Instructor Manual for Elementary Statistics: Picturing the World, 8th Edition by Larson & Farber

INSTRUCTOR’S SOLUTIONS MANUAL DIACRITECH

E LEMENTARY S TATISTICS P ICTURING THE W ORLD EIGHTH EDITION

RON LARSON

CONTENTS

Chapter 1

Introduction to Statistics

Chapter 2

Descriptive Statistics

Chapter 3

Probability

105

Chapter 4

Discrete Probability Distributions

140

Chapter 5

Normal Probability Distributions

173

Chapter 6

Confidence Intervals

226

Chapter 7

Hypothesis Testing with One Sample

255

Chapter 8

Hypothesis Testing with Two Samples

308

Chapter 9

Correlation and Regression

355

Chapter 10

Chi-Square Tests and the F-Distribution

407

Chapter 11

Nonparametric Tests

466

Chapter 6 Alternate (Online) Version Appendix A

Confidence Intervals

517

Alternative Presentation of the Standard Normal Distribution Normal Probability Plots

546

Appendix C

547

Activities

548

Case Studies

557

Uses and Abuses

584

Real StatisticsíReal Decisions

596

Technology

611

CHAPTER

Introduction to Statistics

1.1 AN OVERVIEW OF STATISTICS 1.1 TRY IT YOURSELF SOLUTIONS 1. The population consists of the responses of all teens in the United States. The sample consists of the responses of the 1516 teens in the survey. The sample data set consists of 1228 teens who said mental health is a significant issue for young people in the U.S. and 288 teens who did not say that. 2a. Population parameter, because the total spent on employees’ salaries, $5,150,694, is based on the entire company. b. Sample statistic, because 47% is based on a subset of the population. 3a. The population consists of the responses of all Internet users, and the sample consists of the responses of the 513 Internet users in the study. b. The part of this study that represents the descriptive branch of statistics involves the statement “97% of the respondents said music is important to them, and 83% of the respondents said they actively look for new music.” c. A possible inference drawn from the study is that a large majority of Internet users find music to be important to them and actively look for new music.

1.1 EXERCISE SOLUTIONS 1. A sample is a subset of a population. 2. It is usually impractical (too expensive and/or time consuming) to obtain all the population data. 3. A parameter is a numerical description of a population characteristic. A statistic is a numerical description of a sample characteristic. 4. The two main branches of statistics are descriptive statistics and inferential statistics. 5. False. A statistic is a numerical description of a sample characteristic. 6. True 7. True 8. False. Inferential statistics involves using a sample to draw conclusions about a population. 9. False. A population is the collection of all outcomes, responses, measurements, or counts that are of interest. 10. False. A sample statistic can differ from sample to sample. 1 Copyright © 2023 Pearson Education, Inc.

2 CHAPTER 1 Ň INTRODUCTION TO STATISTICS

11. Population, because it is a collection of the salaries of each employee at an advertising firm. 12. Population, because it is a collection of the energy collected from all the solar panels on a photovoltaic power plant. 13. Sample, because the collection of the 250 people is a subset of the population of 20,000 members of an organized union. 14. Population, because it is a collection of the revenue of each team in a pro sports league. 15. Sample, because the collection of the 12 people is a subset of the 49 people who escaped a burning building. 16. Population, because it is a collection of the number of Electoral College votes in each states in the U.S. and the District of Columbia. 17. Population, because it is the collection of all the guests in each room of a hotel. 18. Sample, because only the amount spent by every tenth person cashing out is recorded. 19. Population, because it is a collection of the nationalities of every person passing through a customs station. 20. Sample, because the collection of the 15 precipitation amounts is a subset of the population. 21. Population: Parties of registered voters Sample: Parties of registered voters who respond to a survey 22. Population: Meal plan choices of college students Sample: Meal plan choices of freshmen 23. Population: Ages of adults in the United States who own motor vehicles Sample: Ages of adults in the United States who own two-wheeled motor vehicles 24. Population: Incomes of adults in Nevada Sample: Incomes of adults in Nevada who own homes 25. Population: Collections of the responses of all U.S. adults Sample: Collection of the responses of the 1021 U.S. adults surveyed Sample data set: 45% of adults who have a favorable view of Cuba and 55% who do not 26. Population: Collection of the gut microbiomes of all infants Sample: Collection of the gut microbiomes of 227 U.S. infants Sample data set: The gut microbiomes of infants and their health 27. Population: Collection of the U.S. adults who have received vaccines Sample: Collection of responses of 1500 U.S. adults who have received vaccines Sample data set: 59% of U.S. adults who have never had a vaccine reaction and 41% who have

CHAPTER 1 Ň INTRODUCTION TO STATISTICS 3

28. Population: Collection of the responses of U.S. adults Sample: Collection of the responses of 1028 U.S. adults surveyed Sample data set: 7% of adults who have never heard of organ and tissue donation and 93% who have 29. Population: Collection of responses from small business owners in the U.S. Sample: Collection of responses of 2111 U.S. small business owners Sample data set: 54% of small business owners who oppose increasing the minimum wage and 46% who do not 30. Population: Collection of plans after college of all seniors graduating with a Bachelor of Science degree Sample: Collection of plans after college of 214 seniors graduating with a Bachelor of Science degree Sample data set: 15% of those surveyed who plan to obtain entry-level jobs in the health field and 85% who do not 31. Population: Collection of all U.S. adults Sample: Collection of responses of 1001 U.S. adults who were surveyed Sample data set: 47% of respondents who typically feel well rested on weekdays and 53% who do not 32. Population: Collection of the responses of all automobile owners who have purchased extended warranties Sample: Collection of responses of the 366 automobile owners who purchased extended warranties surveyed Sample data set: 44% of those surveyed who never used the warranty and 56% who have 33. Population: Collection of all companies listed in the Standard & Poor’s 500 Sample: Collection of the responses of the 74 Standard & Poor’s 500 companies surveyed Sample data set: Starting salaries of the 74 companies surveyed 34. Population: Collection of responses of members of a local children’s museum Sample: Collection of responses of the 679 members of a local children’s museum surveyed Sample data set: Parenting attitudes of the 679 members surveyed 35. Sample statistic. The value $71,000 is a numerical description of a sample of average salaries. 36. Sample statistic. The value of 89% is a numerical description of a sample of college board members. 37. Population parameter. The 62 surviving passengers out of 97 total passengers is a numerical description of all of the passengers of the Hindenburg that survived. 38. Population parameter. The value of 54% is a numerical description of the total number of governors. 39. Sample statistic. The value of 6% is a numerical description of the sample of automobile owners. 40. Population parameter. The value 47% is a numerical description of the total number of voters. 41. Sample statistic. The value 79% is a numerical description of a sample of U.S. adults. 42. Population parameter. The score 20.2 is a numerical description of the ACT scores for all graduates.

4 CHAPTER 1 Ň INTRODUCTION TO STATISTICS

43. The statement “47% of respondents typically feel well rested on weekdays” is an example of descriptive statistics. Using inferential statistics, you may conclude that an association exists between feeling well rested and the day of the week. 44. The statement “44% never used the warranty” is an example of descriptive statistics. Using inferential statistics, you may conclude that an association exists between those who purchase extended warranties and those who use the warranty. 45. Answers will vary. 46. Answers will vary. 47. No, this inference may incorrectly imply that exercise stimulates T cell production for all ages when it was only shown for senior citizens. 48. No, it is not appropriate to infer this because an association does not prove a causation between weight loss and high blood pressure. 49. (a) The sample is scores on quizzes and midterm exams by the participants in the study. (b) The population is the collection of all of the college students’ scores on quizzes and midterm exams. (c) The statement “participants earned higher scores on quizzes and midterm exams with better sleep” is an example of descriptive statistics. (d) Individuals who sleep better will be more likely to perform better on quizzes and midterm exams than they would with worse sleep.

1.2 DATA CLASSIFICATION 1.2 TRY IT YOURSELF SOLUTIONS 1. The city names are nonnumerical entries, so these are qualitative data. The city populations are numerical entries, so these are quantitative data. 2. (1) Ordinal, because the data can be put in order. (2) Nominal, because no mathematical computations can be made. 3. (1) Interval, because the data can be ordered and meaningful differences can be calculated, but it does not make sense to write a ratio using the temperatures. (2) Ratio, because the data can be ordered, meaningful differences can be calculated, the data can be written as a ratio, and the data set contains an inherent zero.

CHAPTER 1 Ň INTRODUCTION TO STATISTICS 5

1.2 EXERCISE SOLUTIONS 1. Nominal and ordinal 2. Ordinal, interval, and ratio 3. False. Data at the ordinal level can be qualitative or quantitative. 4. False. For data at the interval level, you can calculate meaningful differences between data entries. You cannot calculate meaningful differences at the nominal or ordinal levels. 5. False. More types of calculations can be performed with data at the interval level than with data at the nominal level. 6. False. Data at the ratio level can be placed in a meaningful order. 7. Qualitative, because nationalities are nonnumerical entries. 8. Qualitative, because zip codes are labels. 9. Quantitative, because age is counted. 10. Quantitative, because capacities are numerical measurements. 11. Qualitative, because flower types are labels. 12. Qualitative, because names are labels. 13. Quantitative, because distances are numerical measurements. 14. Quantitative, because response times are numerical measurements. 15. Interval. Data can be ordered and meaningful differences can be calculated, but it does not make sense to say one year is a multiple of another. 16. Ordinal. Data can be arranged in order, but the differences between data entries are not meaningful. 17. Ratio. A ratio of two data values can be formed, so one data value can be expressed as a multiple of another. 18. Nominal. No mathematical computations can be made, and data are categorized using numbers. 19. Ordinal. Data can be arranged in order, but the differences between data entries are not meaningful. 20. Interval. Data can be ordered and meaningful differences can be calculated, but it does not make sense to say one time is a multiple of another. 21. Horizontal: Nominal; Vertical: Ratio 22. Horizontal: Ratio; Vertical: Ratio

6 CHAPTER 1 Ň INTRODUCTION TO STATISTICS

23. Horizontal: Nominal; Vertical: Ratio 24. Horizontal: Interval; Vertical: Ratio 25. (a) Interval

(b) Nominal

(d) Ordinal

26. (a) Interval

(b) Nominal

(d) Ratio

27. Qualitative. Ordinal. Data can be arranged in order, but the differences between data entries make no sense. 28. Qualitative. Nominal. No mathematical computations can be made, and data are categorized by the type of coffee bean. 29. Qualitative. Nominal. No mathematical computations can be made and data are categorized by region. 30. Quantitative. Interval. Data can be ordered and meaningful differences can be calculated, but it does not make sense to say that one score is a multiple of another. 31. Qualitative. Ordinal. Data can be arranged in order, but the differences between data entries are not meaningful. 32. Quantitative. Ratio. A ratio of two data entries can be formed, so one data entry can be expressed as a multiple of another. 33. An inherent zero is a zero that implies “none.” Answers will vary. 34. Answers will vary.

1.3 DATA COLLECTION AND EXPERIMENTAL DESIGN 1.3 TRY IT YOURSELF SOLUTIONS 1. This is an observational study. 2. There is no way to tell why the people quit smoking. They could have quit smoking as a result of either chewing the gum or watching the DVD. The gum and the DVD could be confounding variables. To improve the study, two experiments could be done, one using the gum and the other using the DVD. Or just conduct one experiment using either the gum or the DVD. 3. Sample answer: Assign numbers 1 to 79 to the employees of the company. Use the table of random numbers and obtain 63, 7, 40, 19, and 26. The employees assigned these numbers will make up the sample. 4. (1) The sample was selected by using the students in a randomly chosen class. This is cluster sampling.

CHAPTER 1 Ň INTRODUCTION TO STATISTICS 7

(2) The sample was selected by numbering each student in the school, randomly choosing a starting number, and selecting students at regular intervals from the starting number. This is systematic sampling.

1.3 EXERCISE SOLUTIONS 1. In an experiment, a treatment is applied to part of a population and responses are observed. In an observational study, a researcher measures characteristics of interest of a part of a population but does not change existing conditions. 2. A census includes the entire population; a sampling includes only a portion of the population. 3. In a random sample, every member of the population has an equal chance of being selected. In a simple random sample, every possible sample of the same size has an equal chance of being selected. 4. Replication is the repetition of an experiment under the same or similar conditions. Replication is important because it enhances the validity of the results. 5. False. A placebo is a fake treatment. 6. False. A double-blind experiment is used to decrease the placebo effect. 7. False. Using stratified sampling guarantees that members of each group within a population will be sampled. 8. False. A convenience sample is not representative of a population. 9. False. To select a systematic sample, a population is ordered in some way and then members of the population are selected at regular intervals. 10. True 11. Experiment. The study applies a treatment (periods of wakeful rest or periods of distraction) to the subjects. 12. Observational study. The study does not apply a treatment to those surveyed. 13. Observational study. The study does not apply a treatment to the adults. 14. Experiment. The study applied a treatment (songs with body-positive and body-negative messages) to the subjects. 15. Answers will vary. Sample answer: Starting at the left-most number in row 6: 28/70/35/17/09/94/45/64/83/96/73/78/ The numbers would be 28, 70, 35, 17, 9, 94, 45, 64, 83, 96, 73, 78. 16. Answers will vary. Sample answer: Starting with the left-most number in row 10: 421/030/278/173/920/562/977/267/812/249/252/ The numbers would be 421, 30, 278, 173, 920, 562, 267, 812, 249, 252.

