Solutions Manual For Seeing Through Statistics 5th Edition By Jessica Utts (All Chapters 1-27

Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025 9780357757550; Chapter 1: The Benefits and Risks of Using Statistics

Solution and Answer Guide

UTTS, SEEING THROUGH STATISTICS, 5E, 2025 9780357757550; CHAPTER 1: THE BENEFITS AND RISKS OF USING STATISTICS

TABLE OF CONTENTS Answers to Exercise Questions ..........................................................................................1 Answers to Reflection Questions ...................................................................................... 9 Answers to Mini Projects ................................................................................................. 10

ANSWERS TO EXERCISE QUESTIONS 1. Explain why the relationship shown in Table 1.1, concerning the use of aspirin and heart attack rates, can be used as evidence that aspirin.

Answer: Because the men were randomly assigned to the two conditions (i.e., because it was a randomized experiment), the only substantial difference between the two groups should have been whether they took aspirin or a placebo.

Therefore, the observed reduction in heart attack rates for those taking aspirin can be attributed to the difference in treatments.

2. “People who often attend cultural activities, such as movies, sports events and concerts, are more likely than their less cultured cousins to survive the next eight to nine years, even when education and income are taken into account, according to a survey by the University of Umea in Sweden” (American Health, April 1997, p. 20). a. Can this claim be tested by conducting a randomized experiment? Explain. b. On the basis of the study that was conducted, can we conclude that attending cultural events makes people more likely to live longer? Explain. c. The article continued, “No one’s sure how Mel Gibson and Mozart help health, but the activities may enhance immunity or coping skills.” Comment on the validity of this statement.

d. The article notes that education and income were taken into account. Give two examples of other factors about the people surveyed that you think should also have been taken into account. (Hint: These should be factors that might differ for those who do and don’t attend cultural events, and also could be related to longevity.)

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025 9780357757550; Chapter 1: The Benefits and Risks of Using Statistics

Answer: a. No. People cannot be randomly assigned to attend cultural events over a long period of time or not. b. We cannot conclude that the relationship is causal because this is an observational study. There could be other factors that account for the increased longevity. For instance, maybe those who are already healthier are more likely to attend cultural events. c. The statement implies that there is a causal connection and seeks an explanation. The causal link is not justified. d. Examples are general health at the start of the study and amount of exercise in their daily lives. (Maybe those who are healthier and who have more energy are more likely to exercise and to attend cultural events.)

Another example is level of income. Those with higher income may be more able to afford cultural events, and may also have better access to other resources that increase longevity.

3. Explain why the number of people in a sample is an important factor to consider when designing a study.

Answer: If the measurements of interest are extremely variable in nature, or if the difference between groups is very small, then a large sample will be required in order to detect real differences between groups or treatments.

4. Explain what problems may arise in trying to make conclusions based on a survey sent to the subscribers of a specialty magazine. Find or construct an example.

Answer: First, even if they all responded to the survey, the results would apply only to subscribers to that magazine, and not to some larger population. Second, it is very unlikely that a large proportion will respond, and those who do are most likely to have strong opinions. Therefore, the results cannot even be extended to all subscribers of that magazine.

5. “If you have borderline high blood pressure, taking magnesium supplements may help, Japanese researchers report. Blood pressure fell significantly in subjects who got 400–500 milligrams of magnesium a day for four weeks, but not in those getting a placebo” (USA Weekend, 22–24 May 1998, p. 11). a. Do you think this was a randomized experiment or an observational study?

Explain. b. Do you think the relationship found in this study is a causal one, in which taking magnesium actually causes blood pressure to be lowered? Explain.

Answer: a. It was most likely a randomized experiment. Placebos are not likely to be taken in observational studies. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025 9780357757550; Chapter 1: The Benefits and Risks of Using Statistics

b. The relationship is probably causal. If the participants were randomly assigned to take magnesium or a placebo, other factors between the two groups should be about equal, so any differences are likely to be caused by taking the magnesium versus taking the placebo.

6. A study similar to the one in Example 1.4 was reported in the May 17, 2012, issue of The New England Journal of Medicine. The researchers followed people for an average of 13 years and found that people who consumed two or more cups of coffee a day were less likely to die during the course of the study than those who drank no coffee. a. Was this study a randomized experiment or an observational study? Explain how you know. b. Based on this study, can it be concluded that drinking coffee causes people to live longer? c. The following headlines appeared on news websites reporting these results.

In each case, explain whether or not the conclusion in the headline is justified. i. “Association of coffee drinking with total and cause-specific mortality” (Source: https://www.coffeeandhealth.org/research/association-ofcoffee-drinking-with-total-andcause-specific-mortality) ii. “NIH Study: Coffee Really Does Make You Live Longer, After All” (Source: http://www.theatlantic.com/health/archive/2012/05/nih-study-coffeereally-does-make-you-livelonger- after-all/257302/) Answer: a. It was an observational study. Coffee drinking over many years cannot be randomly assigned. b. No. A cause and effect conclusion cannot be made based on an observational study. There are likely to be other differences between those who drink coffee and those who do not, and perhaps those other differences have an effect on how long someone lives. c. Headline (i) is justified because it does not imply a causal relationship.

Headline (ii) is not justified because it does imply a causal relationship.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025 9780357757550; Chapter 1: The Benefits and Risks of Using Statistics

7. Refer to Case Study 1.1. When Salk measured the results, he divided the babies into three groups based on whether they had low (2510 to 3000 g), medium (3010 to 3500 g), or high (3510 g and over) birthweights. He then compared the infants from the heartbeat and silent nurseries separately within each birth weight group. Why do you think he did that? (Hint: Remember that it would be easier to detect a difference in adults’ and children’s pulse rates if all adults measured 65 beats per minute and all children measured 75 than it would be if both groups were quite diverse.)

Answer: It is easier to detect a difference between two groups if there is little natural variability among the measurements within each group. By dividing the babies into low, medium and high birth weight groups, Salk was lowering the natural variability among weight gain (or loss) amounts within the groups, compared to what the variability would be among the babies in all of the groups combined.

8. A psychology department is interested in comparing two methods for teaching introductory psychology. Four hundred students plan to enroll for the course at 10:00 a.m., and another200 plan to enroll for the course at 4:00 p.m. The registrar will allow the department to teach multiple sections at each time slot and to assign students to any one of the sections taught at the student’s desired time.

Design a study to compare the two teaching methods. For example, would it be a good idea to use one method on all of the 10:00 sections and the other method on all of the 4:00 sections? Explain your reasoning.

Answer: It would be better to try each teaching method at each time period.

Otherwise, factors related with the time, like how tired the students and instructor are or who takes morning versus afternoon classes, could not be separated from the effects of the different teaching methods. Randomly assign 200 of the 400 students at 10:00 to each method, and 100 of the 200 students at 4:00 to each method. Do not allow the students to choose their method, unless that’s how the course would be taught on an ongoing basis.

9. Suppose you have a choice of two grocery stores in your neighborhood.

Because you hate waiting, you want to choose the one for which there is generally a shorter wait in the checkout line. How would you gather information to determine which one is faster? Would it be sufficient to visit each store once and time how long you had to wait in line? Explain.

Answer: You should visit each store a number of times, at varying times of day and days of the week, unless you always do your shopping at the same time. Even then, you should visit for a number of weeks since on any one visit there could be unusual circumstances. Measurements like time waiting in line have natural variability, so you cannot establish a difference between two conditions (stores) on the basis of a single measurement of each condition (store).

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025 9780357757550; Chapter 1: The Benefits and Risks of Using Statistics

10. Universities are sometimes ranked for prestige according to the amount of research funding their faculty members are able to obtain from outside sources. Explain why it would not be fair simply to use total dollar amounts for each university, and describe what should be used instead.

Answer: Dollars per faculty member should be used. Using the total disadvantages smaller schools because they don’t have as many faculty members.

11. Refer to Case Study 1.3, in which two students were accused of cheating because the licensing board determined the odds of such similar answers were less than 1 in 10,000. Further investigation revealed that over 20% of all pairs of students had matches giving low odds like these (Klein, 1992, p. 26). Clearly, something was wrong with the method used by the board. Read the case study and explain what erroneous assumption they made in their determination of the odds. (Hint: Use your own experience with answering multiple-choice questions.)

Answer: The board assumed that when students answered a question wrong each of the other choices were equally likely to be selected. This is generally not true in multiple-choice exams, in which certain answers can be ruled out and others are plausible, although incorrect.

12. Suppose researchers want to know whether smoking cigars increases the risk of esophageal cancer. a. Could they conduct a randomized experiment to test this? Explain. b. If they conducted an observational study and found that cigar smokers had a higher rate of esophageal cancer than those who did not smoke cigars, could they conclude that smoking cigars increases the risk of esophageal cancer? Explain why or why not.

Answer: a. No, it would be unethical to assign participants randomly to smoke cigars for an extended period or not. b. No, they could not conclude that the relationship was causal. There may be other differences between cigar smokers and non-smokers that could be the cause of the different rates of esophageal cancer, such as higher alcohol consumption by cigar smokers.

13. Suppose the officials in the city or town where you live would like to ask questions of a “representative sample” of the adult population. Explain some of the characteristics this sample should have. For example, would it be sufficient to include only homeowners?

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025 9780357757550; Chapter 1: The Benefits and Risks of Using Statistics

Answer: The sample should include a cross-section of ages, occupations, income, type of neighborhood, political party, and so on. It would not be sufficient to include only homeowners because they would not be representative of all ages, incomes, etc. (Choosing a sample such that all residents are equally likely to be included should result in the desired representation, but choosing a sample like that may not be practical.)

14. Give an example of a decision in your own life, such as which route to take to school, for which you think statistics would be useful in making the decision.

Explain how you could collect and process information to help make the decision.

Answer: The example should be something for which information can be repeatedly collected and for which there is variability in the potential outcomes. As an example, suppose you decide to create a budget for yourself for the coming year, and want to know how much you spend on groceries per week, on average. You could do your usual grocery shopping for multiple weeks and record how much you spent.

15. Suppose you have 20 tomato plants and want to know if fertilizing them will help them produce more fruit. You randomly assign 10 of them to receive fertilizer and the remaining 10 to receive none. You otherwise treat the plants in an identical manner. a. Explain whether this would be an observational study or a randomized experiment. b. If the fertilized plants produce 30% more fruit than the unfertilized plants, can you conclude that the fertilizer caused the plants to produce more?

Explain.

Answer: a. Randomized experiment; plants are randomly assigned to receive the treatments b. Yes. Because of the randomization, all other features, such as how many hearty versus sickly plants there are, should be similar for the two groups of plants.

16. Sometimes websites ask readers to click on choices to register their opinions about an issue. Do you think readers who provide their opinions are a representative sample of all readers who visit the website? Explain.

Answer: They are not likely to be a representative sample. People who feel strongly about the issue are more likely to bother to click on a response. The views of those people are almost surely not representative of all visitors to the website.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025 9780357757550; Chapter 1: The Benefits and Risks of Using Statistics

17. Suppose a researcher would like to determine whether one grade of gasoline produces better gas mileage than another grade. Twenty cars are randomly divided into two groups, with 10 cars receiving one grade and 10 receiving the other. After many trips, average mileage is computed for each car. a. Would it be easier to detect a difference in gas mileage for the two grades if the 20 cars were all the same size, or would it be easier if they covered a wide range of sizes and weights? Explain. b. What would be one disadvantage to using cars that were all the same size?

Answer: a. It would be easier to detect a difference if they were all the same.

Otherwise, it might be hard to distinguish the differences due to grades from the natural variability introduced by using different size cars. For instance, suppose one grade gave 2 more miles to the gallon (mpg) than another. If all of the cars got around 33 mpg with one grade and 35 mpg with the other, the difference would be noticed. If instead they ranged from 15 to 40 with one grade and 17 to 42 with the other, it would be harder to notice the difference. b. One disadvantage would be extending the results to cars of other sizes.

There would be no way to know if they could be extended.

18. National polls are often conducted by asking the opinions of a few thousand adults nationwide and using them to infer the opinions of all adults in the nation. Explain who is in the sample and who is in the population for such polls.

Answer: The sample consists of the few thousand who are asked and the population (at least ideally) is all adults in the nation.

19. Suppose the administration at your school wants to know how students feel about a policy banning smoking on campus. Because they can’t ask all students, they must rely on a sample. a. Give an example of a sample they could choose that would not be representative of all students. Explain how you think they could get a representative sample. A news headline read, “Study finds walking a key to good health: Six brisk outings a month cut death risk.” Comment on what type of study you think was done and whether this is a good headline. b. Explain how you think they could get a representative sample.

Answer: a. A sample that would not be representative would be a poll of everyone who happened to be in the library. If the administration chooses any particular group in which group members’ opinions are likely to differ from the student body as a whole, then they will reach an erroneous conclusion. b. They could get a representative sample by randomly selecting students from the list of all students, maintained by the registrar.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025 9780357757550; Chapter 1: The Benefits and Risks of Using Statistics

20. A news headline read, “Study finds walking a key to good health: Six brisk outings a month cut death risk.” Comment on what type of study you think was done and whether this is a good headline.

Answer: The headline is probably based on an observational study, since it is unlikely that they assigned people to walk and then followed them long enough to ascertain their death risk. If it is an observational study, the headline implying there is a causal connection (walking cuts death risk) is not justified. For instance, the causal relationship could be in the other direction. Perhaps healthier people are more likely to take walks regularly.

21. Suppose a study first asked people whether they meditate regularly and then measured their blood pressures. The idea would be to see if those who meditate have lower blood pressure than those who do not. a. Explain whether this would be an observational study or a randomized experiment. b. If it were found that meditators had lower-than-average blood pressures, can we conclude that meditation causes lower blood pressure? Explain.

Answer: a. It’s an observational study, since people are not randomly assigned to meditate or not. b. No. It could be that people who meditate are also people who tend to have lower blood pressure anyway. You cannot establish a causal connection with an observational study.

22. Refer to the definitions of sample and population on page 7. For each of the following, explain who is in the sample, and what population you think is represented. a. Case Study 1.1, testing the effect of a heartbeat sound in a baby nursery b. Case Study 1.2, testing the effect of taking aspirin on heart attack rates.

Answer: a. The sample consists of the babies who were measured for the study. The population they represent is all babies born under similar circumstances. b. The sample consists of the physicians who participated in the experiment, who were all males. The population they represent is all physicians similar to them who would have agreed to participate if asked. It’s possible that the sample represents a larger group, such as all men similar to the ones in the study.

23. In its March 3–5, 1995 issue, USA Weekend magazine asked readers to return a survey with a variety of questions about sex and violence on television. Of the 65,142 readers who responded, 97% were “very or somewhat concerned about violence on TV” (USA Weekend, 2–4 June 1995, p. 5). Based on this survey, can you conclude that about 97% of all U.S. citizens at that time were concerned about violence on TV? Why or why not?

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025 9780357757550; Chapter 1: The Benefits and Risks of Using Statistics

Answer: You cannot conclude that about 97% of all U.S. citizens at that time were concerned with violence on television. People who choose to respond to a survey in a magazine or on a website are likely to feel strongly about the issue, and thus are not representative of a population.

24. Refer to Thought Question 4 on page 4. A survey was sent to 50,000 employers but only 704 responded. The purpose of the survey was to find out how employers feel about the preparation of college graduates to work in their companies. a. Who do you think would be more likely to take the time to respond to such a survey—employers who are satisfied with the preparation of college graduates or employers who are not satisfied? b. One of the findings of the study was that 31% of respondents thought that college graduates were unprepared for a job search, while only 20% thought they were well prepared or very well prepared. (The remaining 49% thought they were “prepared.”) Using your answer to part (a), explain whether or not the 704 survey respondents are likely to represent how all employers feel on this question.

Answer: a. Employers who are not satisfied would be more likely to respond. b. The respondents are not representative of all employers. People who voluntarily respond to surveys are those who feel strongly, and in this case, they are more likely to be dissatisfied.

ANSWERS TO REFLECTION QUESTIONS 1. If researchers wanted to conduct a study to see if there is an association between drinking beer regularly and having high blood pressure, explain whether they should use a randomized experiment or an observational study.

Answer: They would have to use an observational study because it would not be possible to assign people randomly to drink or not drink beer over a long time period.

2. Suppose an observational study found that college students who participated in at least one volunteer activity in high school had higher grades in college.

Could it be concluded that volunteering in high school causes students to do better in college? What other differences in the students might explain the finding?

Answer: A causal conclusion cannot be made with an observational study. An example of another difference could be that students who volunteer in high school don’t have to work to earn money while in high school or in college. They

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025 9780357757550; Chapter 1: The Benefits and Risks of Using Statistics

would therefore have more time to devote to studying in college, and that could lead to better success.

3. Suppose you received a request to fill out a survey about whether smoking should be allowed on campus and you don’t care about that topic. Would you respond to the survey? Who do you think would respond, and how might that create bias in the results?

Answer: If you don’t care about the topic you are unlikely to respond to the survey. In the case of smoking, people who smoke are probably more likely to respond, and they would be biased in favor of allowing smoking on campus.

4. Suppose you wanted to determine if there is a difference in average blood pressure between English majors and biology majors at your school. Suppose your friend wanted to determine if there is a difference in average number of classes taken per semester for English majors and biology majors. Which one of you do you think would need to survey a larger number of people in order to show that there is a difference between the two majors? Explain.

Answer: The more variability there is likely to be in responses, the larger the number of people needed to show that there is a difference between two groups. In this example blood pressure is likely to be much more variable than number of classes students are taking. Most students take the same number of classes, perhaps plus or minus one. But blood pressure has a wide range of possible values.

ANSWERS TO MINI PROJECTS 1. Design and carry out a study to test the proposition that self-identified males have lower resting pulse rates than self-identified females.

Answer: The intent of this project is to introduce the idea of comparing two groups when there is variability in each one.

Thought should be given to: • How to get a sample without introducing obvious biases • Taking enough observations of each sex to make a meaningful comparison • How to present the results

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025 9780357757550; Chapter 1: The Benefits and Risks of Using Statistics

2. Find a news story that discusses a recent study involving statistical methods.

Identify the study as either an observational study or a randomized experiment. Comment on how well the concepts discussed in this chapter have been applied in the study. Comment on whether the news article, including the headline, accurately reports the conclusions that can legitimately be made from the study. Finally, discuss whether any information is missing from the news article that would have helped you answer the previous questions.

Answer: The kinds of information that should be either described or mentioned as missing are: • Whether a randomized experiment or an observational study was done • The number of observations (the size of the sample) • Whether or not the sample was representative of the population to which the results are being extended • Whether or not a cause-and-effect conclusion was made, and whether or not it was justified given the nature of the study

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 2: Chapter Reading The News

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 2: CHAPTER READING THE NEWS

TABLE OF CONTENTS Answers to the Exercise Questions ....................................................................................1 Answers to the Reflection Questions ................................................................................ 7 Answers to the Mini Projects ............................................................................................. 8

ANSWERS TO THE EXERCISE QUESTIONS 1. Suppose that a television network wants to know how daytime television viewers feel about a new soap opera the network is broadcasting. A staff member suggests that just after the show ends they give two phone numbers, one for viewers to call if they like the show and the other to call if they don’t.

Give two reasons why this method would not produce the desired information.

(Hint: The network is interested in all daytime television viewers. Who is likely to be watching just after the show, and who is likely to call in?)

Answer: First, they would not reach all “daytime viewers”; they would reach only those who were watching the show already, surely a biased sample with regard to the show. Second, those who bothered to call would not even be representative of those who were watching the show. People who want the show to remain on the air would be most likely to call.

2. An e-mail sent to faculty and staff by the Office of Research at the University of California, Irvine (January 19, 2023), had the subject line “Recruiting participants for an at-home sleep study.” The body of the e-mail included the two statements “Did you know that hormones can play a huge role in how many zzz’s we catch each night?”and “WOMEN AGED 42 to 56 ARE NEEDED!” It then gave details about compensation ($200) and location (at home) and provided a link to participate. Assuming that the recruitment effort was limited to that e-mail, to what group could the results of this study be applied?

Explain.

Answer: Results could be extended to women in that age group who would respond to a similar ad, and who are comparable to faculty and staff at that campus. In this case, “comparable” means that they are similar in terms of the purpose of

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 2: Chapter Reading The News

the study. For instance if the study is looking at the relationship between hormones and sleep, then the results could be extended to women who have similar hormone levels and sleep habits as those in the study.

3. A prison administration wants to know whether the prisoners think the guards treat them fairly. Explain how each of the following components could be used to produce biased results, versus how each could be used to produce unbiased results: a. Component 2: The researchers who had contact with the participants b. Component 4: The exact nature of the measurements made or questions asked Answer: a. Having the prison guards themselves asking the prisoners would surely bias the results. Hiring outside interviewers would be best. b. A biased version would be something like “You don’t have any complaints about how you are treated, do you?” while an unbiased version would be something like “We are interested in your opinion about your treatment by the guards. Do you think you are treated fairly or unfairly by them?”

4. According to Cynthia Crossen (1994, p. 106): “It is a poller’s business to press for an opinion whether people have one or not. ‘Don’t knows’ are worthless to pollers, whose product is opinion, not ignorance. That’s why so many polls do not even offer a ‘don’t know’ alternative.” a. Explain how this problem might lead to bias in a survey. b. Which of the Seven Critical Components would bring this problem to light?

Answer: a. People who would prefer to answer that they don’t know, but are forced to choose an answer, will use something other than an informed opinion to do so. For example, if a question were to ask “Do you support the new tax bill in Congress?” and people who were unfamiliar with the bill were forced to answer, their answers would probably be more of a reflection of how they feel about taxes or about the current Congress. b. The lack of “don’t know” as a possible response would be uncovered by Component 4, “the exact nature of the measurements made or questions asked.”

5. Many research organizations give their interviewers an exact script to follow when conducting interviews to measure opinions on controversial issues. Why do you think they do so?

Answer: They give an exact script so that the interviewers don’t change the wording slightly to encourage answers that support their own opinions.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 2: Chapter Reading The News

6. The student who conducted the study in “Hypothetical News Article 1” in this chapter collected two pieces of data from each participant. What were the two pieces of data?

Answer: The two pieces of data from each participant were grade point average and major.

7. Refer to Case Study 1.1, “Heart or Hypothalamus?” Discuss each of the following components, including whether you think the way it was handled would detract from Salk’s conclusion: a. Component 3 b. Component 4 c. Component 5 d. Component 6 Answer: a. Component 3: Presumably the babies were all infants born during a certain time in one hospital. There should be nothing particularly different about these infants. b. Component 4: The measurement of interest was weight gain or loss. That seems like a straightforward, objective measurement, so there shouldn’t be any problems with it. c. Component 5: The setting in which the measurements were taking may have been somewhat of a problem because all of the infants were in a common nursery. Therefore, if one infant cried excessively it may have impacted the entire group. It would have been ideal to test each infant in a separate setting, but that was not practical. d. Component 6: There should have been no additional differences in the two groups of infants, since they were born at the same hospital and therefore probably treated equally except for the heartbeat sound. There may be slight differences if they were born at different times of the year, but that information isn’t provided.

8. Suppose a company that produces plant-based burgers (meat substitutes) is planning to fund a study to compare features of its burgers with those of meat-based burgers. The tests will be conducted in a college snack bar by asking students to sample and compare the two products on characteristics such as taste and texture. In each of the following phases of the study, should the company disclose who is funding it? Explain your answer in each case. a. When students are given the samples and asked for their responses b. When the results are reported in the news c. When the interviewers who will interact with the students are trained and told how to conduct the interviews

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 2: Chapter Reading The News

Answer: a. Absolutely not. If students know who is funding the study, they may try to answer to please the interviewer, who they may assume works for that company. In that case, assuming they can tell which one is the plant-based burger, they are more likely to say that they prefer that one. b. Yes. Reports of research results should always reveal the source of their funding. c. No, unless the instructions are very strict about how to conduct the interviews. Even then, the interviewers are likely to try to solicit opinions to please their employer, and might subtly encourage students to rate the plantbased burger more highly.

9. Suppose a study were to find that twice as many users of nicotine patches quit smoking than nonusers. Suppose you are a smoker trying to quit. Which version of an answer to each of the following components would be more compelling evidence for you to try the nicotine patches? Explain. a. Component 3. Version 1 is that the nicotine patch users were lung cancer patients, whereas the nonusers were healthy. Version 2 is that participants were randomly assigned to use the patch or not after answering an advertisement asking for volunteers who wanted to quit smoking. b. Component 7. Version 1 is that 25% of nonusers quit, whereas 50% of users quit. Version 2 is that 1% of nonusers quit, whereas 2% of users quit.

Answer: a. Version 2 would be more compelling. Lung-cancer patients would be highly motivated to quit already. b. Version 1 would be more compelling, because there isn’t much incentive to try the patches if only 2% of users quit, but there is if 50% quit.

10. In most studies involving human participants, researchers are required to disclose the purpose of the study to the participants. Do you think people should always be informed about the purpose before they participate? Explain.

Answer: The purpose should not always be disclosed in advance, unless it would be unethical for people to participate without knowing the purpose, such as if the purpose was dangerous or harmful. Otherwise, knowing the purpose might bias their answers or behaviors to try to satisfy that purpose.

11. Suppose a study were to find that drinking coffee raised cholesterol levels.

Further, suppose you drink two cups of coffee a day and have a family history of heart problems related to high cholesterol. Pick three of the Seven Critical Components and discuss why knowledge of them would be useful in terms of deciding whether to change your coffee-drinking ha habits based on the results of the study.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 2: Chapter Reading The News

Answer: Components 3 to 7 would all qualify. Probably most important would be 7, knowing by how much it raised cholesterol. Component 3 would be important in order to know if results extend to people of your age, gender and medical type; from component 4 you would learn how “coffee drinking” was defined; from component 5 you might learn about other changes in the environment of the participants, like if they were part of a larger study about healthier diets; from component 6 you would learn whether or not the non-coffee drinkers were different from the coffee drinkers in ways that could affect cholesterol, like amount of exercise or fat in their diets. (A discussion of any three of these is sufficient.)

12. Explain why news reports should give the extent or size of the claimed effects or differences from a study instead of just reporting that an effect or difference was found.

Answer: A statistical difference may not be a meaningful one. Exercise 9b provides a good example. Learning that twice as many nicotine patch users quit smoking than nonusers would mean more if the difference was 50% versus 25% than if it was 2% versus 1%.

13. Refer to the definition of data on page 18. In Hypothetical News Article 3 on page 25, what two pieces of data were collected on each dog? For each one, explain whether it was collected separately for each dog or for the dogs as a group.

Answer: One piece of data was whether the dog was given the drug (yes or no) and this was measured on each dog individually. The other piece of data was amount of barking and this was measured for the dogs as a group, not for each individual dog.

14. Holden (1991, p. 934) discusses the methods used to rank high school math performance among various countries. She notes, “According to the International Association for the Evaluation of Educational Achievement, Hungary ranks near the top in 8th-grade math achievement. But by the 12th grade, the country falls to the bottom of the list because it enrolls more students than any other country—50%—in advanced math. Hong Kong, in contrast, comes in first, but only 3% of its 12th graders take math.” Explain why answers to Components 3 and 6 would be most useful when interpreting the results of rankings of high school math performance in various countries, and describe how your interpretation of the results would be affected by knowing the answers.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 2: Chapter Reading The News

Answer: In comparing test results across countries, states, years, etc., we should learn about exactly who was included each time (Component 3), and what additional differences there might be among the groups taking the tests (Component 6).

Often, differences in results can be explained by the fact that the groups taking them are different on other measures. In this example we would have learned that we were trying to compare the top 50% of students in one country with the top 3% in another, hardly a fair comparison.

15. Moore and Notz (2014, p. 25) reported the following contradictory evidence: “The advice columnist Ann Landers once asked her readers, ‘If you had it to do over again, would you have children? She received nearly 10,000 responses, almost 70% saying ‘No!’. . . A professional nationwide random sample commissioned by Newsday . . . polled 1373 parents and found that 91% would have children again.” Using the most relevant one of the Seven Critical Components, explain the contradiction in the two sets of answers.

Answer: The contradiction is resolved by examining Component 3, the individuals and how they were selected. In the first case, respondents were not a random sample; they were readers who felt strongly enough to write a letter about the issue. In the Newsday study, a random sample was used so the results should not contain a bias in favor of those with strong feelings.

16. An advertisement for a cross-country ski machine, NordicTrack, claimed, “In just 12 weeks, research shows that people who used a NordicTrack lost an average of 18 pounds.” Explain how each of the following components should have been addressed if the research results are fair and unbiased: a. Component 3: The individuals or objects studied and how they were selected b. Component 4: The exact nature of the measurements made or questions asked c. Component 5: The setting in which the measurements were taken. d. Component 6: Differences in the groups being compared, in addition to the factor of interest.

Answer: a. The participants should have been randomly selected from among typical NordicTrack users. b. The participants should have been weighed by the researchers rather than reporting weight loss themselves. c. The NordicTrack should have been used in a natural setting, at home, rather than in a lab where someone else would be making sure participants used it regularly.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 2: Chapter Reading The News

d. The advertisement would be misleading if there were major differences between the group tested and the typical user. For example, if the users in the study started out grossly overweight or if they were on a lean diet at the same time they would not be representative of all users.

17. In a court case similar to those described in Case Study 2.2, the American Basketball Association (ABA) sued AMF Voit, Inc. for manufacturing basketballs with the same red, white, and blue coloring as the ABA basketballs. The ABA commissioned a survey that found 61% of those asked associated the coloring with ABA basketballs. The sample consisted of males aged 12 to 23 who had played basketball within the past year. Do you think the court accepted the survey results as good evidence that the population of people likely to buy basketballs knew the coloring of ABA basketballs? Explain. (If you are curious, do a web search of American Basketball Association v. AMF Voit. The case ultimately went to the Supreme Court.)

Answer: No. Males aged 12 to 23 who played basketball within the past year are not representative of all people who are likely to buy basketballs.

18. Is it necessary that “data” consist of numbers? Explain.

Answer: No, data can be non-numerical information, like college major.

ANSWERS TO THE REFLECTION QUESTIONS 1. Suppose you read a brief news story about a statistical study on a topic that is very important to you. How could you find out more details about the study?

Answer: To find out more details about the study you could consult the original journal article on which the news story was based, if there is one. In most cases there is enough information in the news story for you to find the original source through an Internet search.

2. Suppose a news story reported that people with blood type A are more likely to get a certain disease than are people with other blood types. If you have blood type A, explain what additional information you would want to know about the difference.

Answer: You probably would want to know how much more likely people with blood type A are to get the disease than other blood types, as well as the actual risk of getting it for blood type A.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 2: Chapter Reading The News

3. Suppose an “exit poll” is taken by interviewing voters when they come out of their voting place. Do you think the appearance of the interviewer might make a difference in how honest the voters are? Explain.

Answer: The appearance of the interviewer could easily make a difference in how honest the voters are. If the appearance gives any information about how the interviewer would have voted, some people are likely to say that’s how they voted even if it was not.

4. Suppose you want to estimate the proportion of students at your school who think that everyone should have to take at least one science class. You conduct a survey by sending an e-mail to everyone in your statistics class asking them to respond to the question, “Do you agree that everyone should have to take at least one science class?” Using the material in this chapter, explain three reasons why the results of your survey are not likely to represent the entire school.

Answer: First, the question indicates the desired answer by using the phrase “Do you agree.” Second, students taking a statistics class probably are not representative of all students on campus for this particular issue. They are likely to think science is more important than are some other groups of students. Third, volunteer response means that only those who feel strongly about the issue are likely to respond, leading to a biased sample.

ANSWERS TO THE MINI PROJECTS 1. Scientists publish their findings in technical magazines called journals. Most university libraries have hundreds of journals available for browsing, most of them accessible electronically. Find out where the medical journals are located at your school. Browse the shelves or electronic journals until you find an article with a study that sounds interesting to you. (The New England Journal of Medicine and the Journal of the American Medical Association often have articles of broad interest, but there are also numerous specialized journals on medicine, psychology, environmental issues, and so on.) Read the article and write a report that discusses each of the Seven Critical Components for that particular study. Argue for or against the believability of the results on the basis of your discussion. Be sure you find an article discussing a single study and not a collection or “meta-analysis” of numerous studies.

Answer: The purpose of this project is to get acquainted with asking the right questions when reading the results of research. Make sure all of the components are discussed, and that obvious biases or problems are identified. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 2: Chapter Reading The News

2. Explain how you would design and carry out a study to find out how students at your school feel about an issue of interest to you. Be explicit enough that someone would actually be able to follow your instructions and implement the study. Be sure to consider each of the Seven Critical Components when you design and explain how to do the study.

Answer: The purpose of this project is to learn that there are many difficult decisions that need to be made when designing a research project. Make sure the following issues are discussed: • How to get an unbiased sample • How to collect measurements or ask questions in an unbiased way • How to actually collect the information to ensure good response • How to measure or determine extraneous differences • How to report the results, to include the magnitude 3. Find an example of a statistical study reported in the news for which information about one of the Seven Critical Components is missing. Write two hypothetical reports addressing the missing component that would lead you to two different conclusions about the applicability of the results of the study.

Answer: The purpose of this project is to realize that news reports almost always leave out valuable information, and the interpretation of the results may depend on knowing that information. For instance, news reports about polls often omit the exact wording of the question, yet studies have shown that subtle changes in wording can produce drastic changes in results. Reports about observational studies often omit a description of the group used for comparison, or do not explain the extraneous differences between the groups.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 3: MEASUREMENTS , MISTAKES, AND MISUNDERSTANDINGS

TABLE OF CONTENTS Answers to the Exercise Questions ....................................................................................1 Answers to the Reflection Questions .............................................................................. 16 Answers to the Mini Projects ........................................................................................... 16

ANSWERS TO THE EXERCISE QUESTIONS 1. Give an example of a survey question that is a. Deliberately biased b. Unintentionally biased c. Unnecessarily complex d. Likely to cause respondents to lie Answer: a. “Don’t you agree that our whole tax system, which is far too complicated for anyone to understand, should be overhauled?” (The key here is to include a leading question like “Don’t you agree that ...” or to include emotional wording like “Because abortion involves killing innocent babies...” b. Questions that are unintentionally biased usually involve ambiguous wording. For example, asking “Do you exercise at least three times a week?” may mean something different to young college students than it does to older adults. For instance, the former group might not consider riding a bicycle for transportation as exercise, while the latter group does.

Terms need to be clearly defined. c. Unnecessarily complex questions often are ones that ask more than one question but allow only one answer, like “Do you support doing away with subsidized school lunches and giving food vouchers to families with low incomes instead?” d. Questions that require knowledge people aren’t likely to have, or questions that have socially desirable or even illegal answers are likely to cause respondents to lie. For example, “How many times in the past year have you used illegal drugs?”

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

2. Here is a potential survey question: “Do you agree that marijuana should be legal?” a. Explain which one of the seven pitfalls listed in the box on page 38 of Section 3.2 applies to this question. b. Reword the question so that it avoids the seven pitfalls.

Answer: a. Pitfall 1, deliberate bias, is present. The form of the question indicates that the person asking it does think marijuana should be legal. b. Do you think that the use of marijuana should be legal, or not legal?

3. Here is a potential survey question to determine whether parents support shortening the lunch period at a school: “Do you support cutting the lunch period from 60 minutes to 40 minutes, so that teachers have more time to spend teaching?” a. Explain which two of the seven pitfalls listed in the box on page 38 of Section 3.2 applies to this question. b. Reword the question so that it avoids the seven pitfalls.

Answer: a. Pitfalls 1 (deliberate bias) and 5 (unnecessary complexity) both may be present. The question suggests that there is a good reason to support shortening the lunch period, thus encouraging parents to agree that it should be shortened. It also asks a more complicated question than simply whether the respondent supports shortening the lunch period. b. Do you support or not support shortening the lunch period from 60 minutes to 40 minutes?

4. Here is a potential survey question: “Studies have shown that consuming one alcoholic drink daily helps reduce heart disease. How many alcoholic drinks do you consume daily?” a. Explain which two of the seven pitfalls listed in the box on page 38 of Section 3.2 are most problematic for this question. b. Reword the question so that it avoids the seven pitfalls.

Answer: a. Pitfall 3 (desire to please) is probably most problematic, although pitfalls 1 (deliberate bias) and 6 (order of questions; in this case, order of information presented) may also apply. The statement preceding the question indicates that consuming one drink may actually be good for you, so people would be more likely to admit that they drink. b. The second part of the current version can be used: How many alcoholic drinks do you consume daily? It might make sense to present categories like “none, 0 to 1 per day, 2 to 4 per day, 5 or more per day.”

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

5. Refer to Case Study 2.2, “Flawed Surveys in the Courtroom.” Discuss the study conducted by Brooks Shoe Manufacturing Company in the context of the seven pitfalls (listed in the box on page 38 of Section 3.2) which can be encountered when asking questions in a survey.

Answer: As discussed in Case Study 2.2, the wording was biased, perhaps unintentionally so. Also, desire to please would certainly be a factor in a faceto-face interview like this one, especially in a question like “How long have you known about Brooks Running Shoes?” with the clear implication that you should know about them.

6. Schuman and Presser (1981, p. 277) report the results of a classic study, still relevant today, in which one set of respondents was asked question A, and the other set was asked question B. a. Do you think the United States should forbid public speeches against democracy? b. Do you think the United States should allow public speeches against democracy?

For one version of the question, only about one-fifth of the respondents were against such freedom of speech, whereas for the other version almost half were against such freedom of speech. Which question do you think elicited which response? Explain.

Answer: Although only one-fifth favored forbidding public speeches (version A), almost one-half did not want to allow them (version B). Americans appear to be reluctant to forbid this type of free speech (4/5 didn’t want to do so), but they aren’t as reluctant to withhold approval, with almost half willing to do so.

7. Give an example of two questions in which the order in which they are presented would determine whether the responses were likely to be biased.

Answer: The idea is that one question might suggest a topic that is then considered important or informative for the second question, whereas if the wording had been reversed the topic may not have come to mind. As an example, in the study by Brooks Shoes (Case Study 2.2) participants were told “I am going to hand you a shoe. Please tell me what brand you think it is.” Later, they were asked “How long have you known about Brooks Running Shoes?” If they had been asked the second question first, they would surely have thought of Brooks immediately when asked about the brand of the displayed shoe.

8. A Quinnipiac University Poll conducted in May 2022 asked, “In general, do you think that the Supreme Court is mainly motivated by politics or mainly motivated by the law?” Responses were 63% for “politics,” 32% for “the law,” and 5% “unsure” or no answer. This was followed by the question, “Do you

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

support or oppose limiting the number of years a Supreme Court justice can serve on the Supreme Court?.” Sixty-nine percent responded “Support” and only 27% responded “Oppose.” Do you think the order of these two questions influenced the responses to the second question? Explain.

Answer: The order of the questions most likely influenced the answer to the second one. Almost 2/3 of respondents to the first question said that the Supreme Court at the time was mainly motivated by politics instead of the law. After having brought that possibility to their attention, they probably were more likely to be in favor of term limits than if that had not been brought to their attention first.

9. Sometimes medical tests, such as those for detecting HIV, are so sensitive that people do not want to give their names when they take the test. Instead, they are given a number or code, which they use to obtain their results later. Is this procedure anonymous testing or is it confidential testing? Explain.

Answer: The procedure is an example anonymous testing. If it was confidential, the names would be known but not released to anyone.

10. Some airports still allow smoking in special designated areas. Give three versions of a question to determine whether people think smoking should be completely banned in these airports. Word the question as follows: a. Version 1: As unbiased as possible b. Version 2: Likely to get people to respond that smoking should be forbidden c. Version 3: Likely to get people to respond that smoking should not be forbidden Answer: a. Some airports allow smoking in designated areas but are thinking of completely banning all smoking in the airport. Do you agree or disagree with this policy? b. Don’t you agree that because of the known harmful effects of passive smoke, airports should ban all smoking? c. Don’t you agree that in airports when between flights, smokers should have the right to smoke, as long as a separate, well-ventilated section of the airport is provided for them?

11. The “2003 CASA National Survey of American Attitudes on Substance Abuse VIII: Teens and Parents” discussed in Example 3.2 included 77 questions. a. Two questions asked as “open questions” were Question 1, “What is your age?” and Question 11, “Is there something you do – an activity, hobby, or

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

sport – that you are really into? What is that?” Explain which of the two questions was less likely to cause problems in categorizing the answers. b. Refer to part (a). The most common response to Question 11 was “Sports team.” Read the question and explain why this might have been the case.

Answer: a. Question 1 asked the respondent’s age in years, and would be easy to categorize. Question 11 asked about activities, hobbies and sports, and would be harder to categorize because there are so many possible responses. b. Sports was specifically listed as a possible activity.

12. In 2011 there was a dispute between the Governor of Wisconsin and the teacher’s union about whether teachers should have the right to go on strike.

In February of that year, a Rasmussen Poll asked a nationwide sample “Should teachers, fireman and policemen be allowed to go on strike?” a. Do you think this question is a valid way to determine whether people think teachers should be allowed to go on strike? Explain. b. Do you think the responses to this question would be biased in favor of teachers’ right to strike, not supportive of that view, or not biased in either direction? Explain. c. Write a valid question for determining opinions on this issue. (Source: Rasmussen Reports, 2011) Answer: a. No. It does not allow a response about teachers separate from a response about firemen and policemen. b. Responses would probably be biased in favor of not allowing strikes, because it would be dangerous if policemen and firemen went on strike. c. Wording will vary, but the question should be about teachers only, and should not suggest a desired response. For example, “Should teachers be allowed to go on strike, or should they not be allowed to do so?”

13. Give an example of a survey question that is a. Most appropriately asked as an open question b. Most appropriately asked as a closed question Answer: a. Questions most appropriately asked as an open question are those for which you want to allow for unique answers, like “Give an example of a situation for which an experiment would be unethical.” b. Questions most appropriately asked as a closed question are those for which you want to summarize relative responses from a fixed set of choices. For instance, the student government at a university might ask:

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

“Which of the following areas do you think should receive the highest priority for funding with student fees?” then list choices of projects they are thinking about funding.

14. An advertiser of a certain brand of aspirin (let’s call it Brand B) claims that it is the preferred painkiller for headaches, based on the results of a survey of headache sufferers. The choices given to respondents were: Tylenol, ExtraStrength Tylenol, Brand B aspirin, Advil. a. Is this an open or closed form question? Explain. b. Comment on the variety of choices given to respondents. c. Comment on the advertiser’s claim.

Answer: a. It is a closed-form question; four choices of response were given. b. There is only one choice of aspirin, namely Brand B. Further, those who prefer Tylenol would have to choose between plain Tylenol or Extrastrength Tylenol, thus splitting the vote. c. The claim is very misleading and biased in favor of Brand B aspirin. Anyone who prefers aspirin at all would have to choose Brand B aspirin. Further, those who prefer Tylenol have had their votes split into two groups.

Perhaps the number of people who prefer Tylenol in general is higher than the number who prefer aspirin in general, but the Tylenol vote is split while the aspirin vote all goes to Brand B. 15. The U.S. Centers for Disease Control and Prevention (CDC) includes information on how to define “well-being” for the purpose of conducting medical and psychological studies. They list nine different aspects of wellbeing that have been studied by researchers. Four of these aspects are physical, economic, social, and emotional well-being. a. Give two examples of questions that might be used to measure an individual’s economic well-being. b. Give two examples of questions that might be used to measure an individual’s social well-being. c. Suppose two researchers were to create questions independently to measure these four aspects of well-being. For which two of these four aspects of well-being do you think the researchers’ results would be likely to be similar for a given individual? Explain.

Answer: a. Questions about economic well-being would relate to income and financial security. Examples might include “Do you have enough income or other funds to cover all of your daily expenses?” or “Do you have sufficient savings to cover three months of your typical expenses?”

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

b. Questions about social well-being would relate to social contacts and activities. Examples might include “Do you have a social network of friends and family you could rely on in a crisis?” or “Are you satisfied with the amount of social contact you have with others in a typical week?” c. Physical and economic well-being would probably be easier to measure consistently because they are easier to define, and they are not as subjective as social and emotional well-being.

16. Read Original Source 4 on the companion website or the link that can be accessed at https://corporate.dukehealth.org/news/positive-outlook-linkedlonger-life-heart-patients “Duke Health Briefs: Positive Outlook Linked to Longer Life in Heart Patients.” Explain how the researchers measured “positive outlook.”

Answer: Here is the quote from the article that explains it: “The prevalence of positive emotions and positive personality traits were assessed using the Positive Emotions facet of the Extraversion domain for the NEO personality inventory.

Patients reported how often they experienced positive emotions such as the urge to jump for joy, intense joy, optimism, light-heartedness and the ability to laugh easily.” (See https://corporate.dukehealth.org/news/positive-outlooklinked-longer-life-heart-patients) 17. Locate Original Source 11, “Driving impairment due to sleepiness is exacerbated by low alcohol intake.” It is not on the companion website but may be available at http://www.ncbi.nlm.nih.gov/pmc/articles /PMC1740622/ pdf/v060p00689 or through an Internet search for the title of the article. Find the paragraph describing how the researchers measured “subjective sleepiness.” a. Explain how “subjective sleepiness” was measured. b. Was “subjective sleepiness” measured as a nominal, ordinal, or measurement variable? Explain.

Answer: a. Every 200 seconds the men in the study were asked to rate their level of sleepiness on a scale from 1 to 9, called the Karolinksa Sleepiness Scale, where 1 = extremely alert and 9 = very sleepy, great effort to stay awake, fighting sleep. b. It was an ordinal variable.

18. Give an example of a measure that is a. Valid and categorical b. Reliable but biased c. Unbiased but not reliable

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

Answer: a. The country in which a person was born is valid and categorical. (It measures a non-numerical variable, and truly measures that feature.) b. Time, measured on a clock that is always 5 minutes fast is reliable but biased. (Everyone would get the same reading so it is reliable, but it would be consistently off by 5 minutes, so it is biased.) c. Weight on a postal scale that sometimes measures too high and sometimes measures too low, with equal likelihood is unbiased but not reliable. (It is unbiased since there is no systematic bias in one direction, but unreliable because different people measuring the same package would get different answers.)

19. If we were interested in knowing whether the average price of homes in a certain county had gone up or down this year in comparison with last year, would we be more interested in having a valid measure or a reliable measure of sales price? Explain.

Answer: We would be interested in having a reliable measurement. The reported price may consistently deduct factors like the cost of necessary repairs, but we could make the year-to-year comparison as long as the measurements were taken the same way each year (i.e., were reliable).

20. Do you think the crime statistics reported by the police are a valid measure of the amount of crime in a given city? Are they a reliable measure? Discuss.

Answer: Crime statistics reported by the police are not likely to be a valid measure, because many crimes (like rape or domestic violence) go unreported. They are a reliable measure, because we would all agree on how to get the answer, namely, just look at what is in the records of reported crimes.

21. Specify whether each of the following is a categorical or measurement variable. If you think the variable is ambiguous, discuss why. a. Years of formal education b. Highest level of education completed (grade school, high school, college, higher than college Answer: a. Years of formal education is a measurement variable. b. Highest level of education completed, when given those choices, is categorical (and ordinal, but that was not one of the choices given). It may be ambiguous because it could be quantified to within a certain range of years.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

22. Specify whether each of the following is a categorical or measurement variable. If you think the variable is ambiguous, discuss why. a. Brand of car owned b. Price paid for the last car purchased c. Type of car owned (subcompact, compact, mid-size, full-size, sports, pickup) Answer: a. Brand of car owned is a categorical variable. b. Price paid for the last car purchased is a measurement variable. c. Type of car owned, given those choices, is a categorical variable (and possibly ordinal, but that was not one of the choices given).

23. For each of the variables below, specify whether the variable is categorical and nominal, categorical and ordinal, measurement and interval, or measurement and ratio. a. Years of formal education b. Highest level of education completed (grade school, high school, college, higher than college) c. Brand of car owned d. Price paid for the last car purchased e. Type of car owned (subcompact, compact, mid-size, full-size, sports, pickup) Answer: a. Years of formal education is measurement and ratio. It makes sense to talk about someone having twice as many years as someone else. b. Highest level of education is categorical and ordinal. There is a natural order to the categories. c. Brand of car is categorical and nominal. There is no natural ordering. d. Price paid for a car is measurement and ratio. It makes sense to talk about one car costing twice as much as another. e. Type of car owned is categorical and at least partially ordinal. Not all of the categories can be ordered; sports car and pickup don’t fit anywhere in the natural ordering.

24. Specify whether each of the following measurements is discrete or continuous.

If you think the measurement is ambiguous, discuss why. a. The number of floors in a building b. The height of a building measured as precisely as possible Answer: a. The number of floors in a building is a discrete measurement. b. The height of a building is a continuous measurement.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

25. Specify whether each of the following measurements is discrete or continuous.

If you think the measurement is ambiguous, discuss why. a. The number of words in a book b. The weight of a book c. A person’s IQ Answer: a. The number of words in a book is a discrete measurement. b. The weight of a book is a continuous measurement. c. A person’s IQ is a continuous measurement.

26. Explain whether each measurement below is an interval or a ratio variable. a. The number of floors in a building b. The height of a building measured as precisely as possible c. The number of words in a book d. The weight of a book e. A person’s IQ Answer: a. The number of floors in a building is a ratio variable. It makes sense to talk about one building having twice as many floors as another. b. The height of a building is a ratio variable. It makes sense to talk about one building being twice as tall as another. c. The number of words in a book is a ratio variable. It makes sense to talk about one book having twice as many words as another. d. The weight of a book is a ratio variable. It makes sense to talk about one book weighing twice as much as another. e. IQ is an interval variable. There is no meaningful zero, and it doesn’t make sense to talk about someone having twice as high an IQ as someone else.

27. Can a variable be both nominal and categorical? Explain.

Answer: Yes, nominal variables are one type of categorical variable.

28. Explain whether a variable can be both a. Nominal and ordinal b. Interval and categorical c. Discrete and interval Answer: a. No, categorical variables are either nominal or ordinal, although there could be one or more categories that don’t fit in with an ordinal variable An example is if one of the choices is “prefer not to answer” in a survey question for an ordinal variable.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

b. No, interval variables are measurement variables, not categorical variables. c. Yes, an example is year of someone’s birth.

29. Explain the difference between a discrete variable and a categorical variable.

Give an example of each type.

Answer: A discrete variable measures something numerical with a logical ordering, whereas a categorical variable does not measure something numerical and generally has no logical ordering to the categories. An example of a discrete variable is shoe size. An example of a categorical variable is type of shoe (sandal, athletic shoe, and so on).

30. Suppose you were to compare two routes to school or work by timing yourself on each route for five days. Suppose the times on one route were (in minutes) 10, 12, 13, 15, 20, and on the other route they were 10, 15, 16, 18, 21. a. The average times for the two routes are 14 minutes and 16 minutes. Would you be willing to conclude that the first route is faster over the long run, on average, based on these sample measurements? b. Give an example of two sets of times, where the first has an average of 14 minutes and the second an average of 16 minutes, for which you would be willing to conclude that the first route is faster over the long run.

Answer: a. No, there is so much overlap in the times that it would be difficult to make a definite conclusion about which one is really faster over the long run based on only five times for each route. b. If one route always took 14 minutes and the other always took 16 minutes, that would be convincing.

31. Refer to Exercise 30. Suppose you were to compare two routes to school or work by timing yourself on each route for 5 days. Suppose the times on one route were (in minutes) 13, 14, 14, 14, 15 and on the other route, they were 15, 16, 16, 16, 17. Explain how the concept of natural variability would enter into your decision about whether or not it could be concluded that the first route is faster, on average, than the second route over the long run.

Answer: When measurements are extremely variable, as they were for the two sets in exercise 30a, it is hard to determine whether or not a difference between two groups or conditions really exists over the long run. When measurements have almost no variability, like the examples to be constructed in Exercise 30b, a difference is easy to detect.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

32. Airlines compute the percentage of flights that are on time to be the percentage that arrive no later than 15 minutes after their scheduled arrival time. Is this a valid measure of on-time performance? Is it a reliable measure?

Explain.

Answer: It is not completely valid, because people may interpret “on time” to mean “by the scheduled time” and not 15 minutes later. It is reliable, because everyone using that definition would agree on whether or not the flight made it on time.

33. If each of the following measurements were to be taken on a group of 50 college students (once only for each student), it is unlikely that all 50 of them would yield the same value. In other words, there would be variability in the measurements. In each case, explain whether their values are likely to differ because of natural variability across time, natural variability across individuals, measurement error, or some combination of these three causes. a. Systolic blood pressure b. Time on the student’s phone when that student is measured Answer: a. If taken only once on each individual they would vary due to natural variability across individuals. There could also be measurement error. b. Natural variability across time. Because they are not all measured at the same time, their phones will display different times. If two individuals were measured at the same time, their phones should show the same time.

There should be no measurement error.

34. If the following measurements were to be taken once on each of 50 students, it is unlikely that the values would be the same for all of the students. Explain whether their values are likely to differ because of natural variability across time, natural variability across individuals, measurement error, or some combination of these three causes. a. Blood type (A, B, O, AB) b. Number of contacts in the student’s phone Answer: a. Blood type is likely to differ due to natural variability across individuals. It doesn’t change across time, and it can be measured accurately. b. Natural variability across individuals. Individuals are each measured only once, so there is no variability across time, and the measurements should be accurate.

35. The American Community Survey is conducted annually by the U.S. Census Bureau. The following questions are taken from the survey, addressing issues about the respondent’s home. In each case, state whether the response would

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

be a nominal variable, an ordinal variable, or a measurement variable. (Source: U.S. Census Bureau, 2023.) a. Last month, what was the cost of electricity for this house, apartment, or mobile home? $_____.00 b. Is this house, apartment or mobile home (mark one): Owned by you or someone in this household with a mortgage or loan? Owned by you or someone in this household free and clear, without a mortgage or loan?

Rented? Occupied without payment of rent? c. Does the monthly rent include any meals? (Yes or no?) d. How well does [the person being interviewed] speak English? (Mark one.)

Very well, Well, Not well, Not at all.

Answer: a. Measurement b. Nominal c. Nominal d. Ordinal 36. Explain whether there is likely to be variability in the following measurements if they were to be taken on 10 consecutive days for the same student. If so, explain whether the variability most likely would be due to natural variability across time, natural variability across individuals, measurement error, or some combination of these three causes. a. Systolic blood pressure b. Blood type (A, B, O, AB) Answer: a. Natural variability across time and measurement error are both likely to cause variability in blood pressure measurements for the same person across days. b. Blood type doesn’t change over time for a given individual and is measured accurately, so there would be no variability.

37. For each of the following measurements taken on 10 consecutive days for the same student, explain whether there is likely to be variability. If so, explain whether the variability most likely would be due to natural variability across time, natural variability across individuals, measurement error, or some combination of these three causes. a. The student’s actual height to the nearest 1/10th inch b. A measurement of the student’s height to the nearest 1/10th inch using a 12-inch ruler

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

Answer: a. There is no variability in the actual height. The student’s height should not change over that short time period. b. There would almost surely be variability due to measurement error. It would be difficult to measure height to the nearest 1/10 inch consistently using only a 12-inch ruler.

38. In Chapter 1, we discussed Lee Salk’s experiment in which he exposed one group of infants to the sound of a heartbeat and compared their weight gain to that of a group not exposed. Do you think it would be easier to discover a difference in weight gain between the group exposed to the heartbeat and the “control group” if there were a lot of natural variability among babies, or if there were only a little? Explain.

Answer: It would be easier to detect a difference if there were only a little variability, since then the difference due to the heartbeat might completely separate the groups. For instance, if without the heartbeat all babies gained 10 grams, then even a 1-gram difference in the heartbeat group would be detected. If weight gains ranged from 0 to 20 grams, the 1-gram difference might be masked by the large natural variability.

39. Give an example of a characteristic that could be measured as either a discrete or a continuous variable, depending on the types of units used.

Exercises 40 to 44 refer to a “Hangover Symptoms Scale” (HSS) developed by researchers at the University of Missouri–Columbia (Slutske et al., 2003). The HSS consists of 13 items related to hangovers, such as “Experienced a headache.” Respondents are asked to rate how often in the past year they had these symptoms on a morning after drinking, with the ratings ranging from 0 (not at all) to 4 (100% of the time).

Answer: Anything that gets placed into discrete packets, like candy bar consumption, could be measured in terms of ounces (continuous) or number of bars (discrete) consumed. The size of a book can be measured in number of pages (discrete) or by weight. The height of a building can be measured in inches or by number of floors.

40. a. What are the minimum and maximum possible values for the HSS? b. Explain whether the numerical result of this scale for individuals is an ordinal, interval, or ratio variable.

Answer: a. They listed 13 possible hangover symptoms and asked people to rate the percent of times they had experienced each one of them the morning after drinking, on a 5-point scale ranging from 0 (never) to 4 (every time).

Therefore, the minimum score is 0 and the maximum is 13 × 4 = 52.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

b. It is an interval variable. Even though scores start at 0, it isn’t a ratio variable because it doesn’t make sense to say, for instance, that someone with a score of 20 has twice the severity of hangover symptoms as someone with a score of 10.

41. Explain whether the HSS measurement for each individual is likely to be a reliable measure of hangover severity for that individual.

Answer: It is likely to be reliable because people would probably respond about the same way if you were to ask them to do it over again.

42. Explain whether the HSS measurement for each individual is likely to be a valid measure of hangover severity for that individual.

Answer: It is likely to be valid because it measures severity of 13 of the most common hangover symptoms.

43. Having a valid measurement of hangover severity is important for medical research related to alcohol consumption. To test the validity of this scale, Robertson et al. (2012) conducted a study in which they recruited frequent drinkers and had them fill out the HSS, recalling symptoms over the previous year. They then gave them an electronic device to record evening drinking and morning hangover symptoms in real time for a 3-week period. Describe the general idea of how the researchers could use the electronic records and the HSS result from each participant to assess the validity of the HSS. Answer: The measurements taken in real time over the 3-week period are likely to be an accurate assessment of the individual’s hangover symptoms. The researchers could compare that to their HSS score, and consider the HSS to be a valid measure if there is close agreement.

44. Refer to Exercise 43. Discuss whether the HSS or the electronic records for each individual would be a more reliable measure of hangover severity for that individual. If you think there are reasons to argue in both directions, then explain that as well.

Answer: A reliable measurement is one that is likely to be about the same if you take repeated measurements. The HSS score asks about symptoms over the past year, whereas the electronic real-time measurement is only taken for 3 weeks.

Therefore, the HSS is probably more reliable unless the individual has similar drinking and hangover experiences over any 3-week period.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

ANSWERS TO THE REFLECTION QUESTIONS 1. Why is it important to know the actual wording used in the question(s) asked when you read the results of a survey or poll?

Answer: It is important to know the wording so that you can determine whether it is likely to result in biased responses. There numerous ways in which wording can lead to biased responses, such as if the question starts with something like “Do you agree that...” indicating that the desired answer is yes, you agree.

2. If a candidate for a local office wants to know what issues are of most concern to the local residents, would it be better to ask them with an open form question or a closed form question? Explain.

Answer: If the candidate wants to know what issues are of concern in general, an open form question would be better. A closed form question might be appropriate if there is a specific list of concerns the candidate wants feedback about.

3. Give an example of something that is hard to measure on each individual because the concept is not well defined. Give a different example of something that is hard to measure because different definitions of it may exist.

Answer: An example of something that is not well-defined is an emotion like happiness.

An example of something for which different definitions exist is physical strength. Once a definition is in place it might not be hard to measure physical strength, unlike happiness which is more subjective.

4. Consider measuring something sensitive about which people are likely to lie.

Using the definitions in this chapter, are the measurements more likely to be reliable or more likely to be valid? Explain.

Answer: For sensitive topics measurements might be reliable, but not valid. You are likely to get the about the same response each time (so it’s reliable), but it may not be the truth about the topic, so it’s not valid.

ANSWERS TO THE MINI PROJECTS 1. You will need 10 volunteers for this project: five who exercise regularly (Group E) and five who do not (Group N). The purpose of the project is to examine whether regular exercisers have lower pulse rates, on average, than those who do not exercise regularly. Measure the pulse rates of the 10 individuals for

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

1 minute to get beats per minute (bpm). Draw a line to scale, starting at the lowest pulse in your group and ending at the highest pulse, and mark each exerciser with an E and each non-exerciser with an N. It should look something like this:

Explain exactly how you measured the pulse rates, and then answer each of the following: a. Are your measures valid? b. Are your measures reliable? c. How does the variability in the measurements within each group compare to the difference between the two groups? For example, do all of the exercisers have lower pulse rates than all of the non-exercisers? Are they completely intermixed? d. Do you think your measurements provide convincing evidence that, on average, pulse rates are lower for those who exercise than for those who don’t? Explain. Use your answer to part c as part of your explanation.

Answer: This project has two purposes. One is to illustrate that it is not necessarily easy to measure precisely even something as well defined as pulse rate The measurements, if taken properly, should be valid and would probably be fairly reliable. The second purpose of the project is to recognize that establishing a difference between average values for two groups might take many more than five measurements, even though studies have shown that pulse rates are lower on average for those who exercise regularly than for those who don’t.

With only five in each group, the overlap is often large enough that the difference is not obvious at all. The project should address that concern in the context of the values actually obtained. Sometimes it turns out that for the specific individuals chosen, the variability is low within each group and the difference is apparent.

2. Design a survey with three questions to measure attitudes toward something of interest to you. Now design a new version by changing just a few words in each question to make it deliberately biased. Choose 20 people to whom you will administer the survey. Put their names in a hat (or a box or a bag) and draw out 10 names. Administer the first (unbiased) version of the survey to this group and the second (biased) version to the remaining 10 people. Compare the responses and discuss what happened. (Do not let the individuals in the groups hear the questions administered to the other group.)

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 3: Measurements, Mistakes, And Misunderstandings

Answer: The purpose of this project is to demonstrate that a minor change in wording can result in a major change in responses. Make sure that the “unbiased” versions of the questions are truly neutral. The report should include an explanation of why the biased questions elicited the responses they did. For example, were they biased to produce socially acceptable answers, or to express a strong opinion in hopes that respondents would not disagree, or in some other way?

3. Find a study that includes an emotion like “depression” or “happiness” as one of the measured variables. Explain how the researchers measured that emotion. Discuss whether the method of measurement is likely to produce valid measurements. Discuss whether the method of measurement is likely to produce reliable measurements.

Answer: The main purpose of this project is to encourage students to think about how emotions are measured in studies. Most studies use rating scales based on questions about the emotion and its consequences. The validity of the scale depends on whether the questions asked really do measure that type of emotion. The reliability of the scale is determined by whether people would respond the same way in repeated taking of the test. A scale may not be reliable if it’s highly dependent on how someone is feeling at the moment.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 4: How To Get A Good Sample

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 4: HOW TO GET A GOOD SAMPLE

TABLE OF CONTENTS Answers to Exercise Questions ..........................................................................................1 Answers to Reflection Questions .....................................................................................15 Answers to Mini Projects ................................................................................................. 16

ANSWERS TO EXERCISE QUESTIONS 1. Patients who visit a clinic to help them stop smoking are given a choice of two treatments: undergoing hypnosis or applying nicotine patches. The percentages who quit smoking are compared for the two methods. Is this study a survey, an experiment, an observational study, or a case study? Explain your reasoning.

Answer: Observational study, because patients were allowed to choose their method of treatment rather than having it randomly assigned.

2. A large company wants to compare two incentive plans for increasing sales.

The company randomly assigns a number of its sales staff to receive each kind of incentive and compares the average change in sales of the employees under the two plans. Is this study a survey, an experiment, an observational study, or a case study? Explain your reasoning.

Answer: Experiment, because the incentive methods are randomly assigned.

3. A doctor claims to be able to cure migraine headaches. A researcher administers a questionnaire to each of the patients the doctor claims to have cured. Is this study a survey, an experiment, an observational study, or a case study? Explain your reasoning.

Answer: Case study, because only one doctor is studied in depth.

4. The U.S. government uses a multitude of surveys to measure opinions, behaviors, and so on. Yet, every 10 years it takes a census. What can the government learn from a census that it could not learn from a sample survey?

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 4: How To Get A Good Sample

Answer: The major thing they can learn is how many people live in various areas.

5. Each of the following quotes is based on the results of an experiment or an observational study. Explain which was used. If an observational study was used, explain whether an experiment could have been used to study the topic instead. a. “Spicing up your life with frequent doses of chili could halve the risk of dying from a heart attack or stroke, research suggests. The study of almost 23,000 Italians found that regular consumers had 40 percent fewer fatal heart attacks and 61 percent fewer deadly strokes.” (Donnelly, 2019) b. “The study [of Vitamin D and longevity] ran for five years, and involved more than 21,000 Australians aged between 60 and 79 who were given either vitamin D supplements or a placebo in a double-blind system where neither they nor the researchers knew whether they were getting vitamin D or not.” (Layt, 2022) Answer: a. Observational study. An experiment could not have been done because you can’t randomly assign people to eat chili peppers or not over a long time period. b. This study was a randomized experiment, with each person randomly assigned to take either Vitamin D or a placebo.

6. Explain whether a survey or a randomized experiment would be most appropriate to find out about each of the following: a. Who is likely to win the next presidential election. b. Whether the use of nicotine gum reduces cigarette smoking Answer: a. A survey, because we only need to question people about their preferences and not manipulate their environment. b. A randomized experiment, because we need to randomly assign some people to use the nicotine gum and others to a “control group,” preferably using a placebo gum.

7. Explain whether a survey or a randomized experiment would be most appropriate to find out about each of the following: a. Whether there is a relationship between height and happiness b. Whether a public service advertising campaign has been effective in promoting the use of condoms Answer: a. A survey, because we only need to measure the two variables and not manipulate people’s environment.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 4: How To Get A Good Sample

b. This question could be answered in one of two ways. If the public service campaign had already been conducted, a survey would be appropriate.

Ideally, though, a randomized experiment would be done in which various communities were randomly assigned to receive the advertising campaign and were then compared with similar communities that did not receive it.

8. Refer to Case Study 1.1, “Heart or Hypothalamus?” in which Lee Salk exposed newborn infants to the sound of a heartbeat or to silence, and then measured their weight gain at the end of 4 days. a. What is the sample of units for this study? b. What two variables did Salk measure for each unit in the sample? (Hint: He imposed one of these two variables and measured the other one.) c. To what population of units do you think the results apply?

Answer: a. The sample of units consists of the infants in the two nurseries. b. He measured whether they listened to the sound of a human heartbeat or silence (this was the measure he imposed on them), and he measured weight gain at the end of 4 days. c. The babies in the experiment were born in a New York City hospital, so the population to which the results apply should be all babies similar to the ones born in that hospital.

9. Refer to Case Study 1.1, “Heart or Hypothalamus?” in which Lee Salk exposed newborn infants to the sound of a heartbeat or to silence, and then measured their weight gain at the end of 4 days. a. What is the explanatory variable in Salk’s study? b. What is the outcome (response) variable in Salk’s study?

Answer: a. The explanatory variable is whether the infant listened to the sound of a heartbeat or to silence. b. The outcome (response) variable is weight gain at the end of 4 days.

10. Explain the difference between a proportion and a percent as used to present the results of a sample survey. Include an explanation of how you would convert results from one form to the other.

Answer: A proportion is a part of a whole and a percentage is that same part multiplied by 100. For instance, 1/10th of something is a proportion, equal to 0.10, and the equivalent percentage is 10%.

11. The Sacramento Bee (11 February 2001, p. A20) reported on a Newsweek poll that was based on interviews with 1000 adults, asking questions about a variety of issues.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 4: How To Get A Good Sample

a. What is the margin of error for this poll? b. One of the statements in the news story was “a margin of error of plus or minus three percentage points means that the 43 percent of Americans for and the 48 percent of Americans against oil exploration in Alaska’s Arctic National Wildlife Refuge are in a statistical dead heat.” Explain what is meant by this statement.

Answer: a. With 1000 adults the margin of error is 1 / 1000 = 0.032, or about 3%. b. Adding and subtracting the margin of error of 3% to each of the two figures given, 43% and 48%, produces intervals that overlap. The interval for the percentage who are for oil exploration ranges from 40% to 46%. The interval for the percentage who are against oil exploration ranges from 45% to 51%. The range from 45% to 46% is covered by both intervals. These intervals are likely to cover the true population values for and against oil exploration, so it’s possible that the population percentages are very similar.

12. The study described in Examples 3.2and 4.5 asked teens and their parents about issues related to drug and alcohol consumption. An Associated Press news story about the study included the following information: “QEV Analytics surveyed 1,987 children ages 12 to 17 and 504 parents. . . . The margin of error was plus or minus two percentage points for children and plus or minus four percentage points for parents.” (Kerr, 2003) a. Given the information about how many children and how many parents were included in the survey, verify that the margin of error is about 2% for questions answered by the children and about 4% for questions answered by the parents. b. The news story reported that 20% of the children in the study said they could buy marijuana in an hour or less. What is the interval of values that we can be fairly confident includes the percentage of the population of children who would respond that way at the time of the survey? c. The news story reported that one of the findings of the study was that “more than 5 million children ages 12 to 17, or 20 percent, said they could buy marijuana in an hour or less.” Comment on the accuracy of the wording of that statement.

Answer: a. Children: 1/ 1987 = 0.0224 or about 2%; parents: 1/ 504 = 0.0445, rounded to about 4% b. 20% ± 2% = 18% to 22% c. The study did not ask millions of children, so the statement is not accurate.

It is using the sample value of 20% and extrapolating as if all children in the population had been asked.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 4: How To Get A Good Sample

13. The student newspaper at a university in California reported a debate between two student council members, revolving around a survey of students. The newspaper reported that “according to an AS [Associated Students] Survey Unit poll, 52 percent of the students surveyed said they opposed a diversity requirement.” The report said that one council member “claimed that the roughly 500 people polled were not enough to guarantee a statistically sound cross section of the student population.” Another council member countered by saying that “three percent is an excellent random sampling, so there’s no reason to question accuracy.” The 3% figure is based on the fact that there were about 17,000 undergraduate students currently enrolled at that time.

(Source: California Aggie, 8 November 1994, p. 3) a. Comment on the remark attributed to the first council member, that the sample size is not large enough to “guarantee a statistically sound cross section of the population.” Is the size of the sample the relevant issue to address this concern? b. Comment on the remark by the second council member that “three percent is an excellent random sampling, so there’s no reason to question accuracy.” Is the member correct in their use of terminology and conclusion? c. Assuming a random sample was used, produce the approximate 95% confidence interval that almost certainly covers the true percentage of the population of students who opposed the diversity requirement. Use your result to comment on the debate. In particular, do these results allow a conclusion as to whether the majority of students on campus opposed the requirement?

Answer: a. If the sample was randomly chosen, the size of the sample is the relevant issue and the size of the population does not matter. A sample of 500 students would have a margin of error of about 4.5%, so it may or may not be considered large enough depending on how precise a conclusion was needed. b. The concern should not be with the percent of the population that had been sampled; it should be with the actual number of students sampled.

Further, her use of terminology is not correct. She really means that a random sample of 500 students should be enough to provide a reasonably accurate estimate of the population proportion. c. The margin of error is about 4.5% and the sample percentage is 52%, so we can be fairly certain that between 47.5% and 56.4% of all students oppose the diversity requirement. Because the interval covers 50%, we cannot make a conclusion about majority opinion.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 4: How To Get A Good Sample

14. A Washington Post/Kaiser Family Foundation poll, taken from July 25 to August 5, 2012, asked a random sample of 3130 adults: “Do you support or oppose putting a special tax on junk food—that is, things like soda, chips, and candy— and using the money for programs to fight obesity?” They reported that 53% of the respondents said “Support” and that the margin of error for the poll was ± 2 percentage points. (Source: Washington Post/ Kaiser Family Foundation, 2012) a. Verify that the reported margin of error of ± 2 percentage points is correct (rounded to a whole number of percentage points). b. What is the approximate 95% confidence interval that is likely to cover the true percentage of the entire population that would have answered “Support” if asked at that time? c. Refer to the seven pitfalls in Chapter 3 (listed on page 39) that might be encountered when asking questions in a survey. Which one of them might have been a problem for this survey question? Explain.

Answer: a. The margin of error is 1 / 3130 = .018, which is 1.8% or about 2%. b. The interval is 53% ± 2%, or 51% to 55%. c. Pitfall 5, unnecessary complexity, might be a problem. The question actually contains two separate issues – whether people think a special tax should be put on junk food, and whether the money should be used for programs to fight obesity.

15. Refer to Exercise 14. Based on these sample results, are you convinced that a majority of the population (that is, over 50%) at that time supported putting a special tax on junk food? Provide a numerical argument to support your answer.

Answer: Yes, the interval is 51% to 55%, which is entirely above 50%. Therefore, we should be convinced that the population percentage is more than 50%.

16. Make a list of 20 people you know. Go to the website www.randomizer. org and use it to choose a simple random sample of five people from your list. a. Explain what you did, and give your results by listing the numbers corresponding to the people selected. b. Now use the randomizer website to draw a systematic sample of five people from your list of 20.Give your results (by listing the numbers). c. Would the sample you chose in part (a) have been a possible sample in part (b)? Explain. d. Would the sample you chose in part (b) have been a possible sample in part (a)? Explain.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 4: How To Get A Good Sample

Answer: a. Use the randomizer website to choose 5 numbers from the integers 1 to 20.

Choose those 5 people from your list. b. To get a systematic sample of 5 people from a list of 20, you need to divide the list into 5 parts, so each part will have 4 names. Randomly choose a starting point from 1 to 4, then choose every 4th name from there. For instance, if your initial random selection is 3, you would choose names 3, 7, 11, 15 and 19. c. It is unlikely that the sample you got in part (a) would be a possible systematic sample. The only samples that would qualify are 1, 5, 9, 13, 17; 2, 6, 10, 14, 20; 3, 7, 11, 15, 19; and 4, 8, 12, 16, 20. d. Yes, the sample chosen in part (b) would be a possible simple random sample.

17. In a survey conducted in April 2023, college students were asked to rank a list of reasons about what had made it difficult to succeed in one or more college classes. About 55% ranked “A teaching style that did not work for them” as their top choice. The report said that “The Student Voice survey, conducted in collaboration with College Pulse, asked 3,004 two- and four-year students at 128 institutions a series of questions about academic life.” The survey was supposed to be representative of students at all two- and four-year colleges and universities across the 50 United States. (Source: Flaherty,2023) a. For this survey, explain what a unit is, what the population is, and what the sample is. b. Suppose the survey was conducted by choosing a random sample of students at each of the 128 institutions in the survey. What is that type of sample called? c. Suppose that instead of asking only 3004 students at the institutions, the 128 institutions were randomly selected from all eligible colleges and universities, and then all of the first-year students at those institutions were surveyed. What type of sample is that called?

Answer: a. A unit is a student at a two- or four-year college or university in the United States in April 2023, the population is all such students. The sample is the set of 3004 students who were surveyed. b. A stratified random sample, in which the 128 schools were the strata. c. A cluster sample, in which each of the schools chosen is a cluster.

18. The survey in Exercise 17 was actually conducted by College Pulse using the “American College Student Panel that includes over 675,000 undergraduate college student respondents from more than 1500 two- and four-year colleges

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 4: How To Get A Good Sample

and universities in all 50 states” (https://collegepulse.com/). The students for this survey are a representative sample of 3004 students. a. What is the name for this type of sample? b. For the sample of 3004 students, what is the margin of error? c. About 55% of respondents named “A teaching style that didn’t work for them” as the top barrier to success in a college class. What is an approximate 95% confidence interval for the percentage of the population of college students who would respond that way if asked?

Answer: a. Internet panel sample b. 1/ 3004 = 0.0182 or about 2% c. 55% ± 2% or 53% to 57% 19. For the survey discussed in Exercises 17 and 18, there were 3004 respondents representing students from two- and four-year colleges and universities in the United States. One of the outcomes of the survey was the following: “About two-thirds [67%] of students say their professors grade fairly over all, though just 26 percent of students say they usually understand how their professors grade.”(Source: Flaherty, 2023) a. What is an approximate 95% confidence interval for the percentage of all college students who would say their professors grade fairly, if asked? b. Responses to surveys are always voluntary, although often incentives such as small amounts of money are provided to encourage participation.

Although the report for this survey gave the information that 3004 students responded, it did not provide information about how many were asked to respond. Why would it be important in a survey of this type to know whether most of those asked actually responded?

Answer: a. The margin of error is 1/ 3004 = 0.0182 or about 2%, so the interval is 67% ± 2%, or 65% to 69%. b. If only a relatively small percentage of those asked actually responded, the results would likely be biased. Students who felt strongly about the question of whether their professors grade fairly would be more likely to respond.

20. Give an example in which a. A sample would be preferable to a census b. A cluster sample would be the easiest method to use c. A systematic sample would be the easiest to use and would not be biased

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Answer: a. Typical examples are situations in which the units must be destroyed when tested or in which there are simply too many units to make a census feasible in reasonable time. b. Typical examples are those where units are physically clustered and it is easier to sample an entire cluster, like sampling all students in a random sample of classes at a university. c. Typical examples are when there is a physical layout but not a numbered list, like choosing every fourth house on every street in a neighborhood; or when there is a physical list that isn’t numbered and it is easier to measure a distance on the list, like choosing every thirtieth name on a long unnumbered roster of names.

21. For each of the following situations, state which type of sampling plan was used.

Explain whether you think the sampling plan would result in a biased sample. a. To survey the opinions of its customers, an airline company made a list of all its flights and randomly selected 25 flights. All of the passengers on those flights were asked to fill out a survey. b. A pollster interested in opinions on gun control divided a city into city blocks, then surveyed the third house to the west of the southeast corner of each block. If the house was divided into apartments, the westernmost ground floor apartment was selected. The pollster conducted the survey during the day, but left a notice for those who were not at home with a web address asking them to fill out the survey online. c. To learn how its employees felt about higher student fees imposed by the legislature, a university divided employees into three categories: staff, faculty, and student employees. A random sample was selected from each group and they were contacted and asked for their opinions. d. A large store wanted to know if consumers would be willing to pay slightly higher prices to have computers available throughout the store to help them locate items. An interviewer was posted at the door and told to collect a sample of 100 opinions by asking the next person who came in the door each time an interview was finished.

Answer: a. Cluster sampling; the sample is unlikely to be biased if all passengers complied. b. Assuming the starting point of “third house” was chosen randomly from among the starting options, this was systematic sampling. The sample is likely to be biased for a number of reasons. For instance, ground-floor residents may be more susceptible to crime and thus have more extreme views. Those at home during the day may differ from those not at home, and leaving a notice for the others to call would introduce bias because only those with strong opinions would be likely to do so.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 4: How To Get A Good Sample

c. Stratified sample; the sample is not likely to be biased. d. Convenience or haphazard sample; the sample is likely to be biased for a number of reasons. The most obvious is that the people who looked approachable and had time to answer questions would be the ones chosen.

22. Refer to the previous exercise. Specify the population and the sample, being sure to include both units and measurements, for the situation described in a. Exercise 21a b. Exercise 21b c. Exercise 21c d. Exercise 21d Answer: a. The population of units is all customers of that airline; the measurements are their opinions about the airline. Units in the sample are the passengers on the selected 25 flights, and the sample measurements are their opinions. b. The population of units is all residents of the city, and the measurements are their opinions on gun control. The sample of units consists of those who were either at home and were interviewed or phoned later and were interviewed, and the measurements are their expressed opinions on gun control. c. The population of units is all of the university’s employees, and the measurements are their opinions on the fee increase. The sample units are those reached by telephone, and the measurements are their opinions. d. The ideal population of units is all store customers, and the population measurements are their potential answers to the question posed. The sample consists of those who happened to enter the store and get selected by the interviewer, and their opinions are the sample of measurements.

23. What role does natural variability play when trying to determine the population average of a measurement variable from a sample? (Hint: Read the section on stratified sampling.)

Answer: If there is little natural variability in a set of measurements then the average in the sample will more accurately reflect the average for the whole population than if there is extreme natural variability.

24. Is using a convenience sample an example of a probability sampling plan?

Explain why or why not.

Answer: No. Everyone in the population must have a specified chance of making it into the sample in order to have a probability sampling plan. Approaching people on the street (as an example of a convenience sample) does not allow those probabilities to be specified.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 4: How To Get A Good Sample

25. In early September, 2003, California’s Governor Gray Davis approved a controversial law allowing people who were not legal residents to obtain a California state driver’s license. That week the California Field Poll released a survey showing that 59% of registered voters opposed the law and 34% supported it. This part of the survey was based on a random sample of just over 300 people. a. What is the approximate margin of error for the Field Poll results? b. Provide an interval that is likely to cover the true percentage of registered California voters who supported the law.

Answer: a. The approximate margin of error is 1 / 300 = 0.0577, or about 6%. b. Since 34% of the sample supported it, the true percentage is likely to be in the interval 34% ± 6%, or 28% to 40%.

26. Refer to the previous exercise. The same week that the Field Poll was released, the website SFGate.com (http://www.sfgate.com/polls) asked visitors to “Click to vote” on their preferred response to “Agree with new law allowing drivers’ licenses for illegal immigrants?” The choices and the percent who chose them were “Yes, gesture of respect, makes roads safe” 19%, “No, thwarts immigration law, poses security risk” 79%, and “Oh, great, another messy ballot battle” 2%. The total number of votes shown was 2900. a. What type of sample was used for this poll? b. Explain likely reasons why the percent who supported the law in this poll (19%) differed so much from the percent who supported it in the Field Poll (34%).

Answer: a. This is a volunteer (self-selected) sample. b. There are a couple of possible reasons. First, when interviewed by phone for the Field Poll, people may have thought the socially acceptable answer was to support the law and thus responded that way even if they didn’t support it. Second, who felt strongly about the issue were more likely to oppose the law.

27. Refer to Exercises 25 and 26. Which result, the one from the SFGate poll or from the Field Poll, do you think was more likely to represent the opinion of the population of registered California voters at that time? Explain.

Answer: The Field Poll is more likely to represent the population. The SFGate poll was based on a volunteer sample, not likely to represent any larger group.

28. Despite his success in 1936, George Gallup failed miserably in trying to predict the winner of the 1948 U.S. presidential election. His organization, as well as two others, predicted that Thomas Dewey would beat incumbent Harry

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 4: How To Get A Good Sample

Truman. All three used what is called “quota sampling.” The interviewers were told to find a certain number, or quota, of each of several types of people. For example, they might have been told to interview ten women under age 40, with a fixed number specified for each of three ethnicities. Imagine that you are one of their interviewers trying to follow these instructions. Who would you ask?

Now explain why you think these polls failed to predict the true winner and why quota sampling is not a good method.

Answer: The interviewers were told to choose respondents based on specific characteristics, but not to worry about whether or not they were otherwise representative. The interviewers would undoubtedly choose friends and acquaintances with those characteristics, which would be unlikely to produce a representative sample with regard to voting behavior.

29. Explain why the main problem with the Literary Digest poll is described as “low response rate” and not “volunteer sample.”

Answer: The Literary Digest sent surveys to a legitimate random sample. The problem was that only a small fraction responded, thus the problem was with the low response rate. If they had simply put the survey in their magazine and asked people to respond on a volunteer basis, that would have been a volunteer sample.

30. Suppose the administration at your school has hired you to help with a survey to find out how students feel about opening a new recreation center on campus. They provide you with a random sample of 100 students and their email addresses. After 2 days, 70 of the students have filled out the survey, and you send a reminder to the remaining 30. After only 15 of them respond, you decide to contact 15 of your friends and ask them to take the survey to bring the total to 100. Explain which one of the “difficulties and disasters” in sampling you are most likely to encounter and how it could bias your results.

Answer: The most likely difficulty is “not reaching the individuals selected.” Your friends might not be representative of the larger student body in their opinions on the question of interest.

31. Ask anyone who watched television in the 1980s, and they will remember an ad that claimed “twice as many dentists use Ipana [toothpaste] as any other dentifrice” and that more dentists recommended it than any other dentifrice.

Gastwirth (1988, p. 507) describes a court case in which Bristol-Myers was ordered by the Federal Trade Commission to stop advertising that claim.

