Biostatistics for the Biological and Health Sciences, Global Edition, 2nd edition Marc M Triola Solu

Type:

Solution Manual

Biostatistics for the Biological Resource: and Health Sciences, Global Edition Edition:

2nd Edition

Author(s):

Mac Triola Mario Triola Jason Roy

INSTRUCTOR’S SOLUTIONS MANUAL JAMES LAPP Colorado Mesa University

BIOSTATISTICS FOR THE BIOLOGICAL AND HEALTH SCIENCES

S ECOND E DITION G LOBAL E DITION

Marc Triola New York University School of Medicine

Mario F. Triola Dutchess Community College

Jason Roy

CONTENTS Chapter 1: Introduction To Statistics Section 1-1: Statistical and Critical Thinking ............................................................................1 Section 1-2: Types of Data ........................................................................................................2 Section 1-3: Collecting Sample Data .........................................................................................3 Chapter Quick Quiz ...................................................................................................................5 Review Exercises .......................................................................................................................5 Cumulative Review Exercises ...................................................................................................6

Chapter 2: Exploring Data with Tables and Graphs Section 2-1: Frequency Distributions for Organizing and Summarizing Data ..........................7 Section 2-2: Histograms ...........................................................................................................12 Section 2-3: Graphs That Enlighten and Graphs That Deceive ...............................................14 Section 2-4: Scatterplots, Correlation, and Regression ...........................................................17 Chapter Quick Quiz .................................................................................................................18 Review Exercises .....................................................................................................................19 Cumulative Review Exercises .................................................................................................20

Chapter 3: Describing, Exploring, and Comparing Data Section 3-1: Measures of Center ..............................................................................................23 Section 3-2: Measures of Variation .........................................................................................27 Section 3-3: Measures of Relative Standing and Boxplots ......................................................31 Chapter Quick Quiz .................................................................................................................34 Review Exercises .....................................................................................................................35 Cumulative Review Exercises .................................................................................................35

Chapter 4: Probability Section 4-1: Basic Concepts of Probability .............................................................................37 Section 4-2: Addition Rule and Multiplication Rule ...............................................................38 Section 4-3: Complements, Conditional Probability, and Bayes’ Theorem ............................41 Section 4-4: Risks and Odds ....................................................................................................42 Section 4-5: Rates of Mortality, Fertility, and Morbidity ........................................................44 Section 4-6: Counting ..............................................................................................................46 Chapter Quick Quiz .................................................................................................................48 Review Exercises .....................................................................................................................48 Cumulative Review Exercises .................................................................................................49

Chapter 5: Discrete Probability Distributions Section 5-1: Probability Distributions .....................................................................................51 Section 5-2: Binomial Probability Distributions......................................................................52 Section 5-3: Poisson Probability Distributions ........................................................................55 Chapter Quick Quiz .................................................................................................................57 Review Exercises .....................................................................................................................58 Cumulative Review Exercises .................................................................................................59

Chapter 6: Normal Probability Distributions Section 6-1: The Standard Normal Distribution ......................................................................61 Section 6-2: Real Applications of Normal Distributions .........................................................62 Section 6-3: Sampling Distributions and Estimators ...............................................................65 Section 6-4: The Central Limit Theorem .................................................................................69 Section 6-5: Assessing Normality ............................................................................................72 Section 6-6: Normal as Approximation to Binomial ...............................................................75 Chapter Quick Quiz .................................................................................................................77 Review Exercises .....................................................................................................................77 Cumulative Review Exercises .................................................................................................79

Chapter 7: Estimating Parameters and Determining Sample Sizes Section 7-1: Estimating a Population Proportion.....................................................................81 Section 7-2: Estimating a Population Mean.............................................................................85 Section 7-3: Estimating a Population Standard Deviation or Variance ...................................89 Section 7-4: Bootstrapping: Using Technology for Estimates ................................................92 Chapter Quick Quiz .................................................................................................................94 Review Exercises .....................................................................................................................94 Cumulative Review Exercises .................................................................................................96

Chapter 8: Hypothesis Testing Section 8-1: Basics of Hypothesis Testing ..............................................................................97 Section 8-2: Testing a Claim About a Proportion ....................................................................99 Section 8-3: Testing a Claim About a Mean ..........................................................................104 Section 8-4: Testing a Claim About a Standard Deviation or Variance ................................108 Chapter Quick Quiz ...............................................................................................................110 Review Exercises ...................................................................................................................111 Cumulative Review Exercises ...............................................................................................112

Chapter 9: Inferences from Two Samples Section 9-1: Two Proportions ................................................................................................115 Section 9-2: Two Means: Independent Samples ....................................................................124 Section 9-3: Two Dependent Samples (Matched Pairs) ........................................................130 Section 9-4: Two Variances or Standard Deviations .............................................................134 Chapter Quick Quiz ...............................................................................................................136 Review Exercises ...................................................................................................................137 Cumulative Review Exercises ...............................................................................................138

Chapter 10: Correlation and Regression Section 10-1: Correlation .......................................................................................................141 Section 10-2: Regression .......................................................................................................148 Section 10-3: Prediction Intervals and Variation ...................................................................156 Section 10-4: Multiple Regression.........................................................................................158 Section 10-5: Dummy Variables and Logistic Regression ....................................................160 Chapter Quick Quiz ...............................................................................................................162 Review Exercises ...................................................................................................................162 Cumulative Review Exercises ...............................................................................................164

Chapter 11: Goodness-of-Fit and Contingency Tables Section 11-1: Goodness-of-Fit ...............................................................................................167 Section 11-2: Contingency Tables .........................................................................................170 Chapter Quick Quiz ...............................................................................................................174 Review Exercises ...................................................................................................................175 Cumulative Review Exercises ...............................................................................................176

Chapter 12: Analysis of Variance Section 12-1: One-Way ANOVA ..........................................................................................177 Section 12-2: Two-Way ANOVA .........................................................................................178 Chapter Quick Quiz ...............................................................................................................179 Review Exercises ...................................................................................................................180 Cumulative Review Exercises ...............................................................................................180

Chapter 13: Nonparametric Tests Section 13-2: Sign Test ..........................................................................................................183 Section 13-3: Wilcoxon Signed-Ranks Test for Matched Pairs ............................................184 Section 13-4: Wilcoxon Rank-Sum Test for Two Independent samples ...............................185 Section 13-5: Kruskal-Wallis Test for Three or More Samples ............................................188 Section 13-6: Rank Correlation .............................................................................................190 Chapter Quick Quiz ...............................................................................................................191 Review Exercises ...................................................................................................................192 Cumulative Review Exercises ...............................................................................................192

Chapter 14: Survival Analysis Section 14-1: Life Tables .......................................................................................................195 Section 14-2: Kaplan-Meier Survival Analysis .....................................................................197 Chapter Quick Quiz ...............................................................................................................198 Review Exercises ...................................................................................................................199 Cumulative Review Exercises ...............................................................................................200

Section 1-1: Statistical and Critical Thinking

1

Chapter 1: Introduction to Statistics Section 1-1: Statistical and Critical Thinking 1. The respondents are a voluntary response sample or a self-selected sample. Because those with strong interests in the topic are more likely to respond, it is very possible that their responses do not reflect the opinions or behavior of the general population.

2.

a. The sample consists of the 10,000 adults who were surveyed. The population consists of all the adults. b. When asked, respondents might be inclined to avoid the shame of the unhealthy habit of not brushing their teeth at night, so the reported rate of 45% might well be much higher than it is in reality. It is generally better to observe or measure human behavior than to ask subjects about it.

3. Statistical significance is indicated when methods of statistics are used to reach a conclusion that a treatment is effective, but common sense might suggest that the treatment does not make enough of a difference to justify its use or to be practical. Yes, it is possible for a study to have statistical significance, but not practical significance.

4. No. Correlation does not imply causation. The example illustrates a correlation that is clearly not the result of any interaction or cause effect relationship between deaths in swimming pools and power generated from nuclear power plants.

5.

No, there does not appear to be a potential to create a bias.

6.

Yes, there does appear to be a potential to create bias.

7.

No, there does not appear to be a potential to create a bias.

8.

Yes, there does appear to be a potential to create a bias.

9.

The sample is a voluntary response sample and has strong potential to be flawed.

10. The samples are voluntary response samples and have potential for being flawed, but this approach might be necessary due to ethical considerations involved in randomly selecting subjects and somehow imposing treatments on them.

11. The sampling method appears to be sound.

12. The sampling method appears to be sound.

13. With only a 1% chance of getting such results with a program that has no effect, the program appears to have statistical significance. Also, because the average loss of 22 pounds does seem substantial, the program appears to also have practical significance.

14. Because there is a 0.3% chance of getting such results by chance, the increase in scores does appear to have statistical significance. The typical increase of 5 points suggests that the course does have practical significance.

The course does appear to be successful.

15. Because there is a 19% chance of getting that many girls by chance, the method appears to lack statistical significance. The result of 1020 girls in 2000 births (51% girls) is above the approximately 50% rate expected by chance, but it does not appear to be high enough to have practical significance. Not many couples would bother with a procedure that raises the likelihood of a girl from 50% to 51%.

16. Because there is a 15% chance of getting such results with a medication that has no effect, the medication does not appear to have statistical significance. Because the average decrease is only 2 mm Hg, the medication does not appear to have practical significance.

17. Yes. Each column of 8 AM and 12 AM temperatures is recorded from the same subject, so each pair is matched.

18. No. The source is from university researchers who do not appear to gain from distorting the data.

19. The data can be used to address the issue of whether there is a correlation between body temperatures at 8 AM and at 12 AM. Also, the data can be used to determine whether there are differences between body temperatures at 8 AM and at 12 AM.

.

2

Chapter 1: Introduction to Statistics

20. Because the differences could easily occur by chance (with a 64% chance), the differences do not appear to have statistical significance.

21. No. The white blood cell counts measure a different quantity than the red blood cell counts, so their differences are meaningless.

22. The issue that can be addressed is whether there is a correlation, or association, between white blood cell counts and red blood cell counts.

23. No. The National Center for Health Statistics has no reason to collect or present the data in a way that is biased.

24. No. Correlation does not imply causation, so a statistical correlation between white blood cell counts and red blood cell counts should not be used to conclude that higher white blood cell counts are the cause of higher red blood cell counts.

25. It is questionable that the sponsor is the Idaho Potato Commission and the favorite vegetable is potatoes.

26. The sample is a voluntary response sample, so there is a good chance that the results do not reflect the larger population of people who have a water preference.

27. The correlation, or association, between two variables does not mean that one of the variables is the cause of the other. Correlation does not imply causation. Clearly, cheese consumption is not directly related in any way to bedsheet deaths.

28. The sponsor of the poll is an electronic cigarette maker, so the sponsor does have an interest in the poll results.

The source is questionable.

29. a. 0.45 (3014 ) = 1356.3 adults b. No. Because the result is a count of the people among the 3014 who were surveyed, the result must be a whole number. c. 1356 adults d.

1206 = 0.40, or 40% 1206 + 1808

30. a. 0.19 (1543) = 293.17 women b. No. Because the result is a count of women among the 1543 who were surveyed, the result must be a whole number. c. 293 women

219 = 0.15, or 15% 1458 e. Interpretations of a “typical” week and what it means to “kick back and relax” might vary considerably by different survey respondents. The survey might be improved by asking about behavior within “the past seven days” instead of a “typical” week. Instead of “kick back and relax,” respondents might be surveyed about specific behavior, such as reading, taking a nap, watching television, listening to music, or going for a walk. d.

31. The wording of the question is biased and tends to encourage negative responses. The sample size of 20 is too small. Survey respondents are self-selected instead of being randomly selected by the newspaper. If 20 readers respond, the percentages should be multiples of 5, so 87% and 13% are not possible results.

32. All percentages of success should be multiples of 5. The given percentages cannot be correct.

Section 1-2: Types of Data 1. a. The sample is the 1020 adults who were surveyed. The population is all adults in the United States. b. statistic 2. a. ratio c. categorical b. discrete

.

Section 1-3: Collecting Sample Data 3.

a. categorical b. quantitative c. quantitative d. categorical

4.

Only part (a) describes discrete data.

5.

parameter

6.

statistic

7.

parameter

8.

parameter

9.

statistic

3

15. continuous 16. discrete 17. continuous 18. continuous 19. discrete 20. discrete 21. interval 22. ratio 23. nominal 24. ordinal

10. statistic

25. nominal

11. statistic

26. nominal

12. statistic

27. ratio

13. discrete

28. ordinal

14. continuous

29. The numbers are not counts or measures of anything. They are at the nominal level of measurement, and it makes no sense to compute the average (mean) of them.

30. The digits are not counts or measures of anything. They are at the nominal level of measurement and it makes no sense to calculate their average (mean).

31. The temperatures are at the interval level of measurement. Because there is no natural starting point with 0 F representing “no heat,” ratios such as “twice” make no sense, so it is wrong to say that it is twice as warm at the author’s home as it is in Auckland, New Zealand.

32. The ranks are at the ordinal level of measurement. Differences between the universities cannot be determined, so there is no way to know whether the difference between Princeton and Harvard is the same as the difference between Yale and Columbia.

33. a. Continuous, because the number of possible values is infinite and not countable. b. Discrete, because the number of possible values is finite. c. Discrete, because the number of possible values is finite. d. Discrete, because the number of possible values is infinite and countable.

Section 1-3: Collecting Sample Data 1. The study is an experiment because subjects were given treatments.

2. The subjects in the study did not know whether they were taking a placebo or the paracetamol medication, and those who administered the pills also did not know.

3. The group sample sizes of 547, 550, and 546 are all large so that the researchers could see the effects of the paracetamol treatment.

4. The sample appears to be a convenience sample. Given that the subjects were randomly assigned to the three different treatment groups, it appears that the results of the study are good because they are not likely to be distorted from bias, but we should investigate the sample groups to ensure that they are not fundamentally different from the population.

5. The sample appears to be a convenience sample. By e-mailing the survey to a readily available group of Internet users, it was easy to obtain results. Although there is a real potential for getting a sample group that is not representative of the population, indications of which ear is used for cell phone calls and which hand is dominant do not appear to be factors that would be distorted much by a sample bias. .

4

Chapter 1: Introduction to Statistics

6.

The study is an observational study because the subjects were not given any treatment.

7. With 717 responses, the response rate is 14%, which does appear to be quite low. In general, a very low response rate creates a serious potential for getting a biased sample that consists of those with a special interest in the topic.

8.

Answers vary, but the following are good possibilities. a. Obtain a printed copy of the class roster, assign consecutive numbers (integers), then use a computer to randomly generate six of those numbers. b. Select every third student leaving class until six students are chosen. c. Randomly select three males and three females. d. Randomly select a row, and then select the students in that row. (Use only the first six to meet the requirement of a sample of size six.)

9.

systematic

15. random

10. convenience

16. systematic

11. random

17. random

12. cluster

18. cluster

13. stratified

19. convenience

14. stratified

20. systematic

21. Observational study. The sample is a convenience sample consisting of subjects who decided themselves to respond. Such voluntary response samples have a high chance of not being representative of the larger population, so the sample may well be biased. The question was posted in an electronic edition of a newspaper, so the sample is biased from the beginning.

22. Experiment. The sample subjects consist of male physicians only. It would have been better to include females.

Also, it would be better to include male and females who are not physicians.

23. Experiment. This experiment would create an extremely dangerous and illegal situation that has a real potential to result in injury or death. It’s difficult enough to drive in New York City while being completely sober.

24. Observational study. The sample of four males and four females is too small.

25. Experiment. The biased sample created by using subjects from New York City cannot be fixed by using a larger sample. The larger sample will still be a biased sample that is not representative of subjects in the United States.

26. Experiment. Calling the subjects and asking them to report their weights has a high risk of getting results that do not reflect the actual weights. It would have been much better to somehow measure the weights instead of asking the subjects to report them.

27. Observational study. Respondents who have been convicted of felonies are not likely to respond honestly to the second question. The survey will suffer from a “social desirability bias” because subjects will tend to respond in ways that will be viewed favorably by those conducting the survey.

28. Observational study. The number of responses is very small, and the response rate of only 1.52% is far too small. With such a low response rate, there is a real possibility that the sample of respondents is biased and consists only of those with special interests in the survey topic.

29. prospective study

33. matched pairs design

30. retrospective study

34. randomized block design

31. cross-sectional study

35. completely randomized design

32. prospective study

36. matched pairs design

.

Chapter Quick Quiz

5

37. Prospective: The experiment was begun and results were followed forward in time.

Randomized: Subjects were assigned to the different groups through a process of random selection, whereby they had the same chance of belonging to each group.

Double-blind: The subjects did not know which of the three groups they were in, and the people who evaluated results did not know either.

Placebo-controlled: There was a group of subjects who were given a placebo; by comparing the placebo group to the two treatment groups, the effects of the treatments might be better understood.

38. a. Not a simple random sample, but it is a random sample. b. Simple random sample and also a random sample. c. Not a simple random sample and not a random sample.

Chapter Quick Quiz 1. No. The numbers do not measure or count anything.

2.

nominal

7.

no

3.

continuous

8.

statistic

4.

quantitative data

9.

observational study

5.

ratio

10. false

6.

no

Review Exercises 1. a. discrete c. stratified b. ratio d. cluster e. The mailed responses would be a voluntary response sample, so those with strong opinions are more likely to respond. It is very possible that the results do not reflect the true opinions of the population of all customers.

2. The survey was sponsored by the American Laser Centers, and 24% said that the favorite body part is the face, which happens to be a body part often chosen for some type of laser treatment. The source is therefore questionable.

3.

The sample is a voluntary response sample, so the results are questionable.

4.

a. It uses a voluntary response sample, and those with special interests are more likely to respond, so it is very possible that the sample is not representative of the population. b. Because the statement refers to 72% of all Americans, it is a parameter (but it is probably based on a 72% rate from the sample, and the sample percentage is a statistic). c. observational study

5.

a. If they have no fat at all, they have 100% less than any other amount with fat, so the 125% figure cannot be correct. b. 0.58 (1182 ) = 685.56, so 686

6.

c. 331 1182 = 0.28, or 28%

Only part (c) is a simple random sample.

7. Because there is only a 4% chance of getting the results by chance, the method appears to have statistical significance. The result of 112 girls in 200 births is above the approximately 50% rate expected by chance, but it does not appear to be high enough to have practical significance. Not many couples would bother with a procedure that raises the likelihood of a girl from 50% to 56%.

8.

a. random b. stratified

c. nominal d. Statistic, because it is based on a sample.

e. The mailed responses would be a voluntary response sample. Those with strong opinions about the topic would be more likely to respond, so it is very possible that the results would not reflect the true opinions of the population of all adults. .

6

Chapter 1: Introduction to Statistics

9.

a. systematic; It is likely to result in a representative sample. b. random; It is likely to result in a representative sample. c. cluster; It is likely to result in a representative sample. d. stratified; It is likely to result in a representative sample. e. convenience; It is very possible that the sample is not representative.

10. a. 0.52 (1500) = 780 adults b.

345 = 0.23 or 23% 1500

727 773 = 0.485 or 48.5%; Women: = 0.515 or 51.5% 1500 1500 d. No, although this is a subjective judgment. e. No, although this is a subjective judgment. c. Men:

Cumulative Review Exercises

5 + 16 + 17 + 13 + 13 + 14 + 15 + 15 = 13.5 mg. The tar contents all end as integers, suggesting that all 8 of the tar contents are rounded to the nearest integer, so that there are no decimals.

1.

The mean is

2.

0.56 = 0.015625

3.

272 - 176 = 16, which is an unusually high value.

6

4.

5.

8.

9.

98.2 - 98.6 = - 6.64 0.62 106 1.962 ×0.25 0.032

6.

7.

17 - 5 = 3 mg 4

(5 - 13.5)2

= 1067

(98.4 - 98.6)2 + (98.6 - 98.6)2 + (98.8 - 98.6)2 3- 1

= 0.04 = 0.20

0.67 = 0.0279936

10. 1214 = 1, 283,918, 464,548,86 (or about 1, 283,918, 464,549,000) 11. 712 = 13,841, 287, 201 (or about 13,841, 287,000) 12. 0.415 = 0.000001073741824

.

7

= 10.32 mg 2

Section 2-1: Frequency Distributions for Organizing and Summarizing Data

7

Chapter 2: Exploring Data with Tables and Graphs Section 2-1: Frequency Distributions for Organizing and Summarizing Data 1. The table summarizes measurements from 40 subjects. It is not possible to identify the exact values of all of the original cotinine measurements.

2. The classes of 0–100, 100–200, …, 400–500 overlap, so it is not always clear which class we should put a value in. For example, the value of 100 could go in the first class or the second class. The classes should be mutually exclusive.

3.

Cotinine (ng/Ml)

Relative Frequency

0–99 100–199 200–299 300–399 400–499

27.5% 30.0% 35.0% 2.5% 5.0%

4. The sum of the relative frequencies is 125%, but it should be 100%, with a small round off error. All of the relative frequencies appear to be roughly the same, but if they are from a normal distribution, they should start low, reach a maximum, and then decrease.

5.

Class width: 100 Class midpoints: 49.5, 149.5, 249.5, 349.5, 449.5, 549.5 Class boundaries: –0.5, 99.5, 199.5, 299.5, 399.5, 499.5, 599.5

6.

Class width: 90 Class midpoints: 1004.5, 1094.5, 1184.5, 1274.5, 1364.5,1454.5 Class boundaries: 959.5, 1049.5, 1139.5, 1229.5, 1319.5, 1409.5, 1499.5

7.

Class width: 20 Class midpoints: 39.5, 59.5, 79.5, 99.5, 119.5, 139.5, 159.5 Class boundaries: 29.5, 49.5, 69.5, 89.5, 109.5, 129.5, 149.5, 169.5

8.

Class width: 20 Class midpoints: 59.5, 79.5, 99.5, 119.5, 139.5 Class boundaries: 49.5, 69.5, 89.5, 109.5, 129.5, 149.5

9. No. The maximum frequency is in the first class instead of being near the middle, so the frequencies below the maximum do not mirror those above the maximum.

10. No. The frequencies start high and then decrease, so the frequencies below the maximum do not mirror those above the maximum.

11. By symmetry, the last three frequencies would be 20, 9, and 7. The middle frequency would be 147 - 2 (20) - 2 (9) - 2 (7 ) = 75.

12. No. The maximum frequency is in the second class instead of being near the middle, and the frequencies in the first half of the table are much more than those in the second half of the table, so the frequencies below the maximum do not mirror those above the maximum.

.

8

Chapter 2: Exploring Data with Tables and Graphs

13. The pulse rates appear to have a distribution that is approximately normal.

Pulse Rate (Male)

Frequency

40–49 50–59 60–69 70–79 80–89 90–99 100–109

2 23 53 43 25 5 2

14. The pulse rates appear to have a distribution that is approximately normal.

Pulse Rate (Female)

Frequency

30–39 40–49 50–59 60–69 70–79 80–89 90–99 100–109

1 1 17 33 41 37 13 4

15. The verbal IQ scores appear to have a distribution that is approximately normal.

IQ (Verbal) 50–59 60–69 70–79 80–89 90–99 100–109 110–119 120–129

Frequency 3 8 13 26 18 6 2 2

16. The verbal IQ scores appear to have a distribution that is approximately normal.

IQ (Verbal)

Frequency

60–69 70–79 80–89 90–99 100–109

1 7 8 4 1

.

Section 2-1: Frequency Distributions for Organizing and Summarizing Data 17. Yes. The distribution appears to be approximately normal.

Red Blood Cell Count (Males)

Frequency

3.00–3.49 3.50–3.99 4.00–4.49 4.50–4.99 5.00–5.49 5.50–5.99

1 16 29 57 44 6

18. Yes. The distribution appears to be approximately normal.

Red Blood Cell Count (Females)

Frequency

3.00–3.49 3.50–3.99 4.00–4.49 4.50–4.99 5.00–5.49 5.50–5.99 6.00–6.49

1 20 85 33 7 0 1

Weight (kg) in September

Frequency

50–59 60–69 70–79 80–89 90–99

2 12 11 3 4

Weight (kg) in April

Frequency

40–49 50–59 60–69 70–79 80–89 90–99 100–109

2 20 28 8 7 1 1

19.

20.

.

9

10

Chapter 2: Exploring Data with Tables and Graphs

21. The frequency distribution suggests that the reported weights were not rounded since the last digits seem equally distributed.

Last Digit 0 1 2 3 4 5 6 7 8 9

Frequency 5 6 4 4 4 5 4 4 6 5

22. The frequency distribution suggests that the reported heights were rounded with disproportionately many 0s and 5s. This suggests that the results are not very accurate.

Last Digit 0 1 2 3 4 5 6 7 8 9

Frequency 15 3 2 1 5 16 2 1 4 1

23. The two distributions differ substantially. The presence of cotinine appears to be much higher for smokers than for nonsmokers exposed to smoke.

Cotinine (ng/mL)

Smokers

0–99 100–199 200–299 300–399 400–499 500–599

27.5% 30.0% 35.0% 2.5% 5.0% 0%

.

Nonsmokers Exposed to Smoke 85.0% 5.0% 2.5% 2.5% 0% 5.0%

Section 2-1: Frequency Distributions for Organizing and Summarizing Data 24. The frequency distributions show moderate difference. It appears females have slightly higher weights Weight (kg) 30–49 50–69 70–89 90–109 110–129 130–149 150–169

Males 0% 19.6% 43.1% 28.1% 7.9% 1.3% 0%

Females 5.4% 36.1% 33.3% 19.0% 3.4% 2.1% 0.7%

25.

Cotinine (Nonsmokers Exposed Cumulative Frequency to Smoke in ng/mL) Less than 100 34 Less than 200 36 Less than 300 37 Less than 400 38 Less than 500 38 Less than 600 40 26.

Brain Volume (cm3)

Frequency

Less than 1049 Less than 1139 Less than 1229 Less than 1319 Less than 1409 Less than 1499

6 13 16 18 19 20

27. a. The values of 551 and 543 are clearly outliers; the values of 384, 241, 197, and 178 could also be outliers. b. The number of classes increases from six to ten. The outlier can greatly increase the number of classes. If there are too many classes, we might use a larger class width with the effect that the true nature of the distribution may be hidden.

Cotinine (Nonsmokers Exposed to Smoke in ng/mL) 0–99 100–199 200–299 300–399 400–499 500–599 600–699 700–799 800–899 900–999

.

Frequency 34 2 1 1 0 2 0 0 0 1

11

12

Chapter 2: Exploring Data with Tables and Graphs

Section 2-2: Histograms 1. It is easier to see the distribution of the data by examining the graph of the histogram than by examining the numbers in a frequency distribution.

2. Not necessarily. Because the sample subjects themselves chose to be included, the voluntary response sample might not be representative of the population.

3.

With a data set that is so small, the true nature of the distribution cannot be seen with a histogram.

4. No, the term “normal distribution” has a different meaning than the term “normal” that is used in ordinary speech. A normal distribution will have a histogram that is approximately bell shaped. Determining whether a histogram fits the bell shape would be subjective.

5.

approximately 50

6.

Class width: 0.5 mm, lower limit of first class: 2.0 mm, upper limit of first class: 2.4 mm

7.

The largest possible value would be approximately 4.5 mm, which would not be an outlier.

8.

The sample does not appear to be from a normal distribution, since it is not symmetric about the middle class.

9.

The pulse rates of males appear to have a distribution that is approximately normal.

Histogram for Exercise 10

Histogram for Exercise 9 60

40 50

30

Frequency

Frequency

40 30

20

20

10 10 0

30

40

50

60

70

80

0

90 100 110

30

40

50

60

70

80

90 100 110

Pulse Rate (Female)

Pulse Rate (M ale)

10. The pulse rates of females appear to have a distribution that is approximately normal.

11. The IQ scores appear to have a distribution that is approximately normal.

Histogram for Exercise 12

Histogram for Exercise 11

9 25

8 7 6

Frequency

Frequency

20 15 10

5 4 3 2

5

1 0

60

80

100

120

0

60

70

80

90

100

Verbal IQ (High Lead)

Verbal IQ (Low Lead)

12. The IQ scores appear to have a distribution that is approximately normal.

.

110

Section 2-2: Histograms 13 13. Yes. The red blood cell counts appear to have a distribution that is very approximately normal, although some might describe the distribution as being left-skewed instead of normal.

Histogram for Exercise 14

Histogram for Exercise 13

90

60

80 50

70 60

Frequency

Frequency

40 30

50 40 30

20

20 10

10 0

3.0

3.5

4.0

4.5

5.0

5.5

0

6.0

3.0

3.5

Red Blood Cell Count (Female)

4.0

4.5

5.0

5.5

6.0

6.5

Red Blood Cell Count (M ale)

14. Yes. The red blood cell counts appear to have a distribution that is very approximately normal.

15. The histogram is shown below.

Histogram for Exercise 16

Histogram for Exercise 15

20

14 12

15

Frequency

Frequency

10 8 6 4

10

5

2 0

40

50

60

70

80

90

0

100

September Weight (kg) of M ales

50

60

70

80

90

April Weight (kg) of M ales

16. The histogram is shown above.

17. The histogram suggests that the reported weights were not rounded since the last digits seem equally distributed.

.

100

14

Chapter 2: Exploring Data with Tables and Graphs

18. The histogram suggests that the reported heights were rounded with disproportionately many 0s and 5s. This suggests that the results are not very accurate.

19. Only part (c) appears to represent data from a normal distribution. Part (a) has a systematic pattern that is not that of a straight line, part (b) has points that are not close to a straight-line pattern, and part (d) is really bad because it shows a systematic pattern and points that are not close to a straight-line pattern.

Section 2-3: Graphs That Enlighten and Graphs That Deceive 1. The data set is too small for a graph to reveal important characteristics of the data. With such a small data set, it would be better to simply list the data or place them in a table.

2. No. If the sample is a bad sample, such as one obtained from voluntary responses, there are no graphs or other techniques that can be used to salvage the data.

3. No. Graphs should be constructed in a way that is fair and objective. The readers should be allowed to make their own judgments, instead of being manipulated by misleading graphs.

4. Center, variation, distribution, outliers, change in the characteristics of data over time. The time-series graph does the best job of giving us insight into the change in the characteristics of data over time.

5.

The systolic blood pressure of 158 mmHg appears to be an outlier.

6.

The pulse rate of 104 beats per minute appears to be an outlier.

7.

The data are arranged in order from lowest to highest, as 112, 114, 116, and so on.

8.

The middle value is 66 beats per minute.

.

Section 2-3: Graphs That Enlighten and Graphs That Deceive

15

9. There has been a steep jump in the last five years, but the numbers of accidental deaths show a random trend in the initial years.

10. The number of fatalities is decreasing, most likely due to greater enforcement of DUI laws and greater public awareness campaigns.

6.5 6.0

Fatalities

5.5 5.0 4.5 4.0 3.5 3.0 1990

1995

2000

2005

2010

2015

Year

11. Misconduct includes fraud, duplication, and plagiarism, so it does appear to be a major factor.

900

Number of Retractions

800 700 600 500 400 300 200 100 0 Fraud

Error

Duplication

Other

Plagiarism

12. The overwhelming response was that thank-you notes should be sent to everyone who is met during a job interview. Given what is at stake, that seems like a wise strategy.

400

Number

300

200

100

0 Everyone

M ost Senior

M ost Time Best Conversation Don't Send

.

16

Chapter 2: Exploring Data with Tables and Graphs

13.

14.

15. The distribution appears to be roughly bell-shaped, so the distribution is approximately normal.

60 50

Frequency

40 30 20 10 0 34.5

44.5

54.5

64.5

74.5

84.5

94.5

104.5

Pulse Rate (M ale)

16. The distribution appears to be roughly bell-shaped, so the distribution is approximately normal.

40

Frequency

30

20

10

0 24.5

34.5

44.5

54.5

64.5

74.5

84.5

94.5

104.5 114.5

Pulse Rate (Female)

17. Because the vertical scale starts with a frequency of 200 instead of 0, the difference between the “no” and “yes” responses is greatly exaggerated. The graph makes it appear that about five times as many respondents said “no,” when the ratio is actually a little less than 2.5 to 1.

18. The two costs are one-dimensional in nature, but the baby bottles are three-dimensional objects. The $4500 cost isn’t even twice the $2600 cost, but the baby bottles make it appear that the larger cost is about five times the smaller cost.

.

Section 2-4: Scatterplots, Correlation, and Regression

17

Section 2-4: Scatterplots, Correlation, and Regression 1. The term linear refers to a straight line, and r measures how well a scatterplot of the sample paired data fits a straight-line pattern.

2. No. Finding the presence of a statistical correlation between two variables does not justify any conclusion that one of the variables is a cause of the other.

3.

A scatterplot is a graph of paired ( x, y ) quantitative data. It helps us by providing a visual image of the data

plotted as points, and such an image is helpful in enabling us to see patterns in the data and to recognize that there may be a correlation between the two variables.

4.

a. 0 b. –1

c. 1 d. 0

5.

There does appear to be a linear correlation between systolic and diastolic blood pressure.

6.

There does not appear to be a linear correlation between heights of mother and heights of their first daughters.

.

18

Chapter 2: Exploring Data with Tables and Graphs

7.

There does not appear to be a linear correlation between shoe sizes and heights of females.

8. There does appear to be a linear correlation between body temperatures at 12 AM on one day and at 12 AM on the following day.

9. With n = 5 pairs of data, the critical values are ±0.878. Because r = 0.781 is between −0.878 and 0.878, evidence is not sufficient to conclude that there is a linear correlation.

10. With n = 10 pairs of data, the critical values are ±0.632. Because r = −0.177 is between −0.632 and 0.632, evidence is not sufficient to conclude that there is a linear correlation.

11. With n = 7 pairs of data, the critical values are ±0.754. Because r = −0.283 is between −0.754 and 0.754, evidence is not sufficient to conclude that there is a linear correlation.

12. With n = 10 pairs of data, the critical value are ±0.632, . Because r = 0.374 is between −0.632 and 0.632, evidence is not sufficient to conclude that there is a linear correlation.

Chapter Quick Quiz 1. The class width is 0.12 - 0.08 = 0.04.

2.

The class boundaries are 0.075 and 0.115.

3.

No, it is impossible to determine the original values.

4.

16, 17, 18, 18, 19

7.

time-series graph

5.

bell-shaped

8.

scatterplot

6.

variation

9.

Pareto chart

10. A frequency distribution is in the format of a table, but a histogram is a graph. .

Section 2-4: Scatterplots, Correlation, and Regression Review Exercises 1.

Birth Weight (grams) 2000−2499 2500−2999 3000−3499 3500−3999 4000−4499

Frequency 1 4 6 6 3

2. Yes, the data appears to be from a population with a normal distribution because the bars start low, reach a maximum, and then decrease.

3.

By using fewer classes, the histogram does a better job of illustrating the distribution.

4.

The birth weight of 2100 grams appears to be an outlier.

.

19

20

Chapter 2: Exploring Data with Tables and Graphs

5.

Yes. There is a pattern suggesting that there is a relationship.

32.5 30.0

Neck Size (in.)

27.5 25.0 22.5 20.0 17.5 15.0 100 150 200 250 300 350 400 450 Weight (lb)

6.

a. time-series graph b. scatterplot

c. Pareto chart

7. By using a vertical scale that starts at 45% instead of 0%, the difference is greatly exaggerated. The graph creates the false impression that male enrollees outnumber female enrollees by a ratio of about 3:1, but the actual percentages of 53% and 47% are much closer than that.

Cumulative Review Exercises 1.

Grooming Time (min)

Frequency

0–9 10–19 20–29 30–39 40–49

2 3 9 4 2

2. The histogram is approximately bell-shaped. The frequencies increase to a maximum and then decrease, and the left half of the histogram is roughly a mirror image of the right half. The data do appear to be from a population with a normal distribution.

9 8

Frequency

7 6 5 4 3 2 1 0

0

10

20

30

Grooming Time (min)

.

40

50

Review Exercises

21

3.

4. There are disproportionately many last digits of 0 and 5. Fourteen of the 20 times have last digits of 0 or 5. It appears that the subjects reported their results and they tended to round the results. The data do not appear to be very accurate.

Last Digit

Frequency

0 1 2 3 4 5 6 7 8 9

5 0 2 0 1 9 0 2 1 0

5.

a. ratio b. continuous c. No. The grooming times are quantitative data. d. statistic

6.

The scatterplot helps address the issue of whether there is a correlation between heights of mothers and heights of their daughters. The scatterplot does not reveal a clear pattern suggesting that there is a correlation.

71

Daughter’s Height (in.)

70 69 68 67 66 65 58

60

62

64

M other’s Height (in.)

.

66

68

Section 3-1: Measures of Center

23

Chapter 3: Describing, Exploring, and Comparing Data Section 3-1: Measures of Center 1. The term average is not used in statistics. The term mean should be used for the result obtained by adding all of the sample values and dividing the total by the number of sample values.

2. No. The 50 amounts are all weighed equally in the calculation that yields the mean of 19.67, but some states have higher smoking rates, so the mean smoking rate should be calculated using a weighted mean that takes into account the smoking rates in the different states.

3.

They use different approaches for providing a value (or values) of the center or middle of the sorted list of data.

120 + 110 + 115 + 125 + 135 = 121 mmHg and the median is 120 mmHg. 5 120 + 110 + 115 + 125 + 135 + 1000 Using the six values, including the outlier of 10, x = = 267.5 mmHg and the 6 120 + 125 median is = 122.5 mmHg. 2 The outlier caused the mean to change by a substantial amount, but the median changed very little. The median is resistant to the effect of the outlier, but the mean is not resistant.

4.

Using the five values, x =

322 +

+ 755 + 800.45 + 1060 = $775.90.

10 755 + 800.45 The median is = $777.73.

2 The mode is $1060.

322 + 1060 The midrange is = $691.

2 Apart from the fact that the other charges are higher than those given, nothing meaningful can be known about the population of all charges.

5.

The mean is x =

26.8 + 27.0 + 27.1 + 27.3 + 27.4 + 27.7 + 27.9 + 28.1 + 28.2 + 28.3 + 28.3 + 28.4 = 27.71.

12 27.7 + 27.9 The median is = 27.80.

2 There is no mode.

26.8 + 28.4 = 27.60 The midrange is 2 The measures of center do not provide any information about the change of MCAT scores over time.

6.

The mean is x =

7 + 19 + 20 + 25 + 55 + 60 + 81 + 82 + 89 + 91 + 99 = 57.1.

11 The median is 60.0.

There is no mode.

7 + 99 = 53.0.

The midrange is 11 The jersey numbers are nominal data that are just replacements for names, and they do not measure or count anything, so the resulting statistics are meaningless.

7.

The mean is x =

8.

The mean is x =

189 + 190 + 190 + 195 + 202 + 225 + 235 + 252 + 254 + 305 + 305 = 231.1 lb.

11 The median is 225.0 lb.

The modes are 190 lb and 305 lb.

.

24

Chapter 3: Describing, Exploring, and Comparing Data

8.

(continued)

189 + 305 = 247.0 lb.

2 All of the weights are from players on only one of the 32 NFL teams, but it isn’t likely that any of the teams has weights that are substantially different from those of any other team, so it is reasonable to conclude that the weights are somewhat representative of all NFL players.

The midrange is

9.

The mean is x =

1 +1 +1+1+1+1+1+1+1+1+

+ 2 + 2 + 2 + 3+ 3+ 3+ 3+ 3+ 3+ 4 = 1.9.

25

The median is 2.0.

The mode is 1.

1+ 4 The midrange is = 2.5.

2 The mode of 1 correctly indicates that the smooth-yellow peas occur more than any other phenotype, but the other measures of center do not make sense with these data at the nominal level of measurement.

10. The mean is x =

950 + 1150 + 1200 + 1400 +

+ 1600 + 1600 + 1750 + 1800 = 1454.2, or $1454.2.

12

1500 + 1500 = 1500.0, or $1500.0.

2 The mode is $1500.

950 + 1800 The midrange is = 1375.0, or $1375.0.

2 The sample consists of “best buy” TVs, so it is not a random sample and is not likely to be representative of the population. The lowest price is a relevant statistic for someone planning to buy one of the TVs. The median is

0.51 + 0.74 + 0.89 + 1.04 + 1.18 + 1.38 + 1.41 + 1.42 + 1.45 + 1.45 + 1.49 = 1.178 W/kg.

11 The median is 1.380 W/kg.

The mode is 1.45 W/kg.

0.51 + 1.49 = 1.000 W/kg.

The midrange is 2 If concerned about radiation absorption, you might purchase the cell phone with the lowest absorption rate. All of the cell phones in the sample have absorption levels below the FCC maximum of 1.6 W/kg.

11. The mean is x =

12. The mean is x =

0+0+0+

+ 34 + 36 + 38 + 41 + 41 + 41 + 20

+ 53 + 54 + 55

= 32.6 mg.

38 + 41 = 39.5 mg.

2 The mode is 0 mg.

0 + 55 = 27.5 mg.

The midrange is 2 Americans consume some brands much more often than others, but the 20 brands are all weighted equally in the calculations, so the statistics are not necessarily representative of the population of all cans of the same 20 brands consumed by Americans.

The median is

75 + 106 + 108 + 111 + 112 + 127 + 129 + 133 + 171 = 119.11 firefighters.

9 The median is 112 firefighters.

There is no mode.

75 + 171 = 123 firefighters.

The midrange is 2

13. The mean is x =

.

Section 3-1: Measures of Center

25

13. (continued) The data are time series data, but the measures of center do not reveal anything about a trend consisting of a pattern of change over time.

8.6 + 9.1 + 9.1 + 9.1 + 9.3 + 9.3 + 9.4 + 9.8 + 9.9 + 10 + 10.4 = 9.45 in.

11 The median is 9.30 in.

The mode is 9.1 in.

8.6 + 10.4 The midrange is = 9.50 in.

2 Because the measurements were made in 1988, they are not necessarily representative of the current population of all Army women.

14. The mean is x =

53,323.00 +

+ 54,528.00 + 54,653.00 + + 57, 261.00 = $54,862.0.

10 54,528.00 + 54,653.00 The median is = $54,590.5.

2 There is no mode.

53,323.00 + 57, 261.00 The midrange is = $55,292.0.

2 Apart from the fact that all other colleges have tuition and fee amounts less than those listed, nothing meaningful can be known about the population.

15. The mean is x =

16. The mean is x =

0+0+0+0+0+0+

+ 9 + 10 + 10 + 20 + 40 + 50 = 2.8 cigarettes.

50

The median is 0.0 cigarettes.

The mode is 0 cigarettes.

0 + 50 The midrange is = 25.0 cigarettes.

2 Because the selected subjects report the number of cigarettes smoked, it is very possible that the data are not at all accurate. And what about that person who smokes 50 cigarettes (or 2.5 packs) a day? What are they thinking?

17. Systolic:

94 + 106 + 106 + 106 + 118 + 118 + 120 + 136 + 144 + 146 = 119.4 mm Hg. 10 118 + 118 The median is = 118 mm Hg. 2 Diastolic: 58 + 64 + 64 + 64 + 66 + 68 + 72 + 80 + 82 + 98 = 71.6 mm Hg. The mean is x = 10 66 + 68 = 67 mm Hg. The median is 2 Given that systolic and diastolic blood pressures measure different characteristics, a comparison of the measures of center doesn’t make sense. Because the data are matched, it would make more sense to investigate whether there is an association or correlation between systolic blood pressure measurements and diastolic blood pressure measurements.

The mean is x =

18. Weight measure:

64.1 + 64.2 + 71.3 + 73.1 + 79.2 + 83.1 + 86.5 + 96.9 + 98.2 + 98.6 + 108.2 + 118.8 = 86.85 kg.

12 83.1 + 86.5 = 84.80 kg.

The median is 2 The mean is x =

.

26

Chapter 3: Describing, Exploring, and Comparing Data

18. (continued) Height measure:

154.4 + 155.7 + 157.6 + 160.5 + 166.3 + 166.7 + 168.6 + 172.0 + 178.5 + 179.0 + 180.4 + 186.0 12 = 168.51 cm.

166.7 + 168.6 The median is = 167.65 cm.

2 Because the data are matched, it would make more sense to investigate whether there is an association or correlation between weight and height measures.

The mean is x =

19. Males: The mean is x = The median is

59.5 + 64.1 + 71.3 + 71.7 + 72.3 + 73.1 + 75.3 + 75.9 + 76.5 + 78.5 + 81.3 + 83.1 = 73.55 kg.

12

73.1 + 75.3 = 74.20 kg.

2

Females: The mean is x = The median is

54.4 + 55.6 + 57.8 + 57.8 + 60.5 + 64.0 + 64.2 + 66.5 + 67.1 + 69.8 + 70.9 + 71.5 = 63.34 kg.

12

64.0 + 64.2 = 64.10 kg.

2

Males appear to weigh more than females.

20. Single Line: The mean is x = The median is

390 + 396 + 402 + 408 + 426 + 438 + 444 + 462 + 462 + 462 = 429.0 seconds.

10

426 + 438 = 432.0 seconds.

2

Individual Lines: The mean is x =

252 + 324 + 348 + 372 + 402 + 462 + 462 + 510 + 558 + 600 = 429.0 seconds.

10

402 + 462 = 432.0 seconds.

2 Individual lines: same results as with a single line. Although the mean and median are the same, the times with individual lines are much more varied than those with a single line.

The median is

21. Using all values, the mean is x = 53.7 mg/dL and the median is 52.0 mg/dL. The highest value of 138 mg/dL appears to be an outlier. Excluding 138 mg/dL, the mean is x = 53.4 mg/dL and the median is 52.0 mg/dL. Excluding the outlier does not cause much of a change in the mean, and the median remains the same.

22. Using all values, the mean is x = 113.7 mg/dL and the median is 113.0 mg/dL. The highest value of 251 mg/dL appears to be an outlier. Excluding 251 mg/dL, the mean is x = 113.3 mg/dL and the median is 113.0 mg/dL. Excluding the outlier does not cause much of a change in the mean, and the median remains the same.

23. The mean is x = 98.20°F and the median is 98.40 F. These results suggest that the mean is less than 98.6 F. 24. The mean is x = 3152.0 g and the median is 3300 g. All of the weights end in 00, so they are all rounded to the nearest 100 grams. This suggests that the results should be rounded as follows: x = 3150.0 g and the median is 3300 g.

.

Section 3-2: Measures of Variation

27

1(49.5) + 51(149.5) + 90 (249.5) + 10 (349.5) + 0 (449.5) + 0 (549.5 ) +1 (649.5) 1 + 51 + 90 + 10 + 0 + 0 + 1 = 224.0 (1000 cells / mL). The mean from the frequency distribution is quite close to the mean of

25. The mean is x =

= 224.3 (1000 cells / mL ). obtained by using the original list of values.

25 (149.5) + 92 (249.5) + 28 (349.5) + 0 (449.5) + 2 (549.5) = 255.6 (1000 cells / mL ). The mean 25 + 92 + 28 + 0 + 2 from the frequency distribution is quite close to the mean of = 225.1 (1000 cells / mL ). obtained by using the

26. The mean is x =

original list of values.

27. The mean is x =

3 (4 ) + 3 (2 ) + 3 (3) + 4 (4 ) + 1(1) = 3.14, so the student made the dean’s list.

3 + 3 + 3 + 4 +1

æ63 + 91 + 88 + 84 + 79 ö + 0.10 (86) + 0.15 (90) + 0.15 (70) = 81.2, so the student earned a B. 28. The mean is x = 0.60 ç ÷ è ø 5 29. a. The missing value is 5 (78.0) - 82 - 78 - 56 - 84 = 90 beats per minute. b. n - 1 30. The mean ignoring the presidents who are still alive is 15.0 years. The mean including the presidents who are still alive is at least 15.7. years. The results do not differ by a considerable amount.

31. Mean: 113.7 mg/dL; 10% trimmed mean: 112.6 mg/dL; 20% trimmed mean: 112.2 mg/dL. The 10% trimmed mean and 20% trimmed mean are both fairly close, but the untrimmed mean of 113.7 mg/dL differs from them because it is more strongly affected by the outliers.

Section 3-2: Measures of Variation

2264.25 - 1684.89 = 144.84 cm2 , which is close to the exact value of the standard deviation of 174.8 cm2.

4

1.

s»

2.

Significantly low values are less than or equal to 1906.3 - 2 (174.8) = 1556.7 cm2 and significantly high values

are greater than or equal to 1906.3 + 2 (174.8) = 2255.9 cm2. A brain area of 1500 cm2 is significantly low.

3.

(56.125 cm)2 = 3150.0156 cm2

4.

s, s , s 2, s 2

5.

The range is 1060 – 322 = $738.0.

(322 - 775.9)2 +

2

+ (1060 - 775.9) = 49,615.38 dollars 2 .

10 - 1 The standard deviation is s = 49,615.38 = $222.70.

Because only the 10 lowest sample values are used, nothing much can be known about the population of all such charges.

The variance is s 2 =

6.

The range is 28.4 - 26.8 = 1.60.

(26.8 - 27.2)2 +

2

+ (28.4 - 27.2) = 0.324.

12 - 1 The standard deviation is s = 0.324 = 0.57.

The results do not provide any information about the change of MCAT scores over time.

The variance is s 2 =

.

28

Chapter 3: Describing, Exploring, and Comparing Data

7.

The range is 99 - 7 = 92.0.

(7 - 57.1)2 +

2

+ (99 - 57.1) = 1149.5.

11 - 1 The standard deviation is s = 1149.5 = 33.9.

The jersey numbers are nominal data that are just replacements for names, and they do not measure or count anything, so the resulting statistics are meaningless.

The variance is s 2 =

8.

The range is 305 - 189 = 116.0 lb.

(189 - 231.1)2 +

2

+ (305 - 231.1) = 1923.7 lb2.

11 - 1 The standard deviation is s = 1923.7 = 43.9 lb.

All of the weights are from players on only 1 of the 32 teams, but it isn’t very likely that any of the teams has weights that are substantially different from those of any other team, so it is reasonable to conclude that the measures of variation are typical of NFL players.

The variance is s 2 =

9.

The range is 4 - 1 = 3.0.

2

The variance is s 2 =

25 (109) - (47) = 0.9.

25 (25 - 1)

The standard deviation is s = 0.9 = 0.9.

The measures of variation can be found, but they make no sense because the data don’t measure or count anything.

They are nominal data.

10. The range is 1800 - 950 = $850.0.

2

The variance is s 2 =

12 (26,047,500) - (17, 450) = 61,117.4 (dollars)2.

12 (12 - 1)

The standard deviation is s = 61,117.4 = $247.2.

The sample consists of “best buy” TVs, so it is not a random sample and is not likely to be representative of the population. The measures of variation are not likely to be typical of all TVs that are 60 inches or larger.

11. The range is 1.49 - 0.51 = 0.980 W/kg.

2

The variance is s 2 =

11(16.4078) - (12.96) 2 = 0.114 (W/kg ) .

11(12 - 1)

The standard deviation is s = 0.114 = 0.337 W/kg.

No. Some models of cell phones have a much larger market share than others, so the measures from the different models should be weighted according to their size in the population.

12. The range is 55 - 0 = 55.0 mg.

2

The variance is s 2 =

20 (29,045) - (651) = 413.4 mg2.

20 (20 - 1)

The standard deviation is s 2 = 413.4 = 20.3 mg.

Americans consume some brands much more often than others, but the 20 brands are all weighted equally in the calculations, so the statistics are not necessarily representative of the population of all cans of the same 20 brands consumed by Americans.

.

Section 3-2: Measures of Variation

29

13. The range is 171 – 75 = 96.0 firefighters.

2

9 (133090) - (1072) The variance is s = = 675.36 firefighters2.

9 (9 - 1) 2

The standard deviation is s = 675.36 = 25.99 firefighters.

The data are time-series data, but the measures of variation do not reveal anything about a trend consisting of a pattern of change over time.

14. The range is 10.4 - 8.6 = 1.80 in.

2

The variance is s 2 =

11(985.94) - (107) = 0.27 in2 11(11 - 1)

The standard deviation is s = 0.27 = 0.52 in.

Because the measurements were made in 1988, they are not necessarily representative of the current population of all Army women.

15. The range is 57, 261 - 53, 323 = $3938.0.0.

2

The variance is s 2 =

10 (30,113,141,176) - (548,620) = 1,638,970.9 dollars2 .

10 (10 - 1)

The standard deviation is s = 1,638,970.9 = $1280.2.

Because the data include only the 10 highest costs, the measures of variation don’t tell us anything about the variation among costs for the population of all U.S. college tuitions.

16. The range is 50 - 0 = 50.0 cigarettes.

2

The variance is s 2 =

50 (4781) - (139 ) = 89.7 (cigarettes)2.

50 (50 - 1)

The standard deviation is s = 89.7 = 9.5 cigarettes.

Because the selected subjects report the number of cigarettes smoked, it is very possible that the data are not at all accurate, so the results might not reflect the actual smoking behavior of California adults.

17.56 ×100% = 14.71%.

119.4 11.92 Diastolic: x = 71.6, s = 11.92; The coefficient of variation is ×100% = 16.65%.

71.6 The variation is roughly about the same.

17. Systolic: x = 119.4, s = 17.56; The coefficient of variation is

17.53 ×100% = 20.18%.

86.85 10.54 Height measure: x = 168.81, s = 10.54; The coefficient of variation is ×100% = 6.25%.

168.81 Weights appear to vary more than heights.

18. Weight measure: x = 86.85, s = 17.53; The coefficient of variation is

6.67 ×100% = 9.07%.

73.55 6.03 ×100% = 9.52%.

Females: x = 63.34, s = 6.03; The coefficient of variation is 63.34 The variation is roughly about the same.

19. Males: x = 73.55, s = 6.67; The coefficient of variation is

.

30

Chapter 3: Describing, Exploring, and Comparing Data

28.6 ×100% = 6.7%.

429.0 109.3 Individual Lines: x = 429.0, s = 109.3; The coefficient of variation is ×100% = 25.5%.

429.0 The single line has much less variation than with individual lines.

20. Single Line: x = 429.0, s = 28.6; The coefficient of variation is

2

21. Using all data: Range = 112.0 mg/dL, s 2 = 238.3 (mg/dL ) , s = 15.4 mg/dL 2

Excluding possible outlier of 138: Range = 87.0 mg/dL, s 2 = 215.2 (mg/dL ) , s = 14.7 mg/dL The measures of variation change, but not by substantial amounts.

2

22. Using all data: Range = 212.0 mg/dL, s 2 = 1238.3 (mg/dL ) , s = 35.2 mg/dL 2

Excluding possible outlier of 251: Range = 184.0 mg/dL, s 2 = 1179.0 (mg/dL ) , s = 34.3 mg/dL The measures of variation change, but not by substantial amounts.

23. Range = 3.10 F, s 2 = 0.39

2

( F) , s = 0.62 F

24. Range = 4600.0 g, s2 = 480,848.1 g2 , s = 693.4 g; All of the weights end in 00, so they are all rounded to the nearest 100 grams. This suggests that the results should be rounded as follows:

Range = 4600.0 g, s2 = 480,850 g2 , s = 690 g.

25. The rule of thumb standard deviation is s »

138 - 26 = 28.0 mg/dL, which is much larger than s = 15.4 mg/dL 4

found by using all of the data.

26. The rule of thumb standard deviation is s »

251 - 39 = 53.0 mg/dL, which is much larger than s = 35.2 mg/dL 4

found by using all of the data.

27. The rule of thumb standard deviation is s »

99.6 - 96.5 = 0.78 F, which is not substantially different from 4

s = 0.62 F found by using all of the data.

4900 - 300 = 1150.0 g, which differs from s = 693.4 g found by 4 using all of the data by a considerable amount. Several of the lowest weights correspond to premature births, and they cause the range to be larger, with the resulting estimate being larger.

28. The rule of thumb standard deviation is s »

29. Significantly low values are less than or equal to 74.0 - 2 (12.5) = 49.0 beats per minute, and significantly high values are greater than or equal to 74.0 + 2 (12.5) = 99.0 beats per minute. A pulse rate of 44 beats per minute is significantly low.

30. Significantly low values are less than or equal to 69.6 - 2 (11.3) = 47.0 beats per minute, and significantly high values are greater than or equal to 69.6 + 2 (11.3) = 92.2 beats per minute. A pulse rate of 50 beats per minute is neither significantly low or high.

31. Significantly low values are less than or equal to 77.32 - 2 (1.29 ) = 24.74 cm, and significantly high values are greater than or equal to 77.32 + 2 (1.29) = 29.90 cm. A foot length of 30 cm is significantly high.

32. Significantly low values are less than or equal to 98.20 - 2 (0.62) = 96.96 F, and significantly high values are greater than or equal to 98.20 + 2 (0.62) = 99.44 F. A body temperature of 100 F is significantly high. .

Section 3-3: Measures of Relative Standing and Boxplots 31

(

103 1×49.52 + 33. s =

)

+ 1×649.52 - (1×49.5 +

2

+ 1×649.5)

103 (103 - 1)

= 68.4, which is somewhat far from the exact

value of 59.5.

(

147 25 ×149.52 + 34. s =

)

+ 2 ×549.52 - (25 ×149.5 +

2

+ 2 ×549.5)

147 (147 - 1)

= 69.5, which is not substantially far from

the exact value of 65.5.

2

2

2

(9 - 13) + (10 - 13) + (20 - 13) 9 + 10 + 20 = 13 cigarettes and s 2 = = 24.7 cigarettes2 3 3 b. The nine possible samples of two values are the following: {(9, 9), (9, 10), (9, 20), (10, 9), (10, 10), (10, 20), (20, 9), (20, 10), (20,20)} which have the following corresponding sample variances:{0, 0.5, 60.5, 0.5, 0, 50,

35. a. m =

60.5, 50, 0}, that have a mean of s 2 = 2.47 cigarettes2. c. The population variances of the nine samples above are {0, 0.25, 30.25, 0.25, 0, 25, 30.25, 25, 0} that have a mean of s 2 = 12.3 cigarettes2. d. Part (b), because repeated samples result in variances that target the same value (24.7 cigarettes 2) as the population variance. Use division by n - 1. e. No. The mean of the sample variances (24.7 cigarettes2) equals the population variance (24.7 cigarettes2), but the mean of the sample standard deviations (3.5 cigarettes) does not equal the population standard deviation (5.0 cigarettes).

36. The mean absolute deviation of the population is 4.7 cigarettes. With repeated samplings of size 2, the nine different possible samples have mean absolute deviations of 0, 0, 0, 0.5, 0.5, 5, 5, 5.5, 5.5. With many such samples, the mean of those nine results is 2.4 cigarettes, showing that the sample mean absolute deviations tend to center about the value of 2.4 cigarettes instead of the mean absolute deviation of the population, which is 4.7 cigarettes. The sample mean deviations do not target the mean deviation of the population. This is not good. This indicates that a sample mean absolute deviation is not a good estimator of the mean absolute deviation of a population.

Section 3-3: Measures of Relative Standing and Boxplots 1. James’ height is 4.07 standard deviations above the mean.

2.

The minimum height is 155 cm, the first quartile, Q1, is 169.1 cm, the second quartile, Q2, (or the median) is

173.8 cm, the third quartile, Q3, is 179.4 cm, and the maximum height is 193.3 cm.

3.

The bottom boxplot represents weights of women, because it depicts weights that are generally lower.

4. 2.00 should be preferred, because it is 2.00 standard deviations above the mean and would correspond to the highest of the five different possible scores.

5.

a. The difference is 104 - 74 = 30 bpm.

30 = 2.74 standard deviations 12.5 c. z = 2.74 b.

d. The pulse rate of 104 beats per minute is significantly high.

.

32

Chapter 3: Describing, Exploring, and Comparing Data

6.

a. The difference is 36 - 74 = - 38 BPM.