8 CHAPTER 1 Ň INTRODUCTION TO STATISTICS

17. Answers will vary. 18. Answers will vary. 19. (a) The experimental units are the 500 girls ages 13 to 17 who have acne. The treatment is the new cream used to treat acne. (b) Sample answer: A problem with the design is that the sample is not representative of the entire population because only girls ages 13 to 17 were used. To increase validity, use a stratified sample. (c) For the experiment to be double-blind, neither the subjects nor the company would know whether the subjects are receiving the acne cream or the placebo. 20. (a) The experimental units are the 37 teenagers aged 18 to 19 who have social anxiety disorder. The treatment is cannabidiol used to treat social anxiety disorder. (b) A problem with the design is that the sample size is small. The experiment could be replicated to increase validity. (c) In a placebo-controlled, double-blind experiment, neither the subject nor the experimenter knows whether the subject is receiving a treatment or a placebo. The experimenter is informed after all the data have been collected. (d) Divide the subjects into two age categories and then, within each age group, randomly assign subjects to either the treatment group or the control group. 21. Answers will vary. Sample answer: Number the volunteers from 1 to 18. Using the random number table in Appendix B, starting with the left-most number in row 16: 29/55/31/84/32/13/63/00/55/29/02/79/18/10/17/49/02/77/90/31/50/91/20/93/99 23/50/12/26/42/63/08/10/81/91/89/42/06/78/00/55/13/75/47/07/ Treatment group: Jake, Arya, Nyla, Juan, Nia, Bruno, Kayla, Liam, and Kai Control group: Xavier, Shaniece, Chen, Hana, Ansel, Liam, Mei, Zoey, and Sofia 22. Answers will vary. Sample answer: Using a random number generator: Treatment group: 1, 2, 3, 4, 5, 6, 7, 9, 12, 15, 18, 20, 22, 23, 26, 27, 28, 30, 31, 32, 33, 34, 35, 36, 37, 38, 41, 42, 44, 50, 54, 63, 68, 70, 73, 74, 78, 80, 81, 82, 85, 86, 87, 88, 89 Control group: 8, 10, 11, 13, 14, 16, 17, 19, 21, 24, 25, 29, 39, 40, 43, 45, 46, 47, 48, 49, 51, 52, 53, 55, 56, 57, 58, 59, 60, 61, 62, 64, 65, 66, 67, 69, 71, 72, 75, 76, 77, 79, 83, 84, 90. 23. Simple random sampling is used because each student has an equal chance of being contacted, and all samples of 300 students have an equal chance of being selected. A possible source of bias is that the random sample may contain a much greater percentage of students from one major than others. 24. Convenience sampling is used because the students are chosen due to their convenience of location. Bias may enter into the sample because the students sampled may not be representative of the population of students.

CHAPTER 1 Ň INTRODUCTION TO STATISTICS 9

25. Cluster sampling is used because the disaster area is divided into grids, and 30 grids are then entirely selected. A possible source of bias is that certain grids may have been much more severely damaged than others. 26. Systematic sampling is used because every tenth person entering the shopping mall is sampled. It is possible for bias to enter the sample if, for some reason, there is a regular pattern to people entering the shopping mall. 27. Stratified sampling is used because a sample is taken from each one-acre subplot (stratum). 28. Simple random sampling is used because each telephone number has an equal chance of being dialed, and all samples of 1012 phone numbers have an equal chance of being selected. The sample may be biased because telephone sampling only samples those individuals who have telephones, who are available, and who are willing to respond. 29. Census, because it is relatively easy to obtain the grade point averages (GPAs) of 85 students. 30. Sampling, because the population of spectators is too large for their distances traveled to be easily recorded. Systematic sampling would be advised because it would be easy to ask people entering the stadium the distances they traveled and record their responses. 31. The question is biased because it already suggests that eating whole grain foods improves your health. The question might be rewritten as “How does eating whole grain foods affect your health?” 32. The question is biased because it already suggests that text messaging while driving increases the risk of a crash. The question might be rewritten as “Does text messaging while driving affect the risk of a crash?” 33. The survey question is unbiased because it does not imply how much exercise is good or bad. 34. The question is biased because it already suggests that the media influences the opinions of voters. The question could be rewritten as “Does the media influence the opinions of voters?” 35. Answers will vary. 36. Answers will vary. 37. Open Question Advantage: Allows respondent to express some depth and shades of meaning in the answer. Allows for new solutions to be introduced. Disadvantage: Not easily quantified and difficult to compare surveys. Closed Question Advantage: Easy to analyze results. Disadvantage: May not provide appropriate alternatives and may influence the opinion of the respondent. 38. Sample answer: Observational studies may be referred to as natural experiments because they involve observing naturally occurring events that are not influenced by the study.

10 CHAPTER 1 Ň INTRODUCTION TO STATISTICS

CHAPTER 1 REVIEW EXERCISE SOLUTIONS 1. Population: Collection of responses of all U.S. adults Sample: Collection of responses of the 1025 U.S. adults who were sampled Sample data set: Those sampled had visited the library an average of 10.5 times in the past 12 months. 2. Population: Collection of the opinions on healthcare reform of all doctors in the St. Louis area Sample: Collection of the opinions on healthcare reform of the 83 doctors in the St. Louis area who were sampled Sample data set: Doctors in the St. Louis area and their opinions on healthcare reform 3. Population: Collection of responses of all U.S. adults Sample: Collection of the responses of the 23,503 U.S. adults who were sampled Sample data set: 59% of adults who said if they were single, they would look up someone on the Internet before going on a date with them and 41% who would not. 4. Population: Collection of responses of all U.S. adults Sample: Collection of the responses of the 1508 U.S. adults ages 40-64 who were sampled Sample data set: 32% of adults who said they had provided regular support to their parents in the past year and 68% who have not. 5. Population parameter. The value $4.8 billion is a numerical description of the amount of personal protective equipment imported. 6. Sample statistic. The value 24.7% is a numerical description of a sample of household food purchasers. 7. Sample statistic. The value of 39% is a numerical description of a sample of U.S. workers. 8. Population parameter. The value of 52 biology majors is a numerical description of all students receiving Bachelor of Science degrees from a university. 9. The statement “59% said that if they were single, they would look up someone on the Internet before going on a date with them” is an example of descriptive statistics. An inference drawn from the sample is that a majority of adults would look someone up on the Internet before going on a date. 10. The statement “32% had provided regular support to their parents in the past year” is an example of descriptive statistics. An inference drawn from the sample is that a majority of adults ages 40-64 do not provide regular support to their parents. 11. Quantitative, because ages are numerical measurements. 12. Qualitative, because zip codes are labels for customers. 13. Quantitative, because revenues are numerical measures. 14. Qualitative, because marital statuses are attributes.

CHAPTER 1 Ň INTRODUCTION TO STATISTICS 11

15. Interval. The data can be ordered and meaningful differences can be calculated, but it does not make sense to say that 87 degrees is 1.09 times as hot as 80 degrees. 16. Ordinal. The data are qualitative and could be arranged in order of car size. 17. Nominal. The data are qualitative and cannot be arranged in a meaningful order. 18. Ratio. The data are quantitative, and it makes sense to say that $66.9 million is 1.83 times as much as $36.5 million. 19. Experiment. The study applied a treatment (medication used to reduce the risk of cardiac events) to the subjects. 20. Observational study. The study does not attempt to influence the responses of the subjects and there is no treatment. 21. Sample answer: The subjects could be split into male and female and then be randomly assigned to each of the five treatment groups. 22. Sample answer: Number the volunteers and then use a random number generator to assign subjects randomly to one of the treatment groups or the control group. 23. Simple random sampling is used because random telephone numbers were generated and called. A potential source of bias is that telephone sampling only samples individuals who have telephones, who are available, and who are willing to respond. 24. Convenience sampling is used because the student sampled a convenient group of friends. The study may be biased toward the opinions of the student’s friends. 25. Cluster sampling is used because each neighborhood is considered a cluster and every pregnant woman in a selected neighborhood is surveyed. A potential source of bias is that the selected neighborhoods may not be representative of the entire area. 26. Systematic sampling is used because every third car is stopped. A potential source of bias is that the street the law enforcement officials are using may be near a bar. 27. Stratified sampling is used because the population is divided by grade level and then 25 students are randomly selected from each grade level. 28. Convenience sampling is used because of the convenience of surveying people waiting for their baggage. A potential source of bias is that all of the people just got off an airplane. 29. Answers will vary. Sample answer: Sampling, because the population of students at the university is too large for their favorite spring break destinations to be easily recorded. Random sampling would be advised because it would be easy to select students randomly and then record their favorite spring break destination.

12 CHAPTER 1 Ň INTRODUCTION TO STATISTICS

CHAPTER 1 QUIZ SOLUTIONS 1. Population: Collection of grade point averages, SAT scores, and ACT scores of all high school seniors Sample: Collection of grade point averages, SAT scores, and ACT scores of 1622 high school seniors from four public high schools in the northeastern United States. 2. (a) Sample statistic. The value 42% is a numerical description of a sample of U.S. adults. (b) Population parameter. The 90% of members that approved the contract of the new president is a numerical description of all Board of Trustees members. (c) Sample statistic. The value 48% is a numerical description of a sample of small business owners. 3. (a) Qualitative, because debit card personal identification numbers are labels and it does not make sense to find differences between numbers. (b) Quantitative, because final scores are numerical measurements. 4. (a) Ordinal, because badge numbers can be ordered and often indicate seniority of service, but no meaningful mathematical computation can be performed. (b) Ratio, because horsepower of one car can be expressed as a multiple of another. (c) Ordinal, because data can be arranged in order, but the differences between data entries make no sense. (d) Interval, because meaningful differences between years can be calculated, but a zero entry is not an inherent zero. 5. (a) Observational study. The study does not attempt to influence the responses of the subjects and there is no treatment. (b) Experiment. The study applies a treatment (video involving smoking) to the subjects. 6. Randomized block design 7. (a) Convenience sampling is used because all the people sampled are in one convenient location. (b) Systematic sampling is used because every tenth machine part is sampled. (c) Stratified sampling is used because the population is first stratified and then a sample is collected from each stratum. 8. Convenience sampling. People at campgrounds may be strongly against air pollution because they are at an outdoor location.

CHAPTER 1 Ň INTRODUCTION TO STATISTICS 13

CHAPTER 1 TEST SOLUTIONS 1. (a) Sampling, because the population of New Jersey is too large for the most popular type of investment to be easily recorded. Random sampling would be advised because it would be easy to select people from New Jersey randomly and then record their most popular type of investment. (b) Census, because the population is small and it is relatively easy to obtain the ages of the 30 employees. 2. (a) Sample statistic. The value of 27% is a numerical description of a sample of U.S. adults owning a smart watch or fitness tracker. (b) Population parameter. The average evidence-based reading and writing score of 528 is a numerical description of all test takers in a recent year. 3. (a) Stratified sampling is used because the high school students are divided into strata (male and female), and a sample is selected from each stratum. (b) Simple random sampling is used because each customer has an equal chance of being contacted, and all samples of 625 customers have an equal chance of being selected. (c) Convenience sampling is used because a sample is taken from members of a population that are readily available. The sample may be biased because the teachers at that school may not be representative of the population of teachers. 4. (a) Quantitative. Ratio. The number of employees are numerical measurements. A ratio of two data values can be formed, so it makes sense to say that 40 employees are twice as many as 20 employees. (b) Quantitative. Interval. The grade point averages (GPAs) are numerical measurements. Data can be ordered and meaningful differences can be calculated, but it does not make sense to say that a person with a 3.8 GPA is twice as smart as a person with a 1.9 GPA. 5. (a) The survey question is unbiased. (b) The question is biased because it already suggests that the town’s ban on skateboarding in parks is unfair. The question could be written as “What are your thoughts on the town’s ban on skateboarding in parks?” 6. (a) Population: Collection of the responses of all U.S. physicians Sample: Collection of the 17,461 U.S. physicians who were sampled. (b) Both. Gender, location, employment status, specialty, and if they would choose medicine again are qualitative because they are attributes. Age, income, and time spent seeing patients per week are quantitative because they are numerical measurements. (c) Nominal: gender, location, employment status, specialty, would they choose medicine again Ratio: age, income, time spent seeing patients per week (d) Observational study. The study does not attempt to influence the responses of the physicians and there is no treatment. Copyright © 2023 Pearson Education, Inc.

CHAPTER

Descriptive Statistics

2.1 FREQUENCY DISTRIBUTIONS AND THEIR GRAPHS 2.1 TRY IT YOURSELF SOLUTIONS 1. The number of classes is six. Range 55 − 13 = =7 Number of classes 6 Since the class width is a whole number, use the next whole number as the class width. The class width is 8. The minimum data entry is a convenient lower limit for the first class. Then add the class width to get the lower limits of the other classes. The upper limits are one less than the lower limit of the next class.

Min = 13, Max = 55, Class width =

Lower limit 13 21 29 37 45 53

Upper limit 20 28 36 44 52 60

Make a tally mark for each entry in the appropriate class. The number of tally marks for a class is the frequency of that class. Class 13–20 21–28 29–36 37–44 45–52 53–60

Frequency, f 9 16 18 7 4 1

2. Find each midpoint, relative frequency, and cumulative frequency. (Lower class limit) + (Upper class limit) Midpoint = 2 Class frequency f Relative frequency = = Sample size n The cumulative frequency of a class is the sum of the frequencies of that class and all previous classes.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

Class

Midpoint

13–20

21–28

29–36

37–44

45–52

53–60

13 + 20 = 16.5 2 21 + 28 = 24.5 2 29 + 36 = 32.5 2 37 + 44 = 40.5 2 45 + 52 = 48.5 2 53 + 60 = 56.5 2

f = 55

Relative frequency 9 ≈ 0.16 55 16 ≈ 0.29 55 18 ≈ 0.33 55 7 = 0.13 55 4 ≈ 0.07 55 1 ≈ 0.02 55 f ¦ =1 n

Cumulative frequency 9

9 + 16 = 25 25 + 18 = 43 43 + 7 = 50 50 + 4 = 54 54 + 1 = 55

Sample answer: The most common range of points scored by winning teams is 29 to 36. About 9% of the winning teams scored more than 45 points. 3. Find the class boundaries. Because the data entries are integers, subtract 0.5 from each lower limit to find the lower class boundaries and add 0.5 to each upper limit to find the upper class boundaries. Class 13–20 21–28 29–36 37–44 45–52 53–60

Class Boundaries 12.5–20.5 20.5–28.5 28.5–36.5 36.5–44.5 44.5–52.5 52.5–60.5

Frequency, f 9 16 18 7 4 1

Use class midpoints for the horizontal scale and frequency for the vertical scale. (Class boundaries can also be used for the horizontal scale.)

Sample answer: The most common range of points scored by winning teams is 29 to 36. About 9% of the winning teams scored more than 45 points.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

4. To construct the frequency polygon, use the same horizontal and vertical scales that were used in the histogram labeled with the class midpoints in Try It Yourself 3. Then, plot the points that represent the midpoint and frequency of each class and connect the points with line segments. Extend the left side and right side to one class width before the first class midpoint and after the last class midpoint, respectively.

Sample answer: The frequency of points scored increases up to 32.5 points and then decreases. 5. Notice the shape of the relative frequency histogram is the same as the shape of the frequency histogram constructed in Try It Yourself 3. The only difference is that the vertical scale measures the relative frequencies.