Bristol-Myers had based its claim on a survey of 10,000 randomly selected dentists from a list of 66,000 subscribers to two dental magazines. They received 1983 responses, with 621 saying they used Ipana and only 258

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 4: How To Get A Good Sample

reporting that they used the second most popular brand. As for the recommendations, 461 respondents recommended Ipana, compared with 195 for the second most popular choice. a. Specify the sampling frame for this survey, and explain whether you think “using the wrong sampling frame” was a difficulty here, based on what Bristol-Myers was trying to conclude. b. Of the remaining four “difficulties and disasters in sampling” listed in Section 4.6 (other than “using the wrong sampling frame”), which do you think was the most serious in this case? Explain. c. What could Bristol-Myers have done to improve the validity of the results after it had mailed the 10,000 surveys and received 1983 back? Assume the company kept track of who had responded and who had not.

Answer: a. The sampling frame was the list of 66,000 subscribers to the dental magazines. Because Bristol Myers was trying to extend results to all dentists this sampling frame was probably not sufficient. We would need to know whether or not all dentists were likely to subscribe to one of the magazines. b. Having a low response rate was a major problem, since 10,000 dentists received the survey but fewer than 2,000 responded. c. They could have made additional attempts to get the remaining dentists in their sample to respond.

32. Suppose that a gourmet food magazine wants to know how its readers feel about serving beer with various types of food. The magazine sends surveys to 1000 randomly selected subscribers. Explain which one of the “difficulties and disasters” in sampling the magazine is most likely to face.

Answer: They are most likely to face a low response rate, and those who respond are not likely to be representative of all subscribers. For instance, those who prefer beer with their gourmet meals are probably more likely to respond.

33. Explain the difference between a low response rate and a volunteer sample.

Explain which is worse, and why.

Answer: A low response rate refers to a survey given to a legitimate random sample, but for which many don’t respond. A volunteer sample consists of those who respond to a publicly announced survey. A volunteer sample is worse because there is no way to ascertain whether or not it represents any larger group.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 4: How To Get A Good Sample

34. Find a news article describing a survey that is obviously biased. Explain why you think it is biased.

Answer: Answers will differ, but typical examples are surveys with volunteer responses and not random samples. There are other possibilities as well.

35. Construct an example in which a systematic sampling plan would result in a biased sample.

Answer: The example should be one for which there is a systematic difference that would line up with the way the sample is collected.

For Exercises 36 to 38, locate and read the example in the text, and explain what type of sample was used. Then discuss whether you think the results can be applied to any larger population on the basis of the type of sample used, and if so, what population.

36. Case Study 2.1, based on Appendix News Story 7, “Does Eating a Vegan Diet Reduce Stress and Anxiety?”

Answer: This was a volunteer sample of individuals who responded to requests to participate, solicited through physical and social media sites frequented by vegans and vegetarians. The results are likely to represent people who would volunteer through that recruitment method if they were to see it. It may not represent other vegans, vegetarians, and omnivores.

37. The study in Exercise 5a, based on Appendix News Story 13, described as “The study of almost 23,000 Italians found that regular consumers [of chili pepper] had 40 per cent fewer fatal heart attacks and 61 percent fewer deadly strokes.” (Donnelly, 2019) Answer: This was a convenience sample of people who lived in a certain region of Italy.

The results probably can be applied to populations with diets similar to those in that area of Italy.

38. Example 4.6, “Did Dry Eye Symptoms Skyrocket During the COVID-19 Pandemic?”

Answer: The participants were chosen from only the population of people who suffer from dry eyes. Therefore, the results about whether or not dry eye syndrome increased during the COVID-19 pandemic would apply only to that population.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 4: How To Get A Good Sample

ANSWERS TO REFLECTION QUESTIONS 1. Why is it important to learn how the sample was selected in interpreting the results of a survey?

Answer: It is important to know how the sample was selected for a few reasons. First, to extend the results of the survey to a larger population, you need to know what population is represented by the sample. Next, you need to know if the results are likely to be biased based on how the sample was selected. For instance, if the sample relied on volunteers only, it is not likely to represent any larger population.

2. Why is it important to learn who responded to a survey before interpreting the results?

Answer: It is important to know who responded because there could be response bias.

For example, if a survey was sent out to thousands of people but only a few hundred responded, it’s likely that those who responded had stronger opinions about the topic than those who didn’t respond.

3. National surveys used to be almost entirely conducted by contacting a representative sample of households using landlines. What has been the impact of the elimination of landlines on surveys and polls? What new methods for conducting surveys have been developed as a result of this change?

Answer: Given how few people now have landlines, surveys now use other methods to contact people. For instance, Internet panel surveys are done by creating a large and diverse set of people who are willing to respond to surveys if asked.

Then, individual surveys use a representative sample of the larger panel.

4. What is meant by the margin of error for a survey, and how is it used in trying to extend sample results to a population?

Answer: The margin of error represents the amount by which different surveys of the same size from the same population are likely to differ in their results. If a survey is done well, then adding and subtracting the margin of error to the sample result is likely to produce an interval that covers the truth about 95% of the time.

5. Explain the difference between volunteer response and a volunteer sample.

Which one is worse in interpreting the results of a survey?

Answer: Volunteer response refers to a survey in which a legitimate representative sample is contacted and asked to participate, but only some of those contacted actually respond. A volunteer sample is one in which the survey is made available to anyone, and some people volunteer to respond. The latter (volunteer sample) is worse, because it is likely only to represent people who feel strongly about the issue.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 4: How To Get A Good Sample

ANSWERS TO MINI PROJECTS 1. One interesting application of statistics is trying to identify who wrote important historical works that were published using pseudonyms. A classic paper on this topic is “On Sentence-Length as a Statistical Characteristic of Style in Prose: With Application to Two Cases of Disputed Authorship.” (Yule, 1939) Yule identified the length of sentences as a feature that tends to remain consistent across written works by the same author. For this project, you are going to figure out how to estimate the distribution of sentence lengths for a book of your choosing. Find a book that is mainly text and relatively uncluttered with pictures, such as a novel. Choose a random sample of 20 sentences from throughout the book and count how many words they have in them. Use two different sampling methods, chosen from simple random sampling, stratified sampling, cluster sampling, or systematic sampling. In each case, make a table showing how many sentences there were of each length (two words, three words, etc.). a. Explain exactly how you chose your samples. b. Explain which of your two methods was easier to use. c. Of the methods you did not use, explain which of them would have been the most difficult to use. d. Do you think either of your methods produced biased results? Explain. e. Report your results for each of the two methods. Are they generally in agreement with each other? f. Do you think that average sentence length would be as good an indicator of authorship as listing all of the different sentence lengths and the proportion of time each one occurred? Explain.

Answer: This project is designed to illustrate the advantages and disadvantages of various sampling methods in a real setting. Any of the methods should work if done correctly. A simple random sample is probably the most difficult because the sentences would all have to be numbered. A systematic sample would be relatively easy, especially if the first sentence on each of 20 selected pages was used. Divide the pages into 20 segments, randomly choose a starting page in the first segment, then take that same page in each of the other 19 segments. Cluster sampling would be easy, but may not be representative.

Each page of the book could be a cluster. Randomly sample a page and use all of the sentences on that page. If there are fewer than 20 sentences, randomly choose another page, and continue until 20 sentences have been sampled.

Stratified sampling could be used depending on the format of the book. For instance, in a novel divided into chapters, each chapter would be a stratum.

Randomly select a page in the chapter, and then randomly choose sentences within that chapter.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 4: How To Get A Good Sample

2. Go to a large parking lot or a large area where bicycles are parked. Choose a color or a manufacturer. Design a sampling scheme you can use to estimate the percentage of cars or bicycles of that color or model. In choosing the number to sample, consider the margin of error that will accompany your sample result. Now go through the entire area, actually taking a census, and compute the population percentage of cars or bicycles of that type. a. Explain your sampling method and discuss any problems or biases you encountered in using it. b. Construct an interval from your sample that almost surely covers the true population percentage with that characteristic. Does your interval cover the true population percentage you found when you took the census? c. Use your experience with taking the census to name one practical difficulty with taking a census. (Hint: Did all the cars or bicycles stay put while you counted?)

Answer: This project shows the difference between taking a sample and taking a census. It also requires some thought about how sample size relates to margin of error in a real setting. Finally, it should illustrate one of the difficulties of real-world sampling, which is that populations rarely remain fixed in place while you try to sample them. Attention should be paid to getting a representative sample. For example, our university has a parking garage with certain sections for faculty and others for students. Some thought would have to be given to making sure the sample didn’t include cars of just one group, unless that group was the population of interest.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 5: EXPERIMENTS AND OBSERVATIONAL STUDIES

TABLE OF CONTENTS Answers to Exercise Questions ..........................................................................................1 Answers to Reflection Questions .................................................................................... 18 Answers to Mini Projects ................................................................................................. 20

ANSWERS TO EXERCISE QUESTIONS 1. Explain why it may be preferable to conduct a randomized experiment rather than an observational study to determine the relationship between two variables. Support your argument with an example concerning something of interest to you.

Answer: You cannot make a causal conclusion with an observational study, while you often can with a randomized experiment.

2. Suppose a study found that people who drive more than 10 miles to work each day have better knowledge of current events, on average, than people who ride a bicycle to work. a. What is the explanatory variable in this study? b. What is the response variable in this study? c. It was found that people who drive more than 10 miles to work each day also listen to the news on the radio more often than people who ride a bicycle to work. Explain how the variable “how often a person listens to the news on the radio” fits the two properties of a confounding variable (given in the box on page 90) for this study.

Answer: a. From the wording, it follows that the only people included in the study are those who either drive more than 10 miles to work each day and those who ride a bicycle to work. The explanatory variable for each person is which of those methods they use to get to work. b. The response variable is some measure of the person’s knowledge of current events.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

c. Individuals who differ for the explanatory variable (method of getting to work) are also likely to differ on the confounding variable (listening to news on the radio) because those who drive to work are more likely to be listening to new on the radio than are those who ride a bicycle to work. And different values of the confounding variable (listening to news on the radio) are likely to result in different values of the response variable (knowledge of current events). In other words, people who drive a long distance to work have more opportunity to listen to news, and thus are more likely to know current events.

3. A recent study found that people with insomnia are more likely to experience heart problems than people without insomnia. a. What are the explanatory and response variables in this study? b. Explain why amount of caffeine consumed might be a confounding variable in this study by explaining how caffeine consumption fits the two properties of confounding variables given in the box on page 90.

Answer: a. The explanatory variable for each person is whether or not they have insomnia, and the response variable is whether or not they experience heart problems. b. Drinking caffeinated beverages is likely to be related to insomnia (the explanatory variable) because caffeine keeps some people awake, and also because people with insomnia may not sleep enough and may use caffeine to try to stay awake during the day. Caffeine is a stimulant, and therefore may contribute to heart problems.

4. A utility company was interested in knowing if agricultural customers would use less electricity during peak hours if their rates were different during those hours. (Agricultural energy use is substantial, for things like irrigation, lighting, wind turbines to reduce frost damage, and so on.) Customers who volunteered for the study were randomly assigned to continue to get standard rates or to receive the time-of-day rate structure. Special meters were attached that recorded usage during peak and off-peak hours, which the customers could read. The technician who read the meter did not know what rate structure each customer had. a. What was the explanatory variable in this experiment? b. What was the response variable in this experiment? c. Was this experiment single-blind, double-blind, or neither? Explain. d. Did this experiment use matched pairs, blocks, or neither? Explain.

Answer: a. The explanatory variable was the rate structure. b. The response was the amount of electricity used during peak hours.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

c. Single-blind, because customers must (ethically) be told their rate structure, but the meter-reader was blind to the rate structure. d. The experiment did not use matched-pairs or blocks.

5. To test the effects of drugs and alcohol use on driving performance, 20 volunteers were each asked to take a driving test under three conditions: sober; after two drinks; and after smoking marijuana. The order in which they drove the three conditions was randomized. An evaluator watched them drive on a test course and rated their accuracy on a scale from 1 to 10, without knowing which condition they were under each time. a. What was the explanatory variable in this experiment? b. What was the response variable in this experiment? c. Was this experiment single-blind, double-blind, or neither? Explain. d. Did this experiment use matched pairs, blocks, or neither? Explain.

Answer: a. The explanatory variable was the condition of being sober, having drunk alcohol, or having smoked marijuana. b. The response was driving accuracy. c. Single-blind, because drivers knew the condition but the evaluator was blind to it. d. The same drivers were tested under all conditions, so it was a block design with drivers as blocks.

6. To compare four brands of tires, one of each brand was randomly assigned to the four tire locations on each of 50 cars. These tires were specially manufactured without any labels identifying the brand. After the tires had been on the cars for 30,000 miles, the researchers removed them and measured the remaining tread. They were not told which brand was which until the experiment was over. a. What was the explanatory variable in this experiment? b. What was the response variable in this experiment? c. Was this experiment single-blind, double-blind, or neither? Explain. d. Did this experiment use matched pairs, blocks, or neither? Explain.

Answer: a. The explanatory variable was the brand of tire. b. The response variable was the amount of tread remaining on the tire after being driven 30,000 miles. c. Double-blind, because neither the drivers of the cars nor the researchers measuring the tread knew which brand was which. d. It was a block design because all four brands of tires were tested on each car. The cars were the blocks.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

7. Suppose an observational study finds that people who use public transportation to get to work have better knowledge of current affairs than those who drive to work, but that the relationship is weaker for well-educated people. What term from this chapter (for example, response variable) applies to each of the following variables? a. Method of getting to work b. Knowledge of current affairs c. Level of education d. Whether the participant reads the news on their phone every day Answer: a. Explanatory variable b. Response variable c. Interacting variable d. Confounding variable For instance, maybe those taking public transportation used the time to read the news on their phones, and that might have caused the difference in the response variable.

8. Researchers have found that there is an association between taking oral contraceptives (birth control pills) and higher risk of having a heart attack or stroke. They also found that the risk of a heart attack or stroke is substantially higher for smokers. Assume that the proportions who smoke are similar for those who take oral contraceptives and those who don’t. In investigating the relationship between taking oral contraceptives (the explanatory variable) and having a heart attack or stroke (the response variable), explain whether smoking would fit the definition of each of the following: a. A confounding variable b. An effect modifier c. An interacting variable Answer: a. No, smoking is not a confounding variable. The exercise stated that the proportions of smokers were similar in both groups, so there is little or no difference in smoking behavior for the two groups. b. Yes, smoking is an effect modifier because smoking modifies the effect of taking oral contraceptives on the outcome (heart attack or stroke). c. Yes, smoking is an interacting variable, because it changes the relationship between taking oral contraceptives (the explanatory variable) and risk of heart attack or stroke (the response variable).

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

9. A study to see whether birds remember color was done by putting birdseed on a piece of red cloth and letting the birds eat the seed. Later, empty pieces of cloth of varying colors (red, purple, white, and blue) were displayed. The birds headed for the red cloth. The researcher concluded that the birds remembered the color. Using the terminology in this chapter, give an alternative explanation for the birds’ behavior.

Answer: The color of the cloth is a confounding variable. It could be that birds always prefer red to the other colors.

10. Refer to the previous exercise, about whether birds remember colors. Suppose 20 birds were available for an experiment and they could each be tested separately. Suggest a better method for the study than the one used.

Answer: Put the food on different colored cloth. Randomly assign five birds to eat from each of the four colors. Then see which color cloth each bird preferred without food.

11. Refer to Thought Question 5 at the beginning of this chapter. What were the intended explanatory variable and response variables for this study?

Answer: The explanatory variable was whether the person went to college, and the response variable was how happy and healthy they were later in life.

12. Specify the explanatory and response variables for News Story 10 described in the Appendix, “Churchgoers live longer, study finds.”

Answer: The explanatory variable is regular church attendance or not, and response variable is age at death.

13. For each of the following observational studies in the news stories in the Appendix, specify the explanatory and response variables. a. News Story 17: “Drink 2 to 3 cups of coffee a day for longer life, better heart health” b. News Story 16: “More on TV violence” c. Additional News Story 18: “Family: More pets, fewer allergies” Answer: a. The explanatory variable is whether the person drinks 2 to 3 cups of coffee a day, the response variables are length of life and heart health. b. The explanatory variable is how much violent television the participant watched in childhood (1st and 2nd grades or 3rd and 4th grades) and the response variable is violent behavior 15 years later, in adulthood.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

c. The explanatory variable is whether the person lived with a pet dog in infancy, and the response variable is whether they had an allergy to dogs later in childhood.

14. For each of the following news stories in the Appendix, explain whether the study was a randomized experiment or an observational study. If necessary, consult the original source of the study, cited on the website. a. News Story 4: “Happy people can actually live longer” b. News Story 6: “Music as brain builder” Answer: a. Observational study b. Randomized experiment 15. For each of the following news stories summarized in the Appendix, explain whether the study was a randomized experiment or an observational study. a. News Story 11: “Double trouble behind the wheel” b. News Story 15: “Kids’ stress, snacking linked” Answer: a. Randomized experiment b. Observational study 16. Explain why each of the following is used in experiments. You may refer to an example in the chapter if it makes it easier to explain. a. Placebo treatments b. Control groups Answer: a. The power of suggestion may lead to changes in the participants, and those changes would be mistakenly attributed to the treatment or drug. Placebo treatments still elicit the power of suggestion, but do not have any active ingredients. b. Control groups are used to compare the effect of the treatment with what would have happened under similar circumstances without the treatment.

17. a. Explain why blinding is used in experiments. b. Explain why it is not always possible to make an experiment double-blind.

You may refer to an example in the chapter if it makes it easier to explain.

Answer: a. Participants are kept blind so they don’t alter their behavior or outcome to please the experimenter. Those collecting the measurements are kept blind so they don’t inadvertently bias the measurements in the desired direction.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

b. Some treatments are impossible to hide from the people using them, such as when meditation is compared to exercise for the effect on lowering blood pressure. Obviously, people would know if they have been assigned to meditate or to exercise.

18. Refer to each of these news stories in the Appendix or Case Studies in the text, and consult the original source article cited on the companion website if necessary. In each case, explain whether or not a repeated measures design was used. a. News Story 6: “Music as brain builder” b. News Story 11: “Double trouble behind the wheel” c. Case Study 5.3: “Exercise Yourself to Sleep” Answer: a. A repeated measures design was not used. There were separate children in each group. b. A repeated measures design was used. All twelve men drove under all four conditions. c. A repeated measures design was not used. Each participant was assigned to either the exercise group or the control group.

19. Is the “experimenter effect” most likely to be present in a double-blind experiment, a single-blind experiment, or an experiment with no blinding? Explain.

Answer: The experimenter effect is most likely to be present in an experiment with no blinding, because the experimenter can indicate the favored treatment and the participants can respond to that influence. However, it can also be a problem in a single-blind experiment, because the party who isn’t blind to the conditions can try to influence the results.

20. Give an example of a randomized experiment that would have poor ecological validity.

Answer: Any experiment in which the measurements are taken in an artificial setting but the conclusions are supposed to apply to a natural setting would have poor ecological validity.

21. Is it possible to conduct a randomized experiment to compare two conditions using volunteers recruited through a local newspaper? If not, explain why not.

If so, explain how it would be done and explain any “difficulties and disasters” that would be encountered.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

Answer: Yes, in fact many randomized experiments are done in this way. Treatments would be randomly assigned to the volunteers. Generalizing the results to a larger population may be a problem, if the volunteers differ from the larger population in some way.

22. Find the story referenced as News Story 5, “Driving while distracted is common, researchers say,” and consult the first page of the “Executive Summary” in Original Source 5, “Distractions in Everyday Driving” on the website. Explain the extent to which ecological validity may be a problem in this study and what the researchers did to try to minimize the problem.

Answer: A video camera was installed in the car, so drivers may have been on better behavior than they are normally, reducing the ecological validity. Drivers were not told that the purpose of the study was to examine distractions while driving, so that may have helped.

23. Refer to Case Study 5.3, “Exercise Yourself to Sleep.” a. Discuss each of the “difficulties and disasters in experiments” (Section 5.3) as applied to this experiment. b. Explain whether the authors can conclude that exercise actually caused improvements in sleep. c. Draw a picture similar to Figure 5.1 on page 93 illustrating the process used in this experiment.

Answer: a. “Confounding variables” should not be a problem because participants were randomized into the exercise or no exercise conditions; “Interacting variables” may be an issue, for instance, perhaps people who already had sleep problems were more likely to benefit from exercise; “Placebo, Hawthorne and experimenter effects” may be a problem because both groups knew whether or not they were exercising, and results were selfreported, so beneficial effects may have been exaggerated by those who exercised; “Lack of ecological validity and consequent generalization” may be a problem because participants were included in a community exercise program, thus encouraging them more than if people were self-directed in their exercise routine. b. This was a randomized experiment, so confounding factors should have been similar in the two groups. Therefore, a causal conclusion is justified. c. The figure is shown below.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

24. To see if eating just before going to bed causes nightmares, volunteers will be recruited to spend the night in a sleep laboratory. They will be randomly Cassigned to be given a meal before bed or not. Numbers of nightmares will be recorded and compared for the two groups. Explain which of the “difficulties and disasters” is most likely to be a problem in this experiment, and why.

Answer: Ecological validity is likely to be a problem because participants are sleeping in a lab and not at home. Experimenter effects are also possible, because participants know whether or not they ate.

25. Suppose you wanted to know how much money, on average, students at your school spend on food in a week. You consider two different plans for carrying out an observational study.

Plan 1: Ask the participants how much they spent on food for each of the past 5 weeks.

Plan 2: Ask the participants to keep a diary in which they record their food expenditures for the next 5 weeks. a. Which of these plans is a retrospective study? What term is used for the other plan? b. Give one disadvantage of each plan.

Answer: a. Plan 1 is a retrospective study, whereas Plan 2 is a prospective study. b. Plan 1 has the disadvantage that participants may not remember how much they spent. Plan 2 has the disadvantage that participants might change their normal spending behavior, knowing that they were part of a study.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

26. Suppose researchers were interested in determining the relationship, if any, between brain cancer and the use of cell phones. Would it be better to use a randomized experiment or a case control study? Explain.

Answer: A case-control study. First, if there was a suspicion that the phones caused cancer, it would not be ethical to randomly assign some people to use them.

Second, brain cancer is rare, so it would take huge numbers of participants in an experiment to result in even a few cases of brain cancer. And third, people probably would not be willing to be randomly assigned to use a cell phone or not.

27. Read News Story 14 in the Appendix, “Video gaming improves cognitive performance in children.” Answer these questions about the study described in the story. a. Was a case-control study used? If it was, explain how the “cases” and “controls” were chosen. b. Explain whether the research used an observational study or a randomized experiment. c. The title of the news story is “Video gaming improves cognitive performance in children.” Based on how the study was conducted, explain whether the statement in the headline is justified.

Answer: a. Yes, a case-control study was used. Cases and controls were found using the baseline data from the 2019 “Adolescent Brain Cognitive Development” study. Cases were “Video gamers (VGs) defined as those who played video games, while controls were non-video gamers (NVGs) defined as those who put 0 hours into video gaming per week.” b. The research was an observational study. Children were not assigned to either play video games or not. c. The headline claims that video gaming improves cognitive performance, implying a causal relationship. That claim isn’t justified because the research was an observational study.

28. A case-control study claimed to have found a relationship between drinking coffee and pancreatic cancer. The cases were people recently hospitalized with pancreatic cancer, and the controls were people hospitalized for other reasons. When asked about their coffee consumption for the past year, it was found that the cancer cases drank more coffee than the controls. a. Draw a picture illustrating how this study was done, similar to Figure 5.3 on page 102. b. Give a reasonable explanation for this difference in coffee consumption, other than the possibility that drinking coffee causes pancreatic cancer.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

Answer: a. The figure is shown below.

b. People hospitalized for other reasons (the controls) may have been in ill health for the previous year. The ill health may have led them to give up drinking coffee.

29. Explain which of the “difficulties and disasters” in Section 5.5 is most likely to be a problem in each of the following observational studies, and why. a. A study measured the number of writing courses taken by college students and their subsequent scores on the quantitative part of the Graduate Record Exam. The students who had taken the largest number of writing courses scored lowest on the exam, so the researchers concluded that students who want to pursue graduate careers in quantitative areas should not take many writing courses. b. In a study of the relationship between success in high school and success later in life, participants were asked to report their high school grade point average and their salary at age 40.

Answer: a. Confounding variables are the problem. Students who take writing courses are probably likely to excel in verbal abilities rather than quantitative abilities, so it is natural that they would score lower on a quantitative exam. b. Remembering the past would be a problem here. People are not likely to remember their high school grade point average by the time they are age 40.

30. A company wants to know if placing live green plants in workers’ offices will help reduce stress. Employees will be randomly chosen to participate, and plants will be delivered to their offices. One week after they are delivered, all employees will be given a stress questionnaire and those who received plants will be compared with those who did not. Explain which of the “difficulties and disasters” is most likely to be a problem in this experiment, and why.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

Answer: The Hawthorne effect is the biggest problem. Those who received the plants were treated specially. A control group should have been given something like a small figurine in place of a plant.

31. A headline in the Boston Globe (3 April 1998, p. A25) read, “Graduating is good for your health.” The headline was based on a study in which a representative sample of over 400,000 adults in the United States was asked a series of questions, including level of education and on how many of the past 30 days they felt physically and emotionally healthy. Explain how each of the “difficulties and disasters in observational studies” (Section 5.5) applies to this if at all.

Answer: “Confounding variables and the implications of causation” is a major problem, at least as the study was presented in the news headline, since it implies that level of education causes people to feel better. There are lots of possible confounding factors, including the reverse explanation that people who are healthier are more likely to finish college. You can probably think of many other possible confounding factors. “Extending the results inappropriately” is not likely to be a problem, since a large representative sample was used. “Using the past as a source of data” could be a problem because people were asked to recall how they felt for the past 30 days rather than to keep a diary for 30 days.

32. Read the summary of News Story 1 in the Appendix. One of the results reported was that people who participated in the meditation program had better immune system response to a flu vaccine. a. Is a cause-and-effect relationship justified in this situation? Explain. b. Explain the extent to which you think the results can be extended to a population beyond only those who participated in the study.

Answer: a. A cause and effect conclusion is justified because participants were randomly assigned to the meditation or control conditions. b. The participants were volunteers. They were all right-handed and worked for a biotechnology company in Wisconsin. However, the relationship between meditation and immune response would probably not be much different for others, so the sample is likely to represent at least the larger population of people who would be interested in participating in a program like this one.

33. One quote from the news story summarized as News Story 13 in the Appendix was: “Spicing up your life with frequent doses of chili could halve the risk of dying from a heart attack or stroke, research suggests. The study of almost 23,000 Italians found that regular consumers had 40 per cent fewer fatal heart attacks and 61 percent fewer deadly strokes.” (Donnelly, 2019) a. Is a cause-and-effect relationship justified in this situation? Explain. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

b. According to the original research article, participants were “randomly recruited from the general population of Molise, a southern Mediterranean region in Italy” (Bonaccio et al., 2019). Explain the extent to which you think the results can be extended to a population beyond only those who participated in the study.

Answer: a. A cause-and-effect conclusion is not justified because the consumption of chili peppers was not randomly assigned. This was an observational study. b. The participants were randomly selected from a larger population in a region of southern Italy, so results can be extended to at least that population.

34. Find the story referenced as News Story 15 in the Appendix, and consult the original source article cited on the website, if necessary. One of the results reported was that kids with higher levels of stress in their lives were more likely to eat high-fat foods and snacks. a. Is a cause-and-effect relationship justified in this situation? Explain. b. Explain the extent to which you think the results can be extended to a population beyond only those who participated in the study.

Answer: a. A cause-and-effect conclusion is not justified because level of stress cannot be randomly assigned, and there are likely to be confounding variables. b. The participants were children in London, so they probably represent urban children living in a similar culture.

35. Explain why a randomized experiment allows researchers to draw a causal conclusion, whereas an observational study does not.

Answer: Randomized experiments help rule out confounding variables as an alternative explanation for an observed relationship. Randomization works to average out those variables over the different groups or treatments.

36. A headline in the Sacramento Bee (11 December 1997, p. A15) read, “Study: Daily drink cuts death,” and the article began with the statement, “One drink a day can be good for health, scientists are reporting, confirming earlier research in a new study that is the largest to date of the effects of alcohol consumption in the United States.” The article also noted that “most subjects were white, middle-class and married, and more likely than the rest of the U.S. population to be college-educated.” Explain why this study could not have been a randomized experiment.

Answer: It would not be ethical to randomly assign people to either drink alcohol or not.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

37. Refer to Exercise 36, about a report on the relationship between drinking alcohol and likelihood of death, with the headline “Study: Daily drink cuts death.” a. Explain whether you think the headline is justified for this study. b. The study was based on recording drinking habits for the 490,000 participants in 1982 and then noting death rates for the next 9 years. Is this a retrospective or a prospective study? c. Comment on each of the “difficulties and disasters in observational studies” (Section 5.5) as applied to this study.

Answer: a. The headline implies that a daily drink causes better health, but this was an observational study, so a causal conclusion is not justified; see part c. b. A prospective study. It followed people over time (a nine-year period). c. “Confounding variables and the implications of causation” is a major problem because the results are intended to show a causal relationship.

But perhaps the reverse relationship is true, those who are healthier are more likely to allow themselves to drink alcohol, or perhaps those who are more relaxed about life are both healthier (less stressed) and more likely to drink; you can probably think of other confounding factors. “Extending the results inappropriately” would be a problem if the results were applied to any group other than that represented, which was mostly white, middleclass, married residents of the United States. “Using the past as a source of data” is not a problem.

38. Read Summary of News Story 2 in the Appendix, “Benefits of reading books: How it can positively affect your life.” In particular, consider this quote: “A long-term health and retirement study followed a cohort of 3,635 adult participants for a period of 12 years, finding that those who read books survived around 2 years longer than those who either didn’t read or who read magazines and other forms of media” (Stanborough, 2019). Discuss the extent to which each of these problems with observational studies may affect the conclusion based on this study. a. Confounding variables and the implications of causation b. Extending results inappropriately c. Using the past as a source of data Answer: a. This is definitely a problem. There are several possible confounding variables that are related to whether someone is likely to read books and their general health and survival. For example, people who have time to read books may have fewer obligations than those who don’t, and less stress in their lives. b. We don’t know how the sample was selected, so it’s hard to know if the results can be extended at all. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

c. Using the past as a source of data is not particularly likely to be a problem because the implication is that the reading habits of the participants were measured in real time during the 12 years of the study.

Each of the situations in Exercises 39 to 41 contains one of the complications listed as “difficulties and disasters” with designed experiments or observational studies (Sections 5.3 and 5.5). Explain the problem and suggest how it could have been either avoided or addressed. If you think more than one complication could have occurred, mention them all, but go into detail about only the most problematic.

39. (See instructions above.) To study the effectiveness of vitamin C in preventing colds, a researcher recruited 200 volunteers. The researcher randomly assigned 100 of them to take vitamin C for 10 weeks and the remaining 100 to take nothing. The 200 participants recorded how many colds they had during the 10 weeks. The two groups were compared, and the researcher announced that taking vitamin C reduces the frequency of colds.

Answer: The placebo effect is the biggest problem. A look-alike placebo pill should have been given to the control group.

40. (See instructions before Exercise 39.) A researcher was interested in teaching couples to communicate more effectively. There were 20 volunteer couples, 10 of which were randomly assigned to receive the training program and 10 of which were not. After they had been trained (or not), the researcher presented each of the 20 couples with a hypothetical problem situation and asked them to resolve it. Their conversations were recorded. The researcher was blind as to which 10 couples had taken the training program until after the results had been analyzed.

Answer: Ecological validity is a problem. The couples received a hypothetical problem and their communication was recorded, which is not the most realistic setting for determining their natural form of communication. It might have been better to ask them to discuss a real problem in their relationship, but there is no way to avoid having them know they are being watched or recorded.

41. (See instructions before Exercise 39.) Researchers ran an advertisement in a campus newspaper asking for sedentary volunteers who were willing to begin an exercise program. The volunteers were allowed to choose which of three programs they preferred: jogging, swimming, or aerobic dance. After 5 weeks on the exercise programs, weight loss was measured. The joggers lost the most weight, and the researchers announced that jogging was better for losing weight than either swimming or aerobic dance.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

Answer: The problem is that the participants were allowed to choose their method of exercise, yet the results were presented as if jogging was better in general.

There is a confounding variable, namely, preferred form of exercise. Maybe those who prefer jogging are also more enthusiastic about exercise. The researchers should have either randomly assigned the method of exercise or reported the conclusion differently.

42. Refer to Exercise 41, in which volunteers chose their preferred exercise program, and then weight loss was measured after 5 weeks. a. What are the explanatory and response variables in this study? b. Draw a picture similar to Figure 5.1 illustrating the process used in this study.

Answer: a. The explanatory variable is form of exercise and the response variable is weight loss. b. The figure is shown below.

43. For each of the following situations, draw a picture illustrating the process used in the study. The picture should be similar to whichever of Figures 5.1, 5.2, or 5.3 is most appropriate for the situation. a. The experiment described in Example 5.2 on page 94, “Blindly Lowering Cholesterol.” b. The experiment described in Case Study 5.1 on page 95, “Quitting Smoking with Nicotine Patches.” © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

c. The experiment described in Exercise 4 on page 109, in which a utility company randomly assigned volunteers to two different rate plans and compared usage.

Answer: a. The figure is shown below.

b. The figure is shown below.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

c. The figure is shown below.

ANSWERS TO REFLECTION QUESTIONS 1. Suppose a study of college students found a relationship between average study hours per day and resting pulse rate, with students who studied more having higher pulse rates. The researchers speculated that students who study more have less time for exercise, which is known to lower resting pulse rates. a. Explain whether or not this study could have been done as a randomized experiment. b. In this study, explain what the explanatory and response variables are. c. Explain how “amount of exercise” could fit the definition of a confounding variable.

Answer: a. The study could not be a randomized experiment, because you can’t randomly assign students to study different amounts of time. b. The way the question is written, it implies that the explanatory variable is study hours per day and the response variable is resting pulse rate. c. Amount of exercise might differ for those who study lots of hours versus those who study fewer hours, and it also affects resting pulse rate 2. Why is it important to obtain a representative sample in an observational study?

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

Answer: The results of an observational study can be extended to whatever population is represented by those in the sample. So it’s important to get a representative sample from the population for which you want to make conclusions beyond just those in the sample.

3. Why is it important to use random assignment in a randomized experiment?

Answer: In a randomized experiment, using random assignment usually evens out possible confounding variables across the treatment groups. Therefore, the results of an observed relationship between the explanatory and response variables are not confounded with other possible variables that might affect the response variable 4. Create an example of a possible observational study, and use it to explain why a cause-and-effect conclusion cannot generally be made from an observational study.

Answer: Your example should be one in which there are possible confounding variables that can be an alternative explanation for an observed relationship between the explanatory and response variables. For example, an observational study finds that people who smoke more cigarettes have higher resting pulse rates.

It’s likely that those who smoke also exercise less, and exercise is known to lead to lower resting pulse rates.

5. Create an example of a randomized experiment that would not have good ecological validity, and explain why that is the case.

Answer: The example should be one in which participants might behave or react differently because they are in an experiment than they would otherwise. For example, in a study to compare two diets to lower blood pressure, if would be better if blood pressure data was collected by individuals using a monitor at home. It’s well known that blood pressure measurements are often artificially high when taken in a doctor’s office or clinic because of the stress caused by being there.

6. Suppose a randomized experiment found that a new medication helps reduce anxiety compared to a placebo, but only for people with high anxiety at the start of the study. Explain why “anxiety at the start of the study” meets the definition of an interacting variable.

Answer: An interacting variable is one for which the relationship between the explanatory and response variable differs depending on the level of the interacting variable. In the situation described for this question, the effectiveness of the medication depends on initial level of anxiety. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

ANSWERS TO MINI PROJECTS 1. Design an experiment to test something of interest to you. Explain how your design addresses each of the four complications listed in Section 5.3, “Difficulties and Disasters in Experiments.”

Answer: The key feature of an experiment is to be sure proper randomization is used.

Attention should also be paid to trying to use single- or double-blind procedures, if possible.

2. Design an observational study to test something of interest to you. Explain how your design addresses each of the three complications listed in Section 5.5, “Difficulties and Disasters in Observational Studies.”

Answer: For an observational study, it is important to make conclusions that are appropriate to the method used. The most relevant concern to address is the potential for confounding variables to influence the outcome. To the extent possible, confounding variables should be measured.

3. Go to the library or the Internet and locate a journal article that describes a randomized experiment. Explain what was done correctly and incorrectly in the experiment and whether you agree with the conclusions drawn by the authors.

Answer: Most journal articles are careful to address all of the relevant details. Be sure the major concepts of experiments are discussed in the Mini-Project report, including randomization, the use of placebos or control groups, single- or double-blinding and whether or not blocking or pairs were used. Also, make sure that the article describes an experiment and not an observational study.

4. Go to the library or the Internet, and locate a journal article that describes an observational study. Explain how it was done using the terminology of this chapter and whether you agree with the conclusions drawn by the authors.

Answer: Initial topics that should be discussed are whether it was a case-control study or not and whether it was prospective or retrospective. Attention should be paid to whether or not confounding variables were measured and tested for their effects, whether or not any interacting variables were present, and most importantly whether or not the authors addressed the issue of a causal conclusion.

5. Design and carry out a single-blind study using ten participants, as follows.

Your goal is to establish whether people write more legibly with their dominant hand. In other words, do right-handed people write more legibly with their right hand, and do lefthanded people write more legibly with their left hand?

Explain exactly what you did, including how you managed to conduct a single-

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 5: Experiments And Observational Studies

blind study. Mention things such as whether it was an experiment or an observational study and whether you used matched pairs or not.

Answer: The purpose of this project is to illustrate how complex it can be to perform even a simple experiment. The study should be single-blind such that the person evaluating the handwriting does not know which sample was written with which hand. It obviously could not be double-blind, because those providing the writing samples must know which hand was used to produce them. However, they could be kept blind to the purpose of the experiment.

Ideally, this study would be conducted as an experiment in which participants were randomly assigned to write with either their left or their right hand first.

It is a matched-pair experiment, because the same people are supposed to provide samples of writing with each hand.

6. Pick one of the news stories in the Appendix that describes a randomized experiment and that has one or more journal articles accompanying it on the companion website or linked to it. Explain what was done in the experiment using the terminology and concepts in this chapter. Discuss the extent to which you agree with the conclusions drawn by the authors of the study and of the news story. Include a discussion of whether a cause-and-effect conclusion can be drawn for any observed relationships and the extent to which the results can be extended to a larger population.

Answer: The studies that qualify are 1, 3, 6 and 11. The purpose of the project is to learn to read a journal article reporting on a randomized experiment, to identify the important features of the experiment, and to determine what kinds of conclusions can be drawn.

7. Pick one of the news stories in the Appendix that describes an observational study and that has one or more journal articles accompanying it on the companion website or linked to it. Explain what was done in the experiment using the terminology and concepts in this chapter. Discuss the extent to which you agree with the conclusions drawn by the authors of the study and of the news story. Include a discussion of whether a cause-and-effect conclusion can be drawn for any observed relationships and the extent to which the results can be extended to a larger population.

Answer: The studies that qualify are 2, 5, 7, 10, 12, 13 and 20. The purpose of the project is to learn to read a journal article reporting on an observational study, to identify the important features, and to determine what kinds of conclusions can be drawn.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 6: Getting The Big Picture

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 6: GETTING THE BIG PICTURE

TABLE OF CONTENTS Answer to the Reflection Questions...................................................................................1 Answer to the Mini projects ............................................................................................... 2

ANSWER TO THE REFLECTION QUESTIONS 1. In reading a news story about a statistical study, what information would you look for to determine whether it was a sample survey, an observational study, a randomized experiment, or a meta-analysis?

Answer: For a sample survey, participants would simply answer some questions the researchers asked. For an observational study, there would be an explanatory variable and a response variable, but the explanatory variable would be measured and not randomly assigned to participants. For a randomized experiment there would be an explanatory variable and a response variable, and the explanatory variable would be randomly assigned. For a meta-analysis, the story would indicate that the results were a combination of results from a collection of studies rather than a single study.

2. Give an example of how the results of a statistical study might cause you to change something about your lifestyle or beliefs.

Answer: Answers will vary. The example should be something that is part of your life, such as whether amounts and types of exercise, food and drink you might consume, and so on.

3. Suppose a research study were to find that people who own cats are happier, on average, than people who do not own cats. What type of research— observational study or randomized experiment—do you think would have been used? Write two possible headlines for a news story about this research, one that is justified based on the type of study used and one that is not justified.

Answer: Clearly it would be an observational study. You can’t randomly assign people to own cats or not. Here are two headlines: Justified: “Cat owners are happier than non-cat owners, study finds” Not justified: “Want to be happier? Get a cat” © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 6: Getting The Big Picture

ANSWER TO THE MINI PROJECTS 1. Find a news article about a statistical study. Evaluate it using the seven steps on page 118. If all of the required information is not available in the news article, locate the journal article or other source of the research. As part of your analysis, make sure you discuss step 7 with regard to your own life.

Answer: The purpose of this project is to help students learn to utilize studies to make decisions about their own lives. Ideally, the news article chosen will be relevant to the student’s life, and the discussion in the project will include whether or not the results are important and convincing enough to make a change.

2. Choose one of the news stories in the Appendix and the accompanying material provided or cited on the companion website. Evaluate it using the seven steps on page 118. If all of the required information is not available in the news article, locate the journal article or other source of the research. As part of your analysis, make sure you discuss step 7 with regard to your own life.

Answer: The purpose of this project is similar to that of Mini-Project 6.1. Most of the stories can be used to discuss possible lifestyle changes, with a few exceptions such as #18, “Heavier babies become smarter adults, study shows” for which there isn’t much that can be done to change things!

3. Find the journal article in the New England Journal of Medicine on which Case Study 6.5 is based. Evaluate the study using the seven steps on page 118.

Answer: The purpose of this project is to be able to use the seven steps given in the chapter to evaluate a journal article. Most of the steps can be evaluated using the information in the Case Study as presented in the text, but accessing the journal article will provide a few additional details, such as the size of the effect for various subgroups.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 7: SUMMARIZING AND DISPLAYING MEASUREMENT DATA

TABLE OF CONTENTS Answers to the Exercise Questions ....................................................................................1 Answers to the Reflection Questions .............................................................................. 22 Answers to the Mini Project............................................................................................. 23

ANSWERS TO THE EXERCISE QUESTIONS 1. At the beginning of this chapter, the following “oldest ages” were listed, and a stemplot was shown for them in Figure 7.1: 75, 90, 60, 95, 85, 84, 76, 74, 92, 62, 83, 80, 90, 65, 72, 79, 36,78, 65, 98, 70, 88, 99, 60, 82, 65,79,76, 80, 52, 75. a. Create a five-number summary for these ages. b. Create a boxplot using the five-number summary from part (a).

Answer: a. 36, 65, 78, 85, 99 b. The interquartile range is 85 – 65 = 20, so 1.5 × IQR = 30. Therefore, the lines should extend at most to 65 – 30 = 35 at the lower end and 85 + 30 = 115 at the upper end. Because the minimum of 36 is greater than 35 and the maximum of 99 is less than 115, the lines end at those values. The boxplot is shown below.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

2. Refer to the previous exercise, and the stemplot for the “oldest ages” in Figure 7.1. a. Create a stemplot for the oldest ages using each 10s value twice instead of once on the stem. b. Compare the stemplot created in part (a) with the one in Figure 7.1. Are any features of the data apparent in the new stemplot that were not apparent in Figure 7.1? Are any features of the data apparent in Figure 7.1 that are not apparent in the new stemplot? Explain.

Answer: a. 3|6 3| 4| 4| 5|2 5| 6|002 6|555 7|024 7|5566899 8|00234 8|58 9|002 9|589 b. In the new stemplot the low value of 36 stands out as being more removed from the rest of the data, and the value of 52 stands out as well. The shape is more clearly skewed to the left in the new stemplot.

3. Refer to the pulse rate data displayed in the stemplots in Figure 7.2. a. Find the median. b. Create a five-number summary.

Answer: a. The median is the middle value of the ordered list of numbers. There are 25 values, so the median is the value in the 13th position. Counting in to the 13th value from either the upper end or the lower end, the median is 65. b. The low and high values are 54 and 78. The first quartile is 62.5. It is the median of the 12 values below the median, so it’s the average of 6th and 7th values, which are 62 and 63. The upper quartile is 70, which is the average of the 6th and 7th values from the upper end, which are both 70. Therefore, the five number summary is 54, 62.5, 65, 70, 78. Note: Computer software programs may use a more complicated method for finding the quartiles, but it would give the same results for this example.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

4. Suppose the scores on a recent exam in your statistics class were as follows:78, 95, 60, 93, 55, 84, 76, 92, 62, 83, 80, 90, 64, 75, 79, 32, 75, 64, 98, 73, 88, 61, 82, 68, 79, 78, 80, 85. a. Create a stemplot for these test scores using each 10s value once on the stem. b. Create a stemplot for these test scores using each 10s value twice on the stem. c. Compare the stemplots created in parts (a) and (b). Are any features of the data apparent in one of them but not the other one? Explain.

Answer: a. 3|2 4| 5|5 6|024418 7|56598398 8|5430820 9|53208 b. 3|2 3| 4| 4| 5| 5|5 6|02441 6|8 7|3 7|5659898 8|43020 8|58 9|320 9|58 c. The outlier at 32 is much more apparent in part (b). Also, there appears to be a bimodal element with a set of scores in the low 60’s evident in part (b) that is not evident in part (a).

5. Refer to the test scores in the Exercise 4. a. Create a histogram for the test scores. b. Explain how you decided how many intervals to use for the histogram in part (a). c. Comment on the shape of the histogram in part (a).

Answer: a. The histogram is shown below.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

b. The goal is to have approximately 6 to 15 intervals, including the intervals in the gaps between data. So in this situation the easiest method is to have an interval for scores in the 30s, 40s, 50s, up to the 90s. c. The histogram is skewed to the left due to the outlier. The shape depends on what intervals you use. For the histogram shown in part (a), ignoring the outlier, it has a somewhat bell shape ranging from the 50s to 100. If you used intervals with midpoints of 30, 40, 50, etc. you would notice a somewhat bimodal shape.

6. Refer to the test scores in Exercise 4. a. Find the median for the test scores. b. Find the mean for the test scores and compare it to the median. Which one is larger, and why? c. Find the standard deviation for the test scores. If possible, use software or a calculator to do the calculations. d. Do you think the standard deviation would be smaller or larger if the test score of 32 were to be removed from the data? Explain your answer.

Answer: a. The median is 78.5 points. b. The mean is 76.04 points. The median is larger because the outlier of 32 pulls down the mean. c. The standard deviation is 14.19 points. d. The standard deviation would be smaller. The scores would be less spread out.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

7. Refer to the test scores in Exercise 4. a. Create a five-number summary. b. Create a boxplot. c. The test score of 32 should have been identified as an outlier. Which of the three reasons for outliers given on page 135 do you think explains this outlier? Should the outlier be removed when a description of the test scores is presented? Explain your reasoning.

Answer: a. The low and high values are 32 and 98. There are 28 exam scores, so the median is the average of the 14th and 15th values in the ordered list, which are 78 and 79, resulting in a median of 78.5. The quartiles are the medians of the lower 14 values and the upper 14 values, which are 66 and 84.5. So the five-number summary is 32, 66, 78.5, 84.5, 98. Note: Computer software may use a more complicated method for finding the quartiles, resulting in slightly different answers. For instance, Minitab (Version 17) gives the lower and upper quartiles as 65 and 84.75. b. The interquartile range is 84.5 – 66 = 18.5, so 1.5 × IQR = 27.75. Therefore, the lines should extend at most to 66 – 27.75 = 38.25 at the lower end and 84.5 + 27.75 = 112.25 at the upper end. Because the minimum of 32 is less than 38.25, it will be shown as an outlier; the lower line extends to the next lowest value, which is 55. The maximum of 98 is less than 112.25, so the upper line ends at 98. The boxplot is shown below.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

c. The outlier probably is a legitimate test score, and thus represents natural variation in responses (Reason #1). It should not be removed. If it is removed, the natural variability in test scores would be underestimated.

8. Refer to the pulse rates given in the stemplots in Figure 7.2 (page 137). a. Create a five-number summary of the pulse rates (also found in Exercise 3b). b. Create a boxplot of the pulse rates.

Answer: a. The low and high values are 54 and 78. The first quartile is 62.5. It is found as the median of the 12 values below the median, which is the average of 6th and 7th values of 62 and 63. The upper quartile is 70, which the average of the 6th and 7th values from the upper end, which are both 70. Therefore, the five number summary is 54, 62.5, 65, 70, 78. b. The interquartile range is 70 – 62.5 = 7.5, so 1.5 × IQR = 11.25. Therefore, the lines should extend at most to 62.5 – 11.25 = 51.25 at the lower end and 70 + 11.25 = 81.25 at the upper end. Because the minimum of 54 is greater than 51.25 and the maximum of 78 is less than 81.25, the lines end at the minimum and maximum data values (54 and 78). The boxplot is shown below.

9. Give an example for which the median would be more useful than the mean as a measure of center. (Do not reuse an example already discussed in this chapter.)

Answer: Answers will vary, but could be any measurement that is likely to have extreme outliers distorting the mean. Examples include sales prices of used cars, number of people killed in earthquakes of magnitude 4 or higher, salaries at companies with a few high-paid executives, and so on.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

10. Refer to the “oldest ages” given at the beginning of this chapter and in Exercise 1. a. Create a histogram for these ages. b. Explain how you decided how many intervals to use for the histogram in part (a). c. Comment on the shape of the histogram in part (a).

Answer: a. The histogram is shown below

b. The goal is to have approximately 6 to 15 intervals, including the intervals in the gaps between data. So in this situation the easiest method is to have an interval for age in the 30s, 40s, 50s, up to the 90s, creating 7 intervals. c. The histogram is skewed to the left due to the outlier. The shape depends on what intervals you use. For the histogram shown in part (a), ignoring the outlier, it has a somewhat bell shape ranging from the 50s to 90s. If you used intervals with midpoints of 30, 40, 50, etc. you would get a picture skewed to the left with no gap.

11. Construct an example, and draw a histogram for a measurement that you think would be bell-shaped.

Answer: Answers will vary, but include any example of a physical measurement, IQ scores, etc.—for a homogeneous group, such as heights of 21-year-old women.

Figure 7.13b on page 149 shows the kind of histogram you might see.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

12. Construct an example, and draw a histogram for a measurement that you think would be skewed to the right.

Answer: Answers will vary, but include any example with high outliers or with numbers tailing off to the right much more slowly than to the left. Most salary measurements are like that because salary cannot fall below zero. An example would be current salaries of last year’s graduates from your college. Figure 7.7 on page 141 shows the kind of histogram you might see.

13. Construct an example, and draw a histogram for a measurement that you think would be bimodal.

Answer: Answers will vary, but include any example likely to have two clusters of values. For example, temperature at noon in Buffalo recorded for one year would have summer time highs in the 70s to 90s and winter time lows in the 20s to 30s. Figure 7.6 on page 141 shows the kind of histogram you might see.

14. Give one advantage a stemplot has over a histogram and one advantage a histogram has over a stemplot.

Answer: An advantage of a stemplot is that it shows the actual data. An advantage of a histogram is that it can be used for data sets that are too large to allow construction of a stemplot.

15. Which set of data is more likely to have a bimodal shape: daily New York City temperatures at noon for the summer months or daily New York City temperatures at noon for an entire year? Explain.

Answer: Temperatures for the entire year are more likely to be bimodal because there will be clusters of warm temperatures in the summer and cold ones in the winter.

16. Give an example of a set of more than five numbers that has a five-number summary of: Median 40 Quartiles 30 70 Extremes 10 80 Answer: Answers will vary. Here is one example: 10, 11, 30, 38, 39, 41, 65, 70, 72, 80.

17. All of the information contained in the five-number summary for a data set is required for constructing a boxplot. What additional information is required?

Answer: Additional information needed to construct a boxplot is whether or not there are any “outliers” – values more than 1.5 IQRs beyond the quartiles. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

18. Give an example of a measurement for which the mode would be more useful than the median or the mean as an indicator of the “typical” value. (Do not reuse the example of kindergarten children’s ages given on page 135.)

Answer: Answers will vary. Examples include measurements where the numbers have no numerical meaning, or measurements where there are only a few possible values. Example: area code of telephone numbers for students in your class.

19. Three types of pictures were presented in this chapter: stemplots, histograms, and boxplots. Explain the features of a data set for which: a. Stemplots are most useful. b. Histograms are most useful. c. Boxplots are most useful.

Answer: a. A stemplot is useful with one relatively small sample in which it’s useful to see the individual values; information about shape, outliers, and so on is desired. b. A histogram is useful with a moderate to large sample in which there are too many values for a stemplot, but information about shape, outliers, and so on is desired. c. Boxplots are particularly useful with data representing two or more groups for which a comparison is desired, such as hours of sleep for two different age groups.

20. Refer to the data on median household income in Table 7.1; a five-number summary is given in Section 7.3, page 143. What income value would be an outlier at the upper end, using the definition of an outlier on page 145?

Determine if there are any outliers, and if so, which values are outliers.

Answer: Outliers are more than 1.5 interquartile ranges (IQRs) beyond the quartiles. In this case, the IQR = $126,167 – $100,609 = $25,558, so outliers are more than (1.5) × $25,558 = $38,337 beyond the quartiles of $100,609 and $126,167. Thus, an outlier at the lower end is less than $100,609 – $38,337 = $62,272 and at the upper end is more than $126,167 + $38,337 = $164,504. The value of $62,272 is lower than the minimum, so there are no outliers at the lower end.

At the upper end, consulting Table 7.1 reveals that there is one outlier – District of Columbia, with a value of $204,686.

21. Refer to the data on median household income in Table 7.1; a five-number summary is given in Section 7.3, page 143. a. What is the value of the range? b. What is the value of the interquartile range?

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

Answer: a. Range = $204,686 – $82,110 = $122,576 b. IQR = $126,167 – $100,609 = $25,558 22. Refer to the data on median household income in Table 7.1; a five-number summary is given in Section 7.3, page 143. a. Construct a boxplot for this data set. b. Discuss which picture is more useful for this data set: the boxplot from part (a), or the histogram in Figure 7.4.

Answer: a. The boxplot (produced by Minitab Version 17) is shown below

b. The histogram is more useful. In addition to showing the outlier, which is also evident in the boxplot, it illustrates the approximate bell-shape for the remainder of the data.

23. The data on hours of sleep discussed in Example 7.7 was collected from two different introductory statistics classes. One was a statistical literacy class for liberal arts majors, and the other was a standard introductory class for other majors. Here are the separate five-number summaries for “hours of sleep” for the two classes:

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

a. One student in the “Other” class reported sleeping 12 hours. Using this information and the five-number summaries, draw boxplots that allow you to compare the classes on number of hours slept the previous night. Use a format similar to Figure 7.10. b. Based on the boxplots in part (a), describe the similarities and differences between the classes for number of hours slept the previous night.

Answer: a. See the figure below, produced by Minitab (Version 17).

b. The sleep hours were much more variable for the “Other” class than for the liberal arts class, as evidenced by the range and by the interquartile range.

The median was a full hour higher for the liberal arts class than for the other class, at 8 hours and 7 hours, respectively. There were no outliers in the liberal arts class, but there was an outlier of 12 hours in the other class. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

24. Suppose an advertisement reported that the mean weight loss after using a certain exercise machine for 2 months was 10 pounds. You investigate further and discover that the median weight loss was 3 pounds. a. Explain whether it is most likely that the weight losses were skewed to the right, skewed to the left, or symmetric. b. As a consumer trying to decide whether to buy this exercise machine, would it have been more useful for the company to give you the mean or the median? Explain.

Answer: a. It’s most likely that the weight losses were skewed to the right, because the mean is so much higher than the median. There must be some large outliers pulling the mean up. b. The median would be more useful, because half of the users would have lost more and half would have lost less than that amount. The mean is too likely to be distorted by outliers — people who lost much more weight than would typically be expected.

25. Find a set of data of interest to you that includes at least 12 numbers, such as local rents from a website, or scores from a sport of interest. Include the data with your answer. a. Create a five-number summary of the data. b. Create a boxplot of the data. c. Describe the data in a paragraph that would be useful to someone with no training in statistics.

Answer: This requires individual answers.

26. Draw a boxplot illustrating a data set with each of the following features: a. Skewed to the right with no outliers b. Bell-shaped with the exception of one outlier at the upper end c. Values uniformly spread across the range of the data Answer: a. The figure should have a longer line extending from the upper quartile than from the lower quartile, and the box should be wider for the upper half than for the lower half. b. It should look similar to what Figure 7.11 on page 148 (hours of sleep) would look like if it did not have the outlier at the lower end. c. The following four lengths should all be equal: the two whiskers (between the ends of the lines and the ends of the box) and the two halves of the box.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

27. Would outliers more heavily influence the range or the quartiles? Explain.

Answer: The range would be more heavily influenced. The quartiles do not utilize the magnitude of the lower and upper fourth of the data, so an outlier would not have an impact on the quartiles.

28. Suppose you had a choice of two professors for a class in which your grade was very important. They both assign scores on a percentage scale (0 to 100).

You can have access to three summary measures of the last 200 scores each professor assigned. Of the numerical summary measures discussed in this chapter, which three would you choose? Why? Answer: There is no one correct answer to this question. The reasoning is the most important feature; it should show understanding of the measures in question.

For instance, students who expect to get low grades might prefer to know the lower quartile and the minimum grade, whereas better students might prefer to know the upper quartile and the maximum.

29. The students surveyed for the data on hours of sleep in Example 7.7 came from two classes, one for Liberal Arts majors and one for “Other” majors. The students were also asked “How many alcoholic beverages do you consume in a typical week?” Five-number summaries for responses from the two classes are:

a. Draw side-by-side skeletal boxplots for the data. b. Are the values within each set skewed to the right, bell-shaped, or skewed to the left? Explain how you know.

Answer: a. See the figure below, created using Minitab Version 17. (Recall that a skeletal boxplot does not show outliers.)

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

b. The values are clearly skewed to the right. The median and below covers no range at all for the liberal arts class, and a very narrow range for the other class.

30. Refer to the five-number summaries given in Exercise 29 for reported weekly amount of alcohol consumed by students in two statistics classes. For each group, would the mean be higher, lower, or about the same as the median?

Explain how you know.

Answer: In both cases the mean would be higher because of the large outliers.

Exercises 31 to 33 refer to the following scenario. Forty-four people were randomly assigned to two groups. One group was asked, “Do you think the population of Turkey is more than 10 million?” The other group was asked the same question, but “10 million” was replaced by “70 million.” Both groups were then asked to estimate the population of Turkey to the nearest million. The estimates of the population of Turkey (in millions) given by the 22 respondents for each version of the question were as follows, ordered from smallest to largest: “10 million” version: 5, 5, 5, 6, 6, 6, 7, 8, 8, 8, 12, 12, 14, 20, 25, 30, 32, 50, 74, 75,80, 100 “70 million” version: 3, 6, 10, 15, 25, 30, 40, 45, 45, 45, 45, 50, 50, 52, 85, 90, 90, 95, 95, 100, 100, 110 31. a. Create a five-number summary for the “10 million” version. b. Create a five-number summary for the “70 million” version. c. The purpose of this experiment was to test the anchoring effect, discussed in Chapter 16. The anchoring effect occurs when peoples’ estimates of a quantity are influenced by numbers given in prior questions or statements.

Explain whether or not the anchoring effect is evident in this experiment.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

Answer: a. For the “10 million” version, the five-number summary is 5.0, 6.0, 12.0, 32.0, 100.0. Note: Computer software may use a more complicated method to find the quartiles, and they may differ. Minitab Version 17 gives the upper quartile as 36.5 instead of 32.0. b. For the “70 million” version, the five-number summary is 3.0, 30.0, 47.5, 90.0, 110.0. Note: Computer software may use a more complicated method to find the quartiles, and they may differ. For this example, Minitab Version 17 gives the quartiles as 28.75 and 91.25. c. The anchoring effect is clearly present. The values given for the population of Turkey are in general much higher if the question was phrased by asking if it is more than 70 million than by asking if it is more than 10 million.

32. a. Construct a boxplot for the 22 values in the “10 million” version. b. Construct a boxplot for the 22 values in the “70 million” version. c. Describe the information provided by the boxplots, both separately and by comparing them.

Answer: a. The boxplot (created by Minitab Version 17) is shown in the figure below.

b. The boxplot (created by Minitab Version 17) is shown in the figure below.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

c. In both cases the boxplots illustrate that the data values are skewed to the right, but it is more clearly true for the “10 million” version. The anchoring effect is obvious from the boxplots, with the median for the “70 million” version about equal to the upper quartile for the “10 million” version.

33. a. Create a stemplot for the values in the “10 million” version. Describe the shape. (Hint: Use 11 stem values.) b. Create a stemplot for the values in the “70 million” version. Describe the shape. (Hint: Use 12 stem values.) c. For each version separately, explain whether the mean would be higher, lower, or about the same as the median. (You don’t actually need to compute the mean.)

Answer: a. The stemplot is shown below. The values are strongly skewed to the right.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

b. The stemplot is shown below. The pattern is somewhat bimodal, with two “clumps” and a large gap between about 52 million and 85 million.

c. For the 10 million version, the mean would be much higher than the median. For the 70 million version, they would be about the same 34. a. Give an example of a set of five numbers with a standard deviation of 0. b. Give an example of a set of four numbers with a mean of 15 and a standard deviation of 0. c. Is there more than one possible set of numbers that could be used to answer part (a)? Is there more than one possible set of numbers that could be used to answer part (b)? Explain.

Answer: a. Any set in which all five numbers are equal b. 15, 15, 15, 15 c. There are an infinite number of sets that could be used to answer part (a) – any set in which all five numbers are the same. There is only one set of numbers that can be used to answer part (b). If the standard deviation is 0, then all numbers must be the same, and if the mean is 15 then they must all be 15.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

35. Find the mean and standard deviation of the following set of numbers: 10, 20, 25, 30, 40.

Answer: Mean = 25, variance = 125, so standard deviation = 11.18.

36. In each of the following cases, would the mean or the median probably be higher, or would they be about equal? a. Prices of all new cars sold in 1 month in a large city b. Heights of all 7-year-old children in a large city c. Ages of the students graduating from a large university in a given year Answer: a. The mean would be higher. There would probably be a few expensive ones, pulling the mean up. b. The mean and median should be about the same because the heights are likely to be approximately bell-shaped. c. The mean is likely to be higher. Most students would be in their early 20s but there would be a fraction of much older students pulling the mean higher.

37. In each of the following cases, would the mean or the median probably be higher, or would they be about equal? a. Salaries in a company employing 100 factory workers and two highly paid executives b. Ages at which residents of a suburban city die, including everyone from infants to the elderly Answer: a. The mean would be higher because the two executive salaries would pull it up. b. The median would be higher. There would be some very young ages pulling the mean down. In other words, the picture would be skewed to the left.

38. Suppose a set of test scores is approximately bell-shaped, with a mean of 70 and a range of 50. Approximately, what would the minimum and maximum test scores be?

Answer: The bell-shape is symmetric, so we would expect about half of the values below the mean and half above the mean. Therefore, a range of 50 would imply that the minimum is about 25 points below the mean, or about 45 points, and the maximum is about 25 points above, or about 95 points.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

39. What is the variance for the fifth edition of the Stanford-Binet IQ test? (Hint: See Thought Question 4 on page 134.)

Answer: The standard deviation is 15 so the variance is 152 = 225.

40. a. Explain the following statement in words that someone with no training in statistics would understand: Resting pulse rates in a certain age group for non-smokers who exercise regularly are approximately bell-shaped with a mean of 72 and a standard deviation of 5 beats per minute. b. A website reported that the normal resting pulse rate for people like those described in part (a) is 72, but that it could range from about 62 to 82.

Comment on the accuracy of that statement.

Answer: a. Resting pulse rates for this group tend to be clustered around the average, which is 72 beats per minute. Some pulses are higher and some are lower, but the farther you move from the average the fewer people have those values. The average amount by which the pulse rates differ from the mean of 72 is about 5 beats per minute. b. The statement is confusing “average” with “normal.” The average pulse rate is 72. The comment about the range is a good description of the range of “normal” values.

41. The Winters (CA) Express on April 4, 2013, reported that the seasonal rainfall (since July 1, 2012) for the year so far was 16.75 inches, and that the “average to April 3 is 20.09 inches.” Does this mean that the area received abnormally low rainfall in the period from July 1, 2012, to April 4, 2013? Explain and mention any additional information that might help you make this assessment.

Answer: Not necessarily. Normal rainfall is quite variable and the statement about “average to April 3” rainfall indicates that there is year to year variability around that number. To know if 16.75 inches was abnormally low, we would need to know something about the variability from year to year.

42. Suppose you are interested in genealogy, and want to try to predict your potential longevity by using the ages at death of your ancestors. You find out the ages at death for the eight great-grandparents of your biological parents.

Suppose the ages (in numerical order) are as follows: Mother’s great-grandparents: 78, 80, 80, 81, 81, 82, 82, 84 Father’s great-grandparents: 30, 50, 77, 80, 82, 90, 95, 98 a. Compare the medians for the two sets of ages. How do they compare? b. Compare the means for the two sets of ages. How do they compare? c. Find the standard deviation for each set of ages. d. Which set do you think is more useful for predicting longevity in your family? Explain. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

Answer: a. Mother’s median is 81 years and father’s median is also 81 years. b. Mother’s mean is 81 years and father’s mean is 75.25 years. c. Mother’s standard deviation is 1.773 years and father’s standard deviation is 23.55 years. d. Mother’s family is probably more useful because they are much more consistent. Very low ages in father’s family could be due to accidents, death in childbirth, etc.

43. According to the National Weather Service, there is about a 10% chance that total annual rainfall for Sacramento, CA, will be less than 11.1 inches and a 20% chance that it will be less than 13.5 inches. At the upper end, there is about a 10% chance that it will exceed 29.8 inches and a 20% chance that it will exceed 25.7 inches. The average amount is about 19 inches. In the 2022 year (July 1, 2021–June 30, 2022), the area received about 11.2 inches of rain. Write two news reports of this fact, one that conveys an accurate comparison to other years and one that does not.

Refer to Original Source 5 on the companion website, “Distractions in everyday driving.” Notice that on page 86 of the report, when the responses are summarized for the quantitative data in question 8, only the mean is provided.

But for questions 7 and 9 the mean and median are provided. Why do you think the median is provided in addition to the mean for these two questions?

Answer: News story 1: Rainfall in 2022 was 11.2 inches, well within the normal range. In fact, in about one out of ten years, the area receives less than 11.1 inches of rain. So, even though the average amount is 19 inches a year, there is nothing unusual about receiving less rain in any given year.

News story 2: Rainfall in 2022 was only 11.2 inches, well below the normal rainfall for the area of 19 inches. Scientists aren’t sure what caused us to have such a major drought this year.

44. Refer to Original Source 5 on the companion website, “Distractions in everyday driving.” Notice that on page 86 of the report, when the responses are summarized for the quantitative data in question 8, only the mean is provided.

But for questions 7 and 9 the mean and median are provided. Why do you think the median is provided in addition to the mean for these two questions?

Answer: In questions 7 and 9, the mean is much higher than the median. Therefore, it is likely that there were some high outliers. The report writers probably understood that in those circumstances, the mean would not be an accurate reflection of the middle of the data.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

45. Refer to Original Source 20 cited on the companion website, “Organophosphorus pesticide exposure of urban and suburban preschool children with organic and conventional diets.” (Also available at https://ehp.niehs.nih.gov/doi/pdf/10.1289/ehp.5754.) In Table 4 on page 381, information is presented for estimated dose levels of various pesticides for children who eat organic versus conventional produce. Find the data for malathion. Assume the minimum exposure for both groups is zero. Otherwise, all of the information you need for a five-number summary is provided. (Note that the percentiles are the values with that percent of the data at or below the value. For instance, the median is the 50th percentile.) a. Create a five-number summary for the malathion exposure values for each group (organic and conventional). b. Construct side-by-side skeletal boxplots for the malathion exposure values for the two groups. Write a few sentences comparing them. c. Notice that in each case the mean is higher than the median. Explain why this is the case.

Answer: a. Organic: 0, 0.2, 0.2, 0.4, 1.3; conventional: 0, 0.5, 1.4, 3.4, 8.3 b. See the figure below. The malathion values are much more spread out for the conventional group than for the organic group. The median is much lower for the organic group, as are all of the other summary measures except the minimum, which is 0 for both groups of children.

c. The mean is higher than the median in both cases because the values are strongly skewed to the right, and thus the high values are pulling the mean higher.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

ANSWERS TO THE REFLECTION QUESTIONS 1. Suppose you are considering two different cities as a place to live, and you learn that the mean for the temperature at noon for the past 365 days was about 65 degrees for both cities. Explain what additional features you would want to know about the set of daily temperatures before deciding which city you prefer.

Answer: You would want to know about the variability. Is the temperature always close to 65 degrees, or is it much higher in the summer and much lower in the winter?

2. For the employees of a very large company, do you think outliers would be more likely to exist in the list of ages of all employees or in the list of salaries of all employees? Explain.

Answer: Outliers would be more likely to exist in the list of salaries than in the list of ages. Ages can only cover a limited range, probably from about 18 to 80 or so.

Most salaries would be in a relatively small range compared to some possible very large outliers for the highest paid executives. Those very high salaries would be outliers.

3. Three visual displays for quantitative variables are stemplots, histograms, and boxplots. They can be used for investigating data for a single group or for comparing data for two or more groups. What is one advantage of each type of display over the others?

Answer: An advantage of stemplots is that they display the actual data values (or a rounded off version of them). One advantage of histograms is that they can be used for large data sets, unlike stemplots. Both stemplots and histograms can be used to display and compare shapes. One advantage of boxplots is that they can be used to compare multiple groups by displaying one for each group, using the same scale.

4. Suppose a research study investigating sleep habits of college students found that the median amount of sleep per night across all participants was 7.2 hours, and the mean was 8 hours. One news report stated, “Although the average was 7.2 hours, one participant slept 13 hours, almost 6 hours more than normal.” Explain what two errors were made in the news report.

Answer: First, the use of the word “average” when referring to the median is misleading, since that term usually refers to the mean. Next, the word “normal” is used in reference to the median. The normal range is likely to be quite large, and could extend as high as 13 hours. So, referring to 13 hours as being almost 6 hours more than normal is not at all accurate.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 7: Summarizing And Displaying Measurement Data

ANSWERS TO THE MINI PROJECT 1. Find a set of data that has meaning for you. Internet sites are good sources for data. For instance, find ads for something you would like to buy, sports scores, weather data, and so on. Using the methods given in this chapter, summarize and display the data in whatever ways are most useful. Give a written description of interesting features of the data.

Answer: The purpose of this mini-project is to start with a set of numbers and bring meaning to them through useful summaries. The summaries provided should give the reader as much information as he or she would get by closely studying the data. Be sure all features of the data are covered, including some form of average or typical value, variability and shape if appropriate. Any outliers should also be discussed.

2. Measure your pulse rate 25 times over the next few days, but don’t take more than one measurement in any 10-minute period. Record any unusual events related to the measurements, such as if one was taken during exercise or one was taken immediately upon awakening. Create a stemplot and a five-number summary of your measurements. Give a written assessment of your pulse rate based on the data.

Answer: The main purpose of this mini-project is to notice that something as simple as your own pulse has variability, some of which can be explained by events and some of which cannot. Be sure the stemplot has enough branches (three is not enough) and that the written assessment includes features of average, variability, shape and possible outliers.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 8: Bell-Shaped Curves And Other Shapes

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 8: BELL-SHAPED CURVES AND OTHER SHAPES

TABLE OF CONTENTS Answer to the Exercise Questions......................................................................................1 Answer to the Reflection Questions.................................................................................13 Answer to the Mini Projects............................................................................................. 14

ANSWER TO THE EXERCISE QUESTIONS 1. Using Table 8.1, a computer, or a calculator, determine the percentage of the population falling below each of the following standard scores: a. –1.00 b. 1.96 c. 0.84 Answer: a. 16% b. 97.5% c. 80% 2. In each of the following cases, explain how you know that the population of measurements could not be normally distributed. a. The population is all families in the world that have exactly four children, and the measurement is the number of teenagers in the family. b. The population is all college students and the measurement is number of hours per week the student exercises. The mean is 4.5 hours, and the standard deviation is also 4.5 hours.

Answer: a. The only possible values are 0, 1, 2, 3, or 4. b. The values cannot be negative (exercise hours), but 0 is only one standard deviation below the mean. So it is not possible to get any values more than one standard deviation below the mean. That can’t be true for a normally distributed population.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 8: Bell-Shaped Curves And Other Shapes

3. Using Table 8.1, a computer, or a calculator, determine the percentage of the population falling above each of the following standard scores: a. 1.28 b. –0.25 c. 2.33 Answer: a. 1 − 0.9 = 0.1, or 10% b. 1 − 0.4 = 0.6, or 60% c. 1 − 0.99 = 0.01, or 1% 4. Using Table 8.1, a computer, or a calculator, determine the standard score that has the following percentage of the population below it: a. 25% b. 75% c. 45% d. 98% Answer: a. −0.67 b. 0.67 c. −0.13 d. 2.05 5. Give an example of a population of measurements that would not have a normal distribution for each of the following reasons. a. The measurement can only result in a small number of possible values, instead of the continuum over a substantial range that is required for a normal distribution. b. The measurement cannot go below 0, but is likely to be skewed to the right because it is extremely high for a small subset of all individuals.

Answer: a. Answers will vary. One example is the one given in Exercise 8.2a, the number of teenagers in families with four children. b. Answers will vary. One example is the one given in Exercise 8.2b, the number of hours college students exercise.

6. Using Table 8.1, a computer, or a calculator, determine the standard score that has the following percentage of the population above it: a. 2% b. 50% c. 75% d. 10% © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 8: Bell-Shaped Curves And Other Shapes

Answer: a. 2.05 (same as 98% below). b. 0.00 (same as 50% below). c. −0.67 (same as 25% below). d. 1.28 (same as 90% below).

7. Using Table 8.1, a computer, or a calculator, determine the percentage of the population falling between the two standard scores given: a. –1.00 and 1.00 b. –1.28 and 1.75 c. 0.0 and 1.00 Answer: a. 84% − 16% = 68% b. 96% − 10% = 86% c. 84% − 50% = 34% 8. The 84th percentile for the Stanford-Binet IQ test is 115. (Recall that the mean is 100 and the standard deviation is 15.) a. Verify that this is true by computing the standardized score and using Table 8.1. b. Draw pictures of the original and standardized scores to illustrate this situation, similar to the pictures in Figure 8.4.

Answer: a. The standardized score is (115 − 100)/15 = 1.00; by Table 8.1, this is the 84th percentile. b. Both pictures are identical bell-shaped curves. The top picture should be centered at 100 and have a line at 115, with 84% of the area below the line at 115 and 16% of the area above it. The bottom picture should look identical, except it is centered at 0 and has the line at 1 instead of 115. See figures below.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 8: Bell-Shaped Curves And Other Shapes

9. Find the percentile for the observed value in the following situations: a. Quantitative reasoning GRE score of 149 (mean = 154.3, s.d. = 9.6). b. Stanford-Binet IQ score of 97 (mean = 100, s.d. = 15). c. Adult female’s height of 68 inches (mean = 65 inches, s.d. = 2.5 inches). d. Adult male’s height of 68 inches (mean = 70 inches, s.d. = 3 inches).

Answer: a. The standardized score is (149 – 154.3)/9.6, or about −0.55, which is the 29th percentile. b. The standardized score is (97 − 100)/15 = −0.2, which is the 42nd percentile. c. The standardized score is (68 − 65)/2.5 = 1.2. This is approximately the 88.5th percentile. d. The standardized score is (68 – 70)/3 = −0.67, which is the 25th percentile.

10. Draw a picture of a bell-shaped curve with a mean value of 100 and a standard deviation of 10. Mark the mean and the intervals derived from the Empirical Rule in the appropriate places on the horizontal axis. You do not have to mark the vertical axis. Use Figure 8.6 as a guide.

Answer: The picture should show a bell-shaped curve centered on 100. The interval from 90 to 110 should encompass 68% of the area, from 80 to 120 should encompass 95% of the area, and 70 to 130 should encompass 99.7% of the area. See the figure below.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 8: Bell-Shaped Curves And Other Shapes

11. A statistician was consulted by a company called Molegon, which removed unwanted moles from gardens in Britain. The company had kept records showing that mole weights in its region were approximately normally distributed with mean of 150 grams and standard deviation of 56 grams (Stewart, 1994). a. Draw a normal curve showing the distribution of weights. Mark the weights corresponding to the mean, and corresponding to 1, 2, and 3 standard deviations above and below the mean. b. A new regulation was being put into place that restricted the capture of moles weighing less than 68 grams or more than 211 grams. What percentage of the population of moles could be legally captured under this new regulation?

Answer: a. The picture should be a bell-shaped curve centered on the mean of 150.

The interval encompassing 1, 2, and 3 standard deviations above and below the mean are as follows: One standard deviation: 94 to 206 grams; two standard deviations: 38 to 262 grams; three standard deviations: –18 to 318 grams. (Clearly the distribution isn’t exactly correct, because weights cannot be negative.) The mean and the intervals are marked on the horizontal axis in the figure below.

b. The percentage that can be legally captured is the percentage between the values of 68 grams and 211 grams for a normally distributed population with mean and standard deviation of 150 grams and 56 grams, respectively.

Standardized score for 68 grams = (68 – 150)/56 = –1.46. Standardized score for 211 grams = (211 – 150)/56 = 1.09. From Table 8.1, about 86% of weights are below 211 grams and about 7% of weights are below 68 grams.

Therefore, about 86% – 7% = 79% are between 56 grams and 211 grams and could be legally captured. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 8: Bell-Shaped Curves And Other Shapes

12. Every time you have your cholesterol measured, the measurement may be slightly different due to random fluctuations and measurement error. Suppose that for you, the population of possible cholesterol measurements if you are healthy has a mean of 190 and a standard deviation of 10. Further, suppose you know you should get concerned if your measurement ever gets up to the 97th percentile. What level of cholesterol does that represent?

Answer: The 97th percentile corresponds to a standardized score of 1.88 or a cholesterol level of 190 + (1.88)(10) = 208.8.

13. Use Table 8.1 to verify that the Empirical Rule is true. You may need to round off the values slightly.

Answer: The percentage of values within one standard deviation of the mean are the percentage of values between + 1 and −1, 84% – 16% = 68%. The percentage of values within two standard deviations of the mean are those between +2 and −2, or about 97.5% – 2.5% = 95%. The percentage within three standard deviations is about 99.87% − 0.13%, or 99.7%.

14. Remember that the whiskers on a boxplot can extend 1.5 interquartile ranges (IQRs) from either end of the box, and values beyond that range are considered to be outliers. For normally distributed data, the mean and median are equal and are both centered in the middle of the box in a boxplot. a. Draw the “box” part of a boxplot for normally distributed data. What fraction of an IQR is the distance between the mean/median and the ends of the box? (Hint: How many IQRs is the width of the box?) b. For normally distributed data, how many IQRs above or below the mean/median would a data value have to be in order to be considered an outlier? (Hint: Combine the information at the beginning of this exercise and your answer in part (a).

Answer: a. The width of the box is one IQR, so half of the IQR is between the mean/median and each end of the box. b. An outlier is a value that is more than 1.5 IQRs from the ends of the box.

The ends of the box are ½ or 0.5 IQRs below and above the mean/median.

Therefore, an outlier would be a total distance of two IQRs below or above the mean/median.

15. Recall from Chapter 7 that the interquartile range (IQR) covers the middle 50% of the data. a. What range of standardized scores is covered by the interquartile range for normally distributed populations? In other words, what are the standardized scores for the lower quartile (the 25th percentile) and the

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 8: Bell-Shaped Curves And Other Shapes

upper quartile (the 75th percentile)? (Hint: Draw a standard normal curve and locate the 25th and 75th percentiles using Table 8.1.) b. How many standard deviations are covered by the interquartile range for a normally distributed population? (Hint: Use the standardized scores found in part (a) for the lower and upper quartiles.) c. Data values are outliers for normally distributed data if they are more than two IQRs away from the mean. (See Exercise 14.) At what percentiles (at the upper and lower ends) are data values considered outliers for normally distributed data? Use your answers to parts (a) and (b) to help you determine the answer to this question. (Software required.)

Answer: a. The standardized score for the lower quartile (25% below it) is −0.67. The standardized score for the upper quartile (75% below it) is +0.67. So the range is from −0.67 to +0.67. b. Using the answers in part (a), the interquartile range is from −0.67 to +0.67 and covers 1.34 standard deviations. c. From part (a), one IQR is 1.34 standard deviations. So 2  IQR = 2.68 standard deviations. Therefore, two IQRs below the mean is a standardized score of −2.68, which corresponds to about the 0.4 percentile and two IQRs above the mean is +2.68, which corresponds to about the 99.6 percentile.

16. Give an example of a population of measurements that you do not think has a normal curve, and draw its frequency curve.

Answer: There are numerous examples that could be used here. Make sure the horizontal axis covers the range of possible values, and that the height of the curve represents reasonable relative frequencies for those values.

17. A graduate school program in Statistics will admit only students with quantitative reasoning GRE scores in the top 30%. What is the lowest GRE score it will accept? (Recall the mean is 154.3 and the standard deviation is 9.6, and round your answer to the closest whole number.)

Answer: They should accept students with a standardized score of 0.52 or higher, corresponding to a GRE score of 154.3 + (0.52)(9.6) = 159.292. Rounding to a whole number means they accept scores of 159 or higher.

18. Recall that for Stanford-Binet IQ scores the mean is 100 and the standard deviation is 15. a. Use the Empirical Rule to specify the ranges into which 68%, 95%, and 99.7% of Stanford-Binet IQ scores fall. b. Draw a picture similar to Figure 8.6 for Stanford-Binet scores, illustrating the ranges from part (a). © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 8: Bell-Shaped Curves And Other Shapes

Answer: a. 68% of scores are between 85 and 115, 95% are between 70 and 130, and 99.7% are between 55 and 145. b. The picture should be centered on 100 and have the intervals specified in part (a) marked. See the figure below.

68%

95% 99.7% 55

70

85

100

115

130

145

19. For every 1000 live births in the United States, the number that are cesarean deliveries follows, approximately, a normal curve with a mean of 320 and standard deviation of 15. A large hospital is being investigated because 365 of the previous 1000 live births were cesarean deliveries. Is that an unusually high number? Explain.

Answer: The mean is 320 and the standard deviation is 15, so the standardized score corresponding to 365 cesarean deliveries is (365 − 320)/15 = 3.0, which is at the 99.87th percentile. So yes, it is an unusually high number, and 99.87% of the time there would be fewer than that many cesarean deliveries in 1000 births.

20. Suppose a candidate for public office is favored by only 48% of the voters. If a sample survey randomly selects 2500 voters, the percentage in the sample who favor the candidate can be thought of as a measurement from a normal curve with a mean of 48% and a standard deviation of 1%. Based on this information, how often would such a survey show that 50% or more of the sample favored the candidate?

Answer: Notice that 50% is two standard deviations above the mean, or a standardized score of 2.00, which is about the 97.75th percentile. So about 2.25%, or between 2% and 2.5% of such surveys, should show a majority favoring the candidate. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 8: Bell-Shaped Curves And Other Shapes

21. Suppose you record how long it takes you to get to work or school over many months and discover that the times are approximately bell-shaped with a mean of 15 minutes and a standard deviation of 2 minutes. How much time should you allow to get there to make sure you are on time 90% of the time?

Answer: You want your allowed travel time to be at the 90th percentile, which corresponds to a standardized score of 1.28 or travel time of 15 + (1.28)(2) = 17.56 minutes.

22. Give an example of a measurement for which you would prefer your value relative to the population to be: a. Around the 10th percentile b. Around the 50th percentile c. Around the 90th percentile Answer: a. Answers will vary. Examples are situations for which you want your value to be relatively low, such as number of accidents over a lifetime of driving. b. Answers will vary. Examples are situations for which you want your value to be close to average, such as your shoe size. c. Answers will vary. Examples are situations for which you want your value to be relatively high, such as your score on a standardized test.