38 = 3.04 standard deviations 12.5 c. z = - 3.04 b.

d. The pulse rate of 36 beats per minute is significantly low.

7.

a. The difference is 96.50 - 98.20 = - 1.7 F.

1.7 = 2.74 standard deviations 0.62 c. z = - 2.74 b.

d. The temperature of 96.5 F is significantly low.

8.

a. The difference is 98.60 - 98.20 = 0.4 F.

0.4 = 0.65 standard deviations 0.62 c. z = 0.65 b.

d. Q3 is not significantly high.

9.

Significantly low scores are less than or equal to 21.1 - 2 (5.1) = 10.9, and significantly high scores are greater

than or equal to 21.1 + 2 (5.1) = 31.3.

10. Significantly low scores are less than or equal to 25.2 - 2 (6.4 ) = 12.4, and significantly high scores are greater than or equal to 25.2 + 2 (6.4 ) = 38.0.

11. Significantly low weights are less than or equal to 3152.0 - 2 (693.4 ) = 1765.2 g, and significantly high weights are greater than or equal to 3152.0 + 2 (693.4) = 4538.8 g.

12. Significantly low hip breadths are less than or equal to 36.6 - 2 (2.5) = 31.6 cm, and significantly high hip breadths are greater than or equal to 36.6 + 2 (2.5) = 41.6 cm.

251 - 174.12 54.6 - 174.12 = 10.83 and the shortest man’s z score is z = 7.10 7.10 = - 16.83. Chandra Bahadur Dangi has the more extreme height because his z score of –16.83 is farther from the mean than the z score of 10.83 for Sultan Kosen.

13. The tallest man’s z score is z =

14. The female has a higher red blood cell count because her z score is z = number than the z score of z =

5.58 - 4.719 = 1.76 for the male.

0.490

15. The male has a more extreme birth weight because his z score is z = number than the z score of z =

1500 - 3272.8 = - 2.19, which is a lower 660.2

1500 - 3037.1 = - 2.18 for the female.

706.3

16. Julianne Moore had the more extreme age since her z score is z = mean than the z score of z =

5.23 - 4.349 = 2.19, which is a higher 0.402

54 - 36.2 = 1.55, which is farther from the 11.5

33 - 44.1 = - 1.25 for Eddie Redmayne. .

8.9

Section 3-3: Measures of Relative Standing and Boxplots 33

17. For 144 kg,

18 ×100 = 33 , so it is the 33rd percentile.

54

18. For 236 kg,

41 ×100 = 76 , so it is the 76th percentile.

54

19. For 166 kg,

30 ×100 = 56 , so it is the 56th percentile.

54

20. For 365 kg,

50 ×100 = 93 , so it is the 93rd percentile.

54

21. W =

65 ×54 = 35.1 , so P65 = 204 kg. (Tech: Excel: 203.5 kg, Minitab: 203.5 kg) 100

22. W =

150 + 150 50 ×54 = 27 , so P50 = = 150 kg. (Tech: Excel: 150 kg, Minitab: 150 kg) 100 2

23. W =

20 ×54 = 10.8 , so P20 = 76 kg. (Tech: Excel: 76 kg, Minitab: 76 kg) 100

24. W =

25 ×54 = 13.5 , so P25 = 86 kg. (Tech: Excel: 85.5 kg, Minitab: 84.5 kg) 100

25. W =

75 ×54 = 40.5 , so Q3 = P75 = 236 kg. (Tech: Excel: 242.5 kg, Minitab: 242.5 kg) 100

26. W =

70 ×54 = 37.8 , so P70 = 212 kg. (Tech: Excel: 216 kg, Minitab: 216 kg) 100

27. W =

25 ×54 = 13.5 , so Q1 = P25 = 86 kg. (Tech: Excel: 87.75 kg, Minitab: 84.5 kg) 100

28. W =

80 ×54 = 43.2 , so P80 = 316 kg. (Tech: Excel: 316 kg, Minitab: 316 kg) 100

29. The five number summary is 25.1 cm, 26.40 cm, 27.50 cm, 28.60 cm, 29.2 cm.

30. The five number summary is 0.51 W/kg, 0.890 W/kg, 1.380 W/kg, 1.450 W/kg, 1.49 W/kg.

(Excel: Q1 = 0.965 W/kg, Q3 = 1.435 W/kg.)

31. The five number summary is 128 mBq, 140.0 mBq, 150.0 mBq, 158.5 mBq, 172 mBq (Excel: Q1 = 141.0 mBq and Q3 = 157.25 mBq.)

.

34

Chapter 3: Describing, Exploring, and Comparing Data

32. The five number summary is 120 mm Hg, 130.0 mm Hg, 132.5 mm Hg, 140.0 mm Hg, 150 mm Hg.

33. The top boxplot represents BMI values for males. The two boxplots do not appear to be very different, so BMI values of males and females appear to be about the same, except for a few very high BMI values for females that caused the boxplot to extend farther to the right.

34. The low lead level group represented in the top boxplot has much more variation and the IQ scores tend to be higher than the IQ scores from the high lead level group.

35. Bottom boxplot represents females. The two boxplots are not dramatically different. The outliers for females are 48.0 in., 52.6 in., 56.8 in., and 59.0 in. The outliers for males are 43.2 in., 43.7 in., 44.2 in., 45.9 in.

Chapter Quick Quiz

4 +5+6+6+7 +7 +7 +7 +8+8+8+8 = 6.8 hours.

12

1.

The sample mean is x =

2.

The median is

3.

The modes are 7 hours and 8 hours.

4.

The variance is (1.3 hours)2 = 1.7 hours2

5.

Yes, because 0 hours is substantially less than all of the other data values.

6.

z=

7+7 = 7.0 hours.

2

5 - 6.3 = - 0.93 1.4

.

Review Exercises 7.

75% or 60 sleep times

8.

minimum, first quartile Q1, second quartile Q2 (or median), third quartile Q3, maximum 10. x , m, s, s , s2, s 2

10 - 4 = 1.5 hours 4 Review Exercises 9.

s»

1.

a. The mean is x =

35

1497+1538+1550+1571+1642 = 1559.6 mm.

12 b. The median is 1550.0 mm. c. There is no mode.

1497 + 1642 = 1569.5 mm.

2 e. The range is 1642 - 1497 = 145.0 mm. d. The midrange is

f. s =

(1497 - 1559.6)2 + (1538 - 1559.6)2 + (1550 - 1559.6)2 + (1571 - 1559.6)2 + (1642 - 1559.6)2 12 - 1

= 53.4 mm

g. s 2 = 53.42 = 2849.3 mm2

1642 - 1559.6 = 1.54; The eye height is not significantly low or high because its z score is between 2 and 53.4 –2, so it is within 2 standard deviations of the mean.

2.

z=

12 + 14 + 22 + 27 + 40 = 23.0. The numbers don’t measure or count anything. They are used as 5 replacements for the names of the categories, so the numbers are at the nominal level of measurement. In this case the mean is a meaningless statistic.

3.

The mean is x =

3400 - 3273 3200 - 3037 = 0.19 . The female z score is z = = 0.23 . The female has the 660 706 larger relative birth because the female has the larger z score.

4.

The male z score is z =

5. The outlier is 646. The mean and standard deviation with the outlier included are x = 267.8 and s = 131.6.

With the outlier excluded, the values are x = 230.0 and s = 42.0. Both statistics changed by a substantial amount, so here the outlier has a very strong effect on the mean and standard deviation.

6. The minimum value is 119 mm, the first quartile is 128 mm, the second quartile (or median) is 131 mm, the third quartile is 135 mm, and the maximum value is 141 mm.

7.

Significantly low heights are 97.5 - 2 (6.9 ) = 83.7 cm or less; significantly high heights are 111.3 cm or greater.

The height of 97.5 + 2 (6.9 ) = 87.8 cm is not significant, so the physician should not be concerned.

8.

The median would be better because it is not affected much by the one very large income.

Cumulative Review Exercises 1.

Arsenic ( mg )

Frequency

0.0 – 1.9 2.0 – 3.9 4.0 – 5.9 6.0 – 7.9 8.0 – 9.9

1 0 3 7 1

.

36

Chapter 3: Describing, Exploring, and Comparing Data

2.

7 6

Frequency

5 4 3 2 1 0

0.95

2.95

4.95

6.95

8.95

Amount of Arsenic (micrograms/serving)

3.

4.

a. The mean is x = b. The median is

c. s =

1.5 + 4.9 + 5.4 + 5.4 + 6.1 + 6.3 + 6.6 + 6.7 + 6.9 + 7.3 + 7.8 + 8.2 = 6.09 mg.

12

6.3 + 6.6 = 6.45 mg.

2

(1.5 - 6.09)2 + (4.9 - 6.09)2 +

2

2

+ (7.8 - 6.09 ) + (8.2 - 6.09 ) = 1.75 mg 12 - 1

2

d. s 2 = 1.752 = 3.06 ( mg )

e. The range is 8.2 - 1.5 = 6.70 mg.

5.

a. Mode, because the others are numerical measures that require data at the interval or ratio levels of measurement. b. convenience sampling c. More consistency can be achieved by lowering the standard deviation. (It is also important to keep the mean at an acceptable level.)

.

Section 4-1: Basic Concepts of Probability

37

Chapter 4: Probability Section 4-1: Basic Concepts of Probability 1. The probability of a student passing the physical examination is 0.60, or 60%.

2.

The probability of selecting someone who is a post graduate is 0.55.

3.

P ( A) = 0.512, so P( A) = 1 - 0.512 = 0.488

4.

The answers vary, but a high answer in the neighborhood of 0.999 is reasonable.

5.

1, 4/7, 0, 0.461, 0.08

7.

1/9, or 0.111

6.

1/4, or 0.25

8.

{bb, bg, gb, gg}

9.

47 girls is significantly high.

10. 26 girls is neither significantly low nor significantly high.

11. 23 girls is neither significantly low nor significantly high.

12. 5 girls is significantly low.

13. 1 2, or 0.5

17. 1 10, or 0.1

14. 1 5, or 0.2

18. 1 2, or 0.5

15. 1 4, or 0.25

19. 0

16. 0.292

20. 1

5 1 = , or 0.00901; The employer would suffer because it would be at a risk by hiring someone who uses 555 111 drugs.

21.

25 5 = , or 0.0450; The person tested would suffer because he or she would be suspected of using drugs 555 111 when in reality he or she does not use drugs.

22.

23.

50 10 = or 0.0901; This result does appear to be a reasonable estimate of the prevalence rate.

555 111,

25 + 480 505 101 = = , or 0.910; The result does appear to be reasonable as an estimate of the proportion of the 555 555 111 adult population that does not use drugs.

24.

25.

879 , or 0.93; Yes, the technique appears to be effective.

945

26.

239 , or 0.821; Yes, the technique appears to be effective.

291

27.

428 = 0.738; Yes, it is reasonable.

580 c. He already knew. d. 0

28. a. 1 365 b. yes

139 139 = , or 0.130. No, it is not unlikely for someone to never seek medical information online.

933 + 139 1072 Because the responses are from a voluntary response survey, it is very possible that the results are not very good.

29.

.

38

Chapter 4: Probability

30.

83,600 418 = , or 0.0160; Yes, in a passenger car crash, a rollover is unlikely.

83,600 + 5,127, 400 26,055

31. a. brown /brown, brown/blue, blue/brown, blue/blue b. 1 4 c. 3 4 32. In the following, the first letter represents the chromosome contributed by the father and the second letter represents the chromosome contributed by the mother. Let X1 and X2 represent the possible X chromosomes contributed by the mother. a. 0; The possible outcomes are {xX1, xX2, YX1, YX2}, neither son will have the disease. b. 0; The possible outcomes are {xX1, xX2, YX1, YX2}, neither daughter will have the disease. c. 1 2, or 0.5; The possible outcomes are {Xx1, XX2, Yx1, YX2}, one of the two sons will have the disease. d. 0; The possible outcomes are {Xx1, XX2, Yx1, YX2}, neither daughter will have the disease.

33. 3 8, or 0.375

34. 3 8, or 0.375

35. {bbbb, bbbg, bbgb, bbgg, bgbb, bgbg, bggb, bggg, gbbg, gbbb, gbgb, gbgg, ggbb, ggbg, gggb, gggg}; 4 16 = 1 4, or 0.25 36. 2 16 = 1 8, or 0.125 37. The high probability of 0.327 shows that the sample results could have easily occurred by chance. It appears that there is not sufficient evidence to conclude that pregnant women can correctly predict the gender of their baby.

38. The low probability of less than 0.0001 shows that the sample results are not likely to occur by chance. It appears that Tamiflu does have an effect on nausea.

39. The low probability of less than 0.001 shows that the sample results could not have easily occurred by chance.

It appears that OxyContin does have an effect on sleepiness.

40. The high probability of 0.512 shows that the sample results could have easily occurred by chance. It appears that there is not sufficient evidence to conclude that cell phones have an effect on cancer of the brain or nervous system.

Section 4-2: Addition Rule and Multiplication Rule 1. P ( A) represents the probability of selecting an adult with blue eyes, and P( A) represents the probability of selecting an adult who does not have blue eyes.

2.

P ( M | B ) represents the probability of getting a male, given that someone with blue eyes has been selected.

P ( M | B ) is not the same as P ( B | M ).

3. Because the selections are made without replacement, the events are dependent. Because the sample size of 1068 is less than 5% of the population size of 15,524,971, the selections can be treated as being independent (based on the 5% guideline for cumbersome calculations).

4.

It is certain that the selected adult has type B blood or does not have type B blood.

5.

1 - 0.26 = 0.74

6.

1 - 0.0025 = 0.9975

617 617 117 = 1= , or 0.159, where P( H ) is the probability of randomly selecting someone 617 + 117 734 734 who did not get a headache.

7.

P( H ) = 1 -

.

Section 4-2: Addition Rule and Multiplication Rule 8.

P( I ) denotes the probability of screening a driver and finding that he or she is not intoxicated, and P( I ) = 0.99112, or 0.991 when rounded.

Use the following table for Exercises 9–20 O

A

B

AB

Total

+

Rh Rh–

59 9

53 8

12 3

6 2

130 22

Total

68

61

15

8

152

9.

152 - 15 137 = , or 0.901 152 152

10.

130 65 = , or 0.855 152 76

11.

15 22 3 34 17 + = = , or 0.224.The two events are not disjoint.

152 152 152 152 76

12.

8 22 2 28 7 + = = , or 0.184; The two events are not disjoint.

152 152 152 152 38

13. a.

68 67 × = 0.1985. The events are dependent, not independent.

152 151

b.

68 68 × = 0.2001. Yes, the events are independent.

152 152

14. a.

22 22 × = 0.0209; Yes, the events are independent.

152 152

b.

15. a.

22 21 × = 0.0201; The events are dependent, not independent.

152 151 130 130 × = 0.731; Yes, the events are independent.

152 152

b.

130 129 × = 0.731; The events are dependent, not independent.

152 151

16. a.

144 143 × = 0.8972. The events are dependent, not independent.

152 151

b.

144 144 × = 0.8975. Yes, the events are independent.

152 152

17.

61 + 15 22 8 + 3 87 + = , or 0.572 152 152 152 152

19.

61 60 59 × × = 0.0627 152 151 150

18.

68 + 8 130 59 + 12 135 + = , or 0.888 152 152 152 152

20.

22 21 20 × × = 0.00268 152 151 150

.

39

40

Chapter 4: Probability

21. Use the following table for parts (a) and (b).

Positive Test Result

Negative Test Result

Total

False Negative 3 True Negative 154

122

Subject Did Not Use Marijuana

True Positive 119 False Positive 24

Total

143

Subject Used Marijuana

157 154 77 c. = = 0.513 300 150

a. There were a total of 300 subjects in the study. b. 154 subjects had a true negative result.

22.

119 + 3 + 154 23 = = 0.92 300 25

23.

178 300

119 + 24 + 154 99 = = 0.990 300 100

178 89 = = 0.593; No, in the general population, the rate of subjects who do not use marijuana is probably 300 150 much greater than 0.593 or 59.3%.

24.

25. a. 0.03 b. 0.03 ×0.03 = 0.0009 c. 0.03 ×0.03 ×0.03 = 0.000027 d. By using one drive without a backup, the probability of total failure is 0.03, and with three independent disk drives, the probability drops to 0.000027. By changing from one drive to three, the probability of total failure drops from 0.03 to 0.000027, and that is a very substantial improvement in reliability. Back up your data!

26. a. 0.22 ×0.22 = 0.0484 b. 1 - 0.0484 = 0.9516, or 0.952; This probability seems high, but both generators fail about 5% of the time that they are needed. Given the importance of the hospital’s needs, the reliability should be improved.

8330 8329 8328 × × = 0.838; The probability of 0.838 is high, so it is likely that the entire 8834 8833 8832 batch will be accepted, even though it includes many firmware defects.

27. 8834 - 504 = 8330,

407 406 405 × × = 0.218; It is somewhat likely that the entire lot would be accepted. With 676 675 674 about 40% of the transducers not meeting requirements, the probability of 0.218 seems high.

28. 676 - 269 = 407,

29. a.

47,637 47,637 = = 0.299 47,637 + 111,874 159,511 5

b. Using the 5% guideline for cumbersome calculations, (0.299) = 0.00239. Using exact probabilities,

47,637 47,636 47,635 47,634 47,633 × × × × = 0.00238.

159,511 159,510 159,509 159,508 159,507 40

æ47,637 - 188 ö 30. Using the 5% guideline for cumbersome calculations, ç = 0.854. è 47,637 ÷ ø 31. a. 0.985 ×0.985 + 0.985 ×0.015 + 0.015 ×0.985 = 0.999775 b. 0.985 ×0.985 = 0.970225 c. The series arrangement provides better protection.

32.

365 364 363 341 × × ×K × = 0.431 365 365 365 365

.

Section 4-3: Complements, Conditional Probability, and Bayes’ Theorem

41

33. a. P( A or B ) = P( A) + P( B ) - 2 P( A and B ) b.

61 130 æ 53 ö 85 + - 2ç = , or 0.559 è 152 ÷ ø 152 152 152

59 + 9 + 6 + 2 76 1 152 = = , or 0.5, P( A or B ) = , or 1; The results are different. In general, 152 152 152 2 P ( A or B ) is not the same as P( A or B ).

34. P( A or B) =

Section 4-3: Complements, Conditional Probability, and Bayes’ Theorem 1. A is the event of not getting at least 1 defect among the 3 pacemaker batteries, which means that all 3 pacemaker batteries are good.

2.

Parts (a) and (b) are correct.

3. The probability that the test indicates that the subject has glaucoma given that the subject actually does have glaucoma.

4.

Confusion of the inverse is to think that P (Y | G ) = P (G | Y ) or to switch one of those values for the other. That

is, confusion of the inverse is to think that the following two probabilities are equal or to incorrectly use one of them for the other: (1) the probability that the test indicates that the subject has glaucoma given that the subject actually does have glaucoma; (2) probability that the subject actually does have glaucoma given that the test indicates that the subject has glaucoma.

4

5.

8.

æ1 ö 15 1 - ç ÷ = , or 0.9375 è 2 ø 16

6.

1 2, or 0.5

7.

1 - (0.512) = 0.982

6

6

1 - (0.545) = 0.974; The system cannot continue indefinitely because eventually there would be no women to

give birth.

9.

4

1 - (1 - 0.10) = 0.344 10

10. 1 - (1 - 0.20) = 0.893; There is a good chance of continuing.

10

11. 1 - (1 - 0.398) = 0.994; The probability is high enough so that she can be reasonably sure of getting a defective transducer for her work.

5

12. 1 - (1 - 0.25) = 0.763; It is very possible that the researchers will get what they need.

13. a. P (identical twins | having twins) = b. P (identical twins | twin boys ) =

5+5 1 = , or 0.333 30 3

5 1 = , or 0.5 5+5 2

14. a. P (fraternal twins | having twins ) =

20 2 = , or 0.667 30 3

b. P (fraternal twins | one boy and one girl ) = 15. P (one boy and one girl | fraternal twins ) = 16. P ( two girls | fraternal twins ) =

10 , or 1 10

10 1 = , or 0.5 20 2

5 1 = , or 0.25 20 4

.

42

Chapter 4: Probability Use the following table for Exercises 17–20.

Positive Test Result Negative Test Result Total Hepatitis C 335 10 345 No Hepatitis C 2 1153 1155 Total 337 1163 1500 2 17. P (positive result | no hepatitis C) = , or 0.00173; This is the probability of the test making it appear that 1155 the subject has hepatitis C when the subject does not have it, so the subject is likely to experience needless stress and additional testing.

10 , or 0.0290; The subject gets a test result showing that hepatitis C is not 345 present, but it actually is present, so the subject might delay or forego helpful treatment.

18. P (negative result | hepatitis C) =

19. P (hepatitis C | positive result ) =

335 , or 0.994; The very high result makes the test appear to be effective in 337

identifying hepatitis C.

1153 , or 0.991; The very high result makes the test appear to be effective 1163 in identifying subjects not having hepatitis C.

20. P (no hepatitis C | negative result ) =

2

21. a. 1 - (0.03) = 0.9991 3

b. 1 - (0.03) = 0.999973; The usual round-off rule for probabilities would result in a probability of 1.00, which would incorrectly indicate that we are certain to have at least one working hard drive.

3

22. 1 - (0.22) = 0.989; The result is quite high, but there is roughly a 1% chance that in the event of a power failure, none of the backup generators will work. With the possibility of an outage affecting a hospital and all its patients, it would be wise for manufacturers to improve the reliability of the backup generators so that the 22% failure rate is lowered and the probability of 0.989 is increased.

5

23. 1 - (1 - 0.126) = 0.490; The probability is not low, so further testing of the individual samples will be necessary for about 49% of the combined samples.

10

24. 1 - (1 - 0.005) = 0.0489; The probability is quite low, indicating that further testing of the individual samples will be necessary for about 5% of the combined samples.

25. 1 -

365 364 363 341 × × ×K × = 1 - 0.431 = 0.569 365 365 365 365

Section 4-4: Risks and Odds 1. pt is the proportion of the characteristic in the treatment group, and pc is the proportion of the characteristic in the control group.

2. With small differences between the rates, the relative risk may be large, implying a bad situation where the very low incidence rates suggest that there is not much risk in either group.

3. We need to treat 37 subjects with the influenza vaccine in order to prevent one case of influenza. The result applies to a large number of subjects, not every particular group of 37 subjects.

4. In a retrospective (or case-control) study, data are collected from a past time period by going back in time (through examination of records, interviews, and so on). In a prospective (or longitudinal or cohort) study, data are collected in the future from groups that share common factors (such groups are called cohorts).

5.

prospective .

Section 4-4: Risks and Odds 43

6.

P (headache | viagra treatment ) =

117 117 = = 0.159 117 + 617 734

29 117 1 117 = = = 0.040; = 0.159; P (headache | placebo) = 29 + 696 25 117 + 617 734 117 The risk of a headache appears to be higher in the treatment group.

117 + 617

7.

P (headache | viagra treatment ) =

8.Absolute risk reduction =

9.

1 117 29 117 + 617 29 + 696

117 29 = 0.119 117 + 617 29 + 696

= 8.4; 9 when rounded up.

117 117 + 617 = 117 , or 117:617 (roughly 1:5.2) 10. Odds in favor of headache: 617 617 117 + 617 617 617 Odds against headache: 117 + 617 = , or 617:117 (roughly 5.2:1) 117 117 117 + 617 117 pt 117 + 617 2925 11. RR = = = = 3.99, 3.99 or roughly 4. The risk of headaches among Viagra users is roughly 29 pc 734 29 + 696 4 times the risk of headaches for those who take a placebo.

ad 117 ×696 = = 4.551; The odds of having a headache in the treatment group are 4.551 times those of bc 617 ×29 the placebo group.

12. OR =

13. a. P (infection | atorvastatin treatment ) = b. P (infection | placebo ) =

89 = 0.103 89 + 774

27 = 0.100 27 + 243

89 27 = 0.00313; The chance of infection in the atorvastatin 89 + 774 27 + 243 treatment group is slightly higher than for the placebo group. For those in the placebo group, there is a 0.313% reduced chance of infection when compared to the atorvastatin treatment group. c. Absolute risk reduction =

1 = 319.6, or 320 when rounded up. We would need to treat 320 patients with atorvastatin 89 27 89 + 774 27 + 243 (instead of a placebo) to prevent one of the patients from getting an infection.

14.

15. Atorvastatin: 89:774 or roughly 1:9. Placebo: 27:243 or 1:9. There is not much of a difference between these two results.

.

44

Chapter 4: Probability

89 pt 89 + 774 890 ad 89 ×243 16. OR = = = 1.035; RR = = = = 1.031; Since both values are close to 1, it appears 27 bc 27 ×774 pc 863 27 + 243 that atorvastatin does not increase the risk of infection.

26 pt 2103 7241 ad 26 ×(1671 - 22 ) 17. RR = = = = 0.939; OR = = = 0.938; 0.938; The risk of a headache with 22 pc 7711 bc (2103 - 26) ×22 1671 Nasonex treatment is slightly less than the risk of a headache with a placebo.

18. a. It would be unethical to assign subjects to the group that does not use seat belts. b. Answers will vary. One possible answer: The number of subjects that do not use seat belts may be too small.

Section 4-5: Rates of Mortality, Fertility, and Morbidity 1. During a year in Germany, there are 11.6 deaths for every 1000 people in the population.

2. One important advantage of the death rate of 11.6 people (per 1000 people in the population) is that it uses fewer decimal places and is generally easier to use and understand.

3.

About 0.0116 ⋅ 80,722,792 = 936,384 deaths are expected in a year.

4. A disease incidence rate describes the growth of the number of cases while the disease prevalence rate describes the current number of cases.

5.

æ 15,973 ö The neonatal mortality rate is ç 1000 = 4.0, or 4.0 per 1000. è 3,953,590 ÷ ø

6.

26,148 æ ö The fetal mortality rate is ç 1000 = 6.6, or 6.6 per 1000. è 3,953,590 + 26,148 ÷ ø

7.

æ 26,148 + 15,973 ö The perinatal mortality rate is ç 1000 = 10.6, or 10.6 per 1000. è 3,953,590 + 26,148 ÷ ø

8.

æ 3,953,590 ö The crude birth rate is ç 1000 = 12.6, or 12.6 per 1000. è 312,799, 495 ÷ ø

9.

æ 3,953,590 ö The general fertility rate is ç 1000 = 64.3, or 64.3 per 1000 women aged 15–44. è 61, 488, 227 ÷ ø

æ 33,783 ö 10. The motor vehicle death incidence rate is ç 10,000 = 1.1, or 1.1 deaths per 10,000. è 312,799, 495 ÷ ø

æ 1,155,792 ö 11. The HIV infection prevalence rate is ç 1000 = 3.7, or 3.7 per 1000. è 312,799, 495 ÷ ø æ 7683 ö 12. The HIV infection prevalence rate is ç 1000 = 6.6, or 6.6 per 1000. è1,155,792 ÷ ø 13.

8 = 0.008; The rate uses fewer decimal places and is easier to understand.

1000

.

Section 4-5: Rates of Mortality, Fertility, and Morbidity

14. a.

45

6.2 = 0.0062 1000

b. (0.0062)(0.0062)(0.0062) = 0.000000238 c. (1 - 0.0062)(1 - 0.0062)(1 - 0.0062) = 0.982 15. a.

12.4 = 0.0124 1000

b. (0.0124)(0.0124)(0.0124) = 0.00000191 c. 1 - (0.0124)(0.0124)(0.0124) = 0.999998 Using three significant digits would result in a probability of 1.00 which would be misleading because it would incorrectly suggest that it is certain that at least one survives the year.

æ 787650 ö 16. a. ç 1000 = 2.5, or 2.5 per 1000 è 312799495 ÷ ø b.

787650 = 0.00252 312799495

c. (1 - 0.00252 )(1 - 0.00252 )(1 - 0.00252 ) = 0.992

æ 787,650 ö 17. a. ç 100 = 31.3 è 2,515, 458 ÷ ø b. (1 - 0.313)(1 - 0.313)(1 - 0.313) = 0.324

3625 + 39,820 + 129,395 æ ö 18. a. The crude mortality rate for Florida is ç 1000 = 9.1, or 9.1 per 1000. ÷ è 5,716,861 + 9,842,031 + 3,375,303 ø 63, 208 + 619,982 + 1,832, 268 æ ö The crude mortality rate for the United States is ç 1000 = 8.0, or è103,542,603 + 163,960,163 + 45, 296,729 ÷ ø 8.0 per 1000. Florida has a slightly higher mortality rate, but the two rates are close. æ 129395 ö b. The crude mortality rate for Florida (65 and older) is ç 1000 = 6.8, or 6.8 per 1000. è18,934,195 ÷ ø

æ 18,32, 268 ö The crude mortality rate for the United States (65 and older) is ç 1000 = 5.8 or 5.8 per 1000. è 312,799, 495 ÷ ø Florida now has a slightly higher mortality rate, but the two rates are still close.

45, 296,729 3375303 = 0.14, or 14.5% of the United = 0.178, or 17.8% of Florida's is 65 and older and 312,799, 495 18934195 States is 65 and older. Since Florida has a higher percentage of those in the population of 65 years or older than the U.S., the difference in crude mortality rate (6.8 for Florida and 5.8 for the U.S.) would be expected. c.

19. No, the health of the nation is not necessarily declining. The increasing number of deaths each year is probably due to the growing population.

20. The major problem in comparing these rates is that the age distributions of the two nations could be different and that, among other reasons, could be a major factor in the difference.

.

46

Chapter 4: Probability

21. The United States has a population distribution of 33.10190862%, 52.41701653%, and 14.48107485% for the three age categories. If the 18,934,195 Florida residents have that same distribution, the three age groups would have these numbers of people: 6,267,580, 9,924,740, and 2,741,875, respectively. Using the same Florida mortality rates for the three individual age groups and using the new adjusted population sizes for the three Florida age categories, we get these numbers of Florida deaths: 3974, 40,155, and 105,112, respectively. Using the adjusted numbers of deaths and the adjusted population sizes for the different categories, the crude mortality rate for Florida becomes 7.9 per 1000, which is much closer to the U.S. mortality rate of 8.0 per 1000 than the mortality rate of 9.1 per 1000 found for Florida before the adjustments.

Section 4-6: Counting 1. The symbol “!” is the factorial symbol that represents the product of decreasing whole numbers, as in 6! = 6 ×5 ×4 ×3 ×2 ×1 = 720. Six people can be scheduled in 720 different ways.

2.

For permutations, the order of the results matter. For combinations, the order of the results do not matter.

11! = 462. The result of 462 is the number of different combinations that are possible when 6 (11 - 6)!6! items are selected without replacement from 11 different items that are available.

3.

11 C6 =

11! = 332640. The result of 332640 is the number of different permutations that are possible when (11 - 4)!

6 items are selected without replacement from 11 different items that are available.

4.

11 P6 =

5.

1 1 1 1 1 × × × = 10 10 10 10 10,000

7.

There are 19 C2 =

8.

11 C3 =

9.

8! = 40, 320; The probability is 1 40,320.

6.

1 1 1 1 1 1 × × × × = 10 10 10 10 10 100,000

19! = 171 ways to choose a team of physicians. The probability is 1 171, or 0.00585.

(19 - 2)!2!

11! = 165, 3! = 6 (11 - 3)!3!

10. For three additional letters, there are 2 ×26 ×26 ×26 = 35,152 possibilities, for two additional letters, there are 2 ×26 ×26 = 1352 possibilities, for a total of 35,152 + 1352 = 36,504 possibilities.

11. There are 50 P5 =

50! = 254, 251, 200 possible routes. The probability is 1 254, 251, 200.

(50 - 5)!

12.

12! = 3,326, 400 3!4!

13.

1 1 = ; No, there are too many different possibilities.

100 ×100 ×100 ×100 100,000,000

14. 10 C2 =

10! = 10 (10 - 2)!2!

15. The groups of three must be chosen in order, so there are 168,168,000 ways to form the groups.

15!

12!

9!

6!

3!

(15 C3 ) ×(12 C3 ) ×( 9 C3 ) ×( 6 C3 ) ×( 3 C3 ) = (15 - 3)!3! ×(12 - 3)!3! ×(9 - 3)!3! ×(6 - 3)!3! ×(6 - 3)!3! = 168,168,000 16. 7! = 5040 17. 10 ×10 ×10 ×10 ×10 = 100,000; The probability is 1 100,000, or 0.00001.

.

Section 4-6: Counting

18.

47

7! = 2,520 (7 - 5)!

7 P5 =

19. There are 8 ×10 ×10 = 800 possible area codes. There are 8 ×10 ×10 ×8 ×10 ×10 ×10 ×10 ×10 ×10 = 6, 400,000,000 possible area codes. Yes. (With a total population of about 400,000,000, there would be about 16 phone numbers for every adult and child.)

20.

13! = 10,810,800 3!4!2!2!

21. a. 10 P4 =

10! = 5040 (10 - 4)!

b. 10 C4 =

10! = 210 (10 - 4)!4!

c. The probability is 1 210.

22. a. 1 4, or 0.25

3 1 3 × = , or 0.188 4 4 16 c. Trick question. There is no finite number of attempts, because you could continue to get the wrong position every time. b.

23. 10 P4 = 24. a. b.

10! = 5040 (10 - 4)!

1 16

10

1 12

10

1

=

1 10,000,000,000,000,000

=

1 1,000,000,000,000

1 ; The number of possibilities (100,000,000) is still quite large, so there is no reason to 100,000,000 10 worry(unless the information is taken from your credit card or is hacked from the Internet). c.

25.

8

=

15! = 163, 459, 296, 000 2!2!2!

16! = 10, 461,394,944,000; (Most calculators will give a result in scientific notation, so an (16 - 14)! answer such as 10,461,395,000,000 is OK.)

26. a. 16 P14

b. 16 C14

16! = 120 (16 - 14)!14!

c. The probability is 1 120.

27.

1 1 = ; There is a much better chance of being struck by lightning.

258,890,850 75 C5 ×15 2 C1

2 1 = ; Yes, if everyone treated is of one gender while everyone in the placebo group is of the 924 462 12 C6 opposite gender, you would not know if different reactions are due to the treatment or gender.

28.

=

.

48

Chapter 4: Probability

29. There are 2 + 2 ×2 + 2 ×2 ×2 + 2 ×2 ×2 ×2 + 2 ×2 ×2 ×2 ×2 = 2 + 4 + 8 + 16 + 32 = 62 different possible characters. The alphabet requires 26 characters and there are 10 digits, so the Morse code system is more than adequate.

30.

2 C1 8 C4

=

2 1 = 70 35

31. 26 + 26 ×36 + 26 ×362 + 26 ×363 + 26 ×364 + 26 ×365 + 26 ×366 + 26 ×367 = 2,095,681, 645,538 or about 2 trillion.

32. a. 5 C2 = 10

c. 4! = 24

n(n - 1) 2 Chapter Quick Quiz

d. (n - 1)!

b. n C2 =

1.

4 , or 0.8 5

3.

4 1 = 12 3

2.

1 - 0.30 = 0.70

4.

0.6772 = 0.458

5.

Answers will vary, but the probability should be low, such as 0.01.

Use the following table for Exercises 6–10.

Vaccine Treatment Placebo Total

Developed Flu 14 95 109

Did Not Develop Flu 1056 437 1493

6.

109 , or 0.0680 1602

7.

14 + 1056 + 95 1165 109 1070 14 1165 + = = , or , or 0.727 1602 1602 1602 1602 1602 1602

8.

14 7 = = 0.00874 1602 801

9.

10. P (developed flu | given vaccine) =

Total 1070 532 1602

109 108 × = 0.00459 1602 1601

14 7 = , or 0.0131 1070 535

Review Exercises Use the following table for Exercises 1–10.

Splint Treatment Surgery Treatment Total

Successful Treatment 60 67 127

1.

127 , or 0.814 156

2.

P (successful treatment | splinting ) =

3.

P (successful treatment | surgery ) =

4.

127 73 67 133 + = , or 0.853 156 156 156 156

Unsuccessful Treatment 23 6 29

Total 83 73 156

60 , or 0.723 83

67 , or 0.918 73 5.

.

29 83 23 89 + = , or 0.571 156 156 156 156

Section 4-6: Counting

127 127 × = 0.663 156 156

6.

127 126 × = 0.662 (not 0.663) 156 155

8.

A is the event of selecting a patient and getting someone who was not treated with surgery. P( A) = 1 -

= 9.

7.

73 156

83 = 0.532 156

A is the event of selecting a patient and getting someone who did not have a successful treatment.

127 29 P( A) = 1 = = 0.186 156 156

10.

127 126 125 × × = 0.537 156 155 154

11. a. 1 - 0.62 = 0.38, or 38% b. 0.38 ×0.38 ×0.38 = 0.054872 c. Yes, it is unlikely because the probability of 0.054872 shows that the event does not occur quite often.

12. a. 1 365 b. 31 365 c. Answers will vary, but it is probably quite small, such as 0.01 or less. d. yes 10

34 ö æ = 0.0335; No, it is not likely.

13. 1 - ç1 è 10, 000 ÷ ø 14. a. 1 -

1 599 = , or 0.998 600 600 2

359,999 æ1 ö b. 1 - ç = , or 0.999997 è 600 ÷ ø 360, 000 Cumulative Review Exercises 1.

a. The mean is x = b. The median is

0.09 + 0.11 + L + 0.15 + 0.17 + L + 0.23 + 0.35 = 0.165 g dL. 12

0.15 + 0.17 = 0.160 g dL. 2

c. The midrange is

0.09 + 0.35 = 0.220 g dL. 2

d. The range is 0.35 - 0.09 = 0.260 g dL. e. s =

(0.09 - 0.165)2 + (0.11 - 0.165)2 + K + (0.23 - 0.165)2 + (0.35 - 0.165)2 = 0.069 g dL 12 - 1

2

2

f. s 2 = (0.069) = 0.005 (g dL)

.

49

50

Chapter 4: Probability

2.

a. The five number summary is 0.090 g/dL, 0.120 g/dL, 0.160 g/dL, 0.180 g/dL, 0.350 g/dL. The value of 0.350 g/dL is an outlier. b.

c.

3.

2346 = 0.46 = 46% 5100 b. 0.460 a.

c. stratified sample 4.

a. a convenience sample b. If the students at the college are mostly from a surrounding region that includes a large proportion of one ethnic group, the results will not reflect the general population of the United States. c. 0.35 + 0.4 = 0.75 2

d. 1 - (0.6) = 0.64 5. The upward sloping pattern of the points in the scatterplot suggests that there appears to be an association between tar and nicotine content in cigarettes.

.

Section 5-1: Probability Distributions

51

Chapter 5: Discrete Probability Distributions Section 5-1: Probability Distributions 1. The random variable is x, which is the number of girls in four births. The possible values of x are 0, 1, 2, 3, and 4. The values of the random variable x are numerical.

2. The random variable is discrete because the number of possible values is 5, and 5 is a finite number. The random variable is discrete if it has a finite number of values or a countable number of values.

3. The table does describe a probability distribution because the three requirements are satisfied. First, the variable x is a numerical random variable and its values are associated with probabilities. Second, S P( x ) = 0.063 + 0.250 + 0.375 + 0.250 + 0.063 = 1.001, which is not exactly 1 due to round-off error, but is close enough to satisfy the requirement. Third, each of the probabilities is between 0 and 1 inclusive, as required.

4. The probability of 0.136 is relevant; 56 is not significantly high because the probability of 56 or more girls is 0.136, which is not small, such as 0.05 or less. With random chance, it is likely that the outcome could be 56 or more girls.

5.

7.

a. continuous random variable b. not a random variable c. discrete random variable d. continuous random variable e. discrete random variable

6.

Probability distribution with m = 0 ×0.031 + 1 ×156 + 2 ×0.313 + 3 ×313 + 4 ×0.156 + 5 ×0.031 = 2.5

s = (0 - 2.5)2 ×0.031 + (1 - 2.5)2 ×0.156 + 8.

a. not a random variable b. continuous random variable c. discrete random variable d. not a random variable e. discrete random variable

+ (4 - 2.5)2 ×0.156 + (5 - 2.5)2 ×0.031 = 1.1

Probability distribution (The sum of the probabilities is 1.001, but that is due to rounding errors.) with m = 0 ×0.659 + 1 ×0.287 + 2 ×0.050 + 3 ×004 + 4 ×0.001 + 5 ×0 = 0.4

s = (0 - 0.4)2 ×0.659 + (1 - 0.4)2 ×0.287 +

+ (4 - 0.4) 2 ×0.001 + (5 - 0.4) 2 ×0 = 0.6

9. Not a probability distribution because the sum of the probabilities is 0.94, which is not 1 as required. Also, Ted clearly needs a new approach.

10. Probability distribution with m = 0 ×0.0000 + 1×0.0001 + 2 ×0.0006 + 3 ×0.0387 + 4 ×0.9606 = 4.0

s = (0 - 4.0)2 ×0.0000 + (1 - 4.0)2 ×0.0001 + (2 - 4.0)2 ×0.0006 + (3 - 4.0)2 ×0.0387 + (4 - 4.0)2 ×0.9606 = 0.2 11. Probability distribution with m = 0 ×0.4219 + 1×0.4219 + 2 ×0.1406 + 3 ×0.0156 = 0.7

s = (0 - 0.7)2 ×0.4219 + (1 - 0.7)2 ×0.4219 + (2 - 0.7) 2 ×0.1406 + (3 - 0.7)2 ×0.0156 = 0.7 (The sum of the probabilities is 0.999, but that is due to rounding errors.)

12. Not a probability distribution because the sum of the probabilities is 0.986, which is not 1 as required 13. m = 0 ×0.001 + 1×0.010 + 2 ×0.044 +

s =

+ 8 ×0.044 + 9 ×0.010 + 10 ×0.001 = 5.0 girls

(0 - 5.0)2 ×0.001 + (1 - 5.0)2 ×0.010 +

2

2

+ (9 - 5.0) ×0.010 + (10 - 0.10) ×0.001 = 1.6 girls

14. The upper limit is μ + 2σ = 5.0 + 2(1.6) = 8.2 girls. Because 9 girls is greater than or equal to 8.2 girls, it is a significantly high number of girls.

15. The lower limit is μ – 2σ = 5.0 – 2(1.6) = 1.8 girls. Because 1 girl is less than or equal to 1.8 girls, it is a significantly low number of girls.

.

52

Chapter 5: Discrete Probability Distributions

16. a. P( X = 8) = 0.044 b. P( X ³ 8) = 0.044 + 0.010 + 0.001 = 0.055 c. The probability from part (b), since it is the probability of the given or more extreme result. d. No, because the probability of 8 or more girls is 0.055, which is high (more than or equal to 0.05).

17. a. P( X ³ 6) = 0.205 + 0.117 + 0.044 + 0.010 + 0.001 = 0.377 b. P( X = 6) = 0.205 c. The probability from part (b), since it is the probability of the given or more extreme result. d. No, because the probability of 6 or more girls is 0.377, which is not very low (less than or equal to 0.05).

18. a. P( X = 2) = 0.044 b. P( X £ 2) = 0.044 + 0.010 + 0.001 = 0.055 c. The probability from part (b), since it is the probability of the given or more extreme result. d. No, because the probability of 2 or fewer girls is 0.055, which is high (more than or equal to 0.05).

19. m = 0 ×0.172 + 1×0.363 + 2 ×0.306 + 3 ×0.129 + 4 ×0.027 + 5 ×0.002 = 1.5 sleepwalkers

s = (0 - 1.5)2 ×0.172 + (1 - 1.5)2 ×0.363 + ... + (4 - 1.5) 2 ×0.027 + (5 - 1.5)2 ×0.002 = 1.0 sleepwalkers 20. Significantly high numbers of sleepwalkers are greater than or equal to m + 2s = 1.5 + 2(1.0) = 3.5 sleepwalkers. Because 4 sleepwalkers is greater than or equal to 3.5 sleepwalkers, 4 sleepwalkers is a significantly high number.

21. Significantly high numbers of sleepwalkers are greater than or equal to m + 2s = 1.5 + 2(1.0) = 3.5 sleepwalkers. Because 3 sleepwalkers is not greater than or equal to 3.5 sleepwalkers, 3 sleepwalkers is not a significantly high number.

22. a. P ( X = 4) = 0.027 b. P( X ³ 4) = 0.027 + 0.002 = 0.029 c. The probability from part (b), since it is the probability of the given or more extreme result. d. Yes, because the probability of four or more sleepwalkers is 0.029, which is very low (less than or equal to 0.05) 23. a. P( X = 1) = 0.363 b. P ( X £ 1) = 0.172 + 0.363 = 0.535 c. The probability from part (b), since it is the probability of the given or more extreme result. d. No, because the probability of one or fewer sleepwalkers is 0.535, which is not low (less than or equal to 0.05) Section 5-2: Binomial Probability Distributions 1. The given calculation assumes that the first two offspring peas have green pods and the last offspring peas have green pods, but there are other arrangements consisting of offspring peas have green pods and three that do not.

The probabilities corresponding to those other arrangements should also be included in the result.

2.

n = 5, x = 2, p = 0.75, q = 0.25

3. Because the 30 selections are made without replacement, they are dependent, not independent. Based on the 5% guideline for cumbersome calculations, the 30 selections can be treated as being independent. (The 30 selections constitute 3% of the population of 1020 responses, and 3% is not more than 5% of the population.) The probability can be found by using the binomial probability formula, but it would require application of that formula 21 times (or 10 times if we are clever), so it would be better to use technology.

.

Section 5-2: Binomial Probability Distributions

53

4. The 0+ indicates that the probability is a very small positive value. (The actual value is 0.0000508.) The notation of 0+ does not indicate that the event is impossible; it indicates that the event is possible, but very unlikely.

5.

Not binomial; each of the weights has more than two possible outcomes.

6.

binomial

7.

binomial

8.

Not binomial; each of the responses has more than two possible outcomes.

9.

Not binomial; there are more than two possible outcomes.

10. binomial 11. binomial 12. Not binomial; there are more than two possible outcomes.

4 4 1 × × = 0.128 5 5 5 b. {WWC, WCW, CWW}; Each has a probability of 0.128. c. 0.128 ×3 = 0.384

13. a.

14. a. P(EEEN) = 0.53 ×0.53 ×0.53 ×0.47 = 0.0700 b. {EEEN, EENE, ENEE, NEEE }; Each has a probability of 0.0700. c. 0.0700 ×4 = 0.28 15.

6 1 7 C6 ×0.2 ×0.8 = 0.0003584 (Table: 0+)

16.

3 4 4 3 5 2 6 1 7 0 7 C3 ×0.2 ×0.8 + 7 C4 ×0.2 ×0.8 + 7 C5 ×0.2 ×0.8 + 7 C6 ×0.2 ×0.8 + 7 C7 ×0.2 ×0.8 = 0.1480 (Table: 0.1480)

17.

0 7 1 6 2 5 3 4 7 C0 ×0.2 ×0.8 + 7 C1 ×0.2 ×0.8 + 7 C2 ×0.2 ×0.8 + 7 C3 ×0.2 ×0.8 = 0.967 (Table: 0.967)

18. 7 C0 ×0.20 ×0.87 + 7 C1 ×0.21 ×0.86 + 7 C2 ×0.22 ×0.85 + 7 C3 ×0.23 ×0.84 + 7 C4 ×0.24 ×0.83 + 7 C5 ×0.25 ×0.82 = 1.000 (Table: 1.000) 19.

7 0 7 C7 ×0.2 ×0.8 = 0.00001

21. 8 C6 ×0.212 ×0.796 = 0.300

20. 1 - 7 C0 ×0.20 ×0.87 - 7 C1 ×0.21 ×0.86 = 0.423 (Table: 0.423) 23. 10 C3 ×0.213 ×0.797 + 10 C4 ×0.214 ×0.796 +

22.

5 15 20 C5 ×0.21 ×0.79 = 0.184

+ 10 C9 ×0.219 ×0.791 + 10 C10 ×0.2110 ×0.790 = 0.353

24. 10 C0 ×0.210 ×0.7910 + 10 C1 ×0.211 ×0.799 + 10 C2 ×0.212 ×0.798 = 0.647 25. a. m = np = 36 ×0.5 = 18.0 girls, s = np(1 - p ) = 36 ×0.5 ×0.5 = 3.0 girls b. Values of 18.0 - 2(3.0) = 12.0 girls or fewer are significantly low, values of 18.0 + 2(3.0) = 24.0 girls or more are significantly high, and values between 12.0 girls and 24.0 girls are not significant. c. The result is significantly high because the result of 26 girls is greater than or equal to 24.0 girls. A result of 26 girls would suggest that the XSORT method is effective.

26. a. m = np = 16 ×0.5 = 8.0 girls, s = np(1 - p ) = 16 ×0.5 ×0.5 = 2.0 girls b. Values of 8.0 - 2(2.0) = 4.0 girls or fewer are significantly low, values of 8.0 + 2(2.0) = 12.0 girls or more are significantly high, and values between 4.0 girls and 12.0 girls are not significant.

.

54

Chapter 5: Discrete Probability Distributions c. The result is not significant because the result of 11 girls is not greater than or equal to 12.0 girls. A result of 11 girls would not suggest that the XSORT method is effective.

27. a. m = np = 15 ×0.75 = 11.25 peas, s = np(1 - p) = 15 ×0.75 ×0.25 = 1.7 peas b. Values of 11.25 – 2(1.7) = 7.85 peas or fewer are significantly low, values of 11.25 + 2(1.7) = 14.65 peas or more are significantly high, and values between 7.85 peas and 14.65 peas are not significant. c. The result is significant because the result of 2 peas is less than or equal to 7.85 peas.

28. a. m = np = 22 ×0.75 = 16.5 peas, s = np(1 - p) = 22 ×0.75 ×0.25 = 2.0 peas b. Values of 16.5 – 2(2) = 12.5 peas or fewer are significantly low, values of 16.5 + 2(2) = 20.5 peas or more are significantly high, and values between 12.5 peas and 20.5 peas are not significant. c. The result is not significant because the result of 20 peas is less than or equal to 20.5 peas.

29. 1 - 36 C0 ×0.010 ×0.9936 = 0.304; It is not unlikely for such a combined sample to test positive.

30. 1 - 16 C0 ×0.0140 ×0.98616 = 0.202; It is somewhat likely for such a combined sample to test positive.

31. 40 C1 ×0.031 ×0.9739 + 40 C0 ×0.030 ×0.9740 = 0.662; The probability shows that about 2/3 of all shipments will be accepted. With about 1/3 of the shipments rejected, the supplier would be wise to improve quality.

32. About 92% of all shipments will be accepted. Almost all shipments will be accepted, and only 8% of the shipments will be rejected.

33. a. m = np = 14 ×0.5 = 7.0 girls, s = np(1 - p ) = 14 ×0.5 ×0.5 = 1.8 girls; Values between 14.0 - 2(1.8) = 3.4 girls and 14.0 + 2(1.8) = 10.6 girls are not significant (3.3 and 10.7 if using unrounded values). 13 girls lies outside these limits, so it is significant. b. The probability of exactly 13 girls is 14 C13 ×0.513 ×0.51 = 0.000854. c. The probability of 13 or more girls is 14 C13 ×0.513 ×0.51 + 14 C14 ×0.514 ×0.50 = 0.000916. d. The probability from part (c) is relevant. The result of 13 girls is significantly high. e. The results suggest that the XSORT method is effective in increasing the likelihood that a baby is a girl.

34. a. m = np = 40 ×0.6 = 24.0 successes s = np(1 - p ) = 40 ×0.6 ×0.4 = 3.1 successes; Values between 24.0 - 2(3.1) = 17.8 successes and 24.0 + 2(3.1) = 30.2 successes are not significant. The value of 29 successes lies between these boundaries, so it is neither significantly low or high. b. The probability of exactly 29 successes is 40 C29 ×0.629 ×0.411 = 0.0357. c. The probability of 29 or more successes is 40 C29 ×0.629 ×0.411 +

+ 40 C29 ×0.640 ×0.40 = 0.0709.

d. The probability from part (c) is relevant. The result of 29 successes is not significantly high. e. The results do not provide strong evidence for the vaccine’s effectiveness.

35. a. m = np = 47 ×0.75 = 35.3 long stems, s = np(1 - p ) = 14 ×0.75 ×0.25 = 3.0 long stems; Values between 35.3 - 2(3.0) = 29.3 long stems and 35.3 + 2(3.0) = 14.3 long stems are not significant (29.3 and 41.2 if using unrounded values). Because 34 falls between those limits, it is neither significantly low nor significantly high. b. The probability of exactly 34 peas with long stems is 47 C34 ×0.7547 ×0.2513 = 0.118. c. The probability of 34 or more peas with long stems is 47 C34 ×0.7534 ×0.2513 + = 0.390.

+ 47 C47 ×0.7547 ×0.250

d. The probability from part (c) is relevant. The value of 34 peas with long stems is not significantly low. e. The results do not provide strong evidence against Mendel’s claim of 75% for peas with long stems.

.

Section 5-3: Poisson Probability Distributions

55

36. a. m = np = 80 ×0.05 = 4.0 influenza cases, s = np(1 - p ) = 80 ×0.05 ×0.95 = 1.9 influenza cases; Values between 4.0 - 2(1.9) = 0.2 influenza cases and 4.0 + 2(1.9) = 7.8 influenza cases are not significant (0.1 and 7.9 if using unrounded values). 1 influenza case lies between these limits, so it is not significantly low or high. b. The probability of exactly 1 influenza case is 80 C1 ×0.051 ×0.9579 = 0.0695. c. The probability of 1 or fewer influenza cases is 80 C0 ×0.050 ×0.9570 +

+ 80 C1 ×0.051 ×0.9579 = 0.0861.

d. The probability from part (c) is relevant. The value of influenza case is not significantly high. e. The results do not provide strong evidence for the vaccine’s effectiveness.

4

37. P(5) = 0.06 (1 - 0.06) = 0.0468 5

2

3

10! æ 78 ö æ 22 ö æ 21 ö ×ç ÷ ×ç ÷ ×ç ÷ = 0.0485; The expression could not be used without replacement, the outcomes 5!2!3! è121 ø è121 ø è121 ø would no longer be independent.

38.

39. Without replacement: P(3) = 3

4!

16!

(4 + 16)! × ¸ = 0.139 (4 - 3)!3! (16 - 20 + 3)!(20 - 3)! (4 + 16 - 20)!20!

5

( ) ( ) = 0.147

4 With replacement: 8 C3 × 20 × 16 20

Section 5-3: Poisson Probability Distributions 1. m = 31,645 365 = 86.7, which is the mean number of patient admissions per day. x = 85, because we want the probability that a randomly selected day has exactly 85 admissions, and e » 2.71828 which is a constant used in all applications of Formula 5-9.

2.

The mean is m = 31,645 365 = 86.7, patients, the standard deviation is s = 86.7 = 9.3 patients, and the

variance is s 2 = 86.7 patients2.

3. The possible values of x are 0, 1, 2, … (with no upper bound), so x is a discrete random variable. It is not possible to have x = 90.3 patient admissions in a day 4.

P(0) represents the probability of no occurrences of an event during the relevant interval. If x = 0, P (0) = e - m.

5.

P(12) =

6.

P(9) =

7.

P(At least one birth) = 1 - P (0 ) = 1 -

8.

P(At least two births) = 1 - P (0) - P (1) = 1 -

mx ×e- m 11.586312 ×e- 11.5863 = = 0.114 x!

12!

mx ×e- m 11.58639 ×e- 11.5863 = = 0.964 x!

9!

11.58630 ×e- 11.5863 = 0.999991 0!

11.58630 ×e- 11.5863 11.58631 ×e- 11.5863 = 0.99988 0!

1!

0.90 ×e- 0.9 333 = 0.407 (0.402 if using the unrounded mean.) There should be many = 0.9 murders, P(0) = 0!

365 days (roughly 40%) with no murders.

9.

m=

.

56

Chapter 5: Discrete Probability Distributions

10. m =

196 = 9.8 20

a. P(0) =

9.80 ×e- 9.8 = 0.497 0!

c. P(2) =

9.82 ×e- 9.8 = 0.122 2!

b. P(1) =

9.81 ×e- 9.8 = 0.348 1!

d. P(3) =

9.83 ×e- 9.8 = 0.0284 3!

9.84 ×e- 9.8 = 0.00497; The expected frequencies of 139, 97, 34, 8 , and 1.4 compare reasonably well 4! to the actual frequencies, so the Poisson distribution does provide good results. e. P(4) =

0.9292 ×e- 0.929 = 0.170 2! b. The expected number of regions with exactly 2 hits is between 97.9 and 98.2, depending on rounding. c. The expected number of regions with 2 hits is close to 93, which is the actual number of regions with 2 hits.

11. a. P(2) =

12. a. m = 12429 ×0.000011 = 0.1367

0.1370 ×e- 0.137 0.1371 ×e- 0.137 = 0.872 and P(1) = = 0.119; So the probability of 0 or 1 is 0!

1!

0.872 + 0.119 = 0.991.

b. P(0) =

c. 1 - 0.991 = 0.009 d. No, the probability of more than one case is extremely small, so the probability of getting as many as four cases is even smaller.

11.30 ×e- 11.3 33,561 = 0.9999876 or 0.9999877 if using the unrounded mean. = 11.3 fatalities, 1 - P(0) = 1 0!

2969 There is a very high chance (“almost certain”) that at least one fatality will occur.

13. m =

14. a. P(0) =

7.00 ×e- 7.0 = 0.001 0!

b. P(At least one) = 1 - P(0) = 1 -

7.00 ×e- 7.0 = 0.999 0!

c. P(At most two) = P(0) + P(1) + P(2) = 15. a. m =

7.00 ×e- 7.0 7.01 ×e- 7.0 7.02 ×e- 7.0 + + = 0.0296 0!

1!

2!

138 = 10.61538 cases 13

b. P( X = 0) =

10.615380 ×e- 10.61538 = 0.0000245 0!

10.615389 ×e- 10.61538 = 0.116; With 2 of the 13 years having exactly 9 cases of rubella, the 9! probability appears to be 2/13, or 0.154; the Poisson distribution yields a probability of 0.116, which is in the general ballpark, but it is off by a fairly large amount. c. P( X = 9) =

.

Chapter Quick Quiz

16. a. m =

57

56 = 1.647 cases per year 34

b. P( X = 0) =

1.6470 ×e- 1.647 = 0.192 0!

1.6470 ×e- 1.647 + 0! which is significantly high. c. P( X £ 5) =

+

1.6475 ×e- 1.647 =0.993; The probability of more than 5 cases is 0.007, 5!

17. The distribution is skewed to the right.

0.35 0.30

Probability

0.25 0.20 0.15 0.10 0.05 0.00

0

1

2

3

4

5

6

7

Diphtheria Cases in One Year

Chapter Quick Quiz 1. m = 64 ×0.206 = 13.2 males 2.

s = 64 ×0.206 ×0.794 = 3.2 males

3.

parameters

4.

5.

Significantly low values are 13.2 - 2(3.2) = 6.8 or fewer males. Significantly high values are 13.2 + 2(3.2) = 19.6 or more males 3 5 8 C3 ×0.206 ×0.794 = 0.154

6. This is probability distribution because the three requirements are satisfied. First, the variable x is a numerical random variable and its values are associated with probabilities. Second, S P( x ) = 0.762 + 0.213 + 0.024 + 0.001 +0 + 0 = 1. Third, each of the probabilities is between 0 and 1 inclusive, as required.

7.

m = 0 ×0.762 + 1×0.213 + 2 ×0.024 + 3 ×0.001 + 0 ×0 + 0 ×0 = 0.3 male

8.

s 2 = (0.5 male)2 = 0.3 male2 ; (0.3 male2 if calculated using values from table.)

9. 0+ indicates that the probability is a very small positive number. It does not indicate that it is impossible for all five adult males to be heavy drinkers.