6. Use upper class boundaries for the horizontal scale and cumulative frequency for the vertical scale.

Sample answer: The greatest increase in cumulative frequency occurs between 28.5 and 36.5. 7.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

2.1 EXERCISE SOLUTIONS 1. Organizing the data into a frequency distribution may make patterns within the data more evident. Sometimes it is easier to identify patterns of a data set by looking at a graph of the frequency distribution. 2. If there are too few or too many classes, it may be difficult to detect patterns because the data are too condensed or too spread out. 3. Class limits determine which numbers can belong to that class. Class boundaries are the numbers that separate classes without forming gaps between them. 4. Relative frequency of a class is the portion, or percentage, of the data that falls in that class. Cumulative frequency of a class is the sum of the frequencies of that class and all previous classes. 5. The sum of the relative frequencies must be 1% or 100% because it is the sum of all portions or percentages of the data. 6. A frequency polygon displays frequencies or relative frequencies, whereas an ogive displays cumulative frequencies. 7. False. Class width is the difference between the lower (or upper) limits of consecutive classes. 8. True 9. False. The graph of cumulative frequencies always starts at the lower boundary of the first class and increases to the upper boundary of the last class; therefore, always increasing from left to right. 10. True Range 64 − 9 = ≈ 7.9 8 Number of classes 7 Lower class limits: 9, 17, 25, 33, 41, 49, 57 Upper class limits: 16, 24, 32, 40, 48, 56, 64

11. Class width =

Range 88 − 12 = ≈ 12.7 13 Number of classes 6 Lower class limits: 12, 25, 38, 51, 64, 77 Upper class limits: 24, 37, 50, 63, 76, 89

12. Class width =

Range 118 = = 14.75 15 Number of classes 8 Lower class limits: 17, 32, 47, 62, 77, 92, 107, 122 Upper class limits: 31, 46, 61, 76, 91, 106, 121, 136

13. Class width =

Range 93 = = 9.3 10 Number of classes 10 Lower class limits: 154, 164, 174, 184, 194, 204, 214, 224, 234, 244 Upper class limits: 163, 173, 183, 193, 203, 213, 223, 233, 243, 253

14. Class width =

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

15. (a) Class width = 11 − 0 = 11 (b) and

(c)

(Lower class limit) + (Upper class limit) 2 Find the class boundaries. Because the data entries are integers, subtract 0.5 from each lower limit to find the lower class boundaries and add 0.5 to each upper limit to find the upper class boundaries. Midpoint =

Class

Midpoint

0–10 11–21 22–32 33–43 44–54 55–65 66–76

5 16 27 38 49 60 71

Class boundaries −0.5–10.5 10.5–21.5 21.5–32.5 32.5–43.5 43.5–54.5 54.5–65.5 65.5–76.5

16. (a) Class width = 33 − 25 = 8 (b) and

(c)

Class

Midpoint

25௘–௘32 33௘–௘40 41௘–௘48 49௘–௘56 57௘–௘64 65௘–௘72 73௘–௘80

28.5 36.5 44.5 52.5 60.5 68.5 76.5

Class boundaries 24.5௘–௘32.5 32.5௘–௘40.5 40.5௘–௘48.5 48.5௘–௘56.5 56.5௘–௘64.5 64.5௘–௘72.5 72.5௘–௘80.5

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

Class frequency f = Sample size n The cumulative frequency of a class is the sum of the frequencies of that class and all previous classes.

17. Relative frequency =

Class 0–10 11–21 22–32 33–43 44–54 55–65 66–76

Frequency f 188 372 264 205 83 76 32 f = 1220

Midpoint 5 16 27 38 49 60 71

Relative frequency 0.15 0.30 0.22 0.17 0.07 0.06 0.03 ௙ ൌͳ

Cumulative frequency 188 560 824 1029 1112 1188 1220

௡

Class frequency f = Sample size n The cumulative frequency of a class is the sum of the frequencies of that class and all previous classes. Class Frequency, Midpoint Relative Cumulative f frequency frequency 25–32 86 28.5 0.24 86 33–40 39 36.5 0.11 125 41–48 41 44.5 0.11 166 49–56 48 52.5 0.13 214 57–64 43 60.5 0.12 257 65–72 68 68.5 0.19 325 73–80 40 76.5 0.11 365 ௙ f = 365 ൎͳ

18. Relative frequency =

௡

19. (a) Number of classes: 7 (b) Greatest frequency: about 300 Least frequency: about 10 (c) Class width: 10 (d) Sample answer: About half of the employee salaries are between $50,000 and $69,000. 20. (a) Number of classes: 6 (b) Greatest frequency: 37 Least frequency: 1 (c) Class width: 53 (d) Sample answer: The heights of most roller coasters are less than 231 feet.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

21. Identify the highest point and its respective class. Class with greatest frequency: 506 – 510. Identify the lowest point (not including the points on the horizontal axis) and its respective class. Class with least frequency: 474 – 478. 22. Identify the highest point and its respective class. Class with greatest frequency: 3.5 – 4.5 miles. Identify the lowest point (not including the points on the horizontal axis) and its respective class. Class with least frequency: 0.5 – 1.5 miles. 23. (a) Identify the tallest bar and its respective class. Class with greatest relative frequency: 35 – 36 centimeters. Identify the shortest bar and its respective class. Class with least relative frequency: 39 – 40 centimeters. (b) Greatest relative frequency § 0.25 Least relative frequency § 0.01 (c) Sample answer: From the graph, 0.25% or 25% of females have a fibula length between 35 and 36 centimeters. 24. (a) Identify the tallest bar and its respective class. Class with greatest relative frequency: 11 – 12 minutes Identify the shortest bar and its respective class. Class with least relative frequency: 14 – 15 minutes (b) Greatest relative frequency § 38% Least relative frequency § 4% (c) Sample answer: From the graph, about 0.75% or 75% of campus security response times are between 11 and 13 minutes. 25. (a) Locate the cumulative frequency of the highest (right-most) point. The number in the sample is 75. (b) Locate the neighboring points where the pitch between them is the steepest. The greatest increase in frequency is from 158.5 – 201.5 pounds. 26. (a) Locate the cumulative frequency of the highest (right-most) point. The number in the sample is 77. (b) Locate the neighboring points where the pitch between them is the steepest. The greatest increase in frequency is from 68 – 70 inches. 27. (a) Locate 201.5 on the horizontal axis and find the corresponding cumulative frequency at the point on the ogive: 47. (b) Locate 68 on the vertical axis and find the corresponding weight at the point on the ogive: 287.5 pounds. (c) Subtract the cumulative frequency for each weight: 62 – 22 = 40.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

(d) Subtract the cumulative frequency for bears weighing 330.5 pounds from the number in the sample: 75 – 69 = 6. 28. (a) Locate 72 on the horizontal axis and find the corresponding cumulative frequency at the point on the ogive: 71. (b) Locate 15 on the vertical axis and find the corresponding height at the point on the ogive: 68 inches. (c) Subtract the cumulative frequency for each height: 71 – 15 = 56. (d) Subtract the cumulative frequency for adult males who are 70 inches tall from the number in the sample: 77 – 47 = 30. 29. Class width = Class 28ௗ–௘57 58௘–௘87 88௘–௘117 118௘–௘147 148௘–௘177

Range 177 − 28 = = 29.8 30 Number of classes 5

Frequency, f 2 4 7 7 4 ѷf = 24

Midpoint 42.5 72.5 102.5 132.5 162.5

Relative frequency 0.08 0.17 0.29 0.29 0.17 f ≈1 n

Cumulative frequency 2 6 13 20 24

Classes with greatest frequency: 88–117, 118–147 Class with the least frequency: 28–57 30. Class width = Class 30௘–௘113 114௘–௘197 198௘–௘281 282௘–௘365 366௘–௘449 450௘–௘533

Range 530 − 30 = ≈ 83.3 84 Number of classes 6

Frequency, f 5 7 8 3 3 4 ѷf = 30

Midpoint 71.5 155.5 239.5 323.5 407.5 491.5

Relative frequency 0.17 0.23 0.27 0.10 0.10 0.13 f =1 n

Cumulative frequency 5 12 20 23 26 30

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

31. Class width =

Range 7119 − 1000 = ≈ 1019.8 1020 Number of classes 6

Class 1000ௗ௘–௘2019 2020ௗ௘–௘3039 3040௘–௘ௗ4059 4060௘–௘ௗ5079 5080௘–௘ௗ6099 6100௘–௘ௗ7119

Frequency, f 11 3 2 3 1 1 ѷf = 21

Midpoint

Relative frequency 0.52 0.14 0.10 0.14 0.05 0.05 f =1 n

1509.5 2529.5 3549.5 4569.5 5589.5 6609.5

Cumulative frequency 11 14 16 19 20 21

Sample answer: The graph shows that most of the sales representatives at the company sold from $1000 to $2019. 32. Class width =

Range 51 − 32 = = 3.8 4 Number of classes 5

Class

Frequency, f

Midpoint

32–35 36–39 40–43 44–47 48–51

3 9 8 3 1

33.5 37.5 41.5 45.5 49.5

¦ f = 24

Relative frequency 0.1250 0.3750 0.3333 0.1250 0.0417 f =1 n

Cumulative frequency 3 12 20 23 24

Sample answer: The graph shows that most of the pungencies of the peppers were between 36,000 and 43,000 Scoville units.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

33. Class width =

Range 514 − 291 = = 27.875 28 Number of classes 8

Class

Frequency, f

Midpoint

291–318 319–346 347–374 375–402 403–430 431–458 459–486 487–514

5 4 3 5 6 4 1 2

304.5 332.5 360.5 388.5 416.5 444.5 472.5 500.5

¦ f = 30

Relative frequency 0.1667 0.1333 0.1000 0.1667 0.2000 0.1333 0.0333 0.0667 f =1 n

Cumulative frequency 5 9 12 17 23 27 28 30

Sample answer: The graph shows that the most frequent reaction times were between 403 and 430 milliseconds. 34. Class width = Class 2243௘–௘2588 2589௘–௘2934 2935௘–௘3280 3281௘–௘3626 3627௘–௘3972 3973௘–௘4318 4319௘–௘4664 4665௘–௘5010

Range 5008 − 2243 = = 345.625 346 Number of classes 8

Frequency, f 5 4 2 2 4 3 0 1

¦ f = 21

Midpoint 2415.5 2761.5 3107.5 3453.5 3799.5 4145.5 4491.5 4837.5

Relative frequency 0.2381 0.1905 0.0952 0.0952 0.1905 0.1429 0.0000 0.0476 f =1 n

Cumulative frequency 5 9 11 13 17 20 20 21

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

Sample answer: The graph shows that the most frequent finishing times were from 2243 to 2588 seconds. 35. Class width = Class 42௘–௘47 48௘–௘53 54௘–௘59 60௘–௘65 66௘–௘71 72–௘77 78–௘83

Range 78 − 42 = ≈ 5.1 6 Number of classes 7

Frequency, f 6 11 17 8 3 0 1 ¦ f = 46

Midpoint 44.5 50.5 56.5 62.5 68.5 74.5 80.5

Relative frequency 0.13 0.24 0.37 0.17 0.07 0.00 0.02 f =1 n

Cumulative frequency 6 17 34 42 45 45 46

Sample answer: The graph shows that the number of U.S. presidents who were 60 or older at inauguration was less than half as many as those who were 59 and younger. 36. Class width =

Range 18 − 0 = = 3.6 4 Number of classes 5

Class

Midpoint

0௘–௘3 4௘–௘7 8௘–௘11 12௘–௘15 16௘–௘19

Frequency, f 21 18 9 6 2

¦ f = 56

1.5 5.5 9.5 13.5 17.5

Relative frequency 0.3750 0.3214 0.1607 0.1071 0.0357 f ≈1 n

Cumulative frequency 21 39 48 54 56

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

Sample answer: The graph shows that most of the signers of the Declaration of Independence had seven or fewer children. 37. Class width =

Range 10 − 1 = = 1.8 2 Number of classes 5

Class

Midpoint

1௘–௘2 3௘–௘4 5௘–௘6 7௘–௘8 9௘–௘10

Frequency, f 7 8 10 2 9

1.5 3.5 5.5 7.5 9.5

¦ f = 36

Relative frequency 0.19 0.22 0.28 0.06 0.25 f ≈1 n

Cumulative frequency 7 15 25 27 36

Class with greatest relative frequency: 5–6 Class with least relative frequency: 7 – 8 38. Class width = Class 7௘–௘9 10௘–௘12 13௘–௘15 16௘–௘18 19௘–௘21

Range 19 − 7 = = 2.4 3 Number of classes 5

Frequency, f 8 11 8 0 1

Midpoint

¦ f = 28

8 11 14 17 20

Relative frequency 0.2857 0.3929 0.2857 0.0000 0.0357 f ≈1 n

Cumulative frequency 8 19 27 27 28

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

Class with greatest relative frequency: 10–12 Class with least relative frequency: 16–18 Range 75 − 60 = =3 Number of classes 5 Notice that using a class width of 3 is not wide enough to include all the data with five classes. Therefore, use a class width of 4.

39. Class width =

Class 60௘–௘63 64௘–௘67 68௘–௘71 72௘–௘75 76௘–௘79

Frequency ,f 6 7 9 6 0

Midpoint

¦ f = 28

61.5 65.5 69.5 73.5 77.5

Relative frequency 0.2143 0.2500 0.3214 0.2143 0.0000 f ≈1 n

Cumulative frequency 6 13 22 28 28

Class with greatest relative frequency: 68–71 Class with least relative frequency: 76–79 40. Class width =

Range 462 − 138 = = 64.8 65 Number of classes 5

Class

Frequency, f

Midpoint

138௘–௘202 203௘–௘267 268௘–௘332 333௘–௘397 398௘–௘462

13 7 4 1 3

170 235 300 365 430

¦ f = 28

Relative frequency 0.46 0.25 0.14 0.04 0.11 f ≈1 n

Cumulative frequency 13 20 24 25 28

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

Class with greatest relative frequency: 138 – 202 Class with least relative frequency: 333 – 397 41. Class width =

Range 75 − 52 = ≈ 3.8 4 Number of classes 6

Class

Frequency, f

52௘–௘55 56௘–௘59 60௘–௘63 64௘–௘67 68௘–௘71 72௘–௘75

6 4 6 10 5 4

¦ f = 35

Relative frequency 0.1714 0.1143 0.1714 0.2857 0.1429 0.1143

Cumulative frequency 6 10 16 26 31 35 f ≈1 n

Location of the greatest increase in frequency: 64 – 67 42. Class width =

Range 26 − 7 = ≈ 3.2 4 Number of classes 6

Class

Frequency, f

7௘–௘10 11௘–௘14 15௘–௘18 19௘–௘22 23௘–௘26 27௘–௘30

2 11 7 5 3 0

¦ f = 28

Relative frequency 0.0714 0.3929 0.2500 0.1786 0.1071 0.0000 f ≈1 n

Cumulative frequency 2 13 20 25 28 28

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

Location of the greatest increase in frequency: 11 – 14 43. (a) Class width = Class 65–74 75–84 85–94 95–104 105–114 115–124

Range 120 − 65 = ≈ 9.2 10 Number of classes 6

Frequency, f 4 7 4 5 3 1

Midpoint 69.5 79.5 89.5 99.5 109.5 119.5

¦ f = 24

(b)

(c)

(d)

(e)

Relative frequency 0.1667 0.2917 0.1667 0.2083 0.1250 0.0417 f ≈1 N

Cumulative frequency 4 11 15 20 23 24

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

Range 361 − 1 = = 45 Number of classes 8 Since the class width is a whole number, use the next whole number as the class width. The class width is 46.