23. Math SAT scores for students admitted to a university are bell-shaped with a mean of 520 and a standard deviation of 60. a. Draw a picture of these SAT scores, indicating the cutoff points for the middle 68%, 95%, and 99.7% of the scores. See Figure 8.6 for guidance. b. A student had a math SAT score of 490. Find the standardized score for this student and draw where their score would fall on your picture in part (a).

Answer: a. See the figure below.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 8: Bell-Shaped Curves And Other Shapes

68%

95% 99.7% 340

b. z =

400

460

520 Math SAT

580

640

700

490 − 520 − 30 = = −0.5 . It should be drawn halfway between 460 and 60 60

520 on the picture above.

24. According to data obtained by the author from records of a large blood bank, the mean healthy adult temperature is around 97.9°F, not the previously assumed value of 98.6°F. The standard deviation is about 0.7°F and the population of healthy temperatures is bell-shaped. a. What proportion of the population have temperatures at or below the presumed norm of 98.6°F? b. Would it be accurate to say that the normal healthy adult temperature is 97.9°F? Explain.

Answer: a. The standardized score for 98.6 degrees is (98.6 – 97.9)/0.7 = 1.0, so about 84% or 0.84 of the population have temperatures below 98.6 degrees. b. No, that’s the average. Normal temperatures cover a wide range.

25. Remember from Chapter 7 that the range for a data set is found as the Difference between the maximum and minimum values. Explain why it makes sense that for a bell-shaped data set of a few hundred values, the range should be about 4 to 6 standard deviations. (Hint: Use the Empirical Rule.)

Answer: For bell-shaped data, the Empirical Rule tells us that if we had an infinite sample, 99.7% of the data would fall within 3 standard deviations of the mean, and 95% would fall within 2 standard deviations of the mean. For a finite data set, depending on how large it is, it thus makes sense that all of the data will fall within 2 or 3 standard deviations on either side of the mean. Thus, the entire range of the data will be 4 to 6 standard deviations. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 8: Bell-Shaped Curves And Other Shapes

26. Suppose that you were told that scores on an exam in a large class you are taking ranged from 50 to 100 and that they were approximately bell-shaped. a. Estimate the mean for the exam scores. b. For a bell-shaped data set of a few hundred values, the range should be about 4 to 6 standard deviations.

Estimate the standard deviation for the exam scores, using that result and the information in this problem. c. Suppose your score on the exam was 80. Explain why it is reasonable to assume that your standardized score is about 0.5. d. Based on the standardized score in part (c), about what proportion of the class scored higher than you did on the exam?

Answer: a. The mean would be about 75, in the middle of the range (because the bellshape is symmetric). b. The range is 50, so the standard deviation is probably between about 50/4 = 12.5 to 50/6 = 8.33. c. The mean is 75, so your score is 5 points above the mean. From part (b), the standard deviation is around 10, so it is reasonable to assume that your standardized score is around 5/10 = 0.5. d. Using Table 8.1, about 0.31 or 31% 27. Over many years, rainfall totals for Sacramento, CA, in January ranged from a low of about 0.05 inch to a high of about 19.5 inches. The median was about 3.1 inches. Based on this information, explain how you can tell that the distribution of rainfall values in Sacramento in January cannot be bell-shaped.

Answer: The mean is much higher than the median, so the rainfall values must be skewed or have large outliers.

28. Recall that quantitative reasoning GRE scores are approximately bell shaped with a mean of 154.3 and standard deviation of 9.6. The minimum and maximum possible scores on this part of the GRE exam are 130 and 170, respectively. a. What is the range for these GRE scores? b. Based on the Empirical Rule, for a large bell-shaped data set, the range should cover about 4 to 6 standard deviations. Does that make sense for these GRE scores? Explain.

Answer: a. Range = 40 (found as 170 – 130). b. The standard deviation is 9.6, so the range of 40 is 40/9.6 = 4.2 standard deviations. Yes, based on the Empirical Rule the range should be about 4 to 6 standard deviations, and in this case it is 4.2, so that makes sense.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 8: Bell-Shaped Curves And Other Shapes

29. Refer to Example 8.1 on pp. 161–162, and the corresponding Figure 8.3 showing the distribution of sleep hours for college students. The mean is 7.0 hours, and the standard deviation is 1.7 hours. a. Use the Empirical Rule to determine the ranges into which about 68%, 95%, and 99.7% of college students’ sleep hours should fall. b. Draw a picture illustrating the ranges you found in part (a), indicating the cutoff points for the middle 68%, 95%, and 99.7% of sleep hours of college students c. The table below shows some ranges of hours and the number of students out of the 361 in the sample whose sleep hours fell into these ranges.

(Many students reported exactly 7.0 hours, so half of them were assigned to the range 5.3 to 7.0, and half to the range 7.0 to 8.7.) Compare these 361 students to what would be expected if the Empirical Rule held exactly, using the intervals found in part (a).

Answer: a. 68% of sleep hours should be between 5.3 and 8.7 hours, 95% should be between 3.6 and 10.4 hours, and 99.7% should be between 1.9 and 12.1 hours. b. See the figure below.

68%

95% 99.7% 1.9

3.6

5.3

7.0

8.7

10.4

12.1

Sleep hours

c. The table below shows the expected and actual percentages for the three ranges. The actual percentages of 65.1%, 95.3% and 100% are very close to the Empirical Rule percentages of 68%, 95% and 99.7%.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 8: Bell-Shaped Curves And Other Shapes

Range

Expected %

Number (out of 361)

Actual %

5.3 to 8.7

68%

116 + 119 = 235

235/361 is 65.1%

3.6 to 10.4

95%

235 + 58 + 51 = 344

344/361 is 95.3%

1.9 to 12.1

99.7%

344 + 9 + 8 = 361

361/361 is 100%

ANSWER TO THE REFLECTION QUESTIONS 1. What characteristics of a measurement variable would help you decide whether or not the frequency curve for the variable is likely to be bell-shaped?

Answer: To be bell shaped, the range from the mean plus and minus two to three standard deviations must be possible values. Another characteristic that would help you decide is whether measurements are likely to be clustered around the middle, and then become less and less common as you move away from the middle in either direction. Another clue is that the measurement can’t be bell shaped if there are only a few possible values.

2. If you knew the five-number summary for a measurement variable that was not bell shaped, explain whether you could use that information to find the mean and standard deviation.

Answer: You could not use the five-number summary to find the mean and standard deviation. For instance, the five-number summary includes the median, but that is only equal to the mean if the distribution is symmetric. So you would need more information than a five-number summary even to find the mean.

3. For the following variables, explain whether you would prefer the measurement to be in the 10th percentile, the 50th percentile, or the 90th percentile. a. Your yearly income relative to others your age b. Amount you owe for student loans relative to others in school with you c. Temperature on the day of your graduation relative to the 365 daily temperatures for the year Answer: a. You would want your income to be high, so the 90th percentile would be best. b. You would want the amount you own in student loans to be low, so the 10th percentile would be best. c. You would want the temperature to be moderate, so it’s likely that the 50th percentile would be best.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 8: Bell-Shaped Curves And Other Shapes

4. If you took an exam taken by hundreds or thousands of other students and did not know the scale used to grade the exam, would you prefer to be given your raw score or your standardized score? Explain.

Answer: You would learn more by knowing your standardized score than by knowing your raw score. Your raw score would be meaningless unless you know the range of possible values.

ANSWER TO THE MINI PROJECTS 1. An article by researchers in Britain (Linton et al., 2016) reviewed 99 different self-report measures for measuring well-being. An online search should provide access to the research report. (See the References for this chapter for the full citation.) Table 1 in the paper lists the 99 different tests. Find one of them that has at least 20 items. Using online searches and consulting original research articles, find out what you can about the mean, standard deviation, and shape of the scores on that test for any populations to which it was administered.

Answer: The purpose of this Mini-Project is to investigate the characteristics of welldefined assessment tools used by researchers and clinicians. Each of the 99 measures listed in the referenced paper has a research article describing the assessment tool and its characteristics. In addition to learning how to find these research articles, students will learn how to read them to determine the characteristics of the assessment tool.

2. One of the 99 tests identified by Linton et al. (2016) is called the Modified BBC Subjective Well-Being Scale, abbreviated as the BBC-SWB. A research article

by Pontin et al. (2013) investigated the statistical properties of that test. Find that research article. (See the References for this chapter for the full citation.)

Figure 2 in the article provides histograms of the total scores for over 23,000 people and histograms for scores on three of the subscales. The paper also reports these summary statistics: “For the total scale, the mean score for the whole sample was 73.21 (Median = 74; SD = 17.02, Min 24, Max 120, interquartile range 62–86).” a. Use this information to describe the distribution of scores on this test for the population represented by the sample. Include things like a fivenumber summary and the intervals given by the Empirical Rule. b. The research article also gives summary statistics for the three subscales: psychological well-being, physical health and well-being, and relationships.

Find those statistics in the article, and describe each subscale using the same kinds of descriptions as in part (a) for the total scale.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 8: Bell-Shaped Curves And Other Shapes

c. Using an Internet search, find the BBC Well-Being Scale and take the test.

Determine where your own score fits within the distributions you described in part (b).

Answer: The purpose of this project is to investigate in-depth an assessment tool for measuring well-being. In the process of investigating it, students will learn how to read the relevant journal article. In addition, they are asked to find the assessment tool online and take the test themselves. This process will help them connect their own experience to the summary data provided by the researchers.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER: 9 PLOTS, GRAPHS AND PICTURES

TABLE OF CONTENTS Answer to the Exercise Questions......................................................................................1 Answer to the Reflection Questions................................................................................ 16 Answer To the Mini Projects .............................................................................................17

ANSWER TO THE EXERCISE QUESTIONS 1. Use the pie charts in Figure 9.1 to create a bar graph comparing educational attainment for individuals aged 25 to 34 in 1940 and 2021.

Answer: See figure below.

70

Year 1940 2021

60

Percent

50 40 30 20 10 0 Year Degree

1940 2021

1940 2021

1940 2021

1940 2021

No HS degree

HS degree

Some college

Bachelor +

2. Suppose a real estate company in your area sold 100 houses last month, whereas their two major competitors sold 50 houses and 25 houses, respectively. The top company wants to display its better record with a pictogram using a simple two-dimensional picture of a house. Draw two pictograms displaying this information, one of which is misleading and one of which is not. (The horizontal axis should list the three companies and the vertical axis should list the number of houses sold.)

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures

Answer: The misleading pictogram should have realistic looking houses, so that the house with height of 100 has twice the width of the house with height of 50 and four times the area. The one that is not misleading should have distorted pictures of all but one of the houses. For instance, if the one with height of 25 is drawn realistically then the other two will be very tall and thin.

3. Give the name of a type of statistical picture that could be used for each of the following kinds of data. a. One categorical variable b. One measurement variable c. Two categorical variables d. Two measurement variables Answer: a. Pie chart, bar graph or pictogram b. Stemplot, histogram or boxplot c. Bar graph or side by side pie chart. d. Scatter plot or line graph across time (if time is one of the two variables) 4. One method used to compare authors or to determine authorship on unsigned writing is to look at the frequency with which words of different lengths appear in a piece of text. For this exercise, you are going to compare your own writing with that of the author of this book. a. Using the first full paragraph of this chapter (not the Thought Questions), create a pie chart with three segments, showing the relative frequency of words of 1 to 3 letters, 4 to 5 letters, and 6 or more letters in length. (Do not include the numbered list after the paragraph.) b. Find a paragraph of your own writing of at least 50 words. Repeat part (a) of this exercise for your own writing. c. Display the data in parts (a) and (b) of this exercise using a single bar graph that includes the information for both writers. d. Discuss how your own writing style is similar to or different from that of the author of this book, as evidenced by the pictures in parts (a) to (c). e. Name one advantage of displaying the information in two pie charts and one advantage of displaying the information in a single bar graph.

Answer: a. There are 109 words of which 46 (42.2%) are words of 1 to 3 letters, 30 (27.5%) are words of 4 or 5 letters and 33 (30.3%) are 6 or more letters. The pie chart should have three segments, reflecting those percentages. c. The bar graph should have six bars. They should be arranged in pairs, displaying words of 1 to 3 letters for each writer, then 4 or 5 letters for each writer, then 6 or more letters. The heights should be relative

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures

frequencies rather than absolute frequencies because the total number of words are different for the two writers. e. The pie charts are better for getting a picture of how the individual writer uses words, whereas the bar graph is better for comparison.

5. An article in Science (23 January 1998) reported on a “telephone survey of 2600 parents, students, teachers, employers, and college professors” in which people were asked the question, “Does a high school diploma mean that a student has at least learned the basics?” Results are shown in Table 9.2. a. The article noted that “there seems to be a disconnect between the producers [parents, teachers, students] and the consumers [professors, employers] of high school graduates in the United States. Create a bar graph from this study that emphasizes this feature of the data. b. Create a bar graph that deemphasizes the issue raised in part (a).

Answer: a. There are a couple of things you can do to emphasize the distinction.

Probably the most important is to group the parents, teachers and students in 3 adjacent bars, then have a gap, and group the professors and employers in adjacent bars. Use only the percentages who answered “yes.” b. Some ways to mask the differences are to mix up the groups and include bars for both “yes” and “no” next to each other. This would create the effect of having ten bars, with high and low heights intermingled.

6. Figure 9.14a, which displays rising postal rates, is an example of a graph with misleading units because the prices are not adjusted for inflation. The graph actually has another problem as well. Use the checklist in Section 9.6 to determine the problem, then redraw the graph correctly (but still use the unadjusted prices). Comment on the difference between Figure 9.14a and your new picture. (Hint: Look at the scale on the horizontal axis.)

Answer: The problem is that the years on the horizontal axis are not evenly spaced.

They reflect when prices changed. If redrawn correctly, the increases over the years are less even. To be truly accurate, there should be immediate jumps at the times the prices changed, rather than showing gradual change. See the figure below.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures

Rising Cost of First Class Stamp 0.65

Price (dollars)

0.60

0.55

0.50

0.45

0.40 5/07

1/12

4/16

8/21

7/23

Month of price change

7. A report by Pew Research (Fry, 2022) presented data from 1948 to 2022 showing trends in job longevity for young adults (aged 18 to 34) in the United States workforce. Table 9.3 provides a small subset of the data in the report.

For each of the years 2000, 2010, and 2020, the percentage of workers in this age group who had worked for their current employer for less than 12 months and for 3 years or more are shown. The remainder presumably had worked for their current employer for more than 12 months but less than 3 years. a. Draw a bar graph representing the data in Table 9.3. Explain how you decided which variable (year or job tenure) to use for separate sets of bars and which one to use for the individual bars within each set. b. Draw a segmented bar graph representing the data in Table 9.3. Decide whether or not to include the category “More than 12 months but less than 3 years.” Explain your reasoning. c. The “Great Recession” occurred from 2007 to 2009. When a recession occurs, it’s harder to change jobs. Explain how that information helps you understand the numbers in Table 9.3.

Answer: a. Before constructing the graph you should add the missing category of “between 12 months and 3 years.” The graph can be drawn in two ways.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures

You can either display all three categories for 2000 first, followed by 2010 and then 2020. Or you can display all three years together for “Less than 12 months” followed by all three years together for “between 12 months and 3 years” and then “3 years or more.” The decision about which one to use depends on whether you are more interested in comparing job longevity within each time period, or in comparing how things have changed over time. The graph should include the source of the data, and the axes should be clearly labeled and include a descriptive title for the picture. See figure below (created with Minitab Version 17) for an example.

Job Tenure for Young Adults Across Time 50

40

Percent

30

20

10

0 Year

2000 2010 2020

2000 2010 2020

2000 2010 2020

< 12 months

12 months to 3 years

> 3 years

Time in Job

b. You should include the category of “between 12 months and 3 years” so that the total adds to 100% for each bar. See figure below (created with Minitab Version 17).

Job Longevity for Young Adults Length > 3 years 12 months to 3 years < 12 months

100

Percent

80

60

40

20

0 Years

2000

2010

2020

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures

c. As can be seen in the graphs in parts (a) and (b), the year with the highest percentage with job tenure of 3 years or more was 2010, just after the Great Recession. That was also the year with the lowest percentage who had been in their jobs for less than 12 months. These numbers indicate that young adults did not change jobs as much in that time period.

8. Table 9.4 lists the number of aggravated assaults in the United States per 100,000 people for the years 2010 to 2020. (Source: FBI Crime Data Explorer, https://cde.ucr.cjis.gov/) a. Draw two line graphs representing the trend in aggravated assaults over time. Draw the first graph with the vertical axis starting at 0, and the other graph with it starting close to the minimum of the values in the table.

Describe what conclusion you would make about the trend from each of the two graphs. Are the conclusions similar or different? b. Using a subset of consecutive years, draw a line graph showing an ominous increasing trend in aggravated assaults. Using a different subset of consecutive years, draw a line graph showing a steep decrease in aggravated assaults. Do either of the two graphs represent the pattern that is apparent in the full data set? Explain.

Answer: a. The two graphs (created using Minitab Version 17) are shown below. From the graph starting at 0 (on the left) it looks like the rate of assaults has stayed relatively constant across time. The graph starting at 230 makes it clear that the rate of assaults dropped from 2010 to 2014, and then started increasing, ending at a higher value in 2020 than the value in 2010.

Aggravated Assault in the United States across Time

Aggravated Assault in the United States across Time 280

250

Aggravated Assault per 100,000

Aggravated Assault per 100,000

300

200

150

100

50

0

270

260

250

240

230 2010

2012

2014

2016

Year

2018

2020

2010

2012

2014

2016

2018

2020

Year

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures

b. For the graph showing an increasing trend, use the years from 2014 to 2020, shown on the left below. For the graph showing a drop, use the years from 2010 to 2013 or 2014, shown on the right below. Neither of these graphs represent the full picture across all years.

Aggravated Assaults in the United States across Time

Aggravated Assaults in the United States across Time 255

Aggravated Assault per 100,000

Aggravated Assault per 100,000

280

270

260

250

240

250

245

240

235

230

230 2014

2015

2016

2017

2018

2019

2020

2010

2011

Year

2012

2013

2014

Year

9. Table 9.5 displays the percentages of blood types within each race/ethnicity group in the United States. The percentages are based on data from over three million blood-bank donors collected in five cities in the United States from 1991 to 2000 (Garratty et al., 2004). a. Create a pie chart showing the distribution of blood types for all donors combined. Make sure you include all of the features of a good statistical picture. b. Suppose you wanted to create a pie chart showing the breakdown of ethnicities for people who had type O blood. Explain whether you could do it by displaying the percentages shown in the column for type O. If not, what would you need to do to create appropriate percentages?

Answer: a. See the figure below (created with Minitab Version 17).

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures Blood types for all donors AB 4.10%

Category O A B AB

B 12.20%

O 46.60% A 37.10%

b. No. You would need to know what percentage of those with Type O blood fell into each ethnic group. The table only shows what percentage of each ethnic group has Type O blood.

10. As described in Exercise 9, the distribution of blood types in the United States shown in Table 9.5 is based on data from over three million donors to bloodbanks from 1991 to 2000 (Garratty et al., 2004). a. If you wanted to create a stacked bar chart showing all of the data in the table, explain whether each bar would consist of one ethnicity or of one blood type. b. Create a statistical picture incorporating all of the information given.

Answer: a. It would make more sense to show how the blood types are distributed within each race or ethnicity. Therefore, each bar would represent one race or ethnicity. b. See the stacked bar graph below (created using Minitab Version 17).

Blood Type by Race or Ethnicity Blood type AB B A O

100%

Percentage

80%

60%

40%

20%

0

White non-Hispanic Hispanic Black non-Hispanic

Asian

Native American All donors

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures

11. Find an example of a statistical picture in a newspaper, magazine, or on the Internet that has at least one of the problems listed in Section 9.5, “Difficulties and Disasters in Plots, Graphs, and Pictures.” Explain the problem.

If you think anything should have been done differently, explain what and why.

Include the picture with your answer.

Answer: This will differ by individual.

12. Find a graph that does not start at zero. Redraw the picture to start at zero.

Discuss the pros and cons of the two versions.

Answer: This will differ by individual. In general, not starting at zero allows changes to be seen more readily, while starting at zero emphasizes the actual magnitude of the measurements.

13. According to an article in The Seattle Times (Meckler, 2003), living organ donors are most often related to the organ recipient. Table 9.6 gives the percentages of each type of relationship for all 6613 cases in which an organ was transplanted from a living donor in 2002 in the United States. Create a pie chart displaying the relationship of the donor to the recipient, and write a few sentences describing the data.

Answer: The pie chart should have six segments, with almost a third of it labeled as “sibling” then equal parts of almost a fifth each labeled as “child” and “not related” and the remaining parts almost equal, with 13% labeled as “parent,” 11% as “spouse” and 8% as “other relative.” See the figure below.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures

Pie Chart of Relationship

Not related Sibling

Other relativ e

Spouse

C hild Parent

14. A report by the Pew Research Center (DeSilver, 2022) looked at trends in teen summer employment from 1948 to 2021. Table 9.7 shows data for 1948, 1988, and 2021. For each year, the table gives the percentage and the approximate number of teens (aged 16 to 19) who were employed in the summer months (June to August) and during the rest of the year (September to May). If an individual was employed all year, then they are counted in both seasons. a. Create a bar graph for the percentages by placing bars for the two sets of months next to each other for each of the 3 years. Comment on how the percentages compare within each year, and then across years. b. Create a bar graph for the numbers using the same format as in part (a).

Comment on how the numbers compare within each year, and then across years. c. Discuss one advantage of using the percentages and one advantage of using the numbers, for comparing within and across years. d. Would it be appropriate to use a pie chart to display the percentages? If so, explain what the segments would be. If not, explain why not.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures

Answer: a. See figure below. For each year the percentage employed was higher in the summer than in the rest of the year. The percentages for 1948 and 1988 were similar, but they were much lower for 2021, indicating that a smaller percentage of teenagers were working in 2021 than in earlier years, for both seasons.

Percentage of Teenagers Employed 60%

Percent employed

50%

40%

30%

20%

10%

0 Year

Summer Rest of year

Summer Rest of year

Summer Rest of year

1948

1988

2021

b. See the figure below. Within each year, the number of teenagers employed was higher in the summer than during the rest of the year. The total numbers were largest in 1988 and smallest in 1948, for both seasons.

Number of teenagers employed 8

Number (in millions)

7 6 5 4 3 2 1 0

Season

SummerRest of year

SummerRest of year

SummerRest of year

Year

1948

1988

2021

c. The numbers provide information about how many teenagers were employed, which cannot be found from the percentages. However, by looking only at the numbers it is not evident that the percentages were lowest in 2021. That’s because the number of teenagers in the United States has increased over time.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures

d. A pie chart would not be appropriate. A pie chart shows how a total of 100% divides into categories. In this case, we are not breaking a total into parts. A separate pie chart could be created for each season in each year by computing the percentage not employed and having two segments, but that kind of pie chart would not be very informative for comparing across seasons or years.

15. For each of the following time series, do you think the long-term trend would be positive, negative, or nonexistent? a. The cost of a loaf of bread measured monthly from 1960 to the present month b. The temperature in Boston measured at noon on the first day of each month from 1960 to the present month c. The price of a basic computer, adjusted for inflation, measured monthly from 1970 to 2022 d. The number of personal computers sold in the United States measured monthly from 1970 to 2022 Answer: a. Positive, because of inflation b. Nonexistent (ignoring possible global warming) c. Negative. Computer prices have decreased as technology has become more advanced. d. Positive. The first personal computer available was in 1974. Very few people owned computers in the 1970s, and numbers have been increasing ever since.

16. For each of the time series in Exercise 15, explain whether there is likely to be a seasonal component.

Answer: a. No seasonal component in the price of bread b. There is a definite seasonal component in Boston temperatures. c. There is a possible seasonal component for computer prices, with sales at certain times of the year, such as during the December holidays. d. There is a definite seasonal component, with peaks around the December holidays and the beginning of the school year.

17. Global warming is a major concern because it implies that temperatures around the world are going up on a permanent basis. Suppose you were to examine a plot of monthly temperatures in one location for the past 50 years. Explain the role that the three time series components (trend, seasonal, cycles) would play in trying to determine whether global warming was taking place.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures

Answer: A trend would indicate global warming. Seasonal effects would be easy to separate out, but long cycles might lead to an erroneous conclusion that there was global warming. That’s why it’s important to examine data for as many years as possible when making this assessment.

18. If you were to present a time series of the yearly cost of tuition at your local college for the past 30 years, would it be better to first adjust the costs for inflation? Explain.

Answer: It would be better to adjust for inflation to have a fair comparison over the years.

19. The population of the United States rose from about 179 million people in 1960 to about 340 million people in 2023. Suppose you wanted to examine a time series to see if homicides had become an increasing problem over that time period. Would you simply plot the number of homicides versus time, or is there a better measure to plot against time? Explain.

Answer: You should plot the rate of homicides. Divide the number of homicides by the population first. Otherwise, an increasing number of homicides could simply be due to the increase in population. An increase in the rate would reflect a true increase in severity of the problem.

20. Many statistics related to births, deaths, divorces, and so on across time are reported as rates per 100,000 of population rather than as actual numbers.

Explain why those rates may be more meaningful as a measure of change across time than the actual numbers of those events.

Answer: Because the population has been constantly increasing, all statistics related to those kinds of events are also likely to increase. Those increases do not reflect a meaningful change, however, and an increase or decrease in rate is what matters.

21. Discuss which of the three components of a time series (trend, seasonal, and cycles) are likely to be present in each of the following series, reported monthly for the past 10 years: a. Unemployment rates b. Hours per day the average child spends watching television Answer: a. Unemployment rates are likely to have seasonal components and cycles.

Unemployment goes up at certain times of the year, such in May and June when many students graduate, and goes down at other times, such as during the December holiday season when seasonal workers are hired. And there are always long-term cycles in unemployment.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures

b. There are probably trends and seasonal components. For example, children may watch more television during the winter months when it’s too cold to play outside.

22. It has been noted that major wars are good for the economy. Explain which one of the components of an economic time series would be most likely to be influenced by a major war.

Answer: Major wars tend to cause upward cycles in economic time series. In other words, they are good for the economy.

23. Find an example of a statistical picture in a newspaper, magazine, or on the Internet. Answer the 12 questions in Section 9.6 for the picture. In the process of answering the questions, explain what (if any) features you think should have been added or changed to make it a good picture. Include the picture with your answer.

Answer: This differs by individual.

24. Which of the three components (trend, seasonal, and cycles) are likely to be present in a time series reporting interest rates paid by a bank each month for the last 10 years?

Answer: There are likely to be seasonal components and cycles for interest rates paid by banks. For the seasonal, they may offer attractive rates at certain times of the year, like when college students have just moved to the area and may be looking for a bank. But there are also long-term cycles, like there are in most economic time series.

25. Explain why it is important for most economic time series to be seasonally adjusted before they are reported.

Answer: The seasonal fluctuations in many time series would far outweigh the small changes from month to month, but it is those month to month changes that help indicate the health of the economy.

26. The American Association of Colleges and Universities (AACU) surveyed almost 500 executives and hiring managers responsible for hiring and promotions in U.S. companies to find out how they feel about recent college graduates. The survey report (Finley, 2021) presented detailed information about what matters to employers when hiring. One of the questions asked was how effective they thought college graduates are in “communicating the skills and knowledge they have gained in college” (Finley, 2021, p. 14). The results were reported in a figure similar to Figure 9.17.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures

a. Explain why this figure could be misleading. b. Create a more appropriate figure to display these numbers.

Answer: a. The width of the figures appear to represent the numbers in them. But the eye is drawn to the areas of the figures instead of the width. For example, the figure representing 40% is almost four times as wide as the figure representing 11%, but its area is substantially more than four times the area of the 11% figure. b. A bar graph would be an appropriate picture, with the heights of the bars representing the percentages. A pie chart would also be appropriate, because the percentages add to 100%. See the figure below (created using Minitab Version 17).

Effectiveness in Communicating Not at all 2.0%

Category Not at all Not very Somewhat Very

Not very 11.0% Very 40.0% Somewhat 47.0%

27. Refer to Figure 1 on page 691 of Original Source 11, linked to the companion website, “Driving impairment due to sleepiness is exacerbated by low alcohol intake.” a. What type of picture is in the figure? © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures

b. Write a few sentences explaining what you learn from the picture about lane drifting episodes under the different conditions.

Answer: a. It’s a bar graph. b. In all four time periods, there were fewest incidents with no impairment, and the most incidents with the combination of sleep deprivation and alcohol. The number of incidents with either one of those alone was similar for all time periods, with the order of the two changing back and forth.

After 30 minutes, there were a large number of incidents for the combined condition.

28. Refer to Figure 2 on page 691 of Original Source 11, linked to the companion website, “Driving impairment due to sleepiness is exacerbated by low alcohol intake.” a. What type of picture is in the figure? b. Write a few sentences explaining what you learn from the picture about subjective sleepiness ratings under the different conditions.

Answer: a. It’s a line graph across time. b. After about 40 minutes, the self-reported sleepiness leveled off at a very similar value for all three impairment conditions. Sleepiness increased even in the control condition, but always remained below all of the other conditions.

ANSWER TO THE REFLECTION QUESTIONS 1. What type(s) of statistical picture would be appropriate if you wanted to display the breakdown in proportions of adults in a country who fully support, partially support, and don’t support a controversial issue?

Answer: You could use a pie chart showing how the total breaks down into the opinion categories. You could also use a bar graph with the height of each bar showing the percentage in that category.

2. What type(s) of statistical picture would be appropriate if you wanted to compare the proportions who support or do not support an issue across three different countries?

Answer: A bar graph or a segmented bar graph would be most appropriate. A pie chart could be created for each country, but that would make it harder to compare across countries.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures

3. Give two examples of ways to distort a time series plot to make a trend across time appear to be different than it is.

Answer: One way to distort a time series is to show only the part of the series that displays the misleading message you want to convey. Another way is to start the vertical axis at a value that either inflates or deflates the trend.

4. Is it always appropriate for a line graph showing a measurement variable across time to have a line graph with a vertical axis starting at 0? Explain, and provide an example to support your explanation.

Answer: No, it is not always appropriate to start a line graph at 0. For example, if the values range from 1000 to 1500, it would be better to include only that range, or a range slightly larger than that.

5. Why is it important to provide the source of the data when displaying a statistical picture?

Answer: It is important to provide the source of data whenever statistical results are discussed or displayed. The reader can then decide whether the results are likely to be biased.

ANSWER TO THE MINI PROJECTS 1. Collect some categorical data on a topic of interest to you, and represent it in a statistical picture. Explain what you have done to make sure the picture is as useful as possible.

Answer: There are a number of ways to present categorical data, and the purpose of this project is to illustrate their use in an interesting context. If there is just one categorical variable, a pie chart may be most useful, although a pictogram may also be informative. For two variables, a bar graph or pictogram is probably most useful. Be sure the picture is well labeled, including a source for the data and a title for the picture.

2. Collect two measurement variables on each of at least 10 individuals.

Represent them in a statistical picture. Describe the picture in terms of possible outliers, variability, and relationship between the two variables.

Answer: A scatter plot is probably the most useful picture here. Make sure the axes are well labeled and evenly spaced and that the picture itself is labeled. The discussion should include mention of how much variability exists in each one

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter: 9 Plots, Graphs And Pictures

of the measurements, whether or not there are any unusual individuals (outliers on one or both variables) and whether or not a discernible relationship exists between the two measurements.

3. Find some data that represent change over time for a topic of interest to you.

Present a line graph of the data in the best possible format. Explain what you have done to make sure the picture is as useful as possible.

Answer: The line graph should be well-labeled, including a source and title. The axes should be labeled and have constant units. A decision should be made about whether or not to start the labeling of the vertical axis at zero, and an explanation of that feature should be included in the discussion.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 10: Relationships Between Measurement And Variables

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 10: RELATIONSHIPS BETWEEN MEASUREMENT AND VARIABLES

TABLE OF CONTENTS Answer to the Exercise Questions......................................................................................1 Answer to the Reflection Questions.................................................................................12 Answer to the Mini Projects............................................................................................. 14

ANSWER TO THE EXERCISE QUESTIONS 1. For each of the following pairs of variables measured for cities in North America, explain whether the relationship between them would be deterministic or statistical. a. Geographic latitude of the city and average temperature in January for the city b. Average temperature in January for the city in degrees Fahrenheit and average temperature in January for the city in degrees Celsius c. Average temperature in January for the city in degrees Fahrenheit and average temperature in August for the city in degrees Fahrenheit Answer: a. Statistical relationship. Knowing the latitude does not provide exact information about the average temperature. b. Deterministic relationship. If you know the temperature in Fahrenheit, you can convert it to its exact value in Centigrade. c. Statistical relationship. Knowing the average temperature in one month does not provide exact information about the average temperature in a different month.

2. For each of the following pairs of variables measured on college students, explain whether the relationship between them would be deterministic or statistical. a. Hours per day spent studying, on average, and hours per night spent sleeping, on average b. Height in inches and height in centimeters (neither one rounded off) c. Average number of units taken per quarter or semester and GPA

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 10: Relationships Between Measurement And Variables

Answer: a. Statistical relationship. Knowing one of these two variables for an individual does not provide exact information about the other one. b. Deterministic relationship. You can convert inches to centimeters (and vice versa) exactly. c. Statistical relationship. Knowing one of these two variables for an individual does not provide exact information about the other one.

3. Suppose 100 different researchers each did a study to see if there was a relationship between daily coffee consumption and height for adults. Suppose there really is no such relationship in the population. Would you expect any of the researchers to find a statistically significant relationship? If so, approximately how many (using the usual criterion for “small chance” of 5%)?

Explain your answer.

Answer: Yes, about five of the 100 researchers should find a relationship. The definition of statistical significance is such that when there really is no relationship we will erroneously detect one about 5% of the time.

4. Suppose a weak relationship exists between two variables in a population.

Which would be more likely to result in a statistically significant relationship between the two variables: a sample of size 100 or a sample of size 10,000?

Explain.

Answer: The sample of size 10,000 would be more likely. To quote the textbook, “Even a minor relationship will achieve ‘statistical significance’ if the sample is very large.” (p. 203) 5. The relationship between height and weight is a well-established fact.

Suppose you were to sample heights and weights for a small number of your friends, and you failed to find a statistically significant relationship between the two variables. Would you conclude that the relationship doesn’t hold for the population of people like your friends? Explain.

Answer: No, you would conclude that the size of the sample was too small to be able to detect the relationship.

6. A pint of water weighs 1.04 pounds, so 1 pound of water is 0.96 pint. Suppose a merchant sells water in containers weighing 0.5 pound, but customers can fill them to their liking. It is easier to weigh the filled container than to measure the volume of water the customer is purchasing. Define x to be the combined weight of the container and the water and y to be the volume of the water.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 10: Relationships Between Measurement And Variables

a. Write the equation the merchant would use to determine the volume y when the weight x is known. b. Specify the numerical values of the intercept and the slope, and interpret their physical meanings for this example. c. What is the correlation between x and y for this example? d. Draw a picture of the relationship between x and y.

Answer: a. In the container, the water alone weighs (x − 0.5). Every pound of water is 0.96 pints, so multiply the weight of the water by 0.96 to get the number of pints. b. The intercept is −0.48 and it doesn't have a physical meaning. It would be the volume of water corresponding to a weight of zero, but zero is not a possible weight since the container itself weighs 0.5 pounds. The slope is 0.96 and represents the additional volume of water (in pints) when the weight is increased by 1 pound. c. This is a deterministic relationship, not a statistical relationship, so the correlation is 1. d. See the figure below.

7. In Figure 10.2, we observed that the correlation between spouses’ heights, measured in millimeters, was 0.36. Can you determine what the correlation would be if the heights were converted to inches (and not rounded off)?

Explain.

Answer: It would still be 0.36. Correlation does not change when the units of measurement are changed.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 10: Relationships Between Measurement And Variables

8. Are each of the following pairs of variables likely to have a positive correlation or a negative correlation? a. Daily temperatures at noon in New York City and in Boston measured for a year b. Weights of automobiles and their gas mileage in average miles per gallon c. Hours of television watched and GPA for college students d. Years of education and salary Answer: a. A positive correlation. In summer both will be high and in winter both will be low. b. A negative correlation. Heavier cars (higher weight) have lower gas mileage (fewer miles per gallon). c. A negative correlation. Students who watch more TV tend to have lower GPAs. d. A positive correlation. People with more education tend to have higher salaries.

9. Which implies a stronger linear relationship, a correlation of 10.4 or a correlation of 20.6? Explain.

Answer: A correlation of −0.6 implies a stronger relationship. The sign determines whether one measurement increases or decreases as the other one increases, but the number determines the strength of the relationship, and 0.6 is higher than 0.4.

10. Give an example of a pair of variables that are likely to have a positive correlation and a pair of variables that are likely to have a negative correlation.

Answer: The key is that for a positive correlation they increase together, while for a negative correlation one decreases as the other increases. An example of a positive correlation is height and weight. An example of a negative correlation is amount of exercise per week and resting pulse rate.

11. Explain how two variables can have a perfect curved relationship and yet have zero correlation. Draw a picture of a set of data meeting those criteria.

Answer: Correlation measures only how closely the points fall to a straight line. A perfect curved relationship where the best line through the points is flat would have a zero correlation. For example, draw x versus x2 for x = −3, −2, −1, 0, 1, 2, 3.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 10: Relationships Between Measurement And Variables

12. The table below gives the average June and December temperatures (in Fahrenheit) for eight cities in the United States for the years 2000 to 2022.

(Source: https://www.ncei.noaa.gov/access/monitoring/climate-at-aglance/city/time-series)

a. Create a scatter plot for these temperatures, placing June on the horizontal axis and December on the vertical axis. Comment on whether or not it looks like there is a general linear relationship and, if so, whether it is positive or negative. b. Using Excel, a calculator, or other software, find the correlation between average June and December temperatures for these cities. Do the value and the sign (positive or negative) make sense based on the scatterplot from part (a)?

Answer: a. See the figure below. There appears to be a positive linear relationship.

60

December

50

40

30

20 55

60

65

70

75

80

85

90

95

June

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 10: Relationships Between Measurement And Variables

b. The correlation is 0.837. Yes, it makes sense because the scatter plot shows a relatively strong positive linear relationship.

13. The regression line relating verbal SAT scores and GPA for the data exhibited in Figure 9.6 is

Estimate the average GPA for those with verbal SAT scores of 500.

Answer: Average GPA = 0.539 + 0(.00362) (500) = 2.349.

14. Refer to the previous exercise, giving the regression equation relating verbal SAT and GPA. a. Explain what the slope of 0.00362 represents. b. The lowest possible SAT score is 200. Does the intercept of 0.539 have any useful meaning for this example? Explain.

Answer: a. For each increase of 1 in verbal SAT score we would predict an increase of about 0.00362 in GPA. b. No. It would be the GPA for someone with an SAT score of 0, but that's not a possible SAT score.

15. Refer to Case Study 10.2, in which equations are given for maximum heart rate (HRmax) based on age. The equation for all participants combined is given as: Predicted HRmax = 208.852-0.741 (Age) a. Use this equation to predict the maximum heart rate for someone who is 21 years old. b. The formula often recommended is Predicted HRmax 5 220 – Age. Compare the result of that formula to the predicted HRmax from part (a) for someone who is 21 years old.

Answer: a. Predicted maximum heart rate = 208.852 – 0.741 (21) = 193.29 b. 220 – Age = 220 – 21 = 199, slightly higher than the maximum predicted using the equation.

16. Refer to the equation for predicting maximum heart rate given in the previous exercise. a. Does the intercept of 208.852 have a logical physical interpretation in the context of this example? Explain. b. Does the slope of 0.741 have a logical interpretation in the context of this example? Explain.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 10: Relationships Between Measurement And Variables

Answer: a. No, it would be the predicted maximum heart rate for someone who is 0 years old. That age is far outside of the range of data used for the equation. b. Yes, the slope of −0.741 indicates that on average, the predicted maximum heart rate goes down by 0.7416 for each one-year increase in age.

17. Outliers in scatterplots may be within the range of values for each variable individually but lie outside the general pattern when the variables are examined in combination. The mean HRmax for people described in Case Study 10.2 who were between ages 60 and 70 was about 159, with a standard deviation of about 12.4. a. Looking at the values on both axes in Figure 10.9, would someone who was 65 years old and had an HRmax of 200 be considered to be an outlier for either variable (age, HRmax) separately? Explain. b. Would someone with the values given in part (a) be considered to be an outlier when the combination of the two values are used? Explain.

Answer: a. Neither of those values would be an outlier for that variable. Ages range from 10 to 80, so 65 is within that range. Maximum heart rates range from about 140 to 210, so 200 is in this range. b. Although Figure 10.9 doesn’t show the individual data points, it appears to be the case that a maximum heart rate of 200 for someone aged 65 would be an outlier. For people of that age, the predicted maximum heart rate appears to be between 155 and 165.

18. In Chapter 9, we examined a picture of winning time in men’s 500-meter speed skating plotted across time. The data represented in the plot started in 1924 and went through 2022. A regression equation relating winning time and year for 1924 to 2018 is winning time = 266.4 – (0.11526) (year) a. Would the correlation between winning time and year be positive or negative? Explain. b. In 2022, the actual winning time for the gold medal was 34.32 seconds. Use the regression equation to predict the winning time for 2022, and compare the prediction to what actually happened. Was the actual winning time higher or lower than the predicted time? c. Explain what the slope of 20.11526 indicates in terms of how winning times change from year to year.

Answer: a. The correlation would be negative because the slope is negative. b. Predicted time is 266.4 – (0.11526)(2022) = 33.3443; actual time was 0.9759 seconds higher. c. Winning times decrease by 0.11526 seconds per year, on average. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 10: Relationships Between Measurement And Variables

19. Refer to the equation in the previous exercise, relating year to winning time.

The Olympics are held every 4 years. Explain what the slope of 20.11526 indicates in terms of how winning times are predicted to change from one Olympics to the next.

Answer: Winning times should decrease by about 0.11526 × 4 = 0.46104 second per 4-year Olympics.

20. Explain why we should not use the regression equation we found in Exercise 18 for speed-skating time versus year to predict the winning time for the 2032 Winter Olympics.

Answer: We should not extrapolate that far beyond the range of the data. The decreasing trend in winning times cannot continue at the same rate forever, and it may not still apply by the year 2032.

21. In one of the examples in this chapter, we noticed a very strong relationship between spouses’ ages for a sample of 200 British couples, with a correlation of 0.94. Coincidentally, the relationship between putting distance and success rate for professional golfers had a correlation of 20.94, based on 11 data points.

(See Figure 10.4.)

This latter correlation was statistically significant, so we can be pretty sure the observed relationship was not just due to chance. Based on this information, do you think the observed relationship between spouses’ ages is statistically significant? Explain.

Answer: Yes. The strength of the relationship is the same as for the golfers and the sample is much larger.

22. The regression equation relating distance (in feet) and success rate (percent) for professional golfers, based on 11 distances ranging from 5 to 15 feet is success rate = 76.5 – (3.95) (distance) a. What percent success would you expect for these professional golfers if the putting distance is 6.5 feet? b. Explain what the slope of 23.95 means in terms of how success changes with distance.

Answer: a. Predicted success rate = 76.5 − (3.95) (6.5) = 50.825%. b. For each additional foot of distance, success drops by about 3.95%.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 10: Relationships Between Measurement And Variables

23. The original data for the putting success of professional golfers included values beyond those we used in Exercise 22 (5 to 15 feet), in both directions.

At a distance of 2 feet, 93.3% of the putts were successful. At a distance of 20 feet, 15.8% of the putts were successful. a. Use the equation in Exercise 22 to predict success rates for those two distances (2 and 20 feet). Compare the predictions to the actual success rates. b. Use your results from part (a) to explain why it is not a good idea to use a regression equation to predict information beyond the range of values from which the equation was determined. c. Based on the picture in Figure 10.4 and the additional information in this exercise, draw a picture of what you think the relationship between putting distance and success rate would look like for the entire range from 2 to 20 feet. d. Explain why a regression equation should not be formulated for the entire range from 2 to 20 feet.

Answer: a. At 2 feet, predicted success rate = 76.5 − (3.95)(2) = 68.6%. At 20 feet, predicted success rate = 76.5 − (3.95)(20) = −2.5%. These predictions are clearly considerably different from the actual success rates of 93.3% and 15.8%. b. The relationship that held for the distances of 5 to 15 feet obviously does not extend to distances as short as 2 feet or as long as 20 feet. It is not appropriate to assume that relationships continue to hold beyond the range of the data observed. c. The relationship is a curve that looks like an elongated, tilted letter C. d. Regression lines work only for straight line relationships. Clearly, the relationship between success and distance over the entire range from 2 feet to 20 feet is curved, not linear.

24. Suppose a time series across 60 months has a long-term positive trend. Would you expect to find a positive correlation, a negative correlation, or no correlation between the values in the series and the month numbers 1 to 60?

Explain.

Answer: You would expect a positive correlation because the regression line for the trend would have a positive slope.

25. Refer to the temperature data given in the table accompanying Exercise 12. a. Using Excel, a calculator, or other software, find the intercept and slope for the regression equation with x = average June temperature and y = average December temperature for these cities.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 10: Relationships Between Measurement And Variables

b. Use the equation you found in part (a) to predict the average December temperature for a city with an average June temperature of 70 degrees.

What assumption is required about the city in question, in order for this to be a relatively accurate prediction? c. Do you think the equation you found in part (a) could be used to predict average December temperature from June temperature for Honolulu, Hawaii? Explain.

Answer: a. The intercept is −34.2 and the slope is + 0.981. b. The predicted average December temperature is −34.2 + 0.981 (70) = 34.47 degrees. The assumption is that the city in question is similar to those used to generate the equation. c. No. Honolulu has different climate conditions from the cities in the continental United States used to generate the equation.

26. The table below gives the self-reported heights of 10 college women (“Daughter’s height”), along with the heights of their mothers.

a. Draw a scatter plot for these data, placing Mother’s height on the horizontal axis and Daughter’s height on the vertical axis. Comment on whether or not it looks like there is a linear relationship and, if so, whether it is positive or negative. b. Using Excel, a calculator, or other software, find the correlation between the mother’s and daughter’s heights. Do the value and the sign (positive or negative) make sense based on the scatterplot from part (a)? Explain. c. Using Excel, a calculator, or other software, find the intercept and slope for the regression equation with x = Mother’s height and y = Daughter’s height. d. The equation you found in part (c) might be useful for predicting the height of a female from her mother’s height before the daughter is fully grown.

Use the equation to predict the height of the daughter of a mother who is 63 inches tall.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 10: Relationships Between Measurement And Variables

Answer: a. See the figure below. There is a positive linear relationship.

69

Daughter's height (inches)

68 67 66 65 64 63 62 61 60 58

59

60

61 62 63 64 Mother's height (inches)

65

66

67

b. The correlation is 0.739. Yes, it makes sense because the scatter plot shows a moderately strong, positive linear relationship. c. The intercept is 15.2 and the slope is 0.771. d. Predicted daughter’s height is 15.2 + 0.771 (63) = 63.8 inches.

27. Refer to the previous exercise about the relationship between mothers’ and daughters’ heights. Would the intercept of the regression line relating x = Mother’s height to y = Daughter’s height have a useful interpretation in this situation? (You do not need to compute the regression line to answer this question.) Explain.

Answer: No. The intercept would be the predicted daughter’s height for a mother’s height of 0. The intercept would only make sense if a mother’s height of 0 was in the range of possible x values, which it clearly is not.

28. Refer to the temperature data given in Exercise 12. A regression equation could be computed to relate x = average June temperature and y = average December temperature across cities. (This was requested in Exercise 25.)

Answer the following questions without actually computing the regression equation. a. Would the intercept of this regression equation have a useful interpretation in this situation? If you think not, explain why not. If you think it would, give the interpretation. b. Would the slope of this regression equation have a useful interpretation in this situation? If you think not, explain why not. If you think it would, give the interpretation.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 10: Relationships Between Measurement And Variables

Answer: a. No. The intercept would be the predicted average December temperature for a city with an average June temperature of 0, which is far outside the range of June temperatures for the data shown. b. Yes, the slope would have a useful interpretation. It would represent the predicted difference in average December temperature for two cities that differed by one degree in average June temperatures.

29. Refer to News Story 7 in the Appendix and the accompanying journal article, “Vegans report less stress and anxiety than omnivores,” described in Case Study 2.1. One of the measurements on each person was the Depression Anxiety Stress Scales-21, abbreviated DASS-21, which measures “symptoms consistent with depression, anxiety and stress” (Beezhold et al., p. 290). The researchers stated that age was “significantly related” to DASS-21 score, with a correlation of 20.197. a. What two variables were measured for each person to provide this result? b. Explain what a correlation of −0.197 means in the context of these two variables. c. What is meant by the word significantly as it is used in the quote? d. The authors did not provide a regression equation relating the two variables. If a regression equation were to be found for the two variables in the quote, which one do you think would be the logical explanatory variable? Explain.

Answer: a. The two variables measured were DASS-21 score and age. b. The correlation is moderate and negative, which means that in general, people who are older tend to score lower on the DASS-21, the depression, anxiety and stress scale. c. The quote refers to statistical significance. This means that the researchers were convinced that the relationship observed in their sample reflected a real relationship in the population. d. Age would be the logical explanatory variable. The interest would be in predicting the value of the depression, anxiety and stress scale based on someone’s age.

ANSWER TO THE REFLECTION QUESTIONS 1. Suppose x 5 maximum temperature in August in degrees Fahrenheit measured for 20 cities across North America. Give an example of a variable that has a deterministic relationship with x, and an example of a variable that has a statistical, but not deterministic, relationship with x.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 10: Relationships Between Measurement And Variables

Answer: Maximum temperature in August in degrees Celsius (not rounded off) would have a deterministic relationship with the same temperature in degrees Fahrenheit. An example of a variable that has a statistical relationship is the latitude of the city. Cities that are farther north tend to be cooler than cities that are farther south.

2. Suppose that for a certain population the regression equation for the relationship between weight (in pounds) and height (in inches) is weight = –250 + 6(height). a. Explain whether each of the slope and intercept has a meaningful interpretation in this case. If so, explain how it would be interpreted. b. Explain the two uses for this regression equation.

Answer: a. The slope has a meaningful interpretation. It is the predicted difference in weight for two members of the population who differ in height by one inch, with higher predicted weight for the taller individual. The intercept of –250 does not have a meaningful interpretation. It would be the predicted weight for an individual with height = 0, which is not a possible height. b. The equation could be used to predict the weight of an individual of a given height. It could also be used to estimate the average weight for the population of members of a specific height.

3. Using data from 100 donors at a blood bank ranging in age from 17 to 84, a scatterplot showed a linear relationship between x = age and y = body temperature (in Fahrenheit) with a correlation of –0.313. a. Using this information, describe the relationship between age and body temperature. b. Using only this information, is it possible to specify the sign and/or magnitude of the slope of the regression equation relating the two variables? Explain.

Answer: a. There is a moderate negative linear relationship between age and body temperature, with predicted temperature going down as age increases. b. The sign of the slope is the same as the sign of the correlation, so it is negative in this case. The magnitude of the slope cannot be found from the correlation.

4. Suppose a time series for the average price of something taken monthly over 60 months included a trend that showed the price increased on average by 10 cents a month. If you had the list of prices for all of the months in the series, how would you go about creating a plot showing the series with the trend removed? What would go on the horizontal axis, and what would go on the vertical axis? © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 10: Relationships Between Measurement And Variables

Answer: The horizontal axis would be “month” ranging from 1 to 60. The vertical axis would have the original values minus month × 10.

ANSWER TO THE MINI PROJECTS 1. (Computer or statistics calculator required.) It has been claimed that a person’s height and their arm span are almost equal. Arm span is measured as the distance from the tip of one middle finger to the tip of the other one, when your arms are held out horizontally. To test this claim, measure the height and arm span of 10 adults to the nearest 1/2 inch. Draw a scatterplot of the data, with arm span on the horizontal (x) axis and height on the vertical (y) axis. Using software that produces regression equations, find the regression equation for your data. Draw it on your scatter diagram. Now draw a line on your plot that represents the relationship if these two measurements are equal for everyone. Do you believe the claim that they are approximately equal? Finally, use your regression equation to predict the height for someone with an arm span of 65 inches.

Answer: If the claim is correct that height and arm span are almost equal, then a scatter plot of the two variables should show an increasing relationship with a slope of about 1.0. After collecting data, create a scatter plot with arm span as x and height as y. Then draw a line with the equation x = y and see how close that comes to the regression line created from the data.

2. Go to your library or an electronic journal resource and peruse journal articles, looking for examples of scatterplots accompanied by correlations. Find three examples in different journal articles. Present the scatterplots and correlations, and explain in words what you would conclude about the relationship between the two variables in each case.

Answer: The explanation should include an explanation of the two variables, a description about whether the relationship is positive or negative and a discussion about the strength of the relationship. If a regression equation is given, the explanation can also discuss the meaning of the intercept and slope in the context of the example.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 11: Relationships Can Be Deceiving

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 11: RELATIONSHIPS CAN BE DECEIVING

TABLE OF CONTENTS Answers to the Exercise Questions ....................................................................................1 Answers to the Reflection Questions ...............................................................................13 Answers to the Mini Projects ........................................................................................... 14

ANSWERS TO THE EXERCISE QUESTIONS 1. Explain why a strong correlation would be found between weekly sales of firewood and weekly sales of cough drops over a one-year period. Would it imply that fires cause coughs?

Answer: Both firewood sales and cough drop sales are likely to increase during the cold season. No, it does not imply that fires cause coughs!

2. Each of the following headlines reported on an observational study. In each case, explain whether or not you think the headline is warranted. If not, write a more suitable headline. a. “Morning people are happier than night owls, study suggests.” (Welsh, 2012) b. “Breakfast cereals prevent overweight in children.” (World Health, 2013) c. “Do you have high blood pressure? Yogurt may be able to help you, study finds.” (Cutler, 2021) d. “Aspirin tied to lower melanoma risk.” (Pittman, 2014) Answer: a. The headline is justified. It does not make a cause-and-effect statement. b. The headline is not justified. It implies that eating cereal causes a reduction in being overweight. A better headline is “Children who eat breakfast cereal are less likely to be overweight.” c. The headline is not justified. A better headline is “Yogurt linked to lower blood pressure.” d. The headline is justified. It does not make a cause-and-effect statement.

3. Suppose a study of employees at a large company found a negative correlation between weight and distance walked on an average day. In other words, people who walked more weighed less. Would you conclude that walking causes lower weight? Can you think of another potential explanation? © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 11: Relationships Can Be Deceiving

Answer: No, you wouldn’t necessarily conclude that walking causes lower weight.

People who prefer to exercise (walk) may also be people who keep their weight under control, eat less fat, and so on.

4. An article in Science News (1 June 1996, 149, p. 345) claimed that “evidence suggests that regular consumption of milk may reduce a person’s risk of stroke, the third leading cause of death in the United States.” The claim was based on an observational study of 3150 males, and the article noted that the researchers “report strong evidence that men who eschew milk have more than twice the stroke risk of those who drink 1 pint or more daily.” The article concluded by noting that “those who consumed the most milk tended to be the leanest and the most physically active.” Go through the list of seven “reasons two variables may be related,” and discuss each one in the context of this study.

Answer: Reason 1 could be possible but isn’t likely, that milk-drinking alone is responsible for low risk of stroke. Reason 2 is not likely, since it would say that a low risk of stroke causes people to drink milk. People don’t usually even know whether or not they are at risk for strokes. Reason 3 could be possible, that milk-drinking is one of many factors that reduce the risk of a stroke.

Reason 4 is quite likely, that confounding variables exist. One example is general good diet and good health, which leads to reduced stroke risk. Reason 5 is likely, that both reduced stroke risk and increased milk-consumption result from a common cause, such as interest in taking care of one’s health.

Related to reasons 4 and 5, note that the article made a point of saying that the same individuals who consumed the most milk were also “leanest and most physically active.” Reason 6, which is that both variables are changing over time, is not relevant, unless the study did not adjust for age and the men drank less milk as they aged, which would also increase stroke risk. Reason 7, that the association is just a coincidence, is possible. Any single study could find a relationship just by chance.

5. Iman (1994, p. 505) presents data on how college students and experts perceive risks for 30 activities or technologies. Each group ranked the 30 activities. The rankings for the eight greatest risks, as perceived by the experts, are shown in Table 11.5. a. Prepare a scatterplot of the data, with students’ ranks on the vertical axis and experts’ ranks on the horizontal axis. b. The correlation between the two sets of ranks is 0.407. Based on your scatterplot in part (a), do you think the correlation would increase or decrease if X-rays were deleted? Explain. What if pesticides were deleted instead?

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 11: Relationships Can Be Deceiving

c. Another technology listed was nuclear power, ranked first by the students and 20th by the experts. If nuclear power was added to the list, do you think the correlation between the two sets of rankings would increase or decrease? Explain.

Answer: a.

18 X rays 16

Students' rank

14 12 10 8 6 4

Pesticides

2 0 0

1

2

3

4 5 Experts' rank

6

7

8

b. If X-rays were deleted, the correlation would decrease (to 0.13 in case you are interested), because outliers that fall in line with the rest of the data tend to increase correlation. If pesticides were deleted (but not X-rays) correlation would increase (to 0.68), because outliers that do not fit the pattern of the rest of the data tend to decrease correlation.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 11: Relationships Can Be Deceiving

c. Adding nuclear power would decrease the correlation because it would be an extreme outlier, not in line with the data. It also makes sense that it should decrease the correlation because it is an example of a case in which the students and experts are not in agreement at all, and thus weakens the relationship.

6. Give an example of two variables that are likely to be correlated because they are both changing over time.

Answer: Answers will vary. One example is winning time in an Olympic running event and the cost of running shoes, which are likely to be negatively correlated because prices go up and winning times go down over time.

7. Which one of the seven reasons for relationships listed in Section 11.3 is supposed to be ruled out by designed experiments?

Answer: Designed experiments are supposed to rule out Reason 4: “confounding variables may exist.”

8. Nutt (2021) described how a group of experts in the United Kingdom created scores for the harm done by various legal and illegal drugs. Separate scores were created for “harm to user” and “harm to others.” Harm to user included things like mortality, loss of relationships, and dependence on the drug, while harm to others included things like crime, economic cost, and impact on family life. Table 11.6 lists the scores for a subset of the drugs studied. Figure 11.6 shows a scatterplot of the data with two lines. a. Using either Figure 11.6 or Table 11.6, identify which drug is an outlier in the data set. Describe the reason it is an outlier, and explain whether it is a legitimate outlier or not. b. There are two lines shown in Figure 11.6, identified as Line A and Line B. One of them is the regression line without the outlier, and one is the regression line with the outlier. Explain which one is which.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 11: Relationships Can Be Deceiving

Answer: a. The outlier is alcohol. It is not an outlier in the “harm to self” scale, but it is much higher than any of the others on the “harm to others” scale. This is a legitimate outlier, because the experts did in fact conclude that alcohol creates substantial harm to others. The value is not an error. b. Line A includes the alcohol outlier. The value is “pulling” the line towards it.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 11: Relationships Can Be Deceiving

9. The two lines shown in Figure 11.6 represent the regression lines for the data in Table 11.6, with and without the inclusion of an outlier. The equations and correlations for these lines are as follows: Line A: Others=-2.75 + 0.64(User); correlation = 0.5 Line B: Others= -2.16 + 0.45(User); correlation = 0.7 a. Explain why the correlation is lower for Line A than for Line B. b. Do the intercepts in these equations have a meaningful interpretation? If so, interpret the intercept for Line A. If not, explain why not. c. These equations are presented as if “Harm to User” is the explanatory variable and “Harm to Others” is the response variable. Explain whether it makes sense to identify one of the two variables as explanatory and other one as the response.

Answer: a. The correlation is lower for Line A because the outlier (Alcohol) does not fit the linear pattern shown by the rest of the points. The outlier is not used for Line B. b. The intercepts do not have a useful interpretation. They would be the predicted harm to others if the harm to users was 0, but that is outside of the range of the data. c. It does make sense to use “harm to user” as the explanatory variable, because the interesting question is the extent to which users of drugs are harming others for each type of drug.

10. Suppose a study measured total beer sales and number of highway deaths for one month in various cities. Explain why it would make sense to divide both variables by the population of the city before determining whether a relationship exists between them.

Answer: There would be a relationship between the actual numbers, because large cities would have higher numbers for both than small cities. The more meaningful question is whether or not the per capita figures for beer sales and highway deaths are related.

11. Construct an example of a situation in which an outlier inflates the correlation between two variables. Draw a scatterplot.

Answer: An example is square feet versus price for homes sold in a city, where there is one very large mansion that has been sold, and the sales price fits the same pattern as for the other houses. The key is that the outlier fits the same pattern as the rest of the data.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 11: Relationships Can Be Deceiving

12. Construct an example of a situation in which an outlier deflates the correlation between two variables. Draw a scatterplot.

Answer: An example is salary and years of education in a company where the president makes $1 million a year but has an eighth-grade education. The key is that the outlier does not fit the pattern of the rest of the data.

13. According to The Wellness Encyclopedia (University of California, 1991, p.17): “Alcohol consumed to excess increases the risk of cancer of the mouth, pharynx, esophagus, and larynx. These risks increase dramatically when alcohol is used in conjunction with tobacco.” It is obviously not possible to conduct a designed experiment on humans to test this claim, so the causal conclusion must be based on observational studies. Explain three potential additional pieces of information that the authors may have used to lead them to make a causal conclusion.

Answer: First, they may have a biological explanation, like the fact that alcohol and tobacco are irritants on those parts of the body. Second, they may have found that the relationship holds under varying conditions, such as in different cultures, for different types of alcohol, in different age groups and so on. Third, they may have ruled out potential confounding factors (like diet) by measuring them and determining whether or not they also have an impact.

The following information applies to Exercises 14 to 16. Suppose a positive relationship had been found among each of the sets of variables given in Exercises 14 to 16. In Section 11.3, seven potential reasons for such relationships are given. Explain which of the seven reasons is most likely to account for the relationship in each case. If you think more than one reason might apply, mention them all but elaborate on only the one you think is most likely.

14. a. Number of deaths from automobiles and beer sales for each year from 1980 to 2020 b. Number of ski accidents and average wait time for the ski lift for each day during one winter at a ski resort Answer: a. The most likely reason is that both change over time. A secondary reason may be that higher beer sales are contributing to more deaths from automobiles, but that would be indicated by higher rates for both, not by higher raw numbers. b. The most likely reason is that there is a common cause, namely number of people at the resort that day. Obviously, temperature, snow conditions and day of the week all affect attendance, which in turn affects both accidents and waiting time.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 11: Relationships Can Be Deceiving

15. a. Stomach cancer and consumption of barbecued foods, which are known to contain carcinogenic (cancer-causing) substances b. Self-reported level of stress and blood pressure Answer: a. Because the problem mentions that barbecued foods are known to contain cancer-causing substances, there is likely to be a direct causal link. Either direct cause alone, or contributing cause, would be acceptable answers. b. There are a few possibilities here. There may just be a third variable causing both, namely personality type. More high-strung people would both see themselves as being under stress and have high blood pressure. Stress could also be a contributing, but not the sole, cause of high blood pressure.

A weaker explanation is that high blood pressure causes stress, because people worry about having high blood pressure.

16. a. Amount of dietary fat consumed and heart disease b. Twice as many cases of leukemia in a new high school built near a power plant than at the old high school Answer: a. There are too many possible confounding factors to say that there is a direct cause. It could be that those who eat more fat also exercise less, for example.

So, confounding variables make it difficult to determine a causal link. b. Most likely this would be a coincidence. There would be unlikely to be enough cases at each school to conclude that it was anything else.

17. Explain why it would probably be misleading to use correlation to express the relationship between number of acres burned and number of deaths for major fires in the United States.

Answer: Just like for the earthquake example in the text, the number of deaths is highly dependent on the density of houses in the area. Large fires are actually more likely to occur in sparse, wooded areas, where there are fewer people.

Therefore, the relationship may actually be negative, giving the impression that the more acres burned, the fewer the number of deaths. Also, one or two outliers with relatively low acreage but a high number of deaths could completely distort the results.

18. It is said that a higher proportion of drivers of red cars are given tickets for traffic violations than the drivers of any other color car. Does this mean that if you drove a red car rather than a white car, you would be more likely to receive a ticket for a traffic violation? Explain.

Answer: No. There are confounding factors, like more sports cars are red. Also, people who buy red cars may be more likely to speed or drive recklessly.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 11: Relationships Can Be Deceiving

19. Construct an example for which correlation between two variables is masked by grouping over a third variable.

Answer: The key is that there are two clusters of points that don’t fall into the same pattern with each other as the pattern in each cluster. For example, suppose hours spent studying and grade point averages are measured for students at two high schools. Suppose one is a difficult school, where students study a lot but receive moderate grades. The other school doesn’t require much studying and gives very high grades. Then a plot of grades (vertical axis) versus hours spent studying (horizontal axis) may show a cluster in the upper left (low study hours, high grades) and another in the lower right (high study hours, low grades) resulting in a negative correlation between the two. This could happen even though there is a positive correlation between them for each school.

20. An article on the website Market Watch (Fottrell, 2019) had the headline, “This one thing in your marriage increases the risk of divorce by 33%.” Two quotes from the article explain what’s meant by the headline. First, “When wives earn more than their husbands, some men just can’t handle it.” And later in the article, “The risk of divorce is nearly 33% higher when a husband isn’t working full-time.” The headline implies a causal link between wives earning more than their husbands (the explanatory variable) and divorce (the response variable). Choosing from the remaining six possibilities in Section 11.3 (reasons 2 through 7), give three other potential explanations for the observed connection.

Answer: In this example, the explanatory variable is whether the wife earns more than the husband and the response variable is whether the couple gets divorced.

There are a number of possible answers: Reason 2: “The response variable is causing a change in the explanatory variable.” It could be that women who know there is a high risk of divorce are more likely to try to get a high earning job.

Reason 3: “The explanatory variable is a contributing, but not sole, cause of the response variable.” There could be numerous other factors that also contribute to divorce, such as having different interests, but the earning status of each person is also a contributing factor.

Reason 4: “Confounding variables.” There are all sorts of potential confounding variables in this situation. For example, if the wife has a high-powered job she might spend more time away from home, have a work-related group of friends, spend less time with her spouse, and so on.

Reason 5: “Both variables result from a common cause.” One possible common cause is that the wife has a much higher level of education than the husband.

That might lead her to have a higher paying job, and may also lead to a poor marriage. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 11: Relationships Can Be Deceiving

Reasons 6 and 7: “Both variables change over time” does not seem to apply, and “Coincidence” is possible but it’s a less likely explanation than the others.

21. Lave (1990) discussed studies that had been done to test the usefulness of seat belts before and after their use became mandatory. One possible method of testing the usefulness of mandatory seat belt laws is to measure the number of fatalities in a particular region for the year before and the year after the law went into effect and to compare them. If such a study were to find substantially reduced fatalities during the year after the law went into effect, could it be claimed that the mandatory seat belt law was completely responsible? Explain.

(Hint: Consider factors such as weather and the anticipatory effect of the law.)

Answer: No. It is not enough to compare one year with another because there are too many other factors. For instance, if the first winter was very harsh and the second one mild, there would be more fatalities in the first due just to the weather. Second, once people know the law is about to go into effect, they are likely to start getting into the habit of wearing seat belts. It would be better to compare many years before and after the law, although even then there could be confounding factors, like if better roads had been constructed in later years.

22. In Case Study 10.1, we learned how psychologists relied on twins to measure the contributions of heredity to various traits. Suppose a study were to find that identical (monozygotic) twins had highly correlated scores on a certain trait but that pairs of adult friends did not. Why would that not be sufficient evidence to conclude that genetic factors were responsible for the trait?

Answer: The twins would presumably have been raised together, so the correlations could be a result of common environmental experiences rather than genetics.

23. An article in The Wichita Eagle (24 June 2003, p. 4A) read as follows: “Scientists have analyzed autopsy brain tissue from members of a religious order who had an average of 18 years of formal education and found that the more years of schooling, the less likely they were to exhibit Alzheimer’s symptoms of dementia. The study provides the first biological proof of education’s possible protective effect.” Do you agree with the last sentence, that this study provides proof that education has a protective effect? Explain.

Answer: The conclusion is not justified. Clearly this was an observational study, because the participants were not randomly assigned to different years of schooling. There are many possible confounding variables, such as economic status and diet as a child, genetic differences, and so on.

24. One of the features that may suggest a cause-and-effect relationship from an observational study is a “dose-response” relationship. a. Explain what is meant by a dose-response relationship. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 11: Relationships Can Be Deceiving

b. Give an example of a possible dose-response relationship. c. Can a dose-response relationship exist if the explanatory variable is categorical (and not ordinal)? Explain.

Answer: a. There is an association between the quantity of the explanatory variable and the level of the response variable. b. Perhaps the higher the percent fat in someone’s diet the higher their blood pressure is likely to be. c. No. A dose-response relationship requires that the dose has increasing values across groups.

For Exercises 25 to 29, refer to the news stories in the Appendix and/or corresponding original source on the companion website. In each case, identify which of the “Reasons for Relationships between Variables” described in Section 11.3 are likely to apply.

25. News Story 7: “Does veganism help reduce stress and anxiety?”

Answer: The explanatory variable is whether someone follows a vegan diet, and the response variables are measures of stress and anxiety. The news story provides some possible reasons for the relationship. Reason 1 (direct cause and effect) is possible, and the news story gives a possible explanation, that “animal fats may activate inflammatory pathways in the brain that are responsible for stress and mood disorders.” Reason 3 is also possible, that there is a direct effect but that there are other contributing factors. Reason 4 (confounding variables) is a strong possibility, and the news story gives several examples, including that vegans exercise more, spend more time in nature, and so on.

26. News Story 12: “New program for teen drivers with ADHD has significantly reduced crashes.”

Answer: The explanatory variable is whether the teen driver has taken the ADHD training course, and the response variable is frequency of crashes or near crashes. The most likely explanation for the relationship is Reason 1, that there is a direct cause and effect relationship, because the results are based on a randomized experiment.

27. Summary of News Story 15: “Kids’ stress, snacking linked.”

Answer: Reasons 1 to 6 are likely to apply. For instance, for Reason 2, the response variable (unhealthful snacking) may be causing changes in the explanatory variable (stress) because of chemical effects of the poor diet.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 11: Relationships Can Be Deceiving

28. News Story 16: “More on TV Violence” Answer: The explanatory variable is watching TV violence as a child, and the response is aggression in adulthood, so Reasons 2 and 6 cannot apply. Reasons 1, 3, 4 and 5 are all reasonable. Confounding variables may include aggressive behavior in the home when the child is growing up, level of education of parents, and so on. A common cause may be genetic propensity toward violence.

29. News Story 20: “Eating Organic Foods Reduces Pesticide Concentrations in Children.”

Answer: The explanatory variable is type of food eaten (organic or conventional) and the response is pesticide found in the urine of the child. Reasons 1, 3, 4 and 5 are all likely. A common cause may be having parents who are health conscious and who are careful about the child’s surrounding and food intake. A confounding variable may be whether pesticides are used in the home.

Exercises 30 and 31 list the titles of some of the news stories printed or summarized in the Appendix. In each case, determine whether the study was a randomized experiment or an observational study, then discuss whether the title is justified based on the way the study was done.

30. a. News Story 3: “Rigorous veggie diet found to slash cholesterol.” b. News Story 7: “Does veganism help reduce stress and anxiety?”

Answer: a. This was a randomized experiment, so the title is justified. b. This was an observational study, so a cause-and-effect conclusion cannot be made. The title is phrased as a question, but it implies that there is a causal relationship, which is not justified.

31. a. News Story 14: “Video gaming improves cognitive performance in children.” b. News Story 15: “Kids’ stress, snacking linked.” c. News Story 20: “Eating organic foods reduces pesticide concentrations in children.”

Answer: a. This was an observational study, so a cause-and-effect conclusion is not justified. The title is not justified. b. This was an observational study, so a cause-and-effect conclusion is not justified. However, the title does not make one, so the title is justified. c. This was an observational study, so a cause-and-effect conclusion cannot be made. The title is not justified.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 11: Relationships Can Be Deceiving

ANSWERS TO THE REFLECTION QUESTIONS 1. Explain what is meant by a legitimate outlier that results in an illegitimate correlation.

Answer: A legitimate outlier is one that occurs naturally in the data. An outlier produces an illegitimate correlation if it doesn’t fit the pattern of the rest of the data. For example, in the relationship between years of education and annual salary, an outlier that doesn’t fit the pattern could be someone who has very little education but has founded and runs a company for which they receive a very large salary.

2. Explain why a reported correlation should always be accompanied by a scatterplot (or a description of it) showing the data points.

Answer: A scatterplot or a description should always accompany a reported correlation for multiple reasons. First, it is important to see whether the overall relationship is linear, which is required for a correlation to be meaningful.

Second, it is important to see if there are any large outliers distorting the correlation. Third, there could be clusters for which the correlation is different within each cluster than it is for the overall dataset.

3. A national real estate agency is interested in finding a regression line to predict potential sales prices of houses based on the size of the house (in square feet). Give an example of a third variable that is categorical and for which it would make sense to create a separate equation for each category of the variable.

Answer: There are several possibilities for third variables that would affect the price of the house. Examples are whether the house is in an urban or rural setting, whether it has a large yard or not, whether it has a swimming pool or not, and so on.

4. Suppose an observational study found a negative correlation between amount of exercise a person gets per week and their resting pulse rate. a. Explain which one of the two variables is the logical explanatory variable. b. Explain what a negative correlation means in this example. c. Because this is an observational study, a cause-and-effect conclusion may not be warranted. What additional information would help justify such a conclusion?

Answer: a. Explanatory variable is amount of exercise per week and response variable is resting pulse rate. b. As amount of exercise increases, resting pulse rate decreases.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 11: Relationships Can Be Deceiving

c. The reasons given in Section 11.4 that contribute to a cause and effect conclusion would be relevant. For example, it would be useful to see if there is a “dose response” relationship in which as exercise increases, resting pulse rate decreases. It would be useful to know if the medical profession could supply a reasonable explanation for why there would be a cause and effect relationship. It would be helpful to know if the relationship holds for various age groups, ethnic groups, etc.

5. Suppose a strong positive correlation is found between number of new cars sold and per capita average income for the years 1970 to 2020, where each year represents one data point. a. Would you conclude that higher salaries are causing people to buy more cars? Explain. b. What third variable might be at least partially responsible for the observed correlation?

Explain how you could investigate whether your proposed third variable is a possible contributing cause of the correlation.

Answer: a. No, it would not make sense to conclude that higher salaries are causing people to buy more cars. A more reasonable explanation is given in part b. b. Both variables are increasing over time. Number of new cars sold is increasing because the population is increasing. Salaries are increasing because of inflation.

You could use number of new cars per capita instead of overall, and you could adjust the salary amounts for inflation.

ANSWERS TO THE MINI PROJECTS 1. Find a news story or journal article that describes an observational study in which the author’s actual goal is to try to establish a causal connection. Read the article, and then discuss how well the author has made a case for a causal connection. Consider the factors described in Section 11.4, and explain whether they have been addressed by the author. Finally, discuss the extent to which the author has convinced you that there is a causal connection.

Answer: Make sure the article discusses an observational study and not an experiment or other type of study. The factors that should be addressed include possible explanations of cause and effect, whether or not the connection happened under varying conditions and whether or not possible confounding factors were considered. If those points were not addressed, you should not be convinced of a cause-and-effect relationship even if you can think of reasons why one would be plausible.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 11: Relationships Can Be Deceiving

2. Peruse journal articles and find two examples of scatterplots for which the authors have computed a correlation that you think is misleading. For each case, explain why you think it is misleading.

Answer: The scatterplots should either show two or more clusters of dissimilar points or should have outliers distorting the correlation. The explanation given with the project should address which problem occurred and why it leads to an erroneous or misleading conclusion. If the scatter plot does not have any problems, then the example chosen is not appropriate for this project.

3. (Software required.) Find two variables that are both changing over time. A good source is government data available at the website http://Data.gov.

Record the values of the two variables for at least eight time periods (such as years). a. Create a scatterplot of the data, and comment on the relationship. b. Find the correlation between the two variables. Then find the correlation between each variable and year (or whatever time period you are using).

Comment on whether you think there is an important relationship between the two variables, or if they are related only because they have both changed over time.

Answer: The purpose of this mini-project is to explore how relationships can appear to be stronger than they are, when both variables are changing in the same direction over time. It might take some exploration to find two variables that appear to have a linear relationship. If the two variables do not have that kind of relationship, then the purpose of the mini-project will be lost.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 12: RELATIONSHIPS BETWEEN CATEGORICAL VARIABLES

TABLE OF CONTENTS Answer to the Exercise Questions......................................................................................1 Answer to the Reflection Questions.................................................................................17 Answer to the Mini Projects............................................................................................. 19

ANSWER TO THE EXERCISE QUESTIONS 1. Suppose a study on the relationship between housing (own or rent) and politics included 200 homeowners and 200 renters, and found that they included180 Democrats and 220 Republicans. Is that information sufficient for you to construct a contingency table for the study? If so, construct the table.

If not, explain why not.

Answer: No, it is not sufficient. We do not know how many of the homeowners or renters were Democrats versus Republicans.

2. In a survey for a statistics class project, students in the class were asked whether they commute to school by car (Yes or No) and whether they had ever received a parking ticket (Yes or No). Of the 23 students who commuted to school by car, 19 had received a parking ticket. Of the 25 students who did not commute to school by car, seven had received a parking ticket. a. What is the explanatory variable, and what is the response variable in this survey? b. Create a contingency table for these data. c. Find the risk of having ever received a parking ticket for students who commuted by car and the risk for those who did not commute by car. d. Find the relative risk of receiving a parking ticket for those who commuted by car and those who did not. Write a sentence giving this relative risk in words that would be understood by someone with no training in statistics. e. Find the odds of having received a ticket for students who commuted by car, and write the result in a sentence that would be understood by someone with no training in statistics.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

Answer: a. The explanatory variable is whether or not the student commutes to school by car. The response variable is whether or not the student has ever received a parking ticket. b. The table is shown below.

Received parking ticket?

Yes

No

Total

Commute

19

4

23

Doesn’t commute

7

18

25

Total

26

22

48

c. Risk for commuters is 19/23 = 0.826; risk for non-commuters is 7/25 = 0.28. d. Relative risk is 0.826/0.28 = 2.95. The risk of getting a parking ticket is 2.95 times greater for commuters than for non-commuters. e. Odds of having received a ticket (to not receiving one) for commuters are 19 to 4, or 4.75 to 1.

3. In a survey for a statistics class project, 23 students identified as male and 34 identified as female. The students were asked if they had ever been in a traffic accident, including a minor “fender-bender.” Of the 23 males, 16 reported having been in an accident. Of the 34 females, 18 reported having been in an accident. a. What is the explanatory variable, and what is the response variable in this survey? b. Create a contingency table for these data. c. Find the risk of having been in an accident for males and the risk for females. d. Find the relative risk of having been in an accident for males compared to females. Write a sentence giving this relative risk in words that would be understood by someone with no training in statistics. e. Find the odds of having been in an accident for males, and write the result in a sentence that would be understood by someone with no training in statistics.

Answer: a. The explanatory variable is whether the student identifies as male or female and the response variable is whether the student had ever been in a traffic accident.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

b. The table is shown below.

Ever been in an accident?

Yes

No

Total

Male

16

7

23

Female

18

16

34

Total

34

23

57

c. Risk for males is 16/23 = 0.696; risk for females is 18/34 = 0.529. d. Relative risk is 0.696/0.529 = 1.32. The risk of being in an accident is 1.32 times greater for self-identified males than for self-identified females. e. The odds of being in an accident to not being in one for males are 16 to 7, or about 2.29 to 1.

4. According to the U.S. Centers for Disease Control and Prevention, the “ageadjusted” rate of drowning in Florida in recent years was 2.07 deaths per 100,000 population, while in California, it was 1.11 deaths per100,000. (Age adjustment means that these would be the rates assuming the distribution of ages is similar in the two states.) (Source: https://www.cdc.gov/drowning/data/index.html ) a. Express the rate of drowning for Florida as a percentage of the population; then explain why it is better expressed as a rate per 100,000 than as a percentage. b. Find the relative risk of drowning for Florida compared to California. Write your answer in a sentence that would be understood by someone with no training in statistics.

Answer: a. 2.07 per 100,000 is 0.00207% of the population. It is difficult to interpret such a small percentage. The rate of 2.07 per 100,000 is easier to understand. b. The relative risk is 2.07/1.11 = 1.865. Adjusted for age of people in both states, the risk of drowning is 1.865 times greater in Florida than in California.

5. According to the University of California at Berkeley Wellness Letter (February 1994, p. 1), only 40% of all surgical operations at that time required an overnight stay at a hospital. Rewrite this fact as a proportion, as a risk, and as the odds of an overnight stay. In each case, express the result as a full sentence.

Answer: The proportion of surgical operations requiring an overnight hospital stay (at the time of the report) was 0.40. The risk of having to spend the night at the hospital following a surgical procedure (at the time of the report) was 0.40. The odds of having to spend the night at the hospital versus getting to go home on the same day (at the time of the report) were 40 to 60, or 4 to 6, or 2 to 3.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

6. Science News (25 February 1995, p. 124) reported a study of 232 people, aged 55 or over, who had heart surgery. The patients were asked whether their religious beliefs give them feelings of strength and comfort and whether they regularly participate in social activities. Of those who said yes to both, about 1 in 50 died within 6 months after their operation. Of those who said no to both, about 1 in 5 died within 6 months after their operation. What is the relative risk of death (within 6 months) for the two groups? Write your answer in a sentence or two that would be understood by someone with no training in statistics.

Answer: To compute the relative risk, use 1/5 divided by 1/50, or 50/5 = 10. In an observational study, patients who had heart surgery were asked if they had religious beliefs giving them strength and comfort and whether they participated in social activities. The study found that patients who had neither of these had a 10 times higher risk of dying within 6 months than patients who had both.

7. Raloff (1995) reported on a study conducted by Dimitrios Trichopolous of the Harvard School of Public Health in which researchers “compared the diets of 820 Greek women with breast cancer and 1548 others admitted to Athens-area hospitals for reasons other than cancer.” One of the results had to do with consumption of olive oil, a staple in many Greek diets. The article reported that “women who eat olive oil only once a day face a 25 percent higher risk of breast cancer than women who consume it twice or more daily.” a. The article states that the increased risk of breast cancer for those who consume olive oil only once a day is 25%. What is the relative risk of breast cancer for those who consume olive oil only once a day, compared to those who consume it twice or more? b. What information is missing from this article that would help individuals assess the importance of the result in their own lives?

Answer: a. Relative risk = 1 + (increased risk/100) = 1 + 0.25 = 1.25. b. The baseline risk is missing.

8. The headline in an article in the Sacramento Bee read “Firing someone? Risk of heart attack doubles” (Haney, 1998). The article explained that “between 1989 and 1994, doctors interviewed 791 working people who had just undergone heart attacks about what they had done recently. The researchers concluded that firing someone or having a high-stakes deadline doubled the usual risk of a heart attack during the following week. . . . For a healthy 50-year-old man or a healthy 60-year-old woman, the risk of a heart attack in any given hour without any trigger is about 1 in a million.” Assuming the relationship is indeed as stated in the article, write sentences that could be understood by someone with no training in statistics, giving each of the following for this example. a. The odds ratio b. Increased risk c. Relative risk

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

Answer: a. A person who has to fire someone or is subjected to a high-stakes deadline has double the odds of having a heart attack in the following week than someone who has not had to do those things. (Notice that this is not the same thing as saying this person has twice the odds that that same person would have had if they hadn’t fired someone or had the deadline.) b. Although someone who has to fire someone or who has a high-stakes deadline has twice the risk of having a heart attack during the next week as someone who has not done those things, the increased risk is quite small.