10. P( X < 3) = P( X £ 2) = 0.762 + 0.213 + 0.024 = 0.999; Yes, 4 is significantly high.

.

58

Chapter 5: Discrete Probability Distributions

Review Exercises 1.

P( X = 3) = 8 C3 ×0.283 ×0.725 = 0.238

2. P( X ³ 2) = 1 - 8 C0 ×0.280 ×0.728 - 8 C1 ×0.281 ×0.727 = 0.703 No, the eight adults from the same family are not randomly selected from the population of adults. Because they are from the same family, they are likely to share similar diet and genetic factors, so they are not independent.

3.

m = 8 ×0.28 = 2.24 adults, s = 8 ×0.28 ×0.72 = 1.27 adults

4.

No, the limit separating significantly low values is –0.3 adult, and 2 is greater than or equal to −0.3. Also, the

probability that two or fewer adults have high cholesterol is 8 C0 ×0.280 ×0.728 + 8 C1 ×0.281 ×0.727 + 2 6 8 C2 ×0.28 ×0.72 = 0.603, which is not low (less than or equal to 0.05).

5.

Yes, the limit separating significantly high values is 2.24 + 2(1.27) = 4.78 adults and 8 is greater than or equal

to 4.78. Also, the probability that all eight adults have high cholesterol is 8 C8 ×0.288 ×0.720 = 0.00003, which is very low (less than or equal to 0.05).

6.

This is not a probability distribution because the responses are not values of a numerical random variable.

7.

a. This is probability distribution because the three requirements are satisfied. First, the variable x is a numerical random variable and its values are associated with probabilities. Second, SP( x ) = 0.304 + 0.400 + 0.220 + 0.064 + 0.011 + 0.001 + 0 = 1, Third, each of the probabilities is between 0 and 1 inclusive, as required. b. m = 0 ×0.304 + 1 ×0.400 + 2 ×0.220 + 3 ×0.064 + 4 ×0.011 + 5 ×0.001 + 6 ×0 = 1.1 condoms c. s = (0 - 1.1)2 ×0.304 + (1 - 1.1)2 ×0.400 + (2 - 1.1) 2 ×0.220 +

+ (5 - 1.1) 2 ×0.001 + (6 - 1.1)2 ×0 = 0.9 condom

d. Yes, 5 failures is significantly high. A number is significantly high if it is equal to or greater than m + 2s = 1.1 + 2(0.9) = 2.9, and 5 does exceed 2.9. Also, the probability of 5 or more failures is 0.001, which is a low probability. e. Here, the symbol 0+ represents a positive probability that is so small that it is 0.000 when rounded.

8.

a. m = 143 365, or 0.392 death per day b. P( X = 0) =

0.3920 ×e- 0.392 = 0.676 0!

c. P( X > 2) = P( X ³ 2) = 1 - P( X £ 1) = 1 -

0.3920 ×e- 0.392 0.3921 ×e- 0.392 = 0.00750 (0.00749 if using 0!

1!

unrounded value for the mean.) d. No, because the event is so rare. (But it is possible that more than one death occurs, so it might be wise to consider a contingency plan.)

.

Cumulative Review Exercises

59

Cumulative Review Exercises 1.

a. The mean is x =

68 + 69 + 73 + 73 + 73 + 79 + 81 + 83 + 88 + 90 + 92 + 97 = 80.5 manatees.

12

79 + 81 = 80.0 manatees.

2 c. The range is 97 - 68 = 29.0 manatees. b. The median is

d. The standard deviation is s =

(73 - 80.5)2 + (69 - 80.5)2 + + (73 - 80.5)2 + (68 - 80.5)2 = 9.7 manatees.

12 - 1 2

e. The variance is s 2 = (9.7 manatees ) = 94.1 manatees2 (93.9 manatees 2 using unrounded values.) f. The trend of the manatee deaths over time is not addressed by the preceding statistics. g. Significantly low numbers are μ – 2σ = 80.5 – 2(9.7) = 61.1 manatees or lower, and significantly high numbers are μ + 2σ = 80.5 + 2(9.7) = 99.9 manatees or higher. h. None of the listed numbers are significantly low or significantly high. i. ratio j. discrete 2.

a. x =

0 (9) + 1(20) + 2 (14) + 3 (18) + 4 (15) + 5 (15) + 6 (16 ) + 7 (13) + 8 (20) + 9 (13) = 4.6 and 9 + 20 + 14 + 18 + 15 + 15 + 16 + 13 + 20 + 13

(

153 9 ×02 + s=

)

+ 13 ×92 - (9 ×0 + 147 (147 - 1)

2

+ 9 ×9)

= 2.8; They are statistics.

b. The last digits appear to be random. None of the frequencies appear to be substantially different from the others. c. No. The values of x are numerical, but the frequencies are not probabilities, as required.

3. No vertical scale is shown, but a comparison of the numbers shows that 7,066,000 is roughly 1.2 times the number 6,000,000. However, the graph makes it appear that the goal of 7,066,000 people is roughly 3 times the number of people enrolled. The graph is misleading in the sense that it creates the false impression that actual enrollments are far below the goal, which is not the case. Fox News apologized for their graph and provided a corrected graph.

4.

a. 1 - 0.45 = 0.55 b. 0.45 ×0.45 = 0.2025 b. P( X ³ 1) = 1 - P( X = 0) = 1 - 3C0 ×0.450 ×0.553 = 0.166 (Table: 0.162) c. m = 50 ×0.45 = 22.5 adults, s = 50 ×0.45 ×0.55 = 12.375 adults; There are parameters. d. Significantly low numbers are m - 2s = 22.5 - 2 (12.375) = - 2.25 or lower, and significantly high numbers are m + 2s = 22.5 + 2 (12.375) = 47.25 and higher. Because 20 is greater than –2.25, it is not a significantly low or significantly high number of adults with hypertension (among 50).

.

Section 6-1: The Standard Normal Distribution

61

Chapter 6: Normal Probability Distributions Section 6-1: The Standard Normal Distribution 1. The word “normal” has a special meaning in statistics. It refers to a specific bell-shaped distribution that can be described by Formula 6-1. The lottery digits do not have a normal distribution.

2.

3.

The mean and standard deviation have values of m = 0 and s = 1, respectively.

4.

The notation za represents the z score that has an area of a to its right.

5.

P ( x > 3) = 0.2 (5 - 3) = 0.4

7.

P (2 < x < 3) = 0.2 (3 - 2 ) = 0.2

6.

P ( x < 4 ) = 0.2 (4 - 0) = 0.8

8.

P (2.5 < x < 4.5) = 0.2 (4.5 - 2.5) = 0.4

9.

P ( z < 0.44 ) = 0.6700

10. P ( z > - 1.04 ) = 1 - P ( z < - 1.04 ) = 1 - 0.1492 = 0.8508 11. P (- 0.84 < z < 1.28) = P ( z < 1.28) - P ( z < - 0.84 ) = 0.8997 - 0.2005 = 0.6992 (Tech: 0.6993) 12. P (- 1.07 < z < 0.67 ) = P ( z < 0.67 ) - P ( z < - 1.07 ) = 0.7486 - 0.1423 = 0.6063 13. z = 1.23

18. P ( z < - 1.96) = 0.0250

14. z = - 0.51

19. P ( z < 1.28) = 0.8997

15. z = - 1.45 20. P ( z < 2.25) = 0.9878

16. z = 0.82 17. P ( z < - 1.23) = 0.1093 21. P ( z > 0.25) = 1 - P ( z < 0.25) = 1 - 0.5987 = 0.4013 22. P ( z > 0.18) = 1 - P ( z < 0.18) = 1 - 0.5704 = 0.4286 23. P ( z > - 2.55) = 1 - P ( z < - 2.55) = 1 - 0.0054 = 0.9946 24. P ( z > - 3.05) = 1 - P ( z < - 3.05) = 1 - 0.0011 = 0.9989

25. P (2.00 < z < 3.00) = P ( z < 3.00) - P ( z < 2.00 ) = 0.9986 - 0.9772 = 0.0214 (Tech: 0.0215) 26. P (1.50 < z < 2.50) = P ( z < 2.50) - P ( z < 1.50) = 0.9938 - 0.9332 = 0.0606 27. P (- 2.55 < z < - 2.00) = P ( z < - 2.00) - P ( z < - 2.55) = 0.0228 - 0.0054 = 0.0174 28. P (- 2.75 < z < - 0.75) = P ( z < - 0.75) - P ( z < - 2.75) = 0.2266 - 0.0030 = 0.2236

.

62

Chapter 6: Normal Probability Distributions

29. P (- 2.00 < z < 2.00) = P ( z < 2.00) - P ( z < - 2.00) = 0.0228 - 0.9772 = 0.9544 (Tech: 0.9545) 30. P (- 3.00 < z < 3.00) = P ( z < 3.00) - P ( z < - 3.00) = 0.9987 - 0.0013 = 0.9974 (Tech: 0.9973) 31. P (- 1.00 < z < 5.00) = P ( z < 5.00) - P ( z < - 1.00) = 0.9999 - 0.1587 = 0.8412 (Tech: 0.8413) 32. P (- 4.27 < z < 2.34 ) = P ( z < 2.34 ) - P ( z < - 4.27 ) = 0.9904 - 0.0001 = 0.9903 33. P ( z < 4.55) = 0.9999 (Tech: 0.999997)

39. P2.0 = - 2.05 and P98.0 = 2.05

34. P ( z > - 3.75) = 0.9999

40. P6.0 = - 1.56 and P94.0 = 1.56

35. P ( z > 0) = 0.5000

41. z0.10 = 1.28 42. z0.02 = 2.05

36. P ( z < 0) = 0.5000

43. z0.04 = 1.75

37. P99 = 2.33

44. z0.15 = 1.04

38. P15 = - 1.04

45. P (- 1 < z < 1) = P ( z < 1) - P ( z < - 1) = 0.8413 - 0.1587 = 0.6826 = 68.26% (Tech: 68.27%) 46. P (- 2 < z < 2 ) = P ( z < 2 ) - P ( z < - 2 ) = 0.9772 - 0.0228 = 0.9544 = 95.44% (Tech: 95.45%) 47. P (- 3 < z < 3) = P ( z < 3) - P ( z < - 3) = 0.9987 - 0.0013 = 0.9974 = 99.74% (Tech: 99.73%) 48. P (- 3.5 < z < 3.5) = P ( z < 3.5) - P ( z < - 3.5) = 0.9999 - 0.0001 = 0.9998 = 99.98% (Tech: 99.95%) 49. a. P ( z > 2 ) = 1 - P ( z < 2 ) = 1 - 0.9772 = 0.0228 = 2.28% b. P ( z < - 2 ) = 0.0228 = 2.28% c. P (- 2 < z < 2 ) = P ( z < 2 ) - P ( z < - 2 ) = 0.9772 - 0.0228 = 0.9544 = 95.44% (Tech: 95.45%) 50. a. m = 2.5 min. and s = 5

12 = 1.4 min

b. The probability is 1 3 or 0.5774, and it is very different from the probability of 0.6827 that would be obtained by incorrectly using the standard normal distribution. The distribution does affect the results very much.

Section 6-2: Real Applications of Normal Distributions 1. a. m = 0 and s = 1 b. The z scores are numbers without units of measurements.

2.

a. The area equals the maximum probability value of 1. b. The median is the middle value and for normally distributed scores that is also the mean, which is 64.2 inches. c. The mode is also 64.2 inches. d. The variance is the square of the standard deviation which is 5.29 inches 2.

3. The standard normal distribution has a mean of 0 and a standard deviation of 1, but a nonstandard normal distribution has a different value for one or both of those parameters.

4. No. Randomly generated digits have a uniform distribution, but not a normal distribution. The probability of a digit less than 3 is 3/10 = 0.3.

.

Section 6-1: The Standard Normal Distribution

5.

z x =118 =

6.

z x =90 =

63

118 - 100 = 1.2; which has an area of 0.8849 to the left.

15

91 - 100 = - 0.6; which has an area of 0.7257 to the right.

15

79 - 100 133 - 100 = - 1.4; which has an area of = 2.2; which has an area of 0.9861 to the left. z x =79 = 15 15 0.0808 to the left. The area between the two scores is 0.9861 - 0.0808 = 0.9053.

7.

z x =133 =

112 - 100 124 - 100 = 0.8; which has an area of = 1.6; which has an area of 0.9452 to the left. z x =112 = 15 15 0.7881 to the left. The area between the two scores is 0.9452 - 0.7881 = 0.1571.

8.

z x =124 =

9.

z = 2.44; so x = 2.44 ×15 + 100 = 136

10. z = 1; so x = 1×15 + 100 = 115 11. z = - 2.07; so x = - 2.07 ×15 + 100 = 69 12. z = 1.33; so x = 1.33 ×15 + 100 = 120 13. z x =95 =

95 - 69.6 = 2.25, which has an area of 0.9878 to the left.

11.3

14. z x =85 =

85 - 69.6 = 1.36, which has an area of 0.869 to the right.

11.3

65 - 69.6 75 - 69.6 = - 0.41, which has an area of 0.3409 to the left and z x =75 = = 0.48, which has an 11.3 11.3 area of 0.6844 to the left. The area between the two scores is 0.6844 – 0.3409 = 0.3435.

15. z x =65 =

95 - 69.6 75 - 69.6 = 2.25, which has an = 0.48, which has an area of 0.6844 to the left and z x =95 = 11.3 11.3 area of 0.9878 to the left. The area between the two scores is 0.9878 – 0.6844 = 0.3034.

16. z x =75 =

17. P95 = 1.65, so the pulse is 1.65 ⋅ 11.3 + 69.6 = 88.25 BPM. 18. Q3 = 0.67; so the length is 0.67 ⋅ 11.3 + 69.6 = 77.17 BPM. 19. P2 = - 2.05, so the lower length is –2.05 ⋅ 11.3 + 69.6 = 46.4 BPM and P98 = 2.05, so the upper length is 2.05 ⋅ 11.3 + 69.6 = 92.8 BPM, so a pulse rate of 100 BPM is significantly high.

20. P3.3 = - 1.84, so the lower length is –1.84 ⋅ 11.3 + 69.6 = 48.8 BPM and P96.7 = 1.84, so the upper length is 1.84 ⋅ 11.3 + 69.6 = 90.4 BPM, so a pulse rate of 55 BPM is not significantly low.

230.0 - 184.0 = 0.84; so 0.2005, or 20.05% (Tech: 20.15%) of 55.0 people would make eye contact longer than 230 seconds. No, the proportion of schizophrenics is not at all likely to be as high as 0.2005, or about 20%.

21. The z score for eye contact of 230.0 seconds is

22. a. The z score for a temperature of 100.4 is

100.4 - 98.2 = 3.87; which corresponds to a an area of 0.62

1 - 0.9999 = 0.0001 = 0.01% (Tech: 0.02%) to the right, which suggests a cutoff of 100.4 F is appropriate. b. The z score for a probability of 2% is 2.05 which corresponds to a temperature of 2.05 ×0.62 + 98.2

= 99.47 F.

.

64

Chapter 6: Normal Probability Distributions

23. a. The z score for the minimum weight is

2495.0 - 3152.0 = - 0.95; which has an area to the left of 0.1711 693.4

(Tech: 0.1717). b. The z score for 5% to the left is –1.645, which corresponds to a weight of - 1.645 ×693.4 + 3152.0 = 2011.4 g (Tech: 2011.5 g). This is about one pound lower that the criterion of 2495 g. c. Birth weights are significantly low if they are 2011.4 g or less, and they are “low birth weights” if they are 2495 g or less. Birth weights between 2011.4 g and 2495 g are “low birth weights” but they are not significantly low.

308 - 268 = 2.67; which corresponds to a probability of 0.0038 or 15 0.38%. Either a very rare event occurred or the husband is not the father. b. The z score corresponding to 3% is –1.87 which corresponds to a pregnancy of - 1.87 ×15 + 268 = 240 days.

24. a. The z score for a 308 day pregnancy is

25. a. The mean is 71.320 mm Hg and the standard deviation is 11.994 mm Hg. The histogram for the data confirms that the distribution is roughly normal.

35 30

Frequency

25 20 15 10 5 0 40

45

50

55

60

65

70

75

80

85

90

95

100 105

DIASTOLIC (M ale)

b. The z score for the bottom 2.5% is –1.96, which corresponds to a blood pressure of - 1.96 ×11.994 + 71.320 = 47.8 mm Hg. The z score for the top 2.5% is 1.96, which corresponds to a blood pressure of 1.96 ×11.994 + 73.320 = 94.8 mm Hg. 26. a. The mean is 70.163 mm Hg and the standard deviation is 11.220 mm Hg. The histogram for the data confirms that the distribution is roughly normal.

40

Frequency

30

20

10

0 40

45

50

55

60

65

70

75

DIASTOLIC (Female)

.

80

85

90

95

100

Section 6-3: Sampling Distributions and Estimators

65

26. (continued) b. The z score for the bottom 2.5% is –1.96, which corresponds to a blood pressure of - 1.96 ×11.220 + 70.163 = 48.2 mm Hg. The z score for the top 2.5% is 1.95, which corresponds to a blood pressure of 1.96 ×11.220 + 70.163 = 92.2 mm Hg. 27. The z score for Q1 is –0.67, and the z score for Q3 is 0.67. The IQR is 0.67 - (- 0.67 ) = 1.34. 1.5 ×IQR = 2.01, so

Q1 - 1.5 ×IQR = - 0.67 - 2.01 = - 2.68 and Q3 + 1.5 ×IQR = 0.67 + 2.01 = 2.68.

The percentage to the left of –2.68 is 0.0037 and the percentage to the right of 2.68 is 0.0037. Therefore, the percentage of an outlier is 0.0074 (Tech: 0.0070).

Section 6-3: Sampling Distributions and Estimators 1. a. In the long run, the sample proportions will have a mean of 0.512. b. The sample proportions will tend to have a distribution that is approximately normal.

2. a. without replacement b. (1) When selecting a relatively small sample from a large population, it makes no significant difference whether we sample with replacement or without replacement.

(2) Sampling with replacement results in independent events that are unaffected by previous outcomes, and independent events are easier to analyze and they result in simpler calculations and formulas.

3.

sample mean, sample variance, sample proportion

4. No. The data set is only one sample, but the sampling distribution of the mean is the distribution of the means from all samples, not the one sample mean obtained from the one sample in Data Set 4.

5. No, the sample is not a simple random sample from the population of all births worldwide. The proportion of boys born in China is substantially higher than in other countries.

6.

a. The distribution of the sample means is approximately normal. b. The sample means target the mean annual income of all physicians.

7.

a. The population mean is m =

4 +5+9 = 6, and the population variance is 3 (4 - 6)2 + (5 - 6)2 + (9 - 6)2 s 2= = 4.7.

3 b. The possible sample of size 2 are {(4, 4), (4, 5), (4, 9), (5, 4), (5, 5), (5, 9), (9, 4), (9, 5), (9, 9)} which have the following variances {0, 0.5, 12.5, 0.5, 0, 8, 12.5, 8, 0} respectively.

Sample Variance

Probability

0.0 0.5 8 12.5

3/9 2/9 2/9 2/9

3 ×0 + 2 ×0.5 + 2 ×8 + 2 ×12.5 = 4.666 » 4.7.

9 d. Yes. The mean of the sampling distribution of the sample variances (4.7) is equal to the value of the population variance (4.7) so the sample variances target the value of the population variance. c. The sample variance’s mean is

.

66

Chapter 6: Normal Probability Distributions

8.

a. The population standard deviation (using the result from the previous problem) is s = 4.666 = 2.160. b. By taking the square root of the sample variances from the previous problem we get Sample Standard Deviation

Probability

0.000 0.707 2.828 3.536

3/9 2/9 2/9 2/9

3 ×0 + 2 ×0.707 + 2 ×2.828 + 2 ×3.536 = 1.571.

9 d. No. The mean of the sampling distribution of the sample standard deviations is 1.571, and it is not equal to the value of the population standard deviation (2.160), so the sample standard deviations do not target the value of the population standard deviation. c. The mean of the sample standard deviations is

9.

a. The population median is 5. b. The possible sample of size 2 are {(4, 4), (4, 5), (4, 9), (5, 4), (5, 5), (5, 9), (9, 4), (9, 5), (9, 9)}, which have the following medians {4, 4.5, 6.5, 4.5, 5, 7, 6.5, 7, 9} with the following associated probabilities.

Sample Median

Probability

4 4.5 5 6.5 7 9

1/9 2/9 1/9 2/9 2/9 1/9

4 + 2 ×4.5 + 5 + 2 ×6.52 ×7 + 9 = 6.0.

9 d. No. The mean of the sampling distribution of the sample medians is 6.0, and it is not equal to the value of the population median of 5.0, so the sample medians do not target the value of the population median. c. The mean of the sampling distribution of the sampling median is

10. a. The proportion of odd numbers is 2/3, or 0.7. (There are two odd numbers from the population of 4, 5, and 9.) b. The possible sample of size 2 are {(4, 4), (4, 5), (4, 9), (5, 4), (5, 5), (5, 9), (9, 4), (9, 5), (9, 9)}, which have the following proportion of odd numbers {0, 0.5, 0.5, 0.5, 1, 1, 0.5, 1, 1.

Sample Proportion

Probability

0.0 0.5 1.0

1/9 4/9 4/9

0 + 4 ×0.5 + 4 ×1 2 = , or 0.7.

9 3 d. Yes. The mean of the sampling distribution of the sample proportion of odd numbers is 2/3, and it is equal to the value of the population proportion of odd numbers of 2/3, so the sample proportions target the value of the population proportion c. The mean of the sampling distribution of sample proportions is

.

Section 6-3: Real Applications of Normal Distributions

67

11. a. The possible samples of size 2 are {(34, 34), (34, 36), (34, 41), (34, 51), (36, 34), (36, 36), (36, 41), (36, 51), (41, 34), (41, 36), (41, 41), (41, 51), (51, 34), (51, 36), (51, 41), (51, 51)}, which have the following ranges and associated probabilities.

Sample Range

Probability

34 35 36 37.5 38.5 41 42.5 43.5 46 51

1/16 2/16 1/16 2/16 2/16 2/16 2/16 1/16 2/16 1/16

34 + 36 + 41 + 51 = 40.5 and the mean of the sample means is 4 34 + 2 ×35 + 36 + 2 ×37.5 + 2 ×38.5 + 2 ×41 + 2 ×42.5 + 43.5 + 2 ×46 + 51 = 40.5 as well.

16 c. The sample means target the population mean. Sample means make good estimators of population means because they target the value of the population mean instead of systematically underestimating or overestimating it. b. The mean of the population is

12. a. The possible samples of size 2 are {(34, 34), (34, 36), (34, 41), (34, 51), (36, 34), (36, 36), (36, 41), (36, 51), (41, 34), (41, 36), (41, 41), (41, 51), (51, 34), (51, 36), (51, 41), (51, 51)}, which have the following medians and associated probabilities.

b. The median of the population is

Sample Median

Probability

34 35 36 37.5 38.5 41 42.5 43.5 46 51

1/16 2/16 1/16 2/16 2/16 2/16 2/16 1/16 2/16 1/16

38.5 + 41.0 36 + 41 = 40.5. = 38.5, but the median of the sample medians is 2 2

The two values are not equal. c. The sample medians do not target the population median of 38.5, so the sample medians do not make good estimators of the population medians.

.

68

Chapter 6: Normal Probability Distributions

13. a. The possible samples of size 2 are {(34, 34), (34, 36), (34, 41), (34, 51), (36, 34), (36, 36), (36, 41), (36, 51), (41, 34), (41, 36), (41, 41), (41, 51), (51, 34), (51, 36), (51, 41), (51, 51)}, which have the following ranges and associated probabilities.

Sample Range

Probability

0 2 5 7 10 15 17

4/16 2/16 2/16 2/16 2/16 2/16 2/16

b. The range of the population is 51 - 34 = 17.0, but the mean of the sample ranges is 4 ×0 + 2 ×2 + 2 ×5 + 2 ×7 + 2 ×10 + 2 ×15 + 2 ×17 = 7.0. The values are not equal.

16 c. The sample ranges do not target the population range of 17, so sample ranges do not make good estimators of the population range.

14. a. The possible samples of size 2 are {(34, 34), (34, 36), (34, 41), (34, 51), (36, 34), (36, 36), (36, 41), (36, 51), (41, 34), (41, 36), (41, 41), (41, 51), (51, 34), (51, 36), (51, 41), (51, 51)}, which have the following variances and associated probabilities.

Sample Variance (s2)

Probability

0.0 2.0 12.5 24.5 50.0 112.5 144.5

4/16 2/16 2/16 2/16 2/16 2/16 2/16

b. The mean of the sample variances is

4 ×0 + 2 ×2 + 2 ×12.5 + 2 ×24.5 + 2 ×50 + 2 ×112.5 + 2 ×144.5 = 43.25.

16

The two values are equal. c. The sample variances do target the population variance of 43.25, so sample variances do make good estimators of the population variance.

15. The possible birth samples are {(b, b), (b, g), (g, b), (g, g)}.

Proportion of Girls

Probability

0 1/2 2/2

0.25 0.5 0.25

Yes. The proportion of girls in 2 births is 0.5, and the mean of the sample proportions is 0.5. The result suggests that a sample proportion is an unbiased estimator of the population proportion.

16. The possible birth samples are {bbb, bbg, bgb, gbb, ggg, ggb, gbg, bgg}.

Proportion of Girls

Probability

0 1/3 2/3 3/3

1/3 3/8 3/8 1/8

.

Section 6-4: The Central Limit Theorem

69

16. (continued) Yes. The proportion of girls in 3 births is 0.5 and the mean of the sample proportions is 0.5. The result suggests that a sample proportion is an unbiased estimator of the population proportion.

17. The possibilities are: both questions incorrect, one question correct (two choices), both questions correct. a.

Proportion Correct

Probability

0 =0 2

4 4 16 × = 5 5 25

1 = 0.5 2

æ1 4 ö 8 2 ×ç × ÷ = è 5 5 ø 25

2 =1 2

æ1 1 ö 1 = çè × ÷ 5 5 ø 25

16 ×0 + 8 ×0.5 + 1×1 = 0.2.

25 c. Yes, the sampling distribution of the sample proportions has a mean of 0.2 and the population proportion is also 0.2 (because there is 1 correct answer among 5 choices). Yes, the mean of the sampling distribution of the sample proportions is always equal to the population proportion. b. The mean is

18. a. The proportions of 0, 0.5, and 1 have the following probabilities.

Proportion with Yellow Pods

Probability

0 0.5 1

1/25 8/25 16/25

1×0 + 8 ×0.5 + 16 ×1 = 0.8.

25 c. Yes, the population proportion is 0.8 and the mean of the sampling proportions is also 0.8. The mean of the sampling distribution of proportions is always equal to the population proportion. b. The mean is

19. The formula yields P(0) =

1 1 = 0.25 , P(0.5) = = 0.5, and 2(2 - 2 ×0)!(2 ×0)!

2(2 - 2 ×0.5)!(2 ×0.5)!

1 = 0.25, which describes the sampling distribution of the sample proportions. The formula 2(2 - 2 ×1)!(2 ×1)! is just a different way of presenting the same information in the table that describes the sampling distribution.

P(1) =

20. Sample values of the mean absolute deviation (MAD) do not usually target the value of the population MAD, so a MAD statistic is not good for estimating a population MAD. If the population of {4, 5, 9} from Example 5 is used, the sample MAD values of 0, 0.5, 2, and 2.5 have corresponding probabilities of 3/9, 2/9, 2/9, and 2/9. For these values, the population MAD is 2, but the sample MAD values have a mean of 1.1, so the mean of the sample MAD values is not equal to the population MAD. Section 6-4: The Central Limit Theorem 1. The sample must have more than 30 values, or there must be evidence that the population of grade-point averages from statistics students has a normal distribution.

2. No. because the original population is normally distributed, the sample means will be normally distributed for any sample size, not just for n > 30.

3.

mx represents the mean of all sample means and s x which represents the standard deviation of all sample

means. For samples of 64 IQ scores, mx = 100, and s x = 15

.

64 = 1.875.

70

Chapter 6: Normal Probability Distributions

4. No. The sample of annual incomes will tend to have a distribution that is skewed to the right, no matter how large the sample is. If we compute the sample mean, we can consider that value to be one value in a normally distributed population.

5.

a. z x =74 = b. z x =74 =

74.0 - 69.6 = 0.39, which has a probability of 0.6517 to the left.

11.3 74.0 - 69.6 11.3

25

= 1.95, which has a probability of 0.9744 to the left (Tech:0.9728).

c.

Because the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

6.

a. z x =64 = b. z x =64 =

64.0 - 69.6 = 0.50, which has a probability of 1 – 0.3085 = 0.6915 to the right.

11.3 64.0 - 69.6 11.3

36

= - 2.97, which has a probability of 1 – 0.0015 = 0.9985 to the right.

c.

Because the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

7.

82.0 - 69.6 86.0 - 69.6 = 1.10, and z x =86 = = 1.45, which have a probability of 0.9265 – 0.8643 = 11.3 11.3 0.0622 between them (Tech:0.1285). a. z x =82 =

82.0 - 69.6

86.0 - 69.6

= 4.35, which have a probability of 0.9999 – 0.9995 = 11.3 9 11.3 9 0.0004 between them . c.

Because the original population has a normal distribution, the distribution of sample means is normal for any sample size. b. z x =82 =

8.

= 3.29, and z x =86 =

66.0 - 69.6 80.0 - 69.6 = - 0.32, which have a probability of 0.8212 – 0.3745 = = 0.92, and z x =66 = 11.3 11.3 0.4467 between them. a. z x =80 =

80.0 - 69.6

66.0 - 69.6

= - 1.27, which have a probability of 0.9999 – 0.1020 = 11.3 16 11.3 16 0.8979 between them (Tech:0.1018). c.

Because the original population has a normal distribution, the distribution of sample means is normal for any sample size. b. z x =80 =

9.

= 3.68, and z x =66 =

17.7 - 14.7 13.6 - 14.7 = 2.31 and z x=13.6 = = - 0.85 which have a probability of 1.3 1.3 0.9896 - 0.1977 = 0.7919 (Tech: 0.7908) between them.

a. z x =17.7 =

13.6 - 14.7 17.7 - 14.7 = - 7.62 which have a probability of = 20.8 and z x=13.6 = 1.3 9 1.3 9 0.9999 - 0.0001 = 0.9998 (Tech: 0.9944) between them.

b. z x=17.7 =

12.1 - 13.0 15.1 - 13.0 = - 0.69 which have a probability of = 1.62 and z x =12.1 = 1.3 1.3 0.9474 - 0.2451 = 0.7023 (Tech: 0.7025) between them.

10. a. z x =15.1 =

15.1 - 13.0 12.1 - 13.0 = 4.85 and z x=12.1 = = - 2.08 which have a probability of 1.3 9 1.3 9 0.9999 - 0.0188 = 0.9811 between them.

b. z x=15.1 =

.

Section 6-4: The Central Limit Theorem

71

130 - 121.7 = 0.48, which has a probability of 1 – 0.6844 = 0.3156 (Tech: 0.3158) to the right.

17.2

11. a. z x =130 =

130 - 121.7

b. z x =130 =

17.2

9

= 1.45, which has a probability of 1 – 0.9265 = 0.0735 (Tech: 0.0702) to the right.

130 - 124.2 = 0.40, which has a probability of 1 – 0.6554 = 0.3446 (Tech: 0.3449) to the right.

14.4

12. a. z x =130 =

130 - 124.2

b. z x =130 =

14.4

9

= 1.21, which has a probability of 1 – 0.8869 = 0.1131 to the right.

13. a. The z score for the top 2% is 2.05, which corresponds to an IQ score of 2.05 ×15 + 100 = 131.

131 - 100 = 4.13; which has a probability of 1 - 0.9999 = 0.0001 (Tech: 0.0000179) to the right.

15 4

b. z x=131 =

c. No, it is possible that the 4 subjects have a mean of 132 while some of them have scores below the Mensa requirement of 131.

7 - 6.8 9 - 6.8 = 0.14; which have a probability of 0.9418 - 0.5557 = 0.3861 = 1.57 and z x =7 = 1.4 1.4 (Tech: 0.3851) between them.

14. a. z x =9 =

7 - 6.8 9 - 6.8 = 0.32 which have a probability of 0.9999 - 0.6255 = 0.3744 = 3.51 and z x =7 = 1.4 5 1.4 5 (Tech: 0.3745) between them.

15. a. The mean weight of passengers is 3500 25 = 140 lb. b. z x=9 =

b. z x=140 =

140 - 189 = - 6.28; which has a probability of 1 - 0.0001 = 0.9999 (Tech: 0.999999998) to the right.

39 25

c. z x =175 =

175 - 189 = - 1.61; which has a probability of 1 - 0.0537 = 0.9463 (Tech: 0.9458) to the right.

39 20

d. The new capacity of 20 passengers does not appear to be safe enough because the probability of overloading is too high.

16. a. z x=22 =

22 - 18.2 = 3.8; which has a probability of 0.9999. So the percentage is 99.99%.

1

18.5 - 18.2 = 1.8 which has a probability of 0.9641. No, when considering the diameters of 1 36 manholes, we should use a design based on individual men, not samples of 36 men. b. z x=18.5 =

17. a. z x =17 = b. z x=17 =

17.0 - 14.4 = 2.60; which has a probability of 1 - 0.9953 = 0.0047 to the right.

1.0

17.0 - 14.4 = 28.7; which has a probability of 1 - 0.9999 = 0.0001 (Tech: 0.0000) to the right.

1.0 122

c. The result from part (a) is relevant because the seats are occupied by individuals.

211 - 171 140 - 171 = 0.87 and z x =140 = = - 0.67; which have a probability of 0.8078 - 0.2514 46 46 = 0.5564 between them (Tech: 0.5575).

18. a. z x =211 =

.

72

Chapter 6: Normal Probability Distributions

18. (continued)

211 - 171 140 - 171 = 4.35 and z x=140 = = - 3.37; which have a probability of 0.9999 - 0.0004 46 25 46 25 = 0.9995 between them (Tech: 0.9996).

b. z x =211 =

c. Part (a) because the ejection seats will be occupied by individual women, not groups of women.

19. a. z x =72 = b. z x =72 =

72 - 68.6 = 1.21; which has a probability of 0.8869 (Tech: 0.8877).

2.8 72 - 68.6 = 12.1; which has a probability of 0.9999. (Tech: 1.0000 when rounded to four decimal 2.8 100

places.) c. The probability of part (a) is more relevant because it shows that about 89% of male passengers will not need to bend. The result from part (b) gives us useful information about the comfort and safety of individual male passengers. d. Because men are generally taller than women, a design that accommodates a suitable proportion of men will necessarily accommodate a greater proportion of women.

167.6 - 171

= - 0.45, which has a probability of 1 – 0.3264 = 0.6736 to the right. There is a 0.6736 46 37 probability that the aircraft is overloaded. Because the probability is so high, the pilot should take action, such as removing excess fuel and/or requiring that some passengers disembark and take a later flight.

20. z x =167.6 =

21. a. Yes, the sampling is without replacement and the sample size of 50 is greater than 5% of the finite population 16 275 - 50 size of 275. s x = = 2.0504584 50 275 - 1

105 - 95.5 95 - 95.5 = - 0.24; which have a probability of = 4.63 and z x =95.5 = 2.0504584 2.0504584 1 - 0.4053 = 0.5947 between them (Tech: 0.5963).

b. z x =105 =

Section 6-5: Assessing Normality 1. The histogram should be approximately bell-shaped, and the normal quantile plot should have points that approximate a straight-line pattern.

2. Either the points are not reasonably close to a straight-line pattern, or there is some systematic pattern that is not a straight-line pattern.

3. We must verify that the sample is from a population having a normal distribution. We can check for normality using a histogram, identifying the number of outliers, and constructing a normal quantile plot.

4. Because the histogram is roughly bell-shaped, we can conclude that the data are from a population having a normal distribution.

5. Normal; the points are reasonably close to a straight-line pattern and there is no other pattern that is not a straight-line pattern.

6. Normal; the points are reasonably close to a straight-line pattern and there is no other pattern that is not a straight-line pattern.

7. Not normal; the points are not reasonably close to a straight-line pattern and there appears to be a pattern that is not a straight-line pattern.

8. Not normal; the points are not reasonably close to a straight-line pattern and there appears to be a pattern that is not a straight-line pattern.

.

Section 6-5: Assessing Normality 9.

73

not normal Histogram for Exercise 10

Histogram for Exercise 9

400

50

300

Frequency

Frequency

40

30

20

200

100 10 0

1

2

3

4

5

6

0

7

0

10

20

30

40

50

60

70

80

27

30

33

Length of Stay

Petal Length

10. not normal 11. not normal Histogram for Exercise 12

Histogram for Exercise 11

14

5

12 4

Frequency

Frequency

10 3

2

8 6 4

1 0

2 19.5

20.0

20.5

21.0

21.5

22.0

0

22.5

Length of Cuckoo Eggs in Wren Nests

9

12

15

18

21

24

Neck Size

12. normal 13. Not normal, the points are not reasonably close to a straight-line pattern and there appears to be a pattern that is not a straight-line pattern.

.

74

Chapter 6: Normal Probability Distributions

14. Not normal, the points are not reasonably close to a straight-line pattern and there appears to be a pattern that is not a straight-line pattern.

15. normal

16. Normal, but there is a possible outlier.

17. Normal, the points are reasonably close to a straight-line pattern and there is no other pattern that is not a straight-line pattern. The points have coordinates (27.4, –1.13), (30.3, –0.48), (31.4, 0), (34.0, 0.48), and (37.0, 1.13).

18. Not normal, the points are not reasonably close to a straight-line pattern and there appears to be pattern that is not a straight-line pattern. The points have coordinates (1689.80, –1.38), (1709.30, –0.83), (1743.04, –0.46), (1806.31, –0.15), (1850.64, 0.15), (1866.99, 0.46), (2136.37, 0.83) and (2216.40, 1.38).

19. a. Yes, the histogram or normal quantile plot will remain unchanged. b. Yes, the histogram or normal quantile plot will remain unchanged. c. No, the histogram and normal quantile plot will not indicate a normal distribution. .

Section 6-6: Normal as Approximation to Binomial

75

20. The original values are not from a normally distributed population.

After taking the logarithm of each value, the values appear to be from a normally distributed population. The original values are from a population with a lognormal distribution.

Section 6-6: Normal as Approximation to Binomial 1. a. the area below (to the left of) 502.5 b. the area between 501.5 and 502.5 c. the area above (to the right of) 502.5 2. Yes, the circumstances correspond to 100 independent trials of a binomial experiment in which the probability of success is 0.2. Also, with n = 100, p = 0.2, and q = 0.8, the requirements of np = 100 ×0.2 = 20 ³ 5 and nq = 100 ×0.8 = 80 ³ 5 are both satisfied.

3. p = 0.2, q = 0.8, m = 20, s = 4; The value of m = 20 shows that for people who make random guesses for the 100 questions, the mean number of correct answers is 20. For people who make 100 random guesses, the standard deviation of s = 4 is a measure of how much the numbers of correct responses vary.

4. The histogram should be approximately normal or bell-shaped, because sample proportions tend to approximate a normal distribution.

5.

The requirement of np = 7 ⋅ 0.596 = 4.2 ≥ 5 is not satisfied. The normal approximation should not be used.

The requirements for the normal approximation are satisfied with np = 25 ⋅ 0.596 = 14.9 ≥ 5 and nq = 25 ⋅ 0.404 9.5 - 25 ×0.596 = - 2.20, which has a probability of 0.0139 (Tech: 0.0142) to the left. = 10.10 ≥ 5. z x =9.5 = 25 ×0.596 ×0.404

6.

7.

The requirement of np = 20 ×0.2 = 4 ³ 5 is not satisfied. The normal approximation should not be used.

8.

The requirements for the normal approximation are satisfied with np = 50 ×0.2 = 10 ³ 5 and

11.5 - 50 ×0.2 12.5 - 50 ×0.2 = 0.53 and z x =12.5 = = 0.88; which have a probability 50 ×0.2 ×0.8 50 ×0.2 ×0.8 of 0.8106 - 0.7019 = 0.1087 (Tech: 0.1096) between them. nq = 50 ×0.8 = 40 ³ 5. z x =11.5 =

9.

The requirements for the normal approximation are satisfied with np = 100 ×0.35 = 35 ³ 5 and

39.5 - 100 ×0.35 = 0.94; which has a probability of 1 - 0.8264 = 0.1736 (Tech: 100 ×0.35 ×0.65 0.1727) to the to the right. No, 40 people with blue eyes is not significantly high. (Tech: Using the binomial distribution: 0.1724.) nq = 100 ×0.65 = 65 ³ 5. z x =39.5 =

.

76

Chapter 6: Normal Probability Distributions

10. The requirements for the normal approximation are satisfied with np = 100 ×0.35 = 35 ³ 5 and

49.5 - 100 ×0.35 = 3.04; which has a probability of 1 - 0.9988 = 0.0012 to the 100 ×0.35 ×0.65 to the right. Yes, 49 people with blue eyes is significantly high. (Tech: Using the binomial distribution: 0.0027.) nq = 100 ×0.65 = 65 ³ 5. z x =49.5 =

11. The requirements for the normal approximation are satisfied with np = 100 ×0.12 = 12 ³ 5 and

4.5 - 100 ×0.12 = - 2.31; which has a probability of 0.0104 (Tech: 0.0105) to the 100 ×0.12 ×0.88 left. Yes, 4 people with green eyes is significantly low. (Tech: Using the binomial distribution: 0.0053.) nq = 100 ×0.88 = 88 ³ 5. z x =4.5 =

12. The requirements for the normal approximation are satisfied with np = 100 ×0.40 = 40 ³ 5 and

33.5 - 100 ×0.40 = - 1.33; which has a probability of 0.0918 (Tech: 0.0923) to 100 ×0.40 ×0.60 the left. No, 33 people with brown eyes is not significantly low. (Tech: Using the binomial distribution: 0.0913.) nq = 100 ×0.60 = 60 ³ 5. z x =33.5 =

13. a. The requirements for the normal approximation are satisfied with np = 250 ×0.10 = 25 ³ 5 and

17.5 - 250 ×0.1 16.5 - 250 ×0.1 = - 1.58 and z x =16.5 = = - 1.79 = 0.92; which 250 ×0.1 ×0.9 250 ×0.1 ×0.9 have a probability of 0.0571 - 0.0367 = 0.0204 between them. (Tech: Using the binomial distribution: 0.0205.) nq = 250 ×0.90 = 225 ³ 5. z x =17.5 =

b. The requirements for the normal approximation are satisfied with np = 250 ×0.10 = 25 ³ 5 and

17.5 - 250 ×0.1 = - 1.58; which has a probability of 0.0571 to the left.

250 ×0.1×0.9 (Tech: Using the binomial distribution: 0.0513.) nq = 250 ×0.90 = 225 ³ 5. z x =17.5 =

c. The result of 17 cases of nausea is significantly low.

14. a. The requirements for the normal approximation are satisfied with np = 929 ×0.75 = 696.75 ³ 5 and

704.5 - 929 ×0.75 = 0.59; which has a probability of 929 ×0.75 ×0.25 1 - 0.7224 = 0.2776 (Tech: 0.2785) to the right. (Tech: Using the binomial distribution: 0.2799.) nq = 929 ×0.25 = 232.25 ³ 5. z x =704.5 =

b. The result of 705 peas with red flowers is not significantly high. c. The result of 705 peas with red flowers is not strong evidence against Mendel’s assumption that 3 4 of peas will have red flowers.

15. a. The requirements for the normal approximation are satisfied with np = 1480 ×0.292 = 432.16 ³ 5 and

454.5 - 1480 ×0.292 = 1.28; which has a probability of 1480 ×0.292 ×0.708 1 - 0.8997 = 0.1003 (Tech: 0.1008) to the right. (Tech: Using the binomial distribution: 0.1012.) nq = 1480 ×0.708 = 1047.84 ³ 5. z x =454.5 =

b. The result of 455 who have sleepwalked is not significantly high. c. The result of 455 does not provide strong evidence against the rate of 29.2%.

16. a. Using normal approximation: m = 420, 095 ×0.00034 = 142.83, s = 420,095 ×0.000344 ×0.999656

135.5 - 142.83 = - 0.61; which has a probability of 0.2709 (Tech: 420,095 ×0.000344 ×0.999656 0.2697) to the left. (Tech: Using the binomial distribution: 0.2726.) b. Media reports appear to be wrong. = 11.9492, and z x=135.5 =

.

Chapter Quick Quiz 17. (1)

77

The requirements for the normal approximation are satisfied with np = 11 ×0.512 = 5.632 ³ 5 and

7.5 - 11 ×0.512 6.5 - 11×0.512 = 1.13; which = 0.52 and z x =7.5 = 11×0.512 ×0.488 11×0.512 ×0.488 have a probability of 0.8708 - 0.6985 = 0.1723 between them. nq = 11×0.488 = 5.368 ³ 5. z x =6.5 =

(2) 0.1704 (3) 0.1726 No, the approximations are not off by very much.

Chapter Quick Quiz 1.

2.

z = P9 = - 1.34

3.

P ( z > - 2.93) = 1 - P ( z < - 2.93) = 1 - 0.0017 = 0.9983

4.

P (0.87 < z < 1.78) = P ( z < 1.78) - P ( z < 0.87 ) = 0.9625 - 0.8078 = 0.1547 (Tech: 0.1546)

5.

a. m = 0 and s = 1 b. mx represents the mean of all sample means, and s x represents the standard deviation of all sample means.

6 z x =80 = 7.

80.0 - 70.2 = 0.88; which has a probability of 0.8106 (Tech: 0.8092) to the left.

11.2

80.0 - 70.2 60.0 - 70.2 = 0.88; which have a probability of 0.8106 - 0.1814 = - 0.91 and z x =80 = 11.2 11.2 = 0.6292 (Tech: 0.6280) between them. z x=60 =

8. The z score for P90 is 1.28, which corresponds to a diastolic blood pressure of 1.28 ×11.2 + 70.2 = 84.5 mm Hg (Tech: 84.6 mm Hg).

9.

z x=75 =

75.0 - 70.2 = 1.71; which has a probability of 0.9564 (Tech: 0.9568) to the left.

11.2 16

10. The normal quantile plot suggests that diastolic blood pressure levels of women are normally distributed.

Review Exercises 1. a. The probability to the left of a z-score of 1.23 is 0.8907. b. The probability to the right of a z-score of –1.23 is 1 – 0.1093 = 0.8907. c. The probability between z-scores –1.01 and 2.01 is 0.9778 – 0.1562 = 0.8216. d. The z-score for Q3 is 0.67. e. z x =0.50 =

0.50 - 0.0 1 16

= 2.00, which has a probability of 1 – 0.9772 = 0.0228.

.

78

Chapter 6: Normal Probability Distributions

2.

a. z x =54 =

3.

a. z x =70 =

54 - 59.7 = - 2.28; which has a probability of 0.0113 to the left, or 1.13%.

2.5 b. The z score for the lowest 95% is 1.645 which corresponds to a standing eye height of 1.645 ×2.5 + 59.7 = 63.8 in.

70 - 64.3 = 2.19; which has a probability of 1 - 0.9857 = 0.0143 to the right, or 1.43%. (Tech: 2.6

1.42%) b. The z score for the lowest 2% is –2.05 which corresponds to a standing eye height of - 20.05 ×2.6 + 64.3 = 59.0 in.

4.

a. The distribution of samples means is normal. b. mx = 100 c. s x = 15

64 = 1.875

5.

a. An unbiased estimator is a statistic that targets the value of the population parameter in the sense that the sampling distribution of the statistic has a mean that is equal to the mean of the corresponding parameter. b. mean, variance and proportion c. true

6.

a. z x =72 =

7.

a. Because women are generally a little shorter than men, a doorway height that accommodates men will also accommodate women.

72 - 68.6 = 1.21; which has a probability of 0.8869, or 88.69% (Tech: 88.77%) to the left. With 2.8 about 11% of all men needing to bend, the design does not appear to be adequate, but the Mark VI monorail appears to be working quite well in practice. b. The z score for 99% is 2.33 which corresponds to a doorway height of 2.33 ×2.8 + 68.6 = 75.1 in.

b. z x =72 =

72 - 68.6 = 9.41; which has a probability of 0.9999, or 1 when rounded.

2.8 60

c. Because the mean height of 60 men is less than 72 in., it does not follow that the 60 individual men all have heights less than 72 in. In determining the suitability of the door height for men, the mean of 60 heights is irrelevant, but the heights of individual men are relevant. a. No, a histogram is far from bell shaped and a normal quantile plot reveals a pattern of points that is far from a straight-line pattern.

9 8 7 6

Frequency

8.

5 4 3 2 1 0

15

20

25 30 35 40

45

50 55

60

BM I (Female)

.

Cumulative Review Exercises 8.

79

(continued) b. No, the sample size ( n = 13) does not satisfy the condition of n > 30 and the values do not appear to be from a population having a normal distribution.

9.

The requirements for the normal approximation are satisfied with np = 1064 ×0.75 = 798 ³ 5 and

787.5 - 1064 ×0.75 = - 0.74; which has a probability of 0.2296 (Tech: 11064 ×0.75 ×0.25 0.2286) to the left. The occurrence of 787 offspring plants with long stem is not unusually low because its probability is not small. The results are consistent with Mendel’s claimed proportion of 3 4. (Tech: Using the binomial distribution: 0.2278.) nq = 1064 ×0.25 = 266 ³ 5. z z =787.5 =

10. a. z x =70 =

70.0 - 63.7 = 2.17; which has a probability of 1 - 0.9850 = 0.0150 , or 1.50% (Tech: 1.49%) to the 2.9

right. b. The z score for the upper 2.5% is 1.96 which corresponds to a doorway height of 1.96 ×2.9 + 63.7 = 69.4 in.

Cumulative Review Exercises 1.

a. The mean is x = b. The median is

15.9 + 18.7 + 24.2 + 28.7 + 28.8 + 28.9 + 28.9 + 28.9 + 29.0 + 29.1 + 29.3 + 31.4 = 26.82.

12

28.9 + 28.9 = 29.90.

2

c. The standard deviation is s = d. z x =31.4 =

(15.9 - 26.82)2 + ... + (31.4 - 26.82) 2 = 4.76.

12 - 1

31.4 - 26.82 = 0.96 4.76

e. ratio f. continuous 2.

a. Q1: L =

24.2 + 28.7 25 ×12 = 26.45 = 3, so Q1 = 100 2

Q2 : L =

28.9 + 28.9 50 ×12 = 28.90 = 6, so Q2 = 100 2

Q3: L =

75 ×12 29.0 + 29.1 = 9, so Q3 = = 29.05 100 2

b.

c. The sample does not appear to be from a population having a normal distribution.

.

80

Chapter 6: Normal Probability Distributions

3.

a. 0.10 ×0.10 ×0.10 = 0.001 b. P( X ³ 1) = 1 - P( X = 0) = 1 - 3C0 ×0.100 ×0.903 = 0.271 c. The requirement that np > 5 is not satisfied, indicating that the normal approximation would result in errors that are too large. d. m = 50 (0.10) = 5.0 people e. s = 50 ×0.10 ×0.90 = 2.1 people f. Since 8 is not greater than m + 2s = 5.0 + 2(2.1) = 9.2, it is not significantly high.

4.

a. B is the event of selecting someone who does not have blue eyes. b. P( B ) = 1 - 0.35 = 0.65 c. 0.35 ×0.35 ×0.35 = 0.0429 d. The requirements for the normal approximation are satisfied with np = 100 ×0.35 = 35 ³ 5 and

39.5 - 100 ×0.35 = 0.94; which has a probability of 1 - 0.8264 = 0.1736 100 ×0.35 ×0.65 (Tech: 0.1727) to the right. (Tech: Using the binomial distribution: 0.1724.) e. No, 40 people with blue eyes is not significantly high. nq = 100 ×0.65 = 65 ³ 5. z x =39.5 =

5.

a. z x =10 =

10 - 9.6 = 0.80; which has a probability of 0.7881 to the left.

0.5

8.0 - 9.6 11.0 - 9.6 = - 3.20 and z x=11.0 = = 2.80; which have a probability of 0.5 0.5 0.9974 - 0.0007 = 0.9967 (Tech: 0.9968) between them.

b. z x=8.0 =

c. The z score for the top 5% is 1.645, which correspond to the length 1.645 ×0.5 + 9.6 = 10.4 in. d. z x =9.8 =

9.8 - 9.6 = 2.00; which has a probability of 1 - 0.9772 = 0.0228 to the right.

0.5 25

.

Section 7-1: Estimating a Population Proportion

Chapter 7: Estimating Parameters and Determining Sample Sizes Section 7-1: Estimating a Population Proportion 1. The confidence level (such as 95%) was not provided.

2. When using 14% to estimate the value of the population percentage for those who prefer chocolate pie, the maximum likely difference between 14% and the true population percentage is four percentage points, so the interval from 14% - 4% = 10% to 14% + 4% = 18% is likely to contain the true population percentage.

3.

pˆ = 0.14 is the sample proportion; qˆ = 0.86 (found from evaluating 1 - pˆ ); n = 1000 is the sample size; E = 0.04 is the margin of error; p is the population proportion, which is unknown. The value of a is 0.05.

4. The 95% confidence interval will be wider than the 80% confidence interval. A confidence interval must be wider in order for us to be more confident that it captures the true value of the population proportion. (Think of estimating the age of a classmate. You might be 90% confident that she is between 20 and 30, but you might be 99.9% confident that she is between 10 and 40.)

5.

z0.05 = 1.645

7.

z0.0025 = 2.81

6.

z0.005 = 2.576 (Table: 2.575)

8.

z0.025 = 1.96

9.

pˆ =

0.375 + 0.425 0.425 - 0.375 = 0.025; 0.400 ± 0.025 = 0.400 and E = 2 2

10. pˆ =

0.450 - 0.320 0.320 + .0450 = 0.065;0.385 ± 0.065 = 0.385 and E = 2 2

11. pˆ =

0.0780 + 0.162 0.162 - 0.0780 = 0.042; 0.120 - 0.042 < p < 0.120 + 0.42 = 0.120 and E = 2 2

12. 0.070 - 0.021 < p < 0.070 + 0.021, or 0.049 < p < 0.091 13. a. pˆ = 16 227 = 0.0705 16 211 ( 227 )( 227 ) = 0.0333

b. E = za / 2

ˆˆ pq = 1.96 n

c.

pˆ - E < p < pˆ - E

227

0.0705 - 0.0333 < p < 0.0705 + 0.0333 0.0371 < p < 0.104 (Tech: 0.0372 < p < 0.104) d. We have 95% confidence that the interval from 0.0371 to 0.104 actually does contain the true value of the population proportion of subjects treated with OxyContin who experience headaches.

14. a. pˆ = 153 5924 = 0.0258 b. E = za / 2 c.

ˆˆ pq = 2.58 n

153 771 ( 5924 )( 5924 ) = 0.00531

5924

pˆ - E < p < pˆ - E 0.0258 - 0.00531 < p < 0.0258 + 0.00531 0.0205 < p < 0.0311

d. We have 99% confidence that the interval from 0.0205 to 0.0311 actually does contain the true value of the population proportion of all Eliquis users who experience nausea.

.

81

82

Chapter 7: Estimating Parameters and Determining Sample Sizes

15. a. pˆ = 717 5000 = 0.143 b. E = za / 2 c.

ˆˆ pq = 1.645 n

717 4283 ( 5000 )( 5000 ) = 0.00815

5000

pˆ - E < p < pˆ - E 0.143 - 0.00815 < p < 0.143 + 0.00815 0.135 < p < 0.152

d. We have 90% confidence that the interval from 0.135 to 0.152 actually does contain the true value of the population proportion of returned surveys.

16. a. pˆ = 856 1228 = 0.697 b. E = za / 2 c.

ˆˆ pq = 1.96 n

856 372 (1228 )(1228 ) = 0.0257

1228

pˆ - E < p < pˆ - E 0.697 - 0.0257 < p < 0.697 + 0.0257 0.671 < p < 0.723

d. We have 95% confidence that the interval from 0.671 to 0.723 actually does contain the true value of the population proportion of medical malpractice lawsuits that are dropped or dismissed.

( 426 )( 434 )

ˆ ˆ 426 pq = ± 1.96 860 860 Þ 0.462 < p < 0.529; Because 0.512 is contained within the n 860 860 confidence interval, there is not strong evidence against 0.512 as the value of the proportion of boys in all births.

17. 95% CI: pˆ ± za / 2

18. a. 99% CI: pˆ ± za / 2

ˆ ˆ 428 pq = ± 2.58 n 580

428 152 ( 580 )( 580 ) Þ 0.691 < p < 0.785, or 69.1% < p < 78.5%

580

b. No, the confidence interval includes 75%, so the true percentage could easily equal 75%.

19. a. 95% CI: pˆ ± za / 2

ˆˆ pq 52 = ± 1.96 n 227

52 ( 227 )( 175 227 ) Þ 0.174 < p < 0.284, or 17.4% < p < 28.4%

227

b. Because the two confidence intervals overlap, it is possible that the OxyContin treatment group and the placebo group have the same rate of nausea. Nausea does not appear to be an adverse reaction made worse with OxyContin.

20. a. 3005 (0.817 ) = 2455 b. 90% CI: pˆ ± za / 2

ˆˆ (0.817)(0.183) pq = 0.817 ± 1.645 Þ 0.805 < p < 0.829, or 80.5% < p < 82.9% n 3005

c. nothing

ˆˆ (0.000321)(0.999679) pq = 0.000321 ± 1.645 Þ 0.000276 < p < 0.000366, or n 420,095 0.0276% < p < 0.0366%; (Using x = 135: 0.0276% < p < 0.0367%)

21. a. 90% CI: pˆ ± za / 2

b. No, because 0.0340% is included in the confidence interval.

22. Placebo group: 95% CI: pˆ ± za / 2

263 ( 2707 )( 270 ) Þ 0.00697 < p < 0.0449, or 0.697% < p < 4.49%

ˆˆ pq 7 = ± 1.96 n 270

270

.

Section 7-1: Estimating a Population Proportion

83

22. (continued) Treatment group:

( 7 )( 855 )

ˆˆ pq 8 = ± 1.96 863 863 Þ 0.00288 < p < 0.0157, or 0.288% < p < 1.57% n 863 863 Because the two confidence intervals overlap, there does not appear to be a significant difference between the rates of allergic reactions. Allergic reactions do not appear to be a concern for Lipitor users.

95% CI: pˆ ± za / 2

23. XSORT: 95% CI: pˆ ± za / 2

ˆ ˆ 879 pq = ± 1.96 n 945

66 ( 879 945 )( 945 ) Þ 0.914 < p < 0.946, or 91.4% < p < 94.6%

945

( )( 52 ) 239 291

ˆ ˆ 239 pq 291 = ± 1.96 Þ 0.777 < p < 0.865, or 77.7% < p < 86.5% n 291 291 The two confidence intervals do not overlap. It appears that the success rate for the XSORT method is higher than the success rate for the YSORT method, but the big story here is that the XSORT method and the YSORT method both appear to be very effective because the success rates are well above the 50% rates expected with no treatments.

YSORT: 95% CI: pˆ ± za / 2

( 239 )( 52 )

ˆ ˆ 239 pq = ± 1.96 291 291 Þ 0.777 < p < 0.865, or 77.7% < p < 86.5% n 291 291 It appears the YSORT method was very effective because the success rate is well above the 50% rates expected with no treatments.

24. 95% CI: pˆ ± za / 2

(

6062

)(

5938

)

ˆˆ pq 6062 12,000 12,000 = ± 1.96 Þ 0.496 < p < 0.514, or 59.6% < p < 51.4%; No, n 12,000 12,000 because the proportion could easily equal 0.5. The proportion does not appear to be significantly less than 0.5 the week before Thanksgiving.