44. (a) Class width =

Class

Frequency, f

Midpoint

1௘–௘46 47–௘92 93௘–௘138 139–௘184 185௘–௘230 231–௘276 277௘–௘322 323–௘368

24 18 6 3 2 0 0 2 ¦ f = 55

23.5 69.5 115.5 161.5 207.5 253.5 299.5 345.5

(b)

(c)

(d)

(e)

Relative frequency 0.44 0.33 0.11 0.05 0.04 0.00 0.00 0.04 f ¦ =1 n

Cumulative frequency 24 42 48 51 53 53 53 55

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

45. (a) Class width =

Range 104 − 61 = = 5.375 6 Number of classes 8

Class

Frequency, f

Midpoint

61–66 67–72 73–78 79–84 85–90 91–96 97–102 103–108

1 3 6 10 5 2 2 1

63.5 69.5 75.5 81.5 87.5 93.5 99.5 105.5

¦ f = 30

Relative frequency 0.033 0.100 0.200 0.333 0.167 0.067 0.067 0.033 f =1 n

(b) 16.7%, because the sum of the relative frequencies for the last three classes is 0.167. (c) $9700, because the sum of the relative frequencies for the last two classes is 0.10. Range 1500 − 650 = = 85 Number of classes 10 Notice that using a class width of 85 is not wide enough to include all the data with 10 classes. Therefore, use a class width of 86.

46. (a) Class width =

Class

Frequency, f

Midpoint

650–735 736–821 822–907 908–993 994–1079 1080–1165 1166–1251 1252–1337 1338–1423 1424–1509

1 2 3 6 8 8 7 8 4 3

692.5 778.5 864.5 950.5 1036.5 1122.5 1208.5 1294.5 1380.5 1466.5

¦ f = 50

Relative frequency 0.02 0.04 0.06 0.12 0.16 0.16 0.14 0.16 0.08 0.06 f =1 n

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

(b) 64%. The portion of the scores greater than or equal to 1070 is 0.64. (c) A score of 908 or above, because the sum of the relative frequencies of the class starting with 908 and all classes with higher scores is 0.88. 47.

In general, a greater number of classes better preserves the actual values of the data set but is not as helpful for observing general trends and making conclusions. In choosing the number of classes, an important consideration is the size of the data set. For instance, you would not want to use 20 classes if your data set contained 20 entries. In this particular example, as the number of classes increases, the histogram shows more fluctuation. The histograms with 10 and 20 classes have classes with zero frequencies. Not much is gained by using more than five classes. Therefore, it appears that five classes would be best.

2.2 MORE GRAPHS AND DISPLAYS 2.2 TRY IT YOURSELF SOLUTIONS 1. Because the data entries go from a low of 13 to a high of 55, use stem values from 1 to 5. List the stems to the left of a vertical line. For each data entry, list a leaf to the right of its stem. 1 2 3 4 5

3 4 6 6 6 7 Key: 1 | 3 = 13 0 0 0 1 1 1 3 3 4 4 4 4 6 7 7 7 7 7 8 9 0 1 1 1 1 1 1 2 2 3 4 4 4 4 5 5 5 7 8 8 9 1 2 3 6 8 9 2 5

Sample answer: Most of the winning teams scored between 20 and 39 points.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

2. Use the leaves 0, 1, 2, 3, and 4 in the first stem row and the leaves 5, 6, 7, 8, and 9 in the second stem row. 1 1 2 2 3 3 4 4 5 5

3 4 Key: 1 | 3 = 13 6 6 6 7 0 0 0 1 1 1 3 3 4 4 4 4 6 7 7 7 7 7 8 9 0 1 1 1 1 1 1 2 2 3 4 4 4 4 5 5 5 7 8 8 9 1 2 3 6 8 9 2 5

Sample answer: Most of the winning teams scored from 20 to 35 points. 3. Choose the horizontal axis so that each data entry is included in the dot plot. For example, label the horizontal axis from 10 to 55.

Sample answer: Most of the points scored by winning teams cluster between 20 and 40. 4.

Type of degree Associate’s Bachelor’s Master’s Doctoral

f 849 1650 693 159

¦ f = 3351

Relative frequency 0.253 0.492 0.207 0.047 f ≈1 n

Angle 91.1° 177.1° 74.5° 16.9°

¦ ≈ 360°

From 2010 to 2019, as percentages of the total degrees conferred, associate’s degrees increased by 0.2%, bachelor’s degrees increased by 0.2%, master’s degrees decreased by 0.2%, and doctoral degrees decreased by 0.1%. 5.

Cause Roofing contractors Collection agencies Insurance companies New car dealers Internet marketing services

Frequency, f 5987 14,878 10,871 23,848 1687

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

New car dealers are the greatest cause of complaints. 6.

It appears that the longer an employee is with the company, the greater the employee’s salary. 7. Let the horizontal axis represents the years and let the vertical axis represents the number of robberies (in thousands).

Sample answer: The number of robberies decreased between 2009 and 2019.

2.2 EXERCISE SOLUTIONS 1. Quantitative: stem-and-leaf plot, dot plot, histogram, time series chart, scatter plot Qualitative: pie chart, Pareto chart 2. Unlike the histogram, the stem-and-leaf plot still contains the original data values. However, some data are difficult to organize in a stem-and-leaf plot. 3. Both the stem-and-leaf plot and the dot plot allow you to see how data are distributed, determine specific data entries, and identify unusual data values. 4. In a Pareto chart, the height of each bar represents frequency or relative frequency and the bars are positioned in order of decreasing height with the tallest bar positioned at the left.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

5. b

6. d

7. a

8. c

9. 27, 32, 41, 43, 43, 44, 47, 47, 48, 50, 51, 51, 52, 53, 53, 53, 54, 54, 54, 54, 55, 56, 56, 58, 59, 68, 68, 68, 73, 78, 78, 85 Max: 85 Min: 27 10. 12.9, 13.3, 13.6, 13.7, 13.7, 14.1, 14.1, 14.1, 14.1, 14.3, 14.4, 14.4, 14.6, 14.9, 14.9, 15.0, 15.0, 15.0, 15.1, 15.2, 15.4, 15.6, 15.7, 15.8, 15.8, 15.8, 15.9, 16.1, 16.6, 16.7 Max: 16.7 Min: 12.9 11. 13, 13, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 17, 18, 19 Max: 19 Min: 13 12. 214, 214, 214, 216, 216, 217, 218, 218, 220, 221, 223, 224, 225, 225, 227, 228, 228, 228, 228, 230, 230, 231, 235, 237, 239 Max: 239 Min: 214 13. Sample answer: Facebook has the most users, and TikTok has the least. Instagram and WeChat have about the same number of users. 14. Sample answer: The year 2016 had the most automotive sales, and the year 2020 had the least. Automotive sales decreased the most between 2019 and 2020. 15. Sample answer: The Texter is the least popular driver. The Left-Lane Hog is tolerated more than the Tailgater. The Speedster and the Drifter have the same popularity. 16. Sample answer: Dogs are owned by the most amount of households, and horses are owned by the least. Birds and reptiles are the closest for the number of households that own each. 17. Exam Scores

Key: 6 7 = 67

67 8 73 5 5 6 9 80 0 2 3 5 5 7 7 8 90 1 1 1 2 4 5 5 Sample answer: Most grades for the biology midterm were in the 80s and 90s.

18. Hours Worked by Nurses

Key: 2 4 = 24

24 30 2 2 2 3 5 5 6 6 6 6 8 8 9 40 0 0 0 0 0 0 0 8 50 Sample answer: Most nurses work between 30 and 40 hours per week.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

19. Ice Thickness (in centimeters) Key: 4 3 = 4.3

43 9 51 8 8 8 9 64 8 9 9 9 70 0 2 2 2 5 80 1 Sample answer: Most of the ice had a thickness of 5.8 centimeters to 7.2 centimeters. 20. Tomato Prices (in dollars per pound)

15 16 17 18 19 20 21

4 7 Key: 15 | 4 = 1.54 0 1 1 3 4 4 6 8 1 4 7 8 8 9 2 3 6 7 9 1 3 7 8 7 7 8 1 3

Sample answer: Most retail outlets charge $1.60 to $1.79 per pound of tomatoes. 21. Incomes (in millions) of Highest Paid Athletes

3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10

Key: 3 | 7 = 37 7 7 9 0 1 1 2 5 5 5 7 7 8 8 8 9 2 4 6 7 9 1 2 4 4 8 6 4 5 6

Sample answer: Most highly paid athletes have an income around $40 million.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

22. Electoral Votes

0 0 1 1 2 2 3 3 4 4 5 5

3 3 3 3 3 3 3 3 4 4 4 4 4 Key: 0 | 3 = 3 5 5 5 6 6 6 6 6 6 7 7 7 8 8 9 9 9 0 0 0 0 1 1 1 1 2 3 4 5 6 6 8 0 0 9 9 8

Sample answer: Over half the states have fewer than 10 electoral votes. 23. Sample answer: Systolic blood pressure tends to be from 120 to 150 millimeters of mercury. 24.

Sample answer: The life span of a housefly tends to be from 6 to 14 days. 25.

Educational Attainment Not a high school graduate Some college, no degree Bachelor’s degree Doctoral or professional degree High school graduate Associate’s degree Master’s degree

Relative Frequency

Angle

26.6

0.106

38°

45.0

0.180

65°

53.3 7.7

0.213 0.031

77° 11°

70.9

0.283

102°

24.5 22.5

0.098 0.090 f ≈1 n

35° 32° = 360°

¦ f = 250.5

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

Sample answer: The majority of U.S. adults are high school graduates. 26.

Category United States Japan Soviet Union Spain Norway Denmark Portugal Morocco United Kingdom Kenya Mexico Ethiopia

Frequency, f 2 2 1 1 1 1 3 2 6 15 3 4

¦ f = 41

Relative Frequency 0.049 0.049 0.024 0.024 0.024 0.024 0.073 0.049 0.146 0.366 0.073 0.098 f ≈1 n

Angle 18° 18° 9° 9° 9° 9° 26° 18° 53° 132° 26° 35°

¦ ≈ 360°

Sample answer: Most of the London Marathon winners are from Kenya. 27.

Smartphone Manufacturers Apple Huawei Samsung Vivo Xiaomi

Units Sold (in millions) 206.1 189.0 266.7 111.7 147.8

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

Sample answer: Samsung sold the most smartphones out of the five manufacturers and Vivo sold the least. 28. Cause of Death Lower respiratory infections Stroke Chronic obstructive pulmonary disease Ischemic heart disease COVID-19

Numbers 2,971,000 7,809,000 3,910,000

10,890,000 3,068,000

Sample answer: The leading cause of death worldwide in 2020 was ischemic heart disease. 29. Let the horizontal axis represents hours and let the vertical axis represents the hourly wage (in dollars).

Sample answer: It appears that there is no relation between hourly wages and hours worked.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

30. Let the horizontal axis represent the number of students per teacher and let the vertical axis represent the average salary (in thousands of dollars).

Sample answer: It appears that there is no relation between a teacher’s average salary and the number of students per teacher. 31. Let the horizontal axis represents the years and let the vertical axis represents the number of degrees (in thousands).

Sample answer: The number of bachelor’s degrees in engineering conferred in the U.S. has increased from 2011 to 2019. 32. Let the horizontal axis represents the years and let the vertical axis represents the percent.

Sample answer: The percentage of the U.S. gross domestic product (GDP) that comes from the manufacturing sector stayed constant or decreased from 2010 to 2019. 33. Heights (in inches)

7 2 2 4 Key: 7 | 2 = 72 7 5 5 5 5 6 8 1 1 2 2 2 4 8 The dot plot helps you see that the data are clustered from 72 to 76 and 81 to 84, with 75 being the most frequent value. The stem-and-leaf plot helps you see that most values are 75 or greater.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

34.

The stem-and-leaf plot helps you see that most values are from 60 to 69. The dot plot helps you see that the values 55 and 60 occur most frequently. 35.

The pie chart helps you to see the percentages as parts of a whole, with fall being the largest. It also shows that while fall is the largest percentage, it makes up less than half of the pie chart. That means that a majority of U.S. adults ages 18 and older prefer a season other than fall. This means it would not be a fair statement to say that most U.S. adults ages 18 and older prefer fall. The Pareto chart helps you to see the rankings of the seasons. 36.

The Pareto chart helps you see the order from the most favorite to least favorite day. The pie chart helps you visualize the data as parts of a whole and see that about 80% of people say their favorite day is Friday, Saturday, or Sunday. 37. (a) The graph is misleading because the large gap from 0 to 90 makes it appear that the sales for the third quarter are disproportionately larger than the other quarters.

(b)

38. (a) The graph is misleading because the vertical axis has no break. The percent of middle schoolers that responded “yes” appears three times larger than either of the others when the difference is only 10%.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

(b)

39. (a) The graph is misleading because the angle makes it appear as though the third quarter had a larger percent of sales than the others, when the first and third quarters have the same percent.

(b)

40. (a) The graph is misleading because the “non-OPEC countries” bar is wider than the “OPEC countries” bar.

(b)

41. (a) At Law Firm A, the lowest salary was $90,000 and the highest salary was $203,000. At Law Firm B, the lowest salary was $90,000 and the highest salary was $190,000. There are 30 lawyers at Law Firm A and 32 lawyers at Law Firm B.