For a healthy 50-year-old man or 60-year-old woman, the risk in any given hour without a trigger is only about 1 in a million, so the increase is an additional 1 in a million. c. A person who has to fire someone or is subjected to a high-stakes deadline has twice the risk of having a heart attack in the following week as someone who has not had to do those things.

9. The previous exercise described the relationship between recently firing someone and having a heart attack. a. Refer to the types of studies described in Chapter 5. What type of study is described in the previous exercise? b. Refer to the reasons for relationships listed in Section 11.3. Which do you think is the most likely explanation for the relationship found between firing someone and having a heart attack? Do you think the headline for this article was appropriate? Explain.

Answer: a. Observational study. More specifically, it’s a retrospective observational study, and probably case-control, although not enough detail is given to know. b. Reasons 3, 4 and 5 are all possibilities. Reason 3 says that firing someone or having a high-stakes deadline may be a contributor to the heart attack.

Reason 4 says that there may be confounding variables; in this case, overall job responsibility and consequent stress are likely to be involved. Reason 5, that both having to fire someone (or having a deadline) and having a heart attack result from a common cause is possible; they could both indirectly result from having the type of personality that leads one to be driven to succeed, to take high risks, and so on. The headline implies but does not explicitly state that firing someone causes the heart attach rate to double.

That conclusion is not warranted because the results are based on an observational study, so the headline could have stated more clearly that a causal relationship is not appropriate.

10. In a test for extrasensory perception (ESP), described in Case Study 13.1 in the next chapter, people were asked to try to use psychic abilities to describe a hidden photo or video segment being viewed by a “sender.” They were then shown four choices and asked which one they thought was the real answer,

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

based on what they had described. By chance alone, 25% of the guesses would be expected to be successful. The researchers tested 354 people and 122 (about 34.5%) of the guesses were successful. In both parts (a) and (b), express your answer in a full sentence. a. What are the odds of a successful guess by chance alone? b. What were the odds of a successful guess in the experiment?

Answer: a. The odds of a successful guess compared to an unsuccessful guess, by chance alone, are 25 to 75 or 1 to 3. b. In the experiment, the odds of guessing successfully to guessing unsuccessfully were 34 to 66 or 17 to 33 (or about 1 to 2).

11. A newspaper story released by the Associated Press noted that “a study by the Bureau of Justice Statistics shows that a motorist has about the same chance of being a carjacking victim as being killed in a traffic accident, 1 in 5000” [Davis (CA) Enterprise, 3 April 1994, p. A9]. Discuss this statement with regard to your own chances of each event happening to you.

Answer: This does not mean that each, or even any, individual has equal chances of the two events. For instance, someone who drives mainly in cities has a higher chance of being a carjacking victim, while someone who drives mainly on highways has a higher chance of being killed in a traffic accident.

12. The Roper Organization (1992) conducted a study as part of a larger survey to ascertain the number of American adults who had experienced phenomena such as seeing a ghost, “feeling as if you left your body,” and seeing a UFO. A representative sample of adults (18 and over) in the continental United States were interviewed in their homes during July, August, and September of 1991.

The results when respondents were asked about seeing a ghost are shown in Table 12.10. Find numbers for each of the following. a. The percentage of the younger group who reported seeing a ghost b. The proportion of the older group who reported seeing a ghost c. The risk of reportedly seeing a ghost in the younger group d. The odds of reportedly seeing a ghost to not seeing one in the older group.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

Answer: a. 212/1525 = 0.1390 or 13.9% b. 465/4377 = 0.1062 c. This is the same as in part (a), 0.1390 or 13.9%. d. The odds are 465 to 3912, or about 0.12 to 1, 12 to 100, 3 to 25.

13. Refer to the previous exercise and Table 12.10. a. What is the relative risk of reportedly seeing a ghost for one group compared to the other? Write your answer in the form of a sentence that could be understood by someone who knows nothing about statistics. b. Repeat part (a) using increased risk instead of relative risk.

Answer: a. The relative risk is 0.1390/0.1062 = 1.31. We would say: People aged 18 to 29 are 1.31 times as likely to report seeing a ghost as those over 30 (or some similar wording). b. People aged 18 to 29 are 31% more likely to report seeing a ghost than those over 30.

14. Using the terminology of this chapter, what name (for example, odds, risk, relative risk) applies to each of the boldface numbers in the following quotes? a. “Fontham found increased risks of lung cancer with increasing exposure to secondhand smoke, whether it took place at home, at work, or in a social setting. A spouse’s smoking alone produced an overall 30 percent increase in lung-cancer risk” (Consumer Reports, January 1995, p. 28). b. “What they found was that women who smoked had a risk [of getting lung cancer] 27.9 times as great as nonsmoking women; in contrast, the risk for men who smoked regularly was only 9.6 times greater than that for male nonsmokers” (Taubes, 26 November 1993, p. 1375).

Answer: a. Increased risk b. Relative risk, in both cases

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

15. Using the terminology of this chapter, what name (for example, odds, risk, relative risk) applies to the following quote? “One student in five reports abandoning safe-sex practices when drunk” (Newsweek, 19 December 1994, p. 73).

Answer: “One student in five” represents a proportion.

16. For each of the following measures, give the value that implies no difference between the two groups being compared. a. Relative risk b. Odds ratio c. Percent increase in risk Answer: a. A relative risk of 1.0 means there is no difference in risk for the two groups. b. An odds ratio of 1.0 means that the odds are the same for the two groups. c. A 0% increase in risk means that the risk is the same for the two groups.

17. According to Consumer Reports (January 1995, p. 29), “among nonsmokers who are exposed to their spouses’ smoke, the chance of death from heart disease increases by about 30%.” Rewrite this statement in terms of relative risk, using language that would be understood by someone who does not know anything about statistics.

Answer: The increased risk is 0.30 (30%), so the relative risk is 1.30. Nonsmokers who are exposed to their spouses’ smoke have a chance of death from heart disease that is 1.3 times the chance for nonsmokers who are not exposed.

18. Drug use among college students in the United States was particularly heavy during the 1970s. Reporting on a study of drinking and drug use among college students in the United States in 1994, a Newsweek reporter wrote: “Why should college students be so impervious to the lesson of the morning after?

Efforts to discourage them from using drugs actually did work. The proportion of college students who smoked marijuana at least once in 30 days went from one in three in 1980 to one in seven last year [1993]; cocaine users dropped from 7 percent to 0.7 percent over the same period.” (19 December 1994, p. 72) Do you agree with the statement that “efforts to discourage them from using drugs actually did work”? Explain your reasoning.

Answer: No. This is a good example of a relationship being due to changes over time. It could be that drug use declined due to changing priorities, etc., and had nothing to do with the “efforts to discourage them from using drugs.”

19. Refer to the quote in the previous exercise, about marijuana and cocaine use in 1980 and 1993.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

a. What was the relative risk of cocaine use for college students in 1980 compared with college students in 1993? Write your answer as a statement that could be understood by someone who does not know anything about statistics. b. Are the figures given for marijuana use (for example, “one in three”) presented as proportions or as odds? Whichever they are, rewrite them as the other.

Answer: a. A student in 1980 was 10 times as likely to use cocaine as one in 1993. b. They are proportions, although they are expressed differently than proportions are usually expressed. The statement “one in three” means that 1/3 of students smoked in 1980. The odds would be: In 1980, the odds that a student smoked marijuana were 1 to 2. By 1993 they were reduced to 1 to 6. Equivalently, you could say that the odds of not smoking were 2 to 1 in 1980 and 6 to 1 in 1993.

20. A case-control study in Berlin, reported by Kohlmeier, Arminger, Bartolomeycik, Bellach, Rehm, and Thamm (1992) and by Hand and colleagues (1994), asked 239 lung cancer patients and 429 controls (matched to the cases by age and sex) whether they had kept a pet bird during adulthood. Of the 239 lung cancer cases, 98 said yes. Of the 429 controls, 101 said yes. a. Construct a contingency table for the data. b. Compute the risk of lung cancer for bird and non-bird owners for this study. c. Can the risks of lung cancer for the two groups, computed in part (b), be used as baseline risks for the populations of bird and non-bird owners? Explain. d. How much more likely is lung cancer for bird owners than for non-bird owners in this study; that is, what is the increased risk? e. What information about risk would you want, in addition to the information on increased risk in part (d) of this problem, before you made a decision about whether to own a pet bird?

Answer: a. Here is the Table: Lung Cancer

No Lung Cancer

Total

Bird

98

101

199

No Bird

141

328

469

Total

239

429

668

b. For bird owners the risk is 98/199 = 0.4925 = 49.25%. For non-bird owners it is 141/469 = 0.301 = 30.1%. c. No. This was a case-control study, not based on a random sample from the population. Clearly we would not find the risk of lung cancer among bird owners in general to be 0.49. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

d. The relative risk is 0.49/0.30 = 1.63 (or 1.64 if you don’t round off as soon) so the increased risk is 63%. e. You would want to know the baseline risk of lung cancer without owning a bird. If it is very small, then an increased risk of 63% may not mean much to someone who really likes birds.

21. In Example 12.3 and Table 12.4, we presented data from a study examining the relationship between having a dog at home during infancy and having an allergy to dogs at age 7. Refer to the results given in Table 12.4, on page 249. a. What is the risk of having an allergy at age 7 for children who did not have exposure as an infant? b. What is the risk of having an allergy at age 7 for children who had a dog at home as an infant? c. Find the relative risk of having a dog allergy at age 7 under the two infant exposure conditions. Write your answer in a sentence that would be understood by someone with no training in statistics. d. Find the increased risk of having a dog allergy at age 7 for children who did not have infant exposure compared with those who did. Write your answer in a sentence that would be understood by someone with no training in statistics.

Answer: a. The risk is 19/220 = 0.0864 b. The risk is 6/184 = 0.0326. c. The relative risk is 0.0864/0.0326 = 2.65. Children who were not exposed to a dog as an infant are 2.65 times as likely to have a dog allergy at age 7 as children who were exposed to a dog. d. Increased risk = (2.65 – 1) × 100% = 165%. There is a 165% higher risk of dog allergies at age 7 for children who were not exposed to dogs as an infant, compared to those who were.

22. Refer to Exercise 21 and Table 12.4, investigating the risk of having a dog allergy at age 7 based on infant exposure to dogs. a. What are the odds of having an allergy at age 7 to not having one for children with no infant exposure to dogs? b. What are the odds of having an allergy at age 7 to not having one for children who had a dog at home as an infant? c. What is the odds ratio for the two conditions? Write your answer in a sentence that would be understood by someone with no training in statistics.

Answer: a. The odds of having an allergy at age 7 to not having one for children with no infant exposure are 19 to 201, or about 1 to 10.6. b. The odds of having an allergy at age 7 to not having one for children with infant exposure are 6 to 178, or about 1 to 30. c. The odds ratio is (19)(178)/(201)(6) = 2.8. The odds of having an allergy at age 7 were 2.8 times higher for children with no infant dog exposure than for children with infant dog exposure.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

23. Explain the difference between risk and odds. Use the summary on page 250 for guidance.

Answer: Risk compares an outcome to the total; odds compare one outcome to another outcome.

24. A statement quoted in this chapter was, “Use of aspirin for five years or longer was tied to a 30-percent reduction in skin cancer risk [for women aged 50 to 79], according to findings published in the journal Cancer.” (Pittman, 14 March 2013) a. What term from this chapter applies to the phrase “30 percent reduction in skin cancer risk?” b. Based on the information that there was a 30% reduction in skin cancer risk when using aspirin compared to not using aspirin, find the relative risk of skin cancer when using aspirin compared to not using aspirin. c. Is it possible to calculate the baseline risk of skin cancer from the information provided in the quote? If so, calculate it. If not, briefly explain why not.

Answer: a. The statement refers to decreased risk. b. Relative risk = 1 + (increased risk/100), and this case the “decreased risk” is equivalent to a negative “increased risk.” So relative risk = 1 + (−30/100) = 1 – 0.3 = 0.7. c. No. Baseline risk cannot be calculated from either relative risk or increased risk.

25. The data in Table 12.11 are reproduced from Case Study 12.1 and represent employees laid off by the U.S. Department of Labor. a. Compute the odds of being retained to being laid off for each ethnic group. b. Use your results in part (a) to compute the odds ratio and confirm that it is about 3.0, as computed in Case Study 12.1 (where the shortcut method was used).

Answer: a. The odds are 1382 to 130, or about 10.6 to 1 for Blacks and 2813 to 87 or about 32.3 to 1 for Whites. b. Odds ratio is (2813/87)  (1382/130) = 3.04.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

26. One of your teachers tells your class that skipping class often can lead to dire consequences, and that data from previous classes show that students who often skip class have twice the risk of failing compared to students who attend regularly, even though class attendance and participation are not part of the grade. a. What term from this chapter applies to the phrase “twice the risk” (for example, odds ratio, increased risk, etc.)? b. Give two reasons why the information provided by the teacher does not imply that if you skip class regularly you will probably fail the class.

Answer: a. “Twice the risk” refers to relative risk. b. Reason 1: The relative risk is based on an observational study, so it does not imply that skipping class causes people to get lower grades. Reason 2: The teacher did not provide the baseline risk. Perhaps the risk doubles from one student in the class to two students in the class.

27. Kohler (1994, p. 427) reports data on the approval rates and ethnicity for mortgage applicants in Los Angeles in 1990. Of the 4096 Black applicants, 3117 were approved. Of the 84,947 White applicants, 71,950 were approved. a. Construct a contingency table for the data. b. Compute the proportion of each ethnic group that was approved for a mortgage.

Answer: a. Here is the contingency table: Application Approved?

Yes

No

Total

Black

3,117

979

4,096

White

71,950

12,997

84,947

Total

75,067

13,976

89,043

b. 76.1% of blacks were approved and 84.7% of whites.

28. Refer to the previous exercise. a. Compute the ratio of the proportions of each ethnic group that were approved for a mortgage. Would that ratio be more appropriately called a relative risk or a selection ratio? Explain. b. Would the data pass the four-fifths rule used in employment and described in Case Study 12.1? Explain.

Answer: a. The ratio is 76.1/84.7 = 0.898, a selection ratio. b. Yes, it would pass because it is over 80% or 4/5. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

29. A classic and well-publicized example of Simpson’s Paradox, published by Bickel, Hammel, and O’Connell (1975), examined admission rates for men and women who had applied to graduate programs at the University of California at Berkeley. The actual breakdown of data for specific programs is confidential, but the point can be made with similar, hypothetical numbers. For simplicity we will assume there are only two graduate programs. The figures for acceptance to each program are shown in Table 12.12. a. Combine the data for the two programs into one aggregate table. What percentage of all men who applied were admitted? What percentage of all women who applied were admitted? Which gender was more successful? b. What percentage of the men who applied did Program A admit? What percentage of the women who applied did Program A admit? Repeat the question for Program B. Which sex was more successful in getting admitted to Program A? Program B? c. Explain how this problem is an example of Simpson’s Paradox. Provide a potential explanation for the observed figures by guessing what type of programs A and B might have been (in 1975). (Hint: Which program was easier to get admitted to overall? Which program interested men more?

Which program interested women more?)

Answer: a. Here is the combined table: Admit

Deny

Total

Percent Admitted

Men

450

550

1,000

450/1,000 = 45%

Women

175

325

500

175/500 = 35%

Total

625

875

1,500

The last column shows that 45% of the men and 35% of the women were admitted. A higher percentage of men than women were admitted, so the men were more successful. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

b. Program A admitted 400/650 = 61.5% of the men, and 50/75 = 67% of the women. Program B admitted 50/350 = 14.3% of the men and 125/425 = 29.4% of the women. Therefore, it appears that both programs admitted a higher percentage of women than men! c. Simpson’s Paradox occurs when combining groups reverses the direction of the relationship from what it was when the groups were separate. In this case, each program admitted a higher percentage of women, yet overall a lower percentage of women were admitted. What happened was that Program B was harder to get into, and was the one for which the majority of women applied. Of the applicants to Program A, which was relatively easy to get into, only about 10% were women. However, over half of the applicants to Program B, which was hard to get into, were women. These may have been something like Mathematics (Program A) and Psychology (Program B).

For Exercises 30 to 32: Read the summary of News Story 12 in the Appendix and on the companion website, “New program for teen drivers with ADHD has significantly reduced crashes.”

30. What is the statistical term (relative risk, odds, etc.) for “four times” in the following quote? “Teen drivers are four times more likely to be in a crash than adult drivers.”

Answer: Relative risk 31. Identify the statistical term (relative risk, odds, etc.) for the information in bold in each of the following quotes. a. “[The study] found the group that underwent the training had a 40% reduction in crashes and near crashes.” b. “Teens with ADHD are twice as likely to be in a crash compared to teens who do not have the disorder.”

Answer: a. Increased risk, but the term “reduction” means that it was a decreased risk (or a negative increased risk) b. “Twice as likely” is a relative risk.

32. The journal article accompanying the news story (Epstein et al., 2022) provided additional details. The study was a randomized experiment of teens with attention-deficit hyperactivity disorder (ADHD), in which some received driver training to increase focus (the intervention group) and the others did not. For a year following the training, a recording device in their cars measured events that included a certain amount of acceleration or more, and whether a collision or near collision followed each of those events. There were 3213 such events in the intervention group, with 110 (near) crashes. There were 2818 such events in the control group, with 159 (near) crashes.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

a. Create a contingency table for the data, using “group” as the rows and collision or not as the columns. b. The article stated that the relative risk of a (near) collision for the intervention group compared with the control group was 0.6. Show how that number was calculated. c. Calculate the relative risk in the other direction, for the control group compared to the intervention group. d. Using the relative risk from part (c), calculate the increased risk of a (near) collision for the control group. Write a sentence reporting your result that would be understood by someone with no training in statistics.

Answer: a. Here is the contingency table: Near crash or crash?

Group

Yes

No

Total

Intervention

110

3,103

3,213

Control

159

2,659

2,818

Total

269

5,762

6,031

b. The risk for the intervention group is 110/3,213 = 0.0342. The risk for the control group is 159/2,659 = 0.0598, so the relative risk is 0.0342/0.0598 = 0.57, or about 0.6. c. The relative risk in the other direction is 0.0598/0.0342 = 1.75. (If you used the rounded off value of 0.6 as the relative risk instead of 0.57, your answer would be 1/0.6 = 1.67.) d. Increased risk = (Relative risk – 1) × 100% = (1.75 – 1) × 100% = 75%. The group that did not receive the intervention training had a 75% greater risk of a (near) collision than the group that did receive the training.

For Exercises 33 and 34: Refer to the summary of Additional News Story 10 in the Appendix, “‘Keeping the faith’ UC Berkeley researchers links weekly church attendance to longer, healthier life.” Based on the information in the story, identify or calculate a numerical value for each of the following.

33. The increased risk of dying from circulatory diseases for people who attended religious services less than once a week or never, compared to those who attended at least weekly Answer: The increased risk of dying of circulatory disease is reported to be 21%.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

34. a. The relative risk of dying from circulatory diseases for people who attended religious services less than once a week or never, compared to those who attended at least weekly b. The increased risk of dying from digestive diseases for people who attended religious services less than once a week or never, compared to those who attended at least weekly c. The relative risk of dying from digestive diseases for people who attended religious services less than once a week or never, compared to those who attended at least weekly.

Answer: a. The relative risk is 1.21. b. The increased risk of dying of digestive disease is reported to be 99%. c. The relative risk is 1.99.

35. News Story 10, described in the Appendix, is titled “Churchgoers live longer, study finds.” One of the statements in the news story is “women who attend religious services regularly are about 80 percent as likely to die as those not regularly attending.” Discuss the extent to which each of the three “common ways the media misrepresent statistics about risk” from Section 12.3, listed as parts a–c, apply to this quote. a. The baseline risk is missing. b. The time period of the risk is not identified. c. The reported risk is not necessarily your risk.

Answer: a. The baseline risk is indeed missing, so we can’t say what the death rates are for the two groups. Of course it depends on age, so it is not possible to give a baseline risk. Everyone eventually dies! b. The time period of the risk is not identified. We don’t know if it’s in a given year, or in a decade, or since the beginning of the study in 1965. c. The reported risk is not necessarily your risk at all. There are many possible confounding variables in this study, such as social support. Your risk may in fact increase if you attend church, or may not change at all, or may decrease. The reported comparison is an average for everyone.

For Exercises 36 and 37: In Original Source 10, “Religious attendance and cause of death over 31 years” (not available on the companion website), the researchers used a complicated statistical method to assess relative risk by adjusting for factors such as education and income. The resulting numbers are called “relative hazards” instead of relative risks (abbreviated RH), but have the same interpretation as relative risk.

The relative risks and increased risks in Exercises 36 and 37 are from Table 4 of the article. Write a sentence or two that someone with no training in statistics would understand, presenting each of these in context.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

36. The relative risk of dying from all causes for women under age 70, for those who do not attend weekly religious services compared with those who do, is 1.22.

Answer: Women under the age of 70 who attended religious services less than once a week or never were 1.22 times more likely to die during the course of the study than those who attended religious services at least weekly, after adjusting for factors such as education and income.

37. For those who do not attend weekly religious services compared with those who do: a. The increased risk of dying from all causes for men under age 70 is 0.27. b. The relative risk of dying from cancer for men age 701 is 1.00. c. The relative risk of dying from cancer for men under age 70 is 0.74.

Answer: a. Men under the age of 70 who attended religious services less than once a week or never had a 27 percent greater overall risk of dying than men who attended religious services at least weekly, after adjusting for factors such as education and income. b. Men age 70 and older who attended religious services less than once a week or never were no more or less likely to die of cancer than men who attended religious services at least once a week. c. Men under age 70 who attended religious services less than once a week or never were only .74 times (about three-fourths) as likely to die of cancer than men who attended religious services at least weekly, after adjusting for factors such as education and income.

ANSWER TO THE REFLECTION QUESTIONS 1. For two categorical variables with two categories each, suppose you know that 100 individuals fall into each of the two categories for one of the variables, and 200 fall into each category for the other variable. Is that sufficient information to create a contingency table for the relationship between them? If so, create the table. If not, explain what additional information would be needed.

Answer: That is not enough information to create the table. You need to know how the individuals in each category of one variable break down separately into the categories of the other variable 2. Suppose you learn that the relative risk of a certain disease is 3.0 for people who drink a beverage that you like to drink, compared to people who don’t drink it. Explain why that is not sufficient information for you to know whether your risk for the disease is high. What additional information might help you learn about what your risk is?

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

Answer: Knowing the relative risk does not give you any information about the actual risk for each group. You need to know the baseline risk (for the group that does not drink the beverage), or the risk for the group that drinks the beverage.

3. In 2020, shortly after COVID-19 became a worldwide pandemic, various sources of data indicated that individual risk could be related to blood type. Estimates were that individuals with type A blood had a somewhat higher risk of getting the disease than did those with other blood types, with an odds ratio of about 1.20. Explain what this means in words that someone with no training in statistics would understand.

Answer: The odds of getting COVID-19 to not getting it were 1.2 times higher for those with Type A blood than for those with other blood types.

4. Suppose the risk of having an emergency room visit in the next year is computed for people who are left-handed compared to people who are right-handed. Specify whether each of the following outcomes would be possible. If not, explain why not. If so, interpret what it means. a. A relative risk of −0.5 b. An increased risk of −0.5 c. A relative risk of 0 d. An increased risk of 0 Answer: a. A negative relative risk is not possible. b. An increased risk of –0.5 is the same as a decreased risk of 0.5, so it is possible. c. A relative risk of 0 would only be possible if the risk for right-handed people were 0, which is not possible. d. An increased risk of 0 is possible. It means that both groups have the same risk.

5. Explain why Simpson’s Paradox is called a paradox.

Answer: The definition of a paradox is something that seems to be logically contradictory. In Simpson’s Paradox, the contradiction is that a relationship appears to be in one direction for the whole group, but switches to the other direction when broken down into subgroups.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

ANSWER TO THE MINI PROJECTS 1. You are interested in determining whether there is a relationship among data cross-classified by two categorical variables for a given situation. Do not get the data from a book, website or journal; collect it yourself. Be sure to get counts of at least five in each cell and the individuals you use are not related to each other in ways that would influence their data. a. Create a contingency table for the data. b. Compute and discuss the risks and relative risks. Are those terms appropriate for your situation? Explain. c. Write a summary of your findings, including whether a cause-and-effect conclusion could be made if you observed a relationship.

Answer: This project is relatively straightforward in that it involves the same kinds of computations as the exercises. Be careful about collecting the data, to be sure the individuals are independent of each other and the same variables are measured on each one. Some guidelines to check are as follows.

For part (a), make sure the convention of putting the explanatory variable in the rows and the response in the columns is followed, if that distinction can be made.

For part (b) make sure all parts are answered and that relative risk is computed rather than increased risk.

For part (c) remember that a cause-and-effect conclusion can be made only if the data were collected as a randomized experiment.

2. Find a news story that discusses a study showing increased (or decreased) risk of one variable based on another. Write a report evaluating the information given in the article and discussing what conclusions you would reach based on the information in the article. Discuss whether any of the features in Section 12.3, “Misleading Statistics about Risk,” apply to the situation.

Answer: The project is designed to illustrate that relative risk is a common feature in news articles about studies. It should be easy to find such an example. Make sure the report discusses all three of the features listed in Section 12.3.

3. Refer to News Story 17, “Drink 2 to 3 cups of coffee a day for longer life, better heart health” summarized in the Appendix and on the companion website, and the accompanying journal article. Links to the news story and the journal article are provided on the companion website. Write a two- to four-page report describing how the study was done and what was found in terms of hazard ratios, which you can interpret as relative risks. (Note that complicated statistical methods were used that adjusted for things like age and lifestyle

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 12: Relationships Between Categorical Variables

risk factors, but you should still be able to interpret the general results reported in the article.) Discuss shortcomings you think might apply to the results, if any.

Answer: The purpose of this assignment is for students to read both a news story and the original research article from which it was written. Students should interpret the results based on the hazard ratios. They also should recognize that a cause and effect conclusion can’t be made, because clearly the results are based on an observational study.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 13: STATISTICAL SIGNIFICANCE FOR 2 × 2 TABLES

TABLE OF CONTENTS Answer to the Exercise Questions......................................................................................1 Answer to the Reflection Questions................................................................................ 19 Answer to the Mini Projects............................................................................................. 20

ANSWER TO THE EXERCISE QUESTIONS 1. The table below shows the partial results of a survey of 500 middle-aged drivers who were asked whether they had ever had a speeding ticket, and whether the majority of their driving is on surface streets or highways. Specify the number of people who would fall into each cell if there is no relationship between where one mostly drives and getting a speeding ticket. Explain your reasoning.

Answer: We would expect numbers as shown in the table below. Because 4/5 people had a speeding ticket, if there is no relationship we would expect 4/5 of each driver group to have had one. That would be 4/5(200) = 160 of those who mainly drive on surface streets, and 4/5(300) = 240 of those who mainly drive on highways.

Yes

No

Streets

160

40

Highways

240

60

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

2. If there is a relationship between two variables in a population, which is more likely to result in a statistically significant relationship in a sample from that population—a small sample, a large sample, or are they equivalent? Explain.

Answer: A large sample is much more likely to result in a statistically significant relationship. The more data available, the easier it is to rule out chance as an explanation for a relationship observed in the sample.

3. If there is no relationship between two variables in a population, which is more likely to result in a statistically significant relationship in a sample—a small sample, a large sample, or are they equivalent? Explain. (Hint: If there is no relationship in the population, how often will the p-value be less than 0.05?

Does it depend on the size of the sample?)

Answer: They are equivalent; the probability is usually set at 0.05. The result will be statistically significant by chance alone 5% of the time if there is no relationship in the population.

4. Suppose a relationship between two variables is found to be statistically significant. Explain whether each of the following is true in that case. a. There is definitely a relationship between the two variables in the sample. b. There is definitely a relationship between the two variables in the population. c. It is likely that there is a relationship between the two variables in the population.

Answer: a. Yes, there is definitely a relationship in the sample. If there weren’t, the chi-square statistic would be 0 and the p-value would be 1.0. b. There isn’t necessarily a relationship in the population. Using the usual significance level of 5%, in about 5% of tests where there is no relationship in the population, the sample will produce a statistically significant relationship. c. Yes, it is likely that there is a relationship in the population.

5. Are null and alternative hypotheses statements about samples, about populations, or does it depend on the situation? Explain.

Answer: Null and alternative hypotheses are never about samples, they are always about populations. We know whether there is a relationship in the sample because we have the data. We don’t have to make hypotheses about samples.

6. Explain what “expected counts” represent. In other words, under what condition are they “expected”?

Answer: They are expected (on average) if the null hypothesis is true.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

7. For each of the following possible conclusions, state whether it would follow when the p-value is less than 0.05 (assuming a level of 0.05 is desired for the test). a. Reject the null hypothesis. b. Reject the alternative hypothesis. c. Accept the null hypothesis. d. Accept the alternative hypothesis. e. The relationship is not statistically significant. f. The relationship is statistically significant. g. We do not have enough evidence to reject the null hypothesis.

Answer: a. Yes. b. No. (That would be equivalent to accepting the null hypothesis, which is never appropriate, and we do not talk about rejecting the alternative hypothesis in any case.) c. No. (It is never appropriate to accept the null hypothesis.) d. Yes. e. No. f. Yes. g. No. (We do have enough evidence to reject the null hypothesis in this case.)

8. For each of the following possible conclusions, state whether it would follow when the p-value is greater than 0.05 (assuming a level of 0.05 is desired for the test). a. Reject the null hypothesis. b. Reject the alternative hypothesis. c. Accept the null hypothesis. d. Accept the alternative hypothesis. e. The relationship is not statistically significant. f. The relationship is statistically significant. g. We do not have enough evidence to reject the null hypothesis.

Answer: a. No. b. No, we do not talk about rejecting the alternative hypothesis in any case. c. No, the correct terminology is that we fail to reject the null hypothesis. d. No. e. Yes. f. No. g. Yes. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

9. For each of the following situations, would a chi-square test based on a 2 × 2 table using a level of 0.05 be statistically significant? Justify your answer. a. chi-square statistic =1.42 b. chi-square statistic = 14.2 c. p-value =0.02 d. p-value =0.15 Answer: a. No: 1.42 < 3.84. b. Yes: 14.2 > 3.84. c. Yes: 0.02 < 0.05. d. No: 0.15 > 0.05.

10. For each of the following situations, would a chi-square test based on 2 × 2 table using a level of 0.01 be statistically significant? Justify your answer. a. chi-square statistic = 1.42 b. chi-square statistic = 14.2 c. p-value = 0.02 d. p-value = 0.15 Answer: a. No: 1.42 < 6.63. b. Yes: 14.2 > 6.63. c. No: 0.02 > 0.01. d. No: 0.15 > 0.01.

11. Use software (such as Excel), a calculator, or a website to find the p-value for each of the following chi-square statistics calculated from a 2 × 2 table. You may round off your answer to three decimal places. a. chi-square statistic = 3.17 b. chi-square statistic = 5.02 c. chi-square statistic =7.88 d. chi-square statistic =10.81 Answer: a. 0.075 b. 0.025 c. 0.005 d. 0.001

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

12. The chi-square test described in this chapter can be used for tables with more than two rows and/or columns. Use software (such as Excel), a calculator, or a website to find the p-value for each of the following chi-square statistics calculated from a table of the specified number of rows and columns. You may round off your answer to three decimal places. In each case, specify the degrees of freedom you used to find the p-value. (Hint: Remember that in general, degrees of freedom for a table with r rows and c columns are df = (𝑟 – 1)(𝑐 – 1). Then make a conclusion about whether you would reject the null hypothesis for a test with level 0.05. a. Two rows and three columns; chi-square statistic = 7.4 b. Three rows and three columns; chi-square statistic = 11.15 c. Four rows and three columns; chi-square statistic = 7.88 d. Four rows and two columns; chi-square statistic = 12.20 Answer: a. 𝑑𝑓 = 2, 𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 0.025, reject the null hypothesis. b. 𝑑𝑓 = 4, 𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 0.025, reject the null hypothesis. c. 𝑑𝑓 = 6, 𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 0.25, do not reject the null hypothesis. d. 𝑑𝑓 = 3, 𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 0.01, reject the null hypothesis.

13. In each of the following situations, specify the population. Also, state the two categorical variables that would be measured for each unit in the sample and the two categories for each variable. a. Researchers want to know if there is a relationship between having graduated from college or not and voting in the last presidential election, for all registered voters over age 25. b. Researchers want to know if there is a relationship between smoking (either partner smokes) and divorce for people who were married between 1990 and 2000. c. Researchers classify midsize cities in the United States according to whether the city’s median family income is higher or lower than the median family income for the state in which the city is located. They want to know if there is a relationship between that classification and airport availability, defined as whether or not one of the 30 busiest airports in the country is within 50 miles of the city.

Answer: a. Graduated from college (yes, no); (2)Voted in the last presidential election (yes, no). b. The population is people who were married between 1990 and 2000. The two variables and their categories are (1)At least one partner smokes (yes, no) and (2)Got divorced (yes, no).

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

c. The population is midsize cities in the United States. The two variables and their categories are (1)The city’s median family income compared to median family income for the state (higher, lower); (2)One of the 30 busiest airports in the country is within 50 miles of the city (yes, no).

14. Refer to the previous exercise. In each case, state the null and alternative hypotheses.

Answer: a. The null hypothesis is that there is no relationship between having graduated from college and voting in the last presidential election for the population of registered voters over age 25. The alternative hypothesis is that there is a relationship between having graduated from college and voting in the last presidential election for voters over age 25. b. The null hypothesis is that there is no relationship between smoking and getting divorced for people who were married between 1990 and 2000. The alternative hypothesis is that there is a relationship. c. The null hypothesis is that median family income for mid-sized cities is not related to whether there is a major airport within 50 miles. The alternative hypothesis is that median family income is related to whether there is a major airport within 50 miles.

15. A political poll based on a random sample of 1000 likely voters classified them by age group (30 or younger versus over 30) and asked them if they planned to vote for Candidate A or Candidate B in the upcoming election. Results are shown in the accompanying table. a. State the null and alternative hypotheses in this situation. b. Calculate the expected counts. c. Explain in words the rationale for the expected counts in the context of this example. d. Calculate the value of the chi-square statistic. e. Make a conclusion using a level of 0.05. State the conclusion in the context of this situation.

Answer: a. Null hypothesis: There is no relationship between age group and preferred candidate for the population of likely voters. Alternative hypothesis: There is a relationship between age group and preferred candidate for the population of likely voters. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

b. The expected counts are (450)(500)/1000 = 225

(450)(500)/1000 = 225

(550)(500)/1000 = 275

(550)(500)/1000 = 275

c. Half of the people asked said they would vote for Candidate A and half for Candidate B. Therefore, if there is no relationship between age group and preferred candidate, we would expect half of those 30 and under to vote for each candidate and half of those over 30 to vote for each candidate.

Thus, we would expect 225 males and 275 females to vote for each candidate. d. Chi-square statistic = (200 − 225) + (250 − 225) + (300 − 225) + (250 − 275) = 10.1.

2

225

2

225

2

275

2

275

e. Reject the null hypothesis because 10.1 > 3.84. There appears to be a relationship between age group and preferred candidate in the population of likely voters.

16. Refer to Example 13.1, investigating the relationship between taking aspirin and risk of heart attack. As shown in Table 12.1 on page 247, 104 of the 11,037 aspirin takers had heart attacks, whereas 189 of the 11,034 placebo takers had them. The hypotheses for this example are already given in this chapter. a. Calculate the expected counts. b. Calculate the value of the chi-square statistic. c. Make a conclusion using a level of 0.01. State the conclusion in the context of this situation.

Answer: a. The hypotheses are given on page 277. The expected counts and chi-square statistic are shown in the box below. (These were computed using the statistical software Minitab [Version 13], but could have been handcalculated.)

Expected counts are printed below observed counts Heart attack?

Yes No Total Aspirin 104 10933 11037 146.52 10890.48

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

Placebo 189 10845 11034 146.48 10887.52 Total 293 21778 22071 Chi-Sq = 12.339 + 0.166 + 12.343 + 0.166 = 25.014 b. The chi-square statistic is 25.014, as shown in the computer output. c. The chi-square statistic of 25.014 is much larger than the cutoff of 6.63 required for a level of 0.01, so the relationship is statistically significant. We can conclude that there is a relationship between taking aspirin and risk of a heart attack in the population.

17. In Exercise 12 of Chapter 12 results were given for a Roper Poll in which people were classified according to age and were asked if they had ever seen a ghost.

The results from asking Minitab to compute the chi-square statistic are shown in Figure 13.5. (The format is slightly different from the format shown in Figures 13.1 to 13.3 because a different version of Minitab was used here.) What can you conclude about the relationship between age group and reportedly seeing a ghost?

Answer: The decision can be made based on either the reported chi-square statistic of 11.967, or the reported p-value of 0.001. Because the chi-squared statistic of 11.967 exceeds even the value of 6.63 required for a level of 0.01, we can say that there is a statistically significant relationship between age group and reportedly seeing a ghost using a level of 0.01. From the p-value, we make the same conclusion except that we can now say that it holds even if we use a level as low as 0.001.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

18. In a national survey, 1500 randomly selected adults will be asked if they favor or oppose a ban on texting while driving and if they have personally texted while driving during the previous month. Write null and alternative hypotheses about the relationship between the two variables in this situation. Make your hypotheses specific to this situation.

Answer: Null hypothesis: There is no population relationship between opinion about banning texting while driving and personally having texted while driving.

Alternative hypothesis: There is a population relationship between opinion about banning texting while driving and personally having texted while driving.

19. This is a continuation of Exercise 20 in Chapter 12. A case-control study in Berlin, reported by Kohlmeier et al. (1992) and Hand et al. (1994), asked 239 lung cancer patients and 429 controls (matched to the cases by age and sex) whether they had kept a pet bird during adulthood. Of the 239 lung cancer cases, 98 said yes. Of the 429 controls, 101 said yes. a. State the null and alternative hypotheses for this situation. b. Construct a contingency table for the data. c. Calculate the expected counts. d. Calculate the value of the chi-square statistic. e. Make a conclusion about statistical significance using a level of 0.05. State the conclusion in the context of this situation.

Answer: a. Null hypothesis: There is no relationship in the population between having a pet bird during adulthood and getting lung cancer. Alternative hypothesis: There is a relationship in the population between having a pet bird during adulthood and getting lung cancer. b. Here is the contingency table: Lung Cancer

No Lung Cancer

Total

Bird

98

101

199

No Bird

141

328

469

Total

239

429

668

c. Here are the expected counts, with the calculations shown: Lung Cancer

No Lung Cancer

Total

Bird

(239)(199)/668 = 71.2

199 − 71.2 = 127.8

199

No Bird

239 − 71.2 = 167.8

469 − 167.8 = 301.2

469

Total

239

429

668

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

d. The chi-squared statistic is (98 − 71.2)2/71.2 + (101 − 127.8)2/127.8 + (141 − 167.8)2/167.8 + (328 − 301.2)2/301.2 = 10.09 + 5.62 + 4.28 + 2.39 = 22.37 e. Because the chi-squared statistic is (much) larger than 3.84, the relationship is statistically significant. There is indeed a population relationship between bird ownership and lung cancer that is not likely to be due to chance.

20. If a relationship has practical significance, does it guarantee that statistical significance will be achieved in every study that examines it? Explain.

Answer: No. If the sample size of a study is small, there may not be enough information to declare statistical significance, even if the relationship has practical significance. Also, in some studies there may be an unfortunate “luck of the draw” in that sample results may not be consistent with the truth in the population.

21. A Gallup poll asked the same question in multiple years. The essence of the question was whether respondents thought that increases in Earth’s temperature in the past century were due to “human activities” or “natural changes.” (For exact wording, see https://pollingreport.com/enviro.htm.) In 2010, 519 respondents said “human activities” and 520 said “natural changes” or “unsure.” In 2019, 686 said “human activities” and 353 said “natural changes” or “unsure.” a. Create a contingency table for this example, with years as the rows and response as the columns. b. Gallup uses representative samples of adults in the United States, so the sample proportions for each response are estimates of the population proportions who would have responded that way in 2010 and in 2019. Write the null and alternative hypotheses that would be used to test whether the proportion of the population who would have responded “human activity” changed from 2010 to 2019. c. The chi-square test statistic for this example is 55.1. Write a conclusion for the test in statistical language and in words that would be understood by someone with no training in statistics.

Answer: a.

Year 2010 2019

Human 519 686

Natural/unsure 520 353

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

b. Null hypothesis: The proportion of adults in the United States who thought global warming was caused by human activities was the same in 2010 and 2019. Alternative hypothesis: The proportion of adults in the United States who thought global warming was caused by human activities was different in 2010 and 2019. c. Reject the null hypothesis at the 0.05 level of significance because 55.1 > 3.84. The proportion of adults in the United States who thought global warming was caused by human activities was different in 2019 than it had been in 2010.

22. This is a continuation of Exercise 25 in Chapter 12. The data (shown in the accompanying table) are reproduced from Case Study 12.1 and represent employees laid off by the U.S. Department of Labor.

Minitab computed the chi-square statistic as 66.595. Explain what this means about the relationship between the two variables. Include an explanation that could be understood by someone with no knowledge of statistics. Make the assumption that these employees are representative of a larger population of employees.

Answer: The data provide overwhelming evidence that there is a relationship between ethnic group and whether or not someone is laid off. The results from this sample are so strong that we can clearly conclude that there is a relationship in the population of such workers.

Exercises 23 to 25 are based on a Pew Center Research study that surveyed adults living in urban, suburban, and rural areas in the United States (Source: Pew Research Center, 2018). One of the questions asked respondents if they were satisfied with the quality of life in their community. For these exercises, the numbers responding “very satisfied” will be compared for those living in urban areas and in rural areas. The data and chi-square test are shown in the Minitab output in Figure 13.6

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

23. Assuming that these respondents represent all adults who live in urban and rural communities, write the appropriate null and alternative hypotheses that can be tested using these results.

Answer: Null hypothesis: The proportion of adults in the United States who are very satisfied with the quality of life in their community is the same for the population living in urban areas as for the population living in rural areas.

Alternative hypothesis: The proportion of adults in the United States who are very satisfied with the quality of life in their community is different for those living in urban areas and those living in rural areas. An equivalent way to write the null hypotheses is to say that there is no relationship between type of community (urban population or rural population) and being very satisfied with quality of life in one’s community The alternative hypothesis would be that there is a relationship.

24. Using the results produced by Minitab, make a statistical conclusion using 0.05 as the significance level.

Answer: The p-value is 0.054, which is just slightly greater than 0.05. Do not reject the null hypothesis.

25. As shown in Figure 13.6, the p-value for the test in this example is 0.054.

Suppose a reporter wrote, “A new study by the Pew Research Center has resulted in the conclusion that there is no difference in the proportions of urban and rural dwellers who are very satisfied with the quality of life in their communities.” Do you agree with the reporter’s conclusion? If so, explain why.

If not, rewrite it to represent what you think is a more accurate conclusion.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

Answer: The conclusion is not correct. The reporter has made the mistake of accepting the null hypothesis, when the correct conclusion is that there isn’t enough evidence to reject it. A more accurate conclusion would be something like the following: A new study by the Pew Research Center found that 23% of 1,239 urban dwellers and 26% of 2,085 rural dwellers surveyed said they were very satisfied with the quality of life in their community. This difference in percentages was too small to conclude that there is a difference in the percentages of the populations of urban and rural dwellers who would answer that way.

26. Is it harder to find statistical significance using a chi-square test with level 0.05 or a chi-square test with level 0.01? In other words, would a test that is statistically significant using 0.05 always be statistically significant using 0.01, would it be the other way around, or does it depend on the situation? Explain your answer.

Answer: A test that is statistically significant using 0.01 would also be statistically significant for any level greater than 0.01, including 0.05. It requires stronger evidence in favor of the alternative hypothesis to conclude statistical significance using a level of 0.01 than using a level of 0.05 27. Explain whether each of the following is possible. a. A relationship exists in the observed sample but not in the population from which the sample was drawn. b. A relationship does not exist in the observed sample but does exist in the population from which the sample was drawn. c. A relationship does not exist in the observed sample, but an analysis of the sample shows that there is a statistically significant relationship, so it is inferred that there is a relationship in the population.

Answer: a. Yes. By the luck of random sampling, the observed sample might show a relationship when none actually exists in the larger population. b. Yes. By the luck of random sampling, the observed sample might not show a relationship when one actually exists in the larger population. c. No. If there is no relationship in the observed sample the chi-square statistic will be 0 or near 0, so the sample could not show a statistically significant relationship.

Exercises 28 to 32 are based on News Story 12 (summarized in the Appendix), “New program for teen drivers with ADHD has significantly reduced crashes,” and the accompanying Original Source. The purpose of the study was to see if a driver training software program called FOCAL could help teens with ADHD become more focused when driving. Drivers aged 16 to 19 with ADHD were randomly assigned to be trained using either FOCAL or an unrelated control. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

28. According to the research report, “Teens 16 to 19 years of age with ADHD who had a valid driver’s license were recruited at the Cincinnati Children’s Hospital Medical Center by means of radio, social media, and print advertisements” (Epstein et al., 2022). What population do you think is represented by the sample for this study? Explain.

Answer: The population represented is probably all teens with ADHD who live in areas similar to Cincinnati, and who would see ads in the places mentioned.

29. For a year following the training, a recording device in their cars measured events that included a certain amount of acceleration or more, and whether a collision or near collision followed each of those events. The Minitab (Version 17) output in Figure 13.7 provides the data and the results of a chi-square test.

Write the null and alternative hypotheses being tested with these results.

Answer: a. Here is one way to write the hypotheses. Null hypothesis: If all teens with ADHD were given the intervention training, there would be no difference in collision rates than if the teens had not been given the training. Alternative hypothesis: The likelihood of a collision would be different for the population of teens with ADHD if they were given the intervention training than if they were not given the training.

30 a. If the null hypothesis is true, the proportion (0.0446) should be the same for both groups. b. Null hypothesis: In the population similar to those in this experiment, the proportion of times a near collision would occur during an acceleration event would be the same for those with and without the training program.

Alternative hypothesis: In the population similar to those in this experiment, the proportion of times a near collision during an acceleration event would occur would differ for those with and without the training program. Chi-square = 17.343; reject null hypothesis. There is convincing evidence that the proportions would differ. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

c. In the population similar to those in this experiment, the proportion of times there would be a (near) collision during acceleration events would differ for those with and without the training.

Answer: Refer to the previous exercise and the Minitab output in Figure 13.7. a. Of all of the events, 269/6031 = 0.0446 resulted in a collision or near collision. Explain why multiplying this number by the total of 3213 would give the number of (near) collisions that would be expected for the intervention group if the null hypothesis is true. b. Carry out the steps of the hypothesis test, including a statistical conclusion. c. Write the conclusion of the test in a few sentences that convey the results to someone with no training in statistics.

31. The original research report (Epstein et al., 2022) included this quote: “Motor vehicle collisions or near-collisions occurred during 110 of the 3213 g-force events (3.4%) among teens in the intervention group, as compared with during 159 of the 2818 g-force events (5.6%) among teens in the control group (relative risk, 0.60).” A news report about this study (CBS Miami Team, 2022) included the statement, “A small study published in the New England Journal of Medicine this week has found that a new program geared to teen drivers who have attention-deficit/ hyperactivity disorder, or ADHD, has significantly reduced crashes and near crashes.” Explain whether the word “significantly” in this quote refers to statistical significance, practical significance, or both.

Answer: The word “significantly” probably applies to both statistical significance and practical significance. As noted in the previous exercise, the difference is statistically significant, with a p-value of 0 to three decimal places. But the relative risk of only 0.60 has practical significance as well, indicating that the risk of a (near) collision for the group with the training was only 0.6 times the risk for the group without training.

32. The researchers in this study analyzed the number of acceleration events over the course of a year that had a collision or near collision for both the intervention and control groups. There were over 6000 such events. The difference in the percentage of collisions was small, with about 3.42% in the intervention group and 5.64% in the control group. Yet the p-value for the test was 0.000 (0 to 3 decimal places). There were 72 participants in each group who were followed over a year to collect the data. Another option would be to analyze the number of participants in each group who had at least one collision or near collision during the course of the year, rather than the number of events. Why do you think the researchers chose to analyze collisions for the events instead of for the individuals? (Hint: Would they have been likely to find a statistically significant difference if they had used the individuals as the units?)

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

Answer: It is much easier to attain statistical significance with a large sample size than with a small one. If the researchers had used the 72 individuals instead of the thousands of events, even a fairly large difference in the population might not have been enough to achieve statistical significance for that sample size.

Exercises 33 to 37 are based on News Story 5 (summarized in the Appendix), “Driving while distracted is common, researchers say,” and the accompanying Original Source 5, “Distractions in Everyday Driving.”

33. Refer to Table 8 on page 37 of Original Source 5 on the companion website.

Notice that there is a footnote to the table that reads:“∗ p < .05 and ∗∗ p < .01, based on chi-square test of association with sex.” The footnote applies to “Grooming**” and “External distraction*.” a. Explain what null and alternative hypotheses are being tested for “Grooming**.” (Notice that the definition of “grooming” is given on page 41 of the report.) b. Explain what the footnote means.

Answer: a. Null hypothesis: In the population of drivers similar to these, there is no relationship between sex and being distracted by grooming. Alternative hypothesis: There is a relationship. b. The footnote for grooming means that there was a statistically significant relationship between sex and being distracted by grooming, and the associated p-value was less than 0.01.

34. Figure 13.8 provides Minitab output for testing for a relationship between sex and “External distraction,” but the expected counts have been removed for you to fill in. a. Fill in the expected counts. b. State the null and alternative hypotheses being tested. c. Give the value of the chi-square statistic and the p-value, and make a conclusion. State the conclusion in statistical terms and in the context of the situation. d. Explain how this result confirms the footnote given for the table in connection with “External distraction.”

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

Answer: a. The expected count for “Male, Yes” is (35)(60)/70 = 30, for “Male, No” it’s 5, for “Female, Yes” it’s 30 and for “Female, No” it’s 5. b. The null hypothesis is that there is no relationship between sex and external distractions for the population of drivers similar to those in this study. The alternative hypothesis is that there is a relationship. c. Chi-square statistic = 4.20, p-value is 0.04. Reject the null hypothesis.

Conclude that there is a relationship between sex and external distractions while driving, for the population of drivers similar to those in the study. d. The footnote says that the p-value is less than 0.05. The actual p-value is given as 0.04, which is indeed less than 0.05.

35. Table 13.6 shows participants categorized by age group (Under 50, 501) and by whether they were observed grooming (yes, no). What is the expected count for “Under 50, Yes?” Find the remaining expected counts by subtraction.

Answer: The expected count for “Under 50, Yes” is (42)(32)/70 = 19.2. Therefore, for “Under 50, No” it’s 42 – 19.2 = 22.8, for “50 or Older, Yes” it’s 32 – 19.2 = 12.8 and for “50 or Older, No” it’s 38 – 22.8 = 15.2.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

36. Refer to the previous exercise and Table 13.6, categorizing people by age group and whether they were observed grooming. a. State the null and alternative hypotheses that can be tested with this table. b. The chi-square statistic is 0.777 and the p-value is 0.378. Is the relationship statistically significant at the 0.05 level? Give two different methods for determining your answer. c. Make a conclusion in the context of the situation.

Answer: a. The null hypothesis is that there is no relationship between age group and being observed grooming for the population of drivers similar to those in this study. The alternative hypothesis is that there is a relationship. b. The relationship is not statistically significant. The p-value of 0.378 > 0.05, and the chi-square statistic of 0.777 is less than the value of 3.84 required for statistical significance. c. The chi-square statistic is much less than 3.84, therefore, the null hypothesis cannot be rejected. There is not sufficient evidence to conclude that there is a relationship between age group and being observed grooming while driving, for the population of drivers similar to those in the study.

37. Refer to Exercises 35 and 36, in which the relationship was examined between age group (under 50 or 50 and older) and whether the person was observed grooming while driving. a. Calculate the relative risk and the increased risk of being observed grooming for the younger age group compared to the older age group. b. Write a sentence stating the increased risk in words that would be understood by someone with no training in statistics. Explain whether or not you think this increased risk is large enough to be of practical importance. c. The chi-square statistic for the relationship is 0.777, so the relationship is not statistically significant. Suppose the sample size had been 10 times as large as it was, but the proportions in each category remained the same.

What conclusion would be made in that case? Explain. d. Explain how this example illustrates the difference between statistical significance and practical importance.

Answer: a. Risk for the “Under 50” group is 21/42 = 0.5. Risk for the “50 or Older” group is 11/28 = 0.39. Therefore, the relative risk (younger versus older) is 0.5/0.39 = 1.28. Therefore, the increased risk is (1.28 – 1.0) × 100% = 28%. b. “A study of 70 drivers found that for the drivers under 50 there was a 28% higher risk of being observed grooming while driving than there was for drivers aged 50 and older.” Answers may vary on whether this increased © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

risk is large enough to be of practical importance. The reasoning is what counts. However, if the context had been something life-threatening, such as risk of a serious disease, a 28% increased risk certainly would be large enough to be of concern. c. If the sample sizes were 10 times larger, the chi-square statistic would be 10 times as large, at 7.77. Since 7.77 > 3.84, the null hypothesis would be rejected, whereas with the current sample size it was not rejected. (The actual p-value is about 0.005.) d. Even though the increased risk is 28%, the original sample size did not lead to the conclusion that there is a statistically significant increased risk. The larger sample size would have led to a strongly statistically significant increase in risk.

ANSWER TO THE REFLECTION QUESTIONS 1. For hypothesis testing, explain why it is important to have a representative sample from a population and to be able to describe what that population is.

Answer: It is important to have a representative sample for a population because the whole point of hypothesis testing is to make conclusions about a population based on a sample. If the sample does not represent any population, then no such conclusions can be made. It is important to be able to describe the population so that the conclusion can be stated specific to that population.

2. Suppose two independent researchers are studying the relationship between age group (child or adult) and a certain ability. One of them conducts a large study and finds that children are 10% more likely to have the ability, with a pvalue of 0.04. The other researcher conducts a small study and finds that children are twice as likely to have the ability, with a p value of 0.2. Discuss the difference between statistical significance and practical significance for this example.

Answer: The researcher with the larger sample has achieved statistical significance, but the increased likelihood of having the ability of only 10% might not have much practical significance. On the other hand, the researcher with the smaller sample size has a result with great practical significance, given the finding that children are twice as likely to have the ability as adults. But the small sample size was not enough to achieve statistical significance.

3. Part of conducting a chi-square test is to calculate expected counts. Suppose the two variables being tested and their categories are age (child/adult) and having a certain ability (yes/no).

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

a. What is meant by the “expected count” for the cell “child and yes”? b. Suppose the null hypothesis is true. Would the calculated expected counts for each cell be what’s actually observed in that cell when a sample is collected? Explain.

Answer: a. The expected count for the cell “child and yes” is the number that would be expected for that combination, on average, for a sample of this size if there is no difference in having the ability for children and adults in the population. b. The expected counts are not actually what would be likely in any given sample. They represent the average over the long run, for multiple repetitions of the situation with samples of the same size.

4. Which is more informative when you are presented with the results of a hypothesis test: the numerical p-value or a conclusion about whether the result is statistically significant? Or are they equally informative? Explain.

Answer: The numerical p-value is much more informative than just being told whether the results achieved statistical significance. If you know the p-value, you can determine whether statistical significance holds for any specific level you choose.

ANSWER TO THE MINI PROJECTS 1. Carefully collect data cross-classified by two categorical variables for which you are interested in determining whether there is a relationship. Do not get the data from a book, website, or journal; collect it yourself. Be sure to get counts of at least five in each cell and be sure the individuals you use are not related to each other in ways that would influence their data. a. Create a contingency table for the data. b. Compute and discuss the risks and relative risks. Are those terms appropriate for your situation? Explain. c. Determine whether there is a statistically significant relationship between the two variables. d. Discuss the role of sample size in making the determination in part (c). e. Write a summary of your findings.

Answer: This project is relatively straightforward in that it involves the same kinds of computations as the exercises. Be careful about collecting the data, to be sure the individuals are independent of each other and the same variables are measured on each one. Some guidelines to check are as follows. For part a, make sure the convention of putting the explanatory variable in the rows and © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 13: Statistical Significance For 2 × 2 Tables

the response in the columns is followed, if that distinction can be made. For part b make sure all parts are answered and that relative risk is computed rather than increased risk. For part c remember that the chi-squared statistic must exceed 3.84 to achieve statistical significance. For part d it is likely that the sample for the project will have been too small to achieve statistical significance, and this fact should be discussed. If the sample did achieve significance and the sample was small, it should be noted that the relationship is probably very strong.

2. Find a poll that has been conducted at two different time periods or by two different sources. For instance, many polling organizations ask opinions about certain issues on an annual or other regular basis. (A good source for recent polls is http://www.pollingreport.com.) a. Create a 2 × 2 table in which “time period” is one categorical variable and “response to poll” is the other, categorized into just two choices, such as “favor” and “do not favor” for an opinion question. b. State the null and alternative hypotheses for comparing responses across the two time periods. Be sure you differentiate between samples and populations. c. Carry out a chi-square test to see if opinions are different for the two time periods. d. Write a few sentences giving your conclusion in words that someone with no training in statistics could understand.

Answer: This project is relatively straightforward. Make sure the hypotheses are stated in terms of populations at the times the polls are taken.

3. Find a journal article that uses a chi-square test based on a contingency table. a. State the hypotheses being tested. b. Write the contingency table. c. Give the value of the chi-square statistic and the p-value as reported in the article. d. Write a paragraph or more (as needed) explaining what was tested and what was concluded, as if you were writing for a news story for a general audience.

Answer: The purpose of this project is to get experience with reading journal articles and recognizing familiar procedures. Make sure the hypotheses are written about the population and not the sample, and that the population is specified.

The paragraph written for part d should also be clear about the distinction between the sample results and the conclusions that can be made about the population.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 14: Understanding Probability And Long-Term Expectations

Solution and Answer Guide

UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 14: UNDERSTANDING PROBABILITY AND LONG-TERM EXPECTATIONS

TABLE OF CONTENTS Answer to the Exercise Questions......................................................................................1 Answer to the Reflection Questions................................................................................ 14 Answer to the Mini Projects..............................................................................................15

ANSWER TO THE EXERCISE QUESTIONS 1. Recall that there are two interpretations of probability: relative frequency and personal probability. a. Which interpretation applies to this statement: “The probability that I will get the flu this winter is 30%.”? Explain. b. Which interpretation applies to this statement: “The probability that a randomly selected adult in America will get the flu this winter is 30%.”?

Explain. (Assume it is known that the proportion of adults who get the flu each winter remains at about 30%.).

Answer: a. This statement is an example of a personal probability because it is based on an individual’s belief about their chances of getting the flu. b. The relative frequency interpretation applies here because this probability is based on repeated observations of the frequency of flu cases each winter.

2. Explain which of the following more closely describes what it means to say: that the probability of a tossed coin landing with heads up is 1/2: Explanation 1: After more and more tosses, the fraction of heads will get closer and closer to 1/2.

Explanation 2: The number of heads will always be about half the number of tosses.

Answer: Explanation 1 is closer to what it means. The number of heads may be far from half for a small number of tosses, so Explanation 2 is incorrect.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 14: Understanding Probability And Long-Term Expectations

3. Explain why probabilities cannot always be interpreted using the relativefrequency interpretation. Give an example of when that interpretation would not apply.

Answer: Not all events are repeatable. An example is the probability that the Earth will be uninhabitable by the year 2050.

4. Suppose you wanted to test your extrasensory perception (ESP) ability using an ordinary deck of 52 cards, which has 26 red and 26 black cards. You have a friend shuffle the deck and draw cards at random, replacing the card and reshuffling after each guess. You attempt to guess the color of each card. a. What is the probability that you guess the color correctly by chance? b. Is the answer in part (a) based on the relative-frequency interpretation of probability, or is it a personal probability? c. Suppose another friend has never tried the experiment but believes they have ESP and can guess correctly with probability 0.60. Is the value of 0.60 a relative-frequency probability or a personal probability? Explain. d. Suppose another friend guessed the color of 1000 cards and got 600 correct. The friend claims she has ESP and has a 0.60 probability of guessing correctly. Is the value of 0.60 a relative-frequency probability or a personal probability? Explain.

Answer: a. Because half the cards are red and half are black, in the long run half the cards your friend draws will be black and half will be red. So whether you guess black or red you have a probability of 1/2 of being right by chance. b. It is a relative-frequency probability because it is based on knowledge of the physical situation, which can be repeated numerous times. c. This would be a personal probability because it is based neither on physical assumptions nor repeated observations, but on a person’s belief about their own guessing ability. d. In this case, the value of 0.60 is a relative frequency probability because it was the friend’s observed correct proportion over the long run.

5. Suppose you are able to obtain a list of the names of everyone in your school and you want to determine the probability that someone randomly selected from your school has the same first name as you. a. Assuming you had the time and energy to do it, how would you go about determining that probability? (Assume all names listed are fully spelled out.) b. Using the method you described in part (a), would your result be a relativefrequency probability or a personal probability? Explain.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 14: Understanding Probability And Long-Term Expectations

Answer: a. You could go through the whole list and count the portion of who had your first name. You could get an approximate answer by repeatedly randomly select a name on the list and noting the proportion of the time that the first name matched yours. b. By either method the resulting probability would be a relative-frequency probability. The resulting probability would be based either on physical assumptions (how many listings you know to have your first name) or on repeated observations.

6. In Section 14.2 (page 305), you learned two ways in which relative-frequency probabilities can be determined. Explain which method you think was used to determine the following statement: The probability that a particular flight from New York to San Francisco will be on time is 0.78.

Answer: This probability was most likely determined by observing the relative frequency. Flights from New York to San Francisco were observed and their punctuality over a long period of time was recorded. It wouldn’t make sense to determine this probability by making an assumption about the physical world.

7. Which of the two methods for determining relative-frequency probabilities (given in Section 14.2, page 305) was used to determine each of the following? a. On any given day, the probability that a randomly selected American adult will read a book for pleasure is 0.33. b. The probability that a five-card poker hand contains “four of a kind” is 0.00024.

Answer: a. This probability was determined by repeated observations of how often Americans read a book for pleasure. b. This probability was determined by making an assumption about the physical world. It was made by assuming all possible poker hands are equally likely and determining the proportion of them that result in four of a kind.

8. Use your own particular expertise to assign a personal probability to something, such as the probability that a certain sports team will win next week. Now assign a personal probability to another related event. Explain how you determined each probability, and explain how your assignments are coherent.

Answer: Answers will differ. Here is an example provided by a student: I think that the probability that I will finish my math homework by the due date is 70%. I also believe that I will receive an A with a probability of 60%. I will need to turn in the assignment on time in order to get an A, so getting an A is a subset of turning in the assignment on time. This prediction is coherent because 60% is smaller than 70%. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 14: Understanding Probability And Long-Term Expectations

9. For each of the following situations, state whether the relative-frequency interpretation or the personal probability interpretation is appropriate. If it is the relative-frequency interpretation, specify which of the two methods for finding such probabilities (given on page 305) would apply. a. If a spoon is tossed 10,000 times and lands with the rounded head face up on 3000 of those times, we would say that the probability of the rounded head landing face up for that spoon is about 0.30. b. In a debate with you, a friend says that they think there is a 50/50 chance that God exists.

Answer: a. The relative frequency interpretation is appropriate, based on observing the relative frequency over many repetitions. b. Your friend’s statement is based on personal probability.

10. On February 13, 2013, the Wall Street Journal reported, “The odds that . . . an asteroid impact would make us the last generation of human civilization are no lower than the odds of an average American dying in an earthquake (about 0.001%).” (Lu and Rees, 2013) Do you think this probability is based on relative frequency? Explain.

Answer: The probability cannot be based on relative frequency, so it must be based on personal probability. The event of an asteroid impact that wipes out human civilization is not something that could be observed many times!

11. Use the probability rules in this chapter to solve each of the following: a. According to the U.S. Department of Agriculture (Hoy et al., 2021), 33% of children in the United States eat one vegetable in a given day, and an additional 16% eat at least two vegetables in a given day. What is the probability that a randomly selected child does not eat any vegetables in a given day? b. In 2010 in the United States, the probability that a birth would result in twins was 0.0312, and the probability that a birth would result in triplets or more was 0.0008. What was the probability that a birth in 2010 resulted in a single child? (Source: http://www.cdc.gov/nchs/fastats/multiple.htm .)

Answer: a. The probability that a randomly selected child eats one vegetable a day is 0.33, and the probability that the child eats two or more is 0.16. Because these two possibilities are mutually exclusive, the probabilities can be added using Rule 2, so the probability that the child eats at least one vegetable is 0.49. By Rule 1, the probability that the child does not eat any vegetables is 1 – 0.49 = 0.51.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 14: Understanding Probability And Long-Term Expectations

b. Note that the probability of a birth resulting in more than one child is 0.0312 + 0.0008 = 0.032. So by Rule 1, the probability that a birth in 2010 resulted in only one child was 1 − 0.032 = 0.968.

12. Figure 9.1 (page 179) illustrates that in 2021, 26.5% of youth (aged 25 to 34) had some college education, and an additional 40.9% had a bachelor’s degree or higher. a. What is the probability that a randomly selected youth in 2021 would have some college, but not a bachelor’s degree? b. What is the probability that a randomly selected youth in 2021 would have a bachelor’s degree or higher? c. What is the probability that a randomly selected youth in 2021 would have at least some college? d. What is the probability that a randomly selected youth in 2021 would not have any college education?

Answer: a. The percentage of 26.5% is equivalent to a probability of 0.265. b. The percentage of 40.9% is equivalent to a probability of 0.409. c. Having at least some college includes having some college only, and having at least a bachelor’s degree. Therefore, by Rule 2, the probability of having at least some college is 0.265 + 0.409 = 0.674. (Having some college only and having at least a bachelor’s degree are mutually exclusive.) d. Using Rule 1, the probability of having at least some college education and not having any must sum to 1. Therefore, the probability of not having any college education is 1 – 0.674 = 0.326.

13. Example 14.3 states that the probability that a piece of checked luggage will be lost, delayed, damaged, or pilfered on a flight with a U.S. airline is about 5.5/1000 or 0.0055. Interpret that statement, using the appropriate interpretation of probability.

Answer: This is a relative-frequency probability. It means that over the long run, the proportion of times luggage will be temporarily lost, delayed, damaged or pilfered is 5.5/1000, or 0.0055.

14. There is something wrong in each of the following statements. Explain what is wrong. a. The probability that a randomly selected driver will be wearing a seat belt is 0.75, whereas the probability that they will not be wearing one is 0.30. b. The probability that a randomly selected car is red is 1.20. c. The probability that a randomly selected car is red is 0.20, whereas the probability that a randomly selected car is a red sports car is 0.25.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 14: Understanding Probability And Long-Term Expectations

Answer: a. This contradicts Rule 1. There are only two possibilities: a person either is wearing a seat belt or is not. So the two probabilities should sum to 1. b. A probability cannot exceed 1. c. This contradicts Rule 4. Red sports cars are a subset of red cars, so the probability that a car is red cannot be less than the probability that it is both red and a sports car.

15. A small business performs a service and then bills its customers. From past experience, 90% of the customers pay their bills within a week. a. What is the probability that a randomly selected customer will not pay within a week? b. The business has billed two customers this week. What is the probability that neither of them will pay within a week? What assumption did you make to compute that probability? Is it a reasonable assumption?

Answer: a. By Rule 1, it must be 1 − 0.9 = 0.1. This is equivalent to saying that 10% will not pay within a week. b. By Rule 3, the probability that customer A and customer B will not pay their bills in one week is (0.1)(0.1)= 0.01, or 1%. The assumption has been made that customer A not paying in one week is independent of whether or not customer B pays in a week. Yes, it seems reasonable to assume that there will, in general, be no connection between these two events.

16. According to Krantz (1992, p. 111), the probability of being born on a Friday the 13th is about 1/214. a. What is the probability of not being born on a Friday the 13th? b. In any particular year, Friday the 13th can occur once, twice, or three times.

Is the probability of being born on Friday the 13th the same every year?

Explain. c. Explain what it means to say that the probability of being born on Friday the 13th is 1/214.

Answer: a. The probability of not being born on Friday the 13th is 1 − 1/214 = 213/214, by Rule 1. b. No. For instance, the probability of being born on Friday the 13th if you are born in a year with two is roughly 2/365, whereas if there are three in the year it would be about 3/365. c. “The probability of being born on Friday the 13th is 1/214” means that over the long run (many years), about 1 out of every 214 people will be born on a Friday the 13th.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 14: Understanding Probability And Long-Term Expectations

17. Suppose the probability that you get an interesting piece of mail on any given weekday is 1/20. Is the probability that you get at least one interesting piece of mail during the week (Monday to Friday) equal to 5/20? Why or why not?

Answer: No. To say the probability is 5/20, you are assuming that the events are mutually exclusive, meaning that if you got an interesting piece of mail on Monday you could not get one on Tuesday, and so on. But you could easily get an interesting piece of mail on two of these days or all of them. (To find the correct probability you would need to subtract the probability that you would receive an interesting piece of mail on two or more of these days.)

18. You cross a train track on your drive to work or school. If you get stopped by a train you are late. a. Are the events “stopped by train” and “late for work or school” independent events? Explain. b. Are the events “stopped by train” and “late for work or school” mutually exclusive events? Explain.

Answer: a. No they are not independent. If you are stopped by a train it changes the probability of being late to work. In fact, it changes it to 1. b. No they are not mutually exclusive events. If one happens, it does not prevent the other one from happening.

19. On any given day, the probability that a randomly selected adult in the United States drinks water is about 0.86 (86%), and the probability that they drink juice is 0.13 (13%) (Martin et al., 2020). What assumption would we have to make in order to use Rule 3 to conclude that the probability that a person drinks both is(0.86) × (0.13) = 0.112? Do you think that assumption holds in this case? Explain.

Answer: For the probability of 0.112 to be correct, we would have to assume independence, so that if a person drinks water it would not change the probability that they drink juice (and vice versa). This assumption may or may not be valid, and it is the reasoning that counts. For it to be valid, about 13% of the people who drink water would also drink juice, and about 13% of the people who do not drink water would drink juice.

20. A classic study by Kahneman and Tversky (1982, p. 496) asked people the following question: “Linda is 31 years old, single, outspoken, and very bright.

She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in antinuclear demonstrations. Please check off the most likely alternative: a. Linda is a bank teller.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 14: Understanding Probability And Long-Term Expectations

b. Linda is a bank teller and is active in the feminist movement.”

Nearly 90% of the 86 respondents chose alternative B. Explain why alternative B cannot have a higher probability than alternative A. Answer:

By Rule 4, alternative A must have a higher probability than B because bank tellers who are active in the feminist movement are a subset of all bank tellers.

21. Read the definition of “independent events” given in Rule 3. Explain whether each of the following pairs of events is likely to be independent: a. A parent and their adult child live in the same household, and will vote in an upcoming presidential election in the United States. Event A is that the parent votes for the Republican candidate; event B is that the adult child votes for the Republican candidate. b. Event A is that a major earthquake will occur somewhere in the world in the next month; event B is that the Dow Jones Industrial Average will be higher in one month than it is now.

Answer: a. These are not likely to be independent because the political views of people in the same household and family are very likely to influence each other, and they are likely to belong to the same political party. b. These events are probably independent because the Dow Jones Industrial Average is not likely to be affected by earthquakes and earthquakes are clearly not affected by the Dow Jones Industrial Average. You might argue that they are not independent because a major earthquake in a city like New York, Los Angeles or San Francisco would probably play havoc with the stock market. This is an acceptable answer if you provide an explanation.

22. Read the definition of “independent events” given in Rule 3. Explain whether each of the following pairs of events is likely to be independent: a. Event A is that it snows tomorrow; event B is that the high temperature tomorrow is at least 60 degrees Fahrenheit. b. You buy a lottery ticket, betting the same numbers two weeks in a row.

Event A is that you win in the first week; event B is that you win in the second week.

Answer: a. The temperature plays a role in whether or not it snows so these cannot be independent. b. These will be independent because the numbers are chosen randomly each week and knowledge of one week’s numbers will not give you any knowledge of the second week’s numbers.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 14: Understanding Probability And Long-Term Expectations

23. People are surprised to find that it is not all that uncommon for two people in a group of 20 to 30 people to have the same birthday. We will learn how to find that probability in a later chapter. For now, consider the probability of finding two people who have birthdays in the same month. Make the simplifying assumption that the probability that a randomly selected person will have a birthday in any given month is 1/12. Suppose there are three people in a room and you consecutively ask them their birthdays. Your goal, following parts (a–d), is to determine the probability that at least two of them were born in the same calendar month. a. What is the probability that the second person you ask will not have the same birth month as the first person? (Hint: Use Rule 1.) b. Assuming the first and second persons have different birth months, what is the probability that the third person will have yet a different birth month?

(Hint: Suppose January and February have been taken. What proportion of all people will have birth months from March to December?) c. Explain what it would mean about overlap among the three birth months if the outcomes in part (a) and part (b) both happened. What is the probability that the outcomes in part (a) and part (b) will both happen? d. Explain what it would mean about overlap among the three birth months if the outcomes in part (a) and part (b) did not both happen. What is the probability of that occurring?

Answer: a. Whatever the birth month of the first person, the probability that the second person was born in that same month is 1/12. By Rule 1 the probability of a different month is therefore 1 − 1/12 = 11/12. You could also notice that for the second person, 11 out of the 12 months would not match the first person’s birth month, so the probability of a different month is 11/12. b. Two months have been taken by the first two people, so there are ten left.

If the third person was born in any of those ten months, their birth month would be different. Therefore, the probability is 10/12. c. If the outcomes in parts (a) and (b) both happen, then all three were born in different months. The probability of this happening can be found by multiplying the probabilities: (11/12)(10/12) = 110/144 or 0.76. d. If the outcomes in part (a) and part (b) did not both happen this would mean that at least two people have the same birth month. In part (c) we found the probability of the complement of this to be 110/144. Therefore, by Rule 1 we know that the probability that parts (a) and (b) do not both happen is 1 − 110/144 = 34/144 = 0.236. Hence, the probability that at least two of the people were born in the same month is 34/144 or 0.236.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 14: Understanding Probability And Long-Term Expectations

24. A restaurant server knows that the probability that a customer will order coffee is 0.30, the probability that a customer will order a diet soda is 0.40, and the probability that a customer will request a glass of water is 0.70.

Explain what is wrong with his reasoning in each of the following. a. Using Rule 2, the server concludes that the probability that a customer will order either coffee or request a glass of water is 0.30+ 0.70 = 1.0. b. Using Rule 3, the server concludes that the probability that a customer will order coffee and a diet soda is (0.30)(0.40) = 0.12.

Answer: a. To use Rule 2, the two events must be mutually exclusive. In this situation that would mean that no one would order both coffee and water, which is not reasonable. b. To use Rule 3, the two events must be independent, which means that if we know one happens, it doesn’t change the probability of the other happening. In this case, if someone orders a diet soda there would surely be a much lower probability of them ordering coffee (and vice versa).

25. Suppose you routinely check coin-return slots in vending machines that take cash to see if they have any money in them. You have found that about 10% of the time you find money. a. What is the probability that you do not find money the next time you check? b. What is the probability that the next time you will find money is on the third try? c. What is the probability that you will have found money by the third try?

Answer: a. The probability that you do not find money is 1 − 0.1 = 0.9. b. Assuming the events are independent, the probability of finding money for the first time on the third try would be (0.9)(0.9)(0.1) = 0.081. c. This is the accumulated probability at the third try, which is the sum of the individual probabilities for tries 1, 2 and 3. It is thus (0.1) + (0.9)(0.1) + (0.9)(0.9)(0.1) = 0.1 + 0.09 + 0.081 = 0.271.

26. Lyme disease is a disease carried by ticks, which can be transmitted to humans by tick bites. Suppose the probability of contracting the disease is 1/100 for each tick bite. a. What is the probability that you will not get the disease when bitten once? b. What is the probability that you will not get the disease from your first tick bite and will get it from your second tick bite?

Answer: a. 1 – 1/100 = 99/100 b. (99/100)  (1/100) = .0099. (This assumes that the events are independent from one bite to the next.)

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 14: Understanding Probability And Long-Term Expectations

27. Suppose you play a carnival game that requires you to toss a ball to hit a target. The probability that you will hit the target on each play is 0.2 and is independent from one try to the next. You win a prize if you hit the target by the third try. a. What is the probability that you hit the target on the first try? b. What is the probability that you miss the target on the first try but hit it on the second try? c. What is the probability that you miss the target on the first and second tries but hit it on the third try? d. What is the probability that you win a prize?

Answer: a. 0.2. b. (0.8)(0.2) = 0.16. c. (0.8)(0.8)(0.2) = 0.128 d. The probability of winning a prize is the sum of the probabilities from parts (a) to (c), which is 0.2 + 0.16 + 0.128 = 0.488.

28. According to Krantz (1992, p. 161), the probability of being injured by lightning in any given year is 1/685,000. Assume that the probability remains the same from year to year and that avoiding a strike in one year doesn’t change your probability in the next year. a. What is the probability that someone who lives 80 years will never be struck by lightning? You do not need to compute the answer, but write down how it would be computed. b. According to Krantz, the probability of being injured by lightning over the average lifetime is 1/9100. Show how that probability should relate to your answer in part (a), assuming that average lifetime is about 80 years. c. Do the probabilities given in this exercise apply specifically to you? Explain. d. Over 330 million people live in the United States. In a typical year, assuming Krantz’s figure is accurate, about how many people out of 330 million would be expected to be struck by lightning?

Answer: a. The probability of making it 80 years without being struck by lightning would be (684,999/685,000)80. b. If the average lifetime is 80 years, then Krantz’s probability should be 1 − (684,999/685,000)80, or one minus the probability in part (a), which is about 1/8653, close to Krantz’s estimate. (You get closer to Krantz’s estimate if you assume that the average lifespan is 75 years instead of 80 years.) c. These probabilities do not apply to any particular person. Your chances depend on things like where you live and what you do during storms. d. About (330,000,000)  (1/685,000), or approximately 482 people. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 14: Understanding Probability And Long-Term Expectations

29. Suppose you have to cross a train track on your commute. The probability that you will have to wait for a train is 1/5, or 0.20. If you don’t have to wait, the commute takes 15 minutes, but if you have to wait, it takes 20 minutes. a. What is the expected value of the time it takes you to commute? b. Is the expected value ever the actual commute time? Explain.

Answer: a. The expected commute time is (20 min)(0.2) + (15 min)(0.8) = 16 minutes. b. No, the commute times are always either 15 or 20 minutes. This is just an average over the long run.

30. Suppose the probability that you get an A in any class you take is 0.3, and the probability that you get a B is 0.7. To construct a GPA, an A is worth 4.0, and a B is worth 3.0. What is the expected value for your GPA? Would you expect to have this GPA separately for each quarter or semester? Explain.

Answer: Your expected grade point average would be (4.0)(0.3) + (3.0)(0.7) = 3.3. You would not necessarily get this in any one quarter or semester; expected value gives you only a long-term average. Even four or five years of college may not be enough to reach this long-term average exactly.

31. Remember that the probability that a single birth is assigned male is about 0.51. You offer a bet to an unsuspecting friend. Each week you will call the local hospital and find out how many births in the previous week were assigned male and how many were not. For each non-male birth, you will give your friend $1, and for each male birth, your friend will give you $1. a. Suppose that in a given week there are three births. What is the probability that you lose $3 in that week? What is the probability that your friend loses $3? b. Notice that your net profit is $1 for births assigned male and -$1 for births assigned female. What is the expected value of your profit for each birth? c. Using your answer in part (b), how much can you expect to make after 1000 births?

Answer: a. You lose $3 if the births are all assigned female, so the probability that you lose $3 is (0.49)(0.49)(0.49) = 0.117649. Your friend loses $3 if they are all assigned male, so the probability is (0.51)(0.51)(0.51) = 0.132651. b. Expected value = (+$1)(0.51) + (−$1)(0.49) = 0.51 − 0.49 = 0.02 or 2 cents for each birth. c. Making 2 cents on each birth, after 1,000 births you make about ($0.02)(1,000) = $20.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 14: Understanding Probability And Long-Term Expectations

32. In the “3 Spot” version of the former California Keno lottery game, the player picked three numbers from 1 to 40. Ten possible winning numbers were then randomly selected. It cost $1 to play. The accompanying table shows the possible outcomes. Compute the expected value, the net “gain,” for this game.

Interpret what it means.

Answer: It costs $1 to play, so the possible outcomes are $19, $1 and −$1. The expected value is ($19)(0.012) + ($1)(0.137) + (−$1)(0.851) = −$0.486. So your expected loss would be about 49 cents for each dollar played.

33. We have seen many examples for which the term expected value seems to be a misnomer. Construct an example of a situation in which the term expected value would not seem to be a misnomer for what it represents.

Answer: Answers will vary. The best examples are for discrete data in which the expected value is also the mode. For instance, suppose a small box of raisins (sultanas) contains 19 pieces with probability 1/10, 20 with probability 8/10 and 21 with probability 1/10. Then the expected value is 20 raisins, which is also the most likely value.

34. According to the U.S. Census Bureau, in 2018, about 70.1% of children in the United States were living with two parents, 25.8% (0.258) were living with one parent, and the rest, about 4.1% (0.041), were living with no parents. (Many of those were living with grandparents.) What is the expected value for the number of parents a randomly selected child was living with? Does the concept of expected value have a meaningful interpretation for this example?

Explain.

Answer: The possible number of parents is 2 with probability 0.701, 1 with probability 0.258, and 0 with probability 0.041. The expected value for the number of parents is (2)(0.701) + (1)(0.258) = 1.66. This is not meaningful for this example because someone cannot live with 1.66 parents. It can only be interpreted as an average over all children in the United States.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 14: Understanding Probability And Long-Term Expectations

35. Find out your yearly car insurance cost. If you don’t have a car, find out the yearly cost for a friend or relative. Now assume you will either have an accident or not, and if you do, it will cost the insurance company $5000 more than the yearly cost you pay. Calculate what yearly accident probability would result in a “break-even” expected value for you and the insurance company.

Comment on whether you think your answer is an accurate representation of your yearly probability of having an accident.

Answer: The probability that results in a break-even situation is (cost)/(cost+$5,000).

To figure this out, call the probability of an accident p. Then the insurance company earns your cost with probability (1−p) and loses $5,000 with probability p. The expected value is: (cost)(1−p) + (−$5,000)(p) = cost − (cost + $5,000)(p).

Setting this to zero and solving for p gives the answer, which is cost/(cost + 5000). For example, if your cost is $1000, then the answer is p = 1/6. In general, you will find that the answer is much higher than your own personal probability of having an accident. (That’s how the insurance companies pay for their expenses and make a profit.)

ANSWER TO THE REFLECTION QUESTIONS 1. Give an example of two events for which the relative frequency interpretation could not be applied. Choose one example from your own life and one example that affects a population.

Answer: Examples are situations that are not repeatable and for which no physical assumptions can be used to compute probabilities. For example, in your own life you might use the probability that you will live to be 100 years old. An example that affects a population is the probability that an earthquake of magnitude 6.0 or higher will occur next year in a particular region.

2. Suppose two events are mutually exclusive, meaning that they cannot both happen. Can each of the following be true? If so, give an example. If not, explain why not. a. The sum of their probabilities is 1. b. They are independent events.

Answer: a. Yes, the sum of their probabilities can be 1, if they cover all possibilities. An example is the event that a particular sports team will win their next game, and the event that they won’t win their next game. The probabilities for those two events must add to 1.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 14: Understanding Probability And Long-Term Expectations

b. Two events that are mutually exclusive cannot be independent. If they are mutually exclusive, knowing that one of them occurs means you know the probability that the other one occurs is 0. If events are independent, knowing that one occurs does not give you any information about the probability that the other one occurs.

3. Specify which of the following two statements has a higher probability and explain how you know.

Statement 1: Two buildings in San Francisco will simultaneously collapse sometime in the next 10 years.

Statement 2: There will be a major earthquake in San Francisco in the next 10 years during which two buildings will simultaneously collapse.