25. 95% CI: pˆ ± za / 2

( 901 )( 111 )

ˆˆ pq 901 = ± 2.56 1012 1012 Þ 0.871 < p < 0.910 or 87.1% < p < 91.0%; Yes, because n 1012 1012 the proportion that oppose cloning appears to be greater than 0.5.

26. 99% CI: pˆ ± za / 2

27. Sustained care:

ˆˆ pq = 0.828 ± 1.96 n x = 164: 77.5% < p < 88.1%) Standard care: 95% CI: pˆ ± za / 2

(0.828)(0.172) 198

Þ 0.775 < p < 0.881, or 77.5% < p < 88.1%;

(Using

ˆˆ (0.628)(0.372) pq = 0.628 ± 1.96 Þ 0.561 < p < 0.695, or 56.1% < p < 69.5% ; The two n 199 confidence intervals do not overlap. It appears that the success rate is higher with sustained care.

95% CI: pˆ ± za / 2

28. Measured:

ˆˆ pq = 0.258 ± 1.96 n (Using x = 51: 19.7% < p < 31.8%) 95% CI: pˆ ± za / 2

.

(0.258)(0.742) 198

Þ 0.197 < p < 0.319, or 19.7% < p < 31.9%;

84

Chapter 7: Estimating Parameters and Determining Sample Sizes

28. (continued) Reported:

ˆˆ (0.409)(0.591) pq = 0.409 ± 1.96 Þ 0.341 < p < 0.477, or 34.1% < p < 47.7%; n 198 (Using x = 81: 34.1% < p < 47.8%) The two confidence intervals do not overlap. It appears that the success rate is higher with sustained care.

95% CI: pˆ ± za / 2

29. a. n =

b. n =

ˆ ˆ [1.96]2 ×0.25 [ za /2 ]2 pq = = 600 (Tech: 601) E2

0.042

ˆ ˆ [1.96]2 ×(0.02)(0.98) [ za /2 ]2 pq = = 47 E2

0.042

c. They don’t change.

30. a. n =

b. n =

ˆ ˆ [2.575]2 ×0.25 [ za /2 ]2 pq = = 1842 (Tech: 1844) E2

0.032

ˆ ˆ [2.575]2 ×(0.15)(0.85) [ za /2 ]2 pq = = 939 E2

0.032

c. Yes, the added knowledge results in a very substantial decrease in the required sample size.

31. a. n =

ˆ ˆ [1.96]2 ×0.25 [ za / 2 ]2 pq = = 385 E2

0.052

2

b. n =

ˆ ˆ [1.96]2 ×(0.40)(0.60) [ za / 2 ] pq = = 369 E2

0.052

c. No, the sample size doesn’t change much.

32. a. n =

ˆ ˆ [2.575]2 ×0.25 [ za / 2 ]2 pq = = 1037 E2

0.042

2

b. n =

33. a. n =

ˆ ˆ [2.575]2 ×(0.26)(0.74) [ za / 2 ] pq = = 798 E2

0.042

ˆ ˆ [1.96]2 ×0.25 [ za / 2 ]2 pq = = 1537 E2

0.0252

2

b. n =

34. a. n =

ˆ ˆ [1.96]2 ×(0.53)(0.47 ) [ za / 2 ] pq = = 1532 E2

0.0252

ˆ ˆ [2.575]2 ×0.25 [ za / 2 ]2 pq = = 4145 (Tech: 4147) E2

0.022

2

b. n =

ˆ ˆ [2.575]2 ×(0.45)(0.55) [ za / 2 ] pq = = 4103 (Tech: 4106)

E2 0.022 c. No, the sample size doesn’t change much.

35. a. n =

ˆ ˆ [1.645]2 ×0.25 [ za / 2 ]2 pq = = 752 E2

0.032

.

Section 7-2: Estimating a Population Mean

85

35. (continued) b. n =

ˆ ˆ [1.645]2 ×(0.53)(0.47 ) [ za / 2 ]2 pq = = 749

E2 0.032 c. No. A sample of the people you know is a convenience sample, not a simple random sample, so it is very possible that the results would not be representative of the population.

36. a. n =

b. n =

ˆ ˆ [2.575]2 ×0.25 [ za / 2 ]2 pq = = 4145 (Tech: 4147) E2

0.022

ˆ ˆ [2.575]2 ×(0.82 )(0.18) [ za / 2 ]2 pq = = 2447 (Tech: 2449)

E2 0.022 c. Randomly selecting adult women would result in an underestimate, because some women will give birth to their first child after the survey was conducted. It will be important to survey women who have completed the time during which they can give birth.

2

37. n =

2

ˆ ˆ [ za / 2 ] Npq 2

ˆ ˆ [ za / 2 ] + ( N - 1) E pq

2

=

2500 (0.82 )(0.18)[2.575] 2

(0.82)(0.18)[2.575] + (2500 - 1) 0.022

= 1237 (Tech: 1238)

38. Because we have 95% confidence that p is less than 0.0991, we can safely conclude that rate of headaches among OxyContin users is less than 10%.

pˆ + za

ˆ ˆ 16 pq = 227 + 1.645 n

16 211 ( 227 )( 227 ) Þ p < 0.0991

227

39. a. The requirement of at least 5 successes and at least 5 failures is not satisfied, so the normal distribution cannot be used. b. 3 40 = 0.075 Section 7-2: Estimating a Population Mean 1. a. 12.855 g/dL < m < 13.391 g/dL b. The best point estimate of m is x =

12.855 + 13.391 = 13.123 g/dL. The margin of error is 2

13.391 - 12.855 = 0.268 g/dL. 2 c Because the sample size of 50 is greater than 30, we can consider the sample mean to be from a population with a normal distribution.

E=

2.

a. df = 100 - 1 = 99 b. 1.9842 (Table: 1.984) c. In general, the number of degrees of freedom for a collection of sample data is the number of sample values that can vary after certain restrictions have been imposed on all data values.

3. We have 95% confidence that the limits of 12.855 g/dL and 13.391 g/dL contain the true value of the mean hemoglobin level of the population of all adult females.

4. When we say that the confidence interval methods of this section are robust against departures from normality, we mean that these methods work reasonably well with distributions that are not normal, provided that departures from normality are not too extreme.

5.

Neither the normal nor the t distribution applies.

6.

ta / 2 = 1.671

7.

.

za / 2 = 2.576 (Table: 2.575)

86

Chapter 7: Estimating Parameters and Determining Sample Sizes

8.

ta / 2 = 1.972

9.

The sample size is greater than 30 and the data appear to be from a population that is normally distributed. s 7.1 = 30.4 ± 1.972 × Þ 29.4 hg < m < 31.4 hg; No, the results do not differ by much. n 205

95% CI: x ± ta / 2

10. The sample size is greater than 30 and the data appear to be from a population that is normally distributed. s 6.6 = 32.7 ± 1.972 × Þ 31.8 hg < m < 36.6 hg; Yes, it appears that birth weights of boys are 95% CI: x ± ta / 2 n 205 substantially greater than birth weights of girls.

11. The sample size is greater than 30 and the data appear to be from a population that is normally distributed. s 0.62 = 98.20 ± 1.983 × Þ 98.08°F < m < 98.32°F; Because the confidence interval does not 95% CI: x ± ta / 2 n 106 contain 98.6°F, it appears that the mean body temperature is not 98.6°F, as is commonly believed.

s 4.8 = 2.1 ± 1.685 × Þ 0.8 lb < m < 3.4 lb; Because the n 40 confidence interval does not include 0 or negative values, it does appear that the weight loss program is effective, with a positive loss of weight. Because the amount of weight lost is relatively small, the weight loss program does not appear to be very practical.

12. The sample size is greater than 30. 90% CI: x ± ta / 2

13. It is assumed that the 16 sample values appear to be from a normally distributed population. s 42.3 = 98.9 ± 2.602 × Þ 71.4 min < m < 126.4 min; The confidence interval includes the 98% CI: x ± ta / 2 n 16 mean of 102.8 min that was measured before the treatment, so the mean could be the same after the treatment. This result suggests that the zopiclone treatment does not have a significant effect.

s 21.0 = 0.4 ± 2.407 × Þ - 6.8 mg/dL < m < 7.6 mg/dL; n 49 Because the confidence interval includes the value of 0, it is very possible that the mean of the changes in LDL cholesterol is equal to 0, suggesting that the garlic treatment did not affect LDL cholesterol levels. It does not appear that garlic has a significant effect in reducing LDL cholesterol.

14. The sample size is greater than 30. 98% CI: x ± ta / 2

s 1.07 = 2.6 ± 2.262 × Þ 1.8 < m < 3.4; n 10 The given numbers are just substitutes for the four DNA base names, so the numbers don’t measure or count anything, and they are at the nominal level of measurement. The confidence interval has no practical use.

15. The sample appears to have a normal distribution. 95% CI: x ± ta / 2

0.346 = 1.14 ± 2.262 × n 10 Þ 0.89 mg < m < 1.39 mg; No, the samples obtained from Chicago might be very different from those obtained from Tokyo.

16. The sample appears to have a normal distribution. 95% CI: x ± ta /2

s

17. The sample data meet the loose requirement of having a normal distribution. s 0.4229 = 0.9382 ± 1.8125 × Þ 0.707 W/kg < m < 1.169 W/kg; Because the confidence 90% CI: x ± ta / 2 n 11 interval is entirely below the standard of 1.6 W/kg, it appears that the mean amount of cell phone radiation is less than the FCC standard, but there could be individual cell phones that exceed the standard.

18. The sample data meet the loose requirement of having a normal distribution. s 6.46 = 11.05 ± 2.262 × Þ 6.43 mg g < m < 15.7 mg g; No, it does not appear that the lead 95% CI: x ± ta / 2 n 10 concentrations are less than 7 mg g since the upper limit of the confidence exceeds that amount.

.

Section 7-2: Estimating a Population Mean

87

s 0.366 = 0.719 ± 3.143 × n 7 Þ 0.284 ppm < m < 1.153 ppm; Using the FDA guideline, the confidence interval suggests that there could be too much mercury in fish because it is possible that the mean is greater than 1 ppm. Also, one of the sample values exceeds the FDA guideline of 1 ppm, so at least some of the fish have too much mercury.

19. The sample appears to have a normal distribution. 98% CI: x ± ta / 2

20. The data appear to have a distribution that is far from normal, so the confidence interval might not be a good s 3.51 = 6.5 ± 1.729 × Þ 5.1 years < m < 7.9 years estimate of the population mean. 95% CI: x ± ta / 2 n 20 (Tech: 4.9 years < m < 8.1 years); The confidence interval does not contain the value of 4 years, but six of the twenty values are 4, so it is common to earn a bachelor’s degree in four years, but the typical college student uses more than 4 years.

21. The presence of five zeros suggests that the sample is not from a normally distributed population, so the normality requirement is violated and the confidence interval might not be a good estimate of the population mean. s 20.33 = 32.6 ± 2.861× Þ 19.5 mg < m < 45.6 mg; People consume some brands much more 99% CI: x ± ta / 2 n 20 often than others, but the 20 brands are all weighted equally in the calculations. This, along with the violation of the normality requirement, means the confidence interval might not be a good estimate of the population mean.

22. It must be assumed the samples have a normal distribution. a. Manual: 95% CI: x ± ta / 2

s 15 = 175 ± 2.262 × Þ 164 BPM < m < 186 BPM n 10

b. Electric: 95% CI: x ± ta / 2

s 18 = 124 ± 2.262 × Þ 111 BPM < m < 137 BPM n 10

c. The upper limit, which shows maximum strain on the heart. d. Although final conclusions about means of populations should not be based on the overlapping of confidence intervals, the intervals do not overlap, so there does appear to be a significant difference between the mean pulse rate of manual snow shoveling and using an electric snow thrower, with manual snow shoveling resulting in higher pulse rates.

23. The samples sizes are both large. a. Echinacea: 95% CI: x ± ta / 2 b. Placebo: 95% CI: x ± ta / 2

s 2.3 = 6.0 ± 1.967 × Þ 5.8 days < m < 6.2 days n 337

s 2.4 = 6.1 ± 1.966 × Þ 5.9 days < m < 6.3 days n 370

c. The two confidence intervals are very similar. The Echinacea treatment group does not appear to fare any better than the placebo group, so the Echinacea treatment does not appear to be effective.

24. The samples sizes are both large. a. Acupuncture: 95% CI: x ± ta / 2

s 1.4 = 1.8 ± 1.977 × Þ 1.6 < m < 2.0 n 142

b. Sham Treatment: 95% CI: x ± ta / 2

s 1.2 = 1.6 ± 1.990 × Þ 1.3 < m < 1.9 n 80

c. Although final conclusions about means of populations should not be based on the overlapping of confidence intervals, the intervals do overlap, so there does not appear to be a significant difference between the acupuncture treatment and the sham treatment.

.

88

Chapter 7: Estimating Parameters and Determining Sample Sizes

25. Both samples appear to have a normal distribution. s 0.916 = 69.58 ± 1.976 × Þ 67.8 bpm < m < 71.4 bpm Males: 95% CI: x ± ta / 2 n 153 s 1.03 = 74.0 ± 1.976 × Þ 72.0 bpm < m < 76.1 bpm Females: 95% CI: x ± ta / 2 n 147 Although final conclusions about means of populations should not be based on the overlapping of confidence intervals, the intervals do not overlap, so adult females appear to have a mean pulse rate that is higher than the mean pulse rate of adult males.

26. Neither sample appears to have a normal distribution, so the normality requirement is violated and the confidence intervals might not be good estimates of the population means. a. King: 95% CI: x ± ta / 2

s 0.233 = 1.26 ± 2.064 × Þ 1.16 mg < m < 1.35 mg n 25

b. 100: 95% CI: x ± ta / 2

s 0.248 = 0.9 ± 2.064 × Þ 0.8 mg < m < 1.0 mg n 25

c. Although final conclusions about means of populations should not be based on the overlapping of confidence intervals, the intervals do not overlap, so the 100 mm cigarettes appear to have a mean nicotine level that is lower than king size cigarettes.

2

2

2

2

é z s ù é 2.575 ×15 ù 27. The sample size is n = ê a / 2 ú = ê = 94. This does appear to be very practical. ú 4 û ë E û ë é z s ù é 2.326 ×15 ù = 136. This does appear to be very practical.

28. The sample size is n = ê a / 2 ú = ê ú 3 û ë E û ë 2

2

éz s ù é 2.575 ×1.0 ù = 7,367 (Tech: 7,368). This does appear to be very 29. The required sample size is n = ê a /2 ú = ê ë 0.03 ú û ë E û practical.

2

2

éz s ù é1.96 ×7.10 ù = 22. No, the assumption does not seem reasonable.

30. The sample size is n = ê a /2 ú = ê ú 3 ë û ë E û 2

2

é z s ù é1.96 ×17.7 ù = 4815 (Tech: 4814). Yes, the assumption seems 31. The required sample size is n = ê a / 2 ú = ê û ë E û ë 0.5 ú reasonable.

2

32. a. s »

2

186 - 88 éz s ù é1.96 ×24.5 ù = 92. = 24.5. The required sample size is n = ê a /2 ú = ê ú 5 4 ë û ë E û 2

2

éz s ù é1.96 ×17.22 ù = 46. b. The required sample size is n = ê a /2 ú = ê ú 5 ë û ë E û c. The result from part (a) is substantially larger than the result from part (b). The result from part (b) is likely to be better because it uses s instead of the estimated σ obtained from the range rule of thumb.

2

33. a. s »

2

168 - 94 éz s ù é1.96 ×18.5 ù = 53. = 18.5. The required sample size is n = ê a /2 ú = ê ú 5 4 ë û ë E û 2

2

éz s ù é1.96 ×14.35 ù = 32. b. The required sample size is n = ê a /2 ú = ê ú 5 ë û ë E û

.

Section 7-2: Estimating a Population Mean

89

c. The result from part (a) is substantially larger than the result from part (b). The result from part (b) is likely to be better because it uses s instead of the estimated σ obtained from the range rule of thumb.

2

34. a. s »

2

99.6 - 96.5 é z s ù é 2.326 ×0.775 ù = 0.775; The required sample size is n = ê a / 2 ú = ê = 326 (Table: 327). ú 0.1 4 û ë E û ë 2

2

é z s ù é 2.326 ×0.62 ù b. The required sample size is n = ê a / 2 ú = ê = 209. ú 0.1 û ë E û ë c. The result from part (a) is substantially larger than the result from part (b). The result from part (b) is likely to be better because it uses s instead of the estimated s obtained from the range rule of thumb.

s 65.4 = 255.1 ± 2.023 × n 40 Þ 234.2 (1000 cells mL ) < m < 276.0 (1000 cells mL )

35. a. Large pop: 95% CI: x ± ta / 2

s N- n 65.4 500 - 40 = 255.1 ± 2.26 × n n- 1 40 400 - 1 Þ 235.0 (1000 cells mL ) < m < 275.2 (1000 cells mL )

b. Finite pop: 95% CI: x ± ta / 2

c. The second confidence interval is narrower, indicating that we have a more accurate estimate when the relatively large sample is selected without replacement from a relatively small finite population.

Section 7-3: Estimating a Population Standard Deviation or Variance 1.

2

2

( ) < s 2 < 33299.8 (cm3 ) Þ 95.0 cm3 < s < 182.5 cm3 . We have 95% confidence that

9027.8 cm3

the limits of 95.0 cm3 and 182.5 cm3 contain the true value of the standard deviation of brain volumes.

2.

The format implies that s = 15.7, but s is given as 14.3. In general, a confidence interval for s does not have s at the center.

3. The dotplot does not appear to depict sample data from a normally distributed population. The large sample size does not justify treating the values as being from a normally distributed population. Because the normality requirement is not satisfied, the confidence interval estimate of s should not be constructed using the methods of this section.

4. The normality requirement for a confidence interval estimate of s has a much stricter normality requirement than the loose normality requirement for a confidence interval estimate of m. Departures from normality have a much greater effect on confidence interval estimates of s than on confidence interval estimates of m.

5.

df = 24, c L2 = 12.401, and c R2 = 39.364

(n - 1) s 2 c R2

(25 - 1) 0.24

<s <

2

<s <

df = 152, c L2 = 119.759, and c R2 = 118.026 (Table: c L2 = 74.222, and c R2 = 129.561)

(n - 1) s 2 c L2

(25 - 1) 0.24

6.

(n - 1) s 2 c R2

2

(153 - 1) 0.016482

<s <

(n - 1) s 2 c L2

(153 - 1) 0.016482 <s < 118.026 119.759 95% CI: 0.0187 g < s < 0.0186 g (Table: 0.0179 g < s < 0.0226 g)

39.364 12.401 95% CI: 0.19 mg < s < 0.33 mg

.

90

Chapter 7: Estimating Parameters and Determining Sample Sizes

7.

df = 137, c L2 = 106.491, and c R2 = 171.294 (Table: c L2 = 74.222, and c R2 = 129.561)

(n - 1) s 2 c R2

(138 - 1)1.22

<s <

<s <

(n - 1) s 2 c L2

(16 - 1) 42.32

(Table: c L2 = 74.222, and c R2 = 129.561)

(n - 1) s c R2

<s <

(164 - 1) 522

(n - 1) s

2

df = 299, c L2 = 239.769, and c R2 = 365.739 c R2

<s <

(300 - 1)12.132

<s <

(n - 1) s 2 c R2

<s <

<s <

(n - 1) s 2 c L2

(49 - 1) 21.02

13. The sample appears to have a normal distribution.

(n - 1) s 2

c L2

(300 - 1)12.132

c R2

(15 - 1) 6.82

<s <

<s <

(n - 1) s 2 c L2

(15 - 1) 6.82

31.319 4.075 99% CI: 4.55 kmph < s < 12.60 kmph

10. df = 40, c L2 = 26.509, and c R2 = 55.758

c R2

(16 - 1) 42.32

df = 14, c L2 = 4.075, and c R2 = 31.319

(n - 1) s 2

<s < 365.739 239.769 99% CI: 10.96 < s < 13.55 (Tech: 10.97 < s < 13.54)

(n - 1) s 2

c L2

76.154 29.707 98% CI: 16.7 mg/dL < s < 26.7 mg/dL (Tech: 16.9 mg/dL < s < 27.4 mg/dL) No, the confidence interval does not indicate whether the treatment is effective.

(164 - 1) 522

(n - 1) s 2

<s <

(n - 1) s 2

12. df = 50, c L2 = 29.707, and c R2 = 76.154

(49 - 1) 21.02

c L2

<s < 200.243 129.543 97% CI: 46.92 kg < s < 58.33 kg (Table: 58.50 kg < s < 77.06 kg) 9.

<s <

30.578 5.229 98% CI: 29.6 min < s < 71.6 min No, the confidence interval does not indicate whether the treatment is effective.

(138 - 1)1.22

df = 163, c L2 = 129.543, and c R2 = 200.243 2

(n - 1) s 2 c R2

171.294 106.491 97% CI: 1.07 < s < 1.36 (Table: 1.23 < s < 1.63) 8.

11. df = 15, c L2 = 5.229, and c R2 = 30.578

(n - 1) s 2 c L2

(40 - 1) 4.82

(40 - 1) 4.82 <s < 55.758 26.509 90% CI: 4.0 lb < s < 5.9 lb (Tech: 4.1 lb < s < 5.8 lb) The confidence interval gives us information about the variation among the amounts of lost weight, but it does not give us information about the effectiveness. of the diet.

.

Section 7-3: Estimating a Population Standard Deviation or Variance 14. The samples must be assumed to have normal distributions.

Single Line

df = 9, c L2 = 2.700, and c R2 = 19.023

(n - 1) s 2 c R2

<s <

Individual Lines

df = 9, c L2 = 2.700, and c R2 = 19.023

(n - 1) s 2

(n - 1) s 2

c L2

c R2

(10 - 1) 28.62

(10 - 1) 28.62 <s < 19.023 2.700 95% CI: 19.7 sec < s < 52.2 sec

91

(n - 1) s 2

<s <

c L2

(10 - 1)109.32

(10 - 1)109.32 <s < 19.023 2.700 95% CI: 75.0 sec < s < 200.0 sec

The variation appears to be significantly higher for individual lines, so a single line appears to be better.

15. The samples must be assumed to have normal distributions. a. Manual Snow Shoveling

df = 9, c L2 = 2.700, and c R2 = 19.023

(n - 1) s 2 c R2

(10 - 1)152

<s <

<s <

b. Electric Snow Thrower

df = 9, c L2 = 2.700, and c R2 = 19.023

(n - 1) s 2

(n - 1) s 2

c L2

c R2

(10 - 1)182

(142 - 1)152

19.023 2.700 95% CI: 10.3 BPM < s < 27.4 BPM

<s <

<s <

(n - 1) s 2 c L2

(142 - 1)182

19.023 2.700 95% CI: 12.4 BPM < s < 32.9 BPM

c. The variation does not appear to be significantly different.

16. The samples must be assumed to have normal distributions. a. Acupuncture

df = 141, c L2 = 110.020, and c R2 = 175.765

(n - 1) s 2 c R2

(142 - 1)1.4

<s <

2

<s <

b. Sham Treatment

df = 79, c L2 = 56.309, and c R2 = 105.473

(n - 1) s 2

(n - 1) s 2

c L2

c R2

(142 - 1)1.4

2

(80 - 1)1.2

<s <

2

<s <

(n - 1) s 2 c L2

(80 - 1)1.22

105.473 56.309 95% CI: 1.0 < s < 1.4

175.765 110.020 95% CI: 1.3 < s < 1.6 c. The variation does not appear to be significantly different.

17. a. Girls: The sample appears to be from a population with a distribution that is far from normal, so the confidence interval estimate might not be very good.

b. Boys: The sample appears to be from a population with a distribution that is far from normal, so the confidence interval estimate might not be very good.

df = 204, c L2 = 166.338, and c R2 = 245.448

df = 79, c L2 = 0.000, and c R2 = 0.000

(n - 1) s 2 c R2

(205 - 1) 9.72

<s <

9 <s <

(n - 1) s 2

(n - 1) s 2

c L2

c R2

(205 - 1) 9.72

(195 - 1) 8.22

245.448 166.338 95% CI: 8.8 days < s < 10.7 days

<s <

<s <

(n - 1) s 2 c L2

(195 - 1)8.22

234.465 157.321 95% CI: 7.5 < s < 9.2

c. The variation does not appear to be significantly different.

.

92

Chapter 7: Estimating Parameters and Determining Sample Sizes

18. a. The sample appears to have a normal distribution.

df = 204, c L2 = 166.338, and c R2 = 245.448 (Using technology.)

(n - 1) s 2 c R2

(n - 1) s 2

<s <

c L2

(205 - 1) 706.32

(205 - 1) 706.32 <s < 245.448 166.338 95% CI: 643.9 g < s < 782.1 g

b. The sample appears to have a normal distribution.

df = 194, c L2 = 157.321, and c R2 = 234.465 (Using technology.)

(n - 1) s 2 c R2

(195 - 1) 660.22

(n - 1) s 2

<s <

c L2

(195 - 1) 660.22

<s <

234.465 157.321 95% CI: 600.5 g < s < 733.1 g c. The amounts of variation are about the same.

19. 19,205 is too large. There are not 19,205 statistics professors in the population, and even if there were, that sample size is too large to be practical.

20. The sample size is 768. The sample size seems practical.

21. The sample size is 1336. Yes, with the majority of the population concentrated near the mean, the distribution will satisfy the requirement of a normal distribution.

22. The sample size is 336.

2 2 1 1 é- za / 2 + 2k - 1ù = é- 1.96 + 2 ×151 - 1ù = 109.565 and û 2ë û 2ë 2 2 1 1 é ù c R2 = éëza / 2 + 2k - 1ù û = 2 ë1.96 + 2 ×151 - 1û = 175.276 2 The approximate values are quite close to the actual critical values.

23. c L2 =

2

24. n =

2

1 æza / 2 ö 1 æ2.326 ö ÷ = 121 (Statdisk yields a sample size of 122.) ç ÷ = çè 2è d ø 2 0.15 ø

Section 7-4: Bootstrapping: Using Technology for Estimates 1. Without replacement, every sample would be identical to the original sample, so the proportions or means or standard deviations or variances would all be the same, and there would be no confidence “interval.”

2. For the given sample, a bootstrap sample is another sample in which six of the numbers of students in a day who required a book from library are randomly selected with replacement.

3. Part (b): (There are too many samples), Part (d): (There are only four elements, not six), Part (e): (9 and 17 are not in the original sample) are not possible bootstrap samples.

4. There is no universal exact number, but there should be at least 1000 bootstrap samples, and the use of 10,000 or more is common.

5. The proportions from the 10 samples (in ascending order) are: 0.25, 0.25, 0.25, 0.25, 0.25, 0.5, 0.5, 0.5, 0.75, and 0.75. P5 = 0.250 and P95 = 0.750, so the 90% interval is 0.250 < p < 0.750.

6.

The proportions from the 10 samples (in ascending order) are: 0, 0, 0, 0, 0, 0.25, 0.25, 0.25, 0.25, and 0.5.

P10 = 0.000 and P90 = 0.375, so the 80% interval is 0.000 < p < 0.375.

.

Section 7-4: Bootstrapping: Using Technology for Estimates 7.

93

a. The means from the 10 samples (in ascending order) are: –0.25, 0.5, 0.5, 0.75, 3, 3, 3, 5, 8.25, and 9.

P10 = 0.125 and P90 = 8.625, so the 80% interval is 0.1 kg < m < 8.6 kg. b. The standard deviations from the 10 samples (in ascending order) are: 1.5, 2.363, 2.886751, 2.887, 4, 5.5, 5.715, 5.715, 5.715476, and 6.976. P10 = 1.931 and P90 = 6.346, so the 80% interval is 1.9 kg < s < 6.3 kg.

8.

a. The means from the 10 samples (in ascending order) are: 62, 70.5, 73.5, 77.5, 78.25, 81, 81, 85.25, 93, and 103.5. P10 = 66.25 and P90 = 98.25, so the 80% interval is 66.3 cW/kg < m < 98.3 cW/kg. b. The standard deviations from the 10 samples (in ascending order) are: 15.5, 17.898, 27.713, 37.621, 42.429, 45, 47.067, 47.067, 51.772, and 57.449. P10 = 16.699 and P90 = 54.610, so the 80% interval is 16.7 cW/kg < s < 54.6 cW/kg.

9.

Answers will vary, but here are typical answers. a. 90% CI: - 0.8 kg < m < 7.8 kg b. 90% CI: 1.2 kg < s < 7.0 kg

10. Answers will vary, but here are typical answers. a. 90% CI: 50.0 cW/kg < m < 118.3 cW/kg b. 90% CI: 9.8 cW/kg < s < 57.3 cW/kg 11. Answers will vary, but here are typical answers. a. 99% CI: 55.3 min < m < 67.3 min; This isn’t dramatically different from 51.9 min < m < 71.4 min. b. 95% CI: 6.2 min < s < 13.7 min; This isn’t dramatically different from 7.7 min < s < 18.4 min.

12. Answers will vary, but here are typical answers. a. 99% CI: 22.0 min < m < 41.6 min; This isn’t dramatically different from 19.6 min < m < 43.3 min using the methods of Section 7-2. b. 95% CI: 8.5 min < s < 18.9 min; This isn’t dramatically different from 10.7 min < s < 23.7 min using the methods of Section 7-3.

13. Answers will vary, but here is a typical result: 95% CI: 0.348% < p < 0.1.62%. This is quite close to the confidence interval of 0.288% < p < 1.57% found in Exercise 22 from Section 7-1.

14. Answers will vary, but here is a typical result: 99% CI: 0.0208 < p < 0.0317. This is quite close to the confidence interval of 0.0205 < p < 0.0311 found in Exercise 14 from Section 7-1.

15. Answers will vary, but here is a typical result: 90% CI: 0.1356 < p < 0.152. The result is essentially the same as the confidence interval of 0.135 < p < 0.152 found in Exercise 15 from Section 7-1.

16. Answers will vary, but here is a typical result: 95% CI: 0.671 < p < 0.722. The result is very close to the confidence interval of 0.671 < p < 0.723 found in Exercise 16 from Section 7-1.

17. Answers will vary, but here is a typical result: 90% CI: 3.69 < m < 4.15. This result is very close to the confidence interval 3.676 < m < 4.17 found by using the methods from Section 7-2.

18. Answers will vary, but here is a typical result: 99% CI: 21.6 mg < m < 43.1 mg. This result is reasonably close to the confidence interval of 19.5 mg < m < 45.6 mg found in Exercise 21 in Section 7-2.

19. Answers will vary, but here is a typical result: 90% CI: 0.712 W kg < m < 1.18 W kg. This result is very close to the confidence interval of 0.707 W kg < m < 1.169 W kg found in Exercise 17 of Section 7-2.

20. Answers will vary, but here is a typical result: 90% CI: 0.286 W kg < m < 0.492 W kg. This result would be very close to the confidence interval 0.313W kg < m < 0.674 W kg using the methods of Section 7-3.

.

94

Chapter 7: Estimating Parameters and Determining Sample Sizes

21. a. Answers will vary, but here is a typical result: 95% CI: 2.5 < s < 3.3. b. 95% CI: 2.4 < s < 3.7 c. The confidence interval from the bootstrap method is not very different from the confidence interval found using the methods of Section 7-3. Because a histogram or normal quantile plot shows that the sample appears to be from a population not having a normal distribution, the bootstrap confidence interval of 2.5 < s < 3.3 would be a better estimate of s .

22. a. Answers will vary, but here is a typical result: 95% CI: 1.6 < m < 3.3. b. 95% CI: 1.6 < m < 3.3 c. The confidence interval found from the bootstrap method is essentially the same as the confidence interval found using the methods of Section 7-2.

23. Answers will vary, but here is a typical result using 10,000 bootstrap samples: 2.5 < s < 3.3. This result is the same as the confidence interval found using 1000 bootstrap samples. In this case, increasing the number of bootstrap samples from 1000 to 10,000 does not have much of an effect on the confidence interval.

24. a. No. A histogram or normal quantile plot would show that the distribution of the sample data is far from normal. b. Yes. The sample means appear to have a distribution that is approximately normal. c. Yes. The sample standard deviations appear to have a distribution that is approximately normal.

Chapter Quick Quiz 1.

pˆ =

0.110 + 0.150 = 0.130 2

2. We have 95% confidence that the limits of 0.110 and 0.150 contain the true value of the proportion of adults in the population who correct their vision by wearing contact lenses.

3.

z0.01/ 2 = 2.576 (Table: 2.575)

4.

3% - 1.0% < p < 3% + 1.0% Þ 2.0% < p < 4.0%

5.

n=

6.

é z s ù é 2.326 ×15 ù n = ê a /2 ú = ê = 136 ú 3 û ë E û ë

7.

There is a loose requirement that the sample values are from a normally distributed population.

ˆ ˆ [1.96]2 (0.25) [ za / 2 ]2 pq = = 601 E2

0.042

2

2

8. The degrees of freedom is the number of sample values that can vary after restrictions have been imposed on all of the values. For the sample data described in Exercise 7, df = 12 - 1 = 11.

9.

t0.05/ 2 = 2.201

10. No, the use of the c 2 distribution has a fairly strict requirement that the data must be from a normal distribution. The bootstrap method could be used to find a 95% confidence interval estimate of s .

Review Exercises

(

723

)(

222,277

)

ˆˆ pq 223,000 223,000 723 ± 1.96 = 223,000 Þ 0.00301 < p < 0.00348; We have 95% confidence n 223,000 that the limits of 0.00301 and 0.00348 contain the value of the population proportion.

1.

95% CI: pˆ ± za / 2

2.

n=

ˆ ˆ [1.645]2 (0.25) [ za / 2 ]2 pq = = 423 E2

0.042

.

Review Exercises 3.

95

a. x = 10.900 mm b. 95% CI: x ± ta /2

s n

0.148 = 10.9 ± 2.201 × Þ 10.806 mm < m < 10.994 mm 12

c. We have 95% confidence that the limits of 10.806 mm and 10.994 mm contain the value of the population mean m.

2

2

4.

é z s ù é 2.575 ×15 ù n = ê a /2 ú = ê = 94 ú 4 û ë E û ë

5.

a. Student t distribution b. normal distribution c. None of the three distributions is appropriate, but a confidence interval could be constructed by using bootstrap methods. d. c 2 (chi-square distribution) e. normal distribution

6.

a. n =

ˆ ˆ [2.575]2 (0.25) [ za /2 ]2 pq = = 1036 E2

0.042

2

2

éz s ù é 2.575 ×12 ù = 60 b. n = ê a /2 ú = ê ú 4 ë û ë E û c. You must take the larger sample of 1036.

7.

The sample meets the loose requirement of being from a normal distribution. s 32.89 = 500 ± 2.201 × Þ 479.1 mg < m < 520.9 mg; The confidence interval limits do contain n 12 the desired amount of 500 mg, so the mean is perfect. The examination of the individual amounts of paracetamol shows that though most of the tablets fall within the confidence interval limits, a couple of tablets have some production problem. Since it is an oral medicine, an overdose may be harmful. So, care needs to be taken to reduce the quantity of paracetamol in such tablets.

95% CI: x ± ta /2

8.

a. The sample is normally distributed.

df = 11, c L2 = 3.816, and c R2 = 21.920

(n - 1) s 2 c R2

<s <

(n - 1) s 2 c L2

(12 - 1) 32.892

(12 - 1) 32.892 <s < 21.920 3.816 95% CI: 23.3 mg < s < 55.8 mg

550 - 450 100 = = 22.5 mg 4 4 c. The desired s = 22.5 mg from part (b) is almost contained within the confidence interval from part (a). It appears that the current production method produces paracetamol tablets with desired variation, and that should be continued. b. s »

9. Answers will vary, but here is a typical result: 482.4 mg < m < 519.5 mg. The result is very close to the confidence interval found in Exercise 7.

.

Commented [CE1]: Is this value (22.5 mg) correct?

96

Chapter 7: Estimating Parameters and Determining Sample Sizes

10. a. 95% CI: pˆ ± za / 2

ˆˆ pq = 0.02 ± 1.96 n

(0.02)(0.98) 1000

Þ 0.0113 < p < 0.0287

b. Answers will vary, but here is a typical result: 0.0120 < p < 0290. c. The confidence intervals are quite close.

Cumulative Review Exercises 1.

x= s=

73.8 +

+ 75.1 + 75.7 + 76.2 + 76.4 + 10

+ 78.2

= 75.78 cm, Q2 =

75.7 + 76.2 = 75.95 cm, 2

(73.8 - 75.78)2 + + (78.2 - 75.78)2 = 1.37 cm, range = 78.2 – 73.8 = 4.40 cm. These results are statistics.

10 - 1

2. Using the range rule of thumb, significantly low values are μ – 2σ = 75.78 – 2(1.37) = 73.04 cm or lower, and significantly high values are μ + 2σ = 75.78 + 2(1.37) = 78.52 cm or higher. Because 72 cm is lower than 73.04 cm, a walking stride of 72 cm is significantly low (or small).

3.

ratio level of measurement; continuous data.

4. A histogram is not very helpful with only 10 data values, but a normal quantile plot shows that the sample data appear to be from a population having a distribution that is approximately normal.

5.

The sample meets the loose requirement of being from a normal distribution. s 1.37 = 75.78 ± 3.250 × Þ 74.37 cm < m < 77.18 cm.

95% CI: x ± ta /2 n 10 2

2

6.

éz s ù é 2.575 ×1.18 ù n = ê a /2 ú = ê = 148 adult females ë 0.25 ú û ë E û

7.

a. z x =25.00 =

25.00 - 24.20 = 0.71; which has a probability of 1 - 0.7611 = 0.2389 (Tech: 0.2375) to the right.

1.12

b. z x =25.00 =

25.00 - 24.20 = 3.57; which has a probability of 1 - 0.9999 = 0.0001 (Tech: 0.0002) to the right.

1.12 25

c. The z score for the lower 95% is 1.645, which correspond to a length of 1.645 ×1.12 + 24.20 = 26.04 cm (Tech: 15.5 min).

( 30 )( 1 )

ˆ ˆ 30 1440 30 pq = 99% CI: pˆ ± za / 2 = 31 ± 2.575 31 31 Þ 0.956 < p < 0.980; With such high limits 1440 + 48 31 n 1488 on the interval, it does appear that the public is well informed about the risks of tattoos and infectious disease.

8.

pˆ =

.

Section 8-1: Basics of Hypothesis Testing

97

Chapter 8: Hypothesis Testing Section 8-1: Basics of Hypothesis Testing 1. Rejection of the claim about aspirin is more serious because it is a drug used for medical treatments. The wrong aspirin dosage could cause more serious adverse reactions than a wrong vitamin C dosage. It would be wise to use a smaller significance level for testing the claim about the aspirin.

2. Estimates and hypothesis tests are both methods of inferential statistics, but they have different objectives. We could use the sample body temperatures to construct a confidence interval estimate of the population mean, but hypothesis testing is used to test some claim made about the value of the mean body temperature.

3.

a. H0 : m = 161.7 cm b. H1: m ¹ 161.7 cm c. Reject the null hypothesis or fail to reject the null hypothesis. d. No, in this case, the original claim becomes the null hypothesis. For the claim that the mean height of women is equal to 161.7 cm, we can either reject that claim or fail to reject it, but we cannot state that there is sufficient evidence to support that claim.

4. The P-value of 0.001 is preferred because it corresponds to the sample evidence that most strongly supports the alternative hypothesis that the method is effective.

5.

a. p > 0.5 (more than a majority)

7.

b. H0 : m = 73 bpm; H1: m ¹ 73 bpm

b. H 0 : p = 0.5; H1: p > 0.5 6.

a. m = 73 bpm

a. p < 0.95

8.

b. H0 : p = 0.95; H1: p < 0.95

a. s = 13 bpm b. H 0 : s = 13 bpm; H1: s > 13 bpm

9. There is sufficient evidence to support the claim that most adults would erase all of their personal information online if they could.

10. There is sufficient evidence to support the claim that fewer than 95% of adults have a cell phone.

11. There is not sufficient evidence to support the claim that the mean pulse rate (in beats per minute) of adult females is 73 bpm.

12. There is not sufficient evidence to warrant rejection of the claim that standard deviation of pulse rates of adult females is less than 13 bpm.

13. z =

14. z =

15. t =

pˆ - p pq n

pˆ - p pq n

x- m s

n

=

0.709 - 0.50 = 13.11 (0.50)(0.50) 983

0.87 - 0.95 = - 12.33 (or z = - 12.38 if using x = 0.87 ×1128 = 981) (0.95)(0.05)

=

1128

=

74 - 73 12.5 147

= 0.970

16. c 2 =

17. a. right-tailed

(147 - 1) s 2 (147 - 1)12.52 s 2

=

132

= 134.985

19. a. two-tailed

b. P-value = P ( z > 1.00) = 0.1587

b. P-value = 2 ×P ( z > 2.01) = 0.0444

c. 0.1587 > 0.05; Fail to reject H 0 .

c. 0.0444 < 0.05; Reject H 0 .

18. a. two-tailed

20. a. left-tailed

b. P-value = 2 ×P ( z > - 2.26) = 0.0119

b. P-value = P ( z < - 1.58) = 0.0571

c. 0.0119 < 0.05. Reject H 0 .

c. 0.0571 > 0.05. Fail to reject H 0 .

Copyright © 2018 Pearson Education, Inc.

98

Chapter 8: Hypothesis Testing 23. a. Critical values: z = ±1.645

21. a. Critical value: z = 1.645 b. 1.00 < 1.645; Fail to reject H 0 .

b. - 2.01 < - 1.645; Reject H 0 .

22. a. Critical values: z = ±1.96

24. a. Critical value: z = - 1.645

b. - 1.96 < - 1.58 < 1.96. Fail to Reject H 0 .

b. - 2.26 < - 1.645. Reject H 0 .

25. a. 0.2678 > 0.05; Fail to reject H 0 . b. There is not sufficient evidence to support the claim that more than 70% do not have hypertension.

26. a. 0.0003 < 0.05; Reject H 0 . b. There is sufficient evidence to support the claim that fewer than 90% of nurses have a cell phone.

27. a. 0.0095 < 0.05; Reject H 0 . b. There is sufficient evidence to warrant rejection of the claim that the mean pulse rate (in beats per minute) of adult males is 72 bpm.

28. a. 0.3045 > 0.05; Fail to reject H 0 . b. There is not sufficient evidence to support the claim that the standard deviation of pulse rates of adult males is more than 11 bpm.

29. Type I error: In reality p = 0.1, but we reject the claim that p = 0.1. Type II error: In reality p ¹ 0.1, but we fail to reject the claim that p = 0.1.

30. Type I error: In reality p = 0.35, but we reject the claim that p = 0.35. Type II error: In reality p ¹ 0.35, but we fail to reject the claim that p ¹ 0.35.

31. Type I error: In reality p = 0.87, but we support the claim that p > 0.87. Type II error: In reality p > 0.87, but we fail to support that conclusion.

32. Type I error: In reality p = 0.25, but we support the claim that p < 0.25. Type II error: In reality p < 0.25, but we fail to support that conclusion.

33. The power of 0.96 shows that there is a 96% chance of rejecting the null hypothesis of p = 0.08 when the true proportion is actually 0.18. That is, if the proportion of Chantix users who experience abdominal pain is actually 0.18, then there is a 96% chance of supporting the claim that the proportion of Chantix users who experience abdominal pain is greater than 0.08.

34. a. From p = 0.5, pˆ = 0.5 + 1.645 From p = 0.65, z =

(0.5)(0.5)

64 0.6028125 - 0.65 (0.65)(0.35)

= 0.6028125

= - 0.791; Power = P ( z > - 0.791) = 0.7852 (Tech: 0.7857)

64

b. Assuming that p = 0.5, as in the null hypothesis, the critical value of z = 1.645 corresponds to pˆ = 0.6028125, so any sample proportion greater than 0.6028125 causes us to reject the null hypothesis, as shown in the shaded critical region of the top graph. If p is actually 0.65, then the null hypothesis of p = 0.5 is false, and the actual probability of rejecting the null hypothesis is found by finding the area greater than pˆ = 0.6028125 in the bottom graph, which is the shaded area. That is, the shaded area in the bottom graph represents the probability of rejecting the false null hypothesis.

35. From p = 0.5, pˆ = 0.5 + 1.645

pˆ = 0.55 - 0.842

(0.5)(0.5) from p = 0.55, since P z > - 0.842 = 0.8000, ; ( ) n

(0.55)(0.45) n Copyright © 2018 Pearson Education, Inc.

Section 8-2: Testing a Claim About a Proportion

99

35. (continued)

So: 0.5 + 1.645

(0.5)(0.5) n

(0.55)(0.45)

= 0.55 - 0.842

n

0.5 n + 1.645 0.25 = 0.55 n - 0.842 0.2475 0.05 n = 1.645 0.25 + 0.842 0.2475 2

æ1.645 0.25 + 0.842 0.2475 ö n=ç ÷ = 617 0.05 è ø Section 8-2: Testing a Claim About a Proportion 1. a. 0.53 ×510 = 270 b. pˆ = 0.53; The symbol p̂ is used to represent a sample proportion.

2.

H 0 : p = 0.5; H1: p > 0.5

3. The method based on a confidence interval is not equivalent to the P-value method and the critical value method.

4.

a. The first requirement is violated because the sample is a voluntary response sample instead of being a simple random sample. Because a requirement is violated, the methods of this section should not be used to test the claim. b. If the P-value is very low (such as less than or equal to 0.05), “the null must go” means that we should reject the null hypothesis. c. The statement that “if the P is high, the null will fly” suggests that with a high P-value, the null hypothesis has been proved or is supported, but we should never make such a conclusion. d. Choosing a significance level with a number like 0.0483 would make it seem like you’re scheming to reach a desired conclusion.

5.

a. left-tailed. b. z = - 4.46

c. P-value = 0.000004 d. H0 : p = 0.1; Reject the null hypothesis.

e. There is sufficient evidence to support the claim that less than 10% of treated subjects experience headaches.

6.

c. P-value = 0.0017

a. right-tailed. b. z = 2.92

d. H0: p = 0.25; Reject the null hypothesis.

e. There is sufficient evidence to support the claim that more than 1/4 of adults feel comfortable in a selfdriving vehicle.

7.

a. two-tailed. b. z = - 1.36

d. H 0 : p = 0.67; Fail to reject the null hypothesis.

c. P-value = 0.174 e. There is not sufficient evidence to warrant rejection of the claim that 92% of adults own cell phones.

8.

a. two-tailed. b. z = 1.33

d. H0 : p = 0.5; Fail to reject the null hypothesis.

c. P-value = 0.1840 e. There is not sufficient evidence to warrant rejection of the claim that half of us say that we should replace passwords with biometric security.

.

100

Chapter 8: Hypothesis Testing

9.

H0: p = 0.5; H1: p > 0.5; Test statistic: z = 291

239 - 0.5

(0.5)(0.5)

= 10.96;

291

P-value = P ( z > 10.96) = 0.0001 (Tech: 0.0000); Critical value: z = 2.33; Reject H 0 . There is sufficient evidence to support the claim that the YSORT method is effective in increasing the likelihood that a baby will be a boy.

153 - 0.03 10. H 0 : p = 0.03; H1: p ¹ 0.03; Test statistic: z = 5924 = - 1.88; (0.03)(0.97) 5924

P-value = 2 ×P ( z < - 1.88) = 0.0602 (Tech: 0.0597); Critical values: z = ±1.96; Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that 3% of Eliquis users develop nausea. The rate of nausea appears to be quite low, so it is not a problematic adverse reaction.

481 - 0.5 11. H0: p = 0.5; H1: p ¹ 0.5; Test statistic: z = 882 = 2.69; (0.5)(0.5) 882

P-value = 2 ×P ( z > 2.69) = 0.0072 (Tech: 0.0071); Critical values: z = ±2.575 (Tech: z = ±2.576); Reject H 0 . There is sufficient evidence to reject the claim that the proportion of those in favor is equal to 0.5. The result suggests that the politician is wrong in claiming that the responses are random guesses equivalent to a coin toss.

12. H 0 : p = 0.06; H1: p > 0.06; Test statistic: z =

72 - 0.06 724 = 4.47; (0.06)(0.94) 724

P-value = P ( z > 1.47 ) = 0.0001 (Tech: 0.0000); Critical value: z = 1.645; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that the rate of nausea is the same as the rate experienced by flu patients given a placebo. It appears the rate of nausea is greater for the Tamiflu patients, which would be of concern.

13. H 0 : p = 0.20; H1: p > 0.20; Test statistic: z =

52 - 0.20 227 = 1.10; (0.20)(0.80) 227

P-value = P ( z > 1.10) = 0.1357 (Tech: 0.1367); Critical value: z = 1.645; Fail to reject H 0 . There is not sufficient evidence to support the claim that more than 20% of OxyContin users develop nausea. However, with pˆ = 0.229, we see that a large percentage of OxyContin users experience nausea, so that rate does appear to be very high.

856 - 0.5 14. H0: p = 0.5; H1: p > 0.5; Test statistic: z = 1228 = 13.81; (0.5)(0.5) 1228

P-value = P ( z > 13.81) = 0.0001 (Tech: 0.0000); Critical value: z = 2.33; Reject H 0 . There is sufficient evidence to support the claim that most medical malpractice lawsuits are dropped or dismissed. This should be comforting to physicians.

717 - 0.15 = - 1.31; 15. H 0 : p = 0.15; H1: p < 0.15; Test statistic: z = 5000 (0.15)(0.85) 5000

P-value = P ( z < - 1.31) = 0.1357 (Tech: 0.1367); Critical value: z = - 2.33; Fail to reject H 0 . There is not sufficient evidence to support the claim that the return rate is less than 15%. .

Section 8-2: Testing a Claim About a Proportion

16. H 0 : p = 0.10; H1: p < 0.10; Test statistic: z =

101

27 - 0.10 300 = - 0.58; (0.10)(0.90) 300

P-value = P ( z < - 0.58) = 0.2810 (Tech: 0.2819); Critical value: z = - 1.645; Fail to reject H 0 . There is not sufficient evidence to support the claim that less than 10% of the test results are wrong. The sample results suggest that the test is wrong too often to be considered very reliable.

17. H0: p = 0.512; H1: p ¹ 0.512; Test statistic: z =

426 - 0.512 860 = - 0.98; (0.512)(0.488) 860

P-value = 2 ×P ( z < - 0.98) = 0.3270 (Tech: 0.3286); Critical values: z = ±1.96; Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that 51.2% of newborn babies are boys. The results do not support the belief that 51.2% of newborn babies are boys; the results merely show that there is not strong evidence against the rate of 51.2%.

18. H 0 : p = 0.25; H1: p ¹ 0.25; Test statistic: z =

152 - 0.25 580 = 0.67; (0.25)(0.25) 580

P-value = 2 ×P ( z > 0.67) = 0.5028 (Tech: 0.5021); Critical values: z = ±2.575 (Tech: z = ±2.576); Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that 25% of offspring peas are yellow. There is not sufficient evidence to conclude that Mendel’s rate of 25% is wrong.

19. H0: p = 0.5; H1: p ¹ 0.5; Test statistic: z =

0.55 - 0.5 = 1.02; ( z = 0.98 using x = 57 ) (0.5)(0.5) 104

P-value = 2 ×P ( z > 1.02 ) = 0.3078 ; (0.3268, Table: 0.3270 using x = 57) Critical value: z = ±1.96; Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that women who guess the gender of their babies have a success rate equal to 50%.

20. H0: p = 0.5; H1: p ¹ 0.5; Test statistic: z =

32 - 0.5 45 = 2.83; (0.5)(0.5) 104

P-value = P ( z > 2.83) = 0.0023; Critical value: z = 2.33; Reject H 0 . There is sufficient evidence to support the claim that women with more than 12 years of education have a proportion of correct predictions that is greater than 0.5. It appears these women do have an ability to correctly predict the gender of their babies.

123 - 0.5 21. H0: p = 0.5; H1: p ¹ 0.5; Test statistic: z = 280 = - 2.03; (0.5)(0.5) 280

P-value = 2 ×P ( z < - 2.03) = 0.0424 (Tech: 0.0422); Critical values: z = ±1.645; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that touch therapists use a method equivalent to random guesses. However, their success rate of 123 280, or 43.9%, indicates that they performed worse than random guesses, so they do not appear to be effective.

.

102

Chapter 8: Hypothesis Testing

123 - 0.5 22. H0: p = 0.5; H1: p ¹ 0.5; Test statistic: z = 280 = - 2.03; (0.5)(0.5) 280

P-value = 2 ×P ( z < - 2.03) = 0.0424 (Tech: 0.0422); Critical values: z = ±2.575 (Tech: z = ±2.576); Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that touch therapists use a method equivalent to random guesses. After changing the significance level from 0.10 to 0.01, the conclusion does change.

135 - 0.000340 23. H0 : p = 0.00034; H0 : p ¹ 0.00034; Test statistic: z = 420,095 = - 0.66; (0.000340)(0.99966) 420,095

P-value = 2 ×P ( z < - 0.66) = 0.5092 (Tech: 0.5122); Critical values: z = ±2.81; Fail to reject H 0 . There is not sufficient evidence to support the claim that the rate is different from 0.0340%. Cell phone users should not be concerned about cancer of the brain or nervous system.

24. H 0 : p = 0.80; H1: p < 0.80; Test statistic: z =

74 - 0.80 98 = - 1.11; (0.80)(0.20) 98

P-value = P ( z < - 1.11) = 0.1335 (Tech: 0.1332); Critical values: z = ±1.96; Fail to reject H 0 . There is not sufficient evidence support the claim that such polygraph results are correct less than 80% of the time. Opinions may vary, but it appears the polygraph tests are somewhat accurate.

39 - 0.5 71 = 0.83; (0.5)(0.5) 71

25. H0: p = 0.5; H1: p > 0.5; Test statistic: z =

P-value = P ( z > 0.83) = 0.2033 (Tech: 0.2031); Critical value: z = 1.645; Fail to reject H 0 . There is not sufficient evidence to support the claim that among smokers who try to quit with nicotine patch therapy, the majority are smoking one year after the treatment. There isn’t sufficient evidence to conclude that the nicotine patch therapy is not effective.

26. H0: p = 0.5; H1: p < 0.5; Test statistic: z =

6062 - 0.5 12,000

(0.5)(0.5)

= 1.13;

12,000

P-value = P ( z > 1.13) = 0.8708 (Tech: 0.8712); Critical value: z = - 1.645; Fail to reject H 0 . There is not sufficient evidence to support the claim that less than 0.5 of the deaths occur the week before Thanksgiving. Based on these results, there is no indication that people can temporarily postpone their death to survive Thanksgiving. (With 50.5% of the deaths occurring before Thanksgiving, there is no way that the claim of a proportion less than 0.5 could be supported.)

27. H 0 : p = 0.80; H1: p ¹ 0.80; Test statistic: z =

0.828 - 0.80 = 0.98 (using x = 164, z = 0.99); (0.80)(0.20) 198

P-value = 2 ×P ( z > 0.98) = 0.3270 (Tech: 0.3246); Critical values: z = ±2.575 (Tech: z = ±2.576); Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that 80% of patients stop smoking when given sustained care. With a success rate around 80%, it appears that sustained care is effective.

.

Section 8-2: Testing a Claim About a Proportion

28. H 0 : p = 0.75; H1: p > 0.75; Test statistic: z =

103

0.817 - 0.75 = 8.48; (0.75)(0.25) 3005

P-value = P ( z > 8.48) = 0.0001 (Tech: 0.0000); Critical value: z = - 2.33; Reject H 0 . There is sufficient evidence to support the claim that more than 3 4 of adults use at least one prescription medication. With more than 3 4 of adults using at least one prescription medication, it appears that prescription use among adults is high.

29. H0: p = 0.5; H1: p ¹ 0.5; Normal approximation: 9 - 0.5 z = 10 = 2.53; P-value = 2 ×P ( z > 2.53) = 0.0114 (0.5)(0.5) 10

Exact:

(

( )( )

(

)( ))

(

)( )) 12 (10C9 (0.59 )(0.51 )) = 0.0117

P-value = 2 × 10 C9 0.59 0.51 + 10 C10 0.510 0.50 = 0.0215 Continuity Correction:

(

( )( )

P-value = 2 × 10 C9 0.59 0.51 + 10 C10 0.510 0.50 -

H 0 : p = 0.4; H1: p ¹ 0.4; Normal approximation: 9 - 0.4 z = 10 = 3.23; P-value = 2 ×P ( z > 3.23) = 0.0012 (0.4)(0.6) 10

Exact:

(

( )( )

(

)( ))

(

)( )) 12 (10 C9 (0.49 )(0.61 )) = 0.0018

P-value = 2 × 10 C9 0.49 0.61 + 10 C10 0.410 0.60 = 0.0034 Continuity Correction:

(

( )( )

P-value = 2 × 10 C9 0.49 0.61 + 10 C10 0.410 0.60 -

H0: p = 0.5; H1: p > 0.5; Normal approximation: 545 - 0.5 z = 1009 = 2.53; P-value = P ( z > 2.53) = 0.0057 (Tech: 0.0054) (0.5)(0.5) 1009

Exact:

(

)(

)

+ 1009 C1009 0.51009 0.50 = 0.0059

)(

)

+ 1009 C1009 0.51009 0.50 -

P-value = 1009 C545 0.5545 0.5464 +

(

)( )

(

)( ) 12 (1009C545 (0.5545 )(0.5464 ))

Continuity Correction:

(

P-value = 1009 C545 0.5545 0.5464 +

= 0.0054 The P-values agree reasonably well with the large sample size of n = 1009. The normal approximation to the binomial distribution works better as the sample size increases.

30. a. H 0 : p = 0.10; H1: p ¹ 0.10; . Test statistic: z =

0.119 - 0.1 = 2.00; Critical values: z = ±1.96; (0.1)(0.9) 1000

Reject H 0 . There is sufficient evidence to warrant rejection of the claim that the proportion of zeros is 0.1.

.

104

Chapter 8: Hypothesis Testing

30. (continued) b. H 0 : p = 0.10; H1: p ¹ 0.10; Test statistic: z =

0.119 - 0.1 = 2.00; (0.1)(0.9) 1000

P-value = 2 ×P ( z > 2.00) = 0.0456 (Tech: 0.0452); Reject H 0 . There is sufficient evidence to warrant rejection of the claim that the proportion of zeros is 0.1.

ˆˆ (0.119)(0.881) pq = 0.119 ± 1.96 Þ 0.0989 < p < 0.139; Because 0.1 is contained n 1000 within the confidence interval, fail to reject H 0 : p = 0.10. There is not sufficient evidence to warrant rejection of the claim that the proportion of zeros is 0.1. d. The traditional and P-value methods both lead to rejection of the claim, but the confidence interval method does not lead to rejection of the claim. c. 95% CI: pˆ ± za / 2

31. a. From p = 0.40, pˆ = 0.4 - 1.645 From p = 0.25, z =

(0.4)(0.6) 50

= 0.286

0.286 - 0.25 = 0.588; Power = P ( z < 0.588) = 0.7224 (Tech: 0.7219) (0.25)(0.75) 50

b. 1 - 0.7224 = 0.2776 (Tech: 0.2781) c. The power of 0.7224 shows that there is a reasonably good chance of making the correct decision of rejecting the false null hypothesis. It would be better if the power were even higher, such as greater than 0.8 or 0.9.

Section 8-3: Testing a Claim About a Mean 1. The requirements are (1) the sample must be a simple random sample, and (2) either or both of these conditions must be satisfied: The population is normally distributed or n > 30 . There is not enough information given to determine whether the sample is a simple random sample. Because the sample size is not greater than 30, we must check for normality, but the value of 583 sec appears to be an outlier, and a normal quantile plot or histogram suggests that the sample does not appear to be from a normally distributed population.