(b) At Law Firm A, the salaries tend to be clustered at the far ends of the distribution range. At Law Firm B, the salaries are spread out. 42. (a)

3:00 P.M. Class

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

8:00 P.M. Class

1 2 3 4 5 6 7 8

8 8 8 8 8 9 9 9 9 9 0 0 0 2 3 4 4 5 5 8 9 9 1 1 9 3 4 4 6 Key: 5 | 3 | 1 = 35-year-old 1 in 3:00 P.M. class and 31-year-old in 8:00 P.M. class

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

(b) In the 3:00 P.M. class, the lowest age is 35 years old and the highest age is 85 years old. In the 8:00 P.M. class, the lowest age is 18 years old and the highest age is 71 years old. There are 26 participants in the 3:00 P.M. class and there are 30 participants in the 8:00 P.M. class. (c) Sample answer: The participants in each class are clustered at one of the ends of their distribution range. The 3:00 P.M. class mostly has participants over 50 years old and the 8:00 P.M. class mostly has participants under 50 years old. 43. (a)

2 3 4 0 5 6 7

6 Key: 2 | 6 = 26 1 0 4 4 5 6 7 9 9 4 4 5 6 7 9 9 5 5 5 3 4 4 0 1 2 2

(b)

(c)

(d)

(e)

Sample answer: The stem-and-leaf plot, dot plot, frequency histogram, and ogive display the data best because the data are quantitative.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

2.3 MEASURES OF CENTRAL TENDENCY 2.3 TRY IT YOURSELF SOLUTIONS 1.

¦ x = 21 + 29 + 17 + 27 + 31 + ... + 13 + 31 + 31 = 456 ¦ x 456 = = 28.5 x= n 16 The mean points scored by the winning teams in the Super Bowls for the National Football League’s 2005 through 2020 seasons is 28.5.

2. Order the data from smallest to largest. Note: The stem-and-leaf plot from Try It Yourself 1 in Section 2.2 may be helpful to in ordering the data. 13, 14, 16, 16, 16, 17, 20, 20, 20, 21, 21, 21, 23, 23, 24, 24, 24, 24, 26, 27, 27, 27, 27, 27, 28, 29, 30, 31, 31, 31, 31, 31, 31, 32, 32, 33, 34, 34, 34, 34, 35, 35, 35, 37, 38, 38, 39, 41, 42, 43, 46, 48, 49, 52, 55 Because there are 55 entries (an odd number), the median is the middle, or 28th entry. So, the median is 31 points. 3. Order the data from smallest to largest. 13, 17, 21, 21, 24, 27, 28, 29, 31, 31, 31, 31, 34, 34, 41, 43 Because there are an even number of entries, the median is the mean of the two middle entries. 29 + 31 60 Median = = = 30 2 2 The median points scored by the winning teams in the Super Bowls for the National Football League’s 2005 through 2020 seasons is 30 points. 4. Look at the ordered data from Try It Yourself 2 13, 14, 16, 16, 16, 17, 20, 20, 20, 21, 21, 21, 23, 23, 24, 24, 24, 24, 26, 27, 27, 27, 27, 27, 28, 29, 30, 31, 31, 31, 31, 31, 31, 32, 32, 33, 34, 34, 34, 34, 35, 35, 35, 37, 38, 38, 39, 41, 42, 43, 46, 48, 49, 52, 55 The mode is the data entry that occurs with the greatest frequency. The entry 31 occurs the most, so the mode is 31. 5. “Not too much” occurs with the greatest frequency (5970). The mode is “not too much.” 6.

¦ x 410 = ≈ 21.6 n 19 Median = 21 Mode = 20 The mean in Example 6 ( x ≈ 23.8 ) was heavily influenced by the entry 65. Neither the median nor the mode was affected as much by the entry 65. x=

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

Final Grade C C D A B B ¦ ( x ⋅ w)

Points, x 2 2 1 4 3 3

Weight, w 3 4 1 3 2 3 w = 16

x·w 6 8 1 12 6 9 ( x ⋅ w) = 42 .

42 ≈ 2.6 ¦w 16 The new weighted mean is about 2.6. x=

Class 13–20 21–28 29–36 37–44 45–52 53–60

Midpoint, x 16.5 24.5 32.5 40.5 48.5 56.5

Frequency, f 9 16 18 7 4 1

x·f 148.5 392 585 283.5 194 56.5

¦ f = 55 = n

¦ ( x ⋅ f ) = 1659.5

¦(x ⋅ f )

1659.5 = ≈ 30.2 n 55 This is very close to the mean found using the original data set. x=

2.3 EXERCISE SOLUTIONS 1. True 2. False. All quantitative data sets have a median. 3. True 4. True 5. Sample answer: 2, 4, 5, 5, 6, 8 6. Sample answer: 1, 2, 2, 3, 3, 4 7. Sample answer: 2, 5, 7, 9, 35 8. Sample answer: 1, 2, 3, 3, 3, 4, 5 9. The shape of the distribution is skewed right because the bars have a “tail” to the right.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

10. The shape of the distribution is symmetric because a vertical line can be drawn down the middle, creating two halves that are approximately the same. 11. The shape of the distribution is uniform because the bars are approximately the same height. 12. The shape of the distribution is skewed left because the bars have a “tail” to the left. 13. (11), because the distribution values range from 1 to 12 and has (approximately) equal frequencies. 14. (9), because the distribution has values in the thousands and is skewed right due to the few vehicles that have much higher mileages than the majority of the vehicles. 15. (12), because the distribution has a maximum value of 90 and is skewed left due to a few students scoring much lower than the majority of the students. 16. (10), because the distribution is approximately symmetric and the weights range from 80 to 160 pounds. ¦ x 209 = ≈ 14.9 n 14 12 12 13 14 14 15 15 15 16 16 16 16 17 18 N

17. x =

median =

15 + 15 = 15 2

Mode = 16 (occurs four times) ¦ x 1205 = ≈ 172.1 n 7 169 169 170 172 174 175 176 median = 172 Mode = 169 (occurs two times) The mode does not represent the center of the data because 169 is the smallest number in the data set.

18. x =

¦ x 522 = ≈ 74.6 n 7 52 55 56 67 83 90 119 median = 67 Mode = none The mode cannot be found because no data entry is repeated.

19. x =

¦ x 3638 = = 363.8 n 10 34 84 192 240 320 352 384 592 672 768

20. x =

320 + 352 median = = 336 2

Mode = none The mode cannot be found because no data entry is repeated.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

¦ x 812 = = 58 n 14 54 54 54 56 57 57 58 59 59 59 60 60 61 64 N

21. x =

58 + 59 = 58.5 2 Mode = 54, 59 (both occur three times) median =

¦ x 2004 = = 200.4 n 10 154 171 173 181 184 188 203 235 240 275

22. x =

184 + 188 = 186 2 Mode = none; The mode cannot be found because no data entry is repeated. median =

¦ x 114 = ≈ 7.1 n 16 11 2 2 3 4 5 6N7 8 9 10 10 10 17 19

23. x =

median =

6+7 = 6.5 2

Mode = 10 (occurs three times) ¦ x 100 = = 14.3 n 7 6, 7, 8, 9, 13, 15, 42 The median is the middle value, 9. Mode = none The mode cannot be found because no data entry is repeated. The mean does not represent the center of the data set because it is influenced by the outlier of 42.

24. x =

¦ x 1242 = ≈ 59.1 n 21 12, 18, 19, 26, 28, 31, 33, 40, 44, 45, 49, 61, 63, 75, 80, 80, 89, 96, 103, 125, 125 The median is the middle value, 49. Mode = 80, 125 The modes do not represent the center of the data set because they are large values compared to the rest of the data.

25. x =

¦ x 467 = ≈ 12.0 n 39 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 14, 14, 14, 14, 14, 14, 14, 16, 16, 19 The median is the middle value, 12. Mode = 10 (occurs 13 times)

26. x =

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

27. x is not possible (nominal data) Median = not possible (nominal data) Mode = “Social media site” The mean and median cannot be found because the data are at the nominal level of measurement. 28. x is not possible (nominal data) Median is not possible (nominal data) Mode = “Mental health,” “Education” The mean and median cannot be found because the data are at the nominal level of measurement. 29. x is not possible (nominal data) Median is not possible (nominal data) Mode = “Junior” The mean and median cannot be found because the data are at the nominal level of measurement. 30. x is not possible (nominal data) Median is not possible (nominal data) Mode = “No commute” The mean and median cannot be found because the data are at the nominal level of measurement. ¦ x 817 = ≈ 29.2 n 28 5 8 10 1113 16 21 23 23 23 26 27 27 30 31 32 34 34 34 35 37 38 43 44 45 46 49 52 N

31. x =

median = Mode = 23, 34 (both occur three times each)

30 + 31 = 30.5 2

¦ x 29.9 = ≈ 2.49 n 12 0.8 1.5 1.6 1.8 2.1 2.3 2.4 2.5 3.0 3.9 4.0 4.0

32. x =

median =

2.3 + 2.4 = 2.35 2

Mode = 4.0 (occurs two times) The mode does not represent the center of the data set because it is the largest entry in the data set. ¦ x 292 = ≈ 19.5 n 15 5 8 10 15 15 15 17 20 21 22 22 25 28 32 37 Median = 20 Mode = 15 (occurs three times)

33. x =

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

¦ x 3160 = ≈ 197.5 n 16 100 160 160 160 160 160 180 200 200 200 200 220 240 260 280 280

34. x =

median =

200 + 200 = 200 2

Mode = 160 (occurs five times) 35. Cluster around 475–525, gap between 225 and 275 36. Cluster around 525–625, gap between 775 and 825 37. Mode, because the data are at the nominal level of measurement. 38. Mean, because the distribution is symmetric and there are no outliers. 39. Mean, because the distribution is symmetric and there are no outliers. 40. Median, because there is an outlier. 41.

Source Homework Quizzes Project/Speech Final exam

¦ ( x ⋅ w) ¦w

Score, x 85 80 100 93

43.

¦ ( x ⋅ w) ¦w

Balance, x $523 $2415 $250

¦ ( x ⋅ w) ¦w

Score, x 100 89 100 92

Weight, w 20% 30% 10% 40% w = 100%

93.5 = 93.5 1

Days, w 24 2 4 w = 30

¦ =

x·w 4.25 28 35 23.25 ( x ⋅ w) = 90.5

90.5 = 90.5 1

42. Source Quizzes Midterm exam Student lecture Final exam

Weight, w 0.05 0.35 0.35 0.25 w =1

18,382 ≈ $612.73 30

x·w 12,552 4830 1000 ( x ⋅ w) = 18,382

x·w 20 26.7 10 36.8 ( x ⋅ w) = 93.5

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

44.

Balance, x $115.63 $637.19 $1225.06 $0 $34.88

45.

¦w

¦ =

¦ ( x ⋅ w) ¦w

Grade A B B C D

¦ ( x ⋅ w) ¦w

Score, x 85 81 90

¦ ( x ⋅ w) ¦w

x·w 765 1053 450 ( x ⋅ w) = 2268

Credits, w 4 3 3 3 2 w = 15

x·w 16 9 9 6 2 ( x ⋅ w) = 42

Weight, w 0.05 0.35 0.35 0.25 w =1

x·w 4.25 28 35 21.25 ( x ⋅ w) = 88.5

42 = 2.8 15

47. Source Homework Quizzes Project/Speech Final exam

Weight, w 9 13 5 w = 27

2268 = 84 27

Points, x 4 3 3 2 1

x·w 1387.56 3823.14 8575.42 0 139.52 ( x ⋅ w ) = 13,925.64

13,925.64 ≈ $449.21 31

Source Engineering Business Math

46.

¦ ( x ⋅ w)

Days, w 12 6 7 2 4 w = 31

Score, x 85 80 100 85

88.5 = 88.5 1

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

48. Grade A A B C D

¦ ( x ⋅ w) ¦w

Points, x 4 4 3 2 1

49. Class 0–29 30–59 60–89 90–119

Credits, w 4 3 3 3 2 w = 15

x·w 16 12 9 6 2 ( x ⋅ w) = 45

45 =3 15

Midpoint, x 14.5 44.5 74.5 104.5

Frequency, f 9 23 12 6

n = 50 x=

¦(x ⋅ f ) n

50. Class 0–19 20–39 40–59 60–79 80–99

2675 = 53.5 minutes 50

Midpoint, x 9.5 29.5 49.5 69.5 89.5

Frequency, f 8 8 15 10 7

n = 48 x=

¦(x ⋅ f ) n

51. Class 0–9 10–19 20–29 30–39 40–49 50–59 60–69 70–79 80–89 90–99

Midpoint, x 4.5 14.5 24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5

¦(x ⋅ f ) n

x·f 76 236 742.5 695 626.5 ( x ⋅ f ) = 2376

2376 = 49.5 minutes 48

Frequency, f 78 97 54 63 69 86 73 53 43 15

n = 631 x=

x·f 130.5 1023.5 894 627 ( x ⋅ f ) = 2675

26,719.5 ≈ 42.3 years old 631

x·f 351 1406.5 1323 2173.5 3070.5 4687 4708.5 3948.5 3633.5 1417.5 ( x ⋅ f ) = 26,719.5

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

52. Class 0–19 20–39 40–59 60–79 80–99 100–119 120–139 140–159

Midpoint, x 9.5 29.5 49.5 69.5 89.5 109.5 129.5 149.5

Frequency, f 46 3 1 1 1 3 0 1

n = 56 x=

¦(x ⋅ f ) n

53. Class width =

x·f 437 88.5 49.5 69.5 89.5 328.5 0 149.5 ( x ⋅ f ) = 1212

1212 ≈ 21.6 thousand 56

Range 297 − 127 = = 34 35 Number of classes 5

Class

Frequency, f

Midpoint

127௘–௘161 162௘–௘196 197௘–௘231 232௘–௘266 267௘–௘301

7 6 3 3 1

144 179 214 249 284

Positively skewed 54. Class width =

Range 382 − 104 = ≈ 46.3 47 Number of classes 6

Class

Frequency, f

Midpoint

104௘–௘150 151௘–௘197 198௘–௘244 245௘–௘291 292௘–௘338 339௘–௘385

3 4 3 2 1 1

127 174 221 268 315 362

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

Positively skewed

55. Class width =

Range 76 − 62 = = 2.8 3 Number of classes 5

Class 62–64 65–67 68–70 71–73 74–76

Midpoint 63 66 69 72 75

Frequency, f 3 7 9 8 3 ¦ f = 30

Shape: Symmetric 56. Class width = Class 1 2 3 4 5 6

Range 6 −1 = = 0.8333 1 Number of classes 6

Frequency, f 6 5 4 6 4 5 ¦ f = 30

Shape: Uniform

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

¦ x 9109 = ≈ 1518.2 n 6 1502 1511 1516 1525 1526 1529

57. (a) x =

median =

1516 + 1525 = 1520.5 2

¦ x 9127 = = 1521.2 n 6 1511 1516 1520 1525 1526 1529

(b) x =

median =

1520 + 1525 = 1522.5 2

(c) The mean was affected more. ¦ x 948.8 = ≈ 52.7 n 18 8.6 12.0 12.8 15.2 15.6 23.8 24.8 26.4 29.5 29.9 31.7 55.4 55.9 56.7 57.3 69.7 112.7 310.8

58. (a) x =

median =

29.5 + 29.9 = 29.7 2

¦ x 638 = ≈ 37.5 n 17 8.6, 12.0, 12.8, 15.2, 15.6, 23.8, 24.8, 26.4, 29.5, 29.9, 31.7, 55.4, 55.9, 56.7, 57.3, 69.7, 112.7 The median is the middle value, 29.5. The mean was affected more.