Answer: Statement 1 must have a higher probability. Statement 2 is a subset of Statement 1, and therefore it cannot have a higher probability by Rule 4.

4. Give an example of a situation where an “expected value” would not be a value that’s ever expected to occur.

Answer: There are many such situations. For instance, if the outcome of an event can only be an integer but the expected value is not an integer, then it would never be expected to occur. An example is number of children in a randomly selected family from a particular town. Suppose there are a total of 6450 children in the town and 3000 households. Then the average number of children per household is 2.15, which is the “expected value” for the number of children in a randomly selected household.

ANSWER TO THE MINI PROJECTS 1. Refer to Exercise 20. Present the question to 10 people, and note the proportion who answer with alternative B. Explain to the participants why it cannot be the right answer, and report on their reactions.

Answer: The purpose of this mini-project is to learn firsthand that the tendency of most people is to ignore Rule 4 in this setting. The report for this project will be fairly short and should explain whether or not it was difficult to convince people of the right answer.

2. Flip a coin 100 times. Stop each time you have done 10 flips (that is, stop after 10 flips, 20 flips, 30 flips, and so on), and compute the proportion of heads using all of the flips up to that point. Plot that proportion versus the number of flips. Comment on how the plot relates to the relative-frequency interpretation of probability.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 14: Understanding Probability And Long-Term Expectations

Answer: Make sure the plot uses all flips accumulated up to that point and not just the previous ten. If done correctly, the value on the vertical axis should start settling down to a constant as the number of flips (on the horizontal axis) increases. If flipping is done in an unbiased way, that constant should be about 0.50. That is what the relative-frequency interpretation of the probability of a head would predict.

3. Pick an event that will result in the same outcome for everyone, such as whether it will rain next Saturday. Ask 10 people to assess the probability of that event, and note the variability in their responses. (Don’t let them hear each other’s answers, and make sure you don’t pick something that would have 0 or 1 as a common response.) At the same time, ask them the probability of getting a heart when a card is randomly chosen from a fair deck of cards. Compare the variability in responses for the two questions, and explain why one is more variable than the other.

Answer: The purpose of this project is to demonstrate that personal probabilities can be quite variable and still all be correct. In contrast, if we all agree on the physical assumptions, then we should all agree on the relative-frequency probability in this project. The probability of a heart should be given as 1/4 by everyone.

4. Find two lottery or casino games that have fixed payoffs and for which the probabilities of each payoff are available. (Almost all lotteries have these listed on their website. Some books about gambling give the payoffs and probabilities for various casino games.) a. Compute the expected value for each game. Discuss what they mean. b. Using both the expected values and the list of payoffs and probabilities, explain which game you would rather play and why.

Answer: Make sure to count the cost of playing when computing the expected value for each game. The interpretation is that this is what it costs per play in the long run. There is no correct answer to part b, although if one game has a substantially higher expected value, it would probably be the desired game.

The psychological impact of potentially winning a large amount must also be considered.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 15: Understanding Uncertainty Through Simulation

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 15: UNDERSTANDING UNCERTAINTY THROUGH SIMULATION

TABLE OF CONTENTS Answer to the Exercise Questions......................................................................................1 Answer to the Reflection Questions.................................................................................12 Answer to the Mini Projects..............................................................................................13

ANSWER TO THE EXERCISE QUESTIONS For Exercises 1 to 4, assume that birthdays are equally likely to occur on all possible days in any given year, so there are no seasonal variations or day of the week variations.

1. Suppose you wanted to simulate the birthdays (month and day, not year) of the five children in one family. For each child, you tell the computer to choose a number from 1 to 366. This covers all possibilities including February 29. a. Would you tell the computer to make all 366 choices equally likely?

Explain. b. How would you assign the results to birthdays? For example, what number would correspond with a birthday of December 31? c. Would it make sense to tell the computer to allow the same number to be chosen twice or not to allow that? Explain.

Answer: a. No. February 29 should be only ¼ as likely as the others. b. The most logical assignment would be to do them in order, with 1 = January 1st up to 366 = December 31st. c. It would make sense to allow the same number to be chosen twice. Two (or more) of the five children could have the same birthday.

2. Suppose you wanted to simulate the birthdays (month and day, not year) of three children in one family by first choosing a month and then choosing a day.

Assume that none of them were born in a leap year. a. What range of numbers would you tell the computer to use to simulate the month? Would you tell it to make all of those choices equally likely?

Explain.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 15: Understanding Uncertainty Through Simulation

b. What range of numbers would you tell the computer to use to simulate the day? Would you tell it to make all of those choices equally likely? Explain. c. In each case (month and day), would it make sense to tell the computer to allow the same number to be chosen twice, or not to allow that? Explain.

Answer: a. Use the numbers 1 to 12. Do not make them equally likely. For each month, use the number of days in that month divided by 365. For instance, the probability of choosing a 1 (for January) should be 31/365. b. Use the numbers 1 to 31. If the result is beyond the number of days in the month chosen, you could throw that one out and try again. For instance, if the month chosen is April and the day chosen is 31, do it again. c. It would make sense to allow the same number to be chosen twice for both the month and the day. Children in the same family could have the same birthdays.

3. Suppose you wanted to simulate the birthdays (month and day, not year) of three children in one family all of whom were born in years that were not leap years. All you have available is a computer website (www.randomizer.org) that weights all of the choices in the range you provide equally. How could you use the website to carry out this simulation?

Answer: Number the days from 1 (January 1st) to 365 (December 31st). Choose three numbers from 1 to 365 with replacement to represent the three birthdays.

4. Suppose you have two siblings, and all three of you were born in the same month of the year. Explain how you could use simulation to estimate the probability that in a family of three children they are all born in the same month. You do not have to provide exact details about how you would actually simulate each birthday.

Answer: Repeatedly simulate the birth months for three children thousands of times.

You could do this by simulating three numbers, each from 1 to 12, with appropriate probabilities (31/365 for January, etc.) Then count what proportion of the simulated sets of three children all had the same birth month.

5. As a promotion, a cereal brand is offering a prize in each box, and there are four possible prizes. You would like to collect all four prizes, but you only plan to buy six boxes of the cereal before the promotion ends. Assume you have a random number generator that weights all numbers equally in a range you provide. a. Explain how you could simulate the prizes found in one set of six boxes of cereal.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 15: Understanding Uncertainty Through Simulation

b. Explain how you could use simulation to estimate the probability of obtaining all four prizes in six boxes of cereal. Assume that all four prizes are equally likely for any given box, and the choice of prize is independent from one box to the next.

Answer: a. Randomly generate 6 numbers, each from 1 to 4, with replacement. b. Randomly generate 6 numbers, each from 1 to 4, with replacement. Repeat this thousands of times, then count what proportion of the sets include all 4 of the numbers 1, 2, 3, 4.

6. Suppose that 55% of the voters in a large city support a particular candidate for mayor and 45% do not support the candidate. A poll of 100 voters will be conducted, and the proportion of them who support the candidate will be found. Assume you have a random number generator that weights all numbers equally in a range you provide. It will not allow you to provide unequal weights. a. Explain how you could use the random number generator to simulate the response (support or oppose) for one randomly chosen voter. b. Explain how you could simulate the 100 responses (support or oppose) for one poll. c. Explain how you could use simulation to estimate the probability that the actual poll will result in less than half (49 or fewer) of those polled supporting the candidate.

Answer: a. Simulate an integer from 1 to 100. A number from 1 to 55 represents “supports candidate,” a number from 56 to 100 represents “does not support candidate.” b. Simulate 100 numbers each from the range 1 to 100 and use the rule in part (a) for each one. c. Simulate many polls of 100 and see what proportion of those polls result in 49 or fewer supporting the candidate.

7. Suppose that there are 38 students in your class. Each day the teacher randomly selects one student to show the others how to do the previous night’s homework. Each day, all students are eligible and equally likely to be chosen, even if they have been chosen that week already. a. On any given day, what is the probability that you are the student selected? b. What is the probability that you are selected on two consecutive days?

(Hint: Use Rule 3 from Chapter 14.) c. What is the probability that the same student is chosen on two consecutive days? d. Number the students from 1 to 38. Assume you have a random number generator that weights all numbers equally in a range you provide. Explain

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 15: Understanding Uncertainty Through Simulation

how you could use simulation to identify the students who are chosen on two consecutive days. (It could be the same student on both days.) e. Explain how you could use simulation to estimate the probability that the same student is chosen on two consecutive days.

Answer: a. 1/38 b. Using Rule 3, the probability is (1/38)(1/38) = 1/1444. c. The probability is 1/38. The easiest way to see this is that no matter which student is chosen on Day 1, the probability that that same student is chosen on the next day is 1/38. d. Use the random number simulator to choose two numbers, each from 1 to 38, with replacement. e. Repeat the simulation in part (d) thousands of times and count the proportion of those times that the same number is chosen twice.

Exercises 8 to 12: Refer to the box on page 330, describing the website www.randomizer.org. Parts (a) to (e) in each of these exercises list the questions asked when you use that website. In each case, specify how you would answer those questions. (See instructions on page 330.)

8. In Exercise 5, the following scenario was presented: As a promotion, a cereal brand is offering a prize in each box and there are four possible prizes. You would like to collect all four prizes, but you only plan to buy six boxes of the cereal before the promotion ends. For this exercise, you want to carry out 1000 simulations of the prizes (#1, 2, 3, or 4) you will find in six boxes of cereal. The proportion of simulations for which all four prizes appear is an estimate of the probability of getting all four prizes in the six boxes you plan to purchase.

Specify what you would answer for each of the following questions to accomplish this simulation using www.randomizer.org. a. How many sets of numbers do you want to generate? b. How many numbers per set? c. Number range (e.g., 1–50) d. Do you wish each number in a set to remain unique? e. Do you wish to sort the numbers that are generated?

Answer: a. 1000 b. 6 (representing the six boxes of cereal you plan to buy) c. 1 to 4 (representing the four possible prizes) d. No; you could obtain the same prize in more than one box. e. It would make it easier to see whether or not all four prizes were included, but sorting is not necessary. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 15: Understanding Uncertainty Through Simulation

9. In Exercise 6, the following scenario was presented. Suppose that 55% of the voters in a large city support a particular candidate for mayor and 45% do not support the candidate. A poll of 100 voters will be conducted, and the proportion of them who support the candidate will be found. For this exercise, you want to simulate 1000 polls of 100 people each, and find the proportion of those polls for which 49 or fewer support the candidate. That will provide an estimate of the probability that the poll shows less than majority support, even though in truth 55% of the population supports the candidate. For parts (a) to (e), specify what you would answer for each of the questions to accomplish this simulation using www. randomizer.org. Remember that the simulator will not allow you to specify different weights for the outcomes, so you need to figure out how to incorporate the .55 and .45 probabilities into your simulation. a. How many sets of numbers do you want to generate? b. How many numbers per set? c. Number range (e.g., 1–50) d. Do you wish each number in a set to remain unique? e. Do you wish to sort the numbers that are generated? f. Once you had the results of the simulation, how would you use them to estimate the desired probability?

Answer: a. 1000 b. 100 (representing the 100 people in the poll) c. 1 to 100; if the outcome is 1 to 55 it represents someone who supports candidate A, and if it is 56 to 100 it represents someone who does not. d. No e. Yes, because you need to know how many of the “voters” support the candidate, which is found by observing how many of the 100 numbers in each sample are from 1 to 55. f. Count how many of the 1000 samples of 100 numbers had 49 or fewer numbers in the range from 1 to 55.

10. The scenario in this exercise is similar to the one in Exercise 7, except the class size is smaller. Suppose that there are 15 students in your class. Each day the teacher randomly selects one student to show the others how to do the previous night’s homework. Each day, all students are eligible and equally likely to be chosen, even if they have been chosen that week already. For this exercise, you want to simulate three students being drawn, mimicking the Monday, Wednesday, Friday class schedule for this class. You want to estimate the probability that the same student will be chosen all three times in the same week. You plan to simulate results for 10,000 three-day weeks. For parts

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 15: Understanding Uncertainty Through Simulation

(a) to (e), specify what you would answer for each of the questions to accomplish this simulation using www.randomizer.org. a. How many sets of numbers do you want to generate? b. How many numbers per set? c. Number range (e.g., 1–50) d. Do you wish each number in a set to remain unique? e. Do you wish to sort the numbers that are generated? f. Suppose that the same student is chosen on all three days in 47 of your simulated weeks. What is your estimate of the probability that the same student will be chosen all three times in the same week? g. Use the probability rules from Chapter 14 to find the probability that the same student would be chosen all three times in one week, and compare your answer to the estimated probability in part (f).

Answer: a. 10,000 b. 3 (representing the students chosen on Monday, Wednesday and Friday) c. 1 to 15 (representing the 15 students in the class) d. No; the same student could be chosen more than once. e. No; the results are supposed to represent MWF in order. (But to answer the probability question posed in this problem it really doesn’t matter if you sort them.) f. The estimated probability is 47/10000 = 0.0047. g. No matter which student is chosen on Monday, the probability that that same student is chosen again on Wed and Friday is (1/15)(1/15) = 0.0044, which is close to the simulated probability.

11. An airline serves lunch in the first-class cabin. Customers are given a choice of either a sandwich or chicken salad. There are 12 customers in first class, and the airline loads seven of each meal onboard. You would like to simulate the probability that there will not be enough of one or the other meal to meet the preferences of the customers. Assume the probability is 0.60 that a customer requests the sandwich and 0.40 that they request the salad and is independent from one person to the next. You plan to simulate the choices for 100 sets of 12 customers.

For parts (a) to (e), specify what you would answer for each of the questions to accomplish this simulation using www.randomizer.org. Remember that the simulator will not allow you to specify different weights for the outcomes, so you need to figure out how to incorporate the 0.40 and 0.60 probabilities into your simulation. a. How many sets of numbers do you want to generate?

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 15: Understanding Uncertainty Through Simulation

b. c. d. e. f.

How many numbers per set?

Number range (e.g., 1–50) Do you wish each number in a set to remain unique?

Do you wish to sort the numbers that are generated?

Explain how you would use your results to estimate the desired probability.

Answer: a. 100 b. 12 (representing the 12 customers) c. 1 to 5, assigning 1, 2 and 3 to mean that the customer chose the sandwich, and 4 and 5 to mean that the customer chose the chicken salad. There are other ways you could do this, such as using 1 to 10, with 1 to 6 representing the sandwich and 7 to 10 representing the salad. d. No; multiple customers can choose the same meal. e. It doesn’t matter, but it would be easier to count how many of each meal were “chosen” if they are sorted. f. There will not be enough of one meal or the other if there are more than seven of one type chosen. For each of the 100 simulations, determine if there are more than seven of the numbers from 1 to 3 or more than seven of the numbers 4 and 5. If so, that represents a flight for which not all customers got what they wanted. Count the proportion of the 100 simulations for which that was the case.

12. You are competing in a swimming race that has eight contestants starting at the same time, one per lane. Two of your friends are in the race as well. The eight lanes are assigned to the eight swimmers at random, but you hope that you and your two friends will be in three adjacent lanes. You plan to simulate 100 races to estimate the probability that you and your friends will be assigned to three adjacent lanes. For parts (a) to (e), specify what you would answer for each of the questions to accomplish this simulation using www.randomizer.org. a. How many sets of numbers do you want to generate? b. How many numbers per set? c. Number range (e.g., 1–50) d. Do you wish each number in a set to remain unique? e. Do you wish to sort the numbers that are generated? f. Once you had the results of the simulation, how would you use them to estimate the desired probability?

Answer: a. 100 b. 3, to represent the lanes you and your two friends are assigned

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 15: Understanding Uncertainty Through Simulation

c. 1 to 8, to represent the 8 possible lanes you could each be assigned d. Yes; no two swimmers can be assigned to the same lane in the same race. e. Yes or no is okay, but it might be easier to see if you are in adjacent lanes if they are sorted. f. The three numbers represent the lanes you and your friends are in. Find the proportion of the samples that consist of three consecutive numbers.

13. An intersection has a four-way stop sign but no traffic light. Currently, about 1200 cars use the intersection a day, and the rate of accidents at the intersection is about one every two weeks. The potential benefit of adding a traffic light was studied using a computer simulation by modeling traffic flow at the intersection if a light were to be installed. The simulation included 100,000 repetitions of 1200 cars using the intersection to mimic 100,000 days of use. Of the 100,000 simulations, an accident occurred in 5230 of them, and no accident occurred in the rest. (There were no simulated days with two or more accidents.) a. Based on the results of the simulation, what is the estimated number of accidents in a two-week period if the traffic light were to be installed? b. Does the simulation show that adding the traffic light would be a good idea? Explain.

Answer: a. The estimated probability of an accident on any given day is 5230/100,000 or 0.0523. So, the estimated number of accidents in two weeks (14 days) is 14(0.0523) = 0.7322. b. Yes. The current rate of accidents is about one every two weeks, and the simulation shows that the rate would be about 0.7322 every two weeks, which is less than one.

14. Suppose that the chi-square statistic for a chi-square test on a table with two rows and two columns was computed to be 5.3. A simulation was run with 10,000 simulated samples, and 217 of them resulted in chi-square statistics of 5.3 or larger. What is the estimated p-value for the test?

Answer: The estimated p-value is the proportion of the time the simulation resulted in a chi-square statistic of 5.3 or larger, which is 217/10,000 = 0.0217.

15. Suppose that the chi-square statistic for a chi-square test on a table with two rows and two columns was computed to be 2.90. A simulation was run with 1000 simulated samples, and 918 of them resulted in chi-square statistics of less than 2.90. What is the estimated p-value for the test?

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 15: Understanding Uncertainty Through Simulation

Answer: 1 – (918/1000) = 0.082. The estimated p-value is the proportion of the time the simulation resulted in a chi-square statistic of 2.90 or more, which is (1000 – 918)/1000 = 82/1000 = 0.082.

16. Remember that the p-value corresponding to a chi-square statistic of 3.84 for a table with two rows and two columns is 0.05. If you were to simulate 10,000 samples for a 2 × 2 table under the assumption that the null hypothesis is true, about how many of them would you expect to result in a chi-square statistics of less than 3.84?

Answer: About 95% of them should result in a chi-square statistic of less than 3.84; so, about 9500 of the 10,000 simulations.

17. Suppose that a randomization distribution resulting from the simulation of a chi-square test had 7% of the values at or above the chi-square statistic observed for the real sample. a. What is the estimated p-value for the test? b. What decision would you make for the test, using a level of 0.05? c. What decision would you make for the test, using a level of 0.10?

Answer: a. 0.07 b. Do not reject the null hypothesis, because the estimated p-value is greater than 0.05. c. Reject the null hypothesis, because the estimated p-value is less than 0.10.

18. Suppose a teacher observed a correlation of 0.38 between age and number of words children could define on a vocabulary test, for a group of nine children aged 7 to 15. The teacher wanted to confirm that there was a correlation between age and vocabulary test scores for the population of children in this age range. She performed a simulation with 1000 randomized orders, scrambling the test scores across the ages, similar to the simulation used in Example 15.3. She was surprised to find that 29% of the simulated samples resulted in correlations of 0.38 or larger, even though there should not have been any correlation between ages and the scrambled test scores. a. What null and alternative hypotheses was the teacher testing? b. What is the estimated p-value for her test? c. Do these results confirm that there is no correlation between age and vocabulary scores for the population of children aged 7 to 15? Explain.

(Hint: Remember the legitimate conclusions for testing hypotheses.) d. Would a simulation with 10,000 randomized orders be much more likely, much less likely, or about equally likely to enable the teacher to reject the null hypothesis, compared to the simulation performed with 1000 randomized orders?

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 15: Understanding Uncertainty Through Simulation

e. The teacher plans to repeat the experiment. What could she do differently to improve the chance of rejecting the null hypothesis?

Answer: a. Null: Correlation between age and vocabulary score in the population is 0.

Alternative: Correlation is greater than 0 in the population. b. 0.29 c. No. The null hypothesis would not be rejected, but that does not mean it would be accepted as true. d. About equally likely e. Include more children in the study. The size of the sample determines how likely a real relationship is to be found to be statistically significant.

19. Suppose that a chi-square test is carried out on a table with two rows and two columns, with a total of 100 individuals in the four cells. The computed chi-square statistic is 2.17, and a simulation of 1000 samples finds that the estimated p-value is 0.15. The researcher is almost sure that there really is a relationship between the two variables in the population and is disappointed that the results of the test did not provide evidence to reject the null hypothesis using a level of 0.05. The researcher plans to redo the study choosing a new sample of individuals.

Explain whether each of the following actions would increase the likelihood that the null hypothesis would be rejected, decrease the likelihood, or leave the likelihood about the same. a. Use 10,000 simulated samples instead of only 1000. b. Collect a sample of 500 individuals instead of only 100 individuals. c. Use a test with level 0.01 instead of level 0.05.

Answer: a. Doing more simulations would leave the likelihood about the same. b. Including more individuals in the sample would increase the likelihood that the null hypothesis would be rejected. c. Using level 0.01 instead of 0.05 would make it harder to reject the null hypothesis, so the likelihood would decrease.

20. In performing a randomization test, explain why the samples are simulated by using the assumption that the null hypothesis is true.

Answer: The p-value is found by assuming the null hypothesis is true, and then finding the probability of obtaining a test statistic as large as the one observed or larger under that assumption.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 15: Understanding Uncertainty Through Simulation

21. Would probabilities estimated using simulation be considered relativefrequency probabilities or personal probabilities? Explain.

Answer: They are relative frequency probabilities.

22. Using the information in Case Study 15.1 as an example, explain why it is a statistical mistake to formulate hypotheses based on looking at the data you will use to test them.

Answer: Any specific set of data has something unusual about it. If you formulate hypotheses by first noting what is unusual about the data set and then asking how unusual it is, of course you will obtain a low p-value.

23. Three engineers and three artists are given 5 minutes to memorize a list of 25 words, and then asked to recall as many of them as possible. The three engineers recalled 10, 12, and 14 of the words, for an average of 12 words; the three artists recalled 11, 14, and 17 of the words, for an average of 14 words. a. In comparing engineers’ and artists’ ability for memorization, what would be reasonable null and alternative hypotheses to test in this situation? b. The difference in means for the two groups in this sample was two words.

Using that as the “test statistic,” explain how you could carry out a permutation test in this situation. c. Give one example of a permutation from the method you explained in part (b), and what the “test statistic” would be for that permutation.

Answer: a. A reasonable null hypothesis is that the mean number of words recalled in this situation would be the same for engineers and artists in the population. The corresponding alternative hypothesis would be that the mean number of words differs for artists and engineers in the population. b. Randomly choose three of the six scores to represent the engineers, and the remaining three to represent the artists. Compute the difference in the averages for the two groups. Find the proportion of different permutations for which the absolute value of the difference in means is at least 2. That would be the estimated p-value. c. You can use any selection of three of the numbers chosen from 10, 11, 12, 14, 14, 17 to represent the engineers, with the remaining three representing the artists. For example, suppose the engineers are 10, 14 and 14 and the artists are 11, 12 and 17. Then the absolute value of the difference in means is |38/3 – 40/3| = 1.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 15: Understanding Uncertainty Through Simulation

ANSWER TO THE REFLECTION QUESTIONS 1. Do you think humans are able to produce a set of “random numbers” in the same way a computer can? For instance, if you were to ask a group of 100 people to each “randomly choose” a number between 1 and 5, do you think that approximately 20 people would choose each of the possibilities? Explain why or why not.

Answer: No; humans seem to have biases away from numbers at the ends and in favor of numbers in the middle of the options.

2. An odd historical fact is that six of the 14 U.S. presidents between the end of World War II and 2024 were left-handed, even though only about 10% of the U.S. population is left-handed. Suppose you had a very biased coin that had “left” on one side and “right” on the other. The probability that “left” lands face up when you flip the coin is 1 in 10, or 10%. a. Explain how you could use this coin to simulate the “handedness” for a list of 14 individuals randomly chosen from the U.S. population. b. Explain how you could use this coin to estimate the probability that at least six out of 14 randomly selected individuals would be left-handed. (It would take a long time to actually flip the coin enough to do this, so just explain how you would do it.)

Answer: a. You could flip the coin 14 times and record which side landed face up. To use simulation to estimate probabilities for the varying number of lefthanded individuals out of 14, you would need to repeat the process many, many times. b. You would need to carry out 14 flips of the coin many, many times, and record the proportion of them that resulted in “left” landing face up at least six times.

3. Could personal probabilities be estimated using simulation? Explain why or why not.

Answer: Personal probabilities cannot be estimated using simulation. They are based on beliefs, not on repeatable situations.

4. When using simulation to estimate the p-value for a chi-square test, why are the simulated samples created using the assumption that the null hypothesis is true?

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 15: Understanding Uncertainty Through Simulation

Answer: P-values are always calculated by assuming that the null hypothesis is true, and then finding the probability of getting a test statistic as extreme as the one observed or more so, under that assumption.

ANSWER TO THE MINI PROJECTS 1. Use a computer, calculator, or website to simulate one of the situations in Exercises 8 to 12. Use at least 100 repetitions. Use your simulation results to estimate the desired probability.

Answer: The purpose of this project is to get practice with actually carrying out a simulation to estimate a probability. The details of how to do each one are given in the solutions.

2. Find a study that was done using simulation to estimate a p-value, probability, or random outcome, similar to the earthquake simulation study described in Case Study 15.1. Or, find a simulation website that allows you to conduct your own simulation study based on expert opinion or existing data. (Putting terms such as “simulation economics” or “simulation weather” in an internet search engine may help you find an appropriate source.) Explain what was done in the simulation, and then present and interpret the results.

Answer: The purpose of this project is to recognize that simulation is commonly used to estimate probability and uncertainty in situations that are too complex to be handled with probability rules. The report should include a discussion of the details of the simulation, and what results were found.

3. Find out how to do a permutation test for the difference in means for independent samples, and carry out the test using the data in Exercise 23.

State the hypotheses you are testing and the results of the test. Explain how the test was done.

Answer: The details for using a permutation test for the means of two independent samples can be found on many websites or in textbooks. The idea is quite simple. Keep the number of individuals in each group the same as in the original data, but randomly choose which of the responses are assigned to each group. Then compute a two-sample t test statistic for each randomized grouping, and count the proportion of them that are larger than the one observed using the original data.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 15: Understanding Uncertainty Through Simulation

4. Simulate the situation described in Thought Questions 3, 4, and 5. Assume that the probability of being assigned male for each birth is 0.51 and the probability of being assigned female is 0.49. Simulate at least 100 families with four children. Find and report the proportion of your simulated families for which no births are assigned male. Using the probability rules from Chapter 14, calculate the exact probability that a family with four children will not have any births assigned as male, assuming that the probability of being assigned male is 0.51 for each birth. Compare the result with the results of your simulation.

Answer: The purpose of this project is to compare a simulated probability with one that can be computed using probability rules. The answer should come out about the same with both methods, although they may differ somewhat if only 100 simulated families are used. The more simulations done, the closer the two should agree.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 16: Psychological Influences On Personal Probability

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 16: PSYCHOLOGICAL INFLUENCES ON P ERSONAL PROBABILITY

TABLE OF CONTENTS Answer to the Exercise Questions......................................................................................1 Answer to the Reflection Questions................................................................................ 10 Answer to the Mini Projects..............................................................................................12

ANSWER TO THE EXERCISE QUESTIONS 1. Explain how the pseudocertainty effect differs from the certainty effect.

Answer: The pseudocertainty effect applies to situations where the risk is reduced to zero for only a subset of possibilities and is not affected at all for the others.

2. Refer to the scenarios in Table 16.1 on page 346. Suppose someone stands to gain $100,000 in a lawsuit, but there is a 5% chance that something will go wrong and they will get nothing. They are offered a settlement of $90,000 if they are willing to drop the lawsuit. a. Using the definition of “expected value” from Chapter 14, what is the expected value for the amount gained under each decision? b. Using the material in this chapter, explain whether they are likely to accept the settlement. c. Now suppose that instead of a single individual, a company encounters lawsuits of this sort quite frequently. Explain why the decision of the company might be different from the decision of the single individual. Use the concept of expected value in your explanation.

Answer: a. The expected value if they take the settlement is $90,000. The expected value if they don’t is (0.95)($100,000) + (0.05)(0) = $95,000. The expected value is higher by not dropping the lawsuit. b. Because of the certainty effect, people are likely to take the settlement.

(See the first situation in Table 16.1 on page 346.) c. The company might come out ahead in the long run by not settling because the expected value of $95,000 is higher than the settlement value of $90,000. A single individual who only encounters this situation once might © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 16: Psychological Influences On Personal Probability

gain nothing by continuing with the lawsuit, but is guaranteed $90,000 by settling. In other words, the long-run average is relevant for the company, but not as relevant for an individual.

3. Suppose a television advertisement were to show viewers a product and then say, “You might expect to pay $25, $30, or even more for this product. But we are offering it for only $16.99.” Explain which of the ideas in this chapter is being used to try to exploit viewers.

Answer: Anchoring is used. You anchor on the higher values and think they are normal, so the price of $16.99 sounds really good.

4. There are many more termites in the world than there are mosquitoes, but most of the termites live in tropical forests. Using the ideas in this chapter, explain why most people would think there were more mosquitoes in the world than termites.

Answer: The availability heuristic is responsible because mosquitoes are more likely to bother people, and many of the termites are in unpopulated areas. So most people have many more encounters with mosquitos than with termites.

5. Suppose a defense attorney is trying to convince the jury that their client’s wallet, found at the scene of the crime, was actually planted there by the client’s gardener. Here are two possible ways the attorney might present this to the jury: Statement A: The gardener dropped the wallet when no one was looking.

Statement B: The gardener hid the wallet in his sock and when no one was looking he quickly reached down and lowered his sock, allowing the wallet to drop to the ground. a. Explain why the second statement cannot have a higher probability of being true than the first statement. b. Based on the material in this chapter, to which statement are members of the jury likely to assign higher personal probabilities? Explain.

Answer: a. The events in the second statement are a subset of the first. From Rule 4 in Chapter 14, we know that the probability of a subset of events cannot be higher than the probability of the full set. (Notice that personal probabilities are relevant here but the rules should still apply.) b. The members of the jury are likely to assign higher probability to statement B because of the excess detail. It is more representative of how people think such things could happen.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 16: Psychological Influences On Personal Probability

6. Explain why you should be cautious when someone tries to convince you of something by presenting a detailed scenario. Give an example.

Answer: Bringing to mind a detailed, plausible scenario will increase your personal probability that it will happen. An example: Someone trying to sell you a car alarm by describing in detail the methods thieves use.

7. A telephone solicitor recently contacted the author to ask for money for a charity in which typical contributions are in the range of $25 to $50. The solicitor said, “We are asking for as much as you can give, up to $300.00.” Do you think the amount people give would be different if the solicitor said, “We typically get $25 to $50, but give as much as you can.” Explain, using the relevant material from this chapter.

Answer: The solicitor was using “anchoring” by trying to anchor donors on the high value of $300. The resulting donations would probably be higher than they would be if the solicitor has said “we typically get donations of $25 to $50, but give as much as you can.”

8. Research by Slovic and colleagues (1982) found that people judged that accidents and diseases cause about the same number of deaths in the United States, whereas, in truth, diseases cause about 16 times as many deaths as accidents. Using the material from this chapter, explain why the researchers found this misperception.

Answer: The reason is the availability heuristic. Deaths from accidents are more readily brought to mind because they get more media attention. Also, younger people are more likely to know people who have died from accidents and most of these studies are done using students.

9. Determine which statement (A or B) has a higher probability of being true and explain your answer. Using the material in this chapter, also explain which statement you think a statistically naive person would think had a higher probability. a. A car traveling 120 miles per hour on a two-lane highway will have an accident. b. A car traveling 120 miles per hour on a two-lane highway will have a major accident in which all occupants are instantly killed.

Answer: Choice A has a higher probability because the ways in which Choice B can happen are a subset of those for Choice A. But many people would think Choice B had a higher probability because it is representative of what people think would happen.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 16: Psychological Influences On Personal Probability

10. Explain how an insurance salesperson might try to use each of the following concepts to sell you insurance: a. Anchoring b. Pseudocertainty c. Availability Answer: a. Anchoring involves fixing a value in your mind first, then discussing the question of interest. So, for example, the salesperson might tell you what other companies could be charging, then give you their rates (lower, of course). Or, the person might ask you something like “Do you think your risk of needing this insurance is higher or lower than 90%?” thus fixing the idea that you are very likely to need insurance, without actually lying to you. b. Pseudocertainty involves the idea of convincing you that your risk of certain problems will be zero, without mentioning that other problems are not covered at all. For example, the salesperson might tell you that you are completely covered for x, y and z mishaps, and doesn’t that give you peace of mind? You are not told that you are not covered at all for a, b and c mishaps. c. Availability is the idea that you will increase your personal-probability estimate of something that can readily be brought to mind. So the salesperson would try to get you to imagine what terrible things could happen to you. It could be done by saying something like, “Did you see that terrible accident in the news last week? It can happen to anybody!” Or, “Do you know anyone who died young and left a young family with no life insurance? Imagine what would happen to your wife and kids!”

11. In the early 1990s, there were approximately 5 billion people in the world.

Plous (1993, p. 5) asked readers to estimate how wide a cube-shaped tank would have to be to hold all of the human blood in the world. The correct answer is about 870 feet, but most people give much higher answers. Explain which of the concepts covered in this chapter leads people to give higher answers. (Hint: Compare 870 𝑡𝑜 8703 .)

Answer: People use the fact that there are 5 billion people in the world as an anchor.

(Even if the estimate is not given they know there are billions of people.) So they picture a very large container and do not adjust for the fact that a cube has the width multiplied by itself three times. Notice that a width of 870 feet results in a volume of 8703 cubic feet, which is about 670 million cubic feet, which holds about 5 billion gallons of blood.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 16: Psychological Influences On Personal Probability

12. Kahneman and Tversky (1979) asked students if they would be likely to buy “probabilistic insurance.” This insurance would cost half as much as regular insurance but would only cover losses with 50% probability. The majority of respondents (80%) indicated that they would not be interested in such insurance. Notice that the expected value for the return on this insurance is the same as on the regular policy. Explain which of the concepts in this chapter is illustrated by this example.

Answer: This example illustrates the certainty effect and the pseudocertainty effect.

People would rather have a sure thing than a gamble (the certainty effect), even if the sure thing is only relevant to some kinds of losses (the pseudocertainty effect).

13. Barnett (1990) examined front page stories in The New York Times for a year, beginning with October 1, 1988, and found four stories related to automobile deaths but 51 related to deaths from flying on a commercial jet. These correspond to 0.08 story per thousand U.S. deaths by automobile and 138.2 stories per thousand U.S. deaths for commercial jets. He also reported a midAugust 1989 Gallup Poll finding that 63% of Americans had lost confidence in recent years in the safety of airlines. Discuss this finding in the context of the material in this chapter.

Answer: The availability heuristic applies to this situation. The extensive media coverage of the few major airline accidents that have occurred raises people’s perception of the risk of flying.

14. Explain how the concepts in this chapter account for each of the following scenarios: a. Most people rate death by shark attacks to be much more likely than death by falling airplane parts, yet the chances of dying from the latter are actually 30 times greater (Plous, 1993, p. 121). b. You are a juror on a case at your school involving an accusation that a student cheated on an exam. The jury is asked to assess the likelihood of the statement, “Even though he knew it was wrong, the student copied from the person sitting next to him because he desperately wants to get into medical school.” The other jurors give the statement a high probability assessment although they know nothing about the accused student. c. Research by Tversky and Kahneman (1982b) has shown that people think that words beginning with the letter k are more likely to appear in a typical segment of text in English than words with k as the third letter. In fact, there are about twice as many words with k as the third letter than words that begin with k. d. A 45-year-old man dies of a heart attack and does not leave a will.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 16: Psychological Influences On Personal Probability

Answer: a. The availability heuristic applies, because deaths by shark attack are big news and have even been the subject of movies. When did you last see a movie in which someone was killed by falling airplane parts? b. The representative heuristic would come into play. It is consistent with most people’s views of the world to think that wanting desperately to get into medical school would lead someone to fight hard to get good grades, including possibly cheating. Being given a possible explanation that is representative of how they see the world might thus lead the jurors to give the explanation high probability. c. Availability is responsible. It is much easier to bring to mind words that start with k, by making the sound and seeing what follows, than it is to think of words with k as the third letter. d. This is caused by being overly optimistic. Most people in their 40s do not think they are about to die.

15. Suppose you go to your doctor for a routine examination, without any complaints of problems. A blood test reveals that you have tested positive for a certain disease. Based on the ideas in this chapter, what should you ask your doctor in order to assess how worried you should be?

Answer: You should ask about the base rate or probability of the disease for people like you, and the probabilities of false positives and false negatives (or the sensitivity and specificity).

16. Give one example of how each of the following concepts has had or might have an unwanted effect on a decision or action in your daily life: a. Conservatism b. Optimism c. Forgotten base rates d. Availability Answer: Answers will vary for reach part, but an example is given in each case. a. Conservatism is the tendency to change previous probabilities more slowly than warranted by new data. It might lead a smoker to ignore accumulating evidence that the smoke from their cigarettes harms those around them. b. Optimism might lead one to drive after drinking, thinking they won’t have an accident or get caught. c. Forgotten base rates might make someone panic when they learn they have tested positive for a rare disease.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 16: Psychological Influences On Personal Probability

d. Availability may lead one to think crime is much more prevalent than it really is, since it is so often featured in the news.

17. In this chapter, we learned that one way to lower personal-probability assessments that are too high is to list reasons why you might be wrong.

Explain how the availability heuristic might account for this phenomenon.

Answer: The availability heuristic tells us that people assign higher probabilities to events they can readily bring to mind. By listing reasons why you might be wrong, you are actually bringing them to mind, and thus are likely to assign a higher probability to the notion that you might be wrong.

18. Which of the concepts in this chapter might contribute to the decision to buy a lottery ticket? Explain.

Answer: Optimism, because people think they will be the lucky winner. The possibility effect applies as well, because it leads people to give more value than warranted to a small increase from a 0 probability to a non-zero probability of a good outcome.

19. Suppose you have a friend who is willing to ask her friends a few questions and then, based on their answers, is willing to assess the probability that those friends will get an A in each of their classes. She always assesses the probability to be either 0.10 or 0.90. She has made hundreds of these assessments and has kept track of whether her friends actually received A’s.

How would you determine if she is well calibrated?

Answer: You could determine if your friend is well-calibrated by noting what percentage of the time a person gets an A in a class when she has predicted it with probability 0.10 and do the same for her predictions with probability 0.90.

If she is perfectly calibrated, then it will occur in 10% of the cases for which she predicted it with probability 0.10 and in 90% of the cases for which she predicted it with probability 0.90.

20. If you ask five people when their birthdays are, what is the probability you will find someone born in the same month as you? For simplicity, assume that the probability that a randomly selected person will have the same birth month you have is 1/12. Now use the material from Chapter 14 to make a table listing the numbers from 1 to 5, and then fill in the probabilities that you will first encounter someone with your birth month by asking that many people.

Determine the accumulated probability that you will have found someone with your birth month by the time you ask the fifth person. How well calibrated was your initial guess?

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 16: Psychological Influences On Personal Probability

Answer: The probability is 0.353, as shown in the table below.

No. of People

Probability of Match

Accumulated Probability

1

1/12

1/12 = 0.0833

2

(11/12)(1/12)

11/144 + 1/12 = 0.16

3

(11/12)2(1/12)

(11/12)2(1/12) + 0.16 = 0.23

4

(11/12)3(1/12)

(11/12)3(1/12) + 0.23 = 0.294

5

(11/12)4(1/12)

(11/12)4(1/12) + 0.294 = 0.353

Exercises 21 to 24: The scenarios in Exercises 21 to 24 are all taken from the book Thinking, Fast and Slow by Daniel Kahneman (2011). Page numbers from the book are given in each exercise.

21. People were asked to give a personal probability assessment of the possible occurrence of each of the following two events (Kahneman, 2011, p. 159): • A massive flood somewhere in North America next year, in which more than 1000 people drown • An earthquake in California sometime next year, causing a flood in which more than 1000 people drown Using the material from this chapter, explain which of the two possible events you think people found more likely and whether that event is, in fact, more likely.

Answer: The earthquake scenario may be perceived as more likely because of the representativeness heuristic. It provides a plausible scenario for how a flood could occur in which more than 1000 people are killed. But the first scenario must be more likely by Rule 4 of Chapter 14. The second scenario is a subset of the first one.

22. “In a famous study, spouses were asked, ‘How large was your personal contribution to keeping the place tidy, in percentages?’ . . . As expected, the self-assessed contributions [from the two spouses] added up to more than 100% . . . . The bias is not necessarily self-serving: spouses also overestimated their contributions to causing quarrels.” (Kahneman, 2011, p. 131) Use the material in this chapter to explain why the percentages given by the two spouses would add up to more than 100%.

Answer: Availability is the reason for this. Each spouse can readily remember their contributions to keeping the place tidy (and to causing quarrels) but cannot as readily remember their spouse’s contributions.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 16: Psychological Influences On Personal Probability

23. “You see a person reading The New York Times on the New York subway. Which of the following is a better bet about the reading stranger?

• She has a PhD. • She does not have a college degree.” (Kahneman, 2011, p. 151) a. Which of the two statements do you think people would state as more likely, and why? b. Which statement do you think actually is most likely? c. Use the material in this chapter to explain why the answers to parts (a) and (b) might differ.

Answer: a. Most people would choose “She has a PhD” because it is more representative of the kind of person they think reads The New York Times. b. The second statement is probably more likely because many more of the people riding the New York subway do not have college degrees than have PhD degrees. Even if all of the people with PhD degrees read The New York Times and (for instance) only one in 10 of those without college degrees do so, the number of readers without college degrees would be higher than those with PhD degrees. c. As noted in the response to Part (a), the representative heuristic would lead people to be more likely to associate someone reading The New York Times with having a PhD than with having no college degree. Therefore, they would give the first statement a higher probability. The response to Part (b) explains why that would be likely to be incorrect.

24. “In 2002, a survey of American homeowners who had remodeled their kitchens found that, on average, they had expected the job to cost $18,658; in fact, they ended up paying an average of $38,769” (Kahneman, 2011, p. 250). The quote is an example of what Kahneman calls the planning fallacy, which describes “plans and forecasts that are unrealistically close to best case scenarios, [and that] could be improved by consulting the statistics of similar cases” (p. 250).

Explain how each of the following concepts from this chapter could have contributed to the planning fallacy for the kitchen remodeling example. a. Optimism b. Availability c. Anchoring Answer: a. Optimism leads people to think that nothing will go wrong in the process of remodeling and/or that they will not change their minds about anything. In reality, both things are quite likely to happen, thus increasing the cost. b. Availability leads people to bring to mind readily the best-case scenario in which everything proceeds according to the original plan. It is harder to imagine all of the ways in which something might go wrong that would increase the cost. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 16: Psychological Influences On Personal Probability

c. Contractors may anchor consumers on the price of the best-case scenario and fail to mention that something almost always deviates from that scenario.

ANSWER TO THE REFLECTION QUESTIONS 1. Suppose you are buying a new phone because your previous phone fell into the water when you leaned over the edge of a bridge to watch a boat go by. Explain how each of the following might lead you to buy insurance on your new phone. a. The possibility effect b. The availability heuristic c. Detailed imagination Answer: a. The possibility effect holds when people are willing to pay to increase the probability of something good happening from 0 to a small value. In this case, if you don’t buy the insurance and you drop your new phone in the water, you will have the benefit of having it replaced. Although there is only a small probability that you will make that mistake again, buying the insurance means that if you do make it, your phone will be covered. b. The availability heuristic is relevant because you have just experienced a concrete example of how you could lose or damage your phone. So when you are offered insurance to cover a loss or damage, you have an available example of why you might need it. c. Detailed imagination is relevant because, similar to the availability heuristic, you have a detailed example of how you could lose or damage your phone and thus benefit from the insurance.

2. You are planning to sell your car, and based on websites that provide estimates of its worth, you decide that it’s worth about $10,000. Explain how each of the following might lead you to list it for sale for $11,000 instead of $10,000. a. The anchoring effect b. Overconfidence Answer: a. The anchoring effect is relevant because if you list it for $11,000, potential buyers will use that as an “anchor” and negotiate down from there. If you listed it for $10,000, potential buyers would be likely to negotiate down from that price. b. Overconfidence applies because most people are likely to think that they have treated their car better than average, and that it should therefore be worth more than the average price given for that model. Additionally, you might be over confident in your ability to convince potential buyers to pay more than it is worth. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 16: Psychological Influences On Personal Probability

3. Give an example of two statements for which the conjunction fallacy would lead most people to give a higher probability to the wrong one.

Answer: The two statements should be such that one is detailed account of how something could happen, but is a subset of the other one. For example, one statement might be “You have a car accident because there is something in the road and you hit it.” The subset statement might be “You have a car accident because you drive around a curve and there is a deer standing in the road.” The second statement is a subset of the first statement, but people might be likely to assign it a higher probability. They might think that they would not be so negligent as to not see something in the road (as given in the first statement), but could easily envision the scenario in the second statement happening to them.

4. Explain how a lawyer could use “detailed imagination” to mislead a jury into thinking a defendant is guilty based on vague evidence. Provide an example.

Answer: The more detailed the scenario presented, the more jurors are likely to envision that it could have happened. As an example, suppose the charge is that someone was driving while intoxicated, and the evidence is that they just barely failed a roadside sobriety test. Now suppose the prosecuting attorney tells the jury “It’s only three miles from a popular bar frequented by people like this defendant and the defendant’s house, and that’s the road where the defendant was apprehended and tested.” The evidence presented does not include that the defendant was actually drinking at that bar, but the jury can imagine that it could have happened that way.

5. Explain how “forgotten base rates” might lead a physician to misdiagnose someone in each of the following situations. a. The patient has a positive test result for a rare disease. b. The patient has symptoms that are consistent with two different diseases, one of which occurs 10 times more often than the other.

Answer: a. For rare diseases for which the test is not 100% perfect, there are likely to be more false positives than true positives. But physicians often forget that fact, and think that because the test is relatively accurate, if the result is positive the patient probably has the disease. b. Because of the availability heuristic, physicians are likely to choose the diagnosis that corresponds to the more common disease and forget that the other one is a possibility.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 16: Psychological Influences On Personal Probability

6. Suppose you have a friend who is a nurse in a maternity clinic and routinely predicts the sex of the baby when pregnant women first come to the clinic, before any information is available. How could you determine whether your friend is well-calibrated?

Answer: You would need to keep track of the accuracy of your friend’s predictions over the long run and see how it corresponds to what happened. You would create a two-way table with prediction as the rows and outcome as the columns. It would not be sufficient if only one of the predictions was always accurate. For instance, if your friend always predicated that the baby would be assigned male at birth, then they would be 100% accurate for babies assigned male at birth. But they would not be at all accurate for babies assigned female. To be well-calibrated, they would need to be accurate for both.

ANSWER TO THE MINI PROJECTS 1. Design and conduct an experiment to try to elicit misjudgments based on one of the phenomena described in this chapter. Explain exactly what you did and your results.

Answer: Make sure the study was carried out using the ideas from Part 1 of the book.

The wording should be as unbiased as possible. The sample size should be large enough to see the desired effect. Beyond that, there is substantial scope for this project.

2. Find and explain an example of a marketing strategy that uses one of the techniques in this chapter to try to increase the chances that someone will purchase something. Do not use an exact example from the chapter, such as “buy one, get one free.”

Answer: One possibility is a form of anchoring in which a store will announce a competitor’s prices. However, the pseudocertainty type of idea applies because they will announce prices only for certain products, namely, the ones for which they charge less. Another possibility is presenting a detailed picture of how you would use the item, such as when trying to sell a swimming pool or hot tub and people are pictured having fun in one. Yet another possibility is to take advantage of optimism, such as when trying to sell an investment opportunity and explaining how rich you could become if you buy.

3. Find a journal article that describes an experiment designed to test the kinds of biases described in this chapter. Summarize the article, and discuss what conclusions can be made from the research. You can find such articles by searching the web using key words from this chapter. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 16: Psychological Influences On Personal Probability

Answer: There are a multitude of possibilities for this project and it is not possible to anticipate all of the responses.

4. Estimate the probability of some event in your life using a personal probability, such as the probability that a person who passes you on the street will be wearing a hat, or the probability that traffic lights will be red when you get to them. Use an event for which you can keep a record of the relative frequency of occurrence over the next week. How well calibrated were you?

Answer: This project is designed to help understand the meaning of a personal probability and of calibration. The event in question should be one for which the relative-frequency probability would not be known in advance, yet for which the event is repeatable throughout the week. To be well calibrated, the proportion of times the event occurred out of the possible opportunities for it to occur should be close to the assigned personal probability. For example, suppose you estimate that you will have to stop about 40% of the time when you encounter a traffic light on your commute. Then you should count the number of times you have to stop, and divide by the number of traffic lights you pass. If that proportion is close to 40%, you are well calibrated.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 17: When Intuition Differs From Relative Frequency

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 17: WHEN INTUITION DIFFERS FROM RELATIVE FREQUENCY

TABLE OF CONTENTS Answer to the Exercise Questions......................................................................................1 Answer to the Reflection Questions................................................................................ 10 Answer to the Mini Projects..............................................................................................12

ANSWER TO THE EXERCISE QUESTIONS 1. Explain why it is not at all unlikely that, in a class of 50 students, two of them will have the same last name.

Answer: There are 50  49 pairs of students for which there could be a match and some names are extremely common.

2. Suppose twin sisters are reunited after not seeing each other since they were 3 years old. They are amazed to find out that they are both married to men named James and that they each have a daughter named Jennifer. Explain why this is not so amazing.

Answer: Any two people who compare a series of facts about themselves are likely to find some in common. Presumably the sisters would have compared notes on lots of factors. The names cited are heard quite frequently, so that adds to the likelihood of having them in common.

3. Suppose you are seated next to a stranger on an airplane, and you start discussing various topics such as where you were born (country or state), what your favorite movie of all time is, your spouse’s occupation, and so on. For simplicity, assume that the probability that your details match for any given topic is 1/50 and is independent from one topic to the next. If you discuss ten topics, how surprising would it be to find that you match on at least one of them? Support your answer with a numerical assessment.

Answer: It would not be surprising to find a match on at least one topic. The probability can be computed using the rules from Chapter 14. For each item the

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 17: When Intuition Differs From Relative Frequency

probability of not matching is 49/50. Using Rule 3 (extended to ten independent events) the probability of not matching on any of them is (49/50)10 = 0.8171. So the probability of matching on at least one topic is 1 – 0.8171 = 0.1829.

4. Why is it not surprising that the night before a major airplane crash several people will have dreams about an airplane disaster? If you were one of those people, would you think that something amazing had occurred?

Answer: As a rough calculation, there are over 330 million people in the United States, so even if the probability of dreaming of an airplane disaster is only 1 in a million, we would expect 330 people to have done so on any given night.

However, if you were one of them and had never had such a dream before, you would probably think something amazing had occurred. The fact is that someone, somewhere will almost surely have such a dream before any crash.

5. Explain why the story about George D. Bryson, reported in Example 17.1, is not all that surprising.

Answer: There are thousands of hotels and motels with hundreds of rooms each. The probability that two people with the same name will sometimes occupy the same room on consecutive days must take into account millions of possibilities for that to occur. Adding the twist about asking for and receiving mail lowers the probability but would still not be too surprising.

6. Find a dollar bill or other item with a serial number. Write down the number. I predict that there is something unusual about it or some pattern to it. Explain what is unusual about it and how I was able to make that prediction.

Answer: There are numerous meanings to the word “pattern” and it is easy to find something that fits at least one of them.

7. A statistics professor once made a big blunder by announcing to the class of about 50 students that it was fairly certain that someone in the room would share the professor’s birthday. We have already learned that there is a 97% chance that there will be two people in a room of 50 with a common birthday.

Given that information, why was the professor’s announcement a blunder? Do you think the professor was successful in finding a match? Explain.

Answer: The probability of someone having a particular birthday, namely that of the professor, is much smaller than the probability that there will be some date in common for some pair in the class. The professor was not likely to be successful, because the probability for each student would be only 1/365. On

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 17: When Intuition Differs From Relative Frequency

average, we could expect 50/365 or 0.14 matches, which means that a match would occur in about 1 in 7 classes. (To be precise, the probability of no match is found by multiplying 364/365 fifty times, which is about 0.87.)

8. If you wanted to pretend that you could do psychic readings, you could perform “cold readings” by inviting people you do not know to allow you to tell them about themselves. You would then make a series of statements like: “I see that there is some distance between you and your mother that bothers you.”

“It seems that you are sometimes less sure of yourself than you indicate.”

“You are thinking of two men in your life, one of whom is sort of lightcomplexioned and the other of whom is slightly darker. Do you know who I mean?”

In the context of the material in this chapter, explain why this trick would often work to convince people that you are indeed psychic.

Answer: There are numerous ways in which each statement could be true, and at least one of them would apply to almost everyone. For instance, the statement about the mother would apply if she was deceased, if she lived in a distant city, if there was friction in the relationship or if the person just wasn’t as close to her as desired.

9. Many people claim that they can often predict who is on the other end of the phone when it rings before looking at who’s calling. Do you think that phenomenon has a normal explanation? Explain.

Answer: Yes, there are limited choices and people know their friends’ calling behaviors.

10. Explain why it would be much more surprising if someone were to flip a coin and get six heads in a row after telling you they were going to do so than simply watching them flip the coin six times and observe six heads in a row.

Answer: Six heads in a row is just as likely as any of the other possibilities, and there are numerous other sequences that might seem unusual. Specifying the sequence in advance means that the event of interest is only six heads in a row, which has probability (0.5)6 or 1/64 of happening.

11. Comment on the following unusual lottery events, including a probability assessment. a. On September 11, 2002, the first anniversary of the 9/11 attack on the World Trade Center, the winning number for the New York State lottery was 911. b. To play the Maryland Pick 4 lottery, players choose four numbers from the digits 0 to 9. The game is played twice every day, at midday and in the evening. In 1999, holiday players who decided to repeat previous winning

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 17: When Intuition Differs From Relative Frequency

numbers got lucky. At midday on December 24, the winning numbers were 7535, exactly the same as on the previous evening. And on New Year’s Eve, the evening draw produced the numbers 9521—exactly the same as the previous evening.

Answer: a. The number 911 has the same probability as any other 3-digit number, which is 1/1000. b. With 4 digits, the probability of any specific 4-digit number is 1/10,000 because the choices are 0000 to 9999. However, any 4-digit number could have been repeated, so there are 1000 different ways to get the same number twice. Thus, the probability of getting the same 4-digit number on consecutive drawings is 1000/10,000 = 1/1000. You can also see this by realizing that no matter what number was drawn first, the probability of getting that number on the next drawing is 1/1000.

12. Give an example of a sequence of events to which the gambler’s fallacy would not apply because the events are not independent.

Answer: Answers will vary. See exercise 15 for an example. The main idea is that the probability of subsequent similar events actually does increase or decrease after observing an event.

13. Although it’s not quite true, suppose the probability of having a child assigned male at birth (M) is equal to the probability of having a child assigned female (F). A woman gives birth to four children. a. Are they more likely to be FFFF or to be MFFM? Explain your answer. b. Which sequence in part (a) of this exercise would a belief in the law of small numbers cause people to say had higher probability? Explain. c. Is a woman who gives birth to four children more likely to have all of them assigned female at birth, or to have two children of each sex? Explain.

Answer: a. Both sets are equally likely, with probability (0.5)(0.5)(0.5)(0.5). b. People would think the sequence MFFM had higher probability because it has 50% of each sex. c. They are more likely to have two of each sex because there is only one sequence resulting in four babies all assigned female (FFFF) but many sequences resulting in two of each sex (MFFM, MFMF, etc.).

14. You are at a casino with a friend, playing a game in which dice are involved.

Your friend has just lost six times in a row. They are convinced that they will win on the next bet because they claim that, by the law of averages, it’s their turn to win. They explain to you that the probability of winning this game is

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 17: When Intuition Differs From Relative Frequency

40%, and because they have lost six times, they have to win four times to make the odds work out. Is your friend right? Explain.

Answer: Your friend is wrong and is suffering from the gambler’s fallacy. Games like this have no memory, and they are not likely to reach the relative frequency of 40% in the short run.

15. Suppose a friend reports that they have just had a string of “bad luck” with their car. They had three major problems in as many months and now have replaced many of the worn parts with new ones. Your friend concludes that it is their turn to be lucky and that they shouldn’t have any more problems for a while. Is your friend using the gambler’s fallacy? Explain.

Answer: No. The probability of having to replace parts in a car in the near future is likely to depend on having to replace other parts. It would not be luck; it would be because your friend has already replaced the parts most likely to wear out first.

16. A fair coin is flipped six times and the sequence of heads and tails is observed. a. Are the sequences HHHHHH and HTHHTT equally likely? Explain. b. Are the events “A = 6 heads in the 6 tosses” and “B = 3 heads and 3 tails in the 6 tosses” equally likely? Explain. c. Would belief in the law of small numbers lead people to think that the two sequences in part (a) are equally likely, or are not equally likely? Would they be correct? Explain. d. Would belief in the law of small numbers lead people to think that the two events in part (b) are equally likely, or are not equally likely? Would they be correct? Explain.

Answer: a. Yes, each sequence has probability (1/2)6. b. No, they are not equally likely because there is only one sequence with 6 heads (HHHHHH) but many different sequences with three heads and three tails (HHHTTT, THTHTH, etc.). c. Based on the law of small numbers people would think that the sequence HTHHTT was more likely than HHHHHH, but in fact they are equally likely. d. Based on the law of small numbers people would think that event B was more likely than event A, and they would be correct.

17. The University of California at Berkeley Wellness Encyclopedia (1991) contains the following statement in its discussion of HIV testing: “In a high-risk population, virtually all people who test positive will truly be infected, but among people at low risk the false positives will outnumber the true positives.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 17: When Intuition Differs From Relative Frequency

Thus, for every infected person correctly identified in a low-risk population, an estimated ten non-carriers [of the HIV virus] will test positive” (p. 360). a. Suppose you have a friend who is part of this low-risk population but who has just tested positive. Using the numbers in the statement, calculate the probability that the person actually carries the virus. b. Your friend is understandably upset and doesn’t believe that the probability of being infected with HIV isn’t really near 1. After all, the test is accurate, and it came out positive. Explain to your friend how the Wellness Encyclopedia statement can be true, even though the test is very accurate both for people with HIV and for people who don’t carry it. If it’s easier, you can make up numbers to put in a table to support your argument. (See Table 17.1.)

Answer: a. The numbers in the statement say that out of 11 who test positive only 1 is infected. Therefore the probability that the person carries the virus is 1/11 or about 9%. b. You should convey the idea that there are a very large number of false positives because of the fact that most people who get tested do not have the disease. Even if there are no false negatives, the number of people who test positive because they have the disease would be low compared to the number who test positive because of a mistake in the test.

18. Using the data in Table 17.1 about a hypothetical population of 100,000 patients tested for breast cancer, find the probability of each of the following events: a. A patient whose test shows a malignant lump actually has a benign lump. b. A patient who actually has a benign lump has a test that shows a malignant lump. c. A patient with unknown status has a test showing a malignant lump.

Answer: a. 9,900/10,700 = 0.925 b. 9,900/99,000 = 0.10 c. 10,700/100,000 = 0.107 19. Suppose a rare disease occurs in about 1 out of 1000 people who are like you.

A test for the disease has sensitivity of 95% and specificity of 90%. Using the technique described in this chapter, compute the probability that you actually have the disease, given that your test results are positive.

Answer: The best way to do this computation is to construct a table showing 100,000 hypothetical people in a population like the one described. Here is how it would look:

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 17: When Intuition Differs From Relative Frequency

Test positive Actually sick

Test negative

Total

95

5

100

Actually healthy

9,990

89,910

99,900

Total

10,085

89,915

100,000

Of the 10,085 testing positive, 95/10,085 = 0.0094 or 0.94% are sick.

20. Using the data in Table 17.1, give numerical values and explain the meaning of the sensitivity and the specificity of the test.

Answer: Sensitivity is the conditional probability of a positive test, given that you have the disease. In this case, it is 800/1000 = 0.80. Specificity is the conditional probability of having a negative test, given that you do not have the disease. In this case, it is 89,100/99,000 = 0.90. These correspond to the probabilities of correctly testing positive and correctly testing negative.

21. Which one of the problems with probability assessments discussed in Chapter 16 contributes to confusion of the inverse? Explain.

Answer: Not taking the base rate into account is what contributes to confusion of the inverse.

22. Suppose the sensitivity of a test is 0.90. Give either the false positive or the false negative rate for the test, and explain which you are providing. Could you provide the other one without additional information? Explain.

Answer: The sensitivity of 0.90 is the probability of a positive test when someone has the disease, so the probability of a negative test when someone actually has the disease, i.e., a false negative, is 0.10. We cannot find the probability of a false positive from the information provided.

23. In financial situations, are businesses or individuals more likely to make use of expected value for making decisions? Explain.

Answer: Using expected values to make decisions makes the most sense for situations that are repeated over the long run. In general, businesses are more likely to have enough transactions to benefit from the “long-run average” and thus would make use of expected values to make decisions. It’s also the case that for businesses their main concern is an overall profit, whereas individuals tend to incorporate emotional and psychological elements into their decisions.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 17: When Intuition Differs From Relative Frequency

24. We learned in this chapter that one idea researchers have tested was that when forced to make a decision, people choose the alternative that yields the highest expected value. a. If that were the case, explain which of the following two choices people would make: Choice A: Accept a gift of $10.

Choice B: Take a gamble with probability 1/1000 of winning $9000 and 999/1000 of winning nothing. b. Using the material from Chapter 16, explain which of the two choices in part (a) you think the majority of people would select.

Answer: a. The expected value for choice A is $10 and for choice B it is $9, so people would choose A. b. The majority of people would be likely to select Choice B because of the possibility effect. (See Table 16.1.)

25. Refer to the previous exercise, about choosing between a gift and a gamble.

Explain how the situation in part (a) resembles the choices people have when they decide whether to buy lottery tickets.

Answer: The choices are similar to the choice involved in buying a lottery ticket because the expected value for buying a lottery ticket is less than the amount of money you keep by not buying the ticket. But by buying the ticket there is a chance of winning a large amount of money.

26. It is time for the end-of-summer sales. One store is offering bathing suits at 50% of their usual cost, and another store is offering to sell you two for the price of one. Assuming the suits originally all cost the same amount, which store is offering a better deal? Explain.

Answer: They are offering equivalent deals in terms of the average cost of each suit.

You may prefer the 50% off deal, because you have to buy only one suit to achieve the savings.

27. Suppose you believe that the probability that your team will win a game is 1/4.

You are willing to bet $1 that your team will win. What amount should you be offered if you win in order to have a break-even expected value?

Answer: You should be offered $3.00, so the expected value is ($3.00)(1/4) + (−1.00) (3/4) = $0.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 17: When Intuition Differs From Relative Frequency

28. Suppose you are trying to decide whether to park illegally while you attend class. If you get a ticket, the fine is $25. If you assess the probability of getting a ticket to be 1/100, what is the expected value for the fine you will have to pay? Under those circumstances, explain whether you would be willing to take the risk and why. (Note that there is no correct answer to the last part of the question; it is designed to test your reasoning.)

Answer: There are two possibilities. You pay $25 with probability 1/100 or pay nothing with probability 99/100, so the expected value is ($25)(0.01) = $0.25 or 25 cents. It will cost an average of 25 cents each time you risk getting a ticket, which seems like a cheap price for parking!

29. You are making a hotel reservation and are offered a choice of two rates. The advanced purchase rate is $100, but your credit card will be charged immediately and there is no refund, even if you don’t use the room. The flexible rate is $140 but you don’t pay anything if you don’t use the room.

Suppose p is the probability that you will end up using the room. a. Suppose p = 0.70, so there is a 70% chance you will use the room. What is the expected value of your cost if you reserve the room with the flexible rate? (Hint: What are the two possible amounts you could pay, and what are their probabilities?) b. No longer assume a specific value for p. In terms of p, what is the expected value of your cost if you reserve the room with the flexible rate? c. What is the expected value of your cost if you choose the advanced purchase rate? (Hint: There is only one possible amount.) d. For what value of p are the expected values you found in parts (b) and (c) the same? e. For what range of values of p are you better off choosing the advanced purchase rate?

Answer: a. The expected value is ($140)(0.7) + ($0)(0.3) = $98.00. b. The expected value is $140p. c. The expected value is the only possible value, which is $100. d. The expected values are the same if $140p = $100, so p = 100/140 = 0.714. e. You are better off choosing the advanced purchase rate if p is at least 0.714. In that case, you will pay $100 but your expected value for choosing the flexible rate would be $100 or more. So you should choose the advanced purchase rate as long as your personal probability of taking the trip is at least 0.714.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 17: When Intuition Differs From Relative Frequency

ANSWER TO THE REFLECTION QUESTIONS 1. A fact many consider to be an amazing coincidence is that John Adams and Thomas Jefferson, the second and third presidents of the United States, both died on July 4, 1826. Not only did they die within hours of each other, but their death date was the 50th anniversary of the adoption of the U.S. Declaration of Independence, which was signed by both men. Discuss whether you think this event should be considered to be an amazing coincidence.

Answer: There is no right or wrong answer to this question; it’s the reasoning that counts. The situation certainly fits the definition of a coincidence given on page 362, since it’s a low probability event perceived as meaningful. However, the probability that something like this would happen regarding two famous people, or even two politicians, is probably quite high because there are so many famous people and so many different events that would seem surprising.

2. As illustrated in Case Study 17.1, there is an ongoing controversy over whether streak shooting occurs in basketball and whether similar streaks occur in other sports. If streaks do occur, this implies that the probability of a success is higher following a previous success than following a failure. Explain whether streak shooting would be possible if all shooting attempts are independent of each other.

Answer: Streak shooting would not be possible if all attempts are independent. If that were the case, then the probability of a success on one throw would not be related to the probability of a success on the previous throw or the next throw.

3. Explain whether a “false-positive” medical test result is more likely to occur for rare diseases or for common diseases.

Answer: A false positive is more likely for rare diseases than for common diseases. For example, suppose a test is accurate 90% of the time for those who have the disease and for those who don’t have the disease but that only 1 in 1000 people have it. Then if 10,000 people are tested, about 10 of them will have the disease and 9 of them will test positive. But of the 9,990 of them don’t have the disease, and (9990) × 0.9 = 8991 of them will have a positive test.

Therefore, of the 8991 + 9 = 9000 who have a positive test, only 9 of them have the disease.

4. When discussing the effectiveness of vaccines for COVID-19, commentators often confused the two conditional probabilities related to being vaccinated and dying of COVID-19. One of these is the probability of dying given that you

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 17: When Intuition Differs From Relative Frequency

have been vaccinated, and the other is the probability of having been vaccinated given that you died. a. Explain which one of these two probabilities would be of most interest to someone in making a decision about whether to get vaccinated. b. Explain which one of these is likely to be quite high for a population in which most people have been vaccinated. c. Use the information from parts (a) and (b) to explain how confusion of the inverse could cause people to be misled about the effectiveness of getting vaccinated.

Answer: a. A person making the decision would want to know the probability of dying, given that you have been vaccinated. b. If most of a population has been vaccinated, then the probability of having been vaccinated, given that a person dies, would be quite high. c. If commentators were quoting the probability of dying, given that you have been vaccinated, people who didn’t understand the difference would be misled. They might misinterpret that high probability to be the probability of dying, given that you were vaccinated.

5. Would the use of expected values for making decisions make more sense for something you plan to do only once or for something you plan to do repeatedly? Given an example of each situation.

Answer: Using expected values to make decisions makes more sense for something you plan to do repeatedly. The expected value is the average over the long term, not the value expected in a single instance. An example where it would not make sense for an individual to use the expected value to make a decision is given in Chapter 16. If someone is offered a court settlement of many thousands of dollars, they are better off taking it even if the expected value is higher by gambling that they will win more by going to court. An example of something you might do repeatedly is given in Exercise 29 of this chapter.

Suppose you routinely book hotel rooms, and have to choose between the advanced purchase rate which you can’t cancel, or the regular rate, which you can cancel if you don’t use the room. Then assuming you have a high probability of going on each trip, you can figure out the break-even point for which the advanced purchase is the better deal.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 17: When Intuition Differs From Relative Frequency

ANSWER TO THE MINI PROJECTS 1. Find out the sensitivity and specificity of a common medical test. Calculate the probability of a true positive for someone who tests positive with the test, assuming the rate in the population is 1 per 100; then calculate the probability assuming the rate in the population is 1 per 1000.

Answer: Make sure a source is given for the information, because finding these numbers is the hardest part of this project. However, they are often available, and part of the purpose of the project is to make that fact known. One complication of this project is that sometimes the probabilities differ for different populations. If that’s the case, the information should be conveyed in the project and one population can be chosen for the computations.

2. Ask four friends to tell you their most amazing coincidence story. Use the material in this chapter to assess how surprising each of the stories is to you.

Pick one of the stories, and try to approximate the probability of that specific event happening to your friend.

Answer: The idea here is to try to approximate the probabilities using relativefrequency ideas and the rules from Chapter 14. Be sure the rules are properly applied. For instance, event probabilities should only be multiplied if the events can reasonably be assumed to be independent.

3. Conduct a survey in which you ask 20 people the two scenarios presented in Thought Question 5 at the beginning of this chapter and discussed in Section 17.5. Record the percentage who choose alternative A over B and the percentage who choose alternative C over D. a. Report your results. Are they consistent with what other researchers have found? (Refer to p. 371.) Explain. b. Explain how you conducted your survey. Discuss whether you overcame the potential difficulties with surveys that were discussed in Chapter 4.

Answer: Researchers have found that preferences are for a sure gain, but a gamble on the loss. Part (a) of this project simply asks for a comparison with those results. In part (b) it is important to explain how the survey was conducted in an unbiased way. The explanation should include the wording used for the introduction, to make sure participants were not given a clue about the “desired” answer.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 18: Understanding The Economic News

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 18: UNDERSTANDING THE ECONOMIC NEWS

TABLE OF CONTENTS Answer to the Exercise Questions......................................................................................1 Answer to the Reflection Questions................................................................................. 11 Answer to the Mini Projects..............................................................................................13

ANSWER TO THE EXERCISE QUESTIONS 1. The price of a first-class stamp in the United States in January 1985 was 20 cents, whereas in January 2022, it was 58 cents. The Consumer Price Index for January 1985 was 105.5, whereas for January 2022, it was 281.148. If the true cost of a first-class stamp did not increase between January 1985 and January 2022, what should it have cost in 2022? In other words, what would a 20-cent stamp in January 1985 cost in January 2022, when adjusted for inflation?

Answer: Cost of stamp in January 2022 = Cost of stamp in January 1985  (CPI in January 2022/CPI in January 2085) = 0.20  (281.148/105.5) = 0.20 (2.665) = 0.533, or about 53 cents.

2. As shown in Table 18.2, when the CPI was computed for 2023, the relative weight for the food and beverages category was 14.4%, whereas for the recreation category, it was only 5.8%. Explain why food and beverage received higher weight than recreation.

Answer: The weights are based on the proportional expenditures people made in each category in the base period. People spent much more on food and beverages than they did on recreation.

3. On average, a paperback novel cost $3.49 in 1981, $6.99 in 1995, and $15.99 in 2023. Compute a “paperback novel price index” for 1995 and 2023 using 1981 as the base year. In words that can be understood by someone with no training in statistics, explain what the resulting numbers mean.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 18: Understanding The Economic News

Answer: For 1995 the index is 100  (6.99/3.49) = 200.20. For 2023 the index is 100  (15.99 /3.49) = 458.17. In 1995 a paperback book cost 200.20% of what it cost in 1981, and by 2023 a paperback book cost 458.17% of what it cost in 1981. You could also say that in 1995 it cost just over twice as much and in 2013 it cost over 4.5 times as much as in 1981.

4. If you wanted to present a time series of the yearly cost of tuition at your local college for the past 30 years, adjusted for inflation, how would you do the adjustment?

Answer: To adjust for inflation, divide the cost each year by the Consumer Price Index for that year.

5. As noted in Example 18.4, the Dow Jones Industrial Average reached a high of $15,056 on May 7, 2013, and another all-time high of $36,952.65 on January 5, 2022. The Consumer Price Index for May 2013 was 232.9; for January 2022, it was 281.2. By what percentage did the high in the DJIA increase from May 7, 2013, to January 5, 2022, after adjusting for inflation?

Answer: Adjusting for inflation, the May 7, 2013 high of 15,056 would be equivalent to 15,056  (281.2/232.9) = 18,178.39 in January 2022. Therefore, the increase from 15,056 to 36,952.65 represents an increase of (36,952.65/18,178.39) × 100% = 203%, after adjusting for inflation. The Dow Jones Industrial Average rose much more than inflation from May 2013 to January 2022.

6. The CPI in July 1977 was 60.9; in July 2023, it was 305.7. a. The salary of the governor of California in July 1977 was $49,100; in July 2023, it was $224,020. Compute what the July 1977 salary would be in July 2023 adjusted for inflation, and compare it with the actual salary in July 2023. b. The salary of the president of the United States in July 1977, was $200,000.

In July 2023, it was $400,000. Compute what the July 1977, salary would be in July 2023, adjusted for inflation, and compare it with the actual salary.

Answer: a. In July 2023, the 1977 salary adjusted for inflation would be $49,100 × (305.7/60.9) = $246,467.50, which is somewhat higher than the actual salary of $224,000. b. In July 2023, the 1977 salary adjusted for inflation would be $200,000 × (305.7/60.9) = $1,003,941, so it certainly doesn’t seem fair that it was only $400,000 in July 2023.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 18: Understanding The Economic News

7. Find out what the tuition and fees were for your school for the previous 4 years and the current year. Using the cost 5 years ago as the base, create a “tuition index” for each year since then. Write a short summary of your results that would be understood by someone who does not know what an index number is.

Answer: Divide the tuition and fees for each year by the cost five years ago, then multiply by 100.

8. As mentioned in this chapter, both the base year and the relative weights used for the Consumer Price Index are periodically updated. a. Why is it important to update the relative weights used for the CPI? b. Explain why the base year is periodically updated.

Answer: a. Relative weights are based on consumer expenditures. Consumers’ patterns change over the years, and they may spend more on certain categories and less on others than they did a few years ago. b. The Consumer Price Index would get ridiculously large if the base year(s) were never updated. Also, the categories used in the CPI need to be updated, and when the base year is updated the market basket can be updated as well.

9. The CPIs at the start of each decade from 1940 to 2020 are shown in Table 18.5. a. Determine the percentage increase in the CPI for each decade. b. During which decade was inflation the highest, as measured by the percentage change in the CPI? c. During which decade was inflation the lowest, as measured by the percentage change in the CPI?

Answer: a. To compute this for the first decade, use (24.1-14.0)/14.0 = 72% increase from 1940 to 1950. Using this same method for each decade, the increases were: 1940-50 1950-60 1960-70 1970-80 1980-90 1990-2000 2000-2010 2010-2020 72.0% 22.8% 31.1% 112.4% 58.6% 31.8% 26.7% 18.7% b. Inflation was by far highest during the 1970s. c. Inflation was lowest during the 2010s.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 18: Understanding The Economic News

10. In addition to the overall CPI, the BLS reports the index for the subcategories.

The overall CPI in August 2023 was 307.026. Following are the values for some of the subcategories that month (rounded to one decimal place), taken from the CPI website: Dairy products 268.3 Fruits, vegetables 350.4 Alcoholic beverages 285.9 Rent (primary residence) 402.2 Apparel 131.3 New vehicles 179.7 Tobacco and smoking products 1440.6 All of these values are based on using 1982–1984 as the base period, the same period used by the overall CPI. a. Find the subcategory in the list that has the least amount of inflation, compared to the base period. Write a few sentences comparing it to the overall CPI that could be understood by someone who does not know what index numbers are. b. Find the subcategory in the list that has the most amount of inflation, compared to the base period. Write a few sentences comparing it to the overall CPI that could be understood by someone who does not know what index numbers are. c. Write a brief report discussing all of the information that could be understood by someone who does not know what index numbers are.

Answer: a. Apparel had the least amount of inflation. While consumer prices in general in August 2023 were 307.026% of what they were in the 1982-84 timeframe, the cost of Apparel in August 2023 was only about 131.3% of what it was during the 1982-84 time period. b. Tobacco and smoking products had the highest amount of inflation. While consumer prices in general in August 2023 were 307.026% of what they were in the 1982-84 timeframe, the prices for tobacco and smoking products in August 2023 were about 1440.6% of what they were during the 1982-84 time period. c. While consumer prices in general in August 2023 were 307.026% of what they were in the 1982-84 time period, the increase in costs of assorted expenditures varied substantially. Certain goods and services increased at a rate slower than inflation, as follows. Apparel cost only 131.3% of what it did in 1982-84, new vehicles cost only 179.7%, dairy products rose at somewhat less than the overall inflation rate, costing 268.3% of what they did in 1982-84, and alcoholic beverages cost 285.9% of what they did in 1982-84. Other expenditures increased faster than inflation, with fruits and vegetables rising just slightly more than inflation, costing 350.4% in August

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 18: Understanding The Economic News

2023 of what they did in the 1982-84 period, rents costing 402.2% of what they did, and tobacco and smoking products costing a whopping 1440.6% of what they cost in 1982-84.

11. Table 18.6 shows the amounts Americans spent for medical care, per capita, for the years ending in 0 from 1960 to 2020 (Centers for Medicare and Medicaid Services, 2023). a. Create a “medical care index” for each of these years, using 1960 as a base. b. Comment on how the cost of medical care changed between 1960 and 2020, relative to the change in the Consumer Price Index, which was 29.6 in 1960 and 258.8 in 2020.

Answer: a. To find the “Medical Care Index” for each year, simply divide the cost for that year by the 1960 cost of $146: Year

1960

1970

1980

1990

2000

2010

2020

MCI

100

242.8

752.7

1941.8

3318.5

5740.4

8624.0

b. The cost of medical care in 2020 was 86.24 times what it was in 1960, while the Consumer Price Index in 2020 was 8.74 times what it was in 1960.

In other words, the cost of medical care between 1960 and 2020 increased almost 10 times more than the rate of inflation, as measured by the Consumer Price Index.

12. Suppose that you gave your niece a check for $50 on her 13th birthday in 2020, when the CPI was 258.8. Your nephew is now about to turn 13. You discover that the CPI is now 307.0. How much should you give your nephew if you want to give him the same amount you gave your niece, adjusted for inflation?

Answer: In 2020 the CPI was 258.8. Therefore, to be comparable, the check for your nephew would be $50  (307.0/258.8) = $59.31. (You should probably round that up to $60.00.)

13. An article in The Sacramento Bee (Stafford, 2003) on July 7, 2003, reported that the current minimum wage at that time was only $5.15 an hour and that it had not kept pace with inflation. The Consumer Price Index at the time (the end of June 2003) was 183.7. a. One of the quotes in the article was “to keep pace with inflation since 1968, the minimum wage should be $8.45 an hour today.” In 1968, the minimum

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 18: Understanding The Economic News

wage was $1.60 an hour, and the Consumer Price Index was 34.8. Explain how the author of the article determined that the minimum wage should be $8.45 an hour. b. The minimum wage was initiated in October 1938, at $0.25 an hour. The Consumer Price Index in 1938 was 14.1, using the 1982–1984 base years. If the minimum wage had kept pace with inflation from its origination, what should it have been at the end of June 2003? Compare your answer to the actual minimum wage of $5.15 an hour in June 2003. c. The minimum wage of $5.15 an hour quoted in the article was set in September 1997, when the Consumer Price Index was 161.2. If it had kept pace with inflation, what should it have been at the end of June 2003? d. Based on your answers to parts (a) to (c), has the minimum wage always kept pace with inflation, never kept pace with inflation, or some combination?

Answer: a. Minimum wage in June 2003 should be minimum wage in 1968  (CPI in June 2003/CPI in 1968) = $1.60  (183.7/34.8) = $1.60  5.279 = $8.45. b. Minimum wage in June 2003 should be minimum wage in 1938  (CPI in June 2003/CPI in 1938) = $0.25  (183.7/14.1) = $0.25  13.03 = $3.25. The minimum wage would be much lower than the actual minimum wage of $5.15 an hour in June 2003. c. Minimum wage in June 2003 should be: Minimum wage in Sept 1997  (CPI in June 2003/CPI in Sept 1997) = $5.15  (183.7/161.2) = $5.15  1.1395 = $5.87. d. It’s a combination. Based on the 1938 minimum wage, it exceeded inflation

by June 2003, but based on the 1968 minimum wage, it was below what it should have been for inflation. In June 2003 it had not increased since September 1997, even though the value of a dollar had decreased because of inflation.

14. Refer to the previous exercise. Find out the current national minimum wage for the United States and the current Consumer Price Index. (These were available as of 2023, at the websites http://www.dol.gov/dol/topic/wages/ and https://www.bls.gov/news.release/cpi.t01.htm,respectively.) Determine what the minimum wage should be at the current time if it had kept pace with inflation from: a. 1950, when the CPI was 24.1 and the minimum wage was $0.75 an hour. b. 1960, when the CPI was 29.6 and the minimum wage was $1.00 an hour. c. 1990, when the CPI was 130.7 and the minimum wage was $3.80 an hour. d. 2013, when the CPI was 183.7 and the minimum wage was $5.15 an hour.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 18: Understanding The Economic News

Answer: The answers to all four parts depend on the current CPI, but the formulas to find the answer are shown in each part, where CPI = current CPI. Compare the answer to the current minimum wage. a. Minimum wage in 1950  (CPI/CPI in 1950) = $.75  (CPI/24.1) b. Minimum wage in 1960  (CPI/CPI in 1960) = $1.00  (CPI/29.6) c. Minimum wage in 1990  (CPI/CPI in 1990) = $3.80  (CPI/130.7) d. Minimum wage in 2013  (CPI/CPI in 2013) = $5.15  (CPI/183.7) 15. In 1950, being a millionaire was touted as a goal that would be achievable by very few people. The CPI in 1950 was 24.1, and in 2002, it was 179.9. How much money would one need to have in 2002 to be the equivalent of a millionaire in 1950, adjusted for inflation? Would it still have seemed like a goal achievable by very few people in 2002?

Answer: 179.9/24.1 = 7.4647. So one would need 7.464730 × $1,000,000 = $7,464,730 in 2002 to be the equivalent of a millionaire in 1950. That does not seem like a goal achievable by very many people.

16. A report released by the U.S. Census Bureau (2023) on September 12, 2023, included this quote: “Real median household income fell by 2.3% from $76,330 in 2021 to $74,580 in 2022. Income estimates are expressed in real or 2022 dollars to reflect changes in the cost of living. Between 2021 and 2022, inflation rose 7.8%; this is the largest annual increase in the cost-of-living adjustment since 1981.”

Another Census Bureau report added the information that “In nominal terms, meaning unadjusted for inflation, the median income for all U.S. households in 2021 was $70,780” (Unrath, 2023). a. Explain the conceptual difference between the 2021 “real median income” of $76,330 from the first quote and the 2021 income “in nominal terms” of $70,780 from the second quote. b. The Census Bureau sometimes uses the terms “constant dollars” and “current dollars” instead of the terms used in these quotes. Which term (constant or current dollars) do you think is equivalent to the term “real dollars,” and which one do you think is equivalent to “nominal” dollars?

Explain your reasoning. c. Using the terminology introduced in part (b) and the values in the initial quotes, fill in the dollar values for the following sentence. Median household income in the United States in 2021 was ________ in current dollars and ________ in constant dollars. d. Give one advantage to reporting median household income across years in current dollars and one advantage to reporting it in constant dollars.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 18: Understanding The Economic News

Answer: a. The “nominal” figure of $70,780 is not adjusted for inflation, so it is the actual median income in 2021. The “real median income” of $76,330 has been adjusted for inflation, so it is what the dollar amount of $70,780 would have been worth in 2022 after adjusting for the inflation of 7.8%.