9 8

Frequency

7 6 5 4 3 2 1 0

0

100

200

300

400

500

600

Duration Time

2.

df denotes the number of degrees of freedom. For the sample of 12 times, df = 12 - 1 = 11.

3. A t test is a hypothesis test that uses the Student t distribution, such as the method of testing a claim about a population mean as presented in this section. The t test methods are much more likely to be used than the z test methods because the t test does not require a known value of s , and realistic hypothesis tests of claims about m typically involve a population with an unknown value of s .

.

Section 8-3: Testing a Claim About a Mean

105

4. For a 0.05 significance level used in a one-tailed test, use a 90% confidence level. The given confidence interval does contain the value of 90 sec, so it is possible that the value of m is equal to 90 sec or some lower value, so there is not sufficient evidence to support the claim that the mean is greater than 90 sec.

5.

P-value = 0.0437 (Table: 0.025 < P-value < 0.05)

6.

P-value = 0.0000 (Table: P-value < 0.01)

7.

P-value = 0.2581 (Table: P-value > 0.10)

8.

P-value = 0.1324 (Table: 0.05 < P-value < 0.10)

H0: m = 98.6 F; H1: m ¹ 98.6 F; Test statistic: t = - 3.8654; P-value = 0.0004; Critical values (with a = 0.05): t = ±2.0262 (Table: ±2.028); Reject H 0 . There is sufficient evidence to warrant rejection of the claim that the mean body temperature is equal

9.

to 98.6 F. 10. H0: m = 98.6°F; H1: m ¹ 98.6°F; Test statistic: t = - 13.272; P-value < 0.0001; Critical values (with a = 0.05): t = ±1.995 (Table: ±1.994); Reject H 0 . There is sufficient evidence to warrant rejection of the claim that the mean body temperature is equal to 98.6 F. 11. H 0 : m = 270 (1000 cells mL ) ; H1: m < 270 (1000 cells mL ); Test statistic: t = - 2.7637; P-value = 0.0032; Critical value (with a = 0.05): t = - 1.665; (Table: - 1.660); Reject H 0 . There is sufficient evidence to support the claim that the population of adult females has a mean platelet count less than 270 (1000 cells mL ).

12. H 0 : m = 220 (1000 cells mL ) ; H1: m > 220 (1000 cells mL ); Test statistic: t = 0.89; P-value = 0.188; Critical value (with a = 0.05): t = 1.665; (Table: 1.660); Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that the mean platelet count for adult males is different than 220 (1000 cells mL ).

13. The data appear to follow the loose definition for a normal distribution and n > 30.

H0 : m = 100 cm; H1: m > 100 cm; Test statistic: t =

98.24 - 100 17.94 147

= - 1.189; Critical value: t = 2.352;

P-value = 0.2364 (Table: P-value > 0.10); Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that the population mean of waist size of females is less than 100 cm.

14. The data appear to follow the loose definition for a normal distribution and n > 30.

H 0 : m = 98 cm; H1: m < 98 cm; Test statistic: t =

100.1 - 98 14.94 153

= - 1.739; Critical value: t = 2.351;

P-value = 0.0421 (Table: P-value < 0.05); Reject H 0 . There is sufficient evidence to support the claim that the population mean of waist size of males is less than 98 cm.

.

106

Chapter 8: Hypothesis Testing

15. The data cannot be verified to follow a normal distribution, but n > 30.

H0: m = 0; H1: m > 0;

0.4 - 0 = - 0.133; Critical value: t = 1.677 (Table: t » 1.676); 21.0 49 P -value = 0.4472 (Table: P-value > 0.10); Fail to reject H 0 . There is not sufficient evidence to support the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. The results suggest that the garlic treatment is not effective in reducing LDL cholesterol levels.

Test statistic: t =

16. The data cannot be verified to follow a normal distribution and n < 30, so proceed with caution.

H0: m = 102.8 min; H1: m < 102.8 min;

98.9 - 102.8 = - 0.369; Critical value (a = 0.05): t = - 1.753; 42.3 16 P -value = 0.3587 (Table: P-value > 0.10);

Test statistic: t =

Fail to reject H 0 . There is not sufficient evidence to support the claim that after treatment with zopiclone, subjects have a mean wake time less than 102.8 min. These results suggest that the zopiclone treatment is not effective.

17. The data cannot be verified to follow a normal distribution, but n > 30.

H0: m = 0; H1: m > 0;

3.0 - 0.0 = 3.872; Critical value: t = 2.426; 4.9 40 P-value = 0.0002 (Table: P-value < 0.005); Reject H 0 . There is sufficient evidence to support the claim that the mean weight loss is greater than 0. Although the diet appears to have statistical significance, it does not appear to have practical significance, because the mean weight loss of only 3.0 lb does not seem to be worth the effort and cost.

Test statistic: t =

18. The data cannot be verified to follow a normal distribution, but n > 30.

H0 : m = 69.5 years; H1: m > 69.5 years;

73.4 - 69.5 = 2.652; Critical value: t = 1.691; 8.7 35 P -value = 0.0060 (Table: 0.005 < P-value < 0.01);

Test statistic: t =

Reject H 0 . There is sufficient evidence to support the claim that the mean life span for conductors is greater than 69.5 years.

19. The data cannot be verified to follow a normal distribution, but n > 30.

H0 : m = 4.00; H1: m ¹ 4.00;

3.97 - 4.00 = - 1.139; Critical values: t = ±1.965; (Table: 1.966); 0.55 436 P -value = 0.2554 (Table: P-value > 0.20);

Test statistic: t =

Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that the population of student course evaluations has a mean equal to 4.00. Because the data are from the University of Texas at Austin, they don’t necessarily apply to a larger population that extends beyond that one institution.

20. The data cannot be verified to follow a normal distribution and n < 30, so proceed with caution.

H0 : m = 0; H1: m > 0;

4.00 - 0 = 8.447; Critical value: t = 2.584; 2.17 21 P-value = 0.0000 (Table: P-value < 0.005);

Test statistic: t =

.

Section 8-3: Testing a Claim About a Mean

107

20. (continued) Reject H 0 . There is sufficient evidence to support the claim that the mean fatigue level is positive. It appears the treatment is effective.

21. The sample data meet the loose requirement of having a normal distribution.

H0 : m = 14 mg g; H1: m < 14 mg g;

11.05 - 14.0 = - 1.444; Critical value: t = - 1.883; 6.46 10 P-value = 0.0913 (Table: P-value > 0.05); Fail to reject H 0 . There is not sufficient evidence to support the claim that the mean lead concentration for all such medicines is less than 14 mg g.

Test statistic: t =

22. The sample data meet the loose requirement of having a normal distribution.

H 0 : m = 60 sec; H1: m ¹ 60 sec;

62.67 - 60.00 = 0.530; Critical values: t = ±2.145; 9.48 15 P -value = 0.6043 (Table: P-value > 0.20);

Test statistic: t =

Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that the times are from a population with a mean equal to 60 seconds. Although some of the individual estimates are off by a large amount, the group of 15 students yielded the mean of 62.7 sec, which is not significantly different from 60 sec, so as a group they appear to be reasonably good at estimating one minute.

23. The data do not appear to follow a normal distribution and n < 30, so proceed with caution.

H0 : m = 1000 hic; H1: m < 1000 hic;

704 - 1000 = - 2.661; Critical value: t = - 3.365; 273 6 P -value = 0.0224 (Table: 0.1 < P-value < 0.025);

Test statistic: t =

Fail to reject H 0 . There is not sufficient evidence to support the claim that the population mean is less than 1000 hic. There is not strong evidence that the mean is less than 1000 hic, and one of the booster seats has a measurement of 1210 hic, which does not satisfy the specified requirement of being less than 1000 hic.

24. The sample data meet the loose requirement of having a normal distribution.

H0: m = 162 cm; H1: m > 162 cm;

117.25 - 162 = 33.082; Critical value: t = 2.602; 1.84 16 P-value = 0.0000 (Table: P-value < 0.005); Reject H 0 . There is sufficient evidence to support the claim that supermodels have heights with a mean that is greater than the mean height of 162 cm for women in the general population. Supermodels appear to be taller than typical women.

Test statistic: t =

25. The data appear to follow a normal distribution and n > 30.

H0 : m = 75 bpm; H1: m < 75 bpm;

74.04 - 75.00 = - 0.927; Critical value: t = - 1.655 (Table: t » - 1.660); 12.54 147 P -value = 0.1777 (Table: P-value > 0.10); Fail to reject H 0 . There is not sufficient evidence to support the claim that the mean pulse rate of adult females is less than 75 bpm.

Test statistic: t =

.

108

Chapter 8: Hypothesis Testing

26. The data appear to follow a normal distribution and n > 30.

H0 : m = 75 bpm; H1: m < 75 bpm;

69.58 - 75 = - 5.91; Critical value: t = - 1.655 (Table: t » - 1.660); 11.33 153 P-value = 0.0000 (Table: P-value < 0.005); Reject H 0 . There is sufficient evidence to support the claim that the mean pulse rate of adult males is less than 75 bpm.

Test statistic: t =

27. The data appear to follow a normal distribution and n > 30.

H0 : m = 90 mm Hg; H1: m < 90 mm Hg;

70.163 - 90 = - 21.435; Critical value: t = - 1.655 (Table: t » - 1.660); 11.22 147 P-value = 0.0000 (Table: P-value < 0.005);

Test statistic: t =

Reject H 0 . There is sufficient evidence to support the claim that the adult female population has a mean diastolic blood pressure level less than 90 mm Hg. The conclusion addresses the mean of a population, not individuals, so we cannot conclude that there are no female adults in the sample with hypertension.

28. The data appear to follow a normal distribution and n > 30.

H0 : m = 90 mm Hg; H1: m < 90 mm Hg;

71.32 - 90 = - 19.265; Critical value: t = - 1.655 (Table: t » - 1.660); 11.994 153 P-value = 0.0000 (Table: P-value < 0.005); Reject H 0 . There is sufficient evidence to support the claim that the adult male population has a mean diastolic blood pressure level less than 90 mm Hg. The conclusion addresses the mean of a population, not individuals, so we cannot conclude that there are no male adults in the sample with hypertension.

Test statistic: t =

2 1.645 (8 ×146 + 3) = 1.647814371 and t = 146 e1.647814371 146 - 1 = - 1.6554; The approximation yields a 8 ×146 + 1 critical t score of - 1.6554, which is the same as the value of - 1.6554 found from Statdisk. The approximation appears to work quite well.

29. A =

(

)

30. The power of 0.4943 shows that the chance of recognizing that m < 7 hours is not very high when in reality m = 6.0 hours. It would be better if the power was higher, such as 0.8 or greater. b = 1 - 0.4943 = 0.5057, so there is better than a 50% chance of failing to recognize that m < 7 hours when in reality m = 6.

Section 8-4: Testing a Claim About a Standard Deviation of Variance 1. The sample must be a simple random sample and the sample must be from a normally distributed population.

The normality requirement for a hypothesis test of a claim about a standard deviation is much more strict, meaning that the distribution of the population must be much closer to a normal distribution.

2.

H0 : s = 0.470 kg; H1: s ¹ 0.470 kg; Test statistic: c 2 =

3.

a. Reject H 0 .

(n - 1) s2 (16 - 1) 0.65732

= = 29.337 (29.339 if using s 2 0.4702 the original data values.); The sampling distribution of the test statistic is the c 2 (chi-square) distribution.

b. Reject the claim that the sample is from a population with a standard deviation equal to 0.470 kg. c. It appears that the vitamin supplement does affect the variation among birth weights.

4. All of the values in the confidence interval are greater than the standard deviation of 0.470 kg, so it appears that birth weights of babies born to mothers taking the supplement have a larger standard deviation.

.

Section 8-4: Testing a Claim About a Standard Deviation of Variance

109

H0: s = 10 bpm; H1: s ¹ 10 bpm; Test statistic: c 2 = 194.0888; P-value = 0.0239; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that pulse rates of men have a standard deviation equal to 10 beats per minute. Using the normal range of 60 to 100 beats per minute is not very good for estimating s in this case.

5.

6.

H0: s = 10 bpm; H1: s ¹ 10 bpm; Test statistic: c 2 = 229.718 or c 2 =

(147 - 1)12.52 102

= 228.125;

P-value < 0.0001; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that pulse rates of women have a standard deviation equal to 10 beats per minute. Using the normal range of 60 to 100 beats per minute is not very good for estimating s in this case.

7.

H0: s = 2.08 F; H1: s < 2.08 F; Test statistic: c 2 =

(n - 1) s2 (106 - 1) 0.622 s 2

=

2.082

= 9.329;

P-value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 74.252 (Table: c 2 » 70.065);

Reject H 0 . There is sufficient evidence to support the claim that body temperatures have a standard deviation less than 2.08 F. It is very highly unlikely that the conclusion in the hypothesis test in Example 5 from Section 8-3 would change because of a standard deviation from a different sample.

8.

H0 : s = 660.2 hg; H1: s ¹ 660.2 hg; Test statistic: c 2 =

(n - 1) s2 (30 - 1)829.52 s 2

=

660.22

= 45.780;

P -value = 0.0493 (Table: P-value > 0.02); Critical values: c 2 = 13.121 and c 2 = 52.336;

Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that birth weights of girls have the same standard deviation as the birth weights of boys.

9.

H0 : s = 15; H1: s < 15; Test statistic: c 2 =

(n - 1) s2 (18 - 1) 9.62

= = 6.963; s 2 152 P-value = 0.0160 (Table: P-value < 0.025); Critical value: c 2 = 8.672; Reject H 0 . There is sufficient evidence to support the claim that IQ scores of physicians have a standard deviation less than 15.

10. H0 : s = 14.1; H1: s < 14.1; Test statistic: c 2 =

(n - 1) s2 (27 - 1) 9.32

= = 11.311; s 2 14.12 P-value = 0.0056 (Table: P-value < 0.01); Critical value: c 2 = 12.198; Reject H 0 . There is sufficient evidence to support the claim that the last class has less variation. The lower variation implies that the scores are closer together, but it does not imply that the scores are higher, so the lower standard deviation does not suggest that the class is doing better.

11. H0 : s = 4.25 g; H1: s < 4.25 g; Test statistic: c 2 =

(n - 1) s2 (71 - 1) 3.872

= = 58.042; s 2 4.252 P -value = 0.1545 (Table: P-value > 0.10); Critical value: c 2 = 51.739; Fail to reject H 0 . There is not sufficient evidence to support the claim that the new process dispenses amounts with a standard deviation less than the standard deviation of 4.25 g for the old process. The new process does not appear to be better in the sense of dispensing amounts that are more consistent.

.

110

Chapter 8: Hypothesis Testing

12. H0 : s = 7460 words; H1: s > 7460 words; Test statistic: c 2 =

(n - 1) s2 (56 - 1) 78712

= = 61.227; s 2 74602 P -value = 0.2625 (Table: P-value > 0.10); Critical value: c 2 = 82.292 (Table: c 2 » 76.145); Fail to reject H 0 . There is not sufficient evidence to support the claim that males have a standard deviation that is greater than the standard deviation of 7460 words for females.

13. H0 : s = 6.0 lb; H1: s ¹ 6.0 lb; Test statistic: c 2 =

(n - 1) s2 (40 - 1) 4.92

= = 26.011; s 2 6.02 P-value = 0.1101 (Table: P-value > 0.10); Critical values: c 2 = 19.996 and c 2 = 65.475 (Table: 20.707, 66.766); Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that the amounts of weight loss have a standard deviation equal to 6.0 lb.

14. H0: s = 1.5 mm Hg; H1: s > 1.5 mm Hg; Test statistic: c 2 =

(n - 1) s2 (11 - 1) 9.652 =

s 2

1.52

= 413.878;

P-value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 18.307;

Reject H 0 . There is sufficient evidence to support the claim that the sample of devices has a standard deviation greater than 1.5 mm Hg. It appears the accuracy of the tested devices is suspect.

15. H0 : s = 109.3 sec; H1: s < 109.3 sec; Test statistic: c 2 =

(n - 1) s2 (10 - 1) 28.6012 s 2

=

109.32

= 0.616;

P-value = 0.001 (Table: P-value < 0.005); Critical value: c 2 = 3.325;

Reject H 0 . There is sufficient evidence to support the claim that with a single waiting line, the waiting times have a standard deviation less than 109.3 sec. Because the variation among waiting times appears to be reduced with the single waiting line, patients are happier because their waiting times are closer to being the same.

16. H0 : s = 0.03 g; H1: s ¹ 0.03 g; Test statistic: c 2 =

(n - 1) s2 (15 - 1) 0.258752 =

s 2

0.032

= 10.415;

P -value = 0.5375 (Table: P-value > 0.20); Critical values: c 2 = 5.629 and c 2 = 26.119;

Reject H 0 . There is sufficient evidence to warrant rejection of the claim that the weights of bumblebee bats have a standard deviation of 0.03 g. It appears the weights of these bats have a different variation than those from the region in Thailand.

2 1 2.33 + 2 ×55 - 1 = 81.54 (or 81.494 if using z = 2.326348 found from technology), which is 2 reasonably close to the value of 22.465 obtained from STATDISK and Minitab.

17. Critical c 2 =

(

)

3

æ æ 2 2 öö 18. Critical c = 55 ç1 + ç2.33 ÷ = 82.360 (or 82.309 if using z = 2.326348 found from 9 ×55 ÷ øø è 9 ×55 è technology), which is very close to the value of 82.292 obtained from STATDISK and Minitab.

2

Chapter Quick Quiz 1. a. t distribution b. normal distribution c. chi-square distribution

2.

.

a. two-tailed b. left-tailed c. right-tailed

Review Exercises 111 3.

a. H 0 : p = 0.5; H1: p > 0.5 b. Test statistic: z =

0.53 - 0.5 = 1.38 (using x = 282, z = 1.39) (0.5)(0.5) 532

c. Fail to reject H 0 . d. There is not sufficient evidence to support the claim that the majority of Internet users aged 18–29 use Instagram.

4.

P-value = 0.0100

6.

false

5.

true

7.

false

8.

No, all critical values of c 2 are always positive.

9.

The t test requires that the sample is from a normally distributed population, and the test is robust in the sense

that the test works reasonably well if the departure from normality is not too extreme. The c 2 (chi-square) test is not robust against a departure from normality, meaning that the test does not work well if the population has a distribution that is far from normal.

10. The only true statement is the one given in part (a).

Review Exercises 1. a. false b. true c. false 2.

d. false e. false

H0: p = 0.5; H1: p > 0.5; Test statistic: z =

0.64 - 0.5 = 40.91 (z = 40.90 using x = 13,661); (0.5)(0.5) 21,346

P-value = P ( z > 40.91) = 0.0001 (Tech: 0.0000); Critical value: z = 2.33; Reject H 0 . There is sufficient evidence to support the claim that most people believe that the Loch Ness monster exists. Because the sample is a voluntary response sample, the conclusion about the population might not be valid.

3.

The data cannot be verified to have a normal distribution, but n > 30.

H0 : m = 8.3 million cells per microliter; H1: m < 8.3 million cells per microliter;

6.889 - 8.3

= - 4.937; Critical value: t = 2.405; 2.021 50 P-value = 0.0000 (Table: P-value < 0.005); Reject H 0 . There is sufficient evidence to support the claim that the sample is from a population with a mean less than 8.3 million cells per microliter. The test deals with the distribution of sample means, not individual values, so the result does not suggest that each of the 50 females has a white blood cell count below 8.3 million cells per microliter.

Test statistic: t =

4.

H0: p = 0.204; H1: p ¹ 0.204; Test statistic: z =

124 - 0.204 572 = 0.76; (0.204)(0.796) 572

P-value = 2 ×P ( z > 0.76) = 0.4472 (Tech: 0.4480); Critical values: z = ±1.96; Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that the rate of smoking by adult males is now the same as in 2008. The smoking rate appears to be about the same.

Copyright © 2018 Pearson Education Ltd.

112 5.

Chapter 8: Hypothesis Testing

The data appear to follow a normal distribution.

H0: m = 25 mg; H1: m ¹ 25 mg;

24.89 - 25 = - 0.744; Critical values: t = ±2.145; 0.590 15 P -value = 0.4694 (Table: P-value > 0.10); Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that the pills come from a population in which the mean amount of atorvastatin is equal to 25 mg.

Test statistic: t =

6.

The sample data meet the requirement of having a normal distribution.

H0 : m = 20.16; H1: m < 20.16;

18.76 - 20.16 = - 3.732; Critical value: t = - 2.821; 1.186 10 P -value = 0.0023 (Table: P-value < 0.005); Reject H 0 . There is sufficient evidence to support the claim that the population of recent winners has a mean BMI less than 0.16. Recent winners appear to be significantly smaller than those from the 1920s and 1930s.

Test statistic: t =

7.

The sample data meet the requirement of having a normal distribution.

H 0 : s = 1.34; H1: s ¹ 1.34; Test statistic: c 2 =

(n - 1) s2 (10 - 1)1.1862

= = 7.053; s 2 1.342 P -value = 0.7368 (Table: P-value > 0.20); Critical values: c 2 = 1.735 and c 2 = 23.589; Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that the recent winners have BMI values with variation different from that of the 1920s and 1930s.

8.

a. A type I error is the mistake of rejecting a null hypothesis when it is actually true. A type II error is the mistake of failing to reject a null hypothesis when in reality it is false. b. Type I error: In reality, the mean BMI is equal to 20.16, but we support the claim that the mean BMI is less than 20.16. Type II error: In reality, the mean BMI is less than 20.16, but we fail to support that claim.

Cumulative Review Exercises 1.

a. x =

23 + 26 + 27 + 28 + 29 + 32 + 34 + 38 + 43 + 44 + 45 + 48 + 51 + 51 = 37.1 deaths 14

b. Q2 =

34 + 38 = 36.0 deaths 2

c. s =

(23 - 37.1)2 + (26 - 37.1)2 + + (51 - 37.1)2 + (51 - 37.1)2 = 9.8 deaths 14 - 1

d. s 2 = 9.82 = 96.8 deaths2 e. range = 51 - 23 = 28.0 deaths f. The pattern of the data over time is not revealed by the statistics. A time-series graph would be very helpful in understanding the pattern over time.

2.

a. ratio b. discrete c. quantitative d. No, the data are from recent and consecutive years, so they are not randomly selected.

s 9.84 = 37.1 ± 3.012 × Þ 29.1 deaths < m < 45.0 deaths; We have 99% confidence that the n 14 limits of 29.1 deaths and 45.0 deaths contain the value of the population mean.

3.

99% CI: x ± ta / 2

.

Cumulative Review Exercises 4.

113

The sample data meet the loose requirement of having a normal distribution.

H0: m = 72.6 deaths; H1: m < 72.6 deaths;

37.07 - 72.6 = - 13.509; Critical value: t = - 2.650; 9.84 14 P-value = 0.0000 (Table: P-value < 0.005); Reject H 0 . There is sufficient evidence to support the claim that the mean number of annual lightning deaths is now less than the mean of 72.6 deaths from the 1980s. Possible factors: Shift in population from rural to urban areas; better lightning protection and grounding in electric and cable and phone lines; better medical treatment of people struck by lightning; fewer people use phones attached to cords; better weather predictions.

Test statistic: t =

5. Because the vertical scale starts at 50 and not at 0, the difference between the number of males and the number of females is exaggerated, so the graph is deceptive by creating the false impression that males account for nearly all lightning strike deaths. A comparison of the numbers of deaths shows that the number of male deaths is roughly 4 times the number of female deaths, but the graph makes it appear that the number of male deaths is around 25 times the number of female deaths.

6.

232 - 0.5

H0: p = 0.5; H1: p > 0.5; Test statistic: z = 287

(0.5)(0.5)

= 10.45;

287

P-value = P ( z > 10.45) = 0.0001 (Tech: 0.0000); Critical value: z = 2.33; Reject H 0 . There is sufficient evidence to support the claim that the proportion of male deaths is greater than

1 2. More males are involved in certain outdoor activities such as construction, fishing, and golf.

( 232 )( 55 )

ˆ ˆ 232 pq = ± 1.96 287 287 Þ 0.762 < p < 0.854; Because the entire confidence interval is n 287 287 greater than 0.5, it does not seem feasible that males and females have equal chances of being killed by lightning.

7.

95% CI: pˆ ± za / 2

8.

a. 0.8 ×0.8 ×0.8 = 0.512 b. 0.2 ×0.2 ×0.2 = 0.008 c. 1 - 0.2 ×0.2 ×0.2 = 0.992

( )( )

d. 5 C3 0.83 0.22 = 0.205 e. m = np = 50 ×0.8 = 40 males; s = npq = 50 ×0.8 ×0.2 = 2.8 males f. Yes, using the range rule of thumb, values above m + 2s = 40.0 + 2(2.8) = 45.6 are considered significantly high. Since 46 is greater than 45.6, 46 males would be a significantly high number in a group of 50.

331 4489 = 1.14; The odds in favor of experiencing migraine headaches is about 1.14 times higher for 358 5487 overweight people than for people with normal weight.

9.

OR =

.

Section 9-1: Two Proportions 115

Chapter 9: Inferences from Two Samples Section 9-1: Two Proportions 1. The samples are simple random samples that are independent. For each of the two groups, the number of successes is at least 5 and the number of failures is at least 5. (Depending on what we call a success, the four numbers are 33, 115, 201,229 and 200,745 and all of those numbers are at least 5.) The requirements are satisfied.

2.

3.

33 = 0.000163992, qˆ1 = 1 - 0.000163992 = 0.999836; 201,299 115 n2 = 200,745, pˆ 2 = = 0.000572866, qˆ2 = 1 - 0.000572866 = 0.999427; 200,745 33 + 115 p= = 0.000368183, q = 1 - 0.000368183 = 0.999632 201, 299 + 200,745 n1 = 201,229, pˆ1 =

a. H 0 : p1 = p2 ; H1: p1 < p2 b. There is sufficient evidence to support the claim that the rate of polio is less for children given the Salk vaccine than it is for children given a placebo. The Salk vaccine appears to be effective.

4.

a. hypothesis test b. The P-value method and the critical value method are equivalent in the sense that they will always lead to the same conclusion, but the confidence interval method is not equivalent to them. c. 0.90, or 90% d. Because the confidence interval limits do not contain 0, there appears to be a significant difference between the two proportions. Because the confidence interval consists of negative values only, it appears that the first proportion is less than the second proportion. There is sufficient evidence to support the claim that the rate of polio is less for children given the Salk vaccine than it is for children given a placebo.

5.

H0 : p1 = p2 ; H1: p1 > p2 ; population1 = vinyl gloves, population2 = latex gloves; Test statistic: z = 12.82; P-value = 0.0000; Critical value: z = 2.33; Reject H 0 . There is sufficient evidence to support the claim that vinyl gloves have a greater virus leak rate than latex gloves.

6.

H0 : p1 = p2 ; H1: p1 > p2 ; population1 = surgery, population2 = splints; Test statistic: z = 3.12; P-value = 0.0009; Critical value: z = 2.33; Reject H 0 . There is sufficient evidence to support the claim that the success rate is better with surgery.

For Exercises 7–22, assume that the data fit the requirements for the statistical methods for two proportions unless otherwise indicated.

7.

a. H0: p1 = p2 ; H1: p1 < p2 ; population1 = ginkgo, population2 = placebo; Test statistic: z = - 1.66; P-value = 0.0484 (Table: 0.0485); Critical value: z = - 2.33; Fail to reject H 0 . There is not sufficient evidence to support the claim that the rate of dementia among those who use ginkgo is less than the rate of dementia among those who use a placebo. There is not sufficient evidence to support the claim that ginkgo is effective in preventing dementia.

246 + 277 523 523 2546 p= = ; q = 1= ; 1545 + 1524 3069 3069 3069 æ 246 277 ö çè ÷- 0 pˆ1 - pˆ 2 ) - ( p1 - p2 ) ( 1545 1524 ø z= = = - 1.66 pq pq æ 523 öæ2546 ö æ 523 öæ2546 ö + çè ÷ç ÷ ç ÷ç ÷ n1 n2 3069 øè 3069 ø è 3069 øè 3069 ø + 1545 1524

.

116 7.

Chapter 9: Inferences from Two Samples (continued) b. 98% CI: - 0.0541 < p1 - p2 < 0.000904; (Table: - 0.0542 < p1 - p2 < 0.000909) Because the confidence interval limits include 0, there does not appear to be a significant difference between dementia rates for those treated with ginkgo and those given a placebo. There is not sufficient evidence to support the claim that the rate of dementia among those who use ginkgo is less than the rate of dementia among those who use a placebo.

There is not sufficient evidence to support the claim that ginkgo is effective in preventing dementia.

( pˆ1 - pˆ 2 ) ± za / 2

æ 246 öæ1299 ö æ 277 öæ1297 ö çè ÷ç ÷ ç ÷ç ÷ pˆ1qˆ1 pˆ 2 qˆ2 æ 246 277 ö 1545 øè1545 ø è1524 øè1524 ø + =ç ± 2.326 + ÷ è1545 1524 ø n1 n2 1545 1524

c. The sample results suggest that ginkgo is not effective in preventing dementia.

8.

a. H0 : p1 = p2 ; H1: p1 ¹ p2 ; population1 = Lipitor, population2 = Placebo; Test statistic: z = - 0.73; P -value = 0.4368 (Table: 0.4645); Critical values: z = ±1.96; Fail to reject H 0 .

There is not sufficient evidence to reject the claim that Lipitor and the placebo have the same rates of infection.

7 + 27 17 17 165 p= = ; q = 1= ; 94 + 270 182 182 182 27 ö æ7 çè ÷- 0 pˆ1 - pˆ 2 ) - ( p1 - p2 ) ( 94 270 ø z= = = - 0.73 pq pq æ 17 öæ165 ö æ 17 öæ165 ö + çè ÷ç ÷ ç ÷ç ÷ n1 n2 182 øè182 ø è182 øè182 ø + 90 270 b. 95% CI: - 0.0895 < p1 - p2 < - 0.0385 (Table: - 0.0890 < p1 - p2 < - 0.0379); Because the confidence interval limits contain 0, there does not appear\ to be a significant difference between the rates of infection for Lipitor and the placebo.

( pˆ1 - pˆ 2 ) ± za / 2

æ 7 öæ87 ö æ 27 öæ243 ö çè ÷ç ÷ ç ÷ç ÷ ˆp1qˆ1 pˆ 2 qˆ2 7 27 94 øè 94 ø è 270 øè 270 ø + = ± 1.96 + n1 n2 94 270 94 270

c. The sample results suggest that Lipitor and the placebo have the same infection rates.

9.

a. H0 : p1 = p2 ; H1: p1 > p2 ; population1 = sustained care, population2 = standard care; Test statistic: z = 2.64; P-value = 0.0041; Critical value: z = 2.33; Reject H 0 . There is sufficient evidence to support the claim that the rate of success for smoking cessation is greater with the sustained care program.

51 + 30 81 81 316 p= = ; q = 1= ; 198 + 199 397 397 397 æ 51 30 ö çè ÷- 0 pˆ1 - pˆ 2 ) - ( p1 - p2 ) ( 198 199 ø z= = = 2.64 pq pq 81 öæ316 ö æ 81 öæ316 ö æ + çè ÷ç ÷ ç ÷ç ÷ n1 n2 397 øè 397 ø è 397 øè 397 ø + 198 199 b. 98% CI: 0.0135 < p1 - p2 < 0.200 (Table: 0.0134 < p1 - p2 < 0.200); Because the confidence interval limits do not contain 0, there is a significant difference between the two proportions. Because the interval consists of positive numbers only, it appears that the success rate for the sustained care program is greater than the success rate for the standard care program.

( pˆ1 - pˆ 2 ) ± za / 2

æ 51 öæ147 ö æ 30 öæ169 ö çè ÷ç ÷ ç ÷ç ÷ pˆ1qˆ1 pˆ 2 qˆ2 æ 51 30 ö 198 øè198 ø è199 øè199 ø + =ç ± 2.33 + ÷ è198 199 ø n1 n2 198 199

.

Section 9-1: Two Proportions 117 9.

(continued) c. Based on the samples, the success rates of the programs are 25.8% (sustained care) and 15.1% (standard care). That difference does appear to be substantial, so the difference between the programs does appear to have practical significance.

10. a. H0: p1 = p2 ; H1: p1 ¹ p2 ; population1 = women, population2 = men; Test statistic: z = 11.35; P-value = 0.0000 (Table: 0.0001); Critical values: z = ±2.576; Reject H 0 . There is sufficient evidence to support the claim that men and women have different rates of developing thyroid disease.

1397 + 436 1833 1833 2258 p= = ; q = 1= ; 2739 + 1352 4091 4091 4091 æ1397 436 ö çè ÷- 0 pˆ1 - pˆ 2 ) - ( p1 - p2 ) ( 2739 1352 ø z= = = 11.35 pq pq æ1833 öæ2258 ö æ1833 öæ2258 ö + çè ÷ç ÷ ç ÷ç ÷ n1 n2 4091 øè 4091 ø è 4091 øè 4091 ø + 2739 1352 b. 95% CI: 0.147 < p1 - p2 < 0.229; Because the confidence interval limits do not contain 0, there is a significant difference between the two proportions. There is sufficient evidence to support the claim that men and women have different rates of developing thyroid disease and it appear the rate among women is higher.

( pˆ1 - pˆ 2 ) ± za / 2

æ1397 öæ1342 ö æ 436 öæ 837 ö çè ÷ç ÷ ç ÷ç ÷ ˆp1qˆ1 pˆ 2 qˆ2 æ1397 436 ö 2739 øè 2739 ø è1273 øè1273 ø + =ç ± 2.576 + è 2739 1352 ÷ ø n1 n2 1273 1273

11. a. H0: p1 = p2 ; H1: p1 > p2 ; population1 = over age 55, population2 = under age 25; Test statistic: z = 6.44; P-value = 0.0000 (Table: 0.0001); Critical value: z = 2.33; Reject H 0 . There is sufficient evidence to support the claim that the proportion of people over 55 who dream in black and white is greater than the proportion of those under 25.

68 + 13 81 81 523 p= = ; q = 1= ; 306 + 298 604 604 604 æ 68 13 ö çè ÷- 0 pˆ1 - pˆ 2 ) - ( p1 - p2 ) ( 306 298 ø z= = = 6.44 pq pq æ 81 öæ523 ö æ 81 öæ523 ö + çè ÷ç ÷ ç ÷ç ÷ n1 n2 604 øè 604 ø è 604 øè 604 ø + 306 298 b. 98% CI: Because the confidence interval limits do not include 0, it appears that the two proportions are not equal. Because the confidence interval limits include only positive values, it appears that the proportion of people over 55 who dream in black and white is greater than the proportion of those under 25.

( pˆ1 - pˆ 2 ) ± za / 2

æ 68 öæ238 ö æ 13 öæ285 ö çè ÷ç ÷ ç ÷ç ÷ pˆ1qˆ1 pˆ 2 qˆ2 æ 68 13 ö 306 øè 306 ø è 298 øè 298 ø + =ç ± 2.33 + è 306 298 ÷ ø n1 n2 306 298

c. The results suggest that the proportion of people over 55 who dream in black and white is greater than the proportion of those under 25, but the results cannot be used to verify the cause of that difference.

12. a. H0: p1 = p2 ; H1: p1 ¹ p2 ; population1 = OxyContin, population 2 = placebo; Test statistic: z = 1.78; P-value = 0.0757 (Table: 0.0750); Critical values: z = ±1.96; Fail to reject H 0 . There is not sufficient evidence to support the claim that there is a difference between the rates of nausea for those treated with OxyContin and those given a placebo.

.

118

Chapter 9: Inferences from Two Samples

12. (continued)

52 + 5 57 57 215 = ; q = 1= ; 227 + 45 272 272 272 5ö æ 52 ç ÷- 0 ˆ ˆ p p p p è ( 227 45 ø 2) ( 1 2) z= 1 = = 1.78 pq pq 57 öæ215 ö æ 57 öæ215 ö æ + çè ÷ç ÷ ç ÷ç ÷ n1 n2 272 øè 272 ø è 272 øè 272 ø + 227 45 p=

b. 95% CI: 0.0111 < p1 - p2 < 0.225; Because the confidence interval limits do not contain 0, there is a significant difference between the two proportions.

( pˆ1 - pˆ 2 ) ± za / 2

æ 52 öæ175 ö æ 5 öæ40 ö çè ÷ç ÷ ç ÷ç ÷ pˆ1qˆ1 pˆ 2 qˆ2 æ 52 5ö 227 øè 227 ø è 45 øè 45 ø + =ç ± 1.96 + ÷ è 227 45 ø n1 n2 227 45

c. The conclusions from parts (a) and (b) are different. The conclusion from part (a) results from a hypothesis test instead of an estimate of the difference between the two rates of nausea, so there does not appear to be sufficient evidence to conclude that there is a difference between the rates of nausea for those treated with OxyContin and those given a placebo.

13. a. H0 : p1 = p2 ; H1: p1 > p2 ; population1 = wearing seatbelt, population2 = not wearing seatbelt; Test statistic: z = 6.11; P-value = 0.0000 (Table: 0.0001); Critical value: z = 1.645; Reject H 0 . There is sufficient evidence to support the claim that the fatality rate is higher for those not wearing seat belts.

31 + 16 47 47 10,541 p= = ; q = 1= ; 2823 + 7765 10,588 10,588 10,588

( pˆ - pˆ 2 ) - ( p1 - p2 ) = z= 1 pq pq + n1 n2

16 ö æ 31 çè ÷- 0 2823 7765 ø æ 47 öæ10,541 ö æ 47 öæ10,541 ö çè10,588 ÷ç øè10,588 ÷ ø çè10,588 ÷ç øè10,588 ÷ ø + 2823 7765

= 6.11

b. 90% CI: 0.00559 < p1 - p2 < 0.0123; Because the confidence interval limits do not include 0, it appears that the two fatality rates are not equal. Because the confidence interval limits include only positive values, it appears that the fatality rate is higher for those not wearing seat belts.

( pˆ1 - pˆ 2 ) ± za / 2

æ 31 öæ2792 ö æ 16 öæ7749 ö çè ÷ç ÷ ç ÷ç ÷ pˆ1qˆ1 pˆ 2 qˆ2 æ 31 16 ö 2823 øè 2823 ø è 7765 øè 7765 ø + =ç ± 1.645 + ÷ è 2823 7765 ø n1 n2 2823 7765

c. The results suggest that the use of seat belts is associated with fatality rates lower than those associated with not using seat belts.

.

Section 9-1: Two Proportions 119 14. a. H0 : p1 = p2 ; H1: p1 ¹ p2 ; population1 = day, population2 = night; Test statistic: z = 18.26; P-value = 0.0000 (Table: 0.0002); Critical values: z = ±2.576 (Table: z = ±2.575); Reject H 0 . There is sufficient evidence to warrant rejection of the claim that the survival rates are the same for day and night.

11,604 + 4139 15,743 15,743 71,005 p= = ; q = 1= ; 58,593 + 28,155 86,748 86,748 86,748

( pˆ - pˆ 2 ) - ( p1 - p2 ) = z= 1 pq pq + n1 n2

æ11,604 4139 ö -0 çè 58,593 - 28,155 ÷ ø æ15,743 öæ71,005 ö æ15,743 ö æ71,005 ö çè 86,748 ÷ç øè 86,748 ÷ ø çè 86,748 ÷ ø çè 86,748 ÷ ø + 58,593 28,155

= 18.26

b. 99% CI: 0.0441 < p1 - p2 < 0.0579; Because the confidence interval limits do not contain 0, there appears to be a significant difference between the two proportions. There is sufficient evidence to warrant rejection of the claim that the survival rates are the same for day and night. pˆ qˆ pˆ qˆ ( pˆ1 - pˆ 2 ) ± za / 2 1 1 + 2 2 = n1 n2

æ11,604 öæ46,989 ö æ 4139 öæ24,016 ö çè 58,593 ÷ç øè 58,593 ÷ ø çè 28,155 ÷ç øè 28,155 ÷ ø æ11,604 4139 ö ± 2.576 + çè 58,593 - 28,155 ÷ ø 58,593 28,155 c. The data suggest that for in-hospital patients who suffer cardiac arrest, the survival rate is not the same for day and night. It appears that the survival rate is higher for in-hospital patients who suffer cardiac arrest during the day.

15. a. H0: p1 = p2 ; H1: p1 ¹ p2 ; population1 = echinacea, population 2 = placebo; Test statistic: z = 0.57; P-value = 0.5720 (Table: 0.5868); Critical values: z = ±1.96; Fail to reject H 0 . There is not sufficient evidence to support the claim that Echinacea treatment has an effect.

40 + 88 32 32 5 p= = ; q = 1= ; 45 + 103 37 37 37 æ40 88 ö çè ÷- 0 pˆ1 - pˆ 2 ) - ( p1 - p2 ) ( 45 103 ø z= = = 0.57 pq pq 32 öæ 5 ö æ32 öæ 5 ö æ + çè ÷ç ÷ ç ÷ç ÷ n1 n2 37 øè 37 ø è 37 øè 37 ø + 45 103 b. 95% CI: - 0.0798 < p1 - p2 < 0.149; Because the confidence interval limits do contain 0, there is not a significant difference between the two proportions. There is not sufficient evidence to support the claim that Echinacea treatment has an effect.

( pˆ1 - pˆ 2 ) ± za / 2

æ40 öæ 5 ö æ 88 öæ15 ö çè ÷ç ÷ ç ÷ç ÷ pˆ1qˆ1 pˆ 2 qˆ2 æ40 88 ö 45 øè 45 ø è103 øè103 ø + =ç ± 1.96 + è 45 103 ÷ ø n1 n2 45 103

c. Echinacea does not appear to have a significant effect on the infection rate. Because it does not appear to have an effect, it should not be recommended.

.

120

Chapter 9: Inferences from Two Samples

16. a. H0: p1 = p2 ; H1: p1 < p2 ; population1 = used bednet, population2 = did not use bednet; Test statistic: z = - 2.44; P-value = 0.0074; Critical value: Reject H 0 . There is sufficient evidence to support the claim that the incidence of malaria is lower for infants who use the bednets.

15 + 27 6 6 85 p= = ; q = 1= ; 343 + 294 91 91 91 27 ö æ 15 ç ÷- 0 ˆ ˆ p p p p è ( 343 294 ø 2) ( 1 2) z= 1 = = - 2.44 pq pq 6 öæ85 ö æ 6 öæ85 ö æ + çè ÷ç ÷ ç ÷ç ÷ n1 n2 91 øè 91 ø è 91 øè 91 ø + 343 294 b. 98% CI: - 0.0950 < p1 - p2 < - 0.00125; Because the confidence interval does not include 0 and it includes only negative values, it appears that the rate of malaria is lower for infants who use the bednets.

( pˆ1 - pˆ 2 ) ± za / 2

æ 15 öæ328 ö æ 27 öæ267 ö çè ÷ç ÷ ç ÷ç ÷ pˆ1qˆ1 pˆ 2 qˆ2 æ 15 27 ö 343 øè 343 ø è 294 øè 294 ø + =ç ± 2.33 + ÷ è 343 294 ø n1 n2 343 294

c. The bednets appear to be effective.

17. a. H0: p1 = p2 ; H1: p1 < p2 ; population1 = used left ear, population2 = used right ear; Test statistic: z = - 7.94; P-value = 0.0000 (Table: 0.0001); Critical value: z = - 2.33; Reject H 0 . There is sufficient evidence to support the claim that the rate of right-handedness for those who prefer to use their left ear for cell phones is less than the rate of right-handedness for those who prefer to use their right ear for cell phones.

166 + 436 301 301 33 p= = ; q = 1= ; 216 + 452 334 334 334 æ166 436 ö çè ÷- 0 pˆ1 - pˆ 2 ) - ( p1 - p2 ) ( 216 452 ø z= = = - 7.94 pq pq æ301 öæ 33 ö æ301 öæ 33 ö + çè ÷ç ÷ ç ÷ç ÷ n1 n2 334 øè 334 ø è 334 øè 334 ø + 216 452 b. 98% CI: - 0.266 < p1 - p2 < - 0.126; Because the confidence interval limits do not contain 0, there is a significant difference between the two proportions. Because the interval consists of negative numbers only, it appears that the claim is supported. The difference between the populations does appear to have practical significance.

( pˆ1 - pˆ 2 ) ± za / 2

æ166 öæ 50 ö æ436 öæ 16 ö çè ÷ç ÷ ç ÷ç ÷ ˆp1qˆ1 pˆ 2 qˆ2 æ166 436 ö 216 øè 216 ø è 452 øè 452 ø + =ç ± 2.33 + è 216 452 ÷ ø n1 n2 216 452

.

Section 9-1: Two Proportions 121 18. a. H0: p1 = p2 ; H1: p1 < p2 ; population1 = single bill, population2 = multiple bills; Test statistic: z = - 1.85; P-value = 0.0324 (Table: 0.0322); Critical value: z = - 1.645; Reject H 0 . There is sufficient evidence to support the claim that when given a single large bill, a smaller proportion of women in China spend some or all of the money when compared to the proportion of women in China given the same amount in smaller bills.

60 + 68 64 64 11 p= = ; q = 1= ; 75 + 75 75 75 75 æ60 68 ö çè ÷- 0 pˆ1 - pˆ 2 ) - ( p1 - p2 ) ( 75 75 ø z= = = - 1.85 pq pq 64 öæ11 ö æ64 öæ11 ö æ + çè ÷ç ÷ ç ÷ç ÷ n1 n2 75 øè 75 ø è 75 øè 75 ø + 75 75 b. 90% CI: - 0.201 < p1 - p2 < - 0.0127; Because the confidence interval does not include 0 and it includes only negative values, it appears that the first proportion is less than the second proportion. There is sufficient evidence to support the claim that when given a single large bill, a smaller proportion of women in China spend some or all of the money when compared to the proportion of women in China given the same amount in smaller bills.

( pˆ1 - pˆ 2 ) ± za / 2

æ60 öæ15 ö æ68 öæ 7 ö çè ÷ç ÷ ç ÷ç ÷ ˆp1qˆ1 pˆ 2 qˆ2 æ60 68 ö 75 øè 75 ø è 75 øè 75 ø + =ç ± 1.645 + è 75 75 ÷ ø n1 n2 75 75

c. Because the P-value = 0.0324 (Table: 0.0322), the difference is significant at the 0.05 significance level, but not at the 0.01 significance level. The conclusion does change.

19. a. H0 : p1 = p2 ; H1: p1 > p2 ; population1 = oxygen, population 2 = placebo; Test statistic: z = 9.97; P-value = 0.0000 (Table: 0.0001); Critical value: z = 2.33; Reject H 0 . There is sufficient evidence to support the claim that the cure rate with oxygen treatment is higher than the cure rate for those given a placebo. It appears that the oxygen treatment is effective.

116 + 29 145 145 153 = ; q = 1= ; 150 + 148 298 298 298 æ116 29 ö çè ÷- 0 pˆ1 - pˆ 2 ) - ( p1 - p2 ) ( 150 148 ø z= = = 9.97 pq pq æ145 öæ153 ö æ145 öæ153 ö + çè ÷ç ÷ ç ÷ç ÷ n1 n2 298 øè 298 ø è 298 øè 298 ø + 150 148 p=

b. 98% CI: 0.467 < p1 - p2 < 0.687; Because the confidence interval limits do not include 0, it appears that the two cure rates are not equal. Because the confidence interval limits include only positive values, it appears that the cure rate with oxygen treatment is higher than the cure rate for those given a placebo. It appears that the oxygen treatment is effective.

( pˆ1 - pˆ 2 ) ± za / 2

æ116 öæ 34 ö æ 29 öæ119 ö çè ÷ç ÷ ç ÷ç ÷ pˆ1qˆ1 pˆ 2 qˆ2 æ116 29 ö 150 øè150 ø è148 øè148 ø + =ç ± 2.33 + è150 148 ÷ ø n1 n2 150 148

c. The results suggest that the oxygen treatment is effective in curing cluster headaches.

.

122

Chapter 9: Inferences from Two Samples

20. a. H0: p1 = p2 ; H1: p1 ¹ p2 ; population1 = aspirin, population2 = placebo; Test statistic: z = - 5.19; P-value = 0.0000 (Table: 0.0002); Critical values: z = ±1.96; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that aspirin has no effect on myocardial infarctions.

139 + 239 18 18 1033 p= = ; q = 1= ; 11,037 + 11,034 1051 1051 1051

( pˆ - pˆ 2 ) - ( p1 - p2 ) = z= 1 pq pq + n1 n2

239 ö æ 139 -0 çè11,037 - 11,034 ÷ ø æ 18 öæ1033 ö æ 18 öæ1033 ö çè ÷ç ÷ ç ÷ç ÷ 1051 øè1051 ø è1051 øè1051 ø + 11,037 11,034

= - 5.19

b. 95% CI: - 0.0125 < p1 - p2 < - 0.00564; Because the confidence interval limits do not contain 0, there is a significant difference between the two proportions. There is sufficient evidence to warrant rejection of the claim that aspirin has no effect on myocardial infarctions.

( pˆ1 - pˆ 2 ) ± za / 2

æ 139 öæ10,898 ö æ 239 öæ10,795 ö çè11,037 ÷ç øè11,037 ÷ ø çè11,034 ÷ç øè11,034 ÷ ø pˆ1qˆ1 pˆ 2 qˆ2 æ 139 239 ö + =ç ± 1.96 + ÷ è11,037 11,034 ø n1 n2 11,037 11,034

c. It appears that aspirin has an effect. Because the treatment group had a lower rate of myocardial infarctions, the aspirin treatment appears to be associated with a lower rate of myocardial infarctions. It should be noted that the treatment group and placebo group included only male physicians, so the results may or may not apply to the general population.

21. a. H0: p1 = p2 ; H1: p1 < p2 ; population1 = male, population2 = female; Test statistic: z = - 1.17; P-value = 0.1214 (Table: 0.1210); Critical value: z = - 2.33; Fail to reject H 0 . There is not sufficient evidence to support the claim that the rate of left-handedness among males is less than that among females.

23 + 65 11 11 84 = ; q = 1= ; 240 + 520 95 95 95 æ 23 65 ö çè ÷- 0 pˆ1 - pˆ 2 ) - ( p1 - p2 ) ( 240 520 ø z= = = - 1.17 pq pq 11 öæ84 ö æ11 öæ84 ö æ + çè ÷ç ÷ ç ÷ç ÷ n1 n2 95 øè 95 ø è 95 øè 95 ø + 240 520 p=

b. 98% CI: - 0.0848 < p1 - p2 < - 0.0264 (Table: - 0.0849 < p1 - p2 < - 0.0265); Because the confidence interval limits include 0, there does not appear to be a significant difference between the rate of left-handedness among males and the rate among females. There is not sufficient evidence to support the claim that the rate of left-handedness among males is less than that among females.

( pˆ1 - pˆ 2 ) ± za / 2

æ 23 öæ217 ö æ 65 öæ455 ö çè ÷ç ÷ ç ÷ç ÷ pˆ1qˆ1 pˆ 2 qˆ2 æ 23 65 ö 240 øè 240 ø è 520 øè 520 ø + =ç ± 2.33 + ÷ è 240 520 ø n1 n2 240 520

c. The rate of left-handedness among males does not appear to be less than the rate of left-handedness among females.

.

Section 9-1: Two Proportions 123 22. a. H0 : p1 = p2 ; H1: p1 > p2 ; population1 = helicopter, population 2 = ground; Test statistic: z = 10.75; P-value = 0.0000 (Table: 0.0001); Critical value: z = 2.33; Reject H 0 . There is sufficient evidence to support the claim that the rate of fatalities is higher for patients transported by helicopter.

7813 + 17,775 25,588 25,588 197,887 p= = ; q = 1= ; 61,909 + 161,566 223, 475 223, 475 223, 475

( pˆ - pˆ 2 ) - ( p1 - p2 ) = z= 1 pq pq + n1 n2

17,775 ö æ 7813 -0 çè 61,909 - 161,566 ÷ ø æ 25,588 öæ197,887 ö æ 25,588 ö æ197,887 ö çè 223, 475 ÷ç øè 223, 475 ÷ ø çè 223, 475 ÷ ø çè 223, 475 ÷ ø + 61,909 161,566

= 10.75

b. 98% CI: 0.0126 < p1 - p2 < 0.0198; Because the confidence interval limits do not contain 0, there is a significant difference between the two proportions. Because the entire range of values in the confidence interval consists of positive numbers, it appears that the rate of fatalities is higher for patients transported by helicopter. pˆ qˆ pˆ qˆ ( pˆ1 - pˆ 2 ) ± za / 2 1 1 + 2 2 = n1 n2

æ 7813 öæ54,096 ö æ17,775 öæ143,791 ö çè 61,909 ÷ç øè 61,909 ÷ ø çè161,566 ÷ç øè161,566 ÷ ø 17,775 ö æ 7813 ± 2.33 + çè 61,909 161,566 ÷ ø 61,909 161,566 c. The fatality rate for the helicopter sample is 0.126, or 12.6%, and the fatality rate for the ground services sample is 0.110, or 11.0%. The large sample sizes result in a significant difference, but it does not appear that the difference has very much practical significance. Also, it is possible that the most serious of the serious traumatic injuries led to helicopter transportation, and that could partly explain the higher rate of fatalities with helicopter transportation.

z2 1.962 23. n = a /22 = = 2135; The samples should include 2135 men and 2135 women.

2E 2 ×0.032 24. a. The method of this section requires that both samples must have at least 5 successes and 5 failures, but the group not exposed to yawning includes a frequency of 4, which violates that requirement. b. P-value = 0.3729 (Table: 0.3745); which is not close to the P-value of 0.5128 from Fisher’s exact test.

H0: p1 = p2 ; H1: p1 > p2 ; population1 = yawning, population2 = not yawning; 10 + 4 7 7 18 p= = ; q = 1= ; 34 + 16 25 25 25 æ10 4 ö çè ÷- 0 ˆ ˆ p p p p ) ) ( ( 1 2 = 34 16 ø 2 z= 1 = 0.32 pq pq 7 öæ18 ö æ 7 öæ18 ö æ + çè ÷ç ÷ ç ÷ç ÷ n1 n2 25 øè 25 ø è 25 øè 25 ø + 34 16 25. a. 95% CI: 0.0227 < p1 - p2 < 0.217; population1 = first sample, population 2 = second sample; Because the confidence interval limits do not contain 0, it appears that p1 = p2 can be rejected.

( pˆ1 - pˆ 2 ) ± za / 2

æ112 öæ 88 ö æ 88 öæ112 ö çè ÷ç ÷ ç ÷ç ÷ pˆ1qˆ1 pˆ 2 qˆ2 æ112 88 ö 200 øè 200 ø è 200 øè 200 ø + =ç ± 1.96 + ÷ è 200 200 ø n1 n2 200 200

.

124

Chapter 9: Inferences from Two Samples

25. (continued) b. First sample: 95% CI: pˆ ± za / 2

ˆ ˆ 112 pq = ± 1.96 n 200

88 ( 112 200 )( 200 ) Þ 0.491 < p < 0.629 1

200

( 88 )( 112 )

ˆˆ pq 88 = ± 1.96 200 200 Þ 0.371 < p2 < 0.509 n 200 200 Because the confidence intervals do overlap, it appears that p1 = p2 cannot be rejected.

Second sample: 95% CI: pˆ ± za / 2

c. H0 : p1 = p2 ; H1: p1 ¹ p2 ; population1 = first sample, population2 = second sample; Test statistic: z = 2.40; P-value = 0.0164; Critical values: z = ±1.96; Reject H 0 . There is sufficient evidence to reject p1 = p2 .

112 + 88 1 1 1 = ; q = 1- = ; 200 + 200 2 2 2 æ112 88 ö çè ÷- 0 pˆ1 - pˆ 2 ) - ( p1 - p2 ) ( 200 200 ø z= = = 2.40 pq pq 1 öæ1 ö æ1 öæ1 ö æ + çè ÷ç ÷ ç ÷ç ÷ n1 n2 2 øè 2 ø è 2 øè 2 ø + 200 200 p=

d. Reject p1 = p2 . The least effective method is using the overlap between the individual confidence intervals.

26. Hypothesis test: H0 : p1 = p2 ; H1: p1 ¹ p2 ; population1 = first sample, population2 = second sample; Test statistic: z = - 1.9615; P -value = 0.0498 (Table: 0.05); Critical values: z = ±1.96; Reject H 0 . There is sufficient evidence to reject p1 = p2 .

10 + 1404 7 7 3 = ; q = 1= ; 20 + 2000 10 10 10 æ10 1404 ö çè ÷- 0 pˆ1 - pˆ 2 ) - ( p1 - p2 ) ( 20 2000 ø z= = = - 1.9615 pq pq 7 öæ 3 ö æ 7 öæ 3 ö æ + çè ÷ç ÷ ç ÷ç ÷ n1 n2 10 øè10 ø è10 øè10 ø + 20 2000 95% CI: - 0.422 < p1 - p2 < 0.0180; population1 = first sample, population 2 = second sample; which suggests that we should not reject p1 = p2 (because 0 is included). p=

( pˆ1 - pˆ 2 ) ± za / 2

æ10 öæ10 ö æ1404 öæ 596 ö çè ÷ç ÷ ç ÷ç ÷ ˆp1qˆ1 pˆ 2 qˆ2 æ10 1404 ö 20 øè 20 ø è 2000 øè 2000 ø + =ç ± 1.96 + è 20 2000 ÷ ø n1 n2 20 2000

The hypothesis test and confidence interval lead to different conclusions about the equality of p1 = p2 .

Section 9-2: Two Means: Independent Samples 1. Only part (b) describes independent samples.

2.

a. Because the confidence interval does not include 0, it appears that there is a significant difference between the mean level of hemoglobin in women and the mean level of hemoglobin in men. b. We have 95% confidence that the interval from - 1.76 g dL to - 1.62 g dL actually contains the value of the difference between the two population means ( m1 - m2 ) . c. 1.62 g dL < m1 - m2 < 1.76 g dL

3.

a. yes b. yes

c. 98%

.

Section 9-2: Two Means: Independent Samples

125

4. The critical value of t = 1.685 are more conservative than t = 1.665 in the sense that rejection of the null hypothesis requires a greater difference between the sample means. The sample evidence must be stronger with t = 1.685.

5. H0 : m1 = m2 ; H1: m1 ¹ m2 ; population1 = low humidity, population2 = high humidity; Test statistic: t = - 0.452; P-value = 0.6521 (Table: P-value > 0.20); Critical values: t = ±1.987 (Table: t = ±2.014); Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that the two groups are from populations with the same mean. The result suggests that the increased humidity does not help in the treatment of croup. ( x - x ) - ( m1 - m2 ) = (0.98 - 1.09 ) - 0 = - 0.452 (df = 45) t= 1 2 s12 s22 1.222 1.112 + + 46 46 n1 n2 6. H0 : m1 = m2 ; H1: m1 ¹ m2 ; population1 = echinacea, population 2 = placebo; Test statistic: t = - 6.794; P-value = 0.0000 (Table: P-value < 0.005); Critical values: t = ±1.964 (Table: t = ±1.960); Reject H 0 . There is sufficient evidence to support the claim that echinacea has an effect on “days of fever.” ( x - x ) - ( m1 - m2 ) = (0.81 - 1.50) - 0 = - 6.794 (df = 336) t= 1 2 s12 s22 1.502 1.162 + + 337 370 n1 n2

7.

a. H0: m1 = m2 ; H1: m1 < m2 ; population1 = exposed to cocaine, population2 = not exposed to cocaine; Test statistic: t = - 2.908; P-value = 0.0019 (Table: P-value < 0.005); Critical value: t = - 1.649 (Table: t » - 1.653); Reject H 0 . There is sufficient evidence to support the claim that the children exposed to cocaine have a lower mean score. ( x - x ) - ( m1 - m2 ) = (7.3 - 8.2) - 0 = - 2.908 (df = 185) t= 1 2 s12 s22 3.02 3.02 + + 190 186 n1 n2 b. 90% CI: - 1.4 < m1 - m2 < - 0.4; The confidence interval consists of negative numbers only and does not include 0. There is sufficient evidence to support the claim that the children exposed to cocaine have a lower mean score.