(b) x =

¦ x 957 = ≈ 50.4 n 19 8.2, 8.6, 12.0, 12.8, 15.2, 15.6, 23.8, 24.8, 26.4, 29.5, 29.9, 31.7, 55.4, 55.9, 56.7, 57.3, 69.7, 112.7, 310.8 The median is the middle value, 29.5. The mean was affected more.

59. The data are skewed right. A = mode, because it is the data entry that occurred most often. B = median, because the median is to the left of the mean in a skewed right distribution. C = mean, because the mean is to the right of the median in a skewed right distribution. 60. The data are skewed left. A = mean, because the mean is to the left of the median in a skewed left distribution. B = median, because the median is to the right of the mean in a skewed left distribution. C = mode, because it is the data entry that occurred most often.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

61. Increase one of the three-credit B classes to an A. The three-credit class is weighted more than the two-credit classes, so it will have a greater effect on the grade point average. ¦ x 3222 = = 358 n 9 147 177 336 360 375 393 408 504 522 Median = 375

62. (a) x =

¦ x 9666 = = 1074 n 9 441 531 1008 1080 1125 1179 1224 1512 1566 Median = 1125

(b) x =

(c) The mean and median in part (b) are three times the mean and median in part (a). (d) If you multiply the mean and median of the original data set by 36, you will get the mean and median of the data set in inches. 63. Car A ¦ x 152 x= = = 30.4 n 5 28 28 30 32 34 Median = 30 Mode = 28 (occurs two times) Car B ¦ x 151 x= = = 30.2 n 5 29 29 31 31 31 Median = 31 Mode = 31 (occurs three times) Car C ¦ x 151 x= = = 30.2 n 5 28 29 30 32 32 Median = 30 Mode = 32 (occurs two times)

(a) Mean should be used because Car A has the highest mean of the three. (b) Median should be used because Car B has the highest median of the three. (c) Mode should be used because Car C has the highest mode of the three. 34 + 28 = 31 2 31 + 29 Car B: Midrange = = 30 2 32 + 28 Car C: Midrange = = 30 2 Car A because the midrange is the largest.

64. Car A: Midrange =

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

¦ x 1477 = ≈ 49.2 n 30 11 13 22 28 36 36 36 37 37 37 38 41 43 44 46 47 51 51 51 53 61 62 63 64

65. (a) x =

72 72 74 76 85 90

median =

46 + 47 = 46.5 2

(b) Key: 3 6 = 36 11 3 22 8 36 6 41 3 51 1 61 2 72 2 85 90

median 6 4 1 3 4

7 7 7 8 6 7 mean 3 4 6

(c) The distribution is positively skewed. 66. (a) Order the data values. 11 13 22 28 36 36 36 37 37 37 38 41 43 44 46 47 51 51 51 53 61 62 63 64 72 72 74 76 85 90 Delete the lowest 10%, smallest three observations (11, 13, 22). Delete the highest 10%, largest three observations (76, 85, 90). Find the 10% trimmed mean using the remaining 24 observations. ¦ x 1180 x= = ≈ 49.2 n 24 10% trimmed mean ≈ 49.2

(b) x ≈ 49.2 Median = 46.5 Mode = 36, 37, 51 90 + 11 Midrange = = 50.5 2 (c) Using a trimmed mean eliminates potential outliers that may affect the mean of all the observations.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

2.4 MEASURES OF VARIATION 2.4 TRY IT YOURSELF SOLUTIONS 1. Min = 33, or $33,000 and Max = 68, or $68,000 Range = Max – Min = 68 – 33 = 35, or $35,000 The range of the starting salaries for Corporation B is 35, or $35,000. This is much larger than the range for Corporation A. 2. μ = 51.5 , or $51,500 Salary, x

x–ȝ

(x – ȝ)2

–1.5

2.25

–18.5

342.25

–0.5

0.25

8.5

72.25

7.5

56.25

–9.5

90.25

–0.5

0.25

–12.5

156.25

10.5

110.25

16.5

272.25

¦ x = 515

¦( x − μ ) = 0

¦ ( x − μ ) = 1102.5

¦( x − μ )

1102.5 ≈ 110.3 N 10 1102.5 σ = σ2 = = 10.5, or $10,500 10 The population variance is about 110.3 and the population standard deviation is 10.5, or $10,500.

σ2 =

¦ x 272 = = 34 n 8

Time, x

x−x

( x − x)

53 25 41 36 19 22 42 34

19 í9 7 2 í15 í12 8 0

361 81 49 4 225 144 64 0

¦ x = 272

¦( x − x) = 0

¦ ( x − x ) = 928 2

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

(

SS x = ¦ x − x

( s =

¦ x−x

) = 928 2

) = 928 ≈ 132.6 2

n −1 7 928 s = s2 = ≈ 11.5 7

4. Enter the data in a computer or a calculator. x ≈ 20.8, s ≈ 6.7 5. Sample answer: 7, 7, 7, 7, 7, 13, 13, 13, 13, 13 Salary, x 7 7 7 7 7 13 13 13 13 13 x = 100

¦ μ=

σ=

x–ȝ –3 –3 –3 –3 –3 3 3 3 3 3 (x − μ) = 0

(x – ȝ)2 9 9 9 9 9 9 9 9 9 9

¦ ( x − μ ) = 90 2

¦ x 100 = = 10 10 N ¦(x − μ ) N

90 = 9 =3 10

6. 66.7 – 64.1 = 2.6 = 1 standard deviation Because 66.7 is one standard deviation above the mean height, the percent of heights between 64.1 inches and 66.7 inches is 34.13%. Approximately 34.13% of women ages 20-29 are between 64.1 and 66.7 inches tall. 7. 39.8 – 2(23.7) = −7.6 Because –7.6 does not make sense for an age, use 0. 39.8 + 2(23.7) = 87.2 1 1 1 1− 2 =1− = 1 − = 0.75 2 4 k ( 2)

At least 75% of the data lie within 2 standard deviations of the mean. At least 75% of the population of Iowa is between 0 and 87.2 years old. Because 80 < 87.2, and age of 80 lies within two standard deviations of the mean. So, the age is not unusual.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

x 0 1 2 3 4 5 6

f 10 19 7 7 5 1 1

n = 50 x=

xf 0 19 14 21 20 5 6 xf = 85

¦ xf 85 = = 1.7 n 50

x−x

( x − x)

–1.7 –0.7 0.3 1.3 2.3 3.3 4.3

2.89 0.49 0.09 1.69 5.29 10.89 18.49

( x − x) f

28.90 9.31 0.63 11.83 26.45 10.89 18.49

¦ ( x − x ) f = 106.5 2

s= 9.

(

¦ x−x

) f = 106.5 ≈ 1.5 2

n −1

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

x 49.5 149.5 249.5 349.5 449.5 650

f 173 129 39 22 8 19

n = 390 x=

xf 8563.5 19,285.5 9730.5 7689 3596 12,350 xf = 61, 214.5

¦ xf 61, 214.5 = ≈ 157.0 n 390

x−x

( x − x)

–107.5 –7.5 92.5 192.5 292.5 493

11,556.25 56.25 8556.25 37,056.25 85,556.25 243,049

( x − x) f

1,999,231.25 7256.25 333,693.75 815,237.5 684,450 4,617,931

¦ ( x − x ) f = 8, 457,799.75 2

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

(

¦ x−x

) f = 8, 457,799.75 ≈ 147.5 2

n −1 389 Both the mean and sample standard deviation increased. 10. Los Angeles: x ≈ 37.1 , s ≈ 15.5 Dallas: x ≈ 20.8 , s ≈ 6.7 s 15.5 Los Angeles: CV = ⋅ 100% = ⋅ 100% ≈ 41.8% 37.1 x s 6.7 Dallas: CV = ⋅ 100% = ⋅ 100% ≈ 32.2% 20.8 x The office rental rates are more variable in Los Angeles than in Dallas.

2.4 EXERCISE SOLUTIONS 1. The range is the difference between the maximum and minimum values of a data set. The advantage of the range is that it is easy to calculate. The disadvantage is that it uses only two entries from the data set. 2. A deviation ( x − μ ) is the difference between an entry x and the mean of the data ȝ. The sum of the deviations is always zero. 3. The units of variance are squared. Its units are meaningless (example: dollars2). The units of standard deviation are the same as the data. 4. The standard deviation is the positive square root of the variance. The standard deviation and variance can never be negative because squared deviations can never be negative. 5. When calculating the population standard deviation, you divide the sum of the squared deviations by N, then take the square root of that value. When calculating the sample standard deviation, you divide the sum of the squared deviations by n − 1 , then take the square root of that value. 6. When given a data set, you would have to determine if it represented the population or if it was a sample taken from the population. If the data are a population, then σ is calculated. If the data are a sample, then s is calculated. 7. Similarity: Both estimate proportions of the data contained within k standard deviations of the mean. Difference: The Empirical Rule assumes the distribution is approximately symmetric and bell-shaped. Chebychev’s Theorem makes no such assumption. 8. You must know that the distribution is approximately symmetric and bell-shaped. 9. Range = Max – Min = 95 – 45 = 50; Approximately 50, or $50,000 10. Range = Max – Min = 98 – 74 = 24

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

11. (a) Range = Max – Min = 38.5 – 20.7 = 17.8

(b) Range = Max – Min = 60.5 – 20.7 = 39.8 12. Changing the maximum value of the data set greatly affects the range. 13. Range = Max – Min = 11.0 – 9.9 = 1.1 ¦ x 103.3 = = 10.33 μ= 10 N

x 10.1 9.9 10.3 10.1 9.9 10.3 11.0 10.9 10.7 10.1 x = 103.3

¦( x − μ )

σ = σ=

N ¦(x − μ ) N

x−μ

( x − μ )2

í0.23 í0.43 í0.03 í0.23 í0.43 í0.03 0.67 0.57 0.37 í0.23 (x − μ) ≈ 0

0.05 0.18 0.00 0.05 0.18 0.00 0.45 0.32 0.14 0.05

¦ ( x − μ ) = 1.42 2

1.42 ≈ 0.14 10

1.42 ≈ 0.4 10

14. Range = Max – Min = 7870 – 0.09 = 7869.91 ¦ x 22,511.5 = = 2251.15 μ= 10 N

x 1.4 2330 2700 7870 1500 970 900 1740 4500 0.09

x−μ

( x − μ )2

−2249.75 78.85 448.85 5618.85 −751.15 −1281.15 −1351.15 −511.15 2248.85 −2251.06 (x − μ) ≈ 0

5,061,375 6217 201,466 31,571,475 564,226 1,641,345 1,825,606 261,274 5,057,326 5,067,271

¦ x = 22,511.5 ¦

¦ ( x − μ ) = 51, 257,584 2

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

σ = σ=

¦( x − μ )

N ¦(x − μ ) N

51, 257,584 ≈ 5,125,758.4 10

5,125,758.4 ≈ 2264.0 10

15. Range = Max – Min = 23 – 17 = 6 ¦ x 380 x= = = 19 n 20

x−x

( x − x)

19 20 17 19 17 21 23 21 17 17 19 19 17 20 23 18 18 18 18 19

0 1 −2 0 −2 2 4 2 −2 −2 0 0 −2 1 4 −1 −1 −1 −1 0

0 1 4 0 4 4 16 4 4 4 0 0 4 1 16 1 1 1 1 0

¦ x = 380 ¦ ( x − x ) = 0

( s =

) = 66 ≈ 3.5

(

) = 66 ≈ 1.9

¦ x−x n −1

20 − 1

¦ ( x − x ) = 66 2

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

16. Range = Max – Min = 299 – 264 = 35 ¦ x 5902 x= = ≈ 281.0 n 21

x−x

( x − x)

277 277 267 291 282 281 295 279 286 280 296 269 268 285 264 278 269 299 291 293 275

−4 −4 −14 10 1 0 14 −2 5 −1 15 −12 −13 4 −17 −3 −12 18 10 12 −6

16 16 196 100 1 0 196 4 25 1 225 144 169 16 289 9 144 324 100 144 36

¦ x = 5902 ¦ ( x − x ) ≈ 0 ¦ ( x − x ) = 2155 2

( s =

) = 2155 ≈ 107.7

(

) = 2155 ≈ 10.4

¦ x−x n −1

21 − 1

17. The data set in (a) has a standard deviation of 2.4 and the data set in (b) has a standard deviation of 5 because the data in (b) have more variability. 18. The data set in (a) has a standard deviation of 24 and the data set in (b) has a standard deviation of 16 because the data in (a) have more variability. 19. Company B; An offer of $43,000 is two standard deviations from the mean of Company A’s starting salaries, which makes it unlikely. The same offer is within one standard deviation of the mean of Company B’s starting salaries, which makes the offer likely. 20. Company C; An offer of $62,000 is three standard deviations from the mean of Company D’s starting salaries, which makes it unlikely. The same offer is within two standard deviations of the mean of Company C’s starting salaries, which makes the offer somewhat likely. Copyright © 2023 Pearson Education, Inc.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

21. (a) Greatest sample standard deviation: (ii) Data set (ii) has more entries that are farther away from the mean. Least sample standard deviation: (iii) Data set (iii) has more entries that are close to the mean.

(b) The three data sets have the same mean but have different standard deviations. (c) Estimates will vary; (i) s § 1.1; (ii) s § 1.3; (iii) s § 0.8 22. (a) Greatest sample standard deviation: (i) Data set (i) has more entries that are farther away from the mean. Least sample standard deviation: (iii) Data set (iii) has more entries that are close to the mean.

(b) The three data sets have the same mean, median, and mode, but have different standard deviations. (c) Estimates will vary; (i) s § 1.6; (ii) s § 2.9; (iii) s § 0.8 23. (a) Greatest sample standard deviation: (i) Data set (i) has more entries that are farther away from the mean. Least sample standard deviation: (iii) Data set (iii) has more entries that are close to the mean.