(The figures aren’t exact due to rounding the inflation rate; $70,780 × 1.078 = $76,301.) b. Current dollars are adjusted for inflation to represent the value in the current year, and are equivalent to “real dollars.” Constant dollar amounts remain constant, no matter what year you are discussing them, and are equivalent to “nominal” dollars. c. Median household income in the United States in 2021 was $74,580 in current dollars and $70,780 in constant dollars. d. Reporting income in current dollars allows us to know how much money people were actually making. Reporting income in constant dollars allows us to compare incomes over the years and readily see whether they have changed, when adjusted for inflation.

17. In the 1990s, there was considerable debate about whether the standard Consumer Price Index (CPI-U) overestimated inflation, leading a congressional commission to study the issue. In explaining why it is a costly mistake to have the CPI overestimate inflation, the Associated Press (20 October 1994) reported, “Every 1 percentage point increase in the CPI raises the federal budget deficit by an estimated $6.5 billion.” Explain why that would happen.

Answer: Many government payments, such as Social Security and the Supplemental Nutrition Assistance Program, have raises tied to the CPI. For instance, if the CPI goes up 3% in a year, then Social Security recipients get a 3% raise. If it goes up 4% instead, it obviously costs the government much more money (about $6.5 billion more according to this report).

18. Explain which of the criticisms of the CPI given in Section 18.3 is supposed to be addressed by the use of the Chained CPI. Answer: Criticism 2: If the price of one item rises, consumers are likely to substitute another.

19. Many U.S. government payments, such as Social Security benefits, are increased each year by the percentage change in the CPI. In 1995, the government started discussions about lowering these increases or changing the way the CPI is calculated. As discussed in this chapter, one result was the Chained CPI. According to an article in The New York Times, “most economists

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 18: Understanding The Economic News

who have studied the issue closely say the current system is too generous to Federal beneficiaries… the pain of lower COLAs [cost of living adjustments] would be unavoidable but nonetheless appropriate” (Gilpin, 1995, p. D19).

Explain in what sense some economists believe the current system (in place in 1995 and continuing in 2023) is too generous.

Answer: For the reasons given in Section 18.4 of the book, the CPI overestimates increases in the cost of living required to maintain a constant level of lifestyle.

Therefore, raises tied to the CPI may actually be higher than needed to maintain a constant level of lifestyle.

20. Remember that the CPI is supposed to measure the change in what it costs to maintain the same standard of living that was in effect during the base year(s).

Using the material in Section 18.4, explain why it may not do so accurately.

Answer: Your answer should incorporate items one to four in the box on page 389 in the text.

21. Most news accounts of the Consumer Price Index report the percentage change in the CPI from the previous month rather than the value of the CPI itself. Why do you think that is the case?

Answer: What matters to consumers, economists, and policy makers is how prices are changing, not how current prices compare with the base year prices. If the CPI changes by more than a few tenths of a percent from one month to the next, that signifies high inflation.

22. Recall from Chapter 9 that time series (such as the CPI) have three nonrandom components. Which of the three nonrandom components (trend, seasonal, or cycles) is likely to contribute the most to the unadjusted Consumer Price Index? Explain.

Answer: The seasonal component is likely contribute the most to the unadjusted CPI because prices of major items like fresh foods and energy tend to be seasonal.

That’s why the CPI is seasonally adjusted. In periods of very high inflation, the trend might contribute more, but the seasonal component is always relatively important.

23. The Bureau of Labor Statistics reports that one use of the Consumer Price Index is to periodically adjust the federal income tax structure, which sets higher tax rates for higher income brackets. According to the BLS, “these adjustments prevent inflation- induced increases in tax rates, an effect called ‘bracket creep’” (U.S. Bureau of Labor Statistics, 2023a).

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 18: Understanding The Economic News

Explain what is meant by “bracket creep” and how you think the CPI is used to prevent it.

Answer: Income tax rates are higher as income increases, so for instance, those with low income might pay 20% of each additional dollar earned, while those with higher income might pay 30% of each additional dollar earned. The range of incomes for which each percentage is charged is called a “tax bracket.” If someone’s salary increases because of inflation, but the real value of the salary stays constant, they might “creep” into a higher tax bracket even though they didn’t have higher real income. To prevent that, the ranges constituting each tax bracket are raised to account for inflation.

24. Two of the economic indicators used by the U.S. government are “Labor costs per unit of output in manufacturing” and “Industrial production.” One of these is designated as a “lagging economic indicator,” and the other is a “coincident economic indicator.” Explain which you think is which and why.

Answer: “Labor costs per unit of output in manufacturing” is lagging, because the fewer orders there are, the more it costs to make each one. This lags behind the general economy, as a weakening economy results in fewer orders. “Industrial production” is coincident, because it should be a reflection of the current economic conditions.

25. One of the components of the Index of Leading Economic Indicators is the “Average consumer expectations for business conditions.” Why do you think this index would be a leading economic indicator?

Answer: If consumer expectations for the economy fall, consumers are likely to buy less and the economy is likely to falter. If consumer expectations are high, the reverse should happen. Therefore, changes in consumer expectations portend changes in spending habits for the near future.

26. In February 1994, the Index of Leading Economic Indicators dropped slightly, and economists blamed it on unusually severe winter weather that month.

Examine the 10 series that make up the Index of Leading Economic Indicators, listed in Table 18.4. Choose at least two of the series to support the explanation that the drop in these indicators in February was partially due to unusually severe winter weather.

Answer: The most likely ones are #2, “Average weekly initial claims for unemployment insurance,” which may rise because certain work cannot be done in severe winter weather, and # 6, “Building permits, new private housing units,” which is likely to fall because it is hard to start a new building in severe winter weather.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 18: Understanding The Economic News

ANSWER TO THE REFLECTION QUESTIONS 1. Suppose you have been in the same job for the past 10 years, and you want to know how your salary increases compare with inflation for each of those 10 years. You have a list of your annual salaries for the 10 years. a. Explain how you would create a price (salary) index number for each of the 10 years. b. You can find the yearly values of the Consumer Price Index (CPI) online.

Explain how you would use those and your salary index numbers to determine whether your salary kept pace, lagged behind, or exceeded inflation for each of the 10 years.

Answer: a. To create a salary index number, you would divide your salary for each year by your salary in the base year, which would be the first year for which you had a salary figure. If you received other benefits that should be included as part of your compensation package, such as employer contributions to a retirement account or an annual holiday bonus, you would need to decide whether to include those. b. You could use the CPI values to create an inflation-adjusted salary for each year, using the formula on page 386. You would divide the CPI for a particular year by the CPI for your base year, and then multiply your salary in the particular year by that amount.

2. Each of the hundreds of items in the CPI “market basket of goods” receives a different weight in the calculation of the CPI. Explain why they simply aren’t all weighted equally.

Answer: The items in the market basket of goods are weighted according the proportion of consumers’ spending on each item. For example, it wouldn’t make sense to give the same weight to an item people buy every week as to an item they only buy every few years.

3. One of the uses of the CPI described in this chapter is that it is used to determine salary adjustments. In other words, the change in the CPI from one year to the next is used to determine cost of living raises for many retirees and workers. Explain one reason why this use of the CPI is controversial. (There are multiple reasons.)

Answer: As noted in the question, there are multiple reasons. They are listed in the box on page 389. One example is that the CPI does not account for all of the substitutions consumers are likely to make when an item becomes more expensive and can be substituted by a different, cheaper item.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 18: Understanding The Economic News

4. One of the controversies about the CPI as a measure of cost of living is that consumers make substitutions. Substitutions within categories are taken into account, but substitutions across categories are not. Give an example of a within-category substitution you might make when shopping and an example of an across-category substitution you might make.

Answer: An example of a within-category substitution is buying ground beef instead of steak. An example of an across-category substitution is buying frozen vegetables instead of fresh vegetables.

5. The CPI can be used to compare prices across years. But one criticism of the CPI is that technological advances in an item that enhance quality of life are not measured in this comparison. One example is that the inflation-adjusted price of a television has fallen over the years, while the quality and capabilities have significantly increased. Do you think this is a legitimate criticism of the CPI? Discuss your reasoning.

Answer: There is no right or wrong answer, it’s the reasoning that counts. If you think that quality of life is closely tied to technological updates, then you would say that it is a legitimate criticism. For example, if you think that having ondemand streaming services and channels that advocate for certain viewpoints are better than having a fixed set of channels watched by everyone at the same time, then you would think that the change in technology for televisions has enhanced the quality of life.

6. One example of an economic indicator that should be seasonally adjusted before being reported is toy sales, because they are expected to rise during holiday seasons even if the economy is holding steady. Give another example of an indicator that should be seasonally adjusted, and explain why.

Answer: Answers will vary, but examples include anything that fluctuates predictably with the seasons. One example would be the sales of outdoor furniture, which would be higher in spring and early summer than in the fall and winter.

7. It is clear that leading economic indicators are useful because they tend to predict whether the economy is likely to get better or worse over the next several months. Why do you think economists are interested in studying lagging economic indicators?

Answer: Lagging economic indicators are useful for various reasons. For example, it can be useful to see how long after a recession it takes for the lagging indicators to come back up to pre-recession levels. They can also be used to study economic trends that might be able to be predicted more readily in the future.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 18: Understanding The Economic News

ANSWER TO THE MINI PROJECTS 1. Numerous economic indicators are compiled and reported by the U.S. government and by private companies. Various sources are available on the Internet to explain these indicators. Write a report on one of the following.

Explain how it is calculated, how often it is reported, whether it is seasonally adjusted, what its uses are, and what limitations it might have. a. The Producer Price Index b. The Gross Domestic Product c. The Dow Jones Industrial Average Answer: This mini-project is designed to give students experience with doing internet research on economic topics. Students should cite their sources, and make sure that they are reliable and not biased. It can help students gain confidence that they can understand economic news.

2. Find a news story that reports on current values for one of the indexes discussed in this chapter. Discuss the news report in the context of what you have learned in this chapter. For example, does the report contain any information that might be misleading to an uneducated reader? Does it omit any information that you would find useful? Does it provide an accurate picture of the current situation?

Answer: Because release of the Consumer Price Index is usually accompanied by news articles, students will most likely choose an article based on it. As of 1995, the CPI is released monthly, about the middle of the month, so you should time the assignment of this project appropriately or inform students so they can find the news reports. The most common problem with reports of the CPI is omitting the CPI itself and giving only the change since the previous month.

That’s fine for determining whether or not inflation is a problem, but it is not helpful if you want to use the CPI for comparison of costs across years.

3. Refer to Example 18.4, discussing why the Dow Jones Industrial Average keeps reaching new highs. In addition to the Dow Jones, there are other indicators of fluctuation in stock prices. Two examples are the New York Stock Exchange Composite Index and the Standard and Poor’s 500. Choose a stock index (other than the Dow Jones), and write a report about it. Include whether it is adjusted for inflation, seasonally adjusted, or both. Give information about its recent performance, and compare it with performance a few decades ago.

Make a conclusion about whether the stock market has gone up or down in that time period, based on the index you are using, adjusted for inflation.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 18: Understanding The Economic News

Answer: Most stock price indexes are not adjusted for inflation or seasonally adjusted.

Adjustments for inflation can be done using the CPI, but seasonal adjustments can’t be done for this project. No matter what index is used, stock prices traditionally have risen faster than inflation. This project will help students understand the stock market and how the related indexes work.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 19: The Diversity Of Samples From The Same Population

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 19: THE DIVERSITY OF SAMPLES FROM THE SAME POPULATION

TABLE OF CONTENTS Answers to the Exercise Questions ....................................................................................1 Answers to the Reflection Questions ...............................................................................12 Answers to the Mini Projects ........................................................................................... 14

ANSWERS TO THE EXERCISE QUESTIONS 1. Suppose you want to estimate the proportion of students at your college who are left-handed. You decide to collect a random sample of 200 students and ask them which hand is dominant. Go through the conditions for which the rule for sample proportions applies (page 407) and explain why the rule would apply to this situation.

Answer: For condition 1, there exists a population (students at the college) with a fixed proportion with a certain trait (left-handed). For condition 2, a random sample of students is selected. For condition 3, the sample size of 200 is large enough that we are likely to see at least ten right-handed and ten left-handed students.

2. Suppose you want to estimate the proportion of students at your college who are left-handed. Now suppose that the truth is that 0.12, or 12%, of the students are left-handed, and you take a random sample of 200 students. Use the Rule for Sample Proportions to draw a picture similar to Figure 19.3, showing the possible sample proportions for this situation.

Answer: The picture is bell-shaped with a mean of 0.12 and standard deviation of (0.12)(0.88) = 0.023. See the figure below.

200

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 19: The Diversity Of Samples From The Same Population

68%

95%

0.051

0.074

0.097 0.12 0.143 Possible sample proportions

0.166

0.189

3. A Gallup Poll found that of 800 randomly selected drivers surveyed, 70% thought they were better-than-average drivers. In truth, in the population, no more than 50% of all drivers can be “better than average,” assuming “average” is equivalent to the median level of skill. a. Draw a picture of the possible sample proportions that would result from samples of 800 people from a population with a true proportion of 0.50. b. Would we be unlikely to see a sample proportion of 0.70, based on a sample of 800 people, from a population with a proportion of 0.50? Explain, using your picture from part (a). c. Which one of the psychological issues discussed in Chapter 16 would help explain the results of this survey?

Answer: a. The picture is bell-shaped with a mean of 0.50 and standard deviation of (0.5)(0.5) = 0.0177. See the figure below.

800

68%

95%

0.447

0.4646

0.4823 0.50 0.5177 Possible sample proportions

0.5354

0.553

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 19: The Diversity Of Samples From The Same Population

b. Yes, it would be very unlikely. The picture does not extend as far as 0.70. In fact, it is virtually impossible. It represents a standardized score of over 10. c. Optimism or overconfidence. People think they are likely to be better than average on most things.

4. Suppose you are interested in estimating the average number of miles per gallon of gasoline your car can get. You calculate the miles per gallon for each of the next nine times you fill the tank. Suppose, in truth, the values for your car are bell-shaped, with a mean of 25 miles per gallon and a standard deviation of 1. Draw a picture of the possible sample means you are likely to get based on your sample of nine observations. Include the intervals into which 68%, 95%, and almost all of the potential sample means will fall.

Answer: Possible sample means are bell-shaped with a mean of 25 and standard deviation of 1/ 9 = 1/3 or about 0.33. So 68% are between 24.67 and 25.33, 95% are between 24.34 and 25.66 and almost all are between 24.01 and 25.99.

See the figure below.

68%

95% Almost all 24.01

24.34

24.67 25 25.33 Possible sample means

25.66

25.99

5. Refer to Exercise 4. Redraw the picture under the assumption that you will collect 100 measurements instead of only nine. Discuss how the picture differs from the one in Exercise 4.

Answer: Now the picture is even more compact. The mean is still 25, but the standard deviation is only 1/10 = 0.1. See the figure below.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 19: The Diversity Of Samples From The Same Population

68%

95% Almost all 24.7

24.8

24.9 25 25.1 Possible sample means

25.2

25.3

6. Give an example of a situation of interest to you for which the Rule for Sample Proportions would apply. Explain how the conditions allowing the rule to be applied are satisfied for your example.

Answer: This will differ for each student. The explanation of how the conditions are satisfied must include: (1) That there is a fixed proportion in the population with a trait, or a repeatable situation with fixed probability of success; (2) That a random sample is drawn or the situation is repeated independently numerous times; and (3) That the size of the sample is such that there should be at least ten with and ten without the desired trait in the sample.

7. Suppose that 35% of the students at a university favor the semester system, 60% favor the quarter system, and 5% have no preference. Is a random sample of 100 students large enough to provide convincing evidence that the quarter system is favored? Explain.

Answer: Yes. The Rule for Sample Proportions tells us that the sample proportions for this situation would be bell-shaped with a mean of 0.60 and standard (.6)(.4)

deviation of √ 100 = 0.049. Therefore, 95% of the time, the sample proportion would be within 2 standard deviations (0.098) of 0.60, or between 0.502 and 0.698. So there is a very small probability (less than 2.5%) that the sample proportion would fall below 0.50. As long as the sample proportion fell above 0.50 (i.e., 50%), there would be convincing evidence that the quarter system is favored. Another way to see this is to compute the standardized score for a © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 19: The Diversity Of Samples From The Same Population

proportion of 0.50. It would be (0.5 − 0.6)/0.049 = −2.04. That lies at the 2nd percentile, meaning that a value below 0.50 would occur only with probability of 0.02 or 2%.

8. In a 2005 Gallup poll based on a random sample of 1002 adults, 20% said that they believed in reincarnation (Moore, 2005). A 2021 Pew Research Center poll based on an Internet panel survey with 6485 respondents found that 33% of respondents said they believed in reincarnation (Pew Research Center, 2021). a. For the 2005 poll based on a random sample of 1002 adults, suppose that the true population proportion who believed in reincarnation at that time was close to the sample proportion in the poll, or about 0.20. Use the Rule for Sample Proportions to describe the shape, mean, and standard deviation of the distribution of possible sample proportions that would result from samples of size 1002 from this population. b. Assuming a proper random sample was used, verify that the sample proportion for the poll taken in 2005 almost certainly represents the population proportion to within about 0.038. In other words, the true population percentage is almost certainly within 3.8% of the 20% reported in the sample. c. Because the 2021 poll used a complex sampling method, the Rule for Sample Proportions does not apply, but the reported margin of error was 0.019, or 1.9%. Based on these results, would you conclude that the proportion of all adults who believe in reincarnation was higher in 2021 than it was in 2005? Explain.

Answer: a. The possible sample proportions for the 2005 poll would be approximately bell-shaped with a mean of 0.2 and a standard deviation equal to the square root of (0.2)(0.8)/1002, or about 0.0126. b. The sample proportion and the population proportion are almost certainly no more than three standard deviations apart, based on the empirical rule.

Therefore, they should be within 3 × (0.0126) = 0.038 of each other, and thus, the population proportion should be within 3.8% of the reported sample value of 20%. c. From Part (b), the population value in 2005 was almost surely within 3.8% of the sample value of 20%, or in the range 16.2% to 23.8%. The value of 33% found in 2021 far exceeds this range, even after subtracting the margin of error of 1.9%, to get a lower bound of about 31.1%. Therefore, it is reasonable to conclude that there was a real increase in the percentage of all adults who believe in reincarnation between 2005 and 2021.

9. Suppose 20% of all subscribers to a streaming service watch a particular program.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 19: The Diversity Of Samples From The Same Population

a. For a random sample of 2500 subscribers measured by a rating agency, describe the frequency curve for the possible sample proportions who watch the program. b. The program will be canceled if the ratings show less than 17% watching in a random sample of subscribers. Given that 2500 subscribers are used for the ratings, is the program in danger of getting canceled? Explain. c. Draw a picture of the possible sample proportions, similar to Figure 19.3.

Illustrate where the sample proportion of 0.17 falls on the picture. Use this to confirm your answer in part (b).

Answer: a. The frequency curve would be bell-shaped with a mean of 0.2 and a standard deviation of (0.2)(0.8) = 0.008.

2500

b. No, it is not in danger. We can be almost certain that the sample proportion will be within 3 standard deviations of 0.2, or between 0.176 and 0.224, corresponding to 17.6% and 22.4%. So, it is not impossible that only 17% of subscribers would be watching, but quite unlikely. The standardized score is (0.17−0.2)/0.008 = −3.75. c. The figure is shown below.

68%

0.17 is way out here

0.176

95%

0.184

0.192 0.20 0.208 Possible sample proportions

0.216

0.224

10. Use the Rule for Sample Means to explain why it is desirable to take as large a sample as possible when trying to estimate a population value.

Answer: The Rule for Sample Means says that the standard deviation of the frequency curve for the samples will be the population standard deviation divided by the square root of the sample size. So, the larger the sample size, the smaller the standard deviation of this frequency curve; therefore, the closer the estimate will be to the population value. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 19: The Diversity Of Samples From The Same Population

11. A survey of 1012 drivers in the United States conducted by the National Sleep Foundation found that 62% said they have driven “when they are so tired they had a hard time keeping their eyes open” (National Sleep Foundation, 2022). a. Conditions 2 and 3 needed to apply the Rule for Sample Proportions are met because this result is based on a large random sample of drivers.

Explain how condition 1 is also met. b. Suppose that in fact 64% of the population of drivers have driven when they are so tired they had a hard time keeping their eyes open. Find the interval in which about 95% of all sample proportions should fall, based on samples of size 1012. Does the result of this survey fall into that interval? c. Presumably, younger drivers would have had fewer years of driving and thus less opportunity to drive while they were very tired. Therefore, the proportion who report doing so might be lower than for the entire adult population. Suppose a survey based on a random sample of 1012 college students was conducted and 58% report having driven when they are so tired they have a hard time keeping their eyes open. Would it be reasonable to conclude that the population proportion of college students who have done so differs from the proportion of all adults who have done so?

Explain.

Answer: a. There exists an actual population of adults, a fixed proportion of whom have driven when they are so tired they had a hard time keeping their eyes open (and technically, are willing to admit it when asked!). b. The standard deviation is the square root of (0.64)(0.36)/1012, or about 0.015. The interval in which about 95% of sample proportions should fall is thus about 0.64  0.03, or 0.61 to 0.67, equivalent to 61% to 67%. So, the observed proportion of 62% is covered by the interval. c. It would be reasonable to conclude that they differ because 58% is not in the interval computed in part (b).

12. In Example 19.12, a study commissioned by the European Aviation Safety Agency (EASA) was discussed (Berg et al., 2022). The purpose of the study was to determine the weights of airline passengers and their luggage, which is useful for establishing air transportation safety. A similar study in 2009 (Berdowski et al., 2009) found that the mean weight of all passengers (with clothing) was 164.2 pounds, with a standard deviation of 43.9 pounds. That study had a sample size of almost 23,000, so the sample mean and standard deviation are excellent estimates of the corresponding population values. Let’s assume they were 164 pounds and 44 pounds, respectively. a. The sample size for the 2022 survey was 4164, much smaller than in 2009.

Suppose that the population of weights in 2022 is similar to 2009, and has a mean of 164 pounds and a standard deviation of 44 pounds. Use the Rule

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 19: The Diversity Of Samples From The Same Population

for Sample Means to describe the shape, mean, and standard deviation for the possible sample means for samples of size 4164 from this population. b. Draw a picture similar to Figure 19.5 illustrating the possible sample means and the intervals into which 68%, 95%, and almost all of them should fall. c. The actual mean weight for the sample in 2022 was 166.7 pounds. Using your results from parts (a) and (b), explain whether that would be a reasonable value for the sample mean if the population mean and standard deviation in 2022 were the same as in 2009, i.e., 164 pounds and 44 pounds.

Answer: a. The distribution of possible sample means is bell-shaped with mean of 164 pounds and standard deviation of 44/√4164 = 0.68 pounds. b. The figure is shown below.

68%

95% Almost all 161.96

162.64

163.32

164

164.68

165.36

166.04

Possible sample means (pounds)

c. It is not likely that the mean and standard deviation were 164 pounds and 144 pounds in 2022. Using those values and the information from part (a), the standardized score for a sample mean of 166.7 would be (166.7 – 164)/0.68 = 3.97. A standardized score that large or larger is very unlikely.

This can be seen in the figure in part (b) as well. The value of 166.7 is beyond the upper range shown in the figure.

13. Suppose the population of IQ scores in the town or city where you live is bellshaped, with a mean of 105 and a standard deviation of 15. Describe the frequency curve for possible sample means that would result from random samples of 100 IQ scores.

Answer: Using the Rule for Sample Means, we would expect the possible sample means to be bell-shaped, with a mean of 105 and a standard deviation of

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 19: The Diversity Of Samples From The Same Population

15/ 100 = 15/10 = 1.5. Therefore, we would find the sample mean between 103.5 and 106.5 about 68% of the time, between 102 and 108 about 95% of the time, between 100.5 and 109.5 about 99.7% of the time.

14. Explain whether each of the following situations meets the conditions for which the Rule for Sample Pro-portions applies. If not, explain which condition is violated. If so, describe the population parameter of interest. a. You are interested in knowing what proportion of days in typical years have rain or snow in the area where you live. For the months of January and February, you record whether there is rain or snow each day, and then you calculate the proportion of those days that had rain or snow. b. A large company wants to deter-mine what proportion of its employees are interested in on-site day care. The company asks a random sample of 100 employees and calculates the sample proportion who are interested.

Answer: a. This sample would violate conditions 1 and 2. The relative frequency of rain or snow changes from the winter months to the summer months, so it is not fixed. And the sample would not be random. b. This sample meets all the conditions. The population parameter of interest is the proportion of the population of the company’s employees who are interested in on-site day care.

15. Explain whether each of the following situations meets the conditions for which the Rule for Sample Pro-portions applies. If not, explain which condition is violated. If so, describe the population parameter of interest. a. Unknown to the government, 10% of all cars in a certain city do not meet appropriate emissions standards. The government wants to estimate that percentage, so they take a random sample of 30 cars and compute the sample proportion that do not meet the standards. b. The Census Bureau would like to estimate what proportion of households have someone at home between 7 P.M. and 7:30 P.M. on weeknights, to determine whether that would be an efficient time to collect census data.

The Bureau surveys a random sample of 2000 households and visits them during that time to see whether someone is at home.

Answer: a. This random sample does not meet the criteria of condition 3, because the sample size is not large enough. Even if the sample did show exactly 10% of the cars did not meet emissions standards, this would be only 3, not the required minimum of 10. b. This meets all the conditions. The population parameter is the proportion of households that have someone home between 7 and 7:30 p.m. on weeknights.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 19: The Diversity Of Samples From The Same Population

16. Explain whether you think the Rule for Sample Means applies to each of the following situations. If it does apply, specify the population of interest and the population parameter of interest. If it does not apply, explain why not. a. A researcher is interested in what the average cholesterol level would be if people restricted their fat intake to 30% of calories. The researcher gets a group of patients who have had heart attacks to volunteer to participate, puts them on a restricted diet for a few months, and then measures their cholesterol. b. A large corporation would like to know the average income of the spouses of its workers. Rather than go to the trouble to collect a random sample, they post someone at the exit of the building at 5 P.M. Everyone who leaves between 5 P.M. and 5:30 P.M. is asked to complete a short questionnaire on the issue; there are 70 responses.

Answer: a. If the goal is to reach a conclusion about people in general, then the rule does not apply because the sample is not random or representative of that population. It is biased by using only heart attack patients. b. This is a biased sample, so the rule does not apply. People who leave during the 5:00 to 5:30 time period are likely to have different spousal incomes than people who work longer hours and leave later.

17. Explain whether you think the Rule for Sample Means applies to each of the following situations. If it does apply, specify the population of interest and the population parameter of interest. If it does not apply, explain why not. a. A university wants to know the average income of its alumni. Staff members select a random sample of 200 alumni and mail them a questionnaire. They follow up with a phone call to those who do not respond within 30 days. b. An automobile manufacturer wants to know the average price for which used cars of a particular model and year are selling in a certain state. They are able to obtain a list of buyers from the state motor vehicle division, from which they select a random sample of 20 buyers. They make every effort to find out what those people paid for the cars and are successful in doing so.

Answer: a. Incomes are not bell-shaped but this is a large sample so the rule applies (as long as the university is careful to contact the particular people chosen to be in the sample). The population under consideration is alumni and the population parameter of interest is the average income for that population. b. The rule would probably not apply here because sales prices are not likely to be bell-shaped and the sample size of 20 is too small for the rule to apply unless the measurements are bell-shaped.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 19: The Diversity Of Samples From The Same Population

18. Suppose the population of grade-point averages (GPAs) for students at the end of their first year at a large university has a mean of 3.1 and a standard deviation of 0.5. Draw a picture of the frequency curve for the mean GPA of a random sample of 100 students, similar to Figure 19.5.

Answer: It is bell-shaped with a mean of 3.1 and standard deviation of 0.5/√100 = 0.05.

See the figure below.

68%

95% Almost all 2.95

3.0

3.05 3.1 3.15 Possible sample means

3.20

3.25

19. Using data from election officials, the University of Florida Election Lab reported that 46.22% of the voting-eligible population in the United States (citizens old enough to vote and not disqualified) actually voted in the 2022 congressional election (McDonald, 2023). As part of the Current Population Survey, the U.S. Census Bureau asked eligible voters whether they had voted in 2022, and 52.2% of respondents said yes (Fabina, 2023). The sample size for the survey was not provided. a. Suppose the true proportion of the voting-eligible population who actually voted was 0.46 and a random sample of 1600 of them was asked whether or not they voted. Describe the shape, mean, and standard deviation of the possible sample proportions who would say yes in this situation. b. Into what interval of values should the sample proportion who would say yes fall 68%, 95%, and almost all of the time? c. Suppose the Census Bureau result was based on a random sample of 1600 voters. Compute a standardized score to correspond to the sample result of 52.2%, i.e., a sample proportion of 0.522, found in the survey. d. Based on the results from parts (a) to (c), do you think all of the respondents in the Census Bureau survey were truthful about whether they had voted? Explain.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 19: The Diversity Of Samples From The Same Population

Answer: a. The distribution of possible sample proportions who would say yes is bellshaped with a mean of 0.46 and a standard deviation of √(0.46)(0.54)/1600 = 0.0125. b. Using the values from part (a), 68% of the time the sample proportion should be between 0.4475 and 0.4725, 95% of the time it should be between 0.435 and 0.485, and almost all the time it should be between 0.4225 and 0.4975. c. Using the values given in part (a), the mean and standard deviation would be 0.46 and 0.0125, respectively, so the standardized score for 0.522 would be (0.522 – 0.46)/0.125 = 4.96. d. If the true proportion of the population was 0.46 as given in part (a), then it is almost impossible that a sample of size 1600 would result in a sample proportion of 0.522, as shown by the standardized score in part (c).

Therefore, some of the respondents in the Census Bureau survey must have lied about whether they had actually voted.

20. The administration of a large university wants to use a random sample of students to measure student opinion of a new food service on campus.

Administrators plan to use a continuous scale from 1 to 100, where 1 is complete dissatisfaction and 100 is complete satisfaction. They know from past experience with such questions that the standard deviation for the responses is going to be about 5, but they do not know what to expect for the mean. They want to be almost sure that the sample mean is within plus or minus 1 point of the true population mean value. How large will their random sample have to be?

Answer: They want three standard deviations to equal 1, where a single standard deviation is 5/√𝑛 (n is the size of the sample). So, they need 15/√𝑛 = 1, or n = 152 = 225 responses.

ANSWERS TO THE REFLECTION QUESTIONS 1. Suppose a medical study is conducted to test whether a new blood pressure medication is effective in reducing systolic blood pressure. For the 100 patients in the study, the average reduction is 8 mm Hg. a. For this example, describe what is meant by the population parameter and the sample statistic. b. If it is possible to provide a numerical value for the population parameter and for the sample statistic, do so. If it is not possible, explain why not.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 19: The Diversity Of Samples From The Same Population

Answer: a. The population parameter is what the mean reduction in systolic blood pressure would be if everyone in the population were to take the medication the same way those in the sample took it. In this case, the relevant population is whatever is represented by the sample of 100 patients in the study. The sample statistic is the sample mean reduction in systolic blood pressure for the 100 patients in the study. b. It is not possible to give a numerical value for the population parameter unless everyone in the population were to take the medication and have their change in blood pressure measured. The sample statistic is 8 mm/Hg. 2. Suppose a polling organization takes a random sample of 1000 voters each month for a year leading up to a presidential election and finds the sample proportion who say they plan to vote for Candidate X. Would the Rule for Sample Proportions apply to the collection of possible sample proportions for all of these polls together? Explain why or why not.

Answer: The Rule for Sample Proportions would not apply because there is not one fixed value for the population proportion. It would change over time, as voters change their preferences leading up to the election.

3. Suppose a random sample of size 50 is going to be taken from the population of grade point averages (GPAs) for all students at your school. a. Explain which would be more spread out—a histogram of the 50 individual GPA values or the frequency curve for the possible sample means. b. If the sample size is 100 instead of 50, would the histogram of the individual values be likely to be more spread out, less spread out, or about the same as the histogram of individual values for a sample of size 50?

Explain. c. If the sample size is 100 instead of 50, would the frequency curve for the possible sample means be more spread out, less spread out, or about the same as for the sample size of 50? Explain.

Answer: a. The individual GPA values would be more spread out. The standard deviation for the 50 individual values would be similar to the population standard deviation for the population of individual GPA values. But the standard deviation for the possible sample means would be the standard deviation for the individual values divided by √50. b. If the sample size is 100 instead of 50, the histogram of individual values would be similar. The standard deviation would still be similar to whatever the standard deviation is for the population of GPA values.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 19: The Diversity Of Samples From The Same Population

c. If the sample size is 100 instead of 50, the frequency curve for the possible sample means would be less spread out. It’s standard deviation would be that of the individual values divided by √100 instead of by √50.

4. Give an example of a situation in which a random sample of size 15 is to be taken, but the Rule for Sample Means would not apply. Specify what would be measured for each individual in the sample, and explain why the Rule would not apply.

Answer: The Rule for Sample Means does not apply for random samples of size 15 unless the original population of measurements is bell-shaped. Therefore, the example should be something for which the population is not bell-shaped. For instance, it could be yearly household incomes for all households in a large city. The population would be skewed to the right, because of some very high incomes.

ANSWERS TO THE MINI PROJECTS 1. The goal of this mini-project is to help you verify the Rule for Sample Proportions firsthand, using a physical simulation. You will use the population represented in Figure 19.1 to do so. It contains 400 individuals, of whom 160 (40%) are —that is, carry the gene for a disease—and the remaining 240 (60%) are ☺—that is, do not carry the gene. You are going to draw 20 samples of size 15 from this population. Here are the steps you should follow: Step 1: Develop a method for drawing simple random samples from this population. One way to do this is to cut up the symbols and put them all into a paper bag, shake well, and draw from the bag. There are less tedious methods, but make sure you actually get random samples. Explain your method.

Step 2: Draw a random sample of size 15 and record the number and percent age who carry the gene.

Step 3: Repeat step 2 a total of 20 times, thus accumulating 20 samples, each of size 15. Make sure to start over each time; for example, if you used the method of drawing symbols from a paper bag, then put the symbols back into the bag after each sample of size 15 is drawn so they are available for the next sample as well.

Step 4: Create a stemplot or histogram of your 20 sample proportions.

Compute the mean.

Step 5: Explain what the Rule for Sample Proportions tells you to expect for this situation.

Step 6: Compare your results with what the Rule for Sample Proportions tells you to expect. Be sure to mention mean, standard deviation, shape, and the in tervals into which you expect 68%, 95%, and almost all of the sample proportions to fall.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 19: The Diversity Of Samples From The Same Population

Answer: Make sure the method is one for which you put all of the individuals back in place before the start of each sample, and for which the proportions are correct. As an example of how to do this, one student used two different colors of cotton swabs, counting out 240 blue and 160 yellow, and placing them in a container. The student then picked 15, counted the proportion of yellow, then put them all back and started over. Although the sample of 15 is taken without replacement, with a population of 400 and a sample of 15 the Rule for Sample Proportions should still apply, at least approximately. It tells us that the sample proportions based on a true proportion of 0.4 and a sample size of 15 are approximately bell-shaped with a mean of 0.40 and standard deviation of 0.126. So, we should see about 68% of our sample proportions (13 or 14 of them) between 0.274 and 0.526, about 95% (19 of them) between .148 and .652 and all 20 between 0.022 and 0.778. You should compare the percentages you found in each of the three intervals to what the rule tells us to expect.

2. The purpose of this mini-project is to help you verify the Rule for Sample Means, using a physical simulation. Suppose you are interested in measuring the average amount of blood contained in the bodies of adults, in ounces.

Suppose, in truth, the population consists of the following listed values. (Each value would be repeated mil lions of times, but in the same proportions as they exist in this list.) The actual mean and standard deviation for these numbers are 110 ounces and 5 ounces, respectively. The values are bellshaped.

Population Values for Ounces of Blood in Adults 97

100

101

102

103

103

104

104

104

105

106

106

106

107

107

108

108

109

109

109

110

110

110

110

110

111

112

112

112

112

113

113

113

113

113

113

114

114

114

114

114

114

115

115

116

116

116

117

118

118

Step 1: Develop a method for drawing simple random samples from this population. One way to do this is to write each number on a slip of paper, put all the slips into a paper bag, shake well, and draw from the bag. If a number occurs multiple times, make sure you include it that many times. Make sure you actually get random samples. Explain your method.

Step 2: Draw a random sample of size 9. Calculate and record the mean for your sample.

Step 3: Repeat step 2 a total of 20 times, thus accumulating 20 samples, each of size 9. Make sure to start over each time; for example, if you drew numbers from a paper bag, put the numbers back after each sample of size 9 so they are available for the next sample as well. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 19: The Diversity Of Samples From The Same Population

Step 4: Create a stemplot or histogram of your 20 sample means. Compute the mean of those sample means.

Step 5: Explain what the Rule for Sample Means tells you to expect for this situation.

Step 6: Compare your results with what the Rule for Sample Means tells you to expect. Be sure to mention mean, standard deviation, shape, and the intervals into which you expect 68%, 95%, and almost all of the sample means to fall.

Answer: Make sure the method is one for which everyone goes back into the population after each sample of 9 is drawn. The Rule for Sample Means tells us to expect the sample means to be bell-shaped with a mean of 110 and standard deviation of 5/3 = 1.67. Therefore, about 68% of the sample means should fall between 108.33 and 111.67 (about 13 or 14 of the sample means), 95% should fall between 106.67 and 113.33 (about 19 of the sample means) and all 20 should fall within 5 of the population mean, or between 105 and 115. Compare your results to these anticipated percentages.

3. Carry out the simulation in Mini-Project #1 using a computer or website instead of physically doing so. Go through the same steps and answer the same questions.

Answer: The simulation can be done using a website like www.randomizer.org. For the sample of 15, choose 15 integers without replacement from the set of 1 to 400.

Use the numbers from 1 to 160 to represent someone who carries the gene, and the numbers from 161 to 400 to represent someone who does not.

4. Carry out the simulation in Mini-Project #2 using a computer or website instead of physically doing so. Go through the same steps and answer the same questions.

Answer: Some software (such as Minitab) allows you to enter the 50 numbers in a column and then randomly sample from that column. If you use a website like www.randomizer.org you will need to take a random sample of size 9 from the integers 1 to 50, with replacement, and then use as your data the corresponding values from the table on the bottom of page 423. For instance, the integer 1 would correspond to a value of 97, and the integer 50 would correspond to a value of 118.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 20: Estimating Proportions With Confidence

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 20: ESTIMATING PROPORTIONS WITH C ONFIDENCE

TABLE OF CONTENTS Answers to the Exercise Questions ....................................................................................1 Answers to the Reflection Questions ...............................................................................15 Answers to the Mini Projects ........................................................................................... 16

ANSWERS TO THE EXERCISE QUESTIONS 1. One of the questions asked in a Gallup poll taken in May 2012, in the United States was: “All in all, if you had your choice, would you want to be rich, or not?” (Newport, 2012). (A previous question had asked respondents if they thought they were rich, and the 2% who said yes were not asked this question.) Of the 1008 people asked this question, 635 said yes, they would want to be rich. a. What is the sample proportion of respondents who said yes, they would want to be rich? b. Find the standard error to accompany the proportion you found in part (a). c. Use your answers from parts (a) and (b) to construct a 95% confidence interval for the population proportion who want to be rich. d. Interpret the confidence interval found in part (c) by writing a few sentences explaining what it means.

Answer: a. The sample proportion is 635/1008 = 0.63. b. The standard error is √(. 63)(.37)/1008 = 0.015. c. A 95% confidence interval for the population proportion who want to be rich is 0.63 ± 2(0.015), which is 0.63 ± 0.03, or 0.60 to 0.66. d. We are 95% confident that between 60% and 66% of the population of people in the United States in May 2012 who did not think they were rich wanted to be rich.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 20: Estimating Proportions With Confidence

2. Another question asked in the Gallup poll in Exercise 1 was: “Do you think that rich people in America today are happier than you, less happy, or about the same?” (Newport, 2012). Of the 1008 respondents, only 11% said “happier,” 27% said “less happy,” 57% said “about the same,” and the rest were “unsure.” a. Find a 95% confidence interval for the proportion of the population who would have answered “less happy” if asked. b. Convert the interval in part (a) into a 95% confidence interval for the percentage of the population who would have responded “less happy” if asked. c. Write a few sentences interpreting the interval you found in part (b).

Answer: a. The sample proportion is 0.27, so the 95% confidence interval is 0.27 ± 2(0.014), which is 0.242 to 0.298. The standard error of 0.014 is found as √(0.27)(0.73)/1008. b. The interval is 24.2% to 29.8% (or about 24% to 30%). c. We are 95% confident that between 24% and 30% of the population of the United States in May 2012 thought that rich people were less happy than them. The survey was based on a sample of 1008 randomly selected people who did not think themselves rich.

3. Does a confidence interval for a proportion estimate a sample proportion, or a population proportion?

Answer: A confidence interval is an estimate of a population proportion. Sample proportions are computed from data and do not need to be estimated.

4. Research!America, a nonprofit healthcare advocacy foundation, commissioned a national survey of 800 adults in 2008, and another national survey of 1004 adults in 2018 (Research!America, 2018). One of the questions asked in both surveys was “How important do you believe it is for parents to have their children vaccinated?” The results for the answer “very important” are as follows.

Year 2008 2018

Sample Size 800 1004

Number Saying “Very Important” 656 713

Percentage “Very Important” 82% 71%

a. The research report for 2018 stated that “this survey has a theoretical sampling error of ±3.1 percentage points.” (That value was found using the multiplier of 1.96 instead of 2.0.) Use this information to find a 95% confidence interval for the percentage of all adults in the United States in 2018 who would have answered “very important” if asked.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 20: Estimating Proportions With Confidence

b. Calculate the conservative margin of error for the 2008 survey. Use it to find a 95% confidence interval for the percentage of all adults in the United States in 2008 who would have answered “very important” if asked. c. The percentage who answered “very important” in the 2018 sample was 11% less than the corresponding percentage in 2008. Do you think that the difference between the results in the two polls can be attributed to chance variation in the samples taken, or does it represent a real difference of opinion in the population for the 2 years? Explain.

Answer: a. For 2018, the 95% confidence interval for the percentage of all adults in the United States who would have answered “very important” is 71% ± 3.1%, or 67.9% to 74.1%. b. The conservative margin of error for the 2008 survey is 1/√800 = 0.035, or 3.5%. Therefore, for 2008, a 95% confidence interval for the percentage of all adults in the United States who would have answered “very important” is 82% ± 3.5%, or 78.5% to 85.5%. c. It probably represents a real difference. The confidence intervals for the two time periods do not overlap.

5. Refer to Example 20.2, in which the association between visiting green spaces (nature) and medication use was discussed. Example 20.2 focused on the results for people who reported that they visit green spaces three to four times a week. The quote from the journal article also included information for those visiting green spaces five or more times a week, as follows: “In the main model, when the reference was <1 green space visit per week, the odds for psychotropic medication use were . . . 0.78 (0.63 to 0.96) for those visiting green spaces at least five times per week” (Turunen et al., 2023, p. 4). a. Write a sentence interpreting the reported odds of 0.78. b. What is the value of p in the following sentence regarding the participants of the study? There was p% lower odds of using mental health medications for those who reported visiting nature at least five times a week, compared with those who reported that they rarely visited nature. c. Using the reported confidence interval of 0.63 to 0.96, what are the values of a and b in the following sentence? We can be 95% confident that for the entire population, the odds of taking mental health medications are between a% and b% lower for those who visit nature at least five times a week compared with those who rarely do so.

Answer: a. Using the same wording as in Example 20.2, the odds ratio represents the odds of taking psychotropic medication to not taking it for those who visit green space at least five times per week, compared to those same odds for those who rarely visit green spaces. Therefore, the odds of taking

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 20: Estimating Proportions With Confidence

medication for those who visit green spaces at least five times a week is only 0.78 times what the odds are for those who rarely do so. b. The value is 22%, found as (1 – 0.78) × 100%. c. The values are a = 4% and b = 37%, found as (1 – 0.96) × 100% and (1 – 0.63) × 100%.

6. What level of confidence would accompany each of the following intervals? a. Sample proportion ± 1.0(SEP) b. Sample proportion ± 1.645(SEP) c. Sample proportion ± 1.96(SEP) d. Sample proportion ± 2.576(SEP) Answer: a. 68% b. 90% c. 95% d. 99% 7. Use the Empirical Rule to specify what level of confidence would accompany each of the following confidence intervals. a. Sample proportion ± 1(SEP) b. Sample proportion ± 2(SEP) c. Sample proportion ± 3(SEP) Answer: a. 68% b. 95% c. 99.7% 8. Explain whether the width of a confidence interval would increase, decrease, or remain the same as a result of each of the following changes: a. The sample size is doubled, from 400 to 800. b. The population size is doubled, from 25 million to 50 million. c. The level of confidence is lowered from 95% to 90%.

Answer: a. Decrease; the sample size is in the denominator of the standard error, so a larger sample size would produce a smaller standard error. b. Remain the same; the population size does not affect the width of a confidence interval (as long as the population is large relative to the sample). c. Decrease; the higher the desired confidence level the wider the interval must be.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 20: Estimating Proportions With Confidence

9. Parade Magazine reported that “nearly 3200 readers dialed a 900 number to respond to a survey in our Jan. 8 cover story on America’s young people and violence” (19 February 1995, p. 20). Of those responding, “63.3% say they have been victims or personally know a victim of violent crime.” Can the results quoted and methods in this chapter legitimately be used to compute a 95% confidence interval for the proportion of Americans who fit that description? If so, compute the interval. If not, explain why not. (Note that calling a 900 number was not a free call at the time; the caller was charged a fee.)

Answer: No. This was clearly not a random sample of the population. It was a selfselected sample, and people who had a personal experience probably would be more likely to respond.

10. Refer to Example 20.6. It is claimed that a 95% confidence interval for the percentage of placebo-patch users who quit smoking by the eighth week covers 13% to 27%. There were 120 placebo-patch users, and 24 quit smoking by the eighth week. Verify that the confidence interval given is correct.

Answer: The sample proportion is 0.2, so the standard error is √(0.2)(0.8)/120 = 0.037.

Therefore the 95% confidence interval is 0.2 ± 0.07. This gives the desired interval of 13% to 27%.

11. Find the results of a poll reported in a magazine, newspaper, or on the Internet in which a margin of error is also reported. (A good source of polls conducted by various organizations is www.pollingreport.com.) Explain what question was asked and what margin of error was reported; then present a 95% confidence interval for the results. Explain in words what the interval means for your example.

Answer: Answers will differ.

12. Suppose that a survey is planned to estimate the proportion of a population that is left-handed. The sample data will be used to form a confidence interval. Explain which one of the following combinations of sample size and confidence level will give the widest interval. (i) n = 400, confidence level = 90% (ii) n = 400, confidence level = 95% (iii) n = 1000, confidence level = 90% (iv) n = 1000, confidence level = 95% Answer: The widest interval will be for (ii) n = 400 and confidence level = 95%; the smaller sample size and greater confidence level both will tend to increase the width of the confidence interval

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 20: Estimating Proportions With Confidence

13. Confirm that the standard deviation for sample proportions is largest when the proportion used to calculate it is 0.50. Do this by using other values above and below 0.50 and comparing the answers to what you would get using 0.50. Try three values above and three values below 0.50.

Answer: Note that the sample size is fixed so we need look only at the numerator of the standard deviation formula and find its maximum value. Taking the proportion to be 0.10, 0.30, and 0.40, yields numerators: 0.30, 0.458, and 0.49.

Using 0.60, 0.80, and 0.90 gives the values of the numerators as: 0.49, 0.40, and 0.30. None of these are larger than the numerator value of 0.50 found when the proportion 0.50 is used, so these values indicate that the standard deviation is largest at 0.50.

14. A university is contemplating switching from the quarter system to the semester system. The administration conducts a survey of a random sample of 400 students and finds that 240 of them prefer to remain on the quarter system. a. Construct a 95% confidence interval for the true proportion of all students who would prefer to remain on the quarter system. b. Does the interval you computed in part (a) provide convincing evidence that the majority of students prefer to remain on the quarter system? Explain. c. Now suppose that only 50 students had been surveyed and that 30 said they preferred the quarter system. Compute a 95% confidence interval for the true proportion that prefers to remain on the quarter system. Does the interval provide convincing evidence that the majority of students prefer to remain on the quarter system? d. Compare the sample proportions and the confidence intervals found in parts (a) and (c). Use these results to discuss the role sample size plays in helping make decisions from sample data.

Answer: a. The sample proportion is 0.6, and standard error is about 0.02. So a 95% confidence interval is approximately 0.56 to 0.64 or 56% to 64%. b. Yes. We can be 95% sure that the true percentage of people who prefer the quarter system is 56% to 64%; any of those values constitute a majority. c. The sample proportion is 0.6, and the standard error is about 0.07. So the confidence interval is 0.46 to 0.74 or 46% to 74%. Because this confidence interval goes below 50%, it does not provide convincing evidence that the majority prefer to remain on the quarter system. d. The larger sample size in part (a) yielded a smaller standard error and thus a smaller confidence interval. In general the larger the sample size the more narrow the 95% confidence interval will be, so the sample size alone can determine whether we accept that the majority of a population indeed has a certain trait. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 20: Estimating Proportions With Confidence

15. The headline of a press release from the National Opinion Research Center (2018) was “New Survey Finds Large Number of People Skipping Necessary Medical Care Because of Cost.” The story reported on a survey the center conducted jointly with the West Health Institute. Of the 1302 respondents, 40% said that they had skipped at least one recommended medical test or treatment because of cost. The press release included the following information: “The nationwide survey was conducted February 15–19, 2018, using the AmeriSpeak® Panel, the probability-based panel of NORC at the University of Chicago. Online and telephone interviews using landlines and cell phones were conducted with 1,302 adults. The overall margin of sampling error is plus or minus 3.8 percentage points.” a. Use the reported margin of error to calculate a 95% confidence interval for the percentage of the population at the time of the survey that would have said that they had skipped at least one medical test or treatment because of cost. b. If the poll had used a simple random sample, what would be the numerical value of the conservative margin of error in percentage points? c. Compare your answer to part (b) to the margin of error reported in the news story. What is a likely reason that the two values differ?

Answer: a. The 95% confidence interval is about 40% +/− 3.8%, or 36.2% to 43.8%. As a proportion, it’s 0.362 to 0.438. b. With a simple random sample, the conservative margin of error would be 1/√1302 = 0.0277, or 2.77%. c. The reported margin of error for the panel is larger because different types of respondents were weighted differently. That adds to the uncertainty of the sample estimate.

16. A CBS News poll taken in March 2018 asked 1223 adults the question, “In your community, do you think teachers are paid too little, too much, or about the right amount?” (DePinto, 2018). The possible responses and percentages who selected them were “too little” (68%), “too much” (5%), “about right” (21%), “unsure” (6%). The reported margin of error was 3%. a. Using the reported margin of error of 3%, calculate a 95% confidence interval for the percentage of the population who would have answered “too little” to this question. Write a sentence interpreting the interval that could be understood by someone who knows nothing about statistics. b. Calculate a 95% confidence interval for the population percentage who would have answered “too little” using the formula in this chapter rather than the reported margin of error. Compare your answer to the answer in part (a). c. Using the formula in this chapter, compute a 95% confidence interval for the true percentage who would answer “too much” to the question. Now compute a 95% confidence interval using the reported margin of error of 3%. Compare your two intervals. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 20: Estimating Proportions With Confidence

d. If you computed your intervals correctly, you would have found that the two intervals in parts (a) and (b) were quite similar to each other, whereas the two intervals in part (c) were not. In part (c), the interval computed using the reported margin of error was wider than the one computed using the formula. Explain why the two methods for computing the intervals agreed more closely for the survey question in parts (a) and (b) than for the survey question in part (c). (Hint: The only thing that differs for the two situations is the sample proportion. What role does that play in the computations?)

Answer: a. A 95% confidence interval is68% ± 3%, or 65% to 71%. This means that we are 95% confidence that the percentage of the population of adults in 2018 who thought teachers were paid too little was between 65% and 71%. b. The standard error is √(0.68)(0.32)/1223 = 0.0133, so the confidence interval is 0.68 ± 2(0.0133) or about 0.65 to 0.71. As a percentage, it’s 65% to 71%, exactly the same as in part a. c. For a sample proportion of only 0.05 (5%), the standard error is √(0.05)(0.95)/1223 = 0.006, or 0.6% as a percentage. Therefore, the 95% confidence interval is about 5% ± 2(0.6) or 3.8% to 6.2%. Using the reported margin of error of 3%, the confidence interval is 2% to 8%, much wider than the one calculated using the formula. d. The intervals in parts (a) and (b) were quite similar because the sample proportion for that question was 0.68, close to the 0.50 used to calculate the reported margin of error. In part (c) the proportion of 0.05 was much farther away from 0.50 and therefore yielded a smaller standard error and a narrower interval.

17. In a 2008 ABC News survey of 1019 randomly selected adults in the United States, 55% answered “does” to the question “Do you believe that some form of intelligent life does or does not exist in outer space?” (ABC News, 2008). For parts (a) and (b), use the desired confidence level to find a confidence interval for the proportion of all adults in 2008 who would have said at the time that they thought there is intelligent life in outer space. a. 90% b. 99% c. Using the 99% confidence interval for part (b), would you conclude with 99% confidence that a majority of adults (more than 50%) in 2008 thought that there is intelligent life in outer space? Explain. d. In 2020, the U.S. government established a task force to investigate reports of UFOs but renamed them “unidentified aerial phenomena” or UAPs, leading many people to take the phenomenon more seriously than they had previously. In June 2021, the Pew Research Center conducted an Internet

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 20: Estimating Proportions With Confidence

panel poll with the question “Just your best guess, do you think that intelligent life exists on other planets?” Of the 10,417 participants, 65% answered yes (Kennedy and Lau, 2021). The reported margin of error was 1.5%.

Based on this information and the 99% confidence interval computed in part (c), do you think the percentage of the population who think intelligent life exists on other planets increased between 2008 and 2021? Explain.

Answer: For parts (a) and (b), the confidence interval is Sample proportion Multiplier  Standard error. The sample proportion is 0.55 and the standard error is (0.55)(0.45) / 1019 = 0.016.

a. 0.55 ± 1.645 × 0.016, or 0.55 ± 0.026, or 0.524 to 0.576 b. 0.55 ± 2.576 × 0.016, or 0.55 ± 0.04, or 0.51 to 0.59 c. Yes. We can be 99% confident that between 51% and 59% of all adults in 2008 thought that intelligent life exists elsewhere. The entire interval is above 50%, so that indicates that a majority thought so. d. A 95% confidence interval for the poll in 2021 is 65% ± 1.5%, or 63.5% to 66.5%. The entire interval is much higher than the 99% confidence interval for 2008, indicating that the percentage of the population who think intelligent life exists on other planets had increased between 2008 and 2021.

18. In Example 20.5, we found a 95% confidence interval for the proportion of successes likely in a certain kind of ESP test. Construct a 99.7% confidence interval for that example. Explain why a skeptic of ESP would prefer to report the 99.7% confidence interval rather than the 95% confidence interval.

Answer: The standard error is about 0.038 so a 99.7% confidence interval is about 0.37 ± 3(0.038), or 0.256 to 0.484, or 25.6% to 48.4%. A skeptic would prefer to use this confidence interval because it almost covers the 25% value that would be expected if there were no ESP, whereas the 95% confidence interval was from 29% to 45%.

19. Refer to the formula for a confidence interval in the Focus on Formulas section. a. Write the formula for a 90% confidence interval for a proportion. b. Refer to Example 20.6. Construct a 90% confidence interval for the proportion of smokers who would quit after 8 weeks using a nicotine patch. c. Compare the 90% confidence interval you found in part (b) to the 95% confidence interval used in the example. Explain which one you would present if your company were advertising the effectiveness of nicotine patches.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 20: Estimating Proportions With Confidence

Answer: a.

pˆ  1.645

pˆ (1 − pˆ ) n

b. The sample proportion is 0.46, so a 90% confidence interval is 0.46  1.645(0.045) or 0.46  0.074 or 0.386 to 0.534; 38.6% to 53.4%. c. The 95% confidence interval is wider, ranging from 37% to 55%; the 90% confidence interval covered 38.6% to 53.4%. The company would probably prefer the 95% interval because they could say that the data show that up to 55% of smokers treated in this way would quit smoking after 8 weeks.

20. One of the questions asked in a Pew Research Center/USA Today poll taken in June 2013, was: “Overall, do you approve or disapprove of the government’s collection of telephone and internet data as part of anti-terrorism efforts?” (http://www.pollingreport .com/terror.htm). Of the 1512 respondents, 48% said “approve,” 47% said “disapprove,” and the rest were unsure. a. What is the conservative margin of error for this poll? b. Use the conservative margin of error to find a 95% confidence interval for the proportion of all adults who would have answered “approve” if asked. c. Repeat part (b) for the answer “disapprove.” d. Compare the confidence intervals in parts (b) and (c). From those intervals, can you conclude that one of the two opinions (approve or disapprove) was held by a majority of the population at that time? Explain.

Answer: a. The conservative margin of error is 1/√1512 = 0.0257. b. A 95% confidence interval for “approve” is 0.48 ± 0.0257 or 0.4543 to 0.5057. c. A 95% confidence interval for “disapprove” is 0.47 ± 0.0257 or 0.4443 to 0.4957. d. We cannot conclude that a majority held either opinion. Neither of the two intervals is entirely above 0.50, which is what would be required to conclude that a majority of the population held that opinion.

21. In each situation, explain why you think that the sample proportion should or should not be used to estimate the population proportion. a. An Internet news organization asks visitors to its website to respond to the question: “Are you satisfied with the president’s job performance?” Of 3500 respondents, 61% say that they are not satisfied with the president’s performance. On the basis of this survey, the organization writes an article saying that a majority of Americans are not satisfied with the president. b. A convenience sample of 400 college students in two classes at the same university is used to estimate the proportion that is left-handed in the nationwide population of college students.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 20: Estimating Proportions With Confidence

Answer: a. The sample proportion should not be used to estimate the population proportion because the sample was a self-selected sample. Such samples usually do not represent any larger population. People dissatisfied with the president may be more likely to respond. b. While the sample is not a random sample, for the question of interest it may provide a suitable estimate of the proportion that is left-handed in the population. It doesn't seem that this sample of 400 students would be biased toward having a different proportion that's left-handed than in the nationwide population of college students.

22. An advertisement for Seldane-D, a (now discontinued) drug prescribed for seasonal allergic rhinitis, reported results of a double-blind study in which 374 patients took Seldane-D and 193 took a placebo (Time, 27 March 1995, p. 18).

Headaches were reported as a side effect by 65 of those taking Seldane-D and 43 of those taking a placebo. a. What is the sample proportion of Seldane-D takers who reported headaches? b. What is the standard error accompanying the proportion computed in part (a)? c. Construct a 95% confidence interval for the population proportion based on the information from parts (a) and (b), and write a few sentences explaining what it means. d. Compute a 95% confidence interval for the true population proportion that would get headaches after taking a placebo. e. Notice that a higher proportion of placebo takers than Seldane-D takers reported headaches. Use that information to explain why it is important to have a group taking placebos when studying the potential side effects of medications.

Answer: a. The sample proportion of Seldane-D takers who reported headaches is 65/374 = 0.17. b. The standard error is √(0.17)(0.83)/374 = 0.019. c. A 95% confidence interval is 0.17 ± 2(0.019) or 0.132 to 0.208 or 13.2% to 20.8%.

We are 95% confident that the population percentage of people who would experience headaches while taking Seldane-D is between 13.2% and 20.8%. d. The sample proportion is 43/193 = 0.22, so the standard error is √(0.22)(0.78)/193 = 0.03. Hence a 95% confidence interval is 0.22 ± 2(0.03), or 0.16 to 0.28, or 16% to 28%. e. It is important to have a placebo group in order to decide whether it was the medication that actually caused the side effect. As noted, in this case the percentage with the side effect was higher for the placebo group than for the group taking the medication. Therefore, it is not likely that the Seldane-D was causing the side effect. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 20: Estimating Proportions With Confidence

23. For the following situations, explain whether the reported values are a sample percentage (a statistic) or a population percentage (a parameter). For sample percentages, describe the population percentage they could be used to estimate. a. The U.S. Census Bureau conducts a census of the population every 10 years.

They reported that in 1910 only 4.6% of the population was over 65 years old, whereas in 2020, 16.8% of the population was over 65 (Caplan, 2023). b. In a randomized experiment to test a new medication for hay fever, symptoms were reduced for 67% of the 100 participants who took the medication and for 33% of the 90 participants who took the placebo. The participants were volunteers who were representative of all people who suffer from hay fever. c. In a survey based of a random sample of 1015 registered voters taken 1 month before a presidential election, 5% said they still had not decided who would get their vote. d. A press release from the University of California, Irvine (UCI News, 2023), reported that 46% of the 8507 students receiving bachelor’s degrees in 2023 were first-generation college students, meaning that they were the first person in their family to get a college degree.

Answer: a. The census includes the entire population, so the percentages reported are population percentages. b. These are sample percentages. They can be used to estimate what percentage of the entire population of people who suffer from hay fever would have their symptoms reduced if they were to take the medication, and if they were to take a placebo.

24. Suppose 200 different researchers all randomly select samples of 400 individuals from a population. Each researcher uses the sample to compute a 95% confidence interval for the proportion that has blue eyes in the population. About how many of the confidence intervals will cover the population proportion? About how many of the intervals will not cover the population proportion? Briefly explain how you determined your answers.

Answer: About 190 of the confidence intervals, which is 95% of the 200 computed intervals, should contain the population proportion and about 10 should not.

Remember that the confidence level describes the confidence in the procedure and gives the percentage of all intervals that will contain the true value.

25. A study titled “Development and initial validation of the Hangover Symptoms Scale: Prevalence and correlates of hangover symptoms in college students” (Slutske et al., 2003) examined the prevalence of hangover symptoms in college students. Table 3 of the journal article reports that 13% of the 1216 students who answered the questions said that they had not experienced any hangover symptoms in the past year. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 20: Estimating Proportions With Confidence

a. Assuming that the participants in this study are a representative sample of college students, find a 95% confidence interval for the proportion of college students who have not experienced any hangover symptoms in the past year. Use the formula in this chapter. b. Write a sentence or two interpreting the interval you found in part (a) that would be understood by someone without training in statistics. c. The journal article also reported that the study originally had 1474 participants, but only 1234 reported drinking any alcohol in the past year.

(Only those who reported drinking in the past year were retained for the hangover symptom questions, and only 1216 answered those questions.)

Use this information to find a 95% confidence interval for the proportion of all students who would report drinking any alcohol in the past year. d. The article reported that “Participants were selected from 1474 college undergraduates enrolled in Introduction to Psychology courses at the University of Missouri-Columbia during the Fall 2001 semester.” Based on that information, to what population of students do you think the intervals in this exercise apply? Explain.

Answer: a. The sample proportion is 0.13 and the sample size is 1216. Therefore, the standard error is the square root of (0.13)(0.87)/1216 = 0.0096. So the interval is 0.13 ± 2(0.0096) or 0.13 ± 0.02 or 0.11 to 0.15. b. A study of college students found that most of them have experienced hangover symptoms in the past year. In a sample of 1216 students, only 13% said they had not experienced such symptoms. Extending this to the population, we can be fairly certain that somewhere between 11% and 15% of all college students have not experienced such symptoms in the past year. c. The sample proportion who reported drinking any alcohol is 1234/1474 = 0.837. The associated standard error is the square root of (0.837)(0.163)/1474 = 0.0096. Therefore, a 95% confidence interval for the proportion who would report drinking any alcohol is 0.837 ± 0.02, or about 0.817 to 0.857, or about 82% to 86%. d. The sample was taken from college students at the University of MissouriColumbia, a university in the Midwestern part of the United States.

Therefore, the results probably apply to only to students similar to the students at that university. You would need to know more about that student body to be more specific.

26. The National Center on Addiction and Substance Abuse at Columbia University (2003) conducted a lengthy telephone survey of 1987 teenagers aged 12 to 17, asking 77 questions about the prevalence of tobacco, alcohol, and illegal drugs in the teens’ lives. The results were published in a detailed report called the “2003 CASA National Survey of American Attitudes on Substance Abuse VIII: Teens and Parents.”

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 20: Estimating Proportions With Confidence

a. The margin of error for the teens was reported to be 2.2%. Verify that this is the value of the conservative margin of error for the sample size of 1987 (rounded to one decimal place). b. A survey was also conducted on 504 parents. Verify that the conservative margin of error for that survey is about 4.4%. c. The 1987 teens in the survey were asked, “How harmful to the health of someone your age is the regular use of alcohol—very harmful, fairly harmful, not too harmful, or not harmful at all?” Forty-nine percent responded that it was very harmful. Using the margin of error given in part (a), find a 95% confidence interval for the proportion of all teens who would respond that way. d. The 504 parents in the survey were asked, “How harmful to the health of a teenager is the regular use of alcohol—very harmful, fairly harmful, not too harmful, or not harmful at all?” Seventy-seven percent responded that it was very harmful. Using the margin of error given in part (b), find a 95% confidence interval for the proportion of all parents (similar to those in this study) who would respond that way. e. Compare the confidence intervals in parts (c) and (d). In particular, do they indicate that there is a difference in the population proportions of teens and parents who think alcohol is very harmful to teens? f. Write a short news story reporting the results you found in parts (c) to (e).

Answer: a. The formula for the conservative margin of error in this case is 1/√1987 ×100% = 2.24%, which was rounded to 2.2% when it was reported. b. The formula for the conservative margin of error in this case is 1/√504 ×100% = 4.45%, which was rounded to 4.4% when it was reported. c. The 95% confidence interval is 49% ± 2.2%, which is 46.8% to 51.2%. d. The 95% confidence interval is 77% ± 4.4%, which is 72.6% to 81.4%. e. There is obviously a big difference between how teens and their parents feel in the population. The two confidence intervals are very far apart. f. A study found a big discrepancy between how teens and their parents feel about whether or not regular use of alcohol is harmful to teens. Only about 49% of the teens in the study felt that alcohol was very harmful, whereas about 77% of their parents felt that way. Taking sampling variability into account, the percent of all teens who feel that way is probably between 47% and 51%, whereas for parents it is probably between 73% and 81%. It is clear that parents and teens have very different perceptions about the harm that can be done when teens drink alcohol.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 20: Estimating Proportions With Confidence

ANSWERS TO THE REFLECTION QUESTIONS 1. The most common confidence level used in reporting confidence intervals is 95%, but other confidence levels are sometimes used, such as 99%. Give one advantage to using a level of 95% and one advantage to using a level of 99%.

Answer: A 95% confidence interval will not be as wide as a 99% confidence interval, but the price to pay is that you can only be 95% confident, instead of 99% confident, that it covers the true value of the population parameter.

2. A professor asked all of the students in a class whether they preferred an inclass final exam or a take-home final exam, and 40% said they preferred an inclass exam. Explain why it would not be appropriate to construct a confidence interval in this situation.

Answer: The professor had information from the entire population, and therefore did not need to estimate anything.

3. If hundreds of 95% confidence intervals are computed, approximately what percentage of them will not cover the true value of the population parameter the interval is supposed to estimate?

Answer: The confidence level of 95% means that over the long run, about 95% of the computed intervals will cover the true value of the population parameter.

Therefore, the remaining 5% will not cover it.

4. Does the margin of error for a confidence interval for a proportion depend on the size of the sample, the size of the population, or both? Explain.

Answer: The margin of error depends on the size of the sample, but it does not depend on the size of the population, as long as the population is relatively large compared to the size of the sample.

5. Explain why it makes sense that a large sample will produce a more narrow confidence interval than a small sample for the same situation.

Answer: The larger the sample, the more information it provides. Therefore, the sample statistic will be closer to the population parameter, on average, for a large sample than for a small sample.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 20: Estimating Proportions With Confidence

ANSWERS TO THE MINI PROJECTS 1. You are going to use the methods discussed in this chapter to estimate the proportion of all cars in your area that are red. Stand on a busy street and count cars as they pass by, or visit a large parking lot and observe the colors of the cars. Count 100 cars and keep track of how many are red. a. Using your data, compute a 95% confidence interval for the proportion of cars in your area that are red. b. Based on how you conducted the survey, are any biases likely to influence your results? Explain.

Answer: A 95% confidence interval for a sample of size 100 should have a margin of error of about 1/10 or 0.1 but may be slightly smaller if your observed percentage is small. Sources of bias would be if you happened to take your sample at a time of day that is unrepresentative, like early morning or late evening when only certain types of drivers are on the road. It could also be the case that certain types of drivers use their cars more frequently and are also more (or less) likely to have red cars, which would introduce bias. The project should show that thought has been given to these issues.

2. Collect data and construct a confidence interval for a proportion for which you already know the answer. Use a sample of at least 100. You can select the situation for which you would like to do this. For example, you could flip a coin 100 times and construct a confidence interval for the proportion of heads, knowing that the true proportion is 0.5. Report how you collected the data and the results you found. Explain the meaning of your confidence interval, and compare it to what you know to be the truth about the proportion of interest.

Answer: If you used a coin and a sample size of 100, the width of the interval should be about ± 0.1. It is possible that the interval will not cover what is known to be the truth; that should happen to about 5% of those who do this project. If it does happen, that explanation should accompany the project. If the interval does cover the known true value, the project should include an explanation of that fact, and that it is to be expected once in a while. The project should also explain that if we did not know the truth, we could be 95% certain that it was covered by the computed interval.

3. Choose a categorical variable for which you would like to estimate the true proportion that fall into a certain category. Conduct an experiment or a survey that allows you to find a 95% confidence interval for the proportion of interest.

Explain exactly what you did, how you computed your results, and how you would interpret the results.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 20: Estimating Proportions With Confidence

Answer: The sample should be as representative as possible, and large enough to follow the Rule for Sample Proportions, which means you should expect at least 10 of each of the two responses. The interval should be interpreted to mean that you can be 95% sure that the truth about the population from which the sample was taken is covered by it.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 21: THE ROLE OF CONFIDENCE INTERVALS IN RESEARCH

TABLE OF CONTENTS Answers to the Exercise Questions ....................................................................................1 Answers to the Reflection Questions .............................................................................. 19 Answers to the Mini Projects ........................................................................................... 20

ANSWERS TO THE EXERCISE QUESTIONS 1. In Chapter 20, we saw that to construct a confidence interval for a population proportion it was enough to know the sample proportion and the sample size.

Is the same true for constructing a confidence interval for a population mean?

That is, is it enough to know the sample mean and sample size? Explain.

Answer: No. You also need the sample standard deviation.

2. Explain the difference between a population mean and a sample mean using one of the studies discussed in the chapter as an example.

Answer: Using Example 21.1, the sample mean weight loss for the 42 men who dieted was 7.2 kg. The corresponding population mean is the hypothetical average weight loss that would be achieved if all men similar to this group were to follow a diet similar to the one used by this group, for the same amount of time.

3. The Baltimore Sun (Haney, 21 February 1995) reported on a study by Dr. Sara Harkness in which she compared the sleep patterns of 6-month-old infants in the United States and the Netherlands. She found that the 36 U.S. infants slept an average of just under 13 hours out of every 24, whereas the 66 Dutch infants slept an average of almost 15 hours. a. The article did not report a standard deviation, but suppose it was 0.5 hour for each group. Compute the standard error of the mean (SEM) for the U.S. babies. b. Continuing to assume that the standard deviation is 0.5 hour, compute an approximate 95% confidence interval for the mean sleep time for 6-monthold babies in the United States.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

c. Continuing to assume that the standard deviation for each group is 0.5 hour, compute an approximate 95% confidence interval for the difference in average sleep time for 6-month-old Dutch and U.S. infants. Interpret the interval. d. Does the interval provide convincing evidence that the mean hours of sleep differed for the populations of Dutch and U.S. infants at the time of the study? Explain.

Answer: a. The standard error of the mean is 0.5/√36 = 0.5/6 = 1/12 or 0.083 hours or 5 minutes. b. A 95% confidence interval is 12 hours 50 minutes to 13 hours 10 minutes, found as 13 hours ± 2(5) minutes. c. The SEM for the Dutch babies is 0.062, so the standard error of the difference (SED) is the square root of (0.083)2 + (0.062)2 = 0.104. A 95% confidence interval for the difference is 2 ± 2(0.104) or 1.79 to 2.21 hours. d. Yes, there is convincing evidence that the population means are different.

The confidence interval for the difference in population means does not cover 0.

4. What is the probability that a 95% confidence interval will not cover the true population value?

Answer: In the long run, 95% of these confidence intervals will cover the true population value, so the probability is 0.05, or 5% that an interval will not cover it. (Technically, once an interval is computed, it either does or does not cover the true value, so this probability applies over the long run.)

5. Suppose a university wants to know the average income of its students who work, and all students supply that information when they register. Would the university need to use the methods in this chapter to compute a confidence interval for the population mean income? Explain. (Hint: What is the sample mean and what is the population mean?)

Answer: No. They would have the necessary information on the entire population so they could compute the desired population mean income directly.

6. Suppose you were given a 95% confidence interval for the difference in two population means. What could you conclude about the population means if a. The confidence interval did not cover zero. b. The confidence interval did cover zero.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

Answer: a. We could be 95% confident that the two population means are not equal. b. The population means could be (but are not necessarily) the same.

7. Suppose you were given a 95% confidence interval for the relative risk of disease under two different conditions. What could you conclude about the risk of disease under the two conditions if a. The confidence interval did not cover 1.0. b. The confidence interval did cover 1.0.

Answer: a. With 95% confidence, we could conclude that the risk of disease was greater under one of the conditions than the other. b. We could not conclude that the risks under the two conditions were different (but we would not go so far as to say they are equal.)

8. Ever since President John F. Kennedy was assassinated in 1963, there has been speculation about whether there was a single assassin or more than one.

Periodically, a Gallup poll asks this question: “Turning now to the assassination of John F. Kennedy in 1963—Do you think that one man was responsible for the assassination of President Kennedy, or do you think that others were involved in a conspiracy?” In a poll taken shortly after Kennedy’s assassination in 1963, 52% of respondents said “others involved.” In a poll taken in October 2023, 65% of respondents said “others involved” (Bernan, 2023). Although the sample size for the 1963 poll has been lost to history, the best guess is that an approximate 95% confidence interval for the difference in population percentages who felt that way in 2023 versus 1963 is 13% ± 6%, with the higher percentage in 2023. a. Compute the endpoints of the interval. b. Write a sentence or two interpreting the interval in words that would be understood by someone with no training in statistics. c. Does the interval indicate that opinion on this issue changed between 1963 and 2023 for the population of U.S. adults? Explain.

Answer: a. The interval is 7% to 19%. b. Surveys of U.S. adults in 1963 and 2023 asked the question “Do you think one man was responsible for the assassination of President Kennedy, or do you think that others were involved in a conspiracy?” Based on the results of those asked, we can say with 95% confidence that the percentage who thought that others were involved increased by somewhere from 7% to 19% from 1963 to 2023.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

c. Yes. The confidence interval does not include 0 and in fact is far removed from 0, so we can be fairly certain that the population opinion changed from 1963 to 2023.

9. On page 453 in this chapter technical details were presented for finding a t-multiplier for constructing a confidence interval for a population mean. Use appropriate software, a calculator, or a website to find the t-multiplier for the following situations. a. 95% confidence interval, n = 200 b. 95% confidence interval, n = 50 c. 90% confidence interval, n = 200 d. 90% confidence interval, n = 50 Answer: a. Degrees of freedom = 199, so t = 1.97 b. Degrees of freedom = 49, so t = 2.01 c. Degrees of freedom = 199, so t = 1.65 d. Degrees of freedom = 49, so t = 1.68 10. On page 453 in this chapter technical details were presented for finding a t-multiplier for constructing a confidence interval for a population mean. Use appropriate software, a calculator, or a website to find a confidence interval for the population mean for the following situations. Use the t-multiplier in each case. a. Refer to Table 21.2, which provided summary statistics for the Peak Fast Motion Sickness (FMS) scale during a simulated 15-minute helicopter ride under different conditions. Find a 95% confidence interval for the population mean Peak FMS score if everyone in the population were in the control condition. Relevant numbers from the table are the sample mean of 4.56, SEM of 0.68, and sample size of 27. b. Repeat part (a) for the population mean Peak FMS score if everyone were to chew ginger gum during the 15-minute simulated ride. Relevant numbers from the table are the sample mean of 2.57, SEM of 0.69, and sample size of 23. c. In Example 21.4, we found with 95% confidence that the difference in population means for the Peak FMS under the two conditions was between 0.05 and 3.93. Based on that interval, we could conclude that the difference was not 0, indicating that chewing ginger gum would reduce motion sickness for the population. Compare the two confidence intervals found in parts (a) and (b). Could the same conclusion be made based on those intervals? Explain.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

Answer: a. The t-multiplier is t(0.025, 26) = 2.06, so the 95% confidence interval for the mean Peak FMS score in the control condition is 4.56 ± 2.06 × 0.68, which is 3.16 to 5.96. b. The t-multiplier is t(0.025, 22) = 2.07, so the 95% confidence interval for the mean Peak FMS score if everyone were to chew ginger gum is 2.57 ± 2.07 × 0.69, which is 1.14 to 4.0. c. The two confidence intervals overlap, so the same conclusion could not be made. There are possible values for the population mean that are included in both intervals.

11. As noted on pp. 455–456, finding a confidence interval for the difference in two means with paired data requires finding the differences first, and then computing a confidence interval for the mean difference. For each of the following situations, identify whether it would be more appropriate to use the procedure for paired data or the procedure for two independent samples to find a confidence interval for the difference in means. a. A researcher wants to estimate the difference between mean scores on a memorization test for the population of people over 60 and the population of adults under 40 years old. Separate samples are tested for each age group. b. Patients who come to a medical clinic are asked how much they weigh, and then their weight is measured. Researchers want to esti-mate the mean difference between what people claim to weigh and what they actually weigh. c. In a double-blind study to measure the effect of a new drug designed to reduce postsurgery pain, patients who have just had surgery are randomly assigned to take the new drug or the current standard drug. Patients report their level of pain an hour after taking the drugs. The researchers want to find a confidence interval for the difference in the mean pain ratings for the two drugs, if everyone in the population were to take each one. d. Researchers want to know if teenagers who have not taken a typing class can type out a paragraph faster on their phones or using a computer keyboard. Fifty volunteers type the paragraph using both methods, with the order randomized for each person. The goal is to find a confidence interval for the difference in mean typing times for the two methods, for the population similar to the volunteers.

Answer: a. The procedure for two independent samples is appropriate. There is no pairing between the individuals in the two age groups. b. The procedure for paired data is appropriate. Each individual provides two measurements, and the mean difference for the population is of interest.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

c. The procedure for independent samples is appropriate. Each individual takes only one of the drugs, and there is no pairing of the individuals in the two groups. d. The procedure for paired data is appropriate. Each individual types the paragraph using both methods.

12. In a study comparing age of death for left- and right-handed baseball players, Coren and Halpern (1991, p. 93) provided the following information: “Mean age of death for strong right-handers was 64.64 years (SD = 15.5, n = 1472); mean age of death for strong left-handers [was] 63.97 years (SD = 15.4, n = 236).”

The term “strong handers” applies to baseball players who both threw and batted with the same hand. The data were actually taken from entries in The Baseball Encyclopedia (6th ed., New York: Macmillan, 1985), but, for the purposes of this exercise, pretend that the data were from a sample drawn from a larger population. a. Compute an approximate 95% confidence interval for the mean age of death for the population of strong right-handers from which this sample was drawn. b. Repeat part (a) for the strong left-handers. c. Compare the results from parts (a) and (b) in two ways. First, explain why one confidence interval is substantially wider than the other. Second, explain whether you would conclude that there is a difference in the mean ages of death for left- and right-handers on the basis of these results. d. Compute an approximate 95% confidence interval for the difference in mean ages of death for the strong right- and left-handers. Interpret the result.

Answer: a. The confidence interval is 64.64 ± 0.81 or 63.83 to 65.45 years of age at death. b. The confidence interval is 63.97 ± 2.0 or 61.97 to 65.97 years of age at death. c. The interval in part b. is wider because the sample size is smaller. We could not conclude that there is a difference in the mean ages because there is overlap in the two intervals. d. The interval is (64.64 − 63.97) ± 2(1.08) or −1.49 to 2.83 years of age at death. This interval does not lie entirely above zero, so it is possible that the difference in these sample means does not represent a real difference in the population of strong left-handers and right-handers.

13. In revisiting Case Study 5.4, we quoted the original journal article as reporting that “for any vertex baldness (i.e., mild, moderate, and severe combined), the age-adjusted RR was 1.4 (95% CI, 1.2 to 1.9)” (Lesko et al., 1993, p. 1000).

Interpret this result. Explain in words that someone with no training in statistics would understand.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

Answer: The statement means that with 95% confidence we can say that men with any vertex baldness are at a higher risk of heart attack than men with no baldness; their risk is 1.2 to 1.9 times greater.

14. In a report titled, “Secondhand Smoke: Is It a Hazard?” (Consumer Reports, January 1995, pp. 27–33), 26 studies linking secondhand smoke and lung cancer were summarized by noting, “those studies estimated that people breathing secondhand smoke were 8 to 150 percent more likely to get lung cancer sometime later” (p. 28). Although it is not explicit, assume that the statement refers to a 95% confidence interval and interpret what this means.

Answer: The study says that with 95% confidence we can say that people breathing secondhand smoke are between 8% and 150% more likely to get lung cancer than those who don't breathe it.

15. In the study described in Example 21.1, the researchers measured many variables on each participant in addition to weight. For instance, the mean drop in total cholesterol over the year for those who dieted was 13.9 mg/dL, while for those who exercised, it was 9.7 mg/dL. Therefore, those who dieted had an average cholesterol drop that was 4.2 mg/dL higher than those who exercised. A 95% confidence interval for the difference in population means is from –5.5 to +14.0 mg/dL. a. Explain why it would be misleading to report “these results show that men who diet for a year will have a greater drop in cholesterol than men who exercise for a year” even though it was true for the men in this study. b. Write a sentence or two interpreting the 95% confidence interval.

Answer: a. The confidence interval covers 0 and negative values, so it's possible that the population means are equal, or even that the exercisers had a higher drop than the dieters. b. We are 95% confident that the difference in mean cholesterol drop would be anywhere from 5.5 mg/dL lower to 14.0 mg/dL higher if everyone in the population dieted than if they exercised for a year.

16. The study described in Example 21.1 included a control group in addition to those who dieted or exercised. Table 21.4 shows the sample size n, and mean ± standard deviation for weight change during the 1 year of the study, as presented in Table 2 of the journal article (Wood et al., 1988, p. 1175). For example, the mean change for exercisers was −4.0, meaning they lost an average of 4.0 kg. a. Compute the standard error of the mean for each group by dividing the standard deviation by the square root of the sample size.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

b. Draw a figure similar to Figure 21.1 illustrating the mean and standard error of the mean for the three groups. Note that your figure will need to include negative values on the vertical axis. c. Based on the figure drawn in part (b), comment on whether you think the three population means are all different or if it is plausible that two of them are the same.

Answer: a. The formula is SEM = standard deviation/√𝑛. For exercisers SEM = 3.9/√47 = 0.57, and for both dieters and the control group SEM = 3.7/√42 = 0.57. b. The figure is shown below.

Mean ± Standard Error of the Mean

Weight gain over one year (kg)

0

-2

-4

-6

-8 Exercisers

Dieters

Control Group

c. It is likely that the population means are all different. There is no overlap in the ranges covered by the three black lines, which indicate the mean ± standard error for the sample. A 95% confidence interval for each population mean would be the mean ± 2 standard errors, but even then they would not overlap.

17. Refer to the previous exercise and the means, standard deviations, and sample sizes shown in Table 21.4. a. Compute a 95% confidence interval for the difference in population means if everyone in the population were to exercise for a year versus if they were to behave like those in the control group. b. Compute a 95% confidence interval for the difference in population means if everyone in the population were to follow the diet used in this study versus if they were to behave like those in the control group. c. As shown in Example 21.2, a 95% confidence interval for the difference in population means if everyone were to follow the diet for a year versus follow the exercise program for a year is 1.6 to 4.8 kg, with higher weight

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

loss for the dieters. Use that information and the results in parts (a) and (b) to write a paragraph summarizing the results of this study.