( x1 - x2 ) ± ta / 2 8.

s12 s22 3.02 3.02 + = (7.3 - 8.2) ± 1.653 + (df = 185) n1 n2 190 186

a. H0: m1 = m2 ; H1: m1 > m2 ; population1 = magnet treatment, population2 = sham treatment; Test statistic: t = 0.132; P -value = 0.4480 (Table: P-value > 0.10); Critical value: t = 1.691 (Table: t = 1.729); Fail to reject H 0 . There is not sufficient evidence to support the claim that the magnets are effective in reducing pain. ( x - x ) - ( m1 - m2 ) = (0.49 - 0.44 ) - 0 = 0.132 (df = 19) t= 1 2 s12 s22 0.962 1.42 + + 20 20 n1 n2 b. 90% CI: - 0.59 < m1 - m2 < 0.69 (Table: - 0.61 < m1 - m2 < 0.71);

( x1 - x2 ) ± ta / 2

s12 s22 0.962 1.42 + = (0.49 - 0.44 ) ± 1.729 + (df = 19) n1 n2 20 20

c. Magnets do not appear to be effective in treating back pain. It is valid to argue that the magnets might appear to be effective if the sample sizes were larger. .

126 9.

Chapter 9: Inferences from Two Samples a. H0 : m1 = m2 ; H1: m1 > m2 ; population1 = king-size, population2 = 100-mm; Test statistic: t = 5.004; P-value = 0.0000 (Table: P-value < 0.005); Critical value: t = 1.678 (Table: t = 1.711); Reject H 0 . There is sufficient evidence to support the claim that the unfiltered king-size cigarettes have a mean nicotine content greater than that of 100-mm cigarettes. The result suggests that the filters are effective in reducing nicotine content, assuming that both types of cigarettes are about the same size. ( x - x ) - (m1 - m2 ) (1.26 - 0.92) - 0 t= 1 2 = = 5.004 (df = 24) s12 s22 0.232 0.252 + + 25 25 n1 n2 b. 90% CI: 0.22 mg < m1 - m2 < 0.46 mg (Table: 0.23 mg < m1 - m2 < 0.46 mg); The confidence interval includes positive numbers only, which suggests that the mean nicotine content of unfiltered king-size cigarettes is greater than the mean for 100-mm cigarettes.

( x1 - x2 ) ± ta /2

s12 s22 0.232 0.252 + = (1.26 - 0.92) ± 1.711 + (df = 24) n1 n2 25 25

10. H0 : m1 = m2 ; H1: m1 ¹ m2 ; population1 = paroxetine, population2 = placebo; Test statistic: t = - 1.334; P -value = 0.1866 (Table: 0.10 < P-value < 0.20); Critical values: t = ±2.159 (Table: t = ±2.023); Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim the paroxetine and placebo groups have the same mean depression scale scores. Paroxetine does not appear to be an effective treatment for depression. ( x - x ) - ( m1 - m2 ) = (20.38 - 21.57) - 0 = - 1.334 (df = 42) t= 1 2 s12 s22 3.912 3.872 + + 33 43 n1 n2 11. a. H0 : m1 = m2 ; H1: m1 ¹ m2 ; population1 = treatment, population2 = placebo; Test statistic: t = 5.045; P-value = 0.0000 (Table: P-value < 0.01); Critical values: t = ±2.058 (Table: t = ±2.080); Reject H 0 . There is sufficient evidence to warrant rejection of the claim that there is a significant different between the treatment and control groups. We cannot conclude that the cause is due to the treatment. ( x - x ) - ( m1 - m2 ) = (4.20 - 1.71) - 0 = 5.045 (df = 21) t= 1 2 s12 s22 2.202 0.722 + + 22 22 n1 n2 b. 95% CI: 1.47 < m1 - m2 < 3.51 (Table: 1.46 < m1 - m2 < 3.52); Because the interval does not include 0, there appears to be a significant difference between the two population means. It does appear that there are significantly more errors made by those treated with alcohol.

( x1 - x2 ) ± ta / 2

s12 s22 2.202 0.722 + = (4.20 - 7.71) ± 2.080 + (df = 21) n1 n2 22 22

12. a. H0 : m1 = m2 ; H1: m1 < m2 ; population1 = heavy users, population2 = light users; Test statistic: t = - 2.790; P -value = 0.0031 (Table: P-value < 0.005); Critical value: t = - 2.357 (Table: t = - 2.390); Reject H 0 . There is sufficient evidence to support the claim that the heavy users of marijuana have a lower mean of items sorted than light marijuana users. ( x - x ) - ( m1 - m2 ) = (51.3 - 53.3) - 0 = - 2.790 (df = 63) t= 1 2 s12 s22 4.52 3.62 + + 65 64 n1 n2

.

Section 9-2: Two Means: Independent Samples

127

12. (continued) b. 98% CI: - 3.96 < m1 - m2 < - 0.310 (Table: - 3.713 < m1 - m2 < - 0.286); Because the confidence interval includes only negative numbers, there is sufficient evidence to support the claim that the heavy users of marijuana have a lower mean of items sorted than light marijuana users.

( x1 - x2 ) ± ta / 2

s12 s22 4.52 3.62 + = (51.3 - 53.3) ± 2.390 + (df = 63) n1 n2 65 64

13. a. H0: m1 = m2 ; H1: m1 > m2 ; population1 = exposed, population2 = not exposed; Test statistic: t = 1.845; P-value = 0.0363 (Table: P-value < 0.05); Critical value: t = 2.419 (Table:

t = 2.426); Reject H 0 . There is not sufficient evidence to support the claim that the nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke. ( x - x ) - (m1 - m2 ) (60.58 - 16.35) - 0 t= 1 2 = = 1.845 (df = 39) s12 s22 138.082 62.532 + + 40 40 n1 n2 b. 98% CI: 0 mg mL < m1 - m2 < 102.21 mg mL (Table: 0 mg mL < m1 - m2 < 102.37 mg mL);

( x1 - x2 ) ± ta /2

s12 s22 138.082 62.532 + = (60.58 - 16.35) ± 2.426 + (df = 39) n1 n2 40 40

c. Exposure to second-hand smoke does not appear to have the effect of being associated with extremely great nicotine than for those not exposed to second hand smoke.

14. a. H0 : m1 = m2 ; H1: m1 ¹ m2 ; population1 = female, population 2 = male; Test statistic: t = 0.674; P -value = 0.5015 (Table: P-value > 0.20); Critical values: t = ±1.979 (Table: t = ±1.995); Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that the females and the males have the same BMI. ( x - x ) - (m1 - m2 ) (29.29 - 28.31) - 0 t= 1 2 = = 0.983 (df = 79) s12 s22 7.292 5.292 + + 80 85 n1 n2 b. 99% CI: - 1.65 < m1 - m2 < 3.61 (Table: - 1.68 < m1 - m2 < 3.62); Because the confidence interval includes 0,there is not sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean.

( x1 - x2 ) ± ta /2

s12 s22 7.292 5.292 + = (29.29 - 28.31) ± 2.64 + (df = 79) n1 n2 80 85

c. Based on the available sample data, it appears that males and females have the same mean BMI, but we can only conclude that there isn’t sufficient evidence to say that they are different.

15. a. H0 : m1 = m2 ; H1: m1 > m2 ; population1 = low lead, population 2 = high lead; Test statistic: t = 1.063; P-value = 0.1502 (Table: P-value > 0.05); Critical value: t = 2.502 (Table:

t = 2.528); Reject H 0 . There is not sufficient evidence to support the claim that the verb IQ score of people with low blood lead levels is higher than the verb IQ score of people with high blood lead levels. ( x - x ) - (m1 - m2 ) (85.14103 - 82.52381) - 0 t= 1 2 = = 1.063 (df = 20) s12 s22 14.686092 8.316362 + + 78 21 n1 n2

.

128

Chapter 9: Inferences from Two Samples

15. (continued) b. 98% CI: - 3.6 < m1 - m2 < 8.8 (Table: - 3.6 < m1 - m2 < 8.7);

( x1 - x2 ) ± ta /2

s12 s22 14.686092 8.316362 + = (85.14103 - 82.52381) ± 2.528 + (df = 20) n1 n2 78 21

c. No, it does not appear that exposure to lead has an effect on verb IQ scores.

16. a. H0 : m1 = m2 ; H1: m1 < m2 ; population1 = wearing seatbelts, population2 = not wearing seatbelts; Test statistic: t = - 2.330 P-value = 0.0102 (Table: P-value < 0.025); Critical value: t = - 1.649; (Table: t = - 1.660); Reject H 0 . There is sufficient evidence to support the claim that children wearing seat belts have a lower mean length of time in an ICU than do children not wearing seat belts. ( x - x ) - ( m1 - m2 ) = (0.83 - 1.39) - 0 = - 2.330 (df = 122) t= 1 2 s12 s22 1.772 3.062 + + 123 290 n1 n2 b. 90% CI: - 0.96 day < m1 - m2 < - 0.16 day;

( x1 - x2 ) ± ta / 2

s12 s22 1.772 3.062 + = (0.83 - 1.39) ± 1.657 + (df = 122) n1 n2 123 290

c. It appears that after motor vehicle crashes, children wearing seat belts spend less time in intensive care units than children who don’t wear seat belts. Children should wear seat belts (except for young children who should use properly installed car seats)!

17. a. H0 : m1 = m2 ; H1: m1 ¹ m2 ; population1 = Disney movies, population2 = other movies; Test statistic: t = 1.520; P -value = 0.15; Critical values: t = ±2.020 (Table: t = ±2.120); Fail to reject H 0 .

There is not sufficient evidence to warrant rejection of the claim that Disney animated children^' s movies and other animated children’s movies have the same mean time showing alcohol use. ( x - x ) - (m1 - m2 ) (40.4 - 17.1) - 0 t= 1 2 = = 1.520 (df = 16) s12 s22 78.32 28.92 + + 33 17 n1 n2 b. 95% CI: - 7.66 sec < m1 - m2 < 54.3 sec (Table: - 9.19 sec < m1 - m2 < 55.8 sec)

( x1 - x2 ) ± ta /2

s12 s22 78.32 28.92 + = (40.4 - 17.1) ± 2.120 + (df = 16) n1 n2 33 17

c. The times appear to be from a population with a distribution that is not normal (the sample is right-skewed), but the methods in this section are robust against departures from normality. (Results obtained by using other methods confirm that the results obtained here are quite good, even though the non-Disney times appear to violate the normality requirement.)

18. a. H0 : m1 = m2 ; H1: m1 > m2 ; population1 = Pennsylvania, population2 = New York; Test statistic: t = 3.265; P-value = 0.0024 (Table: P-value > 0.005); Critical value: t = 1.746 (Table: t = 1.796); Reject H 0 . There is sufficient evidence to support the claim that the mean amount of strontium-90 from Pennsylvania residents is greater than the mean amount from New York residents. ( x - x ) - ( m1 - m2 ) = (147.6 - 136.4) - 0 = 3.265 (df = 11) t= 1 2 s12 s22 10.62 5.22 + + 12 12 n1 n2

.

Section 9-2: Two Means: Independent Samples

129

18. (continued) b. 95% CI: 3.917 mBq g < m1 - m2 < 18.417 mBq g

( x1 - x2 ) ± ta / 2

s12 s22 10.62 5.22 + = (147.6 - 136.4) ± 1.796 + (df = 11) n1 n2 12 12

19. a. H0 : m1 = m2 ; H1: m1 < m2 ; population1 = popes, population 2 = monarchs; Test statistic: t = - 1.810; P-value = 0.0442 (Table: P-value > 0.025); Critical value: t = - 2.574 (Table: t = - 2.650); Fail to reject H 0 . There is not sufficient evidence to support the claim that the mean longevity for popes is less than the mean for British monarchs after coronation ( x - x ) - ( m1 - m2 ) = (13.1 - 11.7) - 0 = - 1.810 (df = 13) t= 1 2 s12 s22 8.962 18.602 + + 24 14 n1 n2

b. 98% CI: - 23.6 years < m1 - m2 < 4.0 years (Table: - 23.6 years < m1 - m2 < 4.4 years)

( x1 - x2 ) ± ta / 2

s12 s22 8.962 18.602 + = (13.1 - 11.7) ± 1.796 + (df = 13) n1 n2 24 14

20. H0 : m1 = m2 ; H1: m1 ¹ m2 ; population1 = easy to difficult, population2 = difficult to easy; Test statistic: t = - 2.657; P -value = 0.0114 (Table: P-value < 0.02); Critical values (a = 0.05): t = ±2.023 (Table: t = ±2.131); The conclusion depends on the choice of the significance level. There is a significant difference between the two population means at the 0.05 significance level, but not at the 0.01 significance level. ( x - x ) - ( m1 - m2 ) = (27.115 - 31.728) - 0 = - 2.657 (df = 15) t= 1 2 s12 s22 6.8572 4.2602 + + 25 16 n1 n2 21. H0 : m1 = m2 ; H1: m1 ¹ m2 ; population1 = female, population 2 = male; Test statistic: t = - 0.863; P -value = 0.3887 (Table: P-value > 0.20); Critical values: t = ±1.968 (Table: t = ±1.984); Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that women and men have the same mean diastolic blood pressure. ( x - x ) - ( m1 - m2 ) = (70.16 - 71.32) - 0 = - 0.863 (df = 146) t= 1 2 s12 s22 11.222 11.992 + + 147 153 n1 n2 22. H0 : m1 = m2 ; H1: m1 < m2 ; population1 = girls, population2 = boys; Test statistic: t = - 3.450; P -value = 0.0003 (Table: P-value < 0.005); Critical value (a = 0.05): t = - 1.649; (Table: t = - 1.653); Reject H 0 . There is sufficient evidence to support the claim that at birth, girls have a lower mean weight than boys. ( x - x ) - ( m1 - m2 ) = (3037.07 - 3272.82) - 0 = - 3.450 (df = 194) t= 1 2 s12 s22 706.2682 660.1542 + + 205 195 n1 n2

.

130

Chapter 9: Inferences from Two Samples

23. a. H0 : m1 = m2 ; H1: m1 > m2 ; population1 = low lead, population 2 = high lead; Test statistic: t = 0.782; P-value = 0.1632 (Table: P-value > 0.05); Critical value: t = 1.661 (Table: t = 1.987); df = 78 + 21 - 2 = 97; Accept H 0 . There is not sufficient evidence to warrant rejection of the claim that the verb IQ score of people with low blood lead levels is higher than the verb IQ score of people with high blood lead levels. (n - 1) s12 + (n2 - 1) s22 (78 - 1)14.686092 + (21 - 1)8.316362 s 2p = 1 = = 185.471; (78 - 1) + (21 - 1) (n1 - 1) + (n2 - 1)

t=

( x1 - x2 ) - (m1 - m2 ) (85.14103 - 82.52381) - 0 s 2p n1

+

=

s 2p n2

185.471 185.471 + 78 21

= 0.782 (df = 97)

b. 98% CI: - 2.94 < m1 - m2 < 8.18;

( x1 - x2 ) ± ta /2

s12 s22 185.471 185.471 + = (85.14103 - 82.52381) ± 1.661 + (df = 97) n1 n2 78 21

c. Yes, it does appear that exposure to lead has an effect on verb IQ scores. With pooling, df increases dramatically to 97, but the test statistic decreases from 1.063 to 0.782 (because the estimated standard deviation increases from 2.563814 to 3.181429), the P-value increases to 0.1632, and the 98% confidence interval becomes wider. With pooling, these results do not show greater significance.

24. df = 38.9884; Using “ df = smaller of n1 - 1 and n2 - 1 ” is a more conservative estimate of the number of degrees of freedom (than the estimate obtained with Formula 9-1) in the sense that the confidence interval is wider, so the difference between the sample means needs to be more extreme to be considered a significant difference.

2

2

(s12 n1 + s22 n2 ) = (6.8572 25 + 4.2602 16) = 38.9884 df = s12 n1 s22 n2 + n1 - 1 n2 - 1

6.8572 25 4.2602 16 + 25 - 1 16 - 1

25. H0 : m1 = m2 ; H1: m1 ¹ m2 ; population1 = treatment, population2 = placebo; Test statistic: t = 15.322; P-value = 0.0000 (Table: P-value < 0.01); Critical values: t = ±2.080; Reject H 0 .

There is sufficient evidence to warrant rejection of the claim that the two populations have the same mean.

(n1 - 1) s12 + (n2 - 1) s22 = (22 - 1) 0.0152 + (22 - 1) 02 = 0.000125; (22 - 1) + (22 - 1) (n1 - 1) + (n2 - 1) ( x - x ) - ( m1 - m2 ) = (0.049 - 0.000) - 0 = 15.322 t= 1 2 s 2p =

s 2p

n1

+

s 2p

n2

0.000125 0.000125 + 22 22

Section 9-3: Two Dependent Samples (Matched Pairs) 1. Only parts (a) and (c) are true.

2.

d = - 1.26°F and sd = 0.63°F; md represents the mean of the differences from the population of paired data.

Temperature ( °F ) at 8 AM Temperature ( °F ) at 12 AM Difference ( °F )

3.

The results will be the same.

4.

df = n - 1 = 5 - 1 = 4; ta / 2 = 4.604

97.0 97.7 −0.7

98.0 98.8 −0.8

.

97.0 98.0 −1.0

96.4 98.0 −1.6

96.1 98.3 −2.2

Section 9-3: Two Dependent Samples (Matched Pairs)

131

H0 : md = 0; H1: md ¹ 0; difference = right - left; Test statistic: t = - 17.339; P-value = 0.0001 (Table: P-value < 0.01); Critical values: t = ±4.604; Reject H 0 .

There is sufficient evidence to support the claim of a difference in measurements between the two arms. The right and left arms should yield the same measurements, but the given data show that this is not happening for this person. d - md - 72.2 - 0 t= = = - 2.609 (df = 4) sd n 9.31128 5

5.

6.

a. H0 : md = 0 cm; H1: md > 0 cm; difference = president - opponent; Test statistic: t = 1.304; P -value = 0.1246 (Table: P-value > 0.10); Critical value: t = 2.105; Fail to reject H 0 .

There is not sufficient evidence to support the claim that for the population of heights of presidents and their main opponents, the differences have a mean greater than 0 cm (or that presidents tend to be taller than their opponents). d - md 3.67 - 0 t= = = 1.304 (df = 5) sd n 6.890 6 b. 90% CI: - 2.0 cm < md < 9.3 cm; The confidence interval includes 0, so it is possible that md = 0.

d ± ta / 2 7.

sd 6.890 = 3.67 ± 2.015 (df = 5) n 6

a. H0: md = 0°F; H1: md ¹ 0°F; difference = 8 AM - 12 AM; Test statistic: t = - 2.830; P-value = 0.0300 (Table: P-value = 0.0340); Critical values: t = ±1.493. Reject

H 0 . There is sufficient evidence to warrant rejection of the claim that there is no difference between body temperatures measured at 8AM and at 12AM. There appears to be a difference. d - md - 0.8 - 0 t= = = - 2.830 (df = 6) sd n 0.748 7 b. 90% CI: - 1.35°F < md < - 0.25°F; The confidence interval consists of negative numbers only and does not include 0. s 0.748 d ± ta /2 d = - 0.8 ± 1.943 (df = 6) n 7 8.

a. H0: md = 0 words; H1: md < 0 words; difference = male - female; Test statistic: t = - 0.472; P -value = 0.3265 (Table: P-value > 0.10); Critical value: t = - 1.895; Fail to reject H 0 . There is not sufficient evidence to support the claim that among couples, males speak fewer words in a day than females. d - md - 1677.75 - 0 t= = = - 0.472 (df = 7) sd n 10,052.87 8 b. 90% CI: - 8411.5 words < md < 5056.0 words (Table: - 8413.0 words < md < 5057.5 words); The confidence interval includes 0 word, so it is possible that md = 0 word.

d ± ta / 2

sd 10,052.87 = - 1677.75 ± 1.895 (df = 7) n 8

.

132

Chapter 9: Inferences from Two Samples

H0 : md = 0 in.; H1: md ¹ 0 in.; difference = father - son; Test statistic: t = 1.267; P-value = 0.2370 (Table: P-value > 0.20); Critical values: t = ±1.833. Fail to reject H 0 .

There is not sufficient evidence to warrant rejection of the claim that there is no difference in heights between fathers and their first sons. d - md - 0.85 - 0 t= = = - 1.267 (df = 9) sd n 2.122 10

9.

10. H0 : md = 0 in.; H1: md ¹ 0 in.; difference = mother - daughter; Test statistic: t = - 2.323; P-value = 0.2370 (Table: P-value > 0.20); Critical values: t = ±1.833. Reject H 0 .

There is not sufficient evidence to warrant rejection of the claim that there is no difference in heights between mothers and their first daughters. d - md - 1.52 - 0 t= = = - 2.323 (df = 9) sd n 2.069 10 11. 95% CI: - 6.5 admissions < md < - 0.2 admissions; Because the confidence interval does not include 0 admission, it appears that there is sufficient evidence to warrant rejection of the claim that when the 13th day of a month falls on a Friday, the numbers of hospital admissions from motor vehicle crashes are not affected. Hospital admissions do appear to be affected. s 3.011 d ± ta / 2 d = - 3.333 ± 2.571 (difference = 6th - 13th, df = 5) n 6 12. H0 : md = 0; H1: md > 0; difference = before - after; Test statistic: t = 6.371; P-value = 0.0000 (Table: P-value < 0.005); Critical value: t = 2.718; Reject H 0 . There is sufficient evidence to support the claim that captopril is effective in lowering systolic blood pressure. d - md 18.58 - 0 t= = = 6.371 (df = 11) sd n 10.103 12 13. 99% CI: - 66.7 cm3 < md < 49.7 cm3 (Table: - 66.8 cm3 < md < 49.8 cm3 ); Because the confidence interval includes the mean of the differences could be equal to 0 cm3 , so there does not appear to be a significant difference. s 56.679 d ± ta / 2 d = - 8.5 ± 3.249 (difference = first born - second born, df = 9) n 10 14. 95% CI: 0.69 < md < 5.66; Because the confidence interval limits do not contain 0 and they consist of positive values only, it appears that the “before” measurements are greater than the “after” measurements, so hypnotism does appear to be effective in reducing pain. s 2.911 d ± ta / 2 d = 3.125 ± 2.365 (difference = before - after, df = 7) n 8 15. 99% CI: - 4.16 in. < md < 2.16 in.; Because the confidence interval limits contain 0, there is not sufficient evidence to support a claim that there is a difference between self-reported heights and measured heights. We might believe that males would tend to exaggerate their heights, but the given data do not provide enough evidence to support that belief. s 3.520 d ± ta / 2 d = - 1.0 ± 3.106 (difference = reported - measured, df = 11) n 12

.

Section 9-3: Two Dependent Samples (Matched Pairs)

133

16. a. H0 : md = 0 cm; H1: md > 0 cm; difference = regular - kiln dried; Test statistic: t = - 1.532; P -value = 0.1565 (Table: P-value > 0.10); Critical values: t = ±2.228; Fail to reject H 0 . There is not sufficient evidence to reject the claim that the mean yields for the two types of seed are different. d - md - 1.091 - 0 t= = = - 1.532 (df = 10) sd n 2.362 11 b. 90% CI: - 2.678 cwt acre < md < 0.496 cwt acre;

d ± ta / 2

sd 2.362 = - 1.091 ± 2.228 (df = 10) n 11

c. It does not appear that either type of seed is better.

17. a. H0: md = 0°F; H1: md ¹ 0°F; difference = 8 AM - 12 AM; Test statistic: t = - 8.485; P-value = 0.0000 (Table: P-value < 0.01); Critical values: t = ±1.996 (Table: t = ±1.994); Reject H 0 . There is sufficient evidence to warrant rejection of the claim of no difference between body temperatures measured at 8 am and at 12 am. There appears to be a difference. d - md - 0.85 + 0 t= = = - 8.485 (df = 68) sd n 0.833 69 b. 95% CI: - 1.05°F < md < - 0.65°F; The confidence interval consists of negative numbers only and does not include 0. s 0.833 d ± ta / 2 d = - 0.85 ± 1.994 (df = 68) n 69 18. H0: md = 0 in.; H1: md ¹ 0 in.; difference = mother - daughter; Test statistic: t = - 4.090; P-value = 0.0001 (Table: P-value < 0.01); Critical values: t = ±1.978 (Table: t » ±1.974); Reject H 0 . There is sufficient evidence to warrant rejection of the claim of no difference in heights between mothers and their first daughters. d - md - 0.93 - 0 t= = = - 4.090 (df = 133) sd n 2.636 134

19. H0 : md = 0 in.; H1: md ¹ 0 in.; difference = father - son; Test statistic: t = - 6.347; P-value = 0.0000 (Table: P-value < 0.01); Critical values: t = ±1.978 (Table: t » ±1.984); Reject H 0 . There is sufficient evidence to warrant rejection of the claim of no difference in heights between fathers and their first sons. d - md - 1.366 - 0 t= = = - 6.347 (df = 133) sd n 2.491 134 20. H0 : md = 0; H1: md > 0; difference = tobacco - alcohol; Test statistic: t = 1.617; P -value = 0.0561 (Table: 0.05 < P-value < 0.10); Critical value: 1.677; Fail to reject

H 0 . There is not sufficient evidence to support the claim that more movie time is devoted to tobacco use than alcohol use. d - md 24.98 - 0 t= = = 1.617 (df = 49) sd n 109.219 50

.

134

Chapter 9: Inferences from Two Samples

21. For the temperatures in degrees Fahrenheit and the temperatures in degrees Celsius, the test statistic of t = 0.124 is the same, the P-value of 0.9023 is the same, the critical values of t = ±2.028 are the same, and the conclusions are the same, so the hypothesis test results are the same in both cases. The confidence intervals are - 0.25°F < md < 0.28°F and - 0.14°C < md < 0.16°C. The confidence interval limits of - 0.14°C and 0.16°C have numerical values that are 5 9 of the numerical values of - 0.25°F and 0.28°F. 22. a. No, it is possible that values of x are not normally distributed and values of y are not normally distributed, but values of x - y are normally distributed. b. Answers vary, but here is a typical result: 95% CI: - 0.24°F < md < 0.26°F; which is reasonably close to the confidence interval found in Exercise 25.

Section 9-4: Two Variances or Standard Deviations 1. a. No, the numerator will always be larger than the denominator in the fraction. b. No, both variances are nonnegative, so their quotient cannot be negative. c. The two samples have standard deviations (or variances) that are very close in value. d. skewed right 2.

a. s12 = 436.20829 pounds2 and s22 = 311.57192 pounds2 b. H0 : s 1 = s 2 ; H1: s 1 ¹ s 2 ; population1 = women, population2 = men c. F = s12 s22 = 20.88562 17.65142 = 1.4000 d. Fail to reject H 0 . There is not sufficient evidence to support the claim that weights of men and weights of women have different variances.

3. No, unlike some other tests that have a requirement that samples must be from normally distributed populations or the samples must have more than 30 values, the F test has a requirement that the samples must be from normally distributed populations, regardless of how large the samples are.

4. The F test is very sensitive to departures from normality, which means that it works poorly by leading to wrong conclusions when either or both of the populations have a distribution that is not normal.

5.

H0: s 1 = s 2 ; H1: s 1 > s 2 ; population1 = treatment, population 2 = placebo;

Test statistic: F = s12 s22 = 2.202 0.722 = 9.3364; P-value = 0.0000; Critical value: F = 2.0842 (Table: 2.0540 < F < 2.0960); Reject H 0 . There is sufficient evidence to support the claim that the treatment group has errors that vary more than the errors of the placebo group.

6.

a. H0 : s 1 = s 2 ; H1: s 1 > s 2 ; population1 = exposed, population 2 = not exposed; Test statistic: F = s12 s22 = 119.502 62.532 = 3.6552; P-value = 0.0000; Critical value: F = 1.7045 (Table: 1.8752 < F < 2.0739); Fail to reject H 0 . There is sufficient evidence to support the claim that the variation of cotinine in smokers is greater than the variation of cotinine in nonsmokers not exposed to tobacco smoke. b. The sample is not from a normally distributed population as required, so the results in part (a) are highly questionable.

7.

H0 : s 1 = s 2 ; H1: s 1 ¹ s 2 ; population1 = Atkins, population 2 = Zone;

Test statistic: F = s12 s22 = 122 92 = 1.7778; P-value = 0.0762; Upper critical value: F = 1.8907 (Table: 1.8752 < F < 2.0739); Reject H 0 . There is not sufficient evidence to warrant rejection of the claim that subjects from both treatment groups have ages with the same amount of variation. If treatment groups have different characteristics, comparisons of treatments become unclear, because differences might be due to the treatments or they might be due to the different group characteristics.

.

Section 9-4: Two Variances or Standard Deviations 8.

135

H0 : s 1 = s 2 ; H1: s 1 > s 2 ; population1 = low lead, population 2 = high lead;

Test statistic: F = s12 s22 = 14.686092 8.316362 = 3.1185; P -value = 0.0029; Critical value: F = 2.1941 (Table:

2.1562 < F < 2.2234); Reject H 0 . There is sufficient evidence to support the claim that verb IQ scores of people with low lead levels vary more than verb IQ scores of people with high lead levels.

9.

H0: s 1 = s 2 ; H1: s 1 > s 2 ; population1 = sham treatment, population2 = magnet treatment;

Test statistic: F = s12 s22 = 1.42 0.962 = 2.1267; P-value = 0.0543; Critical value: F = 2.1682 (Table: 2.1555 < F < 2.2341); Fail to reject H 0 . There is not sufficient evidence to support the claim that those given a sham treatment have pain reductions that vary more than the pain reductions for those treated with magnets.

10. H0 : s 1 = s 2 ; H1: s 1 ¹ s 2 ; population1 = low humidity, population2 = high humidity; Test statistic: F = s12 s22 = 1.222 1.112 = 1.2080; P-value = 0.5288; Upper critical value: F = 2.4300 (Table: 1.6668 < F < 1.8752); Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that the low humidity and high humidity groups have the same variation.

11. H0 : s 1 = s 2 ; H1: s 1 > s 2 ; population1 = filtered, population 2 = non-filtered; Test statistic: F = s12 s22 = 0.312 0.162 = 3.7539; P-value = 0.0400; Critical value: F = 3.4445; Reject H 0 .

There is sufficient evidence to support the claim that king-size cigarettes with filters have amounts of nicotine that vary more than the amounts of nicotine in non-filtered king-size cigarettes.

12. H0 : s 1 = s 2 ; H1: s 1 ¹ s 2 ; population1 = placebo, population 2 = zinc; Test statistic: F = s12 s22 = 7282 6692 = 1.1842; P-value = 0.3738; Upper critical value: F = 2.8621 (Table: 2.7559 < F < 2.8621); Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that the treatment group and placebo group have the same variation.

13. H0 : s 1 = s 2 ; H1: s 1 > s 2 ; population1 = men, population2 = women; Test statistic: F = s12 s22 = 0.622 0.662 = 0.8825; P -value = 0.0346; Critical value: F = 1.6584 (Table:

1.6428 < F < 1.7215); Fail to reject H 0 . There is not sufficient evidence to support the claim that men have body temperatures that vary more than the body temperatures of women.

14. H0 : s 1 = s 2 ; H1: s 1 > s 2 ; population1 = Pennsylvania, population2 = New York; Test statistic: F = s12 s22 = 10.6382 0.5.2132 = 4.1643 P-value = 0.0130; Critical value: F = 2.8179 (Table: 2.7876 < F < 2.8536); Reject H 0 . There is sufficient evidence to support the claim that the amount of strontium90 for Pennsylvania residents has greater variation than the amount of strontium-90 for New York residents.

15. H0 : s 1 = s 2 ; H1: s 1 ¹ s 2 ; population1 = monarchs, population2 = popes; Test statistic: F = s12 s22 = 18.6032 8.9602 = 4.3103; P-value = 0.0023; Upper critical value: F = 2.5308 (Table: 2.4665 < F < 2.5699); Reject H 0 . There is sufficient evidence to warrant rejection of the claim that both populations of longevity times have the same variation.

16. H0 : s 1 = s 2 ; H1: s 1 ¹ s 2 ; population1 = easy to difficult, population2 = difficult to easy; Test statistic: F = s12 s22 = 6.8572 4.2602 = 2.5908; P-value = 0.0599; Upper critical value: F = 2.7006; Fail to reject H 0 . There is not sufficient evidence to support a claim that the two populations of scores have different amounts of variation.

.

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Chapter 9: Inferences from Two Samples

17. a. Calculations not shown. b. c1 = 3, c2 = 0 c. Critical value =

log (0.05 / 2 ) = 7.4569 æ 25 ö log ç è 25 + 16 ÷ ø

d. c1 = 3 < 7.4569; Fail to reject H 0 . There is not sufficient evidence to support a claim that the two populations of scores have different amounts of variation.

18. H0 : m1 = m2 ; H1: m1 ¹ m2 ; population1 = easy to difficult, population2 = difficult to easy; Test statistic: t = 1.403; P -value = 0.1686 (Table: P-value > 0.10); Critical values: t = ±2.024 (Table: t = ±2.131); Fail to reject H 0 . There is not sufficient evidence to support a claim that the two populations of scores have different amounts of variation. ( x - x ) - ( m1 - m2 ) = (4.914 - 3.296) - 0 = 1.403 (df = 15) t= 1 2 s12 s22 22.7152 6.7592 + + 25 16 n1 n2 19. FL =

1 = 0.4103, FR = 2.7006 2.4374

Chapter Quick Quiz 1.

H0: p1 = p2 ; H1: p1 ¹ p2 ; population1 = women, population2 = men;

2.

x1 = 258, x2 = 282, pˆ1 =

3.

P-value = 0.1015 (Table: 0.1010)

4.

a. 95% CI: - 0.0659 < p1 - p2 < 0.00591

258 282 258 + 282 = 0.230, pˆ 2 = = 0.260, p = = 0.245 1121 1084 1121 + 1084

b. The confidence interval includes the value of 0, so it is possible that the two proportions are equal. There is not a significant difference.

5. Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that for the people who were aware of the statement, the proportion of women is equal to the proportion of men.

6.

True, since n > 30.

7.

False, the requirements are np ³ 5 and nq ³ 5.

8.

Because the data consist of matched pairs, they are dependent.

9.

H0 : md = 0; H1: md ¹ 0

10. a. t =

b. t =

d - md sd n

c. z =

( pˆ1 - pˆ 2 ) - ( p1 - p2 )

( x1 - x2 ) - ( m1 - m2 ) s2 d. F = 12 s2

s12 s22 + n1 n2

.

pq pq + n1 n2

Review Exercises

137

Review Exercises

H0 : p1 = p2 ; H1: p1 ¹ p2 ; population1 = blind, population 2 = not blind; Test statistic: z = - 4.20; P-value = 0.0000 (Table: 0.0002); Critical values: z = ±2.576; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that the acceptance rate is the same with or without blinding.

Without blinding, reviewers know the names and institutions of the abstract authors, and they might be influenced by that knowledge.

13, 200 (0.267 ) + 13, 433 (0.290) p= = 0.637; q = 1 - 0.637 = 0.363; 13, 200 + 13, 433 ( pˆ - pˆ 2 ) - ( p1 - p2 ) = (0.267 - 0.290) - 0 z= 1 = - 4.20 pq pq 0.637 )(0.363) (0.637 )(0.363) ( + + n1 n2 13, 200 13, 433

1.

2. 98% CI: - 0.0372 < p1 - p2 < - 0.00892; The confidence interval limits do not contain 0, so it appears that there is a significant difference between the two proportions.

( pˆ1 - pˆ 2 ) ± za / 2

0.267 (0.733) (0.290)(0.710) pˆ1qˆ1 pˆ 2 qˆ2 + = (0.267 - 0.290) ± 2.576 + n1 n2 13, 200 13, 433

3. 95% CI: - 25.33 cm < m1 - m2 < - 7.51 cm (Table: - 25.70 cm < m1 - m2 < - 7.14 cm); With 95% confidence, we conclude that the mean height of women is less than the mean height of men by an amount that is between 7.51 cm and 25.33 cm (Table: 7.14 cm and 25.70 cm).

( x1 - x2 ) ± ta / 2

s12 s22 11.8472 5.3022 + = (162.35 - 178.77 ) ± 2.262 + (df = 9); n1 n2 10 10

population1 = women, population 2 = men 4. H0 : m1 = m2 ; H1: m1 < m2 ; population1 = women, population 2 = men; Test statistic: t = - 4.001; P-value = 0.0008 (Table: P-value < 0.005); Critical value: t = - 2.666 (Table: t = - 2.821); Reject H 0 . There is sufficient evidence to support the claim that women have heights with a mean that is less than the mean height of men. ( x - x ) - ( m1 - m2 ) = (162.35 - 178.77) - 0 = - 4.001 (df = 9) t= 1 2 s12 s22 11.8472 5.3022 + + 10 10 n1 n2 5.

a. H0 : md = 0 kg; H1: md ¹ 0 kg; difference = pre - post; Test statistic: t = 2.301; P-value = 0.0469; Critical values (a = 0.05): t = ±2.262; Reject H 0 . There is sufficient evidence to conclude that there is a difference between pre-training and post-training weights. d - md 2.0 - 0 t= = = 2.301 (df = 9) sd n 2.749 10 b. 95% CI: 0.0 kg < md < 4.0 kg; The confidence interval consists of negative numbers only and does not include 0. s 2.749 d ± ta / 2 d = 2.0 ± 2.262 (df = 9) n 10

6.

H0 : s 1 = s 2 ; H1: s 1 ¹ s 2 ; population1 = women, population2 = men;

Test statistic: F = s12 s22 = 11.8472 5.3022 = 4.9933; P-value = 0.0252; Upper critical value : F = 4.0260; Reject

H 0 . There is sufficient evidence to warrant rejection of the claim that women and men have heights with the same variation.

.

138

Chapter 9: Inferences from Two Samples

Cumulative Review Exercises 1. a. Because the sample data are matched with each column consisting of heights from the same family, the data are dependent. b. x =

64.0 + 68.0 + 70.0 + 71.0 + 71.0 + 71.0 + 71.0 + 71.7 71.0 + 71.0 = 71.0 in.; = 69.7 in.; Q2 = 2 8

range = 71.7 - 64.0 = 7.7 in.; s =

(64.0 - 69.7)2 + + (71.7 - 69.7) 2 = 2.6 in.; s 2 = 6.6 in 2 14 - 1

c. ratio d. continuous 2. There does not appear to be a correlation or association between the heights of fathers and the heights of their sons.

72 71 70

Son

69 68 67 66 65 64 63 65

66

67

68

69

70

71

Father

s 2.570 = 69.7 ± 2.306 × Þ 67.6 in. < m < 71.9 in.; We have 95% confidence that the limits n 8 of 67.6 in. and 71.9 in. actually contain the true value of the mean height of all adult sons.

3.

90% CI: x ± ta / 2

4. H0 : md = 0 in.; H1: md ¹ 0 in.; difference = father - son; Test statistic: t = - 1.712; P -value = 0.1326 (Table: P-value > 0.10); Critical values: t = ±2.365; Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that differences between heights of fathers and their sons have a mean of 0. There does not appear to be a difference between heights of fathers and their sons. d - md - 1.7125 - 0 t= = = - 1.712 (df = 7) sd n 2.847 8 5. Because the points lie reasonably close to a straight-line pattern, and there is no other pattern that is not a straight-line pattern and there are no outliers, the sample data appear to be from a population with a normal distribution.

.

Cumulative Review Exercises

139

6. The shape of the histogram indicates that the sample data appear to be from a population with a distribution that is approximately normal.

16 14

Frequency

12 10 8 6 4 2 0 28

36

44

52

60

68

M ale Braking Reaction Time

7. Because the points are reasonably close to a straight-line pattern and there is no other pattern that is not a straight-line pattern, it appears that the braking reaction times of females are from a population with a normal distribution.

8. Because the boxplots overlap, there does not appear to be a significant difference between braking reaction times of males and females, but the braking reaction times for males appear to be generally lower than the braking reaction times of females.

9. H0 : m1 = m2 ; H1: m1 ¹ m2 ; population1 = male, population 2 = female; Test statistic: t = - 3.259; P-value = 0.0019 (Table: P-value < 0.005); Critical values t = ±2.664 (Table: t = ±2.724); Reject H 0 . There is sufficient evidence to warrant rejection of the claim that males and females have the same mean braking reaction time. Males appear to have lower reaction times. ( x - x ) - ( m1 - m2 ) = (44.361 - 54.278) - 0 = - 3.259 (df = 35) t= 1 2 s12 s22 9.4722 15.6112 + + 36 36 n1 n2 10. a. The sample sizes are greater than 30 and the data appear to be from a populations that meet the loose requirement of being normally distributed. s 9.472 = 44.631 ± 2.724 × Þ 40.1 < m < 48.7 Males: 99%% CI: x ± ta / 2 n 36 s 15.611 = 54.278 ± 2.724 × Þ 47.2 < m < 61.4 Females: 99%% CI: x ± ta / 2 n 36 The confidence intervals overlap, so there does not appear to be significant difference between the mean braking reaction times of males and females.

.

140

Chapter 9: Inferences from Two Samples

10. (continued) b. 99% CI: - 18.0 < m1 - m2 < - 1.8 (Table: - - 18.2 < m1 - m2 < - 1.6); Because the confidence interval consists of negative numbers and does not include 0, there appears to be a significant difference between the mean braking reaction times of males and females.

( x1 - x2 ) ± ta / 2

s12 s22 9.4722 15.6112 + = (44.361 - 54.278) ± 2.724 + (df = 35); n1 n2 36 36

population1 = men, population 2 = women c. The results from part (b) are better.

.

Section 10-1: Correlation

141

Chapter 10: Correlation and Regression Section 10-1: Correlation 1. a. r is a statistic that represents the value of the linear correlation coefficient computed from the paired sample data, and r is a parameter that represents the value of the linear correlation coefficient that would be computed by using all of the paired data in the population of all statistics students. b. The value of r is estimated to be 0, because it is likely that there is no correlation between body temperature and head circumference. c. The value of r does not change if the body temperatures are converted to Fahrenheit degrees.

2. No, with r = 0, there is no linear correlation, but there might be some association with a scatterplot showing a distinct pattern that is not a straight-line pattern.

3. No, a correlation between two variables indicates that they are somehow associated, but that association does not necessarily imply that one of the variables has a direct effect on the other variable. Correlation does not imply causality.

4.

a. –1 b. 0.746 c. 0.268

d. 0.992 e. 1

5. r = 0.963; P-value = 0.000; Critical values: r = ±0.268 (Table: r » ±0.279); Yes, there is sufficient evidence to support the claim that there is a linear correlation between the weights of bears and their chest sizes. It is easier to measure the chest size of a bear than the weight, which would require lifting the bear onto a scale. It does appear that chest size could be used to predict weight.

6. r = 0.765; P-value < 0.01 (Using table A-6); Critical values: r = ±0.497; Yes, there is sufficient evidence to support the claim that there is a linear correlation between calories and sugar in a gram of cereal.

7. r = 0.552; P-value < 0.0001; Critical values: r » ±0.196; Yes, there is sufficient evidence to support the claim that there is a linear correlation between the heights of fathers and the heights of their sons.

8. r = 0.354; P-value < 0.0001; Critical values: r » ±0.196; Yes, there is sufficient evidence to support the claim that there is a linear correlation between the heights of mothers and the heights of their daughters. a.

10 9 8 7

y

9.

6 5 4 3 5.0

7.5

10.0

12.5

15.0

x

b. H0 : r = 0; H1: r ¹ 0; r = 0.816; P-value = 0.002 (Table: P-value < 0.01); Critical values (a = 0.05): r = ±0.602; There is sufficient evidence to support the claim of a linear correlation between the two variables. c. The scatterplot reveals a distinct pattern that is not a straight line pattern.

.

142

Chapter 10: Correlation and Regression

10. a.

13 12 11

y

10 9 8 7 6 5 5.0

7.5

10.0

12.5

15.0

x

b. H0 : r = 0; H1: r ¹ 0; r = 0.816; P-value = 0.002 (Table: P-value < 0.01); Critical values (a = 0.05): r = ±0.602; There is sufficient evidence to support the claim of a linear correlation between the two variables. c. The scatterplot reveals a perfect straight-line pattern, except for the presence of one outlier.

11. a. Answer will vary, but because there appears to be an upward pattern, it is reasonable to think that there is a linear correlation. b. H0 : r = 0; H1: r ¹ 0; r = 0.906; Critical values (a = 0.05): r = ±0.632; P-value = 0.000 (Table: P-value < 0.01); There is sufficient evidence to support the claim of a linear correlation. c. H0 : r = 0; H1: r ¹ 0; r = 0; Critical values (a = 0.05): r = ±0.666; P-value = 1.000 (Table: P-value > 0.05); There is not sufficient evidence to support the claim of a linear correlation. d. The effect from a single pair of values can be very substantial, and it can change the conclusion.

12. a. There does not appear to be a linear correlation. b. There does not appear to be a linear correlation. c. H0 : r = 0; H1: r ¹ 0; r = 0; Critical values (a = 0.05): r = ±0.950; P-value = 1.000 (Table: P-value > 0.05); There does not appear to be a linear correlation. The same results are obtained with the four points in the upper right corner. d. H0 : r = 0; H1: r ¹ 0; r = 0.985; Critical values (a = 0.05): r = ±0.707; P-value = 0.000 (Table: P-value < 0.01); There is sufficient evidence to support the claim of a linear correlation. e. There are two different populations that should be considered separately. It is misleading to use the combined data from women and men and conclude that there is a relationship between x and y.

.

Section 10-1: Correlation

143

13. H0 : r = 0; H1: r ¹ 0; r = - 0.206; P-value = 0.543 (Table: P-value > 0.05); Critical values: r = ±0.602; Fail to reject H 0 . There is not sufficient evidence to support the claim that there is a linear correlation between pulse rates and systolic blood pressure. It does not appear that people with high systolic blood pressure have high pulse rate.

14. H0 : r = 0; H1: r ¹ 0; r = 0.608; P-value = 0.047 (Table: P-value » 0.05); Critical values: r = ±0.602; Fail to reject H 0 . There is sufficient evidence to support the claim that there is a linear correlation between pulse rates and diastolic blood pressure. It appears that people with high diastolic blood pressure have high pulse rate.

15. H0 : r = 0; H1: r ¹ 0; r = 0.769; P-value = 0.837 (Table: P-value > 0.05); Critical values: r = ±0.878; Fail to reject H 0 . There is not sufficient evidence to support the claim that there is a linear correlation between shoe print lengths and heights of females. The given results do not suggest that police can use a shoe print length to estimate the height of a female.

16. H0 : r = 0; H1: r ¹ 0; r = 0.355; P-value = 0.558 (Table: P-value > 0.05); Critical values: r = ±0.878; Fail to reject H 0 . There is not sufficient evidence to support the claim that there is a linear correlation between foot lengths and heights of females. The given results do not suggest that police can use a foot length to estimate the height of females.

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17. H0 : r = 0; H1: r ¹ 0; r = - 0.959; P-value = 0.010; Critical values: r = ±0.878; Reject H 0 . There is sufficient evidence to support the claim that there is a linear correlation between weights of lemon imports from Mexico and U.S. car fatality rates. The results do not suggest any cause-effect relationship between the two variables.

Scatterplot for Exercise 18 1200

15.8

1150 1100

15.6

Chirps/min

Crash Fatality Rate

Scatterplot for Exercise 17 16.0

15.4

1050 1000

15.2

950

15.0

900 850

200 250 300

70

350 400 450 500 550

75

80

85

90

95

Temperature(°F)

Lemon Imports

18. H0 : r = 0; H1: r ¹ 0; r = 0.874; P-value = 0.005 (Table: P-value < 0.01); Critical values: r = ±0.707; Reject H 0 . There is sufficient evidence to support the claim of a linear correlation between the number of cricket chirps and the temperature.

19. H0 : r = 0; H1: r ¹ 0; r = 0.444; P-value = 0.148 (Table: P-value > 0.05); Critical values: r = ±0.576; Fail to reject H 0 . There is not sufficient evidence to support the claim that there is a linear correlation between heights of fathers and heights of their first sons.

20. H0 : r = 0; H1: r ¹ 0; r = - 0.217; P-value = 0.498 (Table: P-value > 0.05); Critical values: r = ±0.576; Reject H 0 . There is not sufficient evidence to support the claim that there is a linear correlation between heights of mothers and heights of their first sons.

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Section 10-1: Correlation

145

21. H0 : r = 0; H1: r ¹ 0; r = 0.948; P -value = 0.004 (Table: P-value < 0.01); Critical values: r = ±0.811; Reject H 0 . There is sufficient evidence to support the claim of a linear correlation between the overhead width of a seal in a photograph and the weight of a seal.

Scatterplot for Exercise 21 260 240 220

Weight

200 180 160 140 120 100 7.0

7.5

8.0

8.5

9.0

9.5

10.0

Overhead Width

22. H0 : r = 0; H1: r ¹ 0; r = 0.896; P-value = 0.001 (Table: P-value < 0.01); Critical values: r = ±0.798; Reject H 0 . There is sufficient evidence to support the claim that there is a linear correlation between numbers of registered pleasure boats and numbers of manatee boat fatalities.

23. H0 : r = 0; H1: r ¹ 0; r = 0.113; P -value = 0.700 (Table: P-value > 0.05); Critical values: r = ±0.532; Fail to reject H 0 . There is not sufficient evidence to support the claim that there is a linear correlation between heights of winning presidential candidates and heights of their main opponents. In an ideal world, voters would focus on important issues and not height or physical appearance of candidates, so there should not be a correlation.

Scatterplot for Exercise 24

Scatterplot for Exercise 23

100

190

90 80 70

Height

Opponent

185

180

60 50 40

175

30 20 175

180

185

190

2

195

4

6

8

10

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Circumference

President

24. H0 : r = 0; H1: r ¹ 0; r = 0.831; P-value = 0.001 (Table: P-value < 0.01); Critical values: r = ±0.576; Reject H 0 . There is sufficient evidence to support the claim of a linear correlation between circumferences and heights of trees. As trees grow, the become taller and wider, so there should be a correlation.

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25. H0 : r = 0; H1: r ¹ 0; r = - 0.063; P-value = 0.791 (Table: P-value > 0.05); Critical values: r = ±0.444; Fail to reject H 0 . There is not sufficient evidence to support the claim that there is a linear correlation between IQ scores and brain volumes.

Scatterplot for Exercise 26

Scatterplot for Exercise 25

120

120

110

110

IQ

130

IQ

130

100

100

90

90

80 900

80 1000 1100

50

1200 1300 1400 1500

Brain Volume

60

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90 100 110 120 130 140

Body Weight

26. H0 : r = 0; H1: r ¹ 0; r = - 0.003; P-value = 0.991 (Table: P-value > 0.05); Critical values: r = ±0.444; Fail to reject H 0 . There is not sufficient evidence to support the claim that there is a linear correlation between IQ scores and body weight.

27. H0 : r = 0; H1: r ¹ 0; r = 0.594; P -value = 0.007 (Table: P-value < 0.01); Critical values: r » ±0.456; Reject H 0 . There is sufficient evidence to support the claim that there is a linear correlation between shoe print lengths and heights of males. The given results do suggest that police can use a shoe print length to estimate the height of a male.

Scatterplot for Exercise 28 200

190

190

HEIGHT

HEIGHT

Scatterplot for Exercise 27 200

180

170

180

170

160

160

150

150 25.0

27.5

30.0

32.5

21

35.0

22 23 24 25 26

27 28 29 30

FOOT LENGTH

SHOE PRINT

28. H0 : r = 0; H1: r ¹ 0; r = 0.706; P-value = 0.001 (Table: P-value < 0.01); Critical values: r » ±0.456; Reject H 0 . There is sufficient evidence to support the claim that there is a linear correlation between foot lengths and heights of males. The given results do suggest that police can use a foot length to estimate the height of a male.

.

Section 10-1: Correlation

147

29. H0 : r = 0; H1: r ¹ 0; r = - 0.16217; P-value = 0.0497; Critical values: r = ±0.16197; Reject H 0 . There is sufficient evidence to support the claim that there is a linear correlation between pulse rates and systolic blood pressures of adult females. (In this exercise, extra decimal places are needed for r and the P-value. Table A-6 is not adequate to determine the critical values and P-value for this exercise.)

Scatterplot for Exercise 30

Scatterplot for Exercise 29

100

110 100

DIASTOLIC (Female)

90

PULSE (Female)

90 80 70 60 50

80 70 60 50

40

40

30 100

120

140

160

180

100

200

120

140

160

180

200

SYSTOLIC (Female)

SYSTOLIC (Female)

30. H0 : r = 0; H1: r ¹ 0; r = 0.321; P-value = 0.000; Critical values: r = ±0.158; Reject H 0 . There is sufficient evidence to support the claim that there is a linear correlation between the diastolic and systolic blood pressures of adult females. (In this exercise, extra decimal places are needed for r and the P-value. Table A-6 is not adequate to determine the critical values and P-value for this exercise.)

31. H0 : r = 0; H1: r ¹ 0; r = 0.748; P-value = 0.000; Critical values: r = ±0.105; Reject H 0 . There is sufficient evidence to support the claim that there is a linear correlation between right ear threshold measurements and left ear threshold measurements. (Table A-6 is not adequate to determine the critical values and P-value for this exercise.)

Scatterplot for Exercise 32

Scatterplot for Exercise 31

200

80

150

60

RIGHT EYE

LEFT THRESHOLD

70

50 40 30

100

50

20 10

0

0 0

10

20

30

40

50

60

70

0

80

50

100

150

200

LEFT EYE

RIGHT THRESHOLD

32. H0 : r = 0; H1: r ¹ 0; r = 0.595; P-value = 0.000; Critical values: r = ±0.113; Reject H 0 . There is sufficient evidence to support the claim of a linear correlation between the visual acuity of the left eye and the right eye.

(Table A-6 is not adequate to determine the critical values and P-value for this exercise.)

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33. a. 0.911 b. 0.787 c. 0.9999 (largest)

y

x

0.3 05 1.3 1.7 2.0

2 3 20 50 95

t

34. r = ±

d. 0.976 e. –0.948

=±

2

t +n- 2

x2 4 9 400 2500 9025

2.485 2.4852 + 27 - 2

log x

1x

x 1.4142 1.7321 4.4721 7.0711 9.7468

0.3010 0.4771 1.3010 1.6990 1.9777

0.5 0.3333 0.05 0.02 0.0105

= ±0.445

Section 10-2: Regression 1. a. yˆ = 4.915 + 0.3585x

b. ŷ represents the predicted value of white blood cell count.

2. The first equation represents the regression line that best fits sample data, and the second equation represents the regression line that best fits all paired data in a population. The values b0 and b1 are statistics; the values

b 0 and b1 are parameters.

3.

a. A residual is a value of y - yˆ, which is the difference between an observed value of y and a predicted value of y. b. The regression line has the property that the sum of squares of the residuals is the lowest possible sum.

4. The value of r and the value of b1 have the same sign. They are both positive or they are both negative or they are both 0. If r is positive, the regression line has a positive slope and rises from left to right. If r is negative, the slope of the regression line is negative and it falls from left to right.

5.

With a significant linear correlation, the best predicted value is y = - 4.898 + 0.3303(120) = 34.656 cm.

6.

With a significant linear correlation, the best predicted value is yˆ = 13.84 + 3.755(14.5) = 40.6 in.

7.

With no significant linear correlation, the best predicted value is yˆ = 32.036 cm.

8.

With a significant linear correlation, the best predicted value is y = 36.36 + 0.800(94.6) = 112.1 cm.

9.

yˆ = 3.00 + 0.500 x; The data have a pattern that is not a straight line.

Scatterplot for Exercise 10

Scatterplot for Exercise 9

13

10

12 9

11 8

10

7

y

y

9 8

6

7

5

6 4

5

3

4 5.0

7.5

10.0

12.5

5.0

15.0

7.5

10.0 x

x

.

12.5

15.0

Section 10-2: Regression

149

10. yˆ = 3.00 + 0.500 x; There is an outlier.

11. a. yˆ = 0.264 + 0.906 x b. yˆ = 2 + 0 x (or yˆ = 2 ) c. The results are very different, indicating that one point can dramatically affect the regression equation.

12. a. yˆ = 0.0846 + 0.985x b. yˆ = 1.5 + 0 x (or yˆ = 1.5 ) c. yˆ = 9.5 + 0 x (or yˆ = 9.5 ) d. The results are very different, indicating that combinations of clusters can produce results that differ dramatically from results within each cluster alone.

13. yˆ = 134.6 - 0.1754 x, r = - 0.206, P-value = 0.543. With no significant linear correlation, the best predicted value is y = 122.91.

14. yˆ = 41.59 + 0.4187 x, r = 0.608, P-value = 0.047. With a significant linear correlation, the best predicted value is y = 41.59 + 0.4187(75) = 73. The best predicted value is close to the actual value of 74.

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Chapter 10: Correlation and Regression

15. yˆ = 58.54 + 4.059 x, r = 0.769, P-value = 0.128. With no significant linear correlation, the best predicted value is y = 166.1 cm. Because the best predicted value is mean height, it would not be helpful to police in trying to obtain a description of the female.

16. yˆ = 99.3 + 2.786 x, r = 0.355, P-value = 0.558. With no significant linear correlation, the best predicted value is y = 166.1 cm. Because the best predicted value is mean height, it would not be helpful to police in trying to obtain a description of the female.

16.0

0.20

15.8

0.15

15.6

0.10

Residual

Crash Fatality Rate

17. yˆ = 16.5 - 0.00282 x; r = - 0.959; P-value = 0.010; With a significant linear correlation, the best predicted value is yˆ = 16.5 - 0.00282(500) = 15.1 fatalities per 100,000 population. Common sense suggests that the prediction doesn’t make much sense.

15.4

0.05 0.00

15.2

-0.05 15.0 200

250

300

350

400

450

500

-0.10 200

550

250

300

350

400

450

Lemon Imports

Lemon Imports

.

500

550

Section 10-2: Regression

151

95

5.0

90

2.5

Residual

Temperature(°F)

18. yˆ = 27.6 + 0.0523x; r = 0.874; P-value = 0.005; With a significant linear correlation, the best predicted value is yˆ = 27.6 + 0.0523(3000) = 185°F. The value of 3000 chirps in one minute is well beyond the scope of the listed sample data, so the extrapolation might be off by a considerable amount, especially if the cricket is dead from such a high temperature.

85 80 75

0.0

-2.5

-5.0

70 850

900

-7.5 850

950 1000 1050 1100 1150 1200

Chirps/min

900

950 1000 1050 1100 1150 1200

Chirps/min

19. yˆ = 20.84 - 0.6915x, r = 0.444, P-value = 0.148. With no significant linear correlation, the best predicted value is y = 68.9 cm.

20. yˆ = 81.51 - 0.1982 x, r = - 0.217, P-value = 0.498. With a linear correlation, the best predicted value is y = 68.9 cm.

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Chapter 10: Correlation and Regression

21. yˆ = - 157 + 40.2 x; r = 0.948; P-value = 0.004; With a significant linear correlation, the best predicted value is yˆ = - 157 + 40.2(2) = - 76.6 kg; The prediction is a negative weight that cannot be correct. The overhead width of 2 cm is well beyond the scope of the sample widths, so the extrapolation might be off by a considerable amount.