(b) The three data sets have the same mean, median, and mode, but have different standard deviations. (c) Estimates will vary; (i) s § 9.6; (ii) s § 9.0; (iii) s § 5.1 24. (a) Greatest sample standard deviation: (iii) Data set (iii) has more entries that are farther away from the mean. Least sample standard deviation: (i) Data set (i) has more entries that are close to the mean.

(b) The three data sets have the same mean and median but have different modes and standard deviations. (c) Estimates will vary; (i) s § 1.5; (ii) s § 1.8; (iii) s § 2.5 25. Sample answer: 3, 3, 3, 7, 7, 7 26. Sample answer: 3, 3, 3, 3, 9, 9, 9, 9 27. Sample answer: 9, 9, 9, 9, 9, 9, 9 28. Sample answer: 5, 5, 5, 9, 9, 9 29.

( 63, 71) → ( 67 − 1( 4 ) , 67 + 1( 4 ) ) → ( x − s, x + s ) 68% of the vehicles have speeds between 63 and 71 mph.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

30. 95% of the data falls between x − 2s and x + 2s . x − 2s = 70 − 2 ( 8) = 54

x + 2s = 70 + 2 ( 8) = 86 95% of the households have monthly utility bills between $54 and $86.

31. (a) n = 75 ; 68%(75) = (0.68)(75) § 51 vehicles have speeds between 63 and 71 mph.

(b) n = 25 ; 68%(25) = (0.68)(25) § 17 vehicles have speeds between 63 and 71 mph. 32. (a) n = 40 ; 95%(40) = (0.95)(40) § 38 households have monthly utility bills between $54 and $86.

(b) n = 20 ; 95%(20) = (0.95)(20) § 19 households have monthly utility bills between $54 and $86. 33. 78, 76, and 82 are unusual; 82 is very unusual because it is more than three standard deviations from the mean. 34. $52 and $98 are unusual; $98 is very unusual because it is more than three standard deviations from the mean. 35.

( x − 2s, x + 2s ) → ( 0, 4 ) are two standard deviations from the mean. 1−

1 1 1 =1− = 1 − = 0.75 At least 75% of the eruption times lie between 0 and 4. 2 2 4 k ( 2)

If n = 40, at least (0.75)(40) = 30 households have between 0 and 4 pets. 36.

( x − 2s, x + 2s ) → (16.18, 186.94 ) are two standard deviations from the mean. 1−

1 1 1 =1− = 1 − = 0.75 At least 75% of the eruption times lie between 16.18 and 186.94 2 2 4 k ( 2)

minutes. If n = 100, at least (0.75)(100) = 75 eruptions will lie between 16.18 and 186.94 minutes. 37.

( x − 4s, x + 4s ) → ( 70, 94 ) are four standard deviations from the mean. 1 1 1 = 1 − 2 = 1 − = 0.9375 2 16 k 4 At least 93.75% of the exam scores are from 70 to 94. 1−

38. 1 −

1 1 1 =1− = 1 − = 0.75 2 2 4 k ( 2)

At least 75% of the runs lie within two standard deviations of the mean. x − 2 s , x + 2 s → (1.7, 8.9 ) → ( 2, 8 )

(

)

At least 75% of the runs per game scored by the Los Angeles Dodgers during the 2020 World Series are from 2 to 8 (note that 1.7 and 8.9 do not make sense in the context of the data).

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

39.

x−x

( x − x )2

( x − x )2 f

0 1 2 3 4 5 6 7 8 9

3 4 3 9 3 3 8 5 6 6 Σ = 50

0 4 6 27 12 15 48 35 48 54 Σ = 249

−5 −4 −3 −2 −1 0 1 2 3 4

25 16 9 4 1 0 1 4 9 16

75 64 27 36 3 0 8 20 54 96 Σ = 383

s= 40.

x 0 1

s= 41.

¦ xf 249 = ≈ 5.0 n 50

(

¦ x−x

) f = 383 ≈ 2.8 2

n −1

30 20 Σ = 50

0 20 Σ = 20

2 2 x − x (x − x ) (x − x ) f −0.4 0.16 4.8 0.6 0.36 7.2 Σ = 12

¦ xf 20 = ≈ 0.4 n 50

(

¦ x−x

) f = 12 ≈ 0.5 2

n −1

Class

x−x

17,000௘–௘19,499 19,500௘–௘21,999 22,000௘–௘24,499 24,500௘–௘26,999 27,000 or more

18,249.5 20,749.5 23,249.5 25,749.5 28,249.5

9 10 16 11 6 n = 52

164,245.5 207,495 371,992 283,244.5 169,497 Σxf = 1,196, 474

−4759.62 −2259.62 240.38 2740.38 5240.38

¦ xf 1,196, 474 = ≈ $23,009.12 n 52

(

¦ x−x

) f = 503, 245,192.31 ≈ $3141.27 2

n −1

( x − x )2

( x − x )2 f

22,653,982.54 203,885,842.9 5,105,882.544 51,058,825.44 57,782.54 924,520.71 7,509,682.54 82,606,507.99 27,461,582.54 164,769,495.27 Σ ( x − x ) 2 f = 503, 245,192.31

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

42.

Class 0–4 5–9 10–14 15–19 20–24 25–29 30+

Midpoint, x 2 7 12 17 22 27 32

f 5 12 24 17 16 11 5

n = 90 x=

xf 10 84 288 289 352 297 160 xf = 1480

¦ xf 1480 = ≈ 16.4 n 90

x−x

( x − x)

( x − x) f

−14.44 −9.44 −4.44 0.56 5.56 10.56 15.56

208.5136 89.1136 19.7136 0.3136 30.9136 111.5136 242.1136

1042.5680 1069.3632 473.1264 5.3312 494.6176 1226.6496 1210.5680

¦ ( x − x ) f = 5522.2240 2

s= 43.

(

¦ x−x

) f = 5522.2240 ≈ 7.9 2

n −1

x−x

1 2 3 4

2 18 24 16 n = 60

2 36 72 64 Σxf = 174

−1.9 −0.9 0.1 1.1

¦ xf 174 = ≈ 2.9 n 60

(

¦ x−x

) f = 41.4 ≈ 0.8

n −1

( x − x )2

( x − x )2 f

3.61 7.22 0.81 14.58 0.01 0.24 1.21 19.36 2 Σ ( x − x ) f = 41.4

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

44.

Midpoint, x 70.5 92.5 114.5 136.5 158.5

f 1 12 25 10 2

n = 50 x=

xf 70.5 1110.0 2862.5 1365.0 317.0 xf = 5725

¦ xf 5725 = = 114.5 n 50

x−x

( x − x)

–44 –22 0 22 44

1936 484 0 484 1936

( x − x) f

1936 5808 0 4840 3872

¦ ( x − x ) f = 16, 456 2

(

¦ x−x

) f = 16, 456 ≈ 18.33 2

n −1

45. Denver: x =

¦ x 665.1 = ≈ 55.4 n 12

( s =

) = 220.87 ≈ 20.08

(

) = 220.87 ≈ 4.48

( s =

) = 239.97 ≈ 21.82

(

) = 239.97 ≈ 4.67

¦ x−x n −1

¦ x−x

n −1 11 s 4.48 CV = ⋅ 100% = ⋅ 100% ≈ 8.1% 55.4 x ¦ x 774.9 Los Angeles: x = = ≈ 64.6 n 12 2

¦ x−x n −1

¦ x−x

n −1 11 s 4.67 CV = ⋅ 100% = ⋅ 100% ≈ 7.2% 64.6 x Salaries for entry-level architects are more variable in Denver than in Los Angeles.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

¦ x 603.2 = ≈ 67.0 n 9

46. Raleigh:

( s =

) = 459.02 ≈ 57.38

(

) = 459.02 ≈ 7.57

( s =

) = 825 ≈ 103.13

(

) = 825 ≈ 10.16

¦ x−x n −1

9 −1

¦ x−x

n −1 8 s 7.57 CV = = ⋅ 100% ≈ 11.3% 67 x ¦ x 621.4 Wichita: x = = ≈ 69.0 n 9 2

¦ x−x n −1

9 −1

8 s 10.16 CV = ⋅ 100% = ⋅ 100% ≈ 14.7% x 69 Salaries for entry-level software engineers are more variable in Wichita than in Raleigh. ¦ x 642 = ≈ 27.91 N 23 2 ¦( x − μ ) 477.83 σ2 = = ≈ 20.78 N 23

47. Ages: μ =

¦(x − μ )

¦( x − μ )

477.83 ≈ 4.56 N 23 σ 4.56 CV = ⋅ 100% = ⋅ 100% ≈ 16.3% μ 27.91 ¦ x 1816 Caps: μ = = ≈ 78.96 N 23 2 ¦( x − μ ) 130,157 2 σ = = = 5659 N 23

σ=

130,157 ≈ 75.23 N 23 σ 75.23 CV = ⋅ 100% = ⋅ 100% ≈ 95.3% μ 78.96 Caps are more variable than ages for all members of the 2021 Women’s U.S. soccer team.

σ=

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

¦ x 792 = = 72 N 11

48. Heights: μ =

¦( x − μ )

σ2 =

N ¦(x − μ )

104 ≈ 9.45 11

104 ≈ 3.07 11 σ 3.07 CV = ⋅ 100% = ⋅ 100% ≈ 4.3% μ 72 ¦ x 1869 Weights: μ = = ≈ 169.91 N 11 2 ¦( x − μ ) 2402.91 2 σ = = ≈ 218.45 N 11

σ=

¦( x − μ )

2402.91 ≈ 14.78 N 11 σ 14.78 CV = ⋅ 100% = ⋅ 100% ≈ 8.7% μ 169.91 Weight is more variable than height for all France national soccer team players.

σ=

49. Male: x =

( s = 2

¦ x 8760 = = 1095 n 8

¦ x−x n −1

) = 359, 400 ≈ 51,342.86 2

8 −1

(

¦ x−x

n −1

) = 359, 400 ≈ 226.6 2

s 226.6 ⋅ 100% = ⋅ 100% ≈ 20.7% x 1095 ¦ x 9120 Female: x = = = 1140 n 8 CV =

( s =

) = 282,000 ≈ 40, 285.71

(

) = 282,000 = 200.7

¦ x−x n −1

8 −1

s 200.7 ⋅100% = ⋅100% ≈ 17.6% x 1140 SAT scores are more variable for males than for females. CV =

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

50. Male: x =

( s = 2

¦ x 32 = = 3.2 n 10

¦ x−x n −1

) = 4.76 ≈ 0.529 2

10 − 1

(

) = 4.76 ≈ 0.727

( s =

) = 5.9 ≈ 0.656

(

) = 5.9 = 0.810

¦ x−x

n −1 9 s 0.727 CV = ⋅ 100% = ⋅ 100% ≈ 22.7% x 3.2 ¦ x 32 Female: x = = = 3.2 n 10 2

¦ x−x n −1

10 − 1

¦ x−x n −1

9 s 0.810 CV = ⋅ 100% = ⋅ 100% ≈ 25.3% x 3.2 Grade point averages are more variable for females than for males. 51. (a) Answers will vary.

(b) Ages of students x = 380 ; x 2 = 7286

¦ x2 −

( ¦ x )2 n

n −1

7286 −

( 380 )2

20 20 − 1

66 ≈ 1.9 19

¦ x − x = 28

¦ x−x

28 = 1.4 n 20 From Exercise 15, the sample standard deviation is 1.9. The mean absolute deviation is less than the sample standard deviation. MAD =

(b) Days: x ≈ 281.0

¦ x − x = 177

MAD =

¦ x−x n

177 ≈ 8.4 21

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

From Exercise 16, the sample standard deviation is 10.4. The mean absolute deviation is less than the sample standard deviation. 53. (a) x ≈ 42.1; s ≈ 5.6

(b) x ≈ 44.3; s ≈ 5.9 (c) 3.5, 3, 3, 4, 4, 2.75, 4.25, 3.25, 3.25, 3.5, 3.25, 3.75, 3.5, 4.17 x ≈ 3.5; s ≈ 0.47

(d) When each entry is multiplied by a constant k, the new sample mean is k · x, and the new sample standard deviation is k · s. 54. (a) x ≈ 41.2, s ≈ 6.0

(b) x ≈ 42.2, s ≈ 6.0 (c) x ≈ 39.2, s ≈ 6.0 (d) Adding a constant k to, or subtracting it from, each entry makes the new sample mean ‫ݔ‬ҧ + k, or ‫ݔ‬ҧ − k, with the sample standard deviation being unaffected. 55. (a) P =

(b) P =

(d) P =

(e) P =

56. 1 −

(

3 x − median s

(

3 x − median s

(

3 x − median s

(

3 x − median s

(

3 x − median s

) = 3(17 − 19) ≈ −2.61; The data are skewed left. 2.3

) = 3(32 − 25) ≈ 4.12; The data are skewed right. 5.1

) = 3( 9.2 − 9.2) = 0; The data are symmetric. 1.8

) = 3( 42 − 40) = 1; The data are skewed right. 6.0

) = 3(155 − 175) = −3; The data are skewed left. 20.0

1 1 1 = 0.99 1 − 0.99 = 2 k 2 = k= 2 0.01 k k

1 = 10 0.01

At least 99% of the data in any data set lie within 10 standard deviations of the mean.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

2.5 MEASURES OF POSITION 2.5 TRY IT YOURSELF SOLUTIONS 1. Order the data from least to greatest. The median (or Q2 ) is 31. This was also found in Section 2.3, Try It Yourself 2. The first quartile is the median of the data entries to the left of Q2 and the third quartile is the median of the data entries to the right of Q2 . Q1 = 23, Q2 = 31, Q3 = 35 About one-quarter of the winning scores were 23 points or less, about one-half were 31 points or less, and about three-quarters were 35 points or less. 2. Enter data Q1 = 27.5, Q2 = 35, Q3 = 51.5 About one-quarter of these universities charge tuition of $27,500 or less, about one-half charge $35,000 or less, and about three-quarters charge $51,500 or less. 3.

Q1 = 23, Q3 = 35 IQR = Q3 − Q1 = 35 − 23 = 12 Q1 − 1.5(IQR) = 23 − 1.5(12) = 5 ; Q3 + 1.5(IQR) = 35 + 1.5(12) = 53 The score 55 is greater than Q3 + 1.5(IQR) . So, 55 is an outlier.