Table 21.4

Mean ± Standard Deviation for Weight Change over 1 Year (kg)

Exercisers (n = 47)

Dieters (n = 42)

Controls (n = 42)

–4.0 ± 3.9

–7.2 ± 3.7

+0.6 ± 3.7

Answer: a. The procedure for finding an approximate confidence interval for the difference in two population means is given in the box on page 454. In this case, the standard error of the mean (SEM) was shown in the previous exercise to be 0.57 for all three conditions. Therefore, the standard error of the difference, SED is the same for all pairs of differences, found as square root of [(0.57)2 + (0.57)2]= 0.8. Therefore, the approximate 95% confidence interval for the difference in means for exercisers – control group is (−4.0 – 0.6) ± (2 × 0.8), or −4.6 ± 1.6, or −6.2 to – 3.0 kilograms. b. Following the same procedure and the SED computed in part (a), and approximate 95% confidence interval for the difference in means for dieters – control group is −7.8 ± 1.6 or –9.4 to −6.2 kilograms. c. Overweight men were randomly assigned to spend a year exercising, dieting, or continuing their previous behavior. The men in the active samples lost averages of 4 kilograms for exercisers and 7.2 kilograms for dieters, but the men in the control group gained an average of 0.6 kilograms. Extending these results to the population, it is estimated that if all similar men were to follow the diet they would lose an average of between 1.6 and 4.8 kilograms more than if they were to exercise. But in both cases they would lose more than if they did neither of these, losing an average of between 6.2 and 3.0 kilograms more by exercising, and an average of 9.4 to 6.2 kilograms more by following the diet.

The following information is for Exercises 18 to 20. Refer to News Story 13 in the Appendix, “Spicing up your life could make it last longer.” The story describes a study that followed 23,000 Italians who filled out various food and lifestyle questionnaires over a period of many years. One of the quotes in the original news story was: “The study of almost 23,000 Italians found that regular consumers [of chili peppers] had 40 percent fewer fatal heart attacks and 61 percent fewer deadly strokes” (Donnelly, 2019).

18. The original journal article on which the news story is based reported that “Regular intake was also inversely associated with cerebrovascular (HR: 0.39; 95% CI: 0.20 to 0.75) death risks” (Bonaccio et al., 2019, p. 3139). The statistic “HR: 0.39” refers to a hazard ratio, which is the relative risk at a given time point of 0.39 or 39%. The article explained that this was the relative risk of

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

death from strokes for people who ate chili peppers at least four times a week compared to those who rarely ate them. a. Explain how the quoted ratio of 0.39 in the journal article is related to the quote in the news story stating that the study found 61% fewer deadly strokes. b. Write a few sentences interpreting the 95% confidence interval given in the quote from the journal article.

Answer: a. The hazard ratio of 0.39 for death from strokes means that the risk of death at any point in time for those who ate chili peppers at least four times a week was only 39% of what it was for those who didn’t eat them.

That’s equivalent to a (100% − 39%) = 61% reduction in death risk. b. The 95% confidence interval given in the quote is from 0.20 to 0.75. This means we can be 95% confident that for people similar to those in the study, the risk of death from a stroke at any point in time for those who eat chili peppers at least four times a week is only between 20% and 75% of the risk for those who never eat them.

19. The original journal article included information on a “Mediterranean Diet Score” (MDS), which is a measure of how closely participants followed what is known as a Mediterranean diet. The group that rarely ate chili peppers included 7689 people. The mean and standard deviation MDS for that group were 4.2 and 1.6. The mean and standard deviation MDS for the 5546 people in the study who ate chili peppers at least four times a week were 4.8 and 1.6 (Bonaccio et al., 2019). a. Using the Empirical Rule and the quoted mean and standard deviation, find an interval into which about 95% of the population of individual MDS scores would fall for the population who rarely eats chili peppers. Write a sentence interpreting this interval. b. A 95% confidence interval for the mean MDS for the population who rarely eats chili peppers is 4.16 to 4.24. Write a sentence interpreting this interval.

Explain why this interval is so narrow compared with the interval you found in part (a). c. A 95% confidence interval for the mean MDS for the population who eats chili peppers at least four times a week is 4.76 to 4.84. Using this information and the confidence interval in part (b) for those who rarely eat chili peppers, write a sentence or two comparing the mean MDS for these two populations. d. The report in the news story that there were 61% fewer deadly strokes was actually based on a statistical analysis that took into account many measurements on each individual in addition to the frequency of eating chili peppers. One factor taken into account was the individual’s Mediterranean Diet Score (MDS). Using the results in this exercise, explain

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

why it was important for the researchers to adjust for MDS before comparing risks for the differing consumption of chili peppers.

Answer: a. About 95% of the individual scores should fall within 2 standard deviations of the mean. So in this case, about 95% of the Mediterranean diet scores (MDS) for those who rarely eat chili peppers would fall in the range 4.2 ± 2(1.6) or 1.0 to 7.4. b. We can be 95% confident that the mean MDS for the population who rarely eat chili peppers is between 4.16 and 4.24. The confidence interval for the mean is much narrower than the interval that covers most of the population because with such a large sample (7689 people) averaging their individual scores leads to a relatively precise estimate of the corresponding population average. c. On average, people who rarely eat chili peppers are also less likely to follow a Mediterranean diet than those who eat them at least four times a week.

We can be 95% confident that the mean Mediterranean diet score for those who eat them often is between 4.76 and 4.84, while for those who rarely eat them it is only between 4.16 and 4.24 (with 95% confidence). d. The Mediterranean diet score (MDS) is confounded with the frequency of eating chili peppers. Because those who eat peppers more often also are more likely to follow the Mediterranean diet, it was important for the researchers to separate out the effects of these two behaviors on stroke risk.

20. The study investigating death risks based on chili pepper consumption followed almost 22,811 people for a median time of 8.2 years (Bonaccio et al., 2019). a. Do you think this research was based on a randomized experiment or an observational study? Explain. b. Could this study have been conducted as a double-blind study? Explain. c. The news story quoted the author of the journal article as stating “someone can follow the healthy Mediterranean diet, someone else can eat less healthily, but for all of them chili pepper has a protective effect” (Donnelly, 2019). Do you agree that this study has shown that eating more chili peppers has a protective effect? In other words, has the study shown that eating chili peppers causes lower mortality? Explain.

Answer: a. The research was based on an observational study. Researchers could not randomly assign 22,811 people to consume specified levels of chili peppers for over 8 years. People wouldn’t do it!

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

b. The study could not be double blind, even if it had been a randomized experiment. People would know how often they were eating chili peppers. c. The research was based on an observational study, so it cannot be concluded that there is a causal relationship. There could be other confounding variables that are related to eating chili peppers, and that contribute to health outcomes.

21. In Example 21.4, a study was described that tested the use of chewing gum to reduce visually induced motion sickness (VIMS), commonly experienced in virtual reality sessions. Two different flavors of chewing gum were tested in the experiment: peppermint and ginger. One of the outcome measures was the nausea portion of a “simulator sickness questionnaire,” or SSQ, filled out by each participant before and after the 15-minute virtual reality session. The purpose of this exercise is to compare the results at the end of the session for the two flavors of gum. The sample sizes n, means, and standard deviations for the participants who chewed each gum type are shown in Table 21.5 (Kaufeld et al., 2022, p. 657). Higher scores indicate greater nausea at the end of the session.

Table 21.5

Nausea Scores for Two Gum Flavors after a 15-Minute Virtual Reality Session Peppermint (n = 27)

Ginger (n = 23)

Mean

22.97

17.00

Standard deviation

22.65

15.48

a. A 95% confidence interval for the difference in population mean SSQ nausea values if everyone in the population were to chew peppermint gum versus if they were to chew ginger gum is from –4.95 to 116.89 points. Write a few sentences interpreting this interval. b. One of the statements in the journal article was “In our study, we found no additional beneficial effect of ginger on VIMS, as compared to the peppermint flavor. This is contradictory to the findings by [previous researcher]” (Kaufeld et al., 2022, p. 659). The quote then explained that previous research found that administering ginger powder reduced levels of nausea. Referring to the confidence interval given in part (a), do you agree that the current study has shown that there would be no additional beneficial effect of ginger on nausea, compared with peppermint, for the population? Explain.

Answer: a. We can be 95% confident that if everyone who did this virtual reality session were to chew peppermint gum, their nausea scores would be between 4.95 points lower and 16.89 points higher than if they were to chew ginger gum.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

b. No. The confidence interval includes values in both directions, but more of them are in the direction of the ginger gum creating more benefits. The sample was too small to conclude that either gum was superior, but this is not the same as finding that there would be no beneficial effect of ginger over peppermint.

22. In Example 21.4, a study was described that tested whether chewing gum could reduce visually induced motion sickness. One outcome measure was the Peak Fast Motion Sickness (FMS) scale. It was shown with 95% confidence that after chewing ginger gum during a 15-minute virtual reality session, the mean Peak FMS would be reduced more than without chewing any gum. This conclusion was based on an approximate 95% confidence interval for the difference in means, found as (1.99 6 2 SEM) where SEM 5 0.97. The interval did not cover 0, leading to the conclusion that there is a difference in population means. a. Find an approximate 99% confidence interval for the difference in population means for Peak FMS if everyone were to chew ginger gum versus no gum. b. Using the interval you computed in part (a), can it be concluded with 99% confidence that there would be a difference in the population mean Peak FMS if everyone were to chew ginger gum versus not chewing any gum?

Explain.

Answer: a. The formula is the same except that the multiplier is changed from 2 to 2.576. Therefore, the interval is 1.99 ± 2.576(0.97), or −0.51 to +4.49. b. No; it cannot be concluded that there would be a difference because the interval includes 0.

The following information is for Exercises 23 to 25: Refer to Original Source 5 on the companion website, “Distractions in Everyday Driving.” Table 14 on page 51 provides 95% confidence intervals for the average percent of time drivers in the population would be observed not to have their hands on the wheel during various activities while the vehicle was moving, assuming they were like the drivers in this study. The confidence intervals were computed using a different method than the one presented in this book because of the type of data available, but the interpretation is the same. (See Appendix D of the report if you are interested in the details.)

23. a. The confidence interval accompanying “Reading/writing” is from 4.24% to 34.39% of the time. Write a few sentences interpreting this interval that would be understood by someone with no training in statistics. Make sure you are clear about the population to which the result applies. (Additional information about the study is provided in News Story 5 in the Appendix.) b. Repeat part (a) for the confidence interval for “Conversing,” which is from 0.92% to 2.38%.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

c. Compare the intervals in parts (a) and (b). Recall that they represent estimates of the mean percent time drivers in the population have their hands off the steering wheel due to the activity. Using your knowledge of these activities, explain why the values in the interval in part (a), for reading and writing, are so much higher than those in part (b), for conversing.

Answer: a. The drivers in the sample sometimes were observed reading or writing.

Sometimes while they were doing so they did not have their hands on the wheel. Taking into account the variability from driver to driver in this sample, we estimate that the average amount of time drivers in the general population spend with their hands not on the wheel when they are reading or writing is probably between 4.24% and 34.39% of the time they are doing that activity. b. The drivers in the sample were often observed conversing, and occasionally were observed to take their hands off the wheel while they were doing so.

Taking into account the variability from driver to driver in this sample, we estimate that the average amount of time drivers in the general population spend with their hands not on the wheel while they are conversing is minimal, probably between 0.92% and 2.38% of all time they are conversing with someone. c. It is difficult to read or write without taking your hands off the wheel, while it is easy to converse with someone in the car without doing so.

24. Refer to the interval in part (a) of Exercise 23. Write it as a 95% confidence interval for the average number of minutes out of an hour that drivers who were reading or writing would be observed to have their hands off of the wheel.

Answer: Multiply the endpoints by 60 minutes to get 2.5 to 20.6 minutes.

25. In Table 14 (on page 51 of Original Source 5), some of the intervals are much wider than others. For instance, the 95% confidence interval for the average percent of time drivers have their hands off the wheel while not reading or writing is from 0.97 to 1.93, compared with the interval from 4.24 to 34.39 while reading or writing. In News Story 5 in the Appendix, it is reported that only 40% of the 70 drivers in the study were observed reading or writing, and only data from those drivers would be used to compute the interval from 4.24% to 34.39%. All 70 would have been observed during time periods when they were not reading or writing. What two features of the data do you think are responsible for some intervals being wider than others? (Hint: What two features of the data determine the width of a 95% confidence interval for a mean? The answer is the same in this situation.)

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

Answer: The two features are the size of the sample and the variability in the measurement of interest. In this example, the sample size was only 28 for those observed reading or writing, while data for all 70 drivers was available for time periods without reading or writing. It’s also likely that there was much more variability in times with hands off the wheel while reading or writing than for time periods with no reading or writing.

26. Original Source 9, “Suicide Rates in Clinical Trials of SSRIs, Other Antidepressants, and Placebo: Analysis of FDA Reports,” is used for this exercise.

Table 1 in that paper (page 791) provides confidence intervals for suicide rates for patients taking three different kinds of medications. For example, the 95% confidence interval for “Placebo” is 0.01 to 0.19. This interval means that we can be 95% confident that between 0.01% and 0.19% of the population of people similar to the ones in these studies would commit suicide while taking a placebo. Notice that 0.01% is about 1 out of 10,000 people and 0.19% is about 1 out of 526 people, so this is a wide interval in terms of the actual suicide rate. Also note that the population from which the samples were drawn consisted of depressed patients who sought medical help. a. From Table 1 in the article, the 95% confidence interval for the suicide rate for patients taking “selective serotonin reuptake inhibitors” (SSRIs) is from 0.1% to 0.2%. Write a sentence interpreting this interval. b. Compare the interval for patients taking placebos (0.01% to 0.19%) with the one of those taking SSRIs (0.10% to 0.20%). What can you conclude about the effectiveness of SSRIs in preventing suicide, based on a comparison between the two intervals? c. The results were based on double-blind randomized experiments. Based on that information, can you conclude that differences that might have been observed in suicide rates for the different groups were caused by the difference in type of drug treatment? Explain.

Answer: a. We can be 95% confident that between 0.10% and 0.20% of the population of people similar to the ones in these studies would commit suicide while taking SSRIs. b. The two intervals have very substantial overlap, so it’s possible that there is no difference in the effectiveness of placebo and SSRIs for preventing suicide in the population. c. The results are based on randomized experiments, so a cause-and-effect conclusion can be made. It appears that SSRIs may not be any more effective than placebos in preventing suicides.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

The following information is for Exercises 27 to 30: Original Source 10 (link provided on the companion website), titled “Religious attendance and cause of death over 31 years,” provides 95% confidence intervals for the relative risk of death by various causes for those who attend religious services less than weekly versus weekly. (The relative risk is called “relative hazard” in the article and adjustments already have been made for some confounding factors such as age.)

27. The 95% confidence interval for “All Causes” is 1.06 to 1.37. Explain what this interval means in a few sentences that would be understood by some-one with no training in statistics. Make sure your explanation applies to the correct population. (Refer to Additional News Story 10 in the Appendix.)

Answer: The interval is 1.06 to 1.37. We can be fairly confident that the risk of dying from all causes during the time this study took place was lower for the population of people attending church at least once a week than for those attending less often or not at all. Based on results from the study, we can be fairly confident that the risk of dying for the population of those not attending church at least once a week ranges from 1.06 to 1.37 times the risk of dying for those who do attend church as least once a week.

28. What value of relative risk would indicate equal risk for those who attend and do not attend religious services at least weekly?

Answer: A relative risk of 1.0 would indicate equal risk.

29. Confidence intervals are given for the relative risk of death for five specific causes of death, as follows: Circulatory (1.02 to 1.45), Cancer (0.83 to 1.37), Digestive (0.98 to 4.03), Respiratory (0.92, 3.02), and External (0.60 to 2.12). For which of these causes can it be concluded that the risk of death in the population is lower for those who attend religious services at least weekly?

Explain what criterion you need to determine your answer. (Hint: Refer to Exercise 28.)

Answer: If the confidence interval for relative risk is completely above 1.0, then this indicates that the risk of death in the population is lower for those who attend religious services at least weekly. This holds for circulatory disease, but not for any of the other causes of death.

30. Read Additional News Story 10 in the Appendix to determine whether the results are based on an observational study or a randomized experiment. Using that information, explain whether it can be concluded that attending religious services at least weekly causes the risk of death to change, for causes of death for which the confidence interval indicates there is such a change.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

Answer: This was an observational study. Thus, it cannot be concluded that attending regular church services causes a change in risk of death.

31. Read the summary of News Story 11 in the Appendix, and refer to the link to Original Source 11 on the companion website, “Driving impairment due to sleepiness is exacer bated by low alcohol intake.” Table 1 of the article presents mean blood alcohol concentrations (BAC) and standard errors of the mean for the participants before driving. The values for the “With sleep restriction” and “No sleep restriction” conditions are shown in Table 21.6. For the sleep restriction condition, 12 men between the ages of 20 and 26, all experienced drivers, slept 5 hours on a given night, and then at lunchtime the next day ate two cheese rolls and drank 75 mL of 37.5% proof vodka. About half an hour after drinking, their blood alcohol concentration (BAC) was measured. The “no sleep restriction” condition was similar (and used the same participants), but without limiting sleep to 5 hours.

Table 21.6

Blood Alcohol Concentrations (mg alcohol/100 mL blood) before Driving With Sleep Restriction

No Sleep Restriction

Mean

38.3

32.4

Standard error of the mean

4.0

4.0

a. Compute approximate 95% confidence intervals for the mean BAC before driving for each of the two conditions. (Note that the sample sizes are smaller than may be appropriate for using the method in this chapter, but that will cause the in tervals to be just slightly too narrow, or alternatively, the confidence to be slightly less than 95%. You can ignore that detail.) b. Interpret the interval for the “With sleep restriction” condition. Make sure your explanation refers to the correct population. c. Ignoring the fact that the sample size is slightly too small, would it be appropriate to compute a confidence interval for the difference in the two means using the method provided in this chapter? If so, compute the interval. If not, explain why not.

Answer: a. The confidence intervals are Mean ± 2 standard errors. In both cases the standard error is given as 4, so the intervals are Mean ± 8. With sleep restriction: 38.3 ± 8 or 30.3 to 46.3 mg alcohol per 100 ml of blood. No sleep restriction: 32.4 ± 8 or 24.4 to 40.4 mg alcohol per 100 ml of blood. b. Twelve men between the ages of 20 and 26, all experienced drivers, slept 5 hours on a given night, then at lunchtime the next day ate two cheese rolls and drank 75 ml of 37.5% proof vodka. About half an hour after drinking, their blood alcohol concentration (BAC) was measured. The average for the

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

12 men in the study was 38.3 mg alcohol per 100 ml of blood. We are 95% confident that the mean BAC for all men similar to these, under the same conditions, would be between 30.3 and 46.3 mg alcohol/100 ml blood. c. It would not be appropriate because the same men were used in both conditions. Therefore, the samples are not independent.

32. A long-term study in Britain followed the cognitive development of a cohort of babies born in March 1946 (Richards et al., 2001). The goal was to determine whether there was an association between birthweight and cognitive function.

Cognitive function was measured starting at age 8 and continuing to age 43.

The researchers divided birthweights into five ranges and then compared those in the middle range with those in each of the other four ranges. They provided 95% confidence intervals for the difference in mean standardized cognitive scores for midrange birthweight babies (3.01 to 3.50 kg) and babies in each of the four other birthweight groups. The comparisons are made at various ages.

For scores at age 8, the intervals are –0.42 to –0.11 for low weight (0 to 2.50 kg at birth); –0.16 to +0.03 for low-normal weight (2.51 to 3.00 kg at birth); 0.05 to 0.21 for high-normal weight (3.51 to 4.00 kg at birth); and –0.08 to +0.14 for very high-normal weight (4.01 to 5.00 kg at birth). a. Write a sentence or two interpreting the interval for the low-weight group.

Make sure your explanation applies to the correct population. b. What difference value would indicate that the two means are equal in the population? c. For which of the birthweight groups is it clear (with 95% confidence) that there is a difference in the population mean standardized cognitive scores for that group and the midrange group? What criterion did you use to decide? In each case, explain whether the difference indicates that the mean standardized cognitive score is higher or lower for that group than for the midrange birthweight group. d. Do the results of this study imply that low birth weight causes lower cognitive scores? Explain.

Answer: a. We are 95% confident that the average cognitive score at age 8 for children who had a low birth weight (0 to 2.50 kg) is between 0.11 and 0.42 lower than the average cognitive score for children whose birth weights were closer to average (3.01 to 3.50 kg). The study was done on babies born in the United Kingdom in 1946, but should extend to babies born under similar conditions elsewhere. b. A difference of 0 would indicate that the population means are equal. c. There is a clear difference for any groups where the confidence interval does not cover 0. The low birth weight group clearly has a lower average

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

score than the midrange group and the high-normal birth weight group clearly has a higher average score than the midrange group. d. A cause and effect conclusion cannot be made because this is obviously an observational study. Birth weight cannot be randomly assigned. There are many potential confounding variables, such as mother’s diet, quality of health care available, and so on.

ANSWERS TO THE REFLECTION QUESTIONS 1. The standard confidence level to use in most applications is 95%, but sometimes other levels are used. a. What is one advantage of a 99% confidence interval compared to a 95% confidence interval? b. What is one advantage of a 95% confidence interval compared to a 99% confidence interval?

Answer: a. You can state the conclusion with 99% confidence instead of only 95% confidence. In other words, in the long run 99% confidence intervals will not cover the true population value only 1% of the time, compared to 5% of the time for 95% confidence intervals. b. To have more confidence, you need to use a larger multiplier. Therefore, a 99% confidence interval will be wider than a 95% confidence interval, giving a bigger range of possibilities for the true population value.

2. Multiple examples in this chapter included an intervention group and a control group. For both groups, the researchers took measurements at baseline (the start of the study) and after an intervention period. What information could be learned by finding a confidence interval for the difference in means for the two groups at baseline?

Answer: If participants are randomly assigned to an intervention group and a control group, the hope is that they represent populations with about the same mean value for the measurement of interest. A confidence interval for the difference in the population means at baseline would indicate whether or not that seems to be the case.

3. In an experiment with an intervention group and a control group, there are multiple ways to determine whether the intervention was successful. One is to compute a confidence interval for the mean change from baseline to after the intervention for the intervention group only. Another is to compute a confidence interval for the difference in means for the two groups at the end of the intervention period.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

a. What information could be learned from the first method that could not be learned from the second method? b. What information could be learned from the second method that could not be learned from the first method?

Answer: a. A confidence interval for the mean change from baseline to after the intervention would tell you how much the population mean would be likely to change during that time if everyone in the population were given the intervention. b. A confidence interval for the difference in means at the end of the intervention period would tell you how much the means for populations represented by those groups differed relative to each other after the intervention period. But it would not tell you whether they could have started off with different means, or how much the mean of intervention group itself was likely to change over that period.

4. In an experiment involving volunteers, to whom would the results of a 95% confidence interval apply?

Answer: The results would apply to the population of individuals similar to those in the study, and who would have volunteered to be in it if asked.

5. If a 95% confidence interval for a difference in means covers 0, can we conclude with 95% confidence that the two population means are equal?

Explain.

Answer: No, if the confidence interval covers 0, that’s just one possible value for the difference in means. Unless the entire interval is the single value of 0, there are lots of other possible values for the difference in population means, for which they are not equal.

ANSWERS TO THE MINI PROJECTS 1. Find a journal article that reports at least one 95% confidence interval. Explain what the study was trying to accomplish. Give the results as reported in the article in terms of 95% confidence intervals. Interpret the results. Discuss whether you think the article accomplished its intended purpose. In your discussion, include potential problems with the study, as discussed in Chapters 4 to 6.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 21: The Role Of Confidence Intervals In Research

Answer: This is a wide-ranging project. Make sure the confidence interval interpretation is correct, and that the interpretation is not made by assuming the interval covers 95% of all population values, which is a common mistake.

The project should also identify whether the study is an experiment, an observational study, or a survey and consider the appropriate "difficulties and disasters" from Chapters 4 and 5.

2. Collect data on a measurement variable for which the mean is of interest to you. Collect at least 30 observations. Using the data, compute an approximate 95% confidence interval for the mean of the population from which you drew your observations. Explain how you collected your sample, and note whether your method would be likely to result in any biases if you tried to extend your sample results to the population. Interpret the 95% confidence interval to make a conclusion about the mean of the population.

Answer: See the cautions about interpreting a confidence interval given in the notes for Mini-Project 21.1. Also, make sure the project includes a discussion of possible sources of bias, including sampling bias and possible measurement bias.

3. Collect data on a measurement variable for which the difference in the means for two conditions or groups is of interest to you. Collect at least 30 observations for each condition or group. Using the data, compute an approximate 95% confidence interval for the difference in the means of the populations from which you drew your observations. Explain how you collected your samples, and note whether your method would be likely to result in any biases if you tried to extend your sample results to the populations. Interpret the 95% confidence interval to make a conclusion about the difference in the means of the populations or conditions.

Answer: Make sure the two groups or conditions are independent of each other and that they are sampled independently. For example, if the two groups consist of children and adults, don't use a child and an adult from the same family (unless you can be sure their measurements would not be correlated). Be sure that the interpretation is done correctly, and that it applies to the difference in the two population means, and not 95% of all possible differences in the measurements themselves. Finally, be sure possible biases are discussed concerning the choice of units to measure and the taking of the measurements themselves.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 22: Rejecting Chance — Testing Hypotheses In Research

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 22: REJECTING CHANCE — TESTING HYPOTHESES IN RESEARCH

TABLE OF CONTENTS Answers to the Exercise Questions ....................................................................................1 Answers to the Reflection Questions ...............................................................................15 Answers to the Mini Projects ........................................................................................... 16

ANSWERS TO THE EXERCISE QUESTIONS 1. In Example 21.4, a study was described that tested the use of chewing gum to reduce visually induced motion sickness (VIMS), commonly experienced in virtual reality sessions (Kaufeld et al., 2022). Two different flavors of chewing gum were tested in the experiment: peppermint and ginger. One of the outcome measures was the nausea portion of a “simulator sickness questionnaire,” or SSQ, filled out by each participant before and after a 15-minute virtual reality session. Suppose we wanted to conduct a hypothesis test to determine whether the mean SSQ values would differ if everyone in the population were to chew peppermint gum versus if they were to chew ginger gum. What would the null and alternative hypotheses be for the test?

Answer: Null hypothesis: The population mean SSQ value if everyone were to chew peppermint gum is the same as the population mean SSQ value if everyone were to chew ginger gum instead.

Alternative hypothesis: The population mean SSQ value if everyone were to chew peppermint gum is different than the population mean SSQ value if everyone were to chew ginger gum.

2. Refer to Exercise 1. If we had conducted the hypothesis test, the resulting p-value would be 0.28. Explain what the p-value represents for this example.

Answer: Suppose the population mean SSQ values would be the same if everyone were to chew peppermint gum and if everyone were to chew ginger gum. Then we would expect to see a sample difference with absolute value as large as or larger than the one we saw in about 28% of all experiments that used the same size samples as this experiment.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 22: Rejecting Chance — Testing Hypotheses In Research

3. Refer to Case Study 21.1, in which participants were randomly assigned to take a week off of social media or to continue to use it as usual. One outcome measure was a well-being test. Researchers wanted to know if population mean scores on the test would be higher after a week with no social media, compared with usual social media use (Lambert et al., 2022). a. What are the null and alternative hypotheses tested in this experiment? b. The researchers concluded that a week off of social media would result in higher mean well-being scores at the end of the week. Which type of error could they have made? c. What would be the consequences of making a type 1 error in this experiment? What would be the consequences of making a type 2 error?

Answer: a. Null hypothesis: Population mean score on the well-being test would be the same with and without taking a week off of social media.

Alternative hypothesis: Population mean score on the well-being test would be higher after taking a week off of social media than if using social media as usual for the week. b. They could have made a type 1 error, concluding taking a week off of social media would result in a higher mean well-being score when it would not. c. The consequence of a type 1 error is that we conclude that taking a week off of social media would improve well-being, when in fact it would not, so people might take a week off when in fact there is no benefit. The consequence of a type 2 error is that we don’t recognize that taking a week off of social media would benefit well-being, thus not reporting on a benefit that could help people’s well-being.

4. The journal article reporting the experiment described in Case Study 2.1 (Beezhold et al., 2015) found that vegans had lower anxiety and stress scores than omnivores. To help isolate the reason, they also tested how the two groups differed on other measures. The measures and the resulting p-values are given in parts (a) to (c). For each of the following comparisons, state the null and alternative hypotheses and the appropriate conclusion: a. Hours outdoors per week, p-value = 0.003 b. The proportion who regularly practice a stress reduction technique such as yoga, p-value = 0.012 c. Average hours of sleep per night, p-value = 0.235 Answer: Notes for all parts: The wording used for each part is slightly different, to illustrate that there are various ways the hypotheses can be worded. Also, technically, for all parts of this question you should add that the populations are people similar to those represented by the participants in this study.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 22: Rejecting Chance — Testing Hypotheses In Research

a. Null hypothesis: The mean number of hours spent outdoors per week is the same for the population of vegans and the population of omnivores.

Alternative hypothesis: The mean number of hours spent outdoors per week differs for the population of vegans and the population of omnivores.

Conclusion: Because the p-value is only 0.003, the null hypothesis is rejected (using a level of 0.05), so we can conclude that the mean number of hours spent outdoors per week differs for the populations of vegans and omnivores. b. Null hypothesis: The proportion of the population of vegans that regularly practices a stress reduction technique is the same as that proportion is for the population of omnivores.

Alternative hypothesis: The proportions of the populations of vegans and omnivores who regularly practice a stress reduction technique are different.

Conclusion: Because the p-value is only 0.012, the null hypothesis is rejected (using a level of 0.05), so we can conclude that the proportions of the populations of vegans and omnivores who regularly practice a stress reduction technique differ. c. Null hypothesis: The population mean for the average hours of sleep someone gets per night is the same for the populations of vegans and omnivores.

Alternative hypothesis: The population means for average hours of sleep per night differ for the populations of vegans and omnivores.

Conclusion: The p-value is 0.235, so do not reject the null hypothesis at the 0.05 level. Conclude that the population means for the average hours of sleep someone gets per night are not significantly different for vegans and omnivores.

5. The news story discussed in Case Study 6.5 (page 128) was headlined “NIH study finds that coffee drinkers have lower risk of death” (Source: http://www.nih.gov). The news story was based on an article published in The New England Journal of Medicine (Freedman et. al., 2012) that followed hundreds of thousands of older adults from 1995 to 2008 and examined the association between drinking coffee and longevity. a. What are the null and alternative hypotheses for this study? b. The authors concluded that those who drank coffee had a statistically significantly lower risk of death (during the time period of the study) than those who did not. Which type of error, type 1 or type 2, could have been made in making this conclusion? c. Explain what a type 1 error and a type 2 error would be in this situation.

Which one do you think is more serious? Explain.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 22: Rejecting Chance — Testing Hypotheses In Research

Answer: a. Null hypothesis: There is no association between drinking coffee and longevity for the population of people similar to those in this study.

Alternative hypothesis: It is not clear whether the researchers initially thought coffee drinking would be associated with an increase in longevity, or just that there would be a difference. The two-sided alternative is: There is an association between drinking coffee and longevity for the population of people similar to those in this study. The one-sided alternative is: Drinking coffee is associated with increased longevity for the population of people similar to those in this study. b. A type 1 error could have been made, concluding that there is an association when in fact there is no association in the population. c. A type 1 error could be made by concluding that there is an association between drinking coffee and longevity in the population, when in fact there is not. A type 2 error would be made by not concluding that there is an association between drinking coffee and longevity, when in fact there is an association in the population. A type 2 error is probably more serious in this case because if coffee drinking really is associated with longevity and it isn’t discovered, an important medical fact would be missed. (Note that either answer is okay as long as the reasoning is sound.)

6. An article in Science News reported on a study to compare treatments for reducing cocaine use. Part of the results are “short-term psychotherapy that offers cocaine abusers practical strategies for maintaining abstinence sparks a marked drop in their overall cocaine use. . . . In contrast, brief treatment with desipramine—a drug thought by some researchers to reduce cocaine cravings—generates much weaker drops in cocaine use” (Bower, 24 and 31 December 1994, p. 421). a. The researchers were interested in comparing the rates of cocaine use following treatment with psychotherapy and treatment with desipramine.

State the null and alternative hypotheses for this situation. b. Explain what the two types of error could be for this situation and what their consequences would be. c. Although no p-value is given, the researchers presumably concluded that the psychotherapy treatment was superior to the drug treatment. Which type of error could they have made?

Answer: a. Null hypothesis: Psychotherapy and desipramine are equally effective in treating cocaine use. Alternative hypothesis: One method is more effective than the other. b. A type 1 error would occur if there is no difference in the treatments, but the conclusion is that they do differ. A type 2 error would occur if one © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 22: Rejecting Chance — Testing Hypotheses In Research

method is more effective than the other treatment but the study fails to make that conclusion. c. They could have made a type 1error, concluding there is a difference when there isn’t.

7. State the null and alternative hypotheses for each of the following potential research questions: a. Does working 5 hours a day or more at a computer contribute to deteriorating eyesight? b. Does placing babies in an incubator during infancy lead to claustrophobia in adult life? c. Does placing plants in an office lead to fewer sick days?

Answer: a. Null hypothesis: Working 5 hours a day or more at a computer does not contribute to deteriorating eyesight.

Alternative hypothesis: Working 5 hours a day or more at a computer does contribute to deteriorating eyesight. b. Null hypothesis: Placing infants in incubators does not lead to claustrophobia in adult life.

Alternative hypothesis: Placing infants in incubators leads to claustrophobia in adult life. c. Null hypothesis: Placing plants in an office does not reduce number of sick days taken by the people in the office.

Alternative hypothesis: Placing plants in an office reduces the number of sick days taken by the people in that office.

8. For each of the situations in Exercise 7, explain the two errors that could be made and what the consequences would be.

Answer: a. Type 1 error is that working 5 or more hours at a computer does not harm eyesight but the research concludes that it is harmful. Millions of workers would have needless concern. Type 2 is that there really is a harmful effect but it goes undetected. Millions of workers who should be warned about this problem would not learn about it. b. Type 1 error is that incubators do not cause claustrophobia but the research finds that they do. As a consequence, incubator use might be reduced or counseling offered to people who were really okay. Type 2 is that incubators do cause a problem but it goes undetected. Those who were put in incubators would wonder why they had claustrophobia and

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 22: Rejecting Chance — Testing Hypotheses In Research

would not be told there was a reason, knowledge that might help them overcome it. c. Type 1 error is that plants have no impact but the research finds that they do; as a consequence companies would needlessly offer plants to their employees. Type 2 error is that plants do help reduce sick days, but the link goes undetected. A useful strategy would go unused.

9. In Case Study 6.1, a study was described that compared scores on a mock GRE exam for students who had learned and practiced meditation and students who had taken a nutrition class (Mrazek et al., 2013). Although the analysis used by the researchers was somewhat complicated, for simplicity in this exercise, suppose the measure of interest was the difference in mean GRE scores for the two groups. The researchers were interested in showing that practicing meditation boosts mean GRE scores. a. The null hypothesis can be stated as “Population mean difference = null value.” What is the null value? b. Write the alternative hypothesis for this situation. c. Describe what a type 1 error would be in this situation. d. Describe what a type 2 error would be in this situation. e. Discuss the consequences of the two types of errors in this situation.

Which error do you think is more serious? (There is no correct answer; it is your reasoning that counts.)

Answer: a. The null value is 0. b. Population mean GRE score if everyone were to practice meditation is higher than the population mean GRE score if everyone were to take a nutrition class. c. Meditation does not increase mean GRE score any more than taking a nutrition class does, but we conclude that it does. d. Meditation does increase mean GRE score any more than taking a nutrition class does, but we fail to make that conclusion. e. It seems like a type 2 error would be more serious because a useful method for increasing test scores would go undiscovered. But you could also argue that a type 1 error would be more serious because people might spend time and money on a method they thought would increase their tests scores, when it would not.

10. Explain why we can specify the probability of making a type 1 error, given that the null hypothesis is true, but we cannot specify the probability of making a type 2 error, given that the alternative hypothesis is true.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 22: Rejecting Chance — Testing Hypotheses In Research

Answer: We do not specify a particular value for the population when we state the alternative hypothesis, so there is no way to determine what to expect by chance as there is when we assume the null hypothesis is true.

11. Compute a 95% confidence interval for the probability of a successful session in the ganzfeld studies reported in Case Study 22.1. There were 355 sessions and 122 were successful.

Answer: The standard deviation is the square root of (0.344)(0.656)/355 = 0.025; a 95% confidence interval is 0.344 ± 2(0.025) or 0.294 to 0.394. The interval corresponds to getting a correct answer between 29.4% and 39.4% of the time.

12. Specify what a type 1 and a type 2 error would be for the ganzfeld studies reported in Case Study 22.1.

Answer: A type 1 error would be that no ESP is present but the study concludes it is being used. A type 2 error would be that ESP is indeed present but the study does not detect it.

13. Given the convention of declaring that a result is statistically significant if the p-value is 0.05 or less, what conclusion would be made concerning the null and alternative hypotheses in each of the following cases? Be explicit about the wording of the conclusion. a. p-value = 0.35 b. p-value = 0.04 Answer: a. You could say either of the following. “Do not reject the null hypothesis” or “There is no statistically significant difference (relationship, effect, etc.).” b. You could say any of the following. “Reject the null hypothesis” or “Accept the alternative hypothesis” or “Conclude that there is a statistically significant difference (relationship, effect, etc.).”

14. For the research described in Case Study 6.2, the goal was to find out if eating breakfast cereal was associated with a reduction in body mass index (BMI) for children (Frantzen et al., 2013). Although the analysis used by the researchers was somewhat complicated, for simplicity in this exercise, suppose the measure of interest was the difference between population mean BMI for children who eat cereal and children who don’t. a. Write the null and alternative hypotheses for this situation. b. Describe what a type 1 error would be in this situation. c. Describe what a type 2 error would be in this situation.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 22: Rejecting Chance — Testing Hypotheses In Research

d. Which type of error do you think is more serious in this situation? (There is no correct answer; it is your reasoning that counts.)

Answer: a. Null hypothesis: There is no difference in mean BMI for the population of children who eat cereal and the population of children who don’t eat cereal.

Alternative hypothesis: The population of children who eat cereal has a lower mean BMI than the population of children who don’t eat cereal. b. A type 1 error would occur if there is no difference in population mean BMI, but the study concludes that the mean is lower for children who eat cereal. c. A type 2 error would occur if the population mean BMI is lower for children who eat cereal than for those who don’t, but the study fails to find a significant difference. d. A type 2 error is probably more serious, especially given the obesity epidemic. If eating cereal is associated with lower BMI, then that would be useful information even though this is an observational study and a cause/effect conclusion cannot be made.

15. In previous chapters, we learned that researchers have discovered a link between vertex baldness and heart attacks in men. a. State the null hypothesis and the alternative hypothesis used to investigate whether there is such a relationship. b. Discuss what would constitute a type 1 error in this situation. c. Discuss what would constitute a type 2 error in this situation.

Answer: a. Null hypothesis: There is no link between vertex baldness and heart attacks in men.

Alternative hypothesis: There is a link between vertex baldness and heart attacks in men. b. A type 1 error would occur if there is no link between the two but the study concludes that there is. c. A type 2 error would occur if there is a link between vertex baldness and heart attacks but the evidence in the sample wasn’t strong enough to conclusively detect it.

16. A report in the Davis (CA) Enterprise (6 April 1994, p. A-11) was headlined, “Highly educated people are less likely to develop Alzheimer’s disease, a new study suggests.” a. State the null and alternative hypotheses the researchers would have used in this study.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 22: Rejecting Chance — Testing Hypotheses In Research

b. What do you think the headline is implying about statistical significance?

Restate the headline in terms of statistical significance. c. Was this a one-sided test or a two-sided test? Explain.

Answer: a. Null hypothesis: Having a high level of education has no relationship with developing Alzheimer’s disease.

Alternative hypothesis: Highly educated people have a lower chance of developing the disease. b. The headline is implying that a study found a statistically significant relationship. “A new study found that the likelihood of highly educated people developing Alzheimer’s disease was statistically significantly lower than the likelihood for others.” (Notice that a causal conclusion is not justified.) c. It was a one-sided test. The researchers specified a direction in the alternative hypothesis, namely, that if there was a relationship between education and developing Alzheimer’s disease, it would be that those with more education were less likely to develop the disease.

17. Suppose that a study is designed to choose between the hypotheses: Null hypothesis: Population proportion is 0.25.

Alternative hypothesis: Population proportion is higher than 0.25.

On the basis of a sample of size 500, the sample proportion is 0.29. The null standard error for the potential sample proportions in this case is about 0.02. a. Compute the standardized score corresponding to the sample pro-portion of 0.29, assuming the null hypothesis is true. b. What is the percentile for the standardized score computed in part (a)? c. What is the p-value for the test? d. Based on the results of parts (a) to (c), make a conclusion. Be explicit about the wording of your conclusion and justify your answer. e. To compute the standardized score in part (a), you assumed the null hypothesis was true. Explain why you could not compute a standardized score under the assumption that the alternative hypothesis was true.

Answer: a. The standardized score is (0.29 − 0.25)/0.02 = 2.0. b. From Table 8.1, the percentile is between 97.5 and 98, so about the 97.75th percentile. Using Excel, it is the 97.725th percentile. c. The corresponding p-value is about 0.023, or to be exact, 0.02275. d. The p-value is 0.023, which is less than 0.05, so the null hypothesis is rejected. We can conclude that the population proportion is higher than 0.25.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 22: Rejecting Chance — Testing Hypotheses In Research

e. The alternative hypothesis contains a whole range of values, from just above 0.25 to 1. Therefore we would not know what to use as the mean and standard deviation.

18. Consider medical tests in which the null hypothesis is that the patient does not have the disease and the alternative hypothesis is that the patient does. a. Give an example of a medical situation in which a type 1 error would be more serious. b. Give an example of a medical situation in which a type 2 error would be more serious.

Answer: a. Answers will vary, but in general a type 1 error would be more serious for a minor disease with serious treatment, like tonsillitis where the treatment is surgery. b. Answers will vary, but in general a type 2 error would be more serious for an infectious or dangerous disease, because it would go undetected. An example is being infected with HIV or having a fast-spreading cancer, but these are not detected by the test.

19. Consider the following quote (hypothetical but based on real data). “In a recent survey of 1000 randomly selected shoppers at grocery stores, 54% reported that they prefer checking out with a cashier rather than using selfcheckout. If there is no difference in the proportions who prefer each type in the population, the chance of getting results that different from 50y50 or more so in a sample this size is about 0.012. Because 0.012 is less than 0.05, we can conclude that there is a statistically significant difference in preference in the population of shoppers.” This quote is giving the results of a hypothesis test for a proportion. Give a numerical value for each of the following. a. The sample proportion b. The null value being tested c. The sample size d. The p-value e. The level of significance Answer: a. The sample proportion is 0.54. It represents the proportion of the sample that prefers checking out with a cashier. b. The null value is 0.5, which would indicate no preference. c. The sample size is 1000. d. The p-value is 0.012. e. The level of significance is given as 0.05.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 22: Rejecting Chance — Testing Hypotheses In Research

20. In Case Study 1.2 and in Chapters 12 and 13, we examined a study showing that there appears to be a relationship between taking aspirin and incidence of heart attack. The null hypothesis in that study would be that there is no relationship between the two variables, and the alternative would be that there is a relationship. Explain what a type 1 error and a type 2 error would be for the study and what the consequences of each type would be for the public.

Answer: Type 1 error would be that there is no relationship between aspirin and heart attack but the study finds one. People would start taking aspirin needlessly.

Type 2 error is that aspirin does help prevent heart attacks but the study does not discover the relationship, so people who could be helped by aspirin are not told to take it.

21. Many researchers decide to reject the null hypothesis as long as the p-value is 0.05 or less. In a testing situation for which a type 2 error is much more serious than a type 1 error, should researchers require a higher or a lower p-value in order to reject the null hypothesis? Explain your reasoning.

Answer: If a type 2 error is very serious, researchers should make it easier to reject the null hypothesis when indeed it should be rejected. They would do that by using a p-value cutoff higher than 0.05 for rejecting the null hypothesis, thus increasing the probability of making a type 1 error and reducing the probability of making a type 2 error.

22. In Case Study 1.1, Lee Salk did an experiment to see if hearing the sound of a human heartbeat would help infants gain weight during the first few days of life. By comparing weight gains for two sample groups of infants, he concluded that it did. One group listened to a heartbeat and the other did not. a. What are the null and alternative hypotheses for this study? b. Was this a one-sided test or a two-sided test? Explain. c. What would a type 1 and type 2 error be for this study? d. Given the conclusion made by Dr. Salk, explain which error he could possibly have committed and which one he could not have committed. e. Rather than simply knowing whether there was a difference in average weight gains for the two groups, what statistical technique would have provided additional information?

Answer: a. Null hypothesis: Listening to the sound of a heartbeat in the first few days of life does not influence weight gain of infants.

Alternative hypothesis: Infants who hear the sound of a heartbeat in the first few days of life gain more weight than those who do not.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 22: Rejecting Chance — Testing Hypotheses In Research

b. This was a one-sided test. Dr. Salk’s research hypothesis was that listening to the heartbeat sound would cause the infants to gain more weight. c. A type 1 error could occur if there really is no benefit from listening to a heartbeat, but the study finds that there is. A type 2 error could occur if there really is a benefit from listening to the heartbeat but the study fails to detect it. d. Because Salk rejected the null hypothesis, he could have only committed a type 1 error. A type 2 error can be committed only when the null hypothesis is not rejected. e. It would have been useful to see a confidence interval for the difference in average weight gain under the two conditions.

23. In recent years, the research community has placed considerable emphasis on the idea that studies should be replicated before they can be considered to be conclusive. One reason for this is that it is easier to get a study published if the results are statistically significant than if they are not. a. If 100 studies are conducted and, in all cases, the null hypothesis is true, approximately how many of them will find statistical significance at the 0.05 level of significance? b. Is a study more likely to get published if the researchers have made a type 1 error or if they have made a type 2 error? Explain. c. Rule 3 from Chapter 14 says that if two events are independent, the probability that they both happen is the product of their individual probabilities. Suppose two researchers independently test a null hypothesis that is in fact true. What is the probability that they will both find statistical significance at the 0.05 level? d. Use the result from part (c) to explain why the research community places an emphasis on replicating significant results.

Answer: a. About 5% of the 100, or 5 studies, will find statistical significance at the 0.05 level. b. A study is more likely to get published if the null hypothesis is rejected, which can happen when a type 1 error is made but not when a type 2 error is made. c. The probability for each one is 0.05, so the probability that they will both find statistical significance is (0.05)(0.05) = 0.0025, or 1/400. d. When null hypotheses are true, a single study will achieve statistical significance with probability 0.05 or 1/20. But two independent studies will both achieve statistical significance with a very low probability of 0.0025, or 1/400. Therefore, true null hypotheses are much less likely to get published if replication is required.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 22: Rejecting Chance — Testing Hypotheses In Research

24. Legends about full moons are abundant, and one of them is that more babies are born during the 24-hour period surrounding a full moon. (A study of this idea would exclude scheduled births such as induced or Cesarean births.) Full moons occur on average every 29.53 days, so if the legend is true, the proportion of births occurring during a full moon period should be about 1/29.53. a. What is the population parameter of interest for testing this legend? b. Write the null and alternative hypotheses that would be used to test this legend. c. Explain how a researcher could collect data to test this legend.

Answer: a. The population parameter is the proportion of babies born during a full moon period. b. Null hypothesis: The proportion of babies born during a full moon period is 1/29.53.

Alternative hypothesis: The proportion of babies born during a full moon period is greater than 1/29.53. c. For a large number of unscheduled births, record whether or not they occurred during a full moon period.

25. For the study described in Example 22.3 in which researchers tested the results of a sniff test that matched identical twins at a higher than chance level, they also tested the same hypothesis for fraternal (dizygotic, or DZ) twins. The result was 35 matches out of 113 tests, which is 0.31, or 31% successes. a. Using Example 22.3 as a guide, state the null and alternative hypotheses. b. The p-value for the appropriate test is 0.089. Using the 0.05 level of significance, what conclusion would be made? Write your answer in statistical terms and in words that would be understood by someone with no training in statistics. c. In their conclusion, the researchers wrote that the results for DZ twins were not significantly better than chance. Why do you think the researchers used that wording, rather than concluding that the results were “no better than chance”?

Answer: a. Null hypothesis: The probability of being able to match the smell of one fraternal twin with that of the other when given four choices is 0.25.

Alternative hypothesis: The probability of being able to match the smell of one fraternal twin with that of the other when given four choices is greater than 0.25.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 22: Rejecting Chance — Testing Hypotheses In Research

b. With a p-value of 0.089 and a 0.05 level of significance, do not reject the null hypothesis. The research showed that people were not able to match the smells of fraternal twins at a level significantly better than chance. c. Just because the null hypothesis was not rejected, that doesn’t mean it’s exactly true. If the researchers had written that the results were “no better than chance” that would imply that the null hypothesis was exactly true.

26. For the study described in Example 22.3 in which researchers tested the results of sniff tests for identical and fraternal twins, they also tested whether the sniff tests could be used to identify odor samples from the same person.

They presented each sniffer with an odor sample from someone, and then with four vials, one of which contained another sample from that same person. Out of 120 tests, there were 41 correct matches. The alternative hypothesis of interest was that, in the population of people who might be asked to do this test, the probability of a correct match is greater than the 25% expected by chance. Go through the four steps of hypothesis testing for this situation.

Answer: Step 1: Null hypothesis: After being presented with an odor sample from someone, the probability of correctly identifying another odor sample from that person, when given four choices, is 0.25. In other words, it is no better than chance.

Alternative hypothesis: The probability described in the null hypothesis is greater than 0.25. In other words, it is better than expected by chance.

Step 2: The sample proportion is 41/120 = 0.34; the standard deviation is the square root of (0.34)(0.66)/120 = 0.0432. The test statistic is standardized score = (0.34 – 0.25)/0.0432 = 2.08.

Step 3: The p-value is about 0.02.

Step 4: Reject the null hypothesis because 0.02 < 0.05. Conclude that people can match the odor sample of an individual to another odor sample from that person at higher than chance when given four choices.

27. An NPR/PBS NewsHour/Marist Poll conducted in July 2019 asked the question “Do you think free college tuition at public colleges or universities is a good idea?” (https://pollingreport.com/ed.htm). Of the 1346 adults who responded, 53% said it is a good idea. The poll was based on a random sample of adults in the United States. Based on this poll, can it be concluded that a majority (more than 50%) of adults in the United States at that time would answer that free college tuition is a good idea? Go through the four steps of hypothesis testing for this situation.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 22: Rejecting Chance — Testing Hypotheses In Research

Answer: Step 1: The null hypothesis is that the population proportion in 2019 who thought free college tuition at public colleges and universities was a good idea was 0.50, and the alternative hypothesis is that the proportion was greater than 0.50.

Step 2: The standard deviation is the square root of (0.5)(0.5)/1346 = 0.0136.

The test statistic is standardized score = (0.53 – 0.5)/0.0136 = 2.20.

Step 3: The p-value is about 0.014.

Step 4: Reject the null hypothesis because 0.014 < 0.05. Conclude that a majority of adults at that time thought free tuition was a good idea.

ANSWERS TO THE REFLECTION QUESTIONS 1. If a researcher is able to measure the entire population, is it necessary to conduct a hypothesis test? Explain.

Answer: If the researcher can measure the entire population, it is not necessary to conduct a hypothesis test. The value of the population parameter(s) can be computed from the data directly.

2. In Chapter 21, some examples were presented in which a confidence interval was used to make a conclusion about a population value. For example, if a 95% confidence interval for relative risk for two groups did not include 1, it was concluded that the risks differed for the two groups in the population. In that case, what null and alternative hypotheses were being tested?

Answer: The null hypothesis in the situation described is that the relative risk for the population is 1.0, indicating that the population risk is the same for the two groups or conditions. The alternative hypothesis is that the risks are not the same for the two populations.

3. The most common level of significance used in hypothesis testing is 0.05, but sometimes researchers use higher or lower values such as 0.01 or 0.10. Explain how the seriousness of making a type 1 error should be used to guide the choice of significance level.

Answer: A type 1 error can only be made when the null hypothesis is true but it is rejected. Therefore, if a type 1 error is serious, a significance level should be chosen that makes it harder to reject the null hypothesis when it is true, such as 0.01 or 0.001. If a type 1 error is not very serious, then a significance level should be chosen that makes it easier to reject the null hypothesis when it is true, such as 0.10.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 22: Rejecting Chance — Testing Hypotheses In Research

4. Suppose you wanted to test whether the proportion of adults who drink coffee is the same in France and in the United States. What would the null and alternative hypotheses be?

Answer: The null hypothesis would be that the proportion of the population of all adults in the United States who drink coffee is the same as the proportion of the population of all adults in France who drink coffee. The alternative hypothesis would be that the two proportions described in the null hypothesis differ from each other.

5. Suppose two researchers each conduct a test for a situation in which the alternative hypothesis is true. One uses a sample of 100 individuals, and the other uses a sample of 1000 individuals. Explain which test would be likely to have higher statistical power.

Answer: The researcher who used 1000 individuals would have higher statistical power, all other things being equal. It’s easier to make the correct conclusion with a larger sample. (This assumes that the alternative hypothesis is true. When the null hypothesis is true, the power is always equal to the level of significance used.)

ANSWERS TO THE MINI PROJECTS 1. Construct a situation for which you can test null and alternative hypotheses for a population proportion. For example, you could see whether you can flip a coin in a manner so as to bias it in favor of heads. Or you could conduct an ESP test in which you ask someone to guess the suits in a deck of cards (hearts, diamonds, spades, or clubs). (To do the latter experiment properly, you must replace the card and shuffle each time so you don’t change the probability of each suit by having only a partial deck, and you must separate the “sender” and “receiver” to rule out normal communication.) Collect data for a sample of at least size 100. Carry out the test. Make sure you follow the four steps given in Section 22.2, and be explicit about your hypotheses, your conclusion, and the reasoning behind your conclusion.

Answer: There are a number of things to check in this project. First, make sure independence holds, and that the probability of a particular outcome remains the same each time. The alternative hypothesis could be one- or two-tailed based on the context, so make sure the p-value is calculated to correspond to the alternative hypothesis. Make sure the hypothesized value and not the sample value is used in constructing the standard deviation. Finally, make sure the conclusion is worded correctly.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 22: Rejecting Chance — Testing Hypotheses In Research

2. Find two news stories reporting on the results of statistical studies with the following characteristics. First, find one that reports on a study that failed to find a relationship. Next, find one that reports on a study that did find a relationship. For each study, state what hypotheses you think the researchers were trying to test. Then explain what you think the results really imply compared with what is implied in the news reports for each study.

Answer: The report about the study that didn’t find a relationship may conclude that there is thus no relationship in the population, which is not a correct conclusion. For the study that did find a relationship, the report might try to draw a causal link, so the discussion should include an assessment of whether or not that is justified. If the study was an observational one, it probably is not justified.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 23: HYPOTHESIS TESTING—EXAMPLES AND CASE STUDIES

TABLE OF CONTENTS Answers to the Exercise Questions ....................................................................................1 Answers to the Reflection Questions ...............................................................................15 Answers to the Mini Projects ........................................................................................... 18

ANSWERS TO THE EXERCISE QUESTIONS 1. Are null and alternative hypotheses statements about populations or samples, or does it depend on the situation?

Answer: Null and alternative hypotheses are always statements about populations.

(Sample values are known – we don’t need to hypothesize about them.)

2. In a February 2013, CBS News/New York Times poll of 250 American Catholics who attend mass at least once a week, 57% answered yes to the question, “Do you favor allowing women to be priests?” (Source: https://archive.nytimes.com /www.nytimes.com/interactive/2013/03/06/us/catholics-america-poll.html) a. Set up the null and alternative hypotheses for deciding whether a majority of American Catholics at that time who attended mass at least once a week favored allowing women to be priests. b. Using Example 23.4 (pages 503–504) as a guide, compute the test statistic for this situation. c. If you have done everything correctly, the p-value for the test is about 0.0127. Based on this, make a conclusion for this situation. Write it in both statistical language and in words that someone with no training in statistics would understand.

Answer: a. Null hypothesis: It is not the case that in February 2013 a majority of American Catholics who attended mass at least once a week favored allowing women to be priests.

Alternative hypothesis: In February 2013 a majority of American Catholics who attended mass at least once a week favored allowing women to be priests.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

b. The null standard error is the square root of (0.5)(0.5)/250 = 0.032, so the standardized score is (0.57 − 0.50)/0.032 = 2.19. c. Reject the null hypothesis at the 0.05 significance level because 0.0127 < 0.05. We can conclude that in February 2013 a majority of American Catholics who attended mass at least once a week favored allowing women to be priests.

3. Refer to Exercise 2. Is the test described there a one-sided or a two-sided test?

Answer: The test is one-sided, because the alternative hypothesis includes values above 0.50 only.

4. Explain the difference between statistical significance and significance as used in everyday language.

Answer: Statistical significance means only that the data in a study are strong enough that the null hypothesis can be rejected. The finding may or may not have any significance in the usual use of the word, meaning that the finding is important. On the other hand, if a study is based on a small sample, an important, significant effect may not achieve statistical significance.

5. Suppose a one-sided test for a proportion resulted in a p-value of 0.03. What would the p-value be if the test were two-sided instead?

Answer: Double the value to get 0.06.

6. Suppose a two-sided test for a difference in two means resulted in a p-value of 0.08. a. Using the usual criterion for hypothesis testing, would we conclude that there was a difference in the population means? Explain. b. Suppose the test had been constructed as a one-sided test instead, and the evidence in the sample means was in the direction to support the alternative hypothesis. Using the usual criterion for hypothesis testing, would we be able to conclude that there was a difference in the population means? Explain.

Answer: a. No, because 0.08 > 0.05. b. Yes, the p-value would be 0.04 and is therefore less than 0.05.

7. Suppose you were given a hypothesized population mean, a sample mean, a sample standard deviation, and a sample size for a study involving a random sample from one population. What formula would you use for the test statistic?

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

Answer: Use z-score = (sample mean - hypothesized mean)/SEM; find SEM using sample the standard deviation divided by the square root of the sample size.

8. In Example 23.4, we showed that the Excel command NORM.S.DIST(z,1) gives the area below the standardized score z. Use Excel or an appropriate calculator, software, website, or table to find the p-value for each of the following examples and case studies, taking into account whether the test is one-sided or two-sided: a. Example 23.3, pages 502–503, z = 2.17, two-sided test (using z, not t) b. Example 22.1, pages 476–478, z = 2.00, one-sided test c. Case Study 22.1 pages 488–490, z = 4.09, one-sided test Answer: a. Using Excel, NORM.S.DIST(–2.17,1) = 0.015, so p-value = 2  0.015 = 0.03. b. Using Excel, NORM.S.DIST(2, 1) = 0.977, so p-value = 1 – 0.977 = 0.023. c. Using Excel, NORM.S.DIST(4.09, 1) = 0.99998, so p-value = 1 – 0.99998 = 0.00002.

9. Suppose you wanted to see whether a training program helped raise students’ scores on a standardized test. You administer the test to a random sample of students from a large university, give them the training program, then readminister the test. For each student, you record the increase (or decrease) in the test score from one time to the next. a. What would the null and alternative hypotheses be for this situation? b. Suppose the mean change for the sample was 10 points and the accompanying standard error was 4 points. What would be the standardized score that corresponded to the sample mean of 10 points? c. Based on the information in part (b), what would you conclude about this situation? (Assume the sample size is large and use z, not t.)

Answer: a. Null hypothesis: The training program has no effect on the mean test score for the population of students. Alternative hypothesis: The population mean test score would be higher after the training program, if students were to take it. b. Standardized score = (10 − 0)/4 = 2.5. c. From Table 8.1, the p-value is between 0.01 and 0.005; using Excel it is 0.006. In either case, the p-value is less than 0.05, so reject the null hypothesis (using a significance level of 0.05) and conclude that scores would be higher after the program.

10. Refer to the previous exercise. a. What explanation might be given for the increased scores, other than the fact that the training program had an impact?

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

b. What would have been a better way to design the study in order to rule out the explanation you gave in part (a)?

Answer: a. The students had already taken the test once, before the program, so when they took it afterward they were familiar with the test already. b. Randomly assign some students to take the program and others not to take it, using matched pairs if possible. Compare scores after the program.

11. An article in The New York Times included this quote: “In a recent clinical trial of emergency-room patients at the Royal University Hospital in Saskatchewan, participants who received a 10-minute visit from a therapy dog reported small but statistically significant decreases in pain, as well as lowered anxiety and depression symptoms” (Wollan, 2022). a. Write the null and alternative hypotheses that were being tested, making sure you specify the appropriate population(s), and explain what the quote tells you about the result of the hypothesis test. b. Consulting the original journal article on which the story was based (Carey et al., 2022) revealed that patients rated their pain on a scale from 0 (none) to 10 (maximum) before and after being visited by a therapy dog. For the sample, the mean reduction in pain was 0.9, and a 95% confidence interval for the population mean reduction was 0.42 to 1.32. The p-value was reported to be 0.004. Discuss whether or not you think these results are consistent with the quote from The New York Times.

Answer: a. The null hypothesis is that a 10-minute visit from a therapy dog would have no effect on pain, anxiety, or depression symptoms for emergency room patients. The alternative is that for the population of emergency room patients, those who have a 10-minute visit with a therapy dog would have a decrease in pain, anxiety, and depression symptoms, on average. (It is not clear if the alternative was one-sided, but the implication is that it was.)

The study found that the null hypothesis could be rejected, although the effect was small. b. The results are consistent with the quote. The results are statistically significant given the p-value of 0.004, but the reduction in pain of 0.9 on a scale from 0 to 10 is relatively small. Even taking the highest endpoint of the confidence interval, a reduction of 1.32 is still somewhat small.

12. Siegel (1993) reported a study in which she measured the effect of pet ownership on the use of medical care by older adults. She interviewed 938 adults age 65 and over. One of her results was reported as: “After demographics and health status were controlled for, subjects with pets had fewer total doctor contacts during the one-year period than those without pets (beta = –0.07, p < 0.05)” (1993, p. 164). © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

a. State the null and alternative hypotheses Siegel was testing. Be careful to distinguish between a population and a sample. b. State the conclusion you would make. Be explicit about the wording.

Answer: a. Null hypothesis: In the population of older adults, there is no relationship between pet ownership and number of doctor contacts, even after controlling for demographics and health status.

Alternative hypothesis: In the population of older adults, even after controlling for demographics and health status, those with pets have fewer doctor contacts on average than those without pets. b. Because we are told that the p-value is less than 0.05, we can reject the null hypothesis. We can conclude that on average, older adults with pets have significantly fewer doctor contacts than those without pets, even after controlling for demographics and health status.

3. Refer to Exercise 12. Here is another of the results reported by Siegel: “For subjects without a pet, having an above-average number of stressful life events resulted in about two more doctor contacts during the study year (10.37 vs. 8.38, p < 0.005). In contrast, the number of stressful life events was not significantly related to doctor visits among subjects with a pet” (1993, p. 164). a. State the null and alternative hypotheses Siegel is testing in this passage.

Notice that two tests are being performed; be sure to cover both. b. Pretend you are a news reporter, and write a short story describing the results reported in this exercise. Be sure you do not convey any misleading information. You are writing for a general audience, so do not use statistical jargon that would be unfamiliar to them.

Answer: a. The same set of hypotheses was tested separately for people without pets and for those with pets.

Null hypothesis: For the population of adults over 65, those with and without an above-average number of stressful life events have equal numbers of doctor visits, on average.

Alternative hypothesis: For the population of adults over 65, those with an above-average number of stressful life events have a higher number of doctor visits than those who don’t, on average. b. Based on a sample of adults over 65, researchers have concluded that for people in this age group who do not have pets, those with an above average number of stressful life events visit the doctor more often than those without above average stress. For people in this age group who have pets, there was no statistically discernable difference in number of doctor visits for those with and without an above average number of stressful life events.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

14. Refer to Example 13.2, in which we tested whether personal injuries from accidents are equally likely at intersections allowing right turn on red and intersections not allowing right turn on red. The alternative hypothesis was that personal injuries from accidents are not equally likely at the two types of intersections. Specify what a type 1 and a type 2 error would be for this example. Explain what the consequences of the two types of error would be in that context.

Answer: A type 1 error would be that personal injury accidents are equally likely at intersections allowing and not allowing right turn on red, but the conclusion is that they are not equally likely. The alternative as stated is two-sided, but assuming the sample showed more personal injuries with right turn on red than without, the consequence might be that fewer intersections are designated as allowing right turn on red when in fact they are just as safe. A type 2 error would be that there is a difference in the probabilities of accidents at intersections allowing and not allowing right turn on red, but the null hypothesis of no difference is not rejected. The consequence might be that the status quo is upheld, with no changes made to allow more or fewer intersections to have right turn on red.

15. A Quinnipiac University poll, conducted in July 2021, asked 1290 adults in the United States, “Are you worried that you or someone in your family might be affected by an extreme weather event, or is that not something you’re worried about?” (Source: https://pollingreport.com/enviro.htm). “Are worried” was chosen by 593 (47%) of the respondents. Suppose an unscrupulous politician wanted to show that climate change was not a pressing issue, and stated “Significantly fewer than half of adults in the United States are worried about extreme weather events.” a. What are the null and alternative hypotheses the politician is implicitly testing in this quote? Make sure you specify the population parameter being tested and the population to which it applies. b. Using the results of the poll, find the value of the standardized score that would be used as the test statistic. c. If you answered parts (a) and (b) correctly, the p-value for the test should be about 0.002. Explain how the politician reached the conclusion stated in the quote. d. Do you think the statement made by the politician is justified? Explain.

Answer: a. Null hypothesis: Half or more of the population of adults at the time of the poll were worried about an extreme weather event affecting them or a member of their family.

Alternative hypothesis: Fewer than half of adults were worried about an extreme weather event. To prepare for computing a test statistic, the

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

hypotheses could be restated by defining p to be the proportion of all adults at the time of the poll who were worried about an extreme weather event. In that case, the hypotheses are Null hypothesis: p ≥ 0.5; Alternative hypothesis: p < 0.5. b. Null standard error = square root of (0.5)(0.5)/1290 = 0.0139 so standardized score is z = (0.46 − 0.5)/0.0139 = 2.87. c. The p-value is about 0.002, so it’s small enough to reject the null hypothesis and accept the alternative. Thus, the observed proportion of 0.46 in the sample is statistically significantly below 0.5. d. The statement is not justified because the phrase “significantly fewer than half of adults…” would lead people to think that much fewer than half of adults thought that way. In other words, people would confuse statistical significance with practical significance, because most people do not know about and/or understand statistical significance.

16. In Example 23.3, we tested whether the average fat lost from 1 year of dieting versus 1 year of exercise was equivalent. The study also measured lean body weight (muscle) lost or gained. The average for the 47 men who exercised was a gain of 0.1 kg, which can be thought of as a loss of –0.1 kg. The standard deviation was 2.2 kg. For the 42 men in the dieting group, there was an average loss of 1.3 kg, with a standard deviation of 2.6 kg. Test to see whether the average lean body mass lost (or gained) would be different for the population of men similar to the ones in this study. Specify all four steps of your hypothesis test. (Use z instead of t.)

Answer: Step 1: Null hypothesis: Average lean body mass lost does not differ for men who diet for a year and men who exercise for a year.

Alternative hypothesis: Average lean body mass lost differs for men who diet for a year and men who exercise for a year.

Step 2: The difference in the sample means (diet – exercise) is 1.3 – (–0.1) = 1.4 kg. The standard error of the difference (SED) is the square root of (2.2)2/47 + (2.6)2/42 or about 0.51. The test statistic is thus 1.4/0.51 = 2.75.

Step 3: Using Excel, the p-value is about 0.006, which is less than 0.05.

Step 4: Since 0.006 < 0.05, reject the null hypothesis and conclude men who dieted would lose more lean body mass than men who exercised.

17. Professors and other researchers use scholarly journals to publish the results of their research. However, only a small minority of the submitted papers is accepted for publication by the most prestigious journals. In many academic fields, there is a debate as to whether submitted papers written by women are

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

treated as well as those submitted by men. In the January 1994 issue of European Science Editing (Maisonneuve, January 1994), there was a report on a study that examined this question. Here is part of that report: “Similarly, no bias was found to exist at JAMA [Journal of the American Medical Association] in acceptance rates based on the gender of the corresponding author and the assigned editor. In the sample of 1851 articles considered in this study female editors used female reviewers more often than did male editors (P < 0.001).”

That quote actually contains the results of two separate hypothesis tests.

Explain what the two sets of hypotheses tested are and what you can conclude about the p-value for each set.

Answer: One set of hypotheses concerns whether acceptance rates are lower (the alternative) or equivalent (the null) when the author or editor is female. The p-value for that test was apparently above 0.05. The other set of hypotheses concerns whether female editors chose females to review the papers more often than (the alternative) or the same amount as (the null) did male editors.

The p-value for that test was reported to be less than 0.001.

18. Use Excel or an appropriate calculator, software, website, or table to find the p-value in each of the following situations. a. Alternative hypothesis is “greater than,” t = +2.5, df = 40. b. Alternative hypothesis is “less than,” t = –2.5, df = 40. c. Alternative hypothesis is “not equal,” t = 2.5, df = 40. d. Alternative hypothesis is “greater than,” z = +1.75. e. Alternative hypothesis is “less than,” z = –1.75. f. Alternative hypothesis is “not equal,” z = –1.75.

Answer: a. 0.008 b. 0.008 c. 0.016 d. 0.04 e. 0.04 f. 0.08 19. Refer to Example 23.4, in which it was concluded that statistically significantly less than a majority of American adults in 1994 thought Bill Clinton had the honesty and integrity they expected in a president. In January 2021, shortly after supporters of former President Trump stormed the Capitol building, a Quinnipiac University poll asked the question, “Do you think that President Trump is mentally stable, or not?” The poll was asked of a national

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

representative sample of 1239 registered voters. The choices and percentage who chose them were: “Think he is” (45%), “Is not” (48%), and “Unsure/no answer” (7%) (https://pollingreport .com/trump_ad.htm). Follow the same steps used in Example 23.4 to test whether statistically significantly less than a majority of American adults in January 2021 would have responded “Think he is” to that question if asked. Go through the four hypothesis-testing steps.

Answer: Step 1: The null hypothesis is that the proportion of the population who would answer “Think he is” is 0.5, and the alternative hypothesis is that it is less than 0.5, i.e. less than a majority would answer that way.

Step 2: The sample proportion is 0.45 and the null standard error is the square root of (0.45)(0.55)/1239 = 0.014. The standardized score is (0.45 – 0.50)/0.014 = −3.57.

Step 3: The p-value is the area below −3.57, which is about 0.0002.

Step 4: Clearly the null hypothesis can be rejected. We can conclude that significantly less than a majority of the population at that time would answer “Think he is” to the question of whether President Trump was mentally stable.

The following information is for Exercises 20 to 22: In Example 23.3 (pages 502–503), we tested to see whether dieters and exercisers had significantly different average fat loss. We concluded that they did because the difference for the samples under consideration was 1.8 kg, with a standard error of 0.83 kg and a standardized score of 2.17. Fat loss was higher for the dieters.

20. Construct an approximate 95% confidence interval for the population difference in mean fat loss. Consider the two different methods for presenting results: (1) the p-value and conclusion from the hypothesis test or (2) the confidence interval. Which do you think is more informative? Explain.

Answer: 1.8 ± 2(0.83) or 0.14 to 1.94kg. The confidence interval is more informative because not only can we see that there is a real difference, which the test also tells us, but the interval tells us the possible magnitude of the difference.

21. Suppose the alternative hypothesis had been that men who exercised lost more fat on average than men who dieted. Would the null hypothesis have been rejected? Explain why or why not. If yes, give the p-value that would have been used.

Answer: No. The data would not have supported that alternative because in the sample the dieters lost more fat. So we could not have rejected the null hypothesis in favor of that alternative hypothesis.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

22. Suppose the alternative hypothesis had been that men who dieted lost more fat on average than men who exercised. Would the null hypothesis have been rejected? Explain why or why not. If yes, give the p-value that would have been used.

Answer: Yes, because the data supported this alternative. The p-value would be between 0.01 and 0.02, using Table 8.1 and the standardized score of 2.17.

Using Excel, the p-value is 0.015.

Exercises 23 to 29 refer to News Story and Original Source 1, “Alterations in Brain and Immune Function Produced by Mindfulness Meditation” (not available on the companion website). For this study, volunteers were randomly assigned to a meditation group (25 participants) or a control group (16 participants). The meditation group received an 8-week meditation training program. Both groups were given influenza shots and their antibodies were measured. This measurement was taken after the meditation group had been practicing meditation for about 8 weeks.

The researchers also measured brain activity, but these exercises will not explore that part of the study.

23. On page 565 of the article, we are told that “Participants were right-handed subjects who were employees of a biotechnology corporation in Madison, Wisconsin.” To what population do you think the results of this study apply?

Answer: The participants were right-handed volunteers in a biotechnology company in Madison, Wisconsin in the Midwestern United States. However, for the relationship studied in this randomized experiment, there should be nothing unusual about them, so the results probably apply to adults who would volunteer for a study like this one.

24. Participants were given psychological tests measuring positive and negative affect (mood) as well as anxiety at three time periods. Time 1 was before the meditation training. Time 2 was at the end of the 8 weeks of training, and Time 3 was 4 months later. One of the results reported is: “There was a significant decrease in trait negative affect with the meditators showing less negative affect at Times 2 and 3 compared with their negative affect at Time 1 [t(20) = 2.27 and t(21) = 2.45, respectively, p < 0.05 for both]. Subjects in the control group showed no change over time in negative affect (t < 1)” (Davidson et al., 2003, p. 565).

The first sentence of the quote reports the results of two hypothesis tests for the meditators. Specify in words the null hypothesis for each of the two tests.

They are the same except that one is for Time 2 and one is for Time 3. State each one separately, referring to what the time periods were. Make sure you don’t confuse the population with the sample and that you state the hypotheses using the correct one.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

Answer: For the first test the null hypothesis is: For volunteers who are trained to meditate, there will be no change on average in their negative affect (mood) from before the training to eight weeks after beginning meditation. For the second test the null hypothesis is: For volunteers who are trained to meditate, there will be no change on average in their negative affect (mood) from before the training to four months after beginning meditation.

25. a. Refer to Exercise 24. State the alternative hypothesis for each test. Explain whether you decided to use a one-sided or a two-sided test and why. b. Write the conclusion for the Time 2 test in statistical language and in plain English.

Answer: a. For the first test the alternative hypothesis is: For volunteers who are trained to meditate, there will be a decrease on average in their negative affect (mood) eight weeks after beginning meditation. For the second test, replace “eight weeks” with “four months.” A one-sided test was used because the researchers speculated in advance that meditation would help reduce negative mood. (This is stated on page 565 of Original Source 1.) b. Reject the null hypothesis. For the population of people who would volunteer for a meditation training program, eight weeks after beginning to meditate there would be a decrease in their trait negative affect (mood) on average.

26. Refer to the quote in the Exercise 24. Explain what is meant by the last sentence, “Subjects in the control group showed no change over time in negative affect (t < 1).” In particular, do you think the sample difference was exactly zero?

Answer: The quote means that the control group did not show a statistically significant change in negative affect. It does not mean that they showed exactly zero change.

27. Another quote in the article is “In response to the influenza vaccine, the meditators displayed a significantly greater rise in antibody titers from the 4 to 8 week blood draw compared with the controls [t(33) = 2.05, p < 0.05, Figure 5]” (Davidson et al., 2003, p. 566).

Specify in words the null hypothesis being tested. Make sure you don’t confuse the population with the sample and that you state the hypothesis using the correct one.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

Answer: Null hypothesis: There would be no difference in antibody response to a flu vaccine for meditators and non-meditators in the population of people who would volunteer for meditation training.

28. Refer to the quote in Exercise 27. Specify in words the alternative hypothesis being tested. Explain whether you decided to use a one-sided or a two-sided test and why.

Answer: Alternative hypothesis: In the population of people who would volunteer for meditation training, people trained to meditate would show a greater rise in antibody response after being given a flu vaccine than would people not trained to meditate. A one-sided test was used because the researchers speculated in advance that meditation would help with immune system response.

29. Refer to the quote in Exercise 27. a. What is the meaning of the word significantly in the quote? b. Explain in plain English what the results of this test mean.

Answer: a. The word “significantly” in the quote is referring to statistical significance, not the usual English use of the word significant. b. Meditation helps increase immune system response, and we can conclude that the increase seen in the sample reflects a non-zero increase in the population, if everyone were to receive the training. However, the magnitude of the increase was not provided, so there is no way to know if it has practical significance.

Exercises 30 to 32 refer to News Story 2 and the journal article Original Source 2, “A chapter a day: Association of book reading with longevity” (not available on the companion website). This long-term study followed a representative sample of 3635 adults over age 50 in the United States for up to 12 years, obtaining information on many variables including their reading habits. During the time of the study, 27.4% of the sample died. All quotes in Exercises 30 to 32 are from Bavishi et al. (2016), with page numbers following the quote.

30. One of the quotes was “Chi-square tests revealed that book readers differed from non-book readers in that a higher proportion was college educated (p < 0.001)” (p. 45). a. The chi-square test for this situation is based on a 2 × 2 table. The row categories are book reader (yes or no). What are the column categories? b. Specify in words the null hypothesis being tested. Make sure to specify the appropriate population as part of the hypothesis.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

c. The article included the information that 49.4% of those who read books had at least some college education, whereas only 31.4% of non-book readers had any college education. Write a paragraph explaining the results of this hypothesis test that would be understood by someone with no training in statistics.

Answer: a. The column categories are whether the individual was college educated – yes or no. b. Null hypothesis: In the population of adults over 50, there is no relationship between whether they read books or not and whether or not they have a college education. c. In a sample of adults over 50, 49.4% of those who read books had at least some college education, whereas only 31.4% of those who did not read books had any college education. This difference is large enough that it can be concluded that there is an association between book reading and having at least some college education in the population of all adults over 50.

31. The article included some complex analyses, but the simplest analysis compared survival for book readers and non-book readers using a hazard ratio.

The result was given as “HR = 0.76, p < 0.001” (p. 46). a. The reported hazard ratio of 0.76 indicates that book readers in the study had a lower risk of death at any given point in time than non-book readers.

Write a few sentences interpreting this hazard ratio in terms that someone with no training in statistics would understand. b. Write the null and alternative hypotheses that are being tested with the result that p < 0.001. Make sure you don’t confuse the population with the sample and that you state the hypotheses using the correct one. c. The hypothesis test result was followed by the statement: “When readers were compared to non-readers at 80% mortality (the time it takes 20% of a group to die), non-book readers lived 85 months (7.08 years), whereas book readers lived 108 months (9.00 years) after baseline. Thus, reading books provided a 23-month survival advantage” (p. 46). Do you think all book readers lived 23 months longer than all non-book readers? If so, explain why. If not, rewrite the statement to clarify what it really means.

Answer: a. A study followed adults over age 50 in the United States for up to 12 years, and compared the risk of death for those who regularly read books and those who don’t. The study found that the risk of dying at any given time during the study for book readers was only 0.76 of what it was for nonbook readers. In other words, there was a 24% reduction in risk of dying for book readers compared to those who didn’t read books.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

b. Null hypothesis: For the population of adults over 50 in the United States, the risk of dying at any given time is equal for those who read books and those who do not read books.

Alternative hypothesis: For the population of adults over 50 in the United States, the risk of dying at any given time is lower for those who read books than for those who don’t. c. No, the survival advantage of 23 months is a comparison of the averages for the two groups. A better way to write the conclusion would be to discuss averages. For instance, “Book readers lived an average of 23 months longer than non-book readers.”

32. In a more complicated analysis than the one described in Exercise 31, the researchers divided the sample into three groups based on how often they reported reading books. They called these groups tertiles. Tertile 1 did not read books, tertile 2 read books but less than 3.5 hours a week, and tertile 3 read books for at least 3.5 hours a week. One of the results reported was “Book reading was associated with significantly greater survival for both the second and third tertiles, with adjustment for all covariates (HRT2 = 0.83, p < 0.001; HRT3 = 0.77, p < 0.001)” (p. 46). The abbreviation “HR” is hazard ratio, and the subscripts T2 and T3 refer to the second and third tertile groups in comparison with the first tertile (the non-book readers). a. Was this research based on an observational study or a randomized experiment? Explain how you know. b. Explain why the article reported that “book reading was associated with significantly greater survival” rather than reporting that book reading leads to significantly greater survival. c. The report noted that the tertiles may have differed in more ways than their frequency of book reading, including health status, education, economic status, levels of depression, age, and more. In this context, explain what is meant by the part of the quote that says “with adjustment for all covariates.” d. The authors wrote that “book reading was associated with significantly greater survival.” Explain whether you think they used the term “significantly” in the statistical sense, in the common language sense of “important,” or both. Include both the reported p-values and the reported hazard ratios in your explanation.

Answer: a. The research clearly was based on an observational study, because you cannot randomly assign people to read books a certain number of hours a week over many years. b. Because the research was based on an observational study, a cause and effect conclusion cannot be made. All we can say is that book reading and survival time are associated or related. We cannot say that book reading causes or leads to greater survival. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

c. The analysis the researchers used took into account the variables mentioned, called covariates, and tried to isolate the relationship of book reading to survival after taking all of those things into account. If they had not done that, then some of the other variables, and not book reading, may have been associated with survival. d. The quote could include both statistical significance and practical importance. The p-values for both tertiles are less than 0.001, indicating strong statistical significance. But the hazard ratios of 0.83 and 0.77 indicate that book readers in the two tertiles had reduced risks of dying at any point, with reductions of 17% and 23%. Those seem like important reductions.

ANSWERS TO THE REFLECTION QUESTIONS 1. A New York Times article described a study testing smart water meters, which notify customers immediately if water flow exceeds set thresholds over a period of time. The article included this quote: “The study found ‘statistically significant’ reductions in the volume of leaks, saving as much as three gallons per meter per day” (Carrns, 2023). Explain what the null and alternative hypotheses would be for this study.

Answer: The null hypothesis would be that having a smart meter does not result in reducing the volume of leaks. The alternative hypothesis would be that having a smart meter leads to reduced volume of leaks.

2. Why do you think journal articles sometimes report p-values using a format such as “p < 0.01” rather than giving the actual numerical value? Explain which version would be more informative.

Answer: One reason is that when lots of tests are done in an analysis, sometimes researchers prefer to identify which ones result in a p-value below a certain threshold. Another reason is that it’s easier to read a table with lots of p-values if they are categorized in that way, so you can easily read down the column of p-values and identify which ones have p < 0.05, p < 0.01, and so on.

But it would be more informative to know the actual p-value. In that case you could make your own conclusion depending on what level of significance you wanted to use. That’s particularly true if the p-value is greater than 0.05, but is only reported as p > 0.05. That result might have a different meaning for you if the p-value is 0.06 than if it is much larger, such as 0.65.

3. Give an example of a hypothesis testing situation for which a one-sided test would be more appropriate, and an example of situation for which a two-sided test would be more appropriate. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

Answer: There are lots of possible answers, but in general, a one-sided test is more appropriate if the researchers have a pre-conceived hypothesis about the direction the difference or effect, if there is one. A two-sided test is appropriate if there is no prior theory about the direction a difference or effect. For example, in Example 23.3 the researchers were testing whether dieting for a year or exercising for a year would lead to greater weight loss.

They did a two-sided test because they had no preconceived theory for which one would be more effective. But if they wanted to compare either of those methods to the control group, it would make sense to use a one-sided test because they speculated in advance that both methods would lead to more weight loss than doing nothing.