Clearly, the predicted negative weight makes no sense.

260

15

240

10 5

200

Residual

Weight

220

180 160

0 -5 -10

140

-15

120

-20

100 7.0

7.5

8.0

8.5

9.0

9.5

7.0

10.0

7.5

8.0

8.5

9.0

9.5

10.0

Overhead Width

Overhead Width

22. yˆ = - 113.6 + 2.287 x , r = 0.896, P-value = 0.001. With a significant linear correlation, the best predicted value is y = - 113.6 + 2.287(84) = 78.5, which happens to be very close to the predicted value of 78 fatalities. Because in this case, the best predicted number of fatalities is calculated on actual predictions, the predicted values are likely to be very good in general.

190

10

185

5

Residual

Opponent

23. yˆ = 162 + 0.0975x; r = 0.113; P -value = 0.700 (Table: P-value > 0.05); With no significant linear correlation, the best predicted value is y = 179.7 cm. Heights of opponents do not appear to be predicted well by using the heights of the presidents.

180

0

175

-5 170 175

180

185

190

175

195

180

185

President

President

.

190

195

Section 10-2: Regression

153

100

25

90

20

80

15

70

10

Residual

Height

24. yˆ = 21.6 + 5.40 x; r = 0.831; P-value = 0.001; With a significant linear correlation, the best predicted value is yˆ = 21.6 + 5.40(5.0) = 48.6 ft. It is much easier to measure the circumference of a tree to predict the height of the tree than having to measure the height of the tree directly.

60 50

5 0

40

-5

30

-10

20

-15 2

4

6

8

10

12

2

14

4

6

8

10

12

14

Circumference

Circumference

25. ŷ = 137.3 - 0.2 x, r = - 0.153, P-value = 0.008. With no significant linear correlation, the best predicted value is y = 123.0.

26. yˆ = 65.3 + 0.07596 x, r = 0.079, P-value = 0.171. With no significant linear correlation, the best predicted value is y = 70.8. The best predicted value is not close to the actual blood pressure of 80.

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Chapter 10: Correlation and Regression

27. yˆ = 93.5 + 2.85x; r = 0.594; P-value = 0.007; With a significant linear correlation, the best predicted value is yˆ = 93.5 + 2.85(31.3) = 183 cm. Although there is a linear correlation, with r = 0.594, we see that it is not very strong, so an estimate of the height of a male might be off by a considerable amount.

15

190

10

180

5

Residual

HEIGHT

200

170

0 -5

160

-10 150 25.0

27.5

30.0

32.5

25.0

35.0

27.5

30.0

32.5

35.0

SHOE PRINT

SHOE PRINT

28. yˆ = 87.1 + 3.50 x; r = 0.706; P-value = 0.001; With no significant linear correlation, the best predicted value is yˆ = 87.1 + 3.50(28.0) = 185 cm. Although there is a linear correlation, with r = 0.706, we see that it is not very close to 1, so an estimate of the height of a male might be off by a considerable amount.

200

10

5

Residual

HEIGHT

190

180

170

0

-5 160

-10 150 21

22

23

24

25

26

27

28

29

21

30

22

23

24

25

26

27

28

29

30

FOOT LENGTH

FOOT LENGTH

29. yˆ = 138 - 0.223x; r = - 0.162; P-value = 0.050; With a significant linear correlation, the best predicted value is yˆ = 138 - 0.223(80) = 120 mm Hg. 200

60 40

160

Residual

SYSTOLIC (Female)

180

140

20 0

120

-20 100

-40 30

40

50

60

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100 110

40

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PULSE (Female)

PULSE (Female)

.

90

100 110

Section 10-2: Regression

155

100

30

90

20 10

80

Residual

DIASTOLIC (Female)

30. yˆ = 44.7 + 0.209 x; r = 0.321; P-value = 0.000; With a significant linear correlation, the best predicted value is yˆ = 44.7 + 0.209(120) = 69.8 mm Hg.

70

0 -10

60

-20 50

-30

40

-40 100

120

140

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180

100

200

120

140

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180

200

SYSTOLIC (Female)

SYSTOLIC (Female)

31. yˆ = 4.67 + 0.758x; r = 0.748; P-value = 0.000; With a significant linear correlation, the best predicted value is yˆ = 4.67 + 0.758(20) = 19.8.

80

LEFT THRESHOLD

70 60 50 40 30 20 10 0 0

10

20

30

40

50

60

70

80

RIGHT THRESHOLD

32. yˆ = 10.7 + 0.678x; r = 0.595; P-value = 0.000; With a significant linear correlation, the best predicted value is yˆ = 10.7 + 0.678(50) = 44.6.

200

200

150 100

Residual

LEFT EYE

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-50 -100

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RIGHT EYE

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RIGHT EYE

.

150

200

156

Chapter 10: Correlation and Regression

33. a. Using yˆ = 3.02 + .0602 x, the sum of squares of the residuals is 12.28. b. Using yˆ = - 3 + 0.0500 x, the sum of squares of the residuals is 15.71, which is larger than 12.28, which is the sum of squares of the residuals for the regression line. x

yˆ = 3.02 + 0.0602 x

( ŷ - y )2

x

yˆ = 3 + 0.0500 x

( ŷ - y )2

56.0 82.0 78.0 86.0 88.0

6.3912 7.9564 7.7156 8.1972 8.3176

0.259 0.021 1.731 3.599 6.669 12.279

56.0 82.0 78.0 86.0 88.0

5.8 7.1 6.9 7.3 7.4

1.21 1 0.25 1 12.25 15.71

Sum:

Sum:

Section 10-3: Prediction Intervals and Variation 1. The value of se = 20.31678 cm is the standard error of estimate, which is a measure of the differences between the observed weights and the weights predicted from the regression equation. It is a measure of the variation of the sample points about the regression line.

2. We have 95% confidence that the limits of 42.9 kg and 123.8 kg contain the value of the weight for a female with a height of 170 cm. The major advantage of using a prediction interval is that it provides us with a range of likely weights, so we have a sense of how accurate the predicted weight is likely to be. The terminology of prediction interval is used for an interval estimate of a variable, whereas the terminology of confidence interval is used for an interval estimate of a population parameter.

3. The coefficient of determination is r2 = 0.060. We know that 6% of the variation in weight is explained by the linear correlation between height and weight, and 94% of the variation in weight is explained by other factors and/or random variation.

4. For the paired weights, se = 0 because there is an exact conversion formula. For a student who weighs 100 lb, the predicted weight is 45.4 kg, and there is no prediction interval because the conversion yields an exact result.

5.

2

r 2 = (0.874) = 0.764; 76.4% of the variation in temperature is explained by the linear correlation between

chirps and temperature, and 23.6% of the variation in temperature is explained by other factors and/or random variation.

6.

2

r 2 = (0.885) = 0.783; 78.3% of the variation in waist size is explained by the linear correlation between

weight and waist size, and 21.7% of the variation in waist size is explained by other factors and/or random variation.

7.

2

r 2 = (0.934) = 0.872; 87.2% of the variation in weights of bears is explained by the linear correlation between

neck size and weight, and 12.8% of the variation in weights is explained by other factors and/or random variation.

8.

2

r 2 = (0.783) = 0.613; 61.3% of the variation in weight is explained by the linear correlation between head

width and weight, and 38.7% of the variation in weight is explained by other factors and/or random variation.

9. r = 0.850; Critical values (a = 0.05): r = ±0.404 (Table: r » ±0.396); Yes, there is sufficient evidence to support a claim of a linear correlation between registered boats and manatee fatalities.

2

10. r 2 = (0.850) = 0.723, or 72.3% 11. The best predicted value is 70.5 manatees.

12. The 95% prediction interval estimate is 50.0 manatees < y < 90.9 manatees. We have 95% confidence that the limits of 50.0 and 90.9 contain the number of manatee fatalities in a year with 850,000 registered boats.

.

Section 10-3: Prediction Intervals and Variation 13. 99% CI: 42.7 manatees < y < 98.3 manatees

yˆ = - 49.049 + 1.406 (85) = 70.46 2

2

n ( x0 - x ) 24 (85 - 85.25) 1 1 E = ta / 2 se 1 + + = 2.819 (9.6605) 1 + + = 27.79 2 2 n n Sx - (Sx ) 24 24 (127822 ) - (2046)2

( )

14. 95% CI: 67.7 manatees < y < 109.8 manatees

yˆ = - 49.049 + 1.406 (98) = 88.74 2

2

n ( x0 - x ) 24 (98 - 85.25) 1 1 E = ta / 2 se 1 + + = 2.074 (9.6605) 1 + + = 20.41 2 2 n n Sx - (Sx ) 24 24 (127822 ) - (2046)2

( )

15. 95% CI: 65.1 manatees < y < 106.8 manatees

yˆ = - 49.049 + 1.406 (96) = 85.93 2

2

n ( x0 - x ) 24 (96 - 85.25) 1 1 E = ta / 2 se 1 + + = 2.074 (9.6605) 1 + + = 20.42 2 2 n n Sx - (Sx ) 24 24 (127822 ) - (2046)2

( )

16. 95% CI: 45.5 manatees < y < 101.1 manatees

yˆ = - 49.049 + 1.406 (87 ) = 73.27 2

E = ta / 2 se

2

n ( x0 - x ) 24 (87 - 85.25) 1 1 1+ + = 2.819 (9.6605) 1 + + = 27.79 2 2 n n Sx - (Sx ) 24 24 (127822 ) - (2046)2

( )

17. a. 652.4986 b. 1039.1005 c. 95% CI: 53.7 mm Hg < y < 101.3 mm Hg

yˆ = 20.3 + 0.477 (120) = 77.5 2

E = ta /2 se

2

n ( x0 - x ) 12 (120 - 125.67 ) 1 1 1+ + = 2.2283 (10.1937 ) 1 + + = 23.8 2 2 n n Sx - (Sx ) 12 12 (192,376) - (1508)2

( )

18. a. 3714.141 b. 1659.426 c. 99% CI: 0.607 ft < y < 85.8 ft; The interval is wide enough to be of no practical use.

yˆ = 21.6 + 5.401(4.0) = 43.2 2

2

n ( x0 - x ) 12 (4.0 - 4.91) 1 1 E = ta / 2 se 1 + + = 3.1693 (12.88187) 1 + + = 42.6 2 2 n n Sx - (Sx ) 12 12 (416.41) - (58.9 )2

( )

19. a. 352.7278 b. 109.3722 c. 90% CI: 71.09°F < y < 88.71° F

yˆ = 27.628 + 0.05227 (1000) = 79.90 2

E = ta / 2 se 1 +

2

n ( x0 - x ) 8 (1000 - 1016.25) 1 1 + = 1.943 (4.2695) 1 + + = 8.81 2 2 n n Sx - (Sx ) 8 8 (8391204 ) - (8130)2

( )

.

157

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Chapter 10: Correlation and Regression

20. a. 8880.12 b. 991.1515 c. 99% CI: 125.0 kg < y < 284.5 kg

yˆ = - 156.879 + 40.182 (9.0) = 204.76 2

2

n ( x0 - x ) 1 1 6 (9.0 - 8.5) E = ta / 2 se 1 + + = 4.604 (15.7413) 1 + + = 79.79 n n Sx 2 - (Sx )2 6 6 (439) - (51)2

( )

21. a. 95% CI: 6.5 (1000cells mL ) < y < 7.2 (1000cells mL )

yˆ = 4.06 + 0.0345 (80) = 6.82 2

E = ta / 2 se

2

n ( x0 - x ) 147 (80.0 - 74.0) 1 1 + = 1.976 (1.92241) + = 0.341 2 2 n n Sx - (Sx ) 147 147 (828,832 ) - (10,884 )2

( )

b. 95% CI: 6.2 (1000cells mL ) < y < 6.8 (1000cells mL )

yˆ = 4.06 + 0.0345 (70) = 6.48 2

E = ta / 2 se

2

n ( x0 - x ) 147 (70.0 - 74.0) 1 1 + = 1.976 (1.92241) + = 0.326 2 2 n n Sx - (Sx ) 147 147 (828,832 ) - (10,884 )2

( )

Section 10-4: Multiple Regression 1. The response variable is neck and the predictor variables are head length and head width.

2. Yes, it is better to use the regression equation with the three predictor variables of head length, head width, and length. The adjusted R2 value of 0.830 is just a little less than 0.837, so in this case it is better to use three predictor variables instead of two.

3.

The unadjusted R 2 increases (or remains the same) as more variables are included, but the adjusted R 2 is

adjusted for the number of variables and sample size. The unadjusted R 2 incorrectly suggests that the best multiple regression equation is obtained by including all of the available variables, but by taking into account the sample size and number of predictor variables, the adjusted R 2 is much more helpful in weeding out variables that should not be included.

4. 83.6% of the variation in necks of bears can be explained by the variables of head length and head width, so 16.4% of the variation in necks can be explained by other factors and/or random variation.

5.

Son = 18.0 + 0.504 Father + 0.277 Mother

6.

a. P-value < 0.0001 b. R 2 = 0.3649 c. adjusted R 2 = 0.3552

7. A P-value less than 0.0001 is low, but the values of R 2 (0.3649) and adjusted R 2 (0.3552) are not high.

Although the multiple regression equation fits the sample data best, it is not a good fit, so it should not be used for predicting the height of a son based on the height of his father and the height of his mother.

8.

Predicted height: Son = 18.0 + 0.504 (70) + 0.277 (60) = 70 in.; This result is not likely to be a good predicted

value because the multiple regression equation is not a good model (based on the results from Exercise 7).

9. WAIST (waist circumference) because it has the highest adjusted R 2 (0.782) and all the P-values are the same (0.000).

.

Section 10-4: Multiple Regression

159

10. WAIST (waist circumference) and ARM (arm circumference) because they have the best combination of small

P-value (0.000) and highest adjusted R 2 (0.878), but WAIST (waist circumference) and HT (height) are almost the same.

11. Some subjective judgment is involved here. A good choice would be to use all three variables because the adjusted R 2 value of 0.939 is considerably higher than any of the other values of adjusted R 2 . With this reasoning, the best regression equation is yˆ = - 147 + 0.632 HT + 0.697 WAIST + 1.58ARM. 12. Predicted city fuel consumption is yˆ = - 147 + 0.632 (186) + 0.697 (107.8) + 1.58 (37.0) = 104.1 kg. (Based on the result from Exercise 11.) The predicted value is a good estimate, given the adjusted R 2 value of 0.939, the P-value of 0.000, and the large sample size.

13. The best regression equation is yˆ = 0.127 + 0.0878x1 - 0.0250 x2 , where x1 represents tar and x2 represents carbon monoxide. It is best because it has the highest adjusted R 2 value of 0.927 and the lowest P-value of 0.000. It is a good regression equation for predicting nicotine content because it has a high value of adjusted R 2 and a low P-value. Possible models:

yˆ = 0.080 + 0.0633x1, adjusted R2 = 0.877 yˆ = 0.328 + 0.0397 x2 , adjusted R2 = 0.437 yˆ = 0.127 + 0.0878x1 - 0.0250x2 , adjusted R2 = 0.927 14. The best regression equation is yˆ = 0.251 + 0.101x1 - 0.0454 x2 , where x1 represents tar and x2 represents carbon monoxide. It is best because it has the highest adjusted R 2 value of 0.908 and the lowest P-value of 0.000. It is a good regression equation for predicting nicotine content because it has a high value of adjusted R 2 and a low P-value. Possible models:

yˆ = 0.139 + 0.0567 x1, adjusted R2 = 0.752 yˆ = 0.385 + 0.0325x2 , adjusted R2 = 0.283 yˆ = 0.251 + 0.101x1 - 0.0454 x2 , adjusted R2 = 0.908 15. The best regression equation is yˆ = 109 - 0.00670 x1, where x1 represents volume. It is best because it has the highest adjusted R 2 value of 0.0513 and the lowest P-value of 0.791. The three regression equations all have adjusted values of R 2 that are very close to 0, so none of them are good for predicting IQ. It does not appear that people with larger brains have higher IQ scores. Possible models: yˆ = 109 - 0.00670 x1, adjusted R2 = 0.0513

yˆ = 101 - 0.00178 x2 , adjusted R2 = 0.0555 yˆ = 108 - 0.00694 x1 + 0.00722 x2 , adjusted R2 = 0.113 16. The best regression equation is yˆ = - 10.0 + 0.567 x1 + 0.532 x2 , where x1 represents verbal IQ score and x2 represents performance IQ score. It is best because it has the highest adjusted R 2 value of 0.999 and the lowest

P-value of 0.000. Because the adjusted R 2 is so close to 1, it is likely that predicted values will be very accurate.

Possible models: yˆ = 11.5 + 0.940 x1, adjusted R2 = 0.762

yˆ = 10.5 + 0.806 x2 , adjusted R2 = 0.814 yˆ = - 10.0 + 0.567 x1 + 0.532 x2 , adjusted R2 = 0.999

.

160

Chapter 10: Correlation and Regression

0.769317 - 0 = 10.813917; P-value < 0.0001; Reject H 0 and conclude that 0.0711414 the regression coefficient of b1 = 0.769 should be kept.

17. For H 0 : b1 = 0, Test statistic: t =

1.009510 - 0 = 29.856; P-value < 0.0001; Reject H 0 and conclude that the 0.0338123 regression coefficient of b2 = 1.01 should be kept.

It appears that the regression equation should include both independent variables of height and waist circumference.

For H 0 : b 2 = 0, Test statistic: t =

18. 0.629 < b1 < 0.910; 0.943 < b 2 < 1.08; Neither confidence interval includes 0, so neither of the two variables should be eliminated from the regression equation.

CI for b1 b1 - E < b1 < b1 + E b1 - ta / 2 sb1 < b1 < b1 + ta / 2 sb1 0.7693 - 1.976(0.0711414) < b1 < 0.7693 + 1.976(0.0711414) 0.629 < b1 < 0.910 CI for b 2 b2 - E < b 2 < b2 + E b2 - ta / 2 sb2 < b 2 < b2 + ta / 2 sb2 1.0095 - 1.976(0.033812) < b 2 < 1.0095 + 1.976(0.033812) 0.943 < b 2 < 1.08 Section 10-5: Dummy Variables and Logistic Regression 1. A dummy variable is a variable having only two values and those values (such as 0 and 1) are used to represent the different categories of a qualitative variable.

2. Linear regression is used to predict the value of a continuous variable while logistic regression is used to predict the probability of an outcome in a binomial event.

3. With simple logistic regression, there is only one predictor variable, but with multiple logistic regression, there are two or more predictor variables.

4.

False, a dummy variable that is a predictor variable does not necessitate the use of logistic regression.

5.

Head length and head width are predictor variables. The response variable is sex, which is a dummy variable.

6. Because the regression equation has the high overall P-value of 0.160, the predicted value is not likely to be very accurate.

7.

The probability that the bear is a female is

e1.50+0.045(14.5) - 0.0444(7.5) 1 + e1.50+0.045(14.5) - 0.0444(7.5)

= 0.235. The probability that the bear is

a male is 1 - 0.235 = 0.765.

8.

The probability that the bear is a female is

e1.50+0.045(16) - 0.0444(9)

1 + e1.50+0.045(16) - 0.0444(9) male is 1 - 0.145 = 0.855. Given its size, the bear is most likely a male.

.

= 0.145. The probability that the bear is a

Section 10-5: Dummy Variables and Logistic Regression 9.

161

yˆ = 3.06 + 82.4 (SEX ) + 2.91(AGE ) a. Female: yˆ = 3.06 + 82.4 (0) + 2.91(20) = 61 lb b. Male: yˆ = 3.06 + 82.4 (1) + 2.91(20) = 144 lb The sex of the bear does appear to have an effect on its weight. The regression equation indicates that the predicted weight of a male bear is about 82 lb more than the predicted weight of a female bear with other characteristics being the same.

10. yˆ = - 56.2 + 0.828 ( HEIGHT ) - 2.37 (SEX ) a. Female: yˆ = - 56.2 + 0.828 (170) - 2.37 (0) = 84.6 kg b. Male: yˆ = - 56.2 + 0.828 (170) - 2.37 (1) = 82.2 kg One would expect a man to weigh more than a woman of the same height, so the results do not make particular sense. Also, the weights seem rather high for a person standing approximately 5 foot 6 inches.

æ p ö 11. ln ç = - 41.2 + 0.250 (Height ) - 0.00856 (Weight ) è1 - p ÷ ø e- 41.2+0.250(170)- 0.00856(90) The probability of a male is = 0.629.

1 + e- 41.2+0.250(170)- 0.00856(90) æ p ö 12. ln ç = - 38.3 + 0.245 ( Height ) - 0.0289 ( Pulse) ; The probability of a male is è1 - p ÷ ø e- 38.3+0.245(170)- 0.0289(60) = 0.834. The negative coefficient for PULSE shows that higher pulse rates make 1 + e- 38.3+0.245(170)- 0.0289(60) males less likely æ p ö 13. ln ç = - 101.5 + 0.191(Foot Length ) + 0.301(Height ) è1 - p ÷ ø e- 101.5+0.191(28)+0.301(190) The probability of a male is = 0.9999.

1 + e- 101.5+0.191(28)+0.301(190) æ p ö 14. ln ç = - 70.8 + 0.993 (Shoe Size ) + 0.350 ( Height ) ; The probability of a male is è1 - p ÷ ø e- 70.8+0.993(9)+0.350(170) = 0.086. The P-value for shoe size in this model is 0.245, whereas the P-value for 1 + e- 70.8+0.993(9)+0.350(170) foot length in Exercise 13 is 0.013, and the model in Exercise 13 has a higher adjusted R 2 value. The predictions using this model would be less accurate.

.

162

Chapter 10: Correlation and Regression

Chapter Quick Quiz 1. Conclude that there is not sufficient evidence to support the claim of a linear correlation between the systolic blood pressure measurements of the right and left arms 2.

None of the given values change when the variables are switched.

3.

The value of r does not change if all values of one of the variables are multiplied by the same constant.

4.

Because r must be between –1 and 1 inclusive, the value of 1.500 is the result of an error in the calculations.

5. The best predicted number of burglaries is 163.2 mm Hg, which is the mean of the five measurements for the left arm.

6.

The best predicted number of burglaries would be yˆ = 43.6 +1.31 (100 ) = 174.6 mm Hg.

7.

r 2 = (0.867) = 0.752

8.

false

9.

2

false

10. r = - 1

Review Exercises 1. a. r = 0.962; P-value = 0.000 (Table: P-value < 0.01); Critical values (a = 0.05): r = ±0.707; There is sufficient evidence to support the claim that there is a linear correlation between the amount of tar and the amount of nicotine. b. (0.962)2 = 0.925, or 92.5% c. yˆ = - 0.758 + 0.0920 x d. The predicted value is yˆ = - 0.758 + 0.0920(23) = 1.358 mg or 1.4 mg rounded, which is close to the actual amount of 1.3 mg. a. The scatterplot (see part c) shows a pattern with nicotine and carbon monoxide both increasing from left to right, but it is a very weak pattern and the points are not very close to a straight-line pattern, so it appears that there is not sufficient sample evidence to support the claim of a linear correlation between amounts of nicotine and carbon monoxide. b. H0 : r = 0; H1: r ¹ 0; r = 0.329; P-value = 0.427 (Table: P-value > 0.05); Critical values (a = 0.05): r = ±0.707; Fail to reject H 0. There is not sufficient evidence to support the claim that there is a linear correlation between amount of nicotine and amount of carbon monoxide. c. yˆ = 14.2 + 1.42 x 2

18

1

Residual

17

CO (mg)

2.

16

15

0

-1

14

-2 1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.0

1.7

1.1

1.2

1.3

1.4

Nicotine (mg)

Nicotine (mg)

d. The predicted value is y = 16.1 mg, which is close to the actual amount of 15 mg.

.

1.5

1.6

1.7

Review Exercises 3.

163

a. The scatterplot shows a pattern with amounts of tar and carbon monoxide both increasing from left to right, but it is a very weak pattern and the points are not very close to a straight-line pattern, so it appears that there is not sufficient sample evidence to support the claim of a linear correlation between amounts of tar and carbon monoxide.

18

CO (mg)

17

16

15

14 20

21

22

23

24

25

26

27

Tar (mg)

b. Three of the original pairs of sample data are the same (20,16) , so those three points are at the same location and they appear to be one point. c. r = 0.437; P -value = 0.279 (Table: P-value < 0.05); Critical values (a = 0.05): r = ±0.707; There is not sufficient evidence to support the claim that there is a linear correlation between amounts of tar and carbon monoxide. d. yˆ = 12.0 + 0.181x e. The predicted value is y = 16.1 mg, which is close to the actual amount of 15 mg.

4.

a. NICOTINE = - 0.443 + 0.0968TAR - 0.0262 CO, or yˆ = - 0.443 + 0.0968x1 - 0.0262 x2 , where x1 represents tar and x2 represents carbon monoxide. b. R 2 = 0.936; adjusted R 2 = 0.910; P-value = 0.001 c. With high values of R 2 and adjusted R 2 and a small P-value of 0.001, it appears that the regression equation can be used to predict the amount of nicotine given the amounts of tar and carbon monoxide. d. The predicted value is yˆ = - 0.443 + 0.0968(23) - 0.0262(15) = 1.39 mg or 1.4 mg rounded, which is close to the actual value of 1.3 mg of nicotine.

.

164

Chapter 10: Correlation and Regression

H0 : r = 0; H1: r ¹ 0; r = 0.450; P -value = 0.192 (Table: P-value < 0.05); Critical values (a = 0.05): r = ±0.632; Fail to reject H 0. There is not sufficient evidence to support the claim that there is a linear correlation between time and population size. Although there is no linear correlation between time and population size, the scatterplot shows a very distinct pattern revealing that time and population size are associated by a function that is not linear.

5.

5

Population (thousands)

4

3

2

1

0 0

5

10

15

20

Time (days)

æ p ö ln ç = - 44.4 + 0.557 (8 AM Temp) - 0.098 (12 AM Temp) ; The probability that the subject smokes is è1 - p ÷ ø e- 44.4+0.557(98.25)- 0.098(98.25) = 0.667. Because the regression equation has the high overall P-value of 0.521, 1 + e- 44.4+0.557(98.25)- 0.098(98.25) the predicted value is not likely to be very accurate.

6.

Cumulative Review Exercises 1.

x=

s=

4 + 14 + 3 + 1 + 4 + 7 + 3 + 4 = 3.3 lb 8

(4 - 3.3)2 + (14 - 3.3)2 + (3 - 3.3)2 + (1 - 3.3)2 + (4 - 3.3)2 + (7 - 3.3)2 + (3 - 3.3)2 + (4 - 3.3)2 8- 1

= 5.7 lb

Before After

183 179

212 198

177 180

209 208

155 159

162 155

167 164

170 166

Difference

4

14

3

1

4

7

3

4

212 - 179.4 = 1.55. The highest weight is 21.0 not significantly high because its z score of 1.55 shows that it is within 2 standard deviations of the mean.

2.

The highest weight before the diet is 212 lb, which converts to z =

3. H0: md = 0; H1: md > 0; difference = Before - After; Test statistic: t = 1.613; P -value = 0.0754 (Table: P-value > 0.05); Critical value: t = 1.895; Fail to reject

H 0 . There is not sufficient evidence to support the claim that the diet is effective. t=

d - md 3.3 - 0 = = 1.613 (df = 7) sd n 5.7 8

.

Cumulative Review Exercises

165

s 21.0 = 179.4 ± 2.365 × Þ 161.8 lb < m < 197.0 lb; We have 95% confidence that the n 8 interval limits of 161.8 lb and 197.0 lb contain the true value of the mean of the population of all subjects before the diet.

4.

95% CI: x ± ta / 2

5.

a. H0 : r = 0; H1: r ¹ 0; r = 0.965; P-value = 0.0001; Critical values (a = 0.05): r = ±0.707; There is sufficient evidence to support the claim that there is a linear correlation between before and after weights. b. r = 1 c. r = 1 d. The effectiveness of the diet is determined by the amounts of weight lost, but the linear correlation coefficient is not sensitive to different amounts of weight loss. Correlation is not a suitable tool for testing the effectiveness of the diet.

6.

a. z x =3500 =

3500 - 3420 = 0.16; which has an area of 1 - 0.5636 = 0.4364, or 43.64% to the right (Tech: 495

43.58%). b. The z score for 10% to the left is –1.28, which corresponds to a weight of - 1.28 ×495 + 3420 = 2786.4 g (Tech: 2785.5 g).

2450 - 3420 4390 - 3420 = - 1.96 and z x =4390 = = 1.96; which has an area of 0.9550 between 495 495 them, so 1 - 0.9500 = 0.0500, or 5.00% require special treatment. It is the case that many of the babies do require special treatment. c. z x =2450 =

7.

a. No, correlation can be used to investigate an association between the two variables, not whether differences between values of the two variables are significant. b. H0 : md = 0; H1: md ¹ 0; difference = Sitting - Supine; Test statistic: t = 1.185; P -value = 0.2663 (Table: P-value > 0.20); Critical values: t = ±2.262; Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that position has no effect. Position does not appear to have a significant effect. d - md 0.140 - 0 t= = = 1.185 (df = 9) sd n 0.380 10 Subject

8.

A

B

C

D

E

F

G

H

I

J

Sitting Supine

4.66 4.63

5.70 6.34

5.37 5.72

3.34 3.23

3.77 3.60

7.43 6.96

4.15 3.66

6.21 5.81

5.90 5.61

5.77 5.33

Difference

0.03

0.64

0.35

0.11

0.17

0.47

0.49

0.40

0.29

0.44

There must be an error, because the rates of 13.7% and 10.6% are not possible with sample sizes of 100.

.

Section 11-1: Goodness-of-Fit

167

Chapter 11: Goodness-of-Fit and Contingency Tables Section 11-1: Goodness-of-Fit 1. a. Observed values are represented by O and expected values are represented by E. b. For the leading digit of 2, O = 62 and E = (317)(0.176) = 55.792. c. For the leading digit of 2, (O - E )2 E = 0.691.

2.

H0 : p1 = 0.301, p2 = 0.176, p3 = 0.125, , p9 = 0.046; H1: At least one of the proportions is not equal to the given claimed value.

3. There is sufficient evidence to warrant rejection of the claim that the leading digits have a distribution that fits well with Benford’s law.

4. Because the leading digits of inter-arrival traffic times do not fit the distribution described by Benford’s law, it appears that those times are not typical. The anomaly could be due to other factors, but there is a good chance that the computer has been hacked. Computer experts should be used to correct the incursion and secure the computer against further incursions.

5.

H0 : pMon = pTue = pWed = pThu = pFri = 1 5; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 0.523; P-value = 0.9712 (Table: P-value > 0.95); Critical value: c 2 = 9.488; Fail to reject

H 0 . There is not sufficient evidence to warrant rejection of the claim that injuries and illnesses occur with equal frequency on the different days of the week. c2= 6.

(23 - 21.4)2 (23 - 21.4)2 (21 - 21.4)2 (21 - 21.4)2 (19 - 21.4)2 107 5

+

107 5

+

107 5

+

107 5

+

107 5

= 0.523 (df = 4)

H0: pSun = pMon = pTue = pWed = pThu = pFri = pSat = 1 7; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 30.017; P -value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 12.592; Reject H 0 .

There is sufficient evidence to warrant rejection of the claim that the different days of the week have the same frequencies of car crash fatalities. This result is due to the higher numbers of fatalities on Fridays, Saturdays, and Sundays. The number of car crash fatalities on Sunday probably result from people returning after a weekend trip.

c2= 7.

(132 - 117)2 (98 - 117)2 819 7

+

819 7

+

+

(133 - 117)2 (159 - 117 )2 819 7

+

819 7

= 30.017 (df = 6)

H 0 : pJan = pFeb = pMar = pApr = pMay = pJun = pJul = pAug = pSep = pOct = pNov = pDec = 1 12;

H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 47.200; P -value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 19.675; Reject H 0 .

There is sufficient evidence to warrant rejection of the claim that motorcycle fatalities occur with equal frequencies in the different months. Fatalities might be lower in winter months when colder weather is associated with substantially less use of motorcycles.

c2=

(6 - 16.67)2 (8 - 16.67)2 200 12

+

200 12

+

+

(10 - 16.67)2 (8 - 16.67)2 200 12

+

.

200 12

= 47.200 (df = 11)

168 8.

Chapter 11: Goodness-of-Fit and Contingency Tables

H0: p1 = p2 = p3 = p4 = 0.25; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 5.714; P-value = 0.187 (Table: P-value > 0.10); Critical value: c 2 = 7.815; Fail to reject

H 0 . There is not sufficient evidence to warrant rejection of the claim that the tires selected by the students are equally likely. It appear that when four students identify a tire, it is not likely that they would all select the same tire. c2= 9.

(20 - 14)2 (12 - 14)2 (16 - 14)2 (8 - 14)2 0.25 ×56

+

0.25 ×56

+

0.25 ×56

+

0.25 ×56

= 5.714 (df = 3)

H 0 : pJan = pFeb = pMar = pApr = pMay = pJun = pJul = pAug = pSep = pOct = pNov = pDec = 1 12;

H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 93.072; P -value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 19.675; Reject H 0 .

There is sufficient evidence to warrant rejection of the claim that American born Major League Baseball players are born in different months with the same frequency. The sample data appear to support Gladwell’s claim.

c2=

(387 - 376.25)2 (329 - 376.25)2 +

4515 12

4515 12

+

+

(398 - 376.25)2 (371 - 376.25)2 4515 12

+

4515 12

= 93.072 (df = 11)

10. H0 : pRE/NW = 9 16, pSE/NW = 3 16, pRE/VW = 3 16, pSE/VW = 1 16;

H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 15.822; P-value = 0.0012 (Table: P-value < 0.005); Critical value: c 2 = 7.815; Reject H 0 .

There is sufficient evidence to warrant rejection of the claim that the actual numbers of genetic results fit the distribution indicated by the proportions listed in the given table.

c

2

(59 - 45)2 (15 - 15)2 (2 - 15)2 (4 - 5)2 = + + + = 15.822 (df = 3) (9 16) ×80 (3 16) ×80 (3 16) ×80 (1 16) ×80

11. H 0 : pbrown = 0.68, pblue = 0.22, pgreen = 0.10;

H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 9.288; P-value = 0.0064 (Table: P-value > 0.005); Critical value: c 2 = 5.991; Reject H 0 .

There is sufficient evidence to warrant rejection of the claim that the actual frequencies correspond to the predicted distribution.

c2=

(128 - 112.88)2 (32 - 36.52)2 (6 - 16.6)2 +

0.68 ×166

0.22 ×166

+

0.10 ×166

= 9.288 (df = 2)

12. H0 : pAA = 0.25, pAa = 0.50, paa = 0.25;

H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 4.400; P-value = 0.0082 (Table: P-value > 0.005); Critical value: c 2 = 5.991; Reject H 0 .

There is no sufficient evidence to warrant rejection of the claim that the actual frequencies correspond to the predicted distribution.

c2=

(50 - 62.50)2 (140 - 125.0)2 (60 - 62.50)2 0.25 ×250

+

0.50 ×250

+

0.25 ×250

= 4.400 (df = 2)

.

Section 11-1: Goodness-of-Fit

169

13. H0 : p1 = 0.756, p2 = 0.091, p3 = 0.108, p4 = 0.038, p5 = 0.007;

H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 524.713; P -value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 13.277; Reject

H 0 . There is sufficient evidence to warrant rejection of the claim that the distribution of clinical trial participants fits well with the population distribution. Hispanics have an observed frequency of 60 and an expected frequency of 391.027, so they are very underrepresented. Also, the Asian/Pacific Islander subjects have an observed frequency of 54 and an expected frequency of 163.286, so they are also underrepresented. c2=

(3855 - 3248.532)2 (60 - 391.027)2 +

0.756 ×4297

0.091×4297

+

+

(54 - 163.286)2 (12 - 30.079)2 0.038 ×4297

+

0.007 ×4297

= 524.713 (df = 4)

14. H0 : pYS = 9 16, pGS = 3 16, pYW = 3 16, pGW = 1 16;

H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 12.455; P-value = 0.023 (Table: P-value < 0.025); Critical value: c 2 = 7.815; Reject H 0 .

There is sufficient evidence to support the claim that the results contradict Mendel’s theory.

c2=

(256 - 236.25)2 (65 - 78.75)2 (86 - 78.75)2 (12 - 26.25)2 +

0.5625 ×420

0.1875 ×420

+

+

0.1875 ×420

0.0625 ×420

= 12.455 (df = 3)

15. H0 : p1 = 0.301, p2 = 0.176, p3 = 0.125,

, p7 = 0.058, p8 = 0.051, p9 = 0.046; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 21.886; P-value = 0.035 (Table: P-value > 0.01); Critical value: c 2 = 20.090; Reject H 0 .

There is sufficient evidence to warrant rejection of the claim that the leading digits are from a population with a distribution that conforms to Benford’s Law. It does not appear that the image has been corrupted.

c2=

(64 - 66.82)2 (28 - 39.07)2 0.301×330

+

+

0.176 ×330

+

(2 - 11.32)2 (0 - 10.21)2 0.051×330

+

0.046 ×330

= 21.886 (df = 8)

16. H0 : p1 = 0.301, p2 = 0.176, p3 = 0.125,

, p7 = 0.058, p8 = 0.051, p9 = 0.046; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 662.917; P-value = 0.000 (Table: P-value < 0.005); Critical value: c 2 = 20.090; Fail to reject H 0 . There is sufficient evidence to warrant rejection of the claim that the leading digits are from a population with a distribution that conforms to Benford’s Law. It appears that the image has been corrupted.

c

2

=

(550 - 270.90)2 (45 - 158.40)2 +

0.301×900

0.176 ×900

+

+

(110 - 45.90)2 (60 - 41.40)2 0.051 ×900

+

0.046 ×900

= 662.917 (df = 8)

17. H0: pSun = pMon = pTue = pWed = pThu = pFri = pSat = 1 7;

H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 9.500; P -value = 0.147 (Table: P-value > 0.10); Critical value: c 2 = 16.812; Fail to reject

H 0 . There is not sufficient evidence to support the claim that births do not occur on the seven different days of the week with equal frequency.

c2=

Day

Sun

Mon

Tue

Wed

Thu

Fri

Sat

Expected Observed

57.14 53

57.14 52

57.14 66

57.14 72

57.14 57

57.14 57

57.14 43

(53 - 57.14)2 (52 - 57.14)2 400 7

+

400 7

+

+

(57 - 57.14)2 (43 - 57.14)2 +

400 7

.

400 7

= 9.500 (df = 6)

170

Chapter 11: Goodness-of-Fit and Contingency Tables

18. H0: pSun = pMon = pTue = pWed = pThu = pFri = pSat = 1 7;

H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 10.760; P -value = 0.096 (Table: P-value > 0.05); Critical value: c 2 = 16.812; Fail to reject

H 0 . There is not sufficient evidence to warrant rejection of the claim that discharges occur on the seven different days of the week with equal frequency.

c2=

Day

Sun

Mon

Tue

Wed

Thu

Fri

Sat

Expected Observed

57.14 65

57.14 50

57.14 47

57.14 48

57.14 73

57.14 64

57.14 53

(65 - 57.14)2 (50 - 57.14)2 400 7

+

400 7

+

+

(64 - 57.14)2 (53 - 57.14)2 +

400 7

400 7

= 10.760 (df = 6)

19. H0 : Heights selected from a normal distribution.

H1: Heights not selected from a normal distribution. a. b.

Height (cm) Frequency Tech: Table:

c.

Tech:

Less than 155.45 26 0.2023 0.2033 0.2023(247)

155.45 – 162.05 46 0.3171 0.3166 0.3171(247)

162.05–168.65 49 0.3046 0.3039 0.3046(247)

Greater than 168.65 26 0.1761 0.1762 0.1761(247)

= 29.7381 0.2033(247) = 29.8851

= 46.6137 0.3166(247) = 46.5402

= 44.7762 0.3039(247) = 44.6733

= 25.8867 0.1762(247) = 25.9014

Table:

d. Test statistic: c 2 = 0.877 (Table: c 2 = 0.831); P-value = 0.831 (Table: P-value > 0.10); Critical value:

c 2 = 11.345; Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that heights were randomly selected from a normally distributed population. The test suggests that we cannot rule out the possibility that the data are from a normally distributed population. c2=

(26 - 29.7381)2 (46 - 46.6137)2 (49 - 44.7762)2 (26 - 25.8867 )2 29.7381

+

46.6137

+

44.7762

+

25.8867

= 0.877 (df = 3)

Section 11-2: Contingency Tables 1.

2.

(16 + 50 + 3)(40 + 3) (69)(43) = = 4.173 436 + 166 + 40 + 16 + 50 + 3 711 b. Because the expected frequency of a cell is less than 5, the requirements for the hypothesis test are not satisfied. a. E =

H 0 : Whether a subject is right-handed or left-handed is independent of ear preference for cell phone use.

H1: Right-left-handedness and ear preference for cell phone use are dependent.

3. Test statistic: c 2 = 64.517; P-value = 0.0000; Reject the null hypothesis of independence between handedness and cell phone ear preference.

4. The test is right-tailed. The test statistic is based on differences between observed frequencies and the frequencies expected with the assumption of independence between the row and column variables. Only large values of the test statistic correspond to substantial differences between the observed and expected values, and such large values are located in the right tail of the distribution.

.

Section 11-2: Contingency Tables 5.

171

H0 : Success is independent of type of treatment.

H1: Success depends on type of treatment.

Test statistic: c 2 = 9.750; P-value = 0.002 (Table: P-value < 0.005); Critical value: c 2 = 6.635; df = (2 - 1)(2 - 1) = 1; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that success is independent of the type of treatment. The results suggest that the surgery treatment is better.

c2= 6.

(60 - 67.6)2 (23 - 15.4)2 (67 - 59.4)2 (6 - 13.6)2 +

67.6

+

15.4

+

59.4

13.6

= 9.750

H0: Texting while driving is independent of driving when drinking alcohol.

H1: Texting while driving depends on driving when drinking alcohol.

Test statistic: c 2 = 576.224; P -value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 3.841; df = (2 - 1)(2 - 1) = 1; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that success is independent of the type of treatment. The results suggest that the surgery treatment is better.

c2= 7.

(731 - 394.7)2 (3054 - 3390.3)2 (156 - 492.3)2 (4564 - 4227.7)2 +

394.7

+

3390.3

+

492.3

4227.7

= 576.224

H0: Adverse health condition is independent of restoration type.

H1: Adverse health condition depends on restoration type.

Test statistic: c 2 = 0.751; P-value = 0.3862 (Table: P-value > 0.10); Critical value: c 2 = 3.841; df = (2 - 1)(2 - 1) = 1; Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim of independence between the type of restoration and adverse health conditions. Amalgam restorations do not appear to affect health conditions.

c2= 8.

(135 - 140)2 (145 - 140)2 (132 - 127)2 (122 - 127)2 +

140

+

140

+

127

127

= 0.751

H0: Sensory disorder is independent of restoration type.

H1: Sensory disorder depends on restoration type.

Test statistic: c 2 = 1.136; P-value = 0.2864 (Table: P-value > 0.10); Critical value: c 2 = 3.841; df = (2 - 1)(2 - 1) = 1; Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim of independence between the type of restoration and sensory disorder. Amalgam restorations do not appear to affect sensory disorder.

c2= 9.

(36 - 32)2 (28 - 32)2 (231 - 235)2 (239 - 235)2 32

+

32

+

235

+

235

= 1.136

H0 : Dog's selection is independent of source of sample.

H1: Dog's selection depends on source of sample.

Test statistic: c 2 = 34.445; P -value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 6.635; df = (2 - 1)(2 - 1) = 1; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that the source of the sample is independent of the dog’s selections. The results suggest that the dogs have some ability to detect bladder cancer, but they did not do well enough for accurate diagnoses.

2

c =

(22 - 7.9)2 (32 - 46.1)2 (32 - 46.1)2 (282 - 267.9)2 7.9

+

46.1

+

46.1

+

267.9

.

= 34.445

172

Chapter 11: Goodness-of-Fit and Contingency Tables

10. H0: Polygraph result is independent of truthfulness of subject.

H1: Polygraph result depends on truthfulness of subject.

Test statistic: c 2 = 25.571; P -value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 3.841; df = (2 - 1)(2 - 1) = 1; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that the results of a polygraph test is independent of the subjects’ truthfulness. The results suggest that polygraphs are effective in distinguishing between truths and lies.

c2=

(15 - 27.3)2 (42 - 29.7)2 (32 - 19.7)2 (9 - 21.3)2 27.3

+

+

29.7

19.7

+

21.3

= 25.571

11. H0: Nausea is independent of treatment or placebo.

H1: Nausea depends on treatment or placebo.

Test statistic: c 2 = 9.854; P -value = 0.0017 (Table: P-value < 0.005); Critical value: c 2 = 6.635; df = (2 - 1)(2 - 1) = 1; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that nausea is independent of whether the subject took a placebo or Chantix. It appears that nausea is more likely to occur among those who use Chantix, so nausea is a concern. However, the rate of nausea among Chantix users is only about 3.7%, so it is not much of a concern.

c2=

(10 - 19.8)2 (30 - 20.2)2 (795 - 785.2)2 (791 - 800.8)2 19.8

+

+

20.2

785.2

+

800.8

= 9.854

12. H0: Gender is independent of whether call is overturned.

H1: Call is overturned depends on gender.

Test statistic: c 2 = 153.462; P -value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 3.841; df = (2 - 1)(2 - 1) = 1; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that developing the flu is independent of whether the subject took a placebo or the vaccine. The rate of developing the flu among the vaccine recipients (1.3%) is much lower than the rate for placebo recipients (17.9%), so it appears the vaccine is effective.

2

c =

(14 - 72.8)2 (1056 - 997.2)2 (95 - 36.2 )2 (437 - 495.8)2 72.8

+

997.2

+

36.2

+

495.8

= 153.462

13. H0 : Texting while driving is independent of irregular seat belt use.

H1: Texting while driving depends on irregular seat belt use.

Test statistic: c 2 = 18.773; P -value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 3.841; df = (2 - 1)(2 - 1) = 1; Reject H 0 . There is sufficient evidence to warrant rejection of the claim of independence between texting while driving and irregular seat belt use. Those two risky behaviors appear to be somehow related.

c2=

(1737 - 1638.6)2 (2048 - 2146.4)2 (1945 - 2043.4 )2 (2775 - 2676.6)2 +

1638.6

2146.4

+

+

2043.4

2676.6

= 18.773

14. H0 : Deaths on a shift is independent of whether Gilbert was working.

H1: Deaths on a shift depends on whether Gilbert was working.

Test statistic: c 2 = 86.481; P -value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 6.635; df = (2 - 1)(2 - 1) = 1; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that deaths on shifts are independent of whether Gilbert was working. The results favor the guilt of Gilbert.

c2=

(40 - 11.6)2 (217 - 245.4)2 (34 - 62.4 )2 (1350 - 1321.6)2 11.6

+

245.4

+

62.4

+

1321.6

.

= 86.481

Section 11-2: Contingency Tables

173

15. H0 : Adverse effect is independent of treatment.

H1: Adverse effect depends on treatment.

Test statistic: c 2 = 42.568; P -value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 9.210; df = (2 - 1)(3 - 1) = 2; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that experiencing an adverse reaction in the digestive system is independent of the treatment group. Treatments with 1332 mg doses of Campral appear to be associated with an increase in adverse effects of the digestive system.

c2=

(344 - 284.1)2 (89 - 143.7 )2 (8 - 13.2 )2 +

284.1

2

+

(1362 - 1421.9 ) 1421.9

+

143.7

13.2 2

+

(774 - 719.3)

+

719.3

(71 - 65.8)2 65.8

= 42.568

16. H0 : Infection is independent of treatment.

H1: Infection depends on treatment.

Test statistic: c 2 = 0.773; P-value = 0.8560 (Table: P-value > 0.10); Critical value: c 2 = 7.815; df = (2 - 1)(4 - 1) = 3; Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that infections are independent of treatment. Based on the given sample data, the atorvastatin treatment does not appear to have an effect on infections.

c2=

(27 - 27.1)2 (89 - 86.6)2 (8 - 7.9 )2 (7 - 9.4 )2 27.1

+

86.6

2

+

(243 - 242.9 ) 242.9

+

2

+

+

7.9

(774 - 776.4 )

+

776.4

9.4

(71 - 71.1)2 (87 - 84.6 )2 71.1

+

84.6

= 0.773

17. H0 : Amount of smoking is independent of seat belt use.

H1: Amount of smoking depends on seat belt use.

Test statistic: c 2 = 1.358; P-value = 0.715 (Table: P-value > 0.10); Critical value (a = 0.05): c 2 = 7.815; df = (2 - 1)(4 - 1) = 3; Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that the amount of smoking is independent of seat belt use. The theory is not supported by the given data.

c2=

(175 - 171.5)2 171.5

+

+

(6 - 7.9)2 (149 - 152.5)2 7.9

+

152.5

+

+

(9 - 7.1)2 7.1

= 1.358

18. H0 : Getting a cold is independent of treatment.

H1: Getting a cold depends on treatment.

Test statistic: c 2 = 2.925; P-value = 0.232 (Table: P-value > 0.10); Critical value: c 2 = 5.991; df = (2 - 1)(3 - 1) = 2; Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that getting a cold is independent of the treatment group. The results suggest that echinacea is not effective for preventing colds.

c2=

(88 - 88.6)2 (48 - 44.7)2 (42 - 44.7)2 (15 - 14.4)2 (4 - 7.3)2 (10 - 7.3)2 88.6

+

44.7

+

44.7

+

14.4

+

7.3

+

7.3

= 2.925

19. H0 : Injuries are independent of helmet color.

H1: Injuries depend on helmet color.

Test statistic: c 2 = 9.971; P-value = 0.041 (Table: P-value < 0.05); Critical value (a = 0.05): c 2 = 9.488; df = (2 - 1)(5 - 1) = 4; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that injuries are independent of helmet color. It appears that motorcycle drivers should use yellow or orange helmets.

c2=

(491 - 509.5)2 509.5

+

+

(55 - 58.6)2 (213 - 194.5)2 58.6

+

194.5 .

+

+

(26 - 22.4)2 22.4

= 9.971

174

Chapter 11: Goodness-of-Fit and Contingency Tables

20. H0 : Months of birth of baseball players are independent of being born in America.

H1: Months of birth of baseball players depends on being born in America.

Test statistic: c 2 = 20.054; P-value = 0.0446 (Table: P-value < 0.05); Critical value: c 2 = 19.675; df = (2 - 1)(12 - 1) = 11; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that months of births of baseball players are independent of whether they are born in America. The data do appear to support Gladwell’s claim.

c2=

(387 - 397.2)2 397.2

+

+

(371 - 368.7)2 (101 - 90.8)2 368.7

+

90.8

+

+

(82 - 84.3)2 84.3

= 20.054

21. From Exercise 5, c 2 = 9.7504 and z = - 3.122560496, so z 2 = c 2 . The critical values are : c 2 = 6.635 and

z 2 = ±2.57583 (Table: ± 2.575), so z 2 = c 2 (approximately).

22. Without Yates’s correction, the test statistic is c 2 = 9.750. With Yates’s correction, the test statistic is

c 2 = 8.574. Yates’s correction decreases the test statistic so that sample data must be more extreme in order to reject the null hypothesis of independence.

Without Yates’s correction:

c2=

(60 - 67.6)2 (23 - 15.4)2 (67 - 59.4)2 (6 - 13.6)2 67.6

+

15.4

+

59.4

+

13.6

= 9.750

With Yates’s Correction: 2 2 2 2 60 - 67.6 - 0.5) ( 23 - 15.4 - 0.5) ( 67 - 59.4 - 0.5) ( 6 - 13.6 - 0.5) ( c = + + + = 8.574 2

67.6

15.4

59.4

13.6

Chapter Quick Quiz 1.

H0 : p0 = p1 = p2 = p3 = p4 = p5 = p6 = p7 = p8 = p9 = 0.1; H1: At least one of the probabilities is different from the others.

2.

O = 27 and E = 0.1×300 = 30

3.

right-tailed

4.

df = n - 1 = 10 - 1 = 9

5. There is not sufficient evidence to warrant rejection of the claim that the last digits are equally likely. Because reported heights would likely include more last digits of 0 and 5, it appears that the heights were measured instead of reported. (Also, most U.S. residents would have difficulty reporting heights in centimeters, because the United States, Liberia, and Myanmar are the only countries that continue to use the Imperial system of measurement.)

6.

H 0 : Surviving the sinking is independent of whether the person is a man, woman, boy, or girl.

H1: Surviving the sinking and whether the person is a man, woman, boy, or girl are somehow related.

7.

chi-squared distribution

8.

right-tailed

9.

df = ( r - 1)( c - 1) = (2 - 1)(4 - 1) = 3

10. There is sufficient evidence to warrant rejection of the claim that surviving the sinking is independent of whether the person is a man, woman, boy, or girl. Most of the women survived, 45% of the boys survived, and most girls survived, but only about 20% of the men survived, so it appears that the rule was followed quite well.

.

Review Exercises

175

Review Exercises 1.

H 0 : pJan = pFeb = pMar = pApr = pMay = pJun = pJul = pAug = pSep = pOct = pNov = pDec = 1 12;

H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 269.147; P -value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 24.725; Reject

H 0 . There is sufficient evidence to warrant rejection of the claim that weather related deaths occur in the different months with the same frequency. The months of May, June, and July appear to have disproportionately more weather-related deaths, and that is probably due to the fact that vacations and outdoor activities are much greater during those months. c2= 2.

(28 - 37.5)2 (17 - 37.5)2 450 12

+

450 12

+

+

(26 - 37.5)2 (25 - 37.5)2 450 12

+

= 269.147 (df = 11)

450 12

H0 : Norovirus is independent of ship.

H1: Norovirus depends on ship.

Test statistic: c 2 = 71.679 P -value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 3.841; df = (2 - 1)(2 - 1) = 1; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that getting norovirus infection is independent of the ship. It appears that an outbreak of norovirus infection has a different effect on different ships.

c2= 3.

(276 - 185.3)2 (1376 - 1466.7)2 (338 - 428.7 )2 (3485 - 3394.3)2 185.3

+

+

1466.7

+

428.7

3394.3

= 71.679

H 0 : pJan = pFeb = pMar = pApr = pMay = pJun = pJul = pAug = pSep = pOct = pNov = pDec = 1 12;

H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 10.375; P-value = 0.4970 (Table: P-value > 0.10); Critical value: c 2 = 19.675; Fail to reject

H 0 . There is not sufficient evidence to warrant rejection of the claim that homicides in New York City are equally likely for each of the 12 months. There is not sufficient evidence to support the police commissioner’s claim that homicides occur more often in the summer when the weather is warmer. c2= 4.

(38 - 42.667)2 (30 - 42.667)2 +

512 12

512 12

+

+

(37 - 42.667)2 (37 - 42.667)2 512 12

+

512 12

= 10.375 (df = 11)

H0 : Left-handedness is independent of parental handedness.

H1: Left-handedness dependents on parental handedness.

Test statistic: c 2 = 784.647; P -value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 11.345; df = (4 - 1)(2 - 1) = 3; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that lefthandedness is independent of parental handedness. It appears that handedness of the parents has an effect on handedness of the offspring, so left-handedness appears to be an inherited trait.

2

c = 5.

(5360 - 6067.9)2 6067.9

+

+

(2736 - 3125.4)2 (741 - 475.2)2 3125.4

+

475.2

+

+

(289 - 341.7)2 341.7

= 784.647

H0 : pUnder 25 = 0.16, p25- 44 = 0.44, p45- 64 = 0.27, pOver 64 = 0.13; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 53.051; P -value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 7.815; Reject H 0 .

There is sufficient evidence to warrant rejection of the claim that the distribution of crashes is the same as the distribution of ages. Drivers under 25 appear to have disproportionately more crashes.

c2=

(36 - 14.08)2 (5985 - 38.72)2 (5985 - 23.76)2 (5985 - 11.44 )2 0.16 ×88

+

0.44 ×88

+

0.27 ×88

0.13 ×88

.

= 53.051 (df = 3)

176

Chapter 11: Goodness-of-Fit and Contingency Tables

Cumulative Review Exercises 1.

x=

29 + 35 + 38 + 42 + 58 + 62 + 64 + 67 + 68 + 74 58 + 62 = 60.0 years; = 53.7 years; Q2 = 2 10

(29 - 53.7)2 + (35 - 53.7) 2 + + (68 - 53.7) 2 + (74 - 53.7) 2 = 16.1 years; s2 = 16.12 = 259.2 years2 (258.9 10 - 1 16 - 53.7 using unrounded values). Because an age of 16 is = - 2.34, or more than 2 standard deviations below the 16.1 mean of 53.7 years, it is significantly low. s=

2.

The sample appears to meet the loose requirement for being normally distributed. s 16.1 = 53.7 ± 2.626 × Þ 42.2 years < m < 65.2 years; Yes, the confidence interval limits do 95% CI: x ± ta / 2 n 10 contain the value of 65.0 years that was found from a sample of 9269 ICU patients.

3.

H0 : Facial injuries received is independent of wearing helmet.

H1: Facial injuries received depends on wearing helmet.

Test statistic: c 2 = 10.708; P-value = 0.0011 (Table: P-value < 0.005); Critical value: c 2 = 3.841; df = (2 - 1)(2 - 1) = 1; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that wearing a helmet has no effect on whether facial injuries are received. It does appear that a helmet is helpful in preventing facial injuries in a crash.

c2= 4.

5.

(30 - 45.1)2 (182 - 166.9)2 (83 - 67.9)2 (236 - 251.1)2 45.1

+

166.9

a.

30 + 83 + 236 349 = , or 0.657 531 531

b.

113 112 6328 × = , or 0.0450 531 530 93,015

c.

182 + 236 418 = , or 0.787 531 531

+

67.9

+

251.1

= 10.708

a. The z score for the bottom 5% is –1.645, which correspond a forward grip reach of - 1.645 ×34 + 686 = 630 mm.

650 - 686 = - 1.06; which has a probability of 0.1448, or 14.48% (Table: 14.46%) to the left. That 34 percentage is too high, because too many women would not be accommodated. b. z x =650 =

680 - 686 = - 0.71; which has a probability of 1 - 0.2401 = 0.7599 (Table: 0.7611) to the right.

34 16 Groups of 16 women do not occupy a driver’s seat or cockpit; because individual women occupy the driver’s seat/cockpit, this result has no effect on the design. c. z x=680 =

Determine whether there is a linear correlation between systolic blood pressure and waist size. r = 0.289; P-value = 0.5791 (Table: P-value > 0.05); Critical values (a = 0.05): r = ±0.8114; There is not sufficient evidence to support the claim that for females, there is a linear correlation between systolic blood pressure and waist size.

6.

.

Section 12-1: One-Way ANOVA

177

Chapter 12: Analysis of Variance Section 12-1: One-Way ANOVA 1. a. The chest deceleration measurements are categorized according to the one characteristic of car size. b. The terminology of analysis of variance refers to the method used to test for equality of the three population means. That method is based on two different estimates of a common population variance.

2. As we increase the number of individual tests of significance, we increase the risk of finding a difference by chance alone (instead of a real difference in the means). The risk of a type I error, finding a difference in one of the pairs when no such difference actually exists, is too high. The method of analysis of variance helps us avoid that particular pitfall (rejecting a true null hypothesis) by using one test for equality of several means, instead of several tests that each compare two means at a time.

3.

The test statistic is F = 3.288, and the F distribution applies.

4. The P-value is 0.061. Because the P-value is greater than the significance level of 0.05, we fail to reject the null hypothesis of equal means. There is not sufficient evidence to warrant rejection of the claim that the different size categories have the same mean chest deceleration. Based on the available data, the size of a car does not appear to affect chest deceleration.