4. Suppose a randomized experiment recruits volunteers who want to learn a foreign language. The purpose of the study is to compare two phone apps designed to teach users the language, with half of the volunteers randomly assigned to use App A and the other half assigned to use App B. After 6 months, a language test is given to all participants, and it is reported that the p-value for comparing the means on the test is 0.04. a. What are the null and alternative hypotheses being tested? Don’t confuse the population and the sample, and be specific about which one is used in the hypotheses. b. Does knowing the p-value tell you which app performed better in the experiment? Explain. c. Write a few sentences interpreting the p-value in this situation. d. Explain why it would be helpful to be given a confidence interval in addition to the p-value.

Answer: a. The null hypothesis is that if everyone in the relevant population were to use App A, the mean language test score after 6 months would be the same as it would be if everyone in the population were to use App B. The alternative hypothesis is that those means would be different. b. No, knowing the p-value does not tell us which app performed better in the experiment. To know that, we would need to know the actual difference in means. That’s because for a two-sided test, the absolute value of the test statistic is used to find the p-value. So the p-value would be the same if the difference was of the same magnitude but in the other direction. c. The p-value of 0.04 tells us that if the two apps are equally effective, then we would only observe a difference in means as large as the one observed or larger (in absolute value) about 4% of the time in experiments of this size.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

d. A confidence interval would not only tell us which app resulted in higher scores, but it would also give us a range that we are 95% confident captures the actual population difference in what the mean scores would be if everyone used App A versus App B. 5. Give an example of the null hypothesis for a real-world application that would be tested with each of the following: a. A t-test for the difference in two means b. A chi-square test c. An ANOVA F test d. A relative risk Answer: Answers will vary for each part, but general guidelines are given. a. A t-test for the difference in two means is used when there are two independent samples for which the same variable is measured, and you want to test whether the population means are equal. Example 23.3, comparing mean weight loss for dieting versus exercise is an example. b. A chi-square test would be used to test whether there is an association between two categorical variables in a population. For example, it could be used when a survey using a random sample asks people if they support or don’t support an issue, and also asks for their political party. The null hypothesis would be that there is no association between support for the issue and political party in the population represented by the sample. c. An ANOVA F test is used to compare more than two population means. For instance, suppose the experiment described in Reflection Question 4, comparing two apps for learning a foreign language, had included a traditional classroom learning situation for a third group, in addition to the two groups using the apps. Then the null hypothesis would be that all three learning methods would work equally well for the population, and the alternative hypothesis would be that at least one of them would result in a different mean test score in the population than one or both of the others.

An ANOVA F test would be used to test those hypotheses. d. Relative risk is used to compare the risk of a certain outcome for two different groups or conditions. Lots of examples have been presented in the text for tests using relative risk. As an example, relative risk is used to test whether the probability (risk) of getting a certain disease is the same for smokers and non-smokers.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 23: Hypothesis Testing—Examples And Case Studies

ANSWERS TO THE MINI PROJECTS 1. Find three separate journal articles that report the results of hypothesis tests.

For each one, do or answer the following: a. State the null and alternative hypotheses. b. Based on the p-value reported, what conclusion would you make? c. What would a type 1 and a type 2 error be for the hypotheses being tested?

Answer: There are so many possibilities here that it is impossible to specify what to look for, except to be sure all parts are answered.

2. Conduct a test for extrasensory perception. You can either create target pictures or use a deck of cards and ask people to guess suits or colors.

Whatever you use, be sure to randomize properly. For example, with a deck of cards you should always replace the previous target and shuffle very well. a. Explain how you conducted the experiment. b. State the null and alternative hypotheses for your experiment. c. Report your results. d. If you do not find a statistically significant result, can you conclude that extrasensory perception does not exist? Explain.

Answer: Make sure the experiment was designed to rule out ordinary cues. Ideally the receiver and sender should be in different rooms. The null hypothesis, in words, is that ESP is not operating. The alternative is that it is. For part (d) the relevant issue is the sample size. Unless hundreds of guesses are used the power for the experiment will be very low. Therefore, unless very strong ESP is present or ordinary biases are contaminating the experiment, it will be difficult to get a statistically significant result.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 24: Significance, Importance, and Undetected Differences

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 24: SIGNIFICANCE , IMPORTANCE , AND UNDETECTED DIFFERENCES

TABLE OF CONTENTS Answers to the Exercise Questions ....................................................................................1 Answers to the Reflection Questions .............................................................................. 14 Answers to the Mini Projects ........................................................................................... 16

ANSWERS TO THE EXERCISE QUESTIONS 1. Explain why it is important to learn what sample size was used in a study for which it was reported that “no difference” was found.

Answer: If “no difference” is found and the sample sizes are small, it is quite possible that the test had very low power and a type 2 error has occurred. It may not mean that there is no difference in the population. In fact even a moderate size difference in the population may not result in statistical significance with small samples.

2. Suppose that you were to read the following news story: “Researchers compared a new drug to a placebo for treating high blood pressure, and it seemed to work. But the researchers were concerned because they found that significantly more people got headaches when taking the new drug than when taking the placebo. Headaches were the only problem out of the 20 possible side effects the researchers tested.” Do you think the researchers are justified in thinking the new drug would cause more headaches in the population than the placebo would? Explain.

Answer: Maybe they are not justified, because 20 side effects were tested so one or a few could appear to be a problem (statistically significant) just by chance even if none of them were a problem. In fact, with 20 independent tests we would expect to find one significant just by chance, on average, if all of the null hypotheses were true. In other words, it is possible that the observed relationship between taking aspirin and having headaches was a false positive and occurred because of multiple testing.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 24: Significance, Importance, and Undetected Differences

3. An advertisement for Claritin, a drug for seasonal nasal allergies, once made the claim that studies showed the incidence of drowsiness from the drug was similar to that from a placebo. The advertisement also reported that 8% of the 1926 Claritin takers and 6% of the 2545 placebo takers reported drowsiness as a side effect. A one-sided test of whether a higher proportion of Claritin takers than placebo takers would experience drowsiness in the population results in a p-value of about 0.005. a. Can you conclude that the incidence of drowsiness for the Claritin takers is statistically significantly higher than that for the placebo takers? b. Does the answer to part (a) contradict the statement in the advertisement that the “incidence of drowsiness was similar to placebo”? Explain.

Answer: a. Yes, because the p-value of 0.005 is less than 0.05. b. No; the magnitude wasn’t much different in a practical sense (6% versus 8%), but the large sample sizes led to a statistically significant difference.

4. Use the example in Exercise 3 to discuss the importance of making the distinction between the common use and the statistical use of the word significant.

Answer: In Exercise 3, we saw that a difference between 6% and 8% experiencing drowsiness was statistically significant. A statistically significant difference indicates that the difference in the population is not zero but does not indicate that it has any practical significance. The meaning should be clarified when the word is used.

5. An article in Time magazine (Gorman, 6 February 1995) reported that an advisory panel recommended that the Food and Drug Administration (FDA) allow an experimental AIDS vaccine to go forward for testing on 5000 volunteers. The vaccine was developed by Jonas Salk, who developed the first effective polio vaccine. The AIDS vaccine was designed to boost the immune system of HIV-infected individuals and had already been tested on a small number of patients, with mixed results but no apparent side effects. a. In making its recommendation to the FDA, the advisory panel was faced with a choice similar to that in hypothesis testing. The null hypothesis was that the vaccine was not effective and therefore should not be tested further, whereas the alternative hypothesis was that it might have some benefit. Explain the consequences of a type 1 and a type 2 error for the decision the panel was required to make. b. The chairman of the panel, Dr. Stanley Lemon, was quoted as saying, “I’m not all that excited about the data I’ve seen . . . [but] the only way the concept is going to be laid to rest . . . is really to try [the vaccine] in a large population” (p. 53). Explain why the vaccine should be tested on a larger group, when it had not proven effective in the initial tests on a small group.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 24: Significance, Importance, and Undetected Differences

Answer: a. A type 1 error is that the vaccine is not effective but the testing goes forward, which would not be too serious. A type 2 error is more serious; the vaccine is effective but testing does not take place and it is shelved. b. The tests on the small group may not have had enough power to detect the effectiveness of the vaccine, if it exists.

6. In Example 12.2, we revisited Case Study 6.3, in which the Supreme Court ruled on whether males were more likely than females to drink and drive. In a roadside survey, 77 out of 481 males (16%) and 16 out of 138 females (11.6%) had been drinking in the 2 hours prior to being stopped. In a test of whether there is a relationship between gender and driving after drinking, we cannot rule out chance as an explanation for the sample data; the chi-square statistic is 1.637, and the p-value is 0.20. Now suppose that the sample contained three times as many drivers, but the proportions of males and females who drank before driving were still 16% and 11.6%, respectively. a. What would be the value of the chi-square statistic for this hypothetical larger sample? (Hint: See Example 24.3.) b. The p-value for the test based on the larger sample would be about 0.03.

State the hypotheses being tested and explain which one you would choose on the basis of this larger hypothetical sample, using the standard 0.05 level of significance. c. In both the original test and the hypothetical one based on the larger sample, the probability of making a type 1 error if there really is no relationship is 5%. Assuming there is a relationship in the population, would the power of the test—that is, the probability of correctly finding the relationship—be higher, lower, or the same for the sample three times larger compared with the original sample size?

Answer: a. It would be 3(1.637) = 4.911. b. Null hypothesis: There is no relationship in the population between gender and driving after drinking. Alternative hypothesis: There is a relationship between them. The alternative would now be chosen because 0.03 is less than 0.05. c. The power would be higher with the larger sample size. If there is a difference or relationship in the population, it is always more likely to be detected with a larger sample.

7. Each of the following statements are similar to quotes that have been made when researchers thought they were reporting a p-value. Unfortunately, some of them are incorrect interpretations. In each case, determine whether the quote is an accurate statement of a p-value.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 24: Significance, Importance, and Undetected Differences

a. We used the data to calculate the probability that there is no difference in the treatments. The probability is 0.03. b. If there is no difference in population proportions of the two groups who would have this opinion, the probability of seeing a sample difference of at least 0.16, which is what we observed, is 0.09. c. Given the data, the odds against chance are 1 in 10,000.

Answer: a. The statement is not accurate. It represents the p-value as the probability that the null hypothesis is true, which is incorrect. b. The statement is an accurate representation of a p-value. c. The statement is not accurate. It implies that the p-value represents the odds of the results being due to chance compared to being due to a real effect, which is not a correct interpretation.

8. You make many decisions on a daily basis, and some of them are similar to hypothesis testing. For example, suppose you need to get to the airport by 10 A.M. to catch a flight. In normal traffic, the ride takes about 30 minutes, but sometimes there is heavy traffic and it can take longer. You must decide when to leave home. The null hypothesis is that traffic will be as usual. The alternative hypothesis is that there will be heavy traffic. a. What action will you take if you reject the null hypothesis? b. What is a type 1 error in this situation and what are the consequences for you? c. What is a type 2 error in this situation and what are the consequences for you? d. Which is more serious, a type 1 error or a type 2 error? Explain.

Answer: a. If you reject the null hypothesis, you will conclude that there will be heavy traffic and you will leave earlier than 10 a.m. b. A type 1 error can be made only if the null hypothesis is true; the error is that you reject it. In this case, it means that traffic will be as usual, but you decide it will be heavier than usual, and you leave earlier than 10 a.m. The only consequence is that you will have some extra time at the airport. c. A type 2 error is that you fail to reject the null hypothesis, when you should have. In this case, it means that you think traffic will be as usual and leave at 10 a.m., when in fact traffic is heavier than usual. The consequence is that you might miss your flight. d. A type 2 error is more serious.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 24: Significance, Importance, and Undetected Differences

9. Suppose a news report quoted the results of a research study as follows: “For people who do not exercise regularly, the risk of a heart attack while shoveling snow was two to six times greater than during periods of inactivity. For those who do exercise regularly, there was no significant increased risk of heart attack while shoveling snow.” a. There is a confidence interval reported in the statement. Explain what it is and what characteristic of the population it is measuring. b. For the part of the quote regarding those who do exercise regularly, what do you think is meant by the term “significant”? Explain.

Answer: a. A confidence interval for the relative risk of heart attack while shoveling snow, for those who do not exercise regularly, is from 2.0 to 6.0. b. The increase in risk, if any, was not statistically significant for those who exercise regularly. The quote does not tell us whether there was a significant difference in the everyday language use of the word.

10. In reporting the results of a study to compare two population means, explain why researchers should report each of the following: a. A confidence interval for the difference in the means b. A p-value for the results of the test, as well as whether it was one-or two-sided c. The sample sizes used d. The number of separate tests they conducted during the course of the research Answer: a. A confidence interval for the difference in means is useful because it provides the range (in which we can have specified confidence) that includes the possible magnitudes for the difference. b. It is useful to know how likely the sample difference, or one more extreme, would be in studies of the same size, if the null hypothesis is true. c. The sample sizes would help us determine the likelihood of a type 2 error or the power of the test. d. If multiple tests were conducted we would expect some to reach statistical significance even if nothing is going on.

11. We have learned that the probability of making a type 1 error when the null hypothesis is true is usually set at 5%. The probability of making a type 2 error when the alternative hypothesis is true is harder to find. Do you think that probability depends on the size of the sample? Explain your answer.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 24: Significance, Importance, and Undetected Differences

Answer: Yes; a type 2 error occurs when the alternative hypothesis is true but the null hypothesis is not rejected. The larger the sample, the bigger the value of the test statistic (holding the effect constant) so the easier it is to reject the null hypothesis with a larger sample. Intuitively this makes sense, because the more data you have, the more likely you are to discover a population difference if there is one.

12. Refer to Case Study 22.1, concerning the ganzfeld procedure for testing ESP. In earlier studies using the ganzfeld procedure, the results were mixed in terms of whether they were statistically significant. In other words, some of the experiments were statistically significant and others were not. Critics used this as evidence that there was really nothing going on, even though more than 5% of the experiments were successful. Typical sample sizes were from 10 to 100 participants. Give a statistical explanation for why the results would have been mixed.

Answer: Even if ESP was present in the experiments, the sample sizes were so small that it would be difficult to find a statistically significant effect. The power was low and the probability of a type 2 error was high.

13. When the Steering Committee of the Physicians’ Health Study Research Group (1988) reported the results of the effect of aspirin on heart attacks, committee members also reported the results of the same aspirin consumption, for the same sample, on strokes. There were 80 strokes in the aspirin group and only 70 in the placebo group. The relative risk was 1.15, with a 95% confidence interval ranging from 0.84 to 1.58. a. What value for relative risk would indicate that there is no relationship between taking aspirin and having a stroke? Is that value contained in the confidence interval? b. Set up the appropriate null and alternative hypotheses for this part of the study. The original report gave a p-value of 0.41 for this test. What conclusion would be made on the basis of that value? c. Compare your results from parts (a) and (b). Explain how they are related. d. There was actually a higher percentage of strokes in the group that took aspirin than in the group that took a placebo. Why do you think this result did not get much press coverage, whereas the result indicating that aspirin reduces the risk of heart attacks did get substantial coverage?

Answer: a. A relative risk of 1.0 indicates equal risk. It is contained in the interval. b. Null hypothesis: There is no relationship between taking aspirin and suffering strokes. Alternative hypothesis: There is a relationship between taking aspirin and suffering from a stroke. Because the p-value is so high

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 24: Significance, Importance, and Undetected Differences

(0.41), we cannot reject the null hypothesis. There is no significant relationship between taking aspirin and having a stroke. c. As noted in part (a), the confidence interval contains the relative risk of 1.0, indicating that it is a plausible value. In part (b), we could not reject the null hypothesis that the relative risk is 1.0, consistent with it being a plausible value as shown in part (a). d. It probably did not get much coverage because the relationship was not statistically significant. It could also be that the physicians wanted to downplay the risks involved with taking aspirin, given the obvious benefit in reducing heart attacks. But in general, studies (or specific results within studies) that do not find statistical significance do not get as much press coverage.

14. In the study by Lee Salk, reported in Case Study 1.1, he found that infants who listened to the sound of a heartbeat in the first few days of life gained more weight than those who did not. In searching for potential explanations, Salk wrote the following. Discuss Salk’s conclusion and the evidence he provided for it. Do you think he effectively ruled out food intake as being important?

“In terms of actual weight gain the heartbeat group showed a median gain of 40 grams; the control group showed a median loss of 20 grams. There was no significant difference in food intake between the two groups. . . .

There was crying 38 percent of the time in the heartbeat group of infants; in the control group one infant or more cried 60 percent of the time. . . .

Since there was no difference in food intake between the two groups, it is likely that the weight gain for the heartbeat group was due to a decrease in crying” (Salk, May 1973, p. 29).

Answer: Food intake is not effectively ruled out just because there is “no difference.”

Salk almost surely did not mean that the difference was exactly 0, rather, he meant that the difference was not statistically significant. It would be helpful to know the magnitude of the difference in food intake for the two groups.

15. The authors of the report in Case Study 24.1, comparing the psychological health of UFO observers and nonobservers, presented a table in which they compared the four groups of volunteers on each of 20 psychological measures.

For each of the measures, the null hypothesis was that there were no differences in population means for the various types of people on that measure. If there truly were no population differences, and if all of the measures were independent of each other, for how many of the 20 measures would you expect the null hypothesis to be rejected, using the 0.05 criterion for rejection? Explain how you got your answer.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 24: Significance, Importance, and Undetected Differences

Answer: If all null hypotheses are true and they are independent, about one in 20 should be rejected by chance using a p-value of 0.05, so we would expect one.

16. An advertisement for an exercise machine claimed that in just 12 weeks, people who used the machine lost an average of 18 pounds. Forgetting about the questions surrounding how such a study might have been conducted, what additional statistical information would you want to know about the study and its results before you could come to a reasonable conclusion?

Answer: You would want to know how many people were tested and a confidence interval for the mean loss, rather than just the sample value. You probably also would want to know the range and distribution of individual weight losses.

17. Explain why it is not wise to accept a null hypothesis.

Answer: All we can determine is how unlikely the sample results would be when the null hypothesis is true. That tells us nothing about whether or not it is true.

Also, the null hypothesis generally contains only one specific value, so unless we measure the entire population there is no way to know if that exact value is the actual population value.

18. Refer to the list of warnings on page 532. Explain which ones should be of concern if the sample size(s) for a test are large.

Answer: Issues 1, 2 and 4 are of particular concern if the sample size is large. As stated in #2, with a large sample size a statistically significant result may not have practical importance. Issue 4 helps to resolve the problem because the magnitude of the effect can be used to determine practical importance.

19. Refer to the list of warnings on page 532. Explain which ones should be of concern if the sample size(s) for a test are small.

Answer: Issue 3 is the most relevant because it addresses the main problem with interpreting results when the sample size is small. Issue 4 is also relevant because the confidence interval can be examined to see if it seems to be centered close to the null value, or is mainly above or below it.

20. Now that you understand the reasoning behind making inferences about populations based on samples (confidence intervals and hypothesis tests), explain why these methods require the use of random, or at least representative, samples instead of convenience samples.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 24: Significance, Importance, and Undetected Differences

Answer: The methods depend on knowledge of what kinds of sample values we expect to get if everyone has an equal chance of being in the sample. In convenience samples those with certain values may have a higher chance than those with other values, so the rules for sample means, proportions, etc., would not apply.

21. Would it be easier to reject hypotheses about populations that had a lot of natural variability in the measurements or a little variability in the measurements? Explain.

Answer: Assuming the alternative hypothesis is true, it is easier to reject hypotheses if there is only a little variability, because in that case the sample values are more accurate estimates of the population values. So, any observed differences would be less likely to be due to random variability.

22. Refer to the sentence in warning 7 on page 532 that states: “If all of the null hypotheses tested are true, then over the long run about 1 in 20 tests should achieve statistical significance just by chance.” Is that statement the same thing as saying that the null hypothesis is likely to be true in about 1 out of 20 tests that have achieved statistical significance? Explain.

Answer: No. Warning 7 says that given that the null hypothesis is true, the probability that a test will achieve statistical significance is 1/20. The statement given in the exercise gives the conditional probability in the other direction – given that a test achieves statistical significance, the probability that the null hypothesis is true is about 1/20, a false statement.

Exercises 23 to 29 refer to News Story and Original Source 1, “Alterations in Brain and Immune Function Produced by Mindfulness Meditation” (not available on the companion website). For this study, volunteers were randomly assigned to a meditation group (25 participants) or a control group (16 participants). The meditation group received an 8-week meditation training program. Both groups were given influenza shots, and their antibodies were measured. This measurement was taken after the meditation group had been practicing meditation for about 8 weeks.

The researchers also measured brain activity, but these exercises will not explore that part of the study.

23. One of the quotes in News Story 1 refers to the results of measuring antibodies to the influenza vaccine. It says: “While everyone who participated in the study had an increased number of antibodies, the meditators had a significantly greater increase than the control group. ‘The changes were subtle, but statistically it was significant,’ says [one of the researchers]” (Kotansky, 2003, p. 35).

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 24: Significance, Importance, and Undetected Differences

a. Do you think it would have been fair reporting if only the first sentence of the quote had been given, without the second sentence? Explain, including what additional information you learned from the second sentence. b. Based on this quote, what null hypothesis do you think was being tested?

Make sure you use correct terminology regarding populations and samples.

Answer: a. It would not have been fair reporting because readers may have interpreted “significantly greater increase” to mean that the increase was very large.

The second sentence makes it clear that the word “significant” is being used in the statistical sense, not the English sense. b. Null hypothesis: For the population of people who would volunteer for meditation training, there would be no difference in immune response to a flu vaccine eight weeks after the training for those trained to meditate and those not trained.

24. Refer to the previous exercise. a. Do you think the alternative hypothesis used by the researchers was one-sided or two-sided? Explain. b. Do you think the researchers would be justified in specifying a one-sided alternative hypothesis in this situation? Explain why or why not.

Answer: a. The alternative hypothesis was one-sided. Before collecting any data, the researchers speculated that meditation would boost immunity. b. Yes, a one-sided alternative hypothesis is justified. Based on folklore and previous studies, the researchers were speculating that meditation would increase immune function response.

25. In Original Source 1, the researchers addressed some limitations with the study. One of them was: “First, there was a relatively small number of subjects who participated and this limited our statistical power. A number of our hypothesized effects were in the predicted direction, but failed to reach significance” (Davidson et al., 2003, p. 569).

Explain what is meant by the first sentence of the quote.

Answer: Because the sample size was small, even if differences exist in the population the study may not provide enough evidence to detect them. The “statistical power” is the probability of being able to detect a difference based on a sample, given that it exists in the population.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 24: Significance, Importance, and Undetected Differences

26. Refer to the quote in Exercise 25. Explain what is meant by the second sentence of the quote.

Answer: The researchers found that differences in the sample of those trained to meditate and those not trained to meditate were in the direction predicted by their one-sided alternative hypotheses. However, the sample differences were not large enough to rule out chance as an explanation. The test statistics were not large enough to result in p-values less than 0.05.

27. One quote from Original Source 1 compared the ages of the participants who were randomly assigned to the two groups. The researchers reported: “Average age of subjects was 36 years and did not differ between [the two] groups” (Davidson et al., 2003, p. 565). Do you think that the average age of the participants in the two groups was exactly the same? Explain.

Answer: They don’t mean that the difference was exactly zero, they mean that it was not statistically significantly different from zero. However, they mention that the average age was 36 for both groups, so if they rounded it to the nearest year, the average age was exactly the same. But if they included partial years, it is almost surely slightly different for the two groups.

28. Rewrite the quote in the previous exercise to convey what you think the researchers meant to say.

Answer: The difference in average ages for the two groups was not statistically significantly different from 0. They could also have said that the average ages rounded to the nearest whole number of years did not differ between the two groups.

29. Refer to items 1 and 2 in Section 24.6 “A Summary of Warnings.” Explain whether or not they are likely to apply to results from this study.

Answer: The warnings are not likely to apply to this study because the sample size was so small. Warnings 1 and 2 are a concern when the sample size is large.

30. Two reasons for doing hypothesis tests are to make a decision leading to action or to explore a question of interest that does not require a decision. The following are examples of null hypotheses. In each case, state whether the purpose is to make a decision leading to action or to explore a question of interest. a. A researcher tests the null hypothesis that there is no such thing as extrasensory perception.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 24: Significance, Importance, and Undetected Differences

b. A pharmaceutical company tests the null hypothesis that there is no difference between the current treatment for a disease and a new treatment. c. A presidential commission on health is charged with making recommendations to the medical community. The commission tests the null hypothesis that there is no relationship between exercise intensity and the resulting benefit to cardiovascular health.

Answer: a. This is to explore a question of interest that does not require a decision.

There is no action associated with the conclusion one way or the other. b. The purpose is to make a decision leading to action. The company would want to bring the new treatment to the medical community if it works. c. The purpose is to make a decision leading to action. The commission is supposed to make recommendations to the medical community based on the outcome. If the null hypothesis is rejected, the commission will make recommendations to improve cardiovascular health based on exercise intensity.

31. Two reasons for doing hypothesis tests are to make a decision leading to action or to explore a question of interest that does not require a decision. The following are examples of null hypotheses. In each case, state whether the purpose is to make a decision or to explore a question of interest that does not require a decision. a. A government regulatory agency is charged with fining companies that pollute the air and water. The agency suspects that a manufacturer is polluting the river that runs by its manufacturing plant. Multiple water samples are taken from a location upriver and a location downriver from the plant. Concentrations of chemical X are measured for each water sample. The null hypothesis is that the mean concentration of chemical X is the same upriver and downriver from the plant. b. A sociologist wants to know if age and opinion on a certain issue are related. The null hypothesis is that they are not related in the population of interest. c. An athlete wants to know if consuming caffeine before her running workout increases her running speed. If she concludes that it does, she will consume caffeine before each of her track meets. Each day for 2 months, she randomly assigns herself to either consume caffeine before her workout or not consume caffeine. She then records her running speed for that day. Her null hypothesis is that caffeine does not change her mean running speed.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 24: Significance, Importance, and Undetected Differences

Answer: a. The purpose is to make a decision leading to action. The company will be fined for polluting the river if the null hypothesis is rejected. b. The purpose is to explore a question of interest that does not require a decision. There is no action associated with the conclusion one way or the other. c. The purpose is to make a decision leading to action; in this case, whether to consume caffeine before her track meets. If the null hypothesis is rejected, the athlete will start consuming caffeine before her track meets.

32. Consider the scenario from Exercise 31(c), repeated here. An athlete wants to know if consuming caffeine before her running workout increases her running speed. If she concludes that it does, she will consume caffeine before each of her track meets. Each day for 2 months, she randomly assigns herself to either consume caffeine before her workout or not consume caffeine. She then records her running speed for that day. Her null hypothesis is that caffeine does not change her mean running speed. a. Explain what a type 1 error would be in this scenario and what the consequence would be for the athlete. b. Explain what a type 2 error would be in this scenario and what the consequence would be for the athlete. c. Which error would be more serious for the athlete: type 1 or type 2?

Explain.

Answer: a. A Type 1 error occurs when the null hypothesis is true, but it is rejected. In this example, it means that the athlete concludes that caffeine increases her running speed when it does not. The consequence is that she would consume caffeine before her track meets when in fact it would not help. b. A Type 2 error occurs when the alternative hypothesis is true, but the null hypothesis is not rejected. In this example, it means that the athlete would not conclude that caffeine increases her running speed when in fact it does. She would not consume caffeine before her track meets, when doing so would have helped increase her speed. c. Assuming there is no harm in consuming caffeine when it does not help, a Type 2 error is more serious. The athlete would not recognize an action that could help her running time in track meets. If the null hypothesis is rejected the athlete will start consuming caffeine before her track meets.

33. Consider the scenario from Exercise 30(c), repeated here. A presidential commission on health is charged with making recommendations to the medical community. The commission tests the null hypothesis that there is no relationship between exercise intensity and the resulting benefit to

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 24: Significance, Importance, and Undetected Differences

cardiovascular health. The conclusions of the report will be widely distributed to the news media. a. Explain what a type 1 error would be in this scenario and what the consequence would be. b. Explain what a type 2 error would be in this scenario and what the consequence would be. c. Which error would be more serious: type 1 or type 2? Explain.

Answer: a. A Type 1 error occurs when the null hypothesis is true, but it is rejected. In this example, it means that the commission concludes that benefit to cardiovascular health depends on exercise intensity when it does not. This erroneous conclusion would be widely reported to the media and to the medical community. b. A Type 2 error occurs when the alternative hypothesis is true, but the null hypothesis is not rejected. In this example, it means that there is a relationship between exercise intensity and benefit to cardiovascular health, but it is not discovered and therefore not reported, so an important health finding goes unreported. c. A Type 2 error is more serious because an important finding that could improve cardiovascular health goes undiscovered and unreported. The consequence of a Type 1 error probably would be that many people increase their exercise intensity, which would probably have other benefits even if it didn’t increase cardiovascular health.

ANSWERS TO THE REFLECTION QUESTIONS 1. When journalists report the results of a research study, explain why it is important that they distinguish between the use of the word “significance” referring to statistical significance and the word “significance” as used in everyday language.

Answer: Most readers don’t know about the difference between statistical significance and the use of the term “significance” in everyday language. Even when readers do know the difference, they won’t know which way the term is being used unless the journalist explains it. Therefore, it is the responsibility of journalists who report on research studies to explain how the term “significant” is being used in their reporting.

2. Suppose a pharmaceutical company does a study to test whether a new vaccine is effective to prevent a disease for which there currently is no vaccine. The p-value for the test is 0.03.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 24: Significance, Importance, and Undetected Differences

a. Is this an example of a hypothesis test being done to make a decision leading to action or simply to explore a question of interest? Explain. b. The pharmaceutical company rejects the null hypothesis and starts a campaign to market the vaccine. Explain which type of error (type 1 or type 2) they could have made and the consequences of the error if it was made. c. Write a sentence or two interpreting the p-value.

Answer: a. This is an example of a test being done to make a decision leading to action. In this case, the action is whether or not to seek approval for the vaccine to be used and then market it. b. If they reject the null hypothesis, they could only have made a type 2 error, not a type 1 error. A type 2 error could mean that the vaccine is not effective, but they conclude it is. The consequences would be that they market an ineffective vaccine, and people get false hope. c. If the truth is that the vaccine is not effective, then the probability of seeing results as strong as or stronger than those seen in this study, for studies of this size, is 0.03.

3. Suppose you have some food in your refrigerator that has been there awhile, and you are not sure if it is safe to eat. You give it the “smell test” to distinguish between two hypotheses. The null hypothesis is that the food is fine. The alternative hypothesis is that it is spoiled. If you don’t reject the null hypothesis, you will eat the food. Explain what a type 1 error and a type 2 error would be in this situation and which one is more serious.

Answer: A type 1 error only can occur when the null hypothesis is true. Therefore, in this case the food is fine but the error is that you think it’s spoiled and you throw it away. A type 2 error would occur if the food is spoiled, but you do not reject the null hypothesis that it is fine, and you eat it. A type 2 error is probably more serious because the food is likely to make you sick.

4. Suppose a public opinion polling agency surveys a representative sample of voters from two political parties to determine whether the corresponding populations differ in the proportion who support an issue. a. What are the null and alternative hypotheses being tested? b. If you are provided with a 95% confidence interval for the difference in population proportions, would you be able to determine whether to reject the null hypothesis at the 0.05 level of significance? Explain. c. If you were given the p-value, would that provide information about the magnitude of the population difference? Explain. d. Using your answers to parts (b) and (c), which do you think is more informative —being given a confidence interval or being given a p-value?

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 24: Significance, Importance, and Undetected Differences

Answer: a. The null hypothesis is that the proportions of the populations of voters in the two parties who support the issue are equal. The alternative hypothesis is that those proportions are not equal. b. Yes; you would reject the null hypothesis if 0 is not in the interval. That would mean that with 95% confidence, the difference is not 0, so the population proportions are not equal. c. If you were given the p-value, it would not give you any information about the magnitude of the population difference, except whether 0 was a possible value with a specified level of confidence. d. It is more informative to be given a confidence interval. In addition to providing information about the likely magnitude of the difference, you can use the interval to decide whether to reject the null hypothesis. But you cannot learn about the magnitude of the difference if you are given only a p-value.

ANSWERS TO THE MINI PROJECTS 1. Find a news story that you think presents the results of a hypothesis test in a misleading way. Explain why you think it is misleading. Rewrite the appropriate part of the article in a way that you consider to be not misleading. If necessary, find the original journal report of the study so you can determine what details are missing from the news account.

Answer: The most common mistakes in news reports are implying a cause-and-effect relationship based on an observational study, and concluding there is no difference or effect when there is no statistically significant one. Most likely the example will fit one of these two scenarios.

2. Find two journal articles, one that reports on a statistically significant result and one that reports on a nonsignificant result. Discuss the role of the sample size in the determination of statistical significance, or lack thereof, in each case. Discuss whether you think the researchers would have reached the same conclusion if they had used a smaller or larger sample size.

Answer: If a statistically significant result is found with a small sample, then the real effect is probably quite strong. The larger the sample the easier it would be to find a significant result, and the project should discuss that idea in the context of the two examples.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 25: Meta-Analysis: Resolving Inconsistencies Across Studies

Solution and Answer Guide

UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 25: META-ANALYSIS: RESOLVING INCONSISTENCIES ACROSS STUDIES

TABLE OF CONTENTS Answers to the Exercise Questions ....................................................................................1 Answers to the Reflection Questions ................................................................................ 9 Answers to the Mini Projects ............................................................................................ 11

ANSWERS TO THE EXERCISE QUESTIONS 1. Explain why the vote-counting method is not a good way to summarize the results in a research area.

Answer: Vote counting refers to doing a meta-analysis by counting the number or proportion of studies on a topic that achieved statistical significance. Whether or not studies are statistically significant depends very much on the sample size used. Vote-counting does not take that into account.

2. Explain why the person deciding which studies to include in a meta-analysis should be told how each study was conducted but not the results of the study.

Answer: Studies should be included or excluded on the basis of quality and methodology, not on the basis of the results. Knowing the results might bias the selection; the temptation might be to include only successful ones, or ones that support the analyst’s desired outcome.

3. When the Cochrane Library (discussed on page 544) was first getting started, they limited their focus to systematic reviews and meta-analyses of randomized controlled trials only. In other words, they did not include observational studies. a. Why do you think they only planned to include “randomized controlled trials” (that is, randomized experiments) and not observational studies? b. Do you think they should include studies that did not find statistically significant results when they do their meta-analyses? Why or why not?

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 25: Meta-Analysis: Resolving Inconsistencies Across Studies

Answer: a. They wanted to be able to make causal conclusions, which cannot be made from observational studies. b. Yes. They provide just as much information about the true size of the effect as the studies that did find significance. Excluding them would bias the results in favor of a strong effect.

4. Refer to Exercise 3. Pick one of the benefits of meta-analysis listed in Section 25.3, and explain how that benefit would apply to reviews in the Cochrane Library that involved only randomized experiments.

Answer: Benefits 1, 2, and 4 are most relevant. The combined results may find useful treatments that were missed because of low power (benefit #1) and may also detect patterns across studies that were not evident in individual studies (benefit #4). The combined results also would lead to more precise estimates of effects and relationships (benefit #2). Benefit #3, determining future research, might sometimes be an outcome.

5. The Cochrane Library includes thousands of meta-analyses to be used by medical practitioners. Each review is accompanied by language that would be understood by practitioners with little training in statistics. Pick one of the possible problems with meta-analysis listed in Section 25.4, and discuss how it might affect the usefulness of these reviews for medical practitioners.

Answer: (1) If results are combined across studies, Simpson’s Paradox could result. Results of individual studies that used different types of individuals could get combined, and mask the fact that the relationship is different for those different types of individuals. (2) Confounding variables shouldn’t be a problem because observational studies are excluded. (3) Subtle differences in treatments by the same name could be a problem, such as the strength of a dose of chemotherapy where a larger dose could be too toxic. (4) The file drawer problem could exist, so the treatments appear more effective than they are. (5) Flawed original studies could produce erroneous summary results. (6) A statistically significant treatment effect might be reported, without emphasizing the size of the effect. Practitioners could confuse statistical significance with practical importance in clinical settings.

(7) False findings of “no difference” may lead practitioners to rule out useful treatments, thinking that the finding is conclusive, or worse, if the practitioners don’t understand statistical concepts they might think it has been proven that there is no benefit or difference.

6. Explain how it is possible that none of the studies included in a meta-analysis found a statistically significant difference in means for the treatment and control group, yet the meta-analysis found a statistically significant difference with a significance level of 0.001. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 25: Meta-Analysis: Resolving Inconsistencies Across Studies

Answer: The meta-analysis combines data across multiple studies, resulting in a larger sample size than any individual study. A larger sample means that there is greater statistical power to discover a difference if one exists.

7. In a meta-analysis, researchers can choose either to combine results across studies to produce a single confidence interval or to report separate confidence intervals for each study and compare them. Give one advantage and one disadvantage of combining the results into one confidence interval.

Answer: An advantage of combining the results into one interval is that the interval will be narrower, thus giving a more precise estimate. A disadvantage is that if the studies used samples from different types of populations, or used subtle treatment differences, then the interval would not represent the truth for any group or treatment and would be misleading. This could be akin to Simpson’s Paradox.

8. One of the statements in the Cochrane Handbook (version 6.1) is “A common mistake is to confuse ‘no evidence of an effect’ with ‘evidence of no effect.’” (https://training .cochrane.org/handbook/archive/v6.1 /chapter-15) Explain what is meant by the quote.

Answer: The quote is saying that if “no evidence of effect” is found, it means that the effect is not statistically significant, which is not the same thing as saying that the effect does not exist. The quote is similar to Warning #3 on page 532 (Chapter 24).

9. Give two reasons why researchers might not want to include all possible studies in a meta-analysis of a certain topic.

Answer: Here are some reasons: Some studies used outdated technology; some are of such poor quality that they would just muddy the results; researchers might want to use experiments only and not include observational studies.

10. Suppose a meta-analysis on the risks of drinking coffee included 100,000 participants of all ages across 80 studies. One of the conclusions was that a confidence interval for the relative risk of heart attack for women over 70 who drank coffee, compared with those who didn’t, was 0.90 to 1.30. The researchers concluded that coffee does not present a problem for women over 70 because a relative risk of 1.0 is included in the interval. Explain why their conclusion is not justified.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 25: Meta-Analysis: Resolving Inconsistencies Across Studies

Answer: Their conclusion is equivalent to accepting the null hypothesis that the relative risk is actually 1.0 for the population. Although the interval covers 1.0, it is wide and predominantly above 1.0, so a larger sample may actually result in an interval entirely above 1.0. This would indicate that there is a difference in risk for the population. An additional issue is that the research is clearly based on observational studies (coffee drinking can’t be randomly assigned) so no causal conclusion can be made in any case.

11. News Story 19, summarized in the Appendix, is titled “Health Benefits of Tea? Here’s What the Evidence Says” (Carroll, 2015). Published in The New York Times, it summarizes the available evidence on the benefits of drinking tea. For instance, it says, “[Drinking tea] does not seem to be linked with a reduced risk of fracture.” The quote included links to two meta-analyses. Results from those analyses are: “The pooled RR of hip fractures for the highest vs. the lowest categories of tea consumption was 0.84 (95% CI 0.66–1.02)” (Sheng et al., 2014).

“In this meta-analysis, tea intake was not associated with the incidence of fractures (OR, 0.89; 95% CI, 0.78-1.04)” (Chen et al., 2014). a. Write a few sentences interpreting the RR (relative risk) of 0.84 and the 95% confidence interval of 0.66 to 1.02 given in the first meta-analysis quote. Write it in terms that would be understood by someone with no training in statistics. b. Do you think the numerical results of the two meta-analyses were similar enough that you would say one replicated the other? Explain. c. Comment on the quote from The New York Times story that, “Drinking tea does not seem to be linked with a reduced risk of fracture.” Do you agree or disagree? Explain.

Answer: a. Based on research combining many studies on the effects of drinking tea, researchers found that the risk of hip fracture for people in these studies who had the highest tea consumption was only 84% of what that risk was for those with the lowest tea consumption. Extending the results to the population, the researchers were 95% confident that the risk for the highest consumption tea drinkers is between 66% and 102% of what the risk is for the lowest consumption tea drinkers. b. The results are similar. Although one is reporting relative risk and the other is reporting an odds ratio, these will be similar for low-risk outcomes like hip fractures. The reported relative risk is 0.84 with a confidence interval from 0.66 to 1.02, while the reported odds ratio is 0.89 with a confidence interval from 0.78 to 1.04. Those are similar enough to say that the results were replicated.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 25: Meta-Analysis: Resolving Inconsistencies Across Studies

c. The statement is not reasonable. In both cases, the confidence interval barely has any values at 1.0 or higher, which is what the population values would have to be if there is no reduced risk. Most of the values in both confidence intervals are less than 1.0, supporting the idea that there is reduced risk of fracture for the highest tea consuming group compared to the lowest tea consuming group. What the quote probably means is that the observed reduction in risk is not statistically significant.

12. The file drawer problem occurs because studies that don’t find statistical significance sometimes aren’t published. a. One way researchers deal with the problem is given on page 553. Explain what it is and how it would help. b. Given your answer in part (a), would the file drawer problem be more likely to present a substantial difficulty in a research field with 50 researchers or in one with 1000 researchers? Explain.

Answer: a. Contact all of the known researchers on a topic and ask if they have any unpublished studies. If they do, then get those studies and include them in the meta-analysis. b. The file drawer problem would be more likely in a research field with 1,000 researchers; 50 could easily be contacted and asked about unpublished studies, and those studies could then be included in the meta-analysis.

13. An article in The New York Times (Dominus, 2023) discussing the effectiveness of therapy quoted several meta-analyses that appear to show that therapy works. Explain what is meant by each of the following quotes from the article. a. “But there are reasons to think that this picture of therapy over-promises.

As is true of much research, studies with less positive or striking results often go unpublished, so the body of scholarly work on therapy may show inflated effects.” b. “Such different interpretations of the data [based on different metaanalyses] persist in part because of some of the field’s particular research challenges, starting with what constitutes a suitable control group.” c. “Most studies do not break down the results of various psychotherapies by type of patient—by gender, for example, or comorbidities, or age of onset of illness. The trials are too small to generate statistically meaningful results for those categories.”

Answer: a. This quote is describing the file drawer problem. Studies that were not published were not likely to be included in the meta-analysis. Studies that were published are likely to be those that show that therapy works.

Therefore, the meta-analysis would include a biased sample of studies, and the overall effect would be overestimated by the meta-analysis.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 25: Meta-Analysis: Resolving Inconsistencies Across Studies

b. In trying to determine whether therapy works, it would be important to compare the outcomes of therapy with what would have happened without therapy. In order to do that, studies would need to have a suitable control group. The quote seems to be saying that the researchers who study therapy do not agree on what would be a suitable control group. c. This quote reflects the fact that these studies do not have enough individuals in various subgroups to have enough statistical power to make appropriate conclusions.

14. One of the quotes in Example 25.2 was “Random effects meta-analyses revealed a large effect size for clinician rated symptoms (g = 0.88, 95% CI 0.55, 1.12), while those for both parent (g = 0.40, 95% CI 0.24, 0.56) and childreported anxiety (g = 0.25, 95% CI 0.06, 0.43) were smaller, but significant.”

Explain what is meant by the last statement, that they were “smaller, but significant.” As part of your explanation, include whether you think the word “significant” is being used in the statistical sense or in the everyday language meaning, or both.

Answer: As noted in Example 25.2, the effect size g given in the results represents the difference in sample means in terms of number of standard deviations. In that example, it was also explained that an effect size of 0.8 is considered to be large, while 0.5 is medium and 0.2 is small. In the quoted results, the effect size for clinician-rated symptoms is large, at 0.88. But the effect size for parent ratings is only 0.40, slightly less than a medium effect. For childreported anxiety, it is only 0.25, a small effect size. In all three cases, the entire confidence interval is above 0, so the intervals represent a statistically significant effect. Subject-matter experts would need to decide whether they have significance in the everyday language meaning of the word.

15. Suppose 10 studies were done to assess the relationship between watching violence on television and subsequent violent behavior in children. Suppose that none of the 10 studies detected a statistically significant relationship. Is it possible for a vote-counting procedure to detect a relationship? Is it possible for a meta-analysis to detect a relationship? Explain.

Answer: Vote-counting could not detect a relationship but meta-analysis could, by increasing the overall power. Vote-counting assesses whether there is a relationship by asking if there are more statistically significant studies than would be expected if there is no effect, so if there are no statistically significant studies, vote counting could not detect a relationship.

16. Refer to the summary of News Story 9 in the Appendix. The story appeared in The Washington Post on May 7, 2002. You may be able to find it through an internet search of the story’s headline, “Against depression, a sugar pill is hard © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 25: Meta-Analysis: Resolving Inconsistencies Across Studies

to beat.” There is a meta-analysis of “96 antidepressant trials” described in the story and the summary in the Appendix. Use that description to answer these questions. a. How did the analyst answer the question “Which studies should be included?” b. How did the analyst answer the question “Should results be compared or combined?” c. One statement in the story is “in 52 percent of them [the studies], the effect of the antidepressant could not be distinguished from that of the placebo.” Do you think this statement is an example of “vote-counting”?

What additional information would you need to determine whether or not it is? d. Refer to part (c). Assuming the statement is an example of vote-counting, should it be used to conclude that the effects of antidepressants and placebos are the same? Explain.

Answer: a. It was reported that the analyst included 96 studies done between 1979 and 1996, including “those that were made public in the medical literature… and those that were not.” b. The analyst combined the studies to find that “in 52 percent of them, the effect of the antidepressant could not be distinguished from that of the placebo.” c. The statement does appear to be using vote-counting. However, we would need to know if the meaning of “could not be distinguished” is that the difference was not statistically significant. If that’s what it means, then this is an instance of vote-counting. d. If the statement is an example of vote-counting, then it could be quite misleading. Depending on the size of a study, a small difference could be statistically significant, or, more importantly in this example, a real difference in the population could go undetected in a study if the sample size isn’t large enough.

17. A meta-analysis of 32 studies on the effectiveness of “self-talk” on sports performance was summarized as follows: “The analysis revealed a positive moderate effect size (ES = 0.48). The moderator analyses showed that self-talk interventions were more effective for tasks involving relatively fine, compared with relatively gross, motor demands, and for novel, compared with well-learned, tasks” (Hatzigeorgiadis et al., 2011, p. 348). a. Read the explanation of a fixed effects model on page 546. Based on the quote above, do you think a fixed effects model would be appropriate for this meta-analysis? Explain.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 25: Meta-Analysis: Resolving Inconsistencies Across Studies

b. Another quote from the article was: “The fail-safe statistic indicated that it would be unlikely that a sufficient number of unpublished studies (102) would exist to reduce the effect to a trivial size” (Hatzigeorgiadis et al., 2011, p. 351). Explain what the “fail-safe statistic” is, and why it is important to include it in the report.

Answer: a. No. To quote from the text, “In a fixed effects model, the assumption is that all of the studies included samples from similar populations, with a fixed but unknown magnitude of the effect being tested.” In this example, the summary mentioned that self-talk interventions were more effective for certain tasks than others, so the same effect was not present across all studies. b. The quote refers to the “fail-safe N” (discussed on page 553), which is the number of undiscovered studies that would need to exist to reduce a statistically significant finding to a non-significant one. In this example, the quote is saying that there would need to be 102 unpublished studies, and the implication is that there are unlikely to be that many unpublished studies. It is important to include this statistic to counter the possible criticism that the results could be explained by the file drawer problem.

18. Explain what is meant by the file drawer problem and why it could distort the results of a meta-analysis.

Answer: The file drawer problem refers to the idea that most studies are only published if they achieve statistical significance. The rest end up in a file drawer. Therefore, if a meta-analysis is based only on published studies, it will be biased in favor of data for which the null hypothesis was rejected.

19. The quality assessment tools available from NIH (https://www.nhlbi .nih.gov/health-topics/study-quality-assessment-tools) include a list of questions for assessing the quality of a meta-analysis. Each of the following parts gives one of those questions. In each case, explain why the question is important when assessing the quality of a meta-analysis. a. Were eligibility criteria for included and excluded studies predefined and specified? b. Did the literature search strategy use a comprehensive, systematic approach? c. Was publication bias assessed?

Answer: a. It is important to specify the eligibility criteria in advance so that the analyst can’t decide which studies to include or omit based on looking at the results of the studies.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 25: Meta-Analysis: Resolving Inconsistencies Across Studies

b. It is important to try to find all studies that have been done on a topic, whether or not the results were statistically significant. c. Publication bias refers to various types of bias that might lead some studies to be published and others not to be published. Examples that might lead some studies to be published and other not to be include whether the results were statistically significant, and whether the results contradict current thinking on the topic.

20. The quality assessment tool available from NIH (https://www.nhlbi.nih .gov/health-topics/study-quality-assessment-tools) lists separate quality assessment question for various types of studies. One of the questions in assessing the quality of observational studies for inclusion in a meta-analysis is “Were all the subjects selected or recruited from the same or similar populations (including the same time period)?” Choose one of the criticisms of meta-analysis listed on page 552 and explain how it could be a problem if the answer to this question is no. Include an example in your answer.

Answer: Criticisms 2 (confounding variables) and 3 (subtle differences in treatments of the same name) are most likely to create problems. For example, suppose a meta-analysis were to examine the relationship between amount of fat in one’s diet and longevity. If some studies were done in countries for which dairy products were the predominant source of fat and others done in countries where olive oil was the predominant source, then type of fat consumed would be a confounding variable. And differences in “treatments of the same name” would apply if “amount of fat in one’s diet” is considered to be the treatment.

ANSWERS TO THE REFLECTION QUESTIONS 1. Suppose a meta-analysis will be done to examine the relationship between regular use of dental floss and tooth decay. Researchers find some observational studies on the topic and some randomized experiments. In the observational studies, participants simply reported their use of dental floss. In the randomized experiments, participants were randomly assigned to use dental floss for a year or not to use it for the year. a. Explain one advantage to considering the observational studies to answer the question and one advantage to using the randomized experiments to answer the question. b. In the meta-analysis, would it be better to combine the data from both types of studies or to analyze them separately? Explain.

Answer: a. The observational studies have ecological validity. In other words, people in those studies use dental floss with the same regularity in the study as they

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 25: Meta-Analysis: Resolving Inconsistencies Across Studies

would in their lives if they were not in the study. However, if it is found that those who regularly use dental floss have less tooth decay in the observational studies, it cannot be concluded that the use of dental floss is the cause of the reduction in tooth decay. There could be confounding variables related to how well someone takes care of their teeth in other ways. With the randomized experiments, participants are randomly assigned to use dental floss or not, so if a relationship is found it can be concluded that using dental floss causes less tooth decay. b. It would be better to analyze the studies separately. As explained in part (a), the kinds of conclusions that can be made are different for the two types of studies.

2. Why is it important for a report on a meta-analysis to identify all of the statistical analyses that were done, rather than just the ones that produced a statistically significant result?

Answer: For any kind of statistical study, it is important to identify all of the analyses that were done, not just for meta-analyses. Otherwise, multiple testing could lead to false positives. Also, it is important for readers to learn what questions were asked and what conclusions were made, even if the conclusions were that there is no significant effect or difference.

3. Consider a collection of 20 individual studies comparing means for a treatment and a control condition, and a meta-analysis combining them. Explain how each of the following could occur. a. None of the individual studies find a statistically significant difference, but the meta-analysis does. b. All of the individual studies find a statistically significant difference using two-sided tests, but the meta-analysis does not find a statistically significant difference.

Answer: a. The meta-analysis would have much higher statistical power than each of the individual studies. With the combined larger sample size, a small effect could be statistically significant even if none of the smaller individual studies made that conclusion. b. It could be that some of the studies in the meta-analysis found a difference in one direction, and others found a difference in the other direction. By combining the studies, these average out to result in a small or 0 overall effect.

4. What effect does the file drawer problem have on the results of a metaanalysis? In particular, is the meta-analysis more likely or less likely to find a statistically significant effect if there is a large file drawer problem? Explain.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 25: Meta-Analysis: Resolving Inconsistencies Across Studies

Answer: The file drawer problem is that studies not finding statistical significance end up in the file drawer rather than getting published. If there is a large file drawer effect but the meta-analysis does not locate and include those studies, then only studies that achieved statistical significance will be included. Those constitute a biased collection of all studies on the topic, with the bias toward studies that found a larger effect in the sample. Therefore, the results of the meta-analysis could overestimate the true effect.

5. One of the criticisms of hypothesis testing is that statistical significance is sometimes confused with practical significance. Is that problem more likely to occur in an individual study or in a meta-analysis combining many studies?

Explain.

Answer: The problem is more likely to occur in a meta-analysis. By combining data from many studies, it is easier to achieve statistical significance even with a very small magnitude for the population difference or effect.

ANSWERS TO THE MINI PROJECTS 1. Find a journal article that presents a meta-analysis. (They are commonly found in medical and psychology journals. You can also find thousands of reports in the Cochrane Library, https://www.cochranelibrary.com/.) a. Give an overview of the article and its conclusions. b. Explain what the researchers used as the criteria for deciding which studies to include. c. Explain which of the benefits of meta-analysis were incorporated in the article. d. Discuss each of the potential criticisms of meta-analysis and how they were handled in the article. e. Summarize your conclusions about the topic based on the article and your answers to parts (b) to (d).

Answer: There is no way to summarize what to expect for this project. Simply make sure all parts were addressed, and that the article actually presented an analytical meta-analysis and not a summary of different studies or a single study.

2. Find a news story reporting on a meta-analysis. Discuss whether important information is missing from the report. Critically evaluate the conclusions of the meta-analysis, taking into consideration the criticisms listed in Section 25.4.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 25: Meta-Analysis: Resolving Inconsistencies Across Studies

Answer: News reports typically leave out important information. Some possibilities are criteria for selecting studies, quality issues, file drawer problems and number of individuals used in the studies. There may be others.

3. Find a news story reporting on a meta-analysis. Then find the original journal article that provides the details for the meta-analysis. Discuss the connection between the news story and the more technical journal report. Include any important information you think is missing from the news story and whether the news story misinterprets any of the results.

Answer: This project is similar to Mini-Project 2, except that in this one students are asked to find the original journal article. By comparing the news story with the original article, they are in a better position to determine whether important information is missing or misinterpreted than they would be if they had only the news story as a source.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 26: Ethics in Statistical Studies

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 26: ETHICS IN STATISTICAL STUDIES

TABLE OF CONTENTS Answers to the Exercise Questions ....................................................................................1 Answers to the Reflection Questions ................................................................................ 8 Answers to the Mini Projects ........................................................................................... 10

ANSWERS TO THE EXERCISE QUESTIONS 1. A classic psychology experiment was conducted by psychology professor Philip Zimbardo in the summer of 1971 at Stanford University. The experiment is described at the website http://www.prisonexp.org/. Visit the website; if it is no longer operative, try an Internet search on “Zimbardo” and “prison” to find information about this study. a. Briefly describe the study and its findings. b. The website has a number of discussion questions related to the study (https://www.prisonexp.org/discussion-questions). Here is one of them: “Was it ethical to do this study? Was it right to trade the suffering experienced by participants for the knowledge gained by the research?”

Discuss these questions. c. Another question asked at the website is, “How do the ethical dilemmas in this research compare with the ethical issues raised by Stanley Milgram’s obedience experiments [described in Example 25.1]? Would it be better if these studies had never been done?” Discuss these questions.

Answer: a. The study recruited college student volunteers to participate in a simulated prison experience. They were randomly assigned to be prisoners or guards.

The simulation was made as real as possible, including a humiliating entry procedure for the prisoners, using numbers instead of names for them, and so on. The experiment was terminated early because the role-playing became too realistic, with the “prisoners” breaking down from the stress and some of the “guards” becoming inhumane and sadistic. b. Answers to this question will vary. The idea is to have students work out these ethical issues for themselves. c. Answers to this question will vary. The idea is to have students work out these ethical issues for themselves. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 26: Ethics in Statistical Studies

2. In the report Principles and Practices for a Federal Statistical Agency (HarrisKojetin and Citro, 2021), one of the recommended practices is “necessary authority and procedures to protect independence.” Each of the following parts gives one of the characteristics that is recommended to help accomplish this. In each case, explain how the recommendation would help ensure a position of independence for the statistical agency. a. “The selection of qualified professional staff, including senior executive career staff, should be determined by the statistical agency” (p. 54). b. “Authority to release statistical information (including press releases) without prior clearance for the statistical content by department policy officials is essential” (p. 55). c. “The head of the statistical agency or unit should be able to meet with congressional staff and members to explain the agency’s statistics and programs” (p. 56).

Answer: a. Selection and promotion of staff should not be tied to politics. Employees should be hired for their professional skills, not for their political views. The leadership in the statistical agencies are the most qualified to know if the people they are hiring have the right professional skills. In most cases, the statistical agencies are part of larger agencies that have political appointees in the higher leadership positions. The higher leadership should not determine who is hired to do the statistical work. b. As noted in part (a), most statistical agencies are part of a larger agency with political appointees in their leadership. Statistical content should stand on its own, not be filtered or censored by those with a political point of view. c. The heads of the statistical agencies must be permitted to present the agency’s findings directly, rather than have them filtered through political appointees before being presented to Congress and others. If they were to be presented by others, they may be distorted to conform to the views of the current administration or other political entities.

3. Refer to Example 26.2 about Janet’s dissertation research. Explain whether each of the following changes would have been a good idea or a bad idea: a. Have someone who is blind to conditions and has not heard the volunteers’ voices before listen to the phone calls to assess whether a firm “no” was given. b. Have Janet flip a coin each time a volunteer comes to her office to decide if they should go into the role-playing group or the control group. c. Use a variety of different solicitors rather than Brad alone.

Answer: a. This would be a good idea. In cases where the response is ambiguous, Janet could frame the response to conform to her desired outcome based on which group she knew they had been in, whereas someone blind to the condition of each volunteer could not do that.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 26: Ethics in Statistical Studies

b. This may be a good idea, as long as Janet was honest about the coin flip and didn’t try to influence it to come out the way she wanted it. She would also have to be sure she knew in advance which coin outcome corresponded to which group. c. It is probably better for one person to do all of the soliciting, for consistency. But, better care should have been taken to make sure Brad did not have any contact with the participants before the phone call.

4. Refer to Example 26.3 about Jordan’s memory experiment. What do you think Jordan should have done regarding the analysis and reporting of it?

Answer: The answer may differ. Certain features are important though. Jordan should have done a two-sided test, as planned. Regarding the two different tests, Jordan should have checked whether the conditions for a t-test were met, and if not, done the Mann-Whitney test. Jordan should have attempted to find the reason for the outlier, and if that proved to be impossible, run the analysis with it and without it and reported both results. Arbitrarily changing the 100 to 70 definitely should not have been done.

5. Find an example of a statistical study reported in the news. Explain whether you think multiple tests were done, and, if so, if they were reported.

Answer: The answer will differ for each student. It is not always possible to determine if multiple tests were done based only on the news story, because sometimes the journalist reports only the most interesting or statistically significant result. If it is not obvious either way, the answer should discuss that possibility.

6. Sarai is a statistician who works for a company that manufactures components for sound systems. Two development teams have each come up with a new method for producing one of the components, and management is to decide which one to adopt based on which produces components that last longer. Sarai is given the data for both and is asked to make a recommendation about which one should be used. Sarai computes 95% confidence intervals for the mean lifetime using each method and finds an interval from 96 hours to 104 hours, centered on 100 hours for one of them, and an interval from 92 hours to 112 hours, centered on 102 for the second one. What should Sarai do? Should Sarai make a clear recommendation? Explain.

Answer: Because the intervals overlap, it is not clear which component is better, so Sarai should not report a firm conclusion. Notice that one of the intervals has a higher center but is more spread out than the other one. Assuming the same number of components was tested for each method, this would mean that one method may produce components that last slightly longer on average, but the other one has more consistent results. Sarai might want to report this observation. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 26: Ethics in Statistical Studies

Use the following scenario for Exercises 7 to 11: Based on the information in Case Study 26.1, describe the extent to which each of the following conditions for reducing the experimenter effect (listed on page 572) was met. If you cannot tell from the description of the experiment, then explain what additional information you would need.

7. Randomization done by a third party with no vested interest in the experiment or at least done by a well-tested computer randomization device.

Answer: This condition was not met. Randomization was done by the 9-year-old experimenter flipping a coin.

8. Automated data recording without intervention from the researcher.

Answer: The Case Study does not specify if this condition was met, but there is no mention of automatic data recording and it is unlikely that something that sophisticated was done in this science project.

9. Double-blind procedures to ensure that no one who has contact with the participants knows which treatment or condition they are receiving.

Answer: This condition was not met. The 9-year-old girl was in immediate proximity with the practitioners, and she knew which hand she was hovering over with her hand.

10. An honest evaluation that what is being measured is appropriate and unbiased for the research question of interest.

Answer: According to those who wrote some of the letters, this condition was not met.

11. A standard protocol for the treatment of all participants that must be strictly followed.

Answer: It is not clear from the details provided if a standard protocol was strictly followed. One of the letter-writers noted that “the procedure was conducted in difference settings with no control of environmental conditions.” If that is the case, then this condition is not met. For instance, different room temperatures, open windows with a breeze and so on, could have affected the results.

Exercises 12 to 21: Explain the main ethical issue of concern in each of the following.

Discuss what, if anything, should have been done differently to address that concern.

12. Example 26.1, Stanley Milgram’s experiment.

Answer: The main problem is that participants were deceived into thinking they were hurting others. But there would have been no way to conduct the experiment if they had been told the truth.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 26: Ethics in Statistical Studies

13. In Example 26.2, Janet’s decision to make Group 2 the control group after the first volunteer came to her office.

Answer: Janet was probably able to influence the results by deciding which treatment that first participant would receive. Assuming she did not have any way to find the information about which should be Group 1, she should have used a randomization method to decide, such as flipping a coin or using a computer to decide randomly.

14. In Example 26.2, Janet’s handling of the fact that her data showed 26 volunteers in the control group when there should have been only 25.

Answer: Janet altered the results by deciding to remove one of the “no” responses from the control group. She should not have removed anyone from the data.

And, of course, she should have been more careful so that she would be able to discover the source of the discrepancy.

15. In Example 20.3 (pages 429–430), the Environmental Protection Agency’s decision to report a 90% confidence interval.

Answer: The problem is that they chose the confidence level to present the outcome they desired. They should have reported the standard 95% confidence interval, or presented both and explained how to interpret them.

16. In Example 26.3, Jordan’s decision to replace the outlier with the median of 70.

Answer: Jordan altered the results by replacing the outlier of 100 with the median of 70. It would have been better to leave the outlier in the data or attempt to find out if it was legitimate by retesting the person who received that score.

17. In Example 26.4, the fact that many media stories mentioned increased risk of breast cancer and coronary heart disease but not any of the other results.

Answer: The issue is selective reporting, which was used to make an interesting news story. The media should have given more comprehensive results so that readers would know that the therapy actually reduced some risks.

18. In Case Study 26.1, the concerns raised in the letter from Freinkel.

Answer: The complaint raised by the letter is that the study was not measuring what it purported to measure. The researchers should have familiarized themselves with what they were intending to study, rather than assuming they knew what it was.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 26: Ethics in Statistical Studies

19. In Case Study 26.1, the concerns raised in the letter from Carpenter and colleagues.

Answer: This letter raises many ethical issues, including poor randomization, lack of adequate controls and not measuring what it purported to measure. The researchers should have done a better job of designing the study to address those issues.

20. In Case Study 26.1, the concerns raised in the letter from Schmidt.

Answer: This letter raises many ethical issues, including incorrect analyses and lack of informed consent. The researchers should have designed the study and analyzed the data more carefully to avoid those issues.

21. In Case Study 26.1, the concerns raised about the experimenter effect.

Answer: The experimenter effect could have been mitigated by using more careful controls, such as better randomization, better data recording, and not allowing the person who did the randomization to apply the treatment.

22. Many scientific journals now have a policy requiring potential authors to specify their planned analyses in advance if the research is to be published in that journal. Refer to the ethical issues described in Sections 26.4 and 26.5 in this chapter. Which one of these issues is addressed by the policy of requiring the specification of planned analyses in advance?

Answer: By specifying the planned analyses in advance, the decisions listed in bullets at the beginning of Section 26.4, on page 573 would be specified in advance rather than after looking at the data. That would avoid data-snooping, which involves trying different analyses to find one that produces the desired results.

23. Refer to Section 8.07(c) of the American Psychological Association’s Ethical Principles, quoted on page 567, just above the heading “Informed Consent.” It explains when psychologists must inform participants about any deception used in the experiment. Now consider Janet’s dissertation research in Example 26.2. a. Was there any deception used in Janet’s experiment? Explain. b. Using the information in Section 8.07(c), explain when Janet should give complete information about her experiment to the participants. c. Would Janet’s experiment have been as effective if she revealed the purpose of it when participants first came to her office? Explain.

Answer: a. Yes, there was deception. Janet deceived the participants by following the training with a solicitation that was not real.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 26: Ethics in Statistical Studies

b. The ethical guidelines on page 567 state that participants should be informed “as early as is feasible, preferably at the conclusion of their participation, but no later than at the conclusion of the data collection.” In Janet’s experiment, she could have informed them immediately after the phone call from Brad. But if she was concerned that the participants might know each other and leak information to other participants, she would want to wait until after all of the phone calls had been made. c. Janet’s experiment would not have been as effective if she had revealed the purpose before Brad made the phone call. If the participants knew it was not a real request for a donation, they would be likely to respond differently.

24. Explain why it is not always wise to tell participants in an experiment the purpose of the experiment before they participate. Give an example.

Answer: It is not always wise to tell participants in advance what the purpose is because they may change their behavior or responses to help create the desired outcome. This may be the outcome desired by them, or by the researcher.

25. Refer to Section 8.07(c) of the American Psychological Association’s Ethical Principles, quoted on page 567, just above the heading “Informed Consent.”

It explains when psychologists must inform participants about any deception used in the experiment. Now consider Jordan’s project described in Example 26.3. a. Was there any deception used in Jordan’s experiment? Explain. b. Using the information in Section 8.07(c), explain when Jordan should have given complete information about the experiment to the participants. c. Jordan told participants that they would be asked to recall the words in their initial list before they were asked to study it. Would it have made Jordan’s experiment less effective, more effective, or the same if Jordan also revealed at that time that the purpose of the experiment was to compare the two majors? Explain.

Answer: a. Although Jordan did not use deception in the experiment, participants were not told that the purpose was to compare students in their major with students in another major. So, participants were not given the full story for why the research was being done. b. If Jordan had told the participants that the purpose was to compare students in their major to another major, the participants might have become more competitive and tried harder to do well. Therefore, it would make sense for Jordan to reveal that fact after the second memory test.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 26: Ethics in Statistical Studies

c. There is no correct answer – it is the reasoning that counts. Jordan’s experiment may have been less effective if the participants were told in advance that the purpose was to compare the two majors, because they might have become more competitive and tried harder to remember the words. On the other hand, if the participants knew it was a competition, they might have done their best, mimicking what would happen in their professional capacity later. Knowing only that it was a student experiment and not that they were being compared to another major, they might not have tried as hard.

ANSWERS TO THE REFLECTION QUESTIONS 1. Suppose you respond to a request for volunteers to participate in a medical study. The goal is to compare an herbal tea with certain ingredients to a placebo tea without those ingredients to see whether drinking the herbal tea daily for 3 months helps reduce cholesterol. What information would you want to have provided to you before you decide whether you consent to participate?

Answer: There are lots of possibilities for what you might want to know. Here are some examples: Will I be paid to participate? Can I drop out at any time if I don’t want to continue? What data will you collect on me, and will it be kept secure? Will I eventually be told which tea I was drinking? Will I be given my individual results at the end of the experiment, as well as the overall results? Are there any possible side effects for either beverage? Who can I contact if I have questions?

2. Statistical studies include multiple steps. Give an example of an ethical issue that is relevant to each of the following steps. a. Designing the study b. Collecting the data c. Analyzing the data d. Reporting the results Answer: a. In designing the study, an example of an ethical issue is making sure that there are enough participants to have a high probability of finding an effect, if there is one of a designated size. In other words, the sample size should be chosen by computing what will be needed to achieve a high level of statistical power for a specified effect size. b. An example of an ethical issue when collecting data is to make sure all of the data values are accurately recorded, and no values are thrown out just because they don’t fit with the others or with the researchers’ theories. c. An example of an ethical issue when analyzing the data is to specify all hypotheses and planned analyses in advance, unless the research was predesignated to be exploratory in nature.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 26: Ethics in Statistical Studies

d. An example of an ethical issue when reporting the results is to make sure that all results are reported, and not just those that achieve statistical significance.

3. The U.S. Bureau of Labor Statistics (BLS) is the federal statistical agency responsible for calculating and reporting the Consumer Price Index (CPI). As described in Chapter 18, the CPI is used to determine the raises given to Social Security recipients, which in turn affects the federal budget. In this context, explain why it is important that the BLS remains independent of political influence from the president and Congress.

Answer: The federal budget always has political implications, with some politicians wanting to cut expenses wherever possible, and others wanting to preserve certain programs even if it increases the bottom line on the budget. Therefore, if politicians were able to influence how the CPI is designed and calculated, those who want to minimize spending might make decisions to reduce the CPI, while those who want social security recipients to keep up with inflation might make decisions differently.

4. Explain why journalists should familiarize themselves with each of the following aspects of a research study before writing a news report about the study. a. Whether the study was a randomized experiment or an observational study b. How many different statistical analyses were done c. The magnitude of effects examined in the study and whether or not they were statistically significant Answer: a. If the study was a randomized experiment, the journalist could write the results using a cause and effect conclusion. But if it was an observational study, the journalist should not state or imply that a causal connection can be made. b. If multiple statistical analyses were done to answer the same question, the researchers could have chosen the one that best supported their hypothesis. If multiple analyses were done to answer different questions, there could be some that result in statistical significance even if all of the null hypotheses are true. c. It is important to report on both the practical significance and the statistical significance of the results, and to explain them so that the readers understand the difference. Knowing the magnitude of the effects determines whether they are of practical significance, and knowing whether the results are statistically significant determines whether it can be concluded that there is a real difference or effect in the population represented by the sample.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 26: Ethics in Statistical Studies

ANSWERS TO THE MINI PROJECTS 1. Find an example of a news story that reports statistical results on a topic of interest to you and that includes enough information for you to find the original research report. a. Read the original research report. Discuss whether the news story excluded anything important and, if so, what. Also discuss whether there were any misleading statements in the news report, based on the information in the original research report. b. Write your own comprehensive and fair news story reporting the results.

Answer: The purpose of this project is to give students experience with looking beyond a news story to get more information. The project will give them experience with critically evaluating reports by journalists, and perhaps learning how to report news stories about research studies themselves. It also will give them practice with putting the ideas and lessons in this book into use in their everyday lives.

2. Visit the websites of one or more professional organizations related to your major. (You may need to ask one of your professors to help with this.) Find one that has a code of ethics. Describe whether the code includes anything about research methods. If not, explain why you think nothing is included. If so, briefly summarize what is included. Include the website address with your answer.

Answer: Almost all professions have a code of ethics, and often it is developed by professional and academic societies related to the profession. This project has two purposes. One is to acquaint students with what professional societies exist in professions related to their major. The other is to learn what ethical issues are of concern to practitioners in jobs related to their major.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 27: Putting What You Have Learned to the Test

Solution and Answer Guide UTTS, SEEING THROUGH STATISTICS, 5E, 2025, 9780357757550; CHAPTER 27: PUTTING WHAT YOU HAVE LEARNED TO THE TEST

TABLE OF CONTENTS Answers to the Reflection Questions for Case Study 27.1................................................1 Answers to the Reflection Questions for Case Study 27.2 .............................................. 2 Answers to the Reflection Questions for Case Study 27.3 .............................................. 2 Answers to the Reflection Questions for Case Study 27.4 .............................................. 3 Answers to the Reflection Questions for Case Study 27.5 .............................................. 4 Answers to the Reflection Questions for Case Study 27.6 .............................................. 4 Answers to the Reflection Questions for Case Study 27.7 .............................................. 5 Answers to the Reflection Questions for Case Study 27.8 .............................................. 6 Answers to the Reflection Questions for Case Study 27.9 .............................................. 7

ANSWERS TO THE REFLECTION QUESTIONS FOR CASE STUDY 27.1 1. Participants in this study were undergraduate students enrolled in an introductory psychology course at a Canadian university. Do you think the results can be extended to a larger population beyond just the students in the study? Explain.

Answer: The extent to which the results can be extended to larger populations depends on whether the types of behaviors mentioned occur throughout different age groups and life circumstances. It is likely that the results can be extended at least to students at other universities, and probably to people in the same general age group as the students in the study.

2. In discussing their work, the authors of the original research noted “Our use of retrospective questionnaires is an important limiting factor. Such questionnaires rely on long-term memory which may be unreliable or confounded by other factors” (Nielsen et al., 2009, p. 1635). Explain what this quote means.

Answer: The study asked participants in groups 1 and 2 to recall what happened when they first woke up over the previous year, and in group 3 the question did not

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 27: Putting What You Have Learned to the Test

specify a time period. It would be surprising if the participants could all remember whether they had the suggested behaviors in the past year, or ever.

It’s possible that the power of suggestion in the questions (for groups 2 and 3) would instead lead them to imagine that they had those experiences.

ANSWERS TO THE REFLECTION QUESTIONS FOR CASE STUDY 27.2 1. Read the last two paragraphs of The New York Times quote. The reporter made it clear that the term “significant” referred to statistical significance. Do you think readers of the news story would be misled into thinking that the results for mild cognitive impairment had greater practical significance than those for probable dementia? Explain.

Answer: Yes, readers are likely to conclude that the results for mild cognitive impairment had greater practical significance for two reasons. First, the entire premise of the news story is that the research offers hope for “staving off dementia” as noted in the headline. Second, the last line of the story gives the magnitude of the effect, by noting that the intensive treatment group had “a 19 percent lower risk of mild cognitive impairment.” That reduction in risk seems like it has practical significance.

2. The lead researcher was quoted in another part of The New York Times story as follows: “Dr. Williamson said the trial has changed how he treats patients, offering those with blood pressure over 130 the intensive treatment. ‘I’ll tell them it will give you a 19 percent lower chance of developing early memory loss,’ he said.” Explain how Dr. Williamson came up with the “19 percent lower chance of memory loss” statistic.

Answer: The “19 percent lower chance of memory loss” is found from the quote given from the original journal article stating that “Intensive BP control significantly reduced the risk of mild cognitive impairment” with a hazard ratio of 0.81. The hazard ratio is similar to a relative risk, and a relative risk of 0.81 means that there is a reduced risk of 0.19, or 19%.

ANSWERS TO THE REFLECTION QUESTIONS FOR CASE STUDY 27.3 1. The news story explains that there were cameras installed in the cars of the participants that recorded what they were doing. Do you think that fact may have altered the participants’ behavior? Explain.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 27: Putting What You Have Learned to the Test

Answer: The cameras probably did alter the participants’ behavior, knowing that their actions would be viewed and analyzed by the researchers. The risky behaviors observed in the study may have underestimated how often drivers engage in those behaviors when they are not being watched.

2. The 95% confidence intervals for the odds ratios associated with dialing a phone are 2.83 to 24.42 for the novice drivers and 1.38 to 4.54 for the experienced drivers. Why do you think the confidence interval is so much wider for the novice drivers than for the experienced drivers?

Answer: There were only 42 novice drivers in the study, compared with 109 experienced drivers. Smaller sample sizes always result in less precise estimates than larger sample sizes.

ANSWERS TO THE REFLECTION QUESTIONS FOR CASE STUDY 27.4 1. Why is it important to know whether this study was observational or a randomized experiment? Your answer should be specific to this study.

Answer: It is important to know whether participants were randomly assigned to “a morning bout of exercise” (a randomized experiment) or if they were observed doing what they would do naturally (an observational study). If they were observed doing what they do naturally, there could be confounding variables that differ for those who prefer to exercise in the morning and those who don’t, that might also be related to cognitive performance throughout the day.

In that case, it cannot be concluded that the morning exercise has a causal relationship to better cognitive performance. But if participants were randomly assigned to morning exercise (or not) then a cause and effect conclusion can be made.

2. Why is it important to know that there were five different aspects of cognition that were investigated, when only two were reported to have significant results at the 0.05 significance level?

Answer: This question addresses the problem of multiple testing in general. It is important to know how many different tests were performed because if all of the null hypotheses are true but multiple tests are performed, some of the null hypotheses may be rejected by chance. In other words, if lots of tests are performed, it’s more likely that some statistically significant results are type 1 errors.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 27: Putting What You Have Learned to the Test

ANSWERS TO THE REFLECTION QUESTIONS FOR CASE STUDY 27.5 1. The original source for this survey reported that 93% of respondents had a Facebook account at the time of the survey in 2018. According to the Pew Research Center (Gramlich, 2021), at that time, about 69% of adults in the United States used Facebook. Explain whether you think the results of this survey were likely to apply to the population of adults in the United States at the time of the survey for all, some, or none of the questions.

Answer: Many of the questions focused on whether people would give up social media in exchange for other activities, and responses to those question would clearly depend on how much people used social media to begin with. Given that a higher percentage of the people in the study than in the whole adult population had a Facebook account, the results for questions related to social media are not likely to apply to the population of all adults. However, some of the questions were not related to social media use, such as willingness to part with possessions. For those questions, it’s possible that the results would be more representative of the adult population.

2. In addition to the 93% of respondents who had a Facebook account, there were 16 other social media outlets represented. For example, 41.6% of respondents said they used Reddit. One of the quotes in the original source was “Our survey found that the average American has accounts on 6.2 different social media platforms and admits to checking them an average of 5 times a day.” Comment on this quote about “the average American.”

Answer: The quote about “the average American” is not accurate if it is based on the participants in this study. The percentages of the participants saying they use social media sites such as Facebook and Reddit were higher in this study than in the general population. It’s almost surely the case that the response to how often people check their social media accounts would be higher for this sample than for the general population as well.

ANSWERS TO THE REFLECTION QUESTIONS FOR CASE STUDY 27.6 1. Coffee consumption for the participants in the study was measured by asking “How many cups of coffee do you drink each day?” and then asking them to choose one type (instant, filtered, decaffeinated, other). Discuss possible problems with the way these questions were asked and how they could have been asked differently. © 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 27: Putting What You Have Learned to the Test

Answer: There are possible problems because not all terms are clearly defined. First, there is no definition given for how large a cup of coffee is defined to be. For some participants, it could be a small two-ounce espresso, while for others it could be a large 12-ounce mug. Another possible problem is that respondents were asked to choose just one type of coffee. For instance, some people might drink filtered coffee in the morning and then have decaffeinated coffee in the afternoon or evening. The survey did not allow them to give that as an answer. A more comprehensive way to ask the question would be to specify how many ounces or milliliters of each type of coffee the person typically consumes in a day.

2. Discuss whether it would be possible to conduct a randomized experiment to investigate the relationship between coffee consumption and mortality.

Answer: It would not be possible to conduct a randomized experiment, at least not using humans. People cannot be randomly assigned to drink specific amounts and types of coffee over a period of many years, which is what would be necessary to study the link with mortality.

ANSWERS TO THE REFLECTION QUESTIONS FOR CASE STUDY 27.7 1. The survey of Australian students quoted in the story asked the students to use a 5-point scale to rate their level of disapproval for each of a list of over 70 actions against animals, such as using animals for testing cosmetics, keeping egg-laying chickens in small cages, and killing mice with a mouse trap.

The final question was “I would be prepared to pay a higher price for meat to cover the cost of more humane methods of rearing animals for slaughter” (Braithwaite and Braithwaite, 1982). Only 20% disagreed. Discuss the extent to which you think the wording of the question and its placement at the end of the survey might affect the responses.

Answer: The wording of the question makes it clear that the researchers would prefer a “yes” answer. The placement of the question at the end of a list of questions about abuse of animals would also indicate what answer the researchers wanted. Additionally, respondents might be more likely to respond with a “yes” answer after reading a long list of ways animals can be abused.

2. Braithwaite and Braithwaite (1982) claim that people’s attitudes and behaviors related to animals are inconsistent. In addition to the pâté example, they note that 89% of students disapproved of “keeping laying chickens in battery cages which are so small that they cannot spread their wings,” yet most students eat eggs produced by those chickens. Discuss the extent to which you think the authors have justified their claim that attitudes and behaviors are inconsistent.

© 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 27: Putting What You Have Learned to the Test

Answer: The claim is not well-justified. Most of the questions about attitudes used biased wording to indicate the desired answer. The two examples given, about force-feeding geese and keeping chickens in a way that they can’t spread their wings illustrate that fact. Therefore, it could be the difference in how the questions about attitudes and behavior were worded that created the discrepancy, and not that the students are inconsistent in their attitudes and behaviors.

ANSWERS TO THE REFLECTION QUESTIONS FOR CASE STUDY 27.8 1. As quoted in Chapter 26, one of the considerations in the ethics code of the American Psychological Association (APA) is “Psychologists do not conduct a study involving deception unless they have determined that the use of deceptive techniques is justified by the study’s significant prospective scientific, educational, or applied value and that effective nondeceptive alternative procedures are not feasible.” Clearly, the experiment described in this case study involved deception. Discuss whether you think it meets this APA criterion for ethical research.

Answer: There is no right or wrong answer to this question; it depends on the extent to which you think the results of the experiment justify the deception. Note that “effective nondeceptive alternative procedures are not feasible” as described at the end of the APA quote. To answer the question the researchers hoped to answer, they needed to use deception. The question is whether the study’s applied value justifies the deception. One could argue that it does, because it is the first study to show that false memories of committing crimes can be implanted, which has implications for criminal prosecutions.

2. Suppose you are on a jury and the defendant has confessed to a crime after being interrogated by the police. The defendant’s lawyer presents the results of the research in this case study and claims that it could mean that the defendant didn’t commit the crime. Discuss the extent to which the distinction between false memories and false beliefs would be relevant to you in assessing the lawyer’s argument.

Answer: The distinction between false memories and false beliefs seems less important than the fact that either one can easily be implanted into someone’s memories. It might matter if the evidence from the suspect was worded in such a way that it clearly represented a memory and not a belief. The research showed that false beliefs are easier to implant than false memories.

© 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Solution and Answer Guide: Utts, Seeing Through Statistics, 5e, 2025, 9780357757550; Chapter 27: Putting What You Have Learned to the Test

ANSWERS TO THE REFLECTION QUESTIONS FOR CASE STUDY 27.9 1. Remember that the concept of “ecological validity” refers to whether the conditions in an experiment mimic the real-world circumstances in which people would receive the “treatment.” Comment on how ecological validity would or would not be possible if researchers were to randomly assign people to spend time in nature or not for a period of a few months.

Answer: Ecological validity in this situation refers to whether the benefits of spending time in nature when you have been randomly assigned to do so would be similar to those benefits if you were spending time in nature by your own choice. One way in which ecological validity could be incorporated into a randomized experiment to some extent would be to allow people to choose where and when to spend the time they were randomly assigned to spend in nature. However, for people randomly assigned to spend less time in nature but who typically like to do so, there could be a problem with ecological validity. They might find other ways to get similar benefits, such as spending time meditating or relaxing in other ways. One solution would be to match people into pairs based on how much time they routinely spend in nature, and randomly assign one person to spend slightly more time and one to spend slightly less.

2. Here are two quotes from the news story that seem to imply a cause and effect relationship: “It’s possible that the solution to some of the stress and anxiety troubling folks of all ages is just a stroll through the neighborhood park away” and “In 2019, researchers from the University of Michigan published findings in Frontiers in Psychology showing that urban nature experiences reduce stress.”

Read the news story and the discussion, and comment on whether a causal connection is justified in each case. (If you need more details for the study leading to the first quote, you can read Example 20.2, pages 428–429.)

Answer: The first quote is based on an observational study, so a causal connection is not justified. It examined the association between time spent in nature and the taking of psychotropic medicines. It’s possible that people who are healthier are more likely to spend time in nature and less likely to take those medications. There are other possible non-causal explanations. The second quote is based on an experiment that measured markers of stress in saliva before and after spending time in nature. A causal connection between time in nature and stress is justified in that case.

© 2025 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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