5. Test statistic: F = 0.39; P-value: 0.677; Fail to reject H0: m1 = m2 = m3. There is not sufficient evidence to warrant rejection of the claim that the three categories of blood lead level have the same mean verbal IQ score.

Exposure to lead does not appear to have an effect on verbal IQ scores.

6. Test statistic: F = 2.3034; P-value: 0.1044; Fail to reject H0: m1 = m2 = m3. There is not sufficient evidence to warrant rejection of the claim that the three categories of blood lead level have the same mean full IQ score. There is not sufficient evidence to conclude that exposure to lead has an effect on full IQ scores.

7. Test statistic: F = 0.161; P-value: 0.852; Reject H0: m1 = m2 = m3. There is not sufficient evidence to warrant rejection of the claim that the three size categories have the same mean head injury measurement. Based on the available data, the size of a car does not appear to affect head injuries.

8. Test statistic: F = 1.1810; P-value: 0.3167; Fail to reject H0 : m1 = m2 = m3 = m4 . There is not sufficient evidence to support the claim that the different hospitals have different mean birth weights. It appears that birth weights are about the same in urban and rural areas.

9. Test statistic: F = 1.304; P-value: 0.275; Fail to reject H0: m1 = m2 = m3. There is not sufficient evidence to warrant rejection of the claim that males from the three age brackets have the same mean pulse rate. It appears that pulse rates of males are not affected by age bracket.

10. Test statistic: F = 7.9338; P-value: 0.0005; Reject H0: m1 = m2 = m3. There is sufficient evidence to warrant rejection of the claim that females from the three age brackets have the same mean pulse rate. It appears that pulse rates of females are affected by age bracket.

11. Test statistic: F = 0.3476; P-value: 0.7111; Fail to reject H0: m1 = m2 = m3. There is not sufficient evidence to warrant rejection of the claim that the three size categories have the same mean pelvis injury measurement. The size of a car does not appear to affect pelvis injuries.

12. Test statistic: F = 22.9477; P-value: 0.000; Reject H0: m1 = m2 = m3. There is sufficient evidence to warrant rejection of the claim that the three different states have the same mean arsenic content in brown rice. Even though Texas has the highest mean, the results from ANOVA do not allow us to conclude that any one specific population mean is different from the others, so we cannot conclude that brown rice from Texas poses the greatest health problem.

13. Test statistic: F = 5.7321; P-value: 0.0071. Reject H0: m1 = m2 = m3. There is sufficient evidence to warrant rejection of the claim that the four treatment categories yield poplar trees with the same mean weight. Although not justified by the results from analysis of variance, the treatment of fertilizer and irrigation appears to be most effective.

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14. Test statistic: F = 2.5411; P-value: 0.0931. Reject H0: m1 = m2 = m3. There is not sufficient evidence to warrant rejection of the claim that the four treatment categories yield poplar trees with the same mean weight.

Although not justified by the results from analysis of variance, the treatment of fertilizer and irrigation appears to be most effective.

15. Test statistic: F = 18.9931; P-value: 0.000; Reject H0: m1 = m2 = m3. There is sufficient evidence to warrant rejection of the claim that the three different types of cigarettes have the same mean amount of nicotine. Given that the king-size cigarettes have the largest mean of 1.26 mg per cigarette, compared to the other means of 0.87 mg per cigarette and 0.92 mg per cigarette, it appears that the filters do make a difference, although this conclusion is not justified by the results from analysis of variance.

16. Test statistic: F = 20.8562; P-value: 0.000; Reject H0: m1 = m2 = m3. There is sufficient evidence to warrant rejection of the claim that the three samples are from populations with the same mean. It appears that cotinine levels are greater with more exposure to tobacco smoke. (The samples do not appear to be from normally distributed populations, but ANOVA is robust against departures from normality. The samples appear to have very different variances, but the largest variance is roughly five times that of the smallest, and given that the sample sizes are all 40, the requirement of equal variances appears to be satisfied.)

17. The Tukey test results show different P-values, but they are not dramatically different. The Tukey results suggest the same conclusions as the Bonferroni test.

18. a. Test statistic: F = 6.1413; P-value: 0.0056; Reject H0: m1 = m2 = m3. There is sufficient evidence to warrant rejection of the claim that the four treatment categories yield poplar trees with the same mean weight. b. The displayed Bonferroni results show that with a P-value of 0.039, there is a significant difference between the mean of the no treatment group (group 1) and the mean of the group treated with both fertilizer and irrigation (group 4). (Also, when comparing group 1 and group 2, there is no significant difference between means, and when comparing group 1 and group 3, there is no significant difference between means.) c. Test statistic : t = - 4.007; P -value: 6(0.001018) = 0.00611; Reject the null hypothesis that the mean weight from the irrigation treatment group is equal to the mean from the group treated with both fertilizer and irrigation.

Section 12-2: Two-Way ANOVA 1. The pulse rates are categorized using two different factors of (1) age bracket and (2) gender.

2. No, to use two individual tests of one-way analysis of variance is to totally ignore the very important feature of the possible effect from an interaction between age bracket and gender. If there is an interaction, it doesn’t make sense to consider the effects of one factor without the other.

3.

a. An interaction between two factors or variables occurs if the effect of one of the factors changes for different categories of the other factor. b. If there is an interaction effect, we should not proceed with individual tests for effects from the row factor and column factor. If there is an interaction, we should not consider the effects of one factor without considering the effects of the other factor. c. Because the lines are far from parallel, the two genders have very different effects for the different age brackets, so there does appear to be an interaction between gender and age bracket.

4.

Yes, the result is a balanced design because each cell has the same number (10) of values.

5. For interaction, the test statistic is F = 9.58 and the P-value is 0.0003, so there is sufficient evidence to warrant rejection of the null hypothesis of no interaction effect. Because there appears to be an interaction between age bracket and gender, we should not proceed with a test for an effect from age bracket and a test for an effect from gender. It appears an interaction between age bracket and gender has an effect on pulse rates. (Remember, these results are based on fabricated data used in one of the cells, so this conclusion does not necessarily correspond to real data.)

.

Chapter Quick Quiz

179

6. For interaction, the test statistic is F = 3.6653 and the P-value is 0.0322, so there is sufficient evidence to warrant rejection of no interaction effect. Because there appears to be an interaction effect, the tests for effects from age bracket and gender are not conducted. There appears to be sufficient evidence to support a claim that weight appears to be affected by an interaction between age bracket and gender.

7. For interaction, the test statistic is F = 1.7970 and the P-value is 0.1756, so there is not sufficient evidence to conclude that there is an interaction effect. For the row variable of age bracket, the test statistic is F = 2.0403 and the P-value is 0.1399, so there is not sufficient evidence to conclude that age bracket has an effect on height. For the column variable of gender, the test statistic is F = 43.4607 and the P-value is less than 0.0001, so there is sufficient evidence to support the claim that gender has an effect on height.

8. For interaction, the test statistic is F = 2.88 and the P-value is 0.076, so there is not sufficient evidence to warrant rejection of no interaction effect. There does not appear to be an interaction between sex and age. For the row variable of sex, the test statistic is F = 2.01 and the P-value is 0.169, so there is not sufficient evidence to warrant rejection of the claim of no effect from gender. For the column variable of age, the test statistic is F = 0.75 and the P-value is 0.482, so there is not sufficient evidence to conclude that age has an effect on cholesterol levels. It appears that cholesterol levels are not affected by an interaction between sex and age, are not affected by sex, and are not affected by age.

9. For interaction, the test statistic is F = 3.7332 and the P-value is 0.0291, so there is sufficient evidence to conclude that there is an interaction effect. The measures of self-esteem appear to be affected by an interaction between the self-esteem of the subject and the self-esteem of the target. Because there appears to be an interaction effect, we should not proceed with individual tests of the row factor (target’s self-esteem) and the column factor (subject’s self-esteem).

10. For interaction, the test statistic is F = 0.6921 and the P-value is 0.4241, so there is not sufficient evidence to warrant rejection of no interaction effect. There does not appear to be an interaction between gender and smoking.

For the row variable of gender, the test statistic is F = 0.3314 and the P-value is 0.5762, so there is not sufficient evidence to warrant rejection of the claim of no effect from gender. For the column variable of smoking, the test statistic is F = 0.2415 and the P-value is 0.6354, so there is not sufficient evidence to conclude that smoking has an effect on body temperatures. It appears that body temperatures are not affected by an interaction between gender and smoking, they are not affected by gender, and they are not affected by smoking.

11. a. Test statistics and P-values do not change. b. Test statistics and P-values do not change. c. Test statistics and P-values do not change. d. An outlier can dramatically affect and change test statistics and P-values.

Chapter Quick Quiz 1. The sample data are partitioned into the three different categories according to the one factor of epoch.

2.

H0: m1 = m2 = m3; H1: At least one of the three population means is different from the others.

3.

Test statistic: F = 4.0497; Larger test statistics result in smaller P-values.

4. Reject H 0 . There is sufficient evidence to support the claim that the different epochs have mean skull breadths that are not all the same.

5.

Right-tailed. Yes, all one-way analysis of variance tests are right-tailed.

6. No. The method of analysis of variance does not justify a conclusion that any particular mean is different from the others.

7. With one-way analysis of variance, data from the different samples are categorized using only one factor, but with two-way analysis of variance, the sample data are categorized into different cells determined by two different factors.

.

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Chapter 12: Analysis of Variance

8. Test statistic: F = 1.41; P-value: 0.281; Fail to reject the null hypothesis of no interaction effect. There is not sufficient evidence to warrant rejection of the claim that head injury measurements are not affected by an interaction between the type of car (foreign, domestic) and size of the car (small, medium, large). There does not appear to be an effect from an interaction between the type of car (foreign or domestic) and whether the car is small, medium, or large.

9. Test statistic: F = 2.25 P-value: 0.159; Fail to reject the null hypothesis of no effect from the type of car.

There is not sufficient evidence to support the claim that whether the car is foreign or domestic has an effect on head injury measurements.

10. Test statistic: F = 0.44; P-value: 0.655; Fail to reject the null hypothesis of no effect from the size of the car.

There is not sufficient evidence to support the claim that whether the car is small, medium, or large has an effect on head injury measurements.

Review Exercises 1. a. one (type of diet) b. one-way analysis of variance c. Because the P-value is high, it appears that the four samples have means that do not differ by significant amounts. It appears that the mean ages of the four treatment groups are about the same. d. A small P-value would indicate that at least one of the treatment groups has a mean age that is significantly different from the others, so we would not know if differences from the diet treatments are due to the diets or to differences in age. A small P-value would undermine the effectiveness of the experiment.

2. Test statistic: F = 18.9941; P-value: 0.000. Reject H0: m1 = m2 = m3. There is sufficient evidence to warrant rejection of the claim that the three different types of cigarettes have the same mean amount of nicotine. Given that the king-size cigarettes have the largest mean of 1.3 mg per cigarette, compared to the other means of 0.9 mg per cigarette and 0.9 mg per cigarette, it appears that the filters do make a difference, although this conclusion is not justified by the results from analysis of variance.

3. For interaction, the test statistic is F = 1.7171 and the P-value is 0.1940, so there is not sufficient evidence to warrant rejection of no interaction effect. There does not appear to be an interaction between femur and car size.

For the row variable of femur, the test statistic is F = 1.3896 and the P-value is 0.2462, so there is not sufficient evidence to conclude that whether the femur is right or left has an effect on load. For the column variable of car size, the test statistic is F = 2.2296 and the P-value is 0.1222, so there is not sufficient evidence to warrant rejection of the claim of no effect from car size. It appears that the crash test loads are not affected by an interaction between femur and car size, they are not affected by femur, and they are not affected by car size.

4. For interaction, the test statistic is F = 2.1018 and the P-value is 0.1731, so there does not appear to be an effect from an interaction between gender and whether the subject smokes. For gender, the test statistic is F = 2.1518 and the P-value is 0.1671, so gender does not appear to have an effect on body temperature. For smoking, the test statistic is F = 1.2110 and the P-value is 0.2921, so there appears to be an effect of smoking on body temperature.

Cumulative Review Exercises 1.

a. Presidents: x = 15.5 years; Popes: x = 13.1 years; Monarchs: x = 22.7 years b. Presidents: s = 9.7 years; Popes: s = 9.0 years; Monarchs: s = 18.6 years c. Presidents: s 2 = 94.5 years2 ; Popes: s 2 = 80.3 years2 ; Monarchs: s2 = 346.1 years2 d. ratio

.

Cumulative Review Exercises

181

2. H0 : m1 = m2 ; H1: m1 ¹ m2 ; population1 = presidents, population 2 = monarchs; Test statistic: t = - 1.383; P -value = 0.1860 (Table: P-value > 0.10); Critical values (a = 0.05): t = ±2.123 Fail to reject H 0 . There is not sufficient evidence to support the claim that there is a difference between the means of the two groups. ( x - x ) - ( m1 - m2 ) = (15.5 - 22.7 ) - 0 = - 1.383 (df = 13) t= 1 2 s12 s22 9.72 18.62 + + 34 14 n1 n2 3. Normal, because the histogram is approximately bell-shaped or the points in a normal quantile plot are reasonably close to a straight-line pattern with no other pattern that is not a straight line pattern.

9 8 7

Frequency

6 5 4 3 2 1 0

0

5

10

15

20

25

30

35

40

Longevity Time (Presidents)

4.

The data appear to fit the loose definition of a normally distribution. s 9.7 = 15.5 ± 2.035 × Þ 12.3 years < m < 18.7 years; We have 95% confidence that the limits 95% CI: x ± ta / 2 n 34 of 12.3 years and 18.7 years contain the true value of the population mean.

5.

a. H0 : m1 = m2 = m3 b. Because the P-value of 0.051 is greater than the significance level of 0.05, fail to reject the null hypothesis of equal means. There is not sufficient evidence to warrant rejection of the claim that the three means are equal.

The three populations do not appear to have means that are significantly different.

6.

a. r = 0.992; Critical values: r = ±0.707; P -value = 0.002; There is sufficient evidence to support the claim that there is a linear correlation between September weights and the subsequent April weights. b. yˆ = 13.09 + 0.844 x c. 13.09 + 0.844(70) = 72.17 kg, which is very close to the actual April weight of 73 kg.

7.

a. z x =345 =

345 - 280 = 1.00; which has a probability of 1 - 0.8413 = 0.1587 to the right.

65

215 - 280 345 - 280 = - 1.00 and z x =345 = = 1.00; which have a probability of 65 65 0.8413 - 0.1587 = 0.6826 (Tech: 0.6827) between them.

b. z x =215 =

c. z x=319 =

319 - 280 = 3.00; which has a probability of 0.9987 to the left.

65 25

d. The z score for the bottom 80% is 0.84, which correspond to a blood platelet count of 0.84 ×65 + 280 = 334.6 (1000 cells mL ). (Tech: 334.7)

.

182 8.

Chapter 12: Analysis of Variance a. 0.20(1000) = 200 b. 95% CI: pˆ ± za / 2

ˆˆ pq = 0.20 ± 1.96 n

(0.20)(0.80)

Þ 0.175 < p < 0.225

1000

c. Yes, the confidence interval shows us that we have 95% confidence that the true population proportion is contained within the limits of 0.175 and 0.225, and 1 4 is not included within that range.

9. Using normal as approximation to binomial: 0.1020. (Exact result using binomial distribution: 0.0995.)

Assuming that one-quarter of all offspring have blue eyes, the probability of getting 19 or fewer offspring with blue eyes is high, so there is not sufficient evidence to conclude that the one-quarter rate is wrong.

19.5 - 100 ×0.25 = - 1.27 Normal approximation: np = 100 ×0.25 = 25 ³ 5 and nq = 100 ×0.75 = 75 ³ 5; z x =19.5 = 100 ×0.25 ×0.75 Binomial distribution: 100 C0 ×0.250 ×0.75100 +

+ 100 C19 ×0.2519 ×0.7581

10. H0 : Injury category is independent of type of weapon.

H1: Injury category depends on type of weapon.

Test statistic: c 2 = 2.909; P-value = 0.2335 (Table: P-value > 0.10); Critical value: c 2 = 5.991; df = (2 - 1)(3 - 1) = 2; Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim of independence between injury category and whether the firearm was a handgun or a rifle or shotgun. The type of injury doesn’t appear to be affected by whether the firearm is a handgun or a rifle or shotgun.

c2=

(31 - 31.8)2 (35 - 30.4)2 (162 - 165.8)2 (13 - 12.2)2 (7 - 11.6)2 (67 - 63.2)2 31.8

+

30.4

+

165.8

+

12.2

.

+

11.6

+

63.2

= 2.909

13-2: Sign Test

183

Chapter 13: Nonparametric Tests 13-2: Sign Test 1. a. The only requirement for the matched pairs is that they constitute a simple random sample. b. There is no requirement of a normal distribution or any other specific distribution. c. The sign test is “distribution-free” in the sense that it does not require a normal distribution or any other specific distribution.

2. There are 4 positive signs, 2 negative signs, no tie, n = 6, and the test statistic is x = 2 (the smaller of 2 and 4).

3.

H 0 : there is no difference between the populations of 8 AM temperatures on Day 1 and the matching 8 AM temperatures on Day 2. H1: there is a difference between the populations of 8 AM temperatures on Day 1 and the matching 8 AM temperature on Day 2. The sample data do not contradict H1 because the number of positive signs (4) and negative signs (2) are not exactly the same.

4. The efficiency of 0.63 indicates that with all other things being equal, the sign test requires 100 sample observations to achieve the same results as 63 sample observations analyzed through a parametric test.

5. The test statistic of x = 4 is not less than or equal to the critical value of 1 (From Table A-7). There is not sufficient evidence to reject the claim of no difference in heights between mothers and their first daughters.

6. The test statistic of x = 4 is not less than or equal to the critical value of 1 (From Table A-7). There is not sufficient evidence to warrant rejection of the claim of no difference. There is not sufficient evidence to support the claim that there is a difference in heights between fathers and their first sons.

7. The test statistic of x = 1 is not less than or equal to the critical value of 0 (from Table A-7). There is not sufficient evidence to warrant rejection of the claim that when the 13th day of a month falls on a Friday, the numbers of hospital admissions from motor vehicle crashes are not affected. Hospital admissions do not appear to be affected.

8. The test statistic of x = 0 is less than or equal to the critical value of 1 (from Table A-7). There is sufficient evidence to warrant rejection of the claim that Captopril affects blood pressure. It appears that Captopril lowers blood pressure.

The test statistic of z =

(401 + 0.5) - 882 2

10. The test statistic of z =

(372 + 0.5) - 1228 2

= - 2.66 results in a P-value of 0.0078, and it is in the critical region 882 2 bounded by z = - 2.575 and 2.575. There is sufficient evidence to warrant rejection of the claim that there is no difference between the proportions of those opposed and those in favor.

9.

= - 13.78 results in a P-value of 0.0000, and it is in the critical region 1228 2 bounded by z = - 2.575 and 2.575. There is sufficient evidence to support the claim that there is a difference between the rate of medical malpractice lawsuits that go to trial and the rate of such lawsuits that are dropped or dismissed.

(426 + 0.5) - 860 2

= - 0.24 results in a P-value of 0.8103, and it is not in the critical 860 2 region bounded by z = - 1.96 and 1.96. There is not sufficient evidence to reject the claim that boys and girls are equally likely.

11. The test statistic of z =

(123 + 0.5) - 280 2

= - 1.97 results in a P-value of 0.0486, and it is not in the critical 280 2 region bounded by z = - 2.575 and 2.575. There is not sufficient evidence to warrant rejection of the claim that that the therapists do better than random guessing. It appears that therapists are not effective at identifying the correct hand.

12. The test statistic of z =

.

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Chapter 13: Nonparametric Tests

13. The test statistic of x = 9 is not less than or equal to the critical value of 5. There is not sufficient evidence to warrant rejection of the claim that the sample is from a population with a median IQ score of 100.

(21 + 0.5) - 772

= - 3.87 results in a P-value of 0.0000, and it is in the critical region 77 2 bounded by z = - 2.575 and 2.575. There is sufficient evidence to warrant rejection of the claim that the median full IQ score of the population with low lead exposure is equal to 100.

14. The test statistic of z =

(117 + 0.5) - 273 2

= - 2.30 results in a P-value of 0.0215 and it is not in the critical 273 2 region bounded by z = - 1.96 and 1.96. There is not sufficient evidence to warrant rejection of the claim that the median is equal to 72 mm Hg.

15. The test statistic of z =

(123 + 0.5) - 300 2

= - 3.06 results in a P-value of 0.0022, and it is in the critical region 300 2 bounded by z = - 1.96 and 1.96. There is sufficient evidence to warrant rejection of the claim that the systolic blood pressure of the population is equal to 125 mm Hg.

16. The test statistic of z =

17. Second approach: The test statistic of z =

(30 + 0.5) - 105 2

= - 4.29 is in the critical region bounded by 105 2 z = - 1.645 , so the conclusions are the same as in Example 4.

(38 + 0.5) - 106 2 = - 2.82 is in the critical region bounded by z = - 1.645 , Third approach: The test statistic of z = 106 2 so the conclusions are the same as in Example 4. The different approaches can lead to very different results; as seen in the test statistics of –4.21, –4.29, and –2.82. The conclusions are the same in this case, but they could be different in other cases.

18. The column entries are *, *, *, *, *, 0, 0, 0.

13-3: Wilcoxon Signed-Ranks Test for Matched Pairs 1. a. The only requirements are that the matched pairs be simple random samples and the population of differences be approximately symmetric. b. There is no requirement of a normal distribution or any other specific distribution. c. The Wilcoxon signed-ranks test is “distribution-free” in the sense that it does not require a normal distribution or any other specific distribution.

2. a. 1.2, –0.2, 0.8, 0.8, 2.0, –0.2, –0.8 d. 21, 7 b. 6, 1.5, 4, 4, 7, 1.5, 4 e. T = 7 c. 6, –1.5, 4, 4, 7, –1.5, -4 f. Critical value of T is 4.

3. The sign test uses only the signs of the differences, but the Wilcoxon signed-ranks test uses the ranks that are affected by the magnitude of the differences.

4. The efficiency of 0.95 indicates that with all other things being equal, the Wilcoxon signed-ranks test requires 100 sample observations to achieve the same results as 95 sample observations analyzed through a parametric test.

5. Test Statistic: T = 7.0; critical value: T = 4. Fail to reject the null hypothesis that the population of differences has a median of 0. There is not sufficient evidence to reject the claim of no difference in heights between mothers and their first daughters. There does not appear to be a difference in heights between mothers and their first daughters.

6. Test Statistic: T = 12; critical value: T = 4. Fail to reject the null hypothesis that the population of differences has a median of 0. There is not sufficient evidence to reject the claim of no difference in heights between fathers and their first sons. There does not appear to be a difference in heights between fathers and their first sons.

.

13-4: Wilcoxon Rank-Sum Test for Two Independent Samples

185

7. Test statistic: T = 1.5; Critical value: T = 1; Fail to reject the null hypothesis that the population of differences has a median of 0. There is not sufficient evidence to warrant rejection of the claim that when the 13th day of a month falls on a Friday, the numbers of hospital admissions from motor vehicle crashes are not affected. Hospital admissions do not appear to be affected.

8. Test statistic: T = 0; Critical value: T = 10; There is sufficient evidence to warrant rejection of the claim that the before and after measurements are the same. there sufficient evidence to support the claim that captopril has an effect on systolic blood pressure 9. Test statistic: T = 98; Critical value: T = 52; There is not sufficient evidence to warrant rejection of the claim that the sample is from a population with a median IQ score of 100.

10. Convert T = 725.5 to the test statistic z = - 3.94. P-value: 0.0000. Critical values: z = ±2.575; There is sufficient evidence to warrant rejection of the claim that the median IQ of the low lead group is 100. n (n + 1) 77 (77 + 1) T725.5 4 4 z= = = - 3.94 n (n + 1)(2n + 1) 77 (77 + 1)(2 ×77 + 1) 24 24 11. Convert T = 16, 236 to the test statistic z = - 1.89 . P-value: 0.0588. Critical values: z = ±1.96; There is not sufficient evidence to warrant rejection of the claim that the sample is from a population with a median diastolic blood pressure level of 72 mm Hg. n (n + 1) 273 (273 + 1) T16, 236 4 4 z= = = - 1.89 n (n + 1)(2n + 1) 273 (273 + 1)(2 ×273 + 1) 24 24 12. Convert T = 17,826 to the test statistic z = - 3.16. P-value: 0.0016. Critical values: z = ±1.96; There is sufficient evidence to warrant rejection of the claim that the times are from a population with a median of 125 mm Hg. n (n + 1) 300 (300 + 1) T17,826 4 4 z= = = - 3.16 n (n + 1)(2n + 1) 300 (300 + 1)(2 ×300 + 1) 24 24 13. a. Smallest: T = 0; Largest: T = 1 + 2 + 3 +

+ 298 + 299 + 300 =

45,150 = 22,575 2 c. 45,150 - 12,345 = 32,805 b.

d.

300 (300 + 1) = 45,150 2 n (n + 1) -k 2

13-4: Wilcoxon Rank-Sum Test for Two Independent Samples 1. The two samples are from populations with the same median.

2.

3.

R = 19.5 + 2 + 5 + 26 + 12.5 + 4 + 9 + 15.5 + 7.5 + 6 + 24.5 + 12.5 + 10.5 + 17.5 = 172 Girl Rank

3500 19.5

800 2

2400 5

4200 26

3100 12.5

2000 4

2900 9

Girl Rank

3300 15.5

2800 7.5

2500 6

4000 24.5

3100 12.5

3000 10.5

3400 17.5

Yes. The samples are independent and both samples have more than 10 values.

.

186

Chapter 13: Nonparametric Tests

4. The efficiency rating of 0.95 indicates that with all other factors being the same, the Wilcoxon rank-sum test requires 100 sample observations to achieve the same results as 95 observations with the parametric t test of Section 9-2, assuming that the stricter requirements of the parametric t -test are satisfied.

R1 = 172; R2 = 179; mR = 189; s R = 19.4422; Test statistic: z = - 0.87; P-value = 0.3843; Critical values: z = ±1.96; Fail to reject the null hypothesis that the populations have the same median. There is not sufficient evidence to warrant rejection of the claim that girls and boys have the same median birth weight. n (n + n + 1) 14 (14 + 12 + 1) mR = 1 1 2 = = 189 2 2

5.

n1n2 (n1 + n2 + 1) 14 ×12 (14 + 12 + 1) = = 19.4422 12 12 R - mR 172 - 189 z= = = - 0.87 sR 19.4422

sR=

R1 = 194.5; R2 = 105.5; mR = 150; s R = 17.321; Test statistic: z = 2.57; P-value = 0.0102; Critical values: z = ±1.96; Reject the null hypothesis that the populations have the same median. There is sufficient evidence to reject the claim that the median amount of strontium-90 from Pennsylvania residents is the same as the median from New York residents. n (n + n + 1) 12 (12 + 12 + 1) mR = 1 1 2 = = 150 2 2

6.

n1n2 (n1 + n2 + 1) 12 ×12 (12 + 12 + 1) = = 17.321 12 12 R - mR 194.5 - 150 z= = = 2.57 sR 17.321

sR=

R1 = 253.5; R2 = 124.5; mR = 182; s R = 20.607; Test statistic: z = 3.47; P-value = 0.0005; Critical values: z = ±1.96; Reject the null hypothesis that the populations have the same median. There is sufficient evidence to reject the claim that for those treated with 20 mg of Lipitor and those treated with 80 mg of Lipitor, changes in LDL cholesterol have the same median. It appears that the dosage amount does have an effect on the change in LDL cholesterol. n (n + n + 1) 13 (13 + 14 + 1) mR = 1 1 2 = = 182 2 2

7.

n1n2 (n1 + n2 + 1) 13 ×14 (13 + 14 + 1) = = 20.607 12 12 R - mR 253.5 - 182 z= = = 3.47 sR 20.607

sR=

R1 = 437; R2 = 424; mR = 525; s R = 37.4166; Test statistic: z = - 2.35; P-value = 0.0188; Critical values: z = ±2.575; Fail to reject the null hypothesis of no difference between the two population medians. There is not sufficient evidence to support the claim that the arrangement of the test items has an effect on the score. n (n + n + 1) 25 (25 + 16 + 1) mR = 1 1 2 = = 525 2 2

8.

n1n2 (n1 + n2 + 1) 25 ×16 (25 + 16 + 1) = = 37.4166 12 12 R - mR 437 - 525 z= = = - 2.35 sR 37.4166

sR=

.

13-4: Wilcoxon Rank-Sum Test for Two Independent Samples

187

9. R1 = 36,531.5; R2 = 46,668.5; mR = 41,102.5; s R = 1155.782; Test statistic: z = - 3.95; P-value = 0.0001 (Table: 0.0002); Critical values: z = ±1.96; Reject the null hypothesis that the populations have the same median.

There is sufficient evidence to warrant rejection of the claim that girls and boys have the same median birth weight. n (n + n + 1) 205 (205 + 195 + 1) mR = 1 1 2 = = 41,102.5 2 2

n1n2 (n1 + n2 + 1) 205 ×195 (205 + 195 + 1) = = 1155.782 12 12 R - mR 36,531.5 - 41,102.5 z= = = - 3.95 sR 1155.782

sR=

10. R1 = 863; R2 = 412; mR = 637.5; s R = 2656.25; Test statistic: z = 4.38; P-value = 0.0001 (Table: 0.0002): Critical value: z = 2.326; Reject the null hypothesis that the populations have the same median. There is sufficient evidence to support the claim that the median amount of nicotine in the nonfiltered king-size cigarettes is greater than the median amount of nicotine in the 100-mm filtered cigarettes. n (n + n + 1) 25 (25 + 25 + 1) mR = 1 1 2 = = 637.5 2 2

n1n2 (n1 + n2 + 1) 25 ×25 (25 + 25 + 1) = = 51.54 12 12 R - mR 863 - 637.5 z= = = 4.38 sR 51.54

sR=

11. R1 = 501; R2 = 445; mR = 484; s R = 41.15823; Test statistic: z = 0.41; P-value = 0.3409; Critical value: z = 1.645; Fail to reject the null hypothesis that the populations have the same median. There is not sufficient evidence to support the claim that subjects with medium lead levels have a higher median of the full IQ scores than subjects with high lead levels. Based on these data, it does not appear that lead level affects full IQ scores.

mR =

n1 (n1 + n2 + 1) 22 (22 + 21 + 1) = = 484 2 2

n1n2 (n1 + n2 + 1) 22 ×21(22 + 21 + 1) = = 41.158 12 12 R - mR 501 - 484 z= = = 0.41 sR 41.15823

sR=

12. R1 = 4178; R2 = 772; mR = 3900; s R = 116.833; Test statistic: z = 2.38; P-value = 0.0087; Critical value: z = 1.645; Reject the null hypothesis that the populations have the same median. There is sufficient evidence to support the claim that subjects with low lead levels have performance IQ scores with a higher median than subjects with high lead levels. It appears that exposure to lead does have an adverse effect. n (n + n + 1) 78 (78 + 21 + 1) mR = 1 1 2 = = 3900 2 2

n1n2 (n1 + n2 + 1) 78 ×21(78 + 21 + 1) = = 116.833 12 12 R - mR 4178 - 3900 z= = = 2.38 sR 116.833

sR=

.

188

Chapter 13: Nonparametric Tests

13. The test statistic is the same value with opposite sign. n (n + 1) 12 (12 + 1) U = n1n2 + 1 1 - R = 12 ×11 + - 123.5 = 86.5 2 2 nn 12 ×11 U- 1 2 86.5 2 2 z= = = 1.26 12 ×11×(12 + 11 + 1) n1n2 (n1 + n2 + 1) 12 12 14. a.

Rank Sum for Treatment A

Rank 1 A A A B B B

2 A B B B A A

3 B A B A A B

4 B B A A B A

3 4 5 7 5 6

b. The R values of 3, 4, 5, 6, 7 have probabilities of 1 6, 1 6, 2 6, 1 6, and 1 6, respectively c. No, none of the probabilities for the values of R would be less than 0.10.

13-5: Kruskal-Wallis Test for Three or More Samples 1.

R1 = 164.5; R2 = 52.5; R3 = 47 Low Lead Level

Medium Lead Level

High Lead Level

70 (1.0) 85 (10.0) 86 (12.5) 76 (5.0) 84 (8.0)

72 (3.0) 90 (14.0) 92 (15.0) 71 (2.0) 86 (12.5) 79 (6.0)

82 (7.0) 93 (16.0) 85 (10.0) 75 (4.0) 85 (10.0)

2.

(Ranks for each value shown in parentheses.)

Yes, the samples are independent simple random samples, and each sample has at least five data values.

3.

n1 = 5, n2 = 6, n3 = 5, and N = 5 + 6 + 5 = 16

4. The efficiency rating of 0.95 indicates that with all other factors being the same, the Kruskal-Wallis test requires 100 sample observations to achieve the same results as 95 observations with the parametric one-way analysis of variance test, assuming that the stricter requirements of the parametric test are satisfied.

5. Test statistic: H = 4.9054; Critical value: c 2 = 5.991; (Tech: P-value = 0.086); Fail to reject the null hypothesis of equal medians. The data do not suggest that larger cars are safer. æR12 R22 R32 ö æ862 972 482 ö 12 12 H= + + - 3 ( N + 1) = + + - 3 (21 + 1) ç ÷ N ( N + 1) è n1 n2 n3 ø 21(21 + 1) çè 7 7 7 ÷ ø

= 4.9054

.

13-5: Kruskal-Wallis Test for Three or More Samples

189

6. Test statistic: H = 2.3503; Critical value: c 2 = 5.991; (Tech: P-value = 0.309); Fail to reject the null hypothesis of equal medians. The data do not suggest that larger cars are safer. æR12 R22 R32 ö æ1422 133.52 189.52 ö 12 12 H= + + 3 N + 1 = + + - 3 (30 + 1) ( ) N ( N + 1) çè n1 n2 n3 ÷ 30 (30 + 1) çè 10 10 10 ÷ ø ø

= 2.3503 7. Test statistic: H = 22.8157; Critical value: c 2 = 9.210; (Tech: P-value = 0.000); Reject the null hypothesis of equal medians. It appears that the three states have median amounts of arsenic that are not all the same. æR12 R22 R32 ö æ1922 116.52 357.52 ö 12 12 H= + + - 3 ( N + 1) = + + - 3 (36 + 1) ç ÷ N ( N + 1) è n1 n2 n3 ø 36 (36 + 1) çè 12 12 12 ÷ ø

= 22.8157 8. Test statistic: H = 2.54; Critical value: c 2 = 5.991; (Tech: P-value = 0.281); Fail to reject the null hypothesis of equal medians. The times appear to increase for later laps. æR12 R22 R32 ö æ342 332 532 ö 12 12 H= + + 3 N + 1 = + + - 3 (15 + 1) ( ) N ( N + 1) çè n1 n2 n3 ÷ 15 (15 + 1) çè 5 5 5 ÷ ø ø

= 2.54 9. Test statistic: H = 59.1546; Critical value: c 2 = 9.210; (Tech: P-value = 0.000); Reject the null hypothesis of equal medians. The data suggest that the amounts of nicotine absorbed by smokers are different from the amounts absorbed by people who don’t smoke. æR12 R22 R32 ö æ3629.52 2393.52 12372 ö 12 12 H= + + 3 N + 1 = + + - 3 (120 + 1) ( ) N ( N + 1) çè n1 n2 n3 ÷ 120 (120 + 1) çè 40 40 40 ÷ ø ø

= 59.1546 10. Test statistic: H = 8.0115; Critical value: c 2 = 9.210; (Tech: P-value = 0.018); Reject the null hypothesis of equal medians. It appears that the three categories of blood lead level have different median performance IQ scores. æR12 R22 R32 ö æ5277.52 11122 991.52 ö 12 12 H= + + - 3 ( N + 1) = + + - 3 (121 + 1) ç ÷ N ( N + 1) è n1 n2 n3 ø 121(121 + 1) çè 78 22 21 ÷ ø

= 8.0115 11. Test statistic: H = 27.9098; Critical value: c 2 = 5.991; (Tech: P-value = 0.000); Reject the null hypothesis of equal medians. There is sufficient evidence to warrant rejection of the claim that the three different types of cigarettes have the same median amount of nicotine. It appears that the filters do make a difference. æR12 R22 R32 ö æ1413.52 650.52 7862 ö 12 12 H= + + 3 N + 1 = + + - 3 (75 + 1) ( ) N ( N + 1) çè n1 n2 n3 ÷ 75 (75 + 1) çè 25 25 25 ÷ ø ø

= 27.9098 12. Test statistic: H = 43.3240; Critical value: c 2 = 5.991; (Tech: P-value = 0.000); Reject the null hypothesis of equal medians. There is sufficient evidence to warrant rejection of the claim that the three different types of cigarettes have the same median amount of nicotine. It appears that the filters do make a difference. æR12 R22 R32 ö æ15342 6202 6962 ö 12 12 H= + + - 3 ( N + 1) = + + - 3 (75 + 1) ç ÷ N ( N + 1) è n1 n2 n3 ø 75 (75 + 1) çè 25 25 25 ÷ ø

= 32.3240

.

190

Chapter 13: Nonparametric Tests

13. There are 10 zeros and 2 ones, so the values of t are 10 and 2. The values of T are 990 and 6, ST = 996, and 996 ö æ = 8.114, which is quite different from the N = 5 + 6 + 7 = 18. The corrected value of H is 6.724 ç1 - 3 è 18 - 18 ÷ ø value of 6.724 found in Example 1. In this case, the large numbers of ties do appear to have a considerable effect on the test statistic H .

Cotinine Level 0 1

t 10 2 SUM

Rank 5.5 11.5

t3 - t 990 6 996

13-6: Rank Correlation 1. The methods of Section 10-2 should not be used for predictions. The regression equation is based on a linear correlation between the two variables, but the methods of this section do not require a linear relationship. The methods of this section could suggest that there is a correlation with paired data associated by some nonlinear relationship, so the regression equation would not be a suitable model for making predictions.

2. Data at the nominal level of measurement have no ordering that enables them to be converted to ranks, so data at the nominal level of measurement cannot be used with the methods of rank correlation.

3. r represents the linear correlation coefficient computed from sample paired data; r represents the parameter of the linear correlation coefficient computed from a population of paired data; rs denotes the rank correlation coefficient computed from sample paired data; r s represents the rank correlation coefficient computed from a population of paired data. The subscript s is used so that the rank correlation coefficient can be distinguished from the linear correlation coefficient r. The subscript does not represent the standard deviation s. It is used in recognition of Charles Spearman, who introduced the rank correlation method.

4. The efficiency rating of 0.91 indicates that with all other factors being the same, rank correlation requires 100 pairs of sample observations to achieve the same results as 91 pairs of observations with the parametric test using linear correlation, assuming that the stricter requirements for using linear correlation are met.

5. rs = 1.000; Critical values are rs = ±0.648. Reject the null hypothesis of r s = 0. There is sufficient evidence to support a claim of a correlation between height and age.

6. rs = - 1.000; Critical values are rs = ±0.648. Reject the null hypothesis of r s = 0. There is sufficient evidence to support a claim of a correlation between service time and number of waiting patients.

7. rs = 0.857; Critical values: rs = ±0.738. Reject the null hypothesis of r s = 0. There is sufficient evidence to conclude that there is a correlation between the number of chirps in 1 min and the temperature.

8. rs = 1.000; Critical values: rs = ±0.886. Reject the null hypothesis of r s = 0. There is sufficient evidence to conclude that there is a correlation between overhead widths of seals from photographs and the weights of the seals.

9. rs = 0.624; Critical values: rs = ±0.587. Reject the null hypothesis of r s = 0. There is sufficient evidence to support a conclusion that there is a correlation between the number of cigarettes smoked and the cotinine level.

10. rs = 0.797; Critical values: rs = ±0.538. Reject the null hypothesis of r s = 0. There is sufficient evidence to support the claim of a correlation between the circumference and height of a tree.

±z ±1.96 = = ±0.159; Reject the null hypothesis of r s = 0. There is n- 1 153 - 1 sufficient evidence to conclude that there is a correlation between the systolic and diastolic blood pressure levels in males.

11. rs = 0.360; Critical values: rs =

.

Chapter Quick Quiz

191

±z ±1.96 = = ±0.450; Fail to reject the null hypothesis of r s = 0. There n- 1 20 - 1 is not sufficient evidence to support the claim of a correlation between brain volumes and IQ scores.

12. rs = 0.106; Critical values: rs =

±z ±1.96 = = ±0.269; Reject the null hypothesis of r s = 0. There is n- 1 54 - 1 sufficient evidence to support the claim of a correlation between chest sizes and weights of bears.

13. rs = 0.984; Critical values: rs =

±z ±1.96 = = ±0.269; Reject the null hypothesis of r s = 0. There is n- 1 54 - 1 sufficient evidence to support the claim of a correlation between head lengths and neck sizes of bears.

14. rs = 0.924; Critical values: rs =

15. rs = ±

1.9757992

=±

1.9782

= ±0.159; (Use either t = 1.975799 from technology or use 1.9757992 + 153 - 2 1.9782 + 153 - 2 interpolation in Table A-3 with 151 degrees of freedom, so the critical value of t is approximately halfway between 1.984 and 1.972, which is 1.978.) The critical values are the same as those found by using Formula 13-1.

Chapter Quick Quiz 1.

Value Rank

1.18 5

1.41 7

1.49 11

1.04 4

1.45 9 + 10 2 = 9.5

0.74 2

0.89 3

1.42 8

1.45 9 + 10 2 = 9.5

0.51 1

1.38 6

2. The efficiency rating of 0.91 indicates that with all other factors being the same, rank correlation requires 100 pairs of sample observations to achieve the same results as 91 pairs of observations with the parametric test for linear correlation, assuming that the stricter requirements for using linear correlation are met.

3.

a. distribution-free test b. The term “distribution-free test” suggests correctly that the test does not require that a population must have a particular distribution, such as a normal distribution. The term “nonparametric test” incorrectly suggests that the test is not based on a parameter, but some nonparametric tests are based on the median, which is a parameter; the term “distribution-free test” is better because it does not make that incorrect suggestion.

4. Rank correlation should be used. The rank correlation test is used to investigate whether there is a correlation between foot length and height.

5.

No, the P-values are almost always different, and the conclusions may or may not be the same.

6. Rank correlation can be used in a wider variety of circumstances than linear correlation. Rank correlation does not require a normal distribution for any population. Rank correlation can be used to detect some (not all) relationships that are not linear.

7. Because the sign test uses only signs of differences while the Wilcoxon signed-ranks test uses ranks of the differences, the Wilcoxon signed-ranks test uses more information about the data and tends to yield conclusions that better reflect the true nature of the data.

8.

Kruskal-Wallis test

9.

two independent samples

10. matched pairs

.

192

Chapter 13: Nonparametric Tests

Review Exercises The test statistic of z =

(907 + 0.5) - 2015 2

= - 4.46 results in a P-value of 0.0000 (Table: 0.0001) and it is less 2015 2 than or equal to the critical value of z = - 1.645. Fail to reject the null hypothesis of p = 0.5. There is sufficient evidence to support the claim that the majority of adults obtain medical information more often from the Internet than a doctor.

1.

2. The test statistic of x = 3 is less than or equal to the critical value of 5 (from Table A-7). There is sufficient evidence to warrant rejection of the claim that the sample is from a population with a median equal to 5 min.

3. The test statistic T = 21 is less than or equal to the critical value of 59. There is sufficient evidence to warrant rejection of the claim that the sample is from a population with a median equal to 5 min.

4. Test statistic: H = 2.5288; (Tech: P -value = 0.2824); Critical value: c 2 = 5.991; Fail to reject the null hypothesis of equal medians. It appears that times of longevity after inauguration for presidents, popes, and British monarchs have the same median. æR12 R22 R32 ö æ1485.52 806.52 6352 ö 12 12 H= + + 3 N + 1 = + + - 3 (76 + 1) ( ) N ( N + 1) çè n1 n2 n3 ÷ 76 (76 + 1) çè 38 24 14 ÷ ø ø

= 2.5289 5. rs = 0.888; The critical values are rs = ±0.618. Reject the null hypothesis of r s = 0. There is sufficient evidence to support the claim of a correlation between chocolate consumption and the rate of Nobel Laureates. It does not make sense to think that there is a cause/effect relationship, so the correlation could be the result of a coincidence or other factors that affect the variables the same way.

6. Test statistic: H = 6.6305; (Tech: P -value = 0.0363); Critical value: c 2 = 5.991; Reject the null hypothesis of equal medians. Interbreeding of cultures is suggested by the data. æR12 R22 R32 ö æ912 112.52 174.52 ö 12 12 H= + + - 3 ( N + 1) = + + - 3 (27 + 1) ç ÷ N ( N + 1) è n1 n2 n3 ø 27 (27 + 1) çè 9 9 9 ÷ ø

= 6.6305 7. R1 = 60; R2 = 111; mR = 85.5; s R = 11.3248; Test statistic: z = - 2.25; (Tech: P -value = 0.0363); Critical values: z = ±1.96; Reject the null hypothesis that the populations have the same median. Skull breadths from 4000 B.C. appear to have a different median than those from A.D. 150.

mR =

n1 (n1 + n2 + 1) 9 (9 + 9 + 1) = = 85.5 2 2

n1n2 (n1 + n2 + 1) 9 ×9 (9 + 9 + 1) = = 11.3248 12 12 R - mR 60 - 85.5 z= = = - 2.25 sR 11.3248

sR=

8. rs = 0.714. Critical values: rs = ±0.738; Fail to reject the null hypothesis of r s = 0. There is not sufficient evidence to support the claim that there is a correlation between the student ranks and the magazine ranks. When ranking colleges, students and the magazine do not appear to agree.

Cumulative Review Exercises 1. x = 14.6 credit hours, Q2 = 15.0 credit hours, s = 1.7 credit hours, s2 = 2.9 (credit hours)2, range = 6.0 credit hours

.

Cumulative Review Exercises 2.

193

a. convenience sample b. Because the sample is from one class of statistics students, it is not likely to be representative of the population of all full-time college students. c. discrete d. ratio

3. The data appear do not appear to follow the loose definition for a normal distribution and n < 30, proceed with caution.

H0 : m = 14 credit hours; H1: m > 14 credit hours;

14.6 - 14 = 1.446; Critical value (a = 0.05): t = 1.729; 1.7 20 P -value = 0.0822 (Table: P-value > 0.05); Fail to reject H 0 . There is not sufficient evidence to support the claim that the mean is greater than 14 credit hours.

Test statistic: t =

4. The test statistic of x = 5 is not less than or equal to the critical value of 4. There is not sufficient evidence to support the claim that the sample is from a population with a median greater than 14 credit hours.

s 1.7 = 14.6 ± 2.093 × Þ 13.8 credit hours < m < 15.3 credit hours; We have 95% n 20 confidence that the limits of 13.8 credit hours and 15.3 credit hours contain the true value of the population mean.

5.

95% CI: x ± ta / 2

ˆˆ (0.039)(0.961) pq = 0.039 ± 1.96 Þ 0.031 < p < 0.047, or 3.1% < p < 4.7%; We have n 2000 95% confidence that the limits of 3.1% and 4.7% actually contain the true percentage of the population of workers who test positive for drugs.

6.

95% CI: pˆ ± za / 2

7.

H 0 : p = 0.03; H1: p > 0.03; Test statistic: z =

0.039 - 0.03 = 2.36; (0.03)(0.97 ) 2000

P-value = P ( z > 2.36) = 0.0091 (Tech: 0.0092); Critical value: z = 1.645; Reject H 0 . There is sufficient evidence to support the claim that the rate of positive drug test results among workers in the United States is greater than 3.0%.

8.

2 ˆ ˆ [1.96]2 ×0.25 za / 2 ] pq [ n= = = 2401

9.

H0: p = 0.5; H1: p > 0.5; Test statistic: z =

E2

0.022

0.54 - 0.5 = 1.36; (0.5)(0.5) 285

P-value = P ( z > 1.36) = 0.0869 (Tech: 0.0865); Critical value: z = 1.645; Fail to reject H 0 . There is not sufficient evidence to support the claim that the majority of the population is not afraid of heights in tall buildings. Because respondents themselves chose to reply, the sample is a voluntary response sample, not a random sample, so the results might not be valid.

10. There must be an error, because the rates of 13.7% and 10.6% are not possible with samples of size 100.

.

Section 14-1: Life Tables

195

Chapter 14: Survival Analysis Section 14-1: Life Tables 1. A cohort life table is a record of the actual mortality experience of a particular group, such as all people born in 1970. This would include data of the mortality of this group at each year. A period life table, on the other hand, the mortality conditions and results for different ages are based on data from one particular year.

2. Period life tables are more practical because the data comes from one year, as opposed to being recorded over the course of decades.

3. The probability of dying during the first interval would remain the same, as would the expected remaining lifetime value. However, the number surviving to the beginning of the interval, number of deaths during the interval, person-years lived and total person-years lived would all be five times the stated values.

4. The period life table assumes that condition affecting mortality remain fixed throughout the lives of everyone in the hypothetical cohort. Over the course of the years since 2000, those conditions may have changed, and so mortality rates may have changed.

5. There were 99,529 white females alive on their first birthday, and there were 38 deaths during the interval. The probability of dying in that interval was 38 99,529 = 0.000382.

6. There were 99,491 white females alive at their second birthday, and the probability of dying between their second and third birthdays was 0.000204. The number of deaths during the interval was 99, 491(0.000204 ) = 20.

7. There were 100,000 white females at birth, and there were 99,491 that survived until their second birthday. The probability of surviving until the second birthday is 99,491 100,000 = 0.99491.

8.

The values would be:

471 + 38 100, 000 = 0.00509

0–2

100,000

509

99,588 + 99,510 = 199,098

8,128,871

81.3

9. From column 2, the probability of dying between the 25th birthday and 26th birthday is 0.000936, so the probability of surviving that same period is 1 – 0.000936 = 0.999064. If 15000 people live to their 26th birthday, we expect that 15000(0.999064) = 14,985.96 will survive to their 26th birthday. Using the values from column 3 92 and 4, 1 = 0.999066 and 15000(0.999066) = 14,985.99.

98483 10. a. The expected remaining lifetime for someone who has just reached their 65th birthday is 19.1 years. For someone reaching their 66th birthday, the expected remaining lifetime is 18.4 years. b. The expected age of death for someone who has just reached their 65th birthday is 84.1 years. For someone reaching their 66th birthday, the expected age of death is 84.4 years. c. The expected age of death is not exactly one year different because the value of total person-years lived does not drop uniformly from one year to the next.

11. There were 99,208 people alive on their 12th birthday and 96,464 people alive on their 42nd birthday. The 96,464

probability of surviving is 99,208 = 0.972341.

12. There were 99,032 people alive on their 18th birthday and 41,497 alive on their 85th birthday. The probability 41,497

of surviving is 99,032 = 0.419026.

13. a. b.

612 + 43 + 27 = 0.006820 100, 000 21 + 16 + 14 = 0.000514 99,318

The results are so different because of the exceptionally high mortality rate at or very near birth.

.

196

Chapter 14: Survival Analysis

14. The values would be:

612 + 43 + 27 = 0.006820 100, 000

0–3 15.

100,000

682

99,318

51

99,465 + 99,366 + 99,331 = 298,162

7,866,027

78.7

The values would be:

21 + 16 + 14 = 0.000514 99,318

3–6 16.

99,307 + 99,289 + 99,274 = 297,870

7,567,864

76.2

The values would be: 0–5

719 = 0.007190 100, 000

100,000

719

496,758

7,866,027

78.7

17. H 0 : p = 0.000936; H1: p > 0.000936. The test statistic is z = 49.6523. The P-value is 0.0000042 (Table : < 0.0001), and the critical value is z = 1.645. Reject H 0 and conclude that there is sufficient evidence to support the claim that the number of deaths is significantly high. [If the probability of dying is calculated from the third column of Table 14-1, use H1: p > 0.00093417 to get a test statistic of z = 1.60, a P-value of 0.0000026 (Table: < 0.00001), and the same critical value and conclusions.]

z=

pˆ - p pq n

=

(

200 15,000

) - 0.000936

(0.000936)(1 - 0.000936)

= 49.6523

15, 000

18. H 0 : p = 0.001663; H1: p > 0.001663. The test statistic is z = 1.10. The P-value is 0.1357 (Table: 0.1357), and the critical value is z = 1.645. Fail to reject H 0 and conclude that there is not sufficient evidence to support the claim that the number of deaths is significantly high.

z=

pˆ - p pq n

=

17 - 0.001633 ( 7844 )

(0.001633)(1 - 0.001633)

= 1.10

7844

99,154 - 95,833 = 0.033493; H1: p > 0.033493. The test statistic is z = 7.29. The P-value is less 99,154 than 0.00001 (Table: < 0.00001), and the critical value is z = - 1.645. Reject H 0 and conclude that there is sufficient evidence to support the claim that the number of deaths is significantly high.

296 - 0.033493 pˆ - p 5842 z= = = 7.29 pq (0.033493)(1 - 0.033493) n 5842

19. H 0 : p =

78, 069 - 23, 619 = 0.697460; H1: p < 0.697460. The test statistic is z = - 109.7. The P-value is less 78, 069 than 0.00001 (Table: < 0.00001), and the critical value is z = 1.645. Reject H 0 and conclude that there is sufficient evidence to support the claim that the number of deaths is significantly low.

20. H 0 : p =

z=

pˆ - p pq n

=

592 - 0.697460 ( 6812 )

(0.697460)(1 - 0.697460)

= - 109.7

6812

.

Section 14-2: Kaplan-Meier Survival Analysis

197

Section 14-2: Kaplan-Meier Survival Analysis 1. A survivor is a subject that successfully lasted throughout a particular time period without reaching some terminating event. A survivor could be a person or an object or some other entity such as a marriage.

2. Censored data are subjects that are removed from the study because they survived past the end of the study or were no longer available to be included in the study.

3. A life table is based on fixed intervals of time, but a table of survival data and Kaplan-Meier calculations is based on times that vary according to the terminating event.

4.

It is a graphical representation of the information generated using Kaplan-Meier calculations.

5. The fast walkers are more likely to be healthier with greater longevity, so they correspond to the top (green) curve. The moderate group corresponds to the middle (black) curve. The slow group of walkers is more likely to have health issues with lower longevity, so they correspond to the bottom (red) curve.

6.

The event is surviving for the given time period. The probability of survival decreases with the passage of time.

7.

a. 4 years

8.

The red graph reaches zero, so none of the slow walkers survived past the study.

b. 1 year

9. The five-year survival rates are 0.2 for the group of slow walkers, 0.5 for the group with moderate walking speeds, and 0.9 for the group of fast walkers. The differences are substantial, and they suggest that after five years, those with faster walking speeds have much greater survival rates, and those with slow walking speeds have much lower survival rates. The data do not necessarily suggest that we can get older people to live longer by somehow getting them to walk faster. It’s very possible that walking speed is one manifestation of overall health status, so longevity and walking speed are likely both affected by one or more other extraneous variables.

10.

Day

Status 0 = Censored 1 = Failed

Number of Patients

Patients Not Requiring Retreatment

Proportion of Patients Not Requiring Retreatment

2 6

0 1

9

8

8 / 9 = 0.889

10

1

8

7

7 / 8 = 0.875

11

1

7

6

6 / 7 = 0.857

0.67 (0.889 ´ 0.875 ´ 0.857)

13

1

6

5

5/ 6 = 0.833

0.56 (0.889 ´ 0.875 ´ 0.857 ´ 0.833)

14 14 14 14 14

0 0 0 0 0

.

Cumulative Proportion of Patients Not Requiring Retreatment 0.89 0.78 (0.889 ´ 0.875)

198

Chapter 14: Survival Analysis

11.

12.

Day

Status 0 = Censored 1 = Failed

Number of Patients

Patients Not Requiring Retreatment

Proportion of Patients Not Requiring Retreatment

2 10

0 1

7

6

6 / 7 = 0.857

0.86

12

1

6

5

5/ 6 = 0.833

0.71 (0.857 ´ 0.833)

0 0 0 0 0

0 0 0 0 0

Cumulative Proportion of Patients Not Requiring Retreatment

13. The graph does not show any information about the six censored survival times. The graph shows information about the two survival times that were not censored.

14. Estimates will vary. Placebo: 0.50; Treatment: 0.80; It appears that the treatment is effective.

Chapter Quick Quiz 1. life table 2.

survival table with Kaplan-Meier calculations

3. A period life table describes mortality and longevity data for a hypothetical group that would have lived with the same mortality conditions throughout their lives.

4.

A cohort life table is a record of the actual observed mortality experience for a particular group.

5.

false

6.

true .

Review Exercises 7.

false

8.

true

9.

The entries are 1, 0, 0, and 0.

10. 1 -

199

35 = 0.999345 98,975

Review Exercises 1.

99,527 - 99, 492 = 0.000352 99,527

2.

99,527 - 99, 492 = 35

3.

99, 492 = 0.99492 100,000

4.

The values would be:

473 + 35 100, 000 = 0.00508

0–2

100,000

508

99,588 + 99,510 = 199,098

8,382,303

83.8

5. 83.8 years; 83.2 years; the second value is less than the first value. As we age, our expected remaining lifetime steadily decreases.

6.

Proportion of Status Number Patients Not Patients Not 0 = Censored of Requiring Requiring Cumulative Proportion of Patients Day 1 = Failed Patients Retreatment Retreatment Not Requiring Retreatment 3

1

4

3

3/ 4 = 0.75

0.75

12

1

3

2

2 / 3 = 0.667

0.50 (0.75 ´ 0.667 )

20

1

2

1

1/ 2 = 0.5

0.25 (0.75 ´ 0.667 ´ 0.5)

30

0

7.

8. The treatment group and the placebo group appear to have approximately the same behavior. The treatment does not appear to be effective.

.

200

Chapter 14: Survival Analysis

Cumulative Review Exercises 1. The table shows that among 100,000 births, 99,492 survived to the second birthday, so the probability of dying is 1 - 0.99492 = 0.00508. Using H1: p < 0.00508, the test statistic is z = - 0.67, the P-value is 0.2501 (Table: 0.2514), and the critical value is z = - 1.645 (assuming a 0.05 significance level), so fail to reject the null hypothesis of p = 0.00508 and conclude that there is not sufficient evidence to support the claim that the proportion of deaths is less than 0.00508. The program does not appear to be effective in reducing the mortality rate.

5 - 0.00508 pˆ - p 1328 z= = = - 0.67 pq (0.00508)(1 - 0.00508) n 1328 2. 95% CI: 0.99294 < p < 0.99953; The confidence interval limits contain 0.99492, so it appears that the mortality rate has not been lowered by a significant amount.

z = pˆ ± za / 2

ˆ ˆ 1323 pq = ± 1.96 n 1328

1323 1 - 1323 (1328 )( 1328 )

1328

3.

(1 - 0.004729)3 = 0.986

4.

1 - (1 - 0.004729) = 0.0188; No, because being in the same family causes the events to be dependent, instead

4

of being independent, as required by the multiplication rule. It is reasonable to expect that four Hispanic females in the same family are more likely to experience similar environmental and hereditary characteristics.

5. The graph is misleading. The vertical scale begins with a frequency of 800 instead of 0, so the difference between the “yes” and “no” responses is greatly exaggerated.

6.

60.0 - 68.6 80.0 - 68.6 = - 3.07 and z x =80 = = 4.07, which have a probability of 0.9999 - 0.0011 2.8 2.8 = 0.9988, or 99.88% (Tech: 99.89%) between them.

a. z x=60 =

b. z x=70 =

70.0 - 68.6 = 1.00, which has a probability of 1 - 0.8413 = 0.1587 to the right.

2.8 4

.