4. Min = 13, Q1 = 23, Q2 = 31, Q3 = 35, Max = 55

About 50% of the winning scores were between 23 and 35 points. About 25% of the winning scores were less than 23 points. About 25% of the winning scores were greater than 35 points. 5. The 65th percentile is 33.5. About 65% of the winning scores were 33.5 or less. 6. 10, 11, 18, 19, 25, 27, 28, 31, 31, 32, 33, 33, 35, 35, 36, 39, 46, 49, 49, 54, 55, 55, 56, 57, 58 Seven data entries are less than 31 number of data entries less than 31 7 Percentile of 31 = ⋅ 100 = ⋅ 100 = 28th percentile total number of data entries 25 The tuition cost of $31,000 is greater than 28% of the other tuition costs. 7.

μ = 70, σ = 8 x − μ 60 − 70 x = 60 : z = = = −1.25 σ 8

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

x−μ

71 − 70 = 0.125 σ 8 x − μ 92 − 70 x = 92 : z = = = 2.75 σ 8 From the z-scores, the utility bill of $60 is 1.25 standard deviations below the mean, the bill of $71 is 0.125 standard deviation above the mean, and the bill of $92 is 2.75 standard deviations above the mean. x = 71: z =

8. 5 feet = 5(12) = 60 inches x − μ 60 − 69.9 x − μ 60 − 64.3 Man: z = = = −3.3 ; Woman: z = = ≈ −1.7 σ σ 3 2.6 The z-score for the 5-foot-tall man is 3.3 standard deviations below the mean. This is an unusual height for a man. The z-score for the 5-foot-tall woman is 1.7 standard deviations below the mean. This is among the typical heights for a woman.

2.5 EXERCISE SOLUTIONS 1. The talk is longer in length than 75% of the lectures in the series. 2. The motorcycle’s fuel efficiency is higher than 90% of the other vehicles in its class. 3. The student’s grade on the Fundamentals of Engineering exam was below the average. 4. The student has a higher IQ score than 91% of the students in the same age group. 5. The interquartile range of a data set can be used to identify outliers because data values that are greater than Q3 + 1.5 ( IQR ) or less than Q1 − 1.5 ( IQR ) are considered outliers. 6. Quartiles are special cases of percentiles. Q1 is the 25th percentile, Q2 is the 50th percentile, and Q3 is the 75th percentile. 7. True 8. False. The second quartile is the median of an ordered data set. 9. False. An outlier is any number above Q3 + 1.5 ( IQR ) or below Q1 − 1.5 ( IQR ) . 10. False. It is possible to have an z-score of zero when the x-value equals the mean. 11. (a) 51 54 56 57 59 60 60 60 60 62 63 63 63 65 80

(b) IQR = Q3 − Q1 = 63 − 57 = 6

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

(c) Q1 − 1.5( IQR ) = 57 − 1.5(6) = 48 ; Q3 + 1.5( IQR) = 63 + 1.5(6) = 72 . The date entry 80 is an outlier. 12. (a) 19 20 20 21 21 21 22 22 22 22 23 23 23 23 24 24 25 25 25 29

Q1 = 21

Q2 = 22.5

Q3 = 24

(b) IQR = Q3 − Q1 = 24 − 21 = 3

(c) Q1 − 1.5( IQR) = 21 − 1.5(3) = 16.5 ; Q3 + 1.5( IQR) = 24 + 1.5(3) = 28.5 . The data entry 29 is an outlier. 13. Min = 0, Q1 = 2, Q2 = 5, Q3 = 8, Max = 10 14. Min = 500, Q1 = 580, Q2 = 605, Q3 = 630, Max = 720 15. (a) 24 26 27 28 30 32 35 35 36 39 39 41 50 51 60

Q1 Q2 Q3 Min = 24, Q1 = 28 , Q2 = 35 , Q3 = 41 , Max = 60 (b)

16. (a)

Min = 150, Q1 = 172 , Q2 = 177 , Q3 = 180 , Max = 182 (b)

17. (a)

Min = 1, Q1 = 4.5 , Q2 = 6 , Q3 = 7.5 , Max = 9 (b)

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

18. (a) 1 1 2 2 2 2 2 3 3 3 3 3 3 4 4 4 5 5 5 5 5 6 7 7 7 8 8 8 9 9 9 9 9 9

Q1 = 3 Q2 = 5 Min = 1, Q1 = 3 , Q2 = 5 , Q3 = 8 , Max = 9

Q3 = 8

(b)

19. None. The data are not skewed or symmetric. 20. Skewed right. Most of the data lie to the left in the box-and-whisker plot. 21. Skewed left. Most of the data lie to the right in the box-and-whisker plot. 22. Symmetric. The data are evenly spaced to the left and to the right of the median. 23. Min = 1, Q1 = 2, Q2 = 3, Q3 = 6.5, Max = 8

24. Min = 0, Q1 = 2, Q2 = 4.5, Q3 = 6.5, Max = 10

25. Min = 1, Q1 = 3, Q2 = 7, Q3 = 12, Max = 45

26. Min = 15.06, Q1 = 23.39, Q2 = 26.19, Q3 = 30.87, Max = 38.24

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

27. (a) 6.5 hours (b) About 50% (c) About 25% 28. (a) $26.19 per hour (b) About 75% (c) About 75% (d) About 25% 29. About 158; About 65% of quantitative reasoning scores on the Graduate Record Examination are less than 158. 30. About 154; About 50% of quantitative reasoning scores on the Graduate Record Examination are less than 154. 31. About 7th percentile; About 7% of quantitative reasoning scores on the Graduate Record Examination are less than 140. 32. About 96th percentile; About 96% of quantitative reasoning scores on the Graduate Record Examination are less than 170. 33. Percentile of 40 =

number of data entries less than 40 3 ⋅ 100 = ⋅ 100 = 10th percentile total number of data entries 30

34. Percentile of 56 =

number of data entries less than 56 21 ⋅ 100 = ⋅ 100 = 70th percentile total number of data entries 30

35. 75th percentile = Q3 = 56 ; Ages over 56 are 57, 57, 61, 61, 65, 66 36. 25th percentile = Q1 = 43 ; Ages below 43 are 28, 35, 38, 40, 41, 41, 42 37.

38. The 50th percentile is about 8 minutes. About 50% of wait times are less than 8 minutes. 39. A wait time of 20 minutes corresponds to about the 85th percentile. 40. The wait times between the 25th and 75th percentiles are about 4.5 minutes to 16 minutes. 41. A z = −1.43 B z = 0 C z = 2.14 The z-score 2.14 is unusual because it is so large.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

42. A z = −1.54 B z = 0.77 C z = 1.54 None of the z-scores are unusual. x−μ

43. Christopher Froome: x = 31 z =

31 − 27.9

≈ 0.91

σ 3.4 Not unusual; The z-score is 0.91, so the age of 31 is about 0.91 standard deviation above the mean. x−μ

24 − 27.9 ≈ −1.15 σ 3.4 Not unusual; The z-score is −1.15, so the age of 24 is about 1.15 standard deviations below the mean.

44. Jan Ullrich: x = 24 z =

x−μ

27 − 27.9 ≈ −0.26 σ 3.4 Not unusual; The z-score is −0.26, so the age of 27 is about 0.26 standard deviation below the mean.

45. Antonin Magne: x = 27 z =

x−μ

36 − 27.9 ≈ 2.38 σ 3.4 Unusual; The z-score is 2.38, so the age of 36 is about 2.38 standard deviations above the mean.

46. Firmin Lambot: x = 36 z =

x−μ

20 − 27.9 ≈ −2.32 σ 3.4 Unusual; The z-score is −2.32, so the age of 20 is about 2.32 standard deviations below the mean.

47. Henri Cornet: x = 20 z =

x−μ

21 − 27.9 ≈ −2.03 σ 3.4 Unusual; The z-score is í2.03, so the age of 21 is about 2.03 standard deviations below the mean.

48. Tadej Pogacar: x = 21 z =

x−μ

34,000 − 35,000 ≈ −0.44 σ 2250 x − μ 37,000 − 35,000 x = 37,000 z = = ≈ 0.89 σ 2250 x − μ 30,000 − 35,000 x = 30,000 z = = ≈ −2.22 2250 σ The tire with a life span of 30,000 miles has an unusually short life span.

49. (a) x = 34,000 z =

x−μ

30,500 − 35,000 = −2 about 2.5th percentile 2250 σ x − μ 37, 250 − 35,000 x = 37, 250 z = = = 1 about 84th percentile σ 2250 x − μ 35,000 − 35,000 x = 35,000 z = = = 0 about 50th percentile σ 2250

(b) x = 30,500 z =

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

x−μ

34 − 33 = 0.25 σ 4 x − μ 30 − 33 x = 30 z = = = −0.75 σ 4 x − μ 42 − 33 x = 42 z = = = 2.25 σ 4 The fruit fly with a life span of 42 days has an unusually long life span.

50. (a) x = 34 z =

x−μ

29 − 33 = −1 about 16th percentile σ 4 x − μ 41 − 33 x = 41 z = = = 2 about 97.5th percentile σ 4 x − μ 25 − 33 x = 25 z = = = −2 about 2.5th percentile σ 4

(b) x = 29 z =

x−μ

59 − 43.8 ≈ 1.75 σ 8.7 x − μ 49 − 50.2 Sam Rockwell: x = 49 z = = ≈ −0.09 σ 13.5 The age of Gary Oldman was between one and two standard deviations above the mean age of Best Actor winners, and the age of Sam Rockwell was less than one standard deviation below the mean age of Best Supporting Actor winners. Neither actor’s age is unusual.

51. Gary Oldman: x = 59 z =

x−μ

37 − 43.8 ≈ −0.78 σ 8.7 x − μ 67 − 50.2 Morgan Freeman: x = 67 z = = ≈ 1.24 σ 13.5 The age of Jamie Foxx was less than one standard deviation below the mean age of Best Actor winners, and the age of Morgan Freeman was between one and two standard deviations above the mean age of Best Supporting Actor winners. Neither actor’s age is unusual.

52. Jamie Foxx: x = 37 z =

x−μ

62 − 43.8 ≈ 2.09 σ 8.7 x − μ 56 − 50.2 Gig Young: x = 56 z = = ≈ 0.43 σ 13.5 The age of John Wayne was more than two standard deviations above the mean age of Best Actor winners, which is unusual. The age of Gig Young was less than one standard deviation above the mean age of Best Supporting Actor winners, which is not unusual.

53. John Wayne: x = 62 z =

x−μ

76 − 43.8 ≈ 3.70 8.7 σ x − μ 77 − 50.2 John Gielgud: x = 77 z = = ≈ 1.99 σ 13.5 The age of Henry Fonda was more than three standard deviations above the mean age of Best Actor winners, which is very unusual. The age of John Gielgud was less than two standard deviations above the mean age of Best Supporting Actor winners, which is not unusual.

54. Henry Fonda: x = 76 z =

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

55.

Midquartile =

Q1 + Q3 3 + 7 = =5 2 2

Midquartile =

Q1 + Q3 28 + 39.5 = = 33.75 2 2

56.

57. (a) The distribution of Concert 1 is symmetric. The distribution of Concert 2 is skewed right. Concert 1 has less variation.

(b) Concert 2 is more likely to have outliers because it has more variation. (c) Concert 1, because 68% of the data should be between ±16.3 of the mean. (d) No, you do not know the number of songs played at either concert or the actual lengths of the songs. 58.

Your distribution is symmetric and your friend’s distribution slightly skewed to the right. 59. (a)

Q1 = 9 , Q2 = 11 , Q3 = 13 IQR = Q3 − Q1 = 13 − 9 = 4 1.5 × IQR = 6 Q1 − (1.5 × IQR ) = 9 − 6 = 3

Q3 + (1.5 × IQR ) = 13 + 6 = 19

Any values less than 6 or greater than 19 are outliers. So, 2 and 24 are outliers. (b)

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

60. (a)

Q1 = 73 , Q2 = 75 , Q3 = 79 IQR = Q3 − Q1 = 79 − 73 = 6 1.5 × IQR = 9 Q1 − (1.5 × IQR ) = 73 − 9 = 64

Q3 + (1.5 × IQR ) = 79 + 9 = 88

Any values less than 64 or greater than 88 are outliers. So, 62 and 95 are outliers. (b)

61. (a) 1, 23, 29, 35, 37, 46, 46, 47, 49, 52, 53, 59, 83 Q1 = 32 , Q2 = 46 , Q3 = 52.5 IQR = Q3 − Q1 = 52.5 − 32 = 20.5 1.5 × IQR = 30.75 Q1 − (1.5 × IQR ) = 32 − 30.75 = 1.25

Q3 + (1.5 × IQR ) = 52.5 + 30.75 = 83.25

Any values less than 1.25 or greater than 83.25 are outliers. So, 1 is an outlier. (b)

62. (a) 19, 27, 30, 36, 38, 47, 47, 48, 50, 50, 53, 54, 56, 60, 62, 90 Q1 = 37 , Q2 = 49 , Q3 = 55 IQR = Q3 − Q1 = 55 − 37 = 18 1.5 × IQR = 27 Q1 − (1.5 × IQR ) = 37 − 27 = 10

Q3 + (1.5 × IQR ) = 55 + 27 = 82

Any values less than 10 or greater than 82 are outliers. So, 90 is an outlier. (b)

63. Answers will vary.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

CHAPTER 2 REVIEW EXERCISE SOLUTIONS

Class width =

Class 26௘–௘31 32௘–௘37 38௘–௘43 44௘–௘49 50௘–௘55

Max − Min 52 − 26 = = 5.2 6 Number of classes 5

Midpoint 28.5 34.5 40.5 46.5 52.5

Class boundaries 25.5௘–௘31.5 31.5௘–௘37.5 37.5௘–௘43.5 43.5௘–௘49.5 49.5௘–௘55.5

Frequency, f 5 4 6 3 2 f ¦ = 20

Relative frequency 0.25 0.20 0.30 0.15 0.10 f ¦ =1 n

Class with greatest relative frequency: 38 – 43 Class with least relative frequency: 50 – 55 3.

Class width =

Max − Min 12.10 − 11.86 = ≈ 0.03 0.04 Number of classes 7

Class

Midpoint

Frequency, f

11.86–11.89 11.90–11.93 11.94–11.97 11.98–12.01 12.02–12.05 12.06–12.09 12.10–12.13

11.875 11.915 11.955 11.995 12.035 12.075 12.115

3 5 8 7 1 0 1

¦ f = 25

Relative frequency 0.12 0.20 0.32 0.28 0.04 0.00 0.04 f =1 n

Cumulative frequency 5 9 15 18 20

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

Class width =

Max − Min 12.10 − 11.86 = ≈ 0.03 0.04 Number of classes 7

Class

Midpoint

Frequency, f

11.86–11.89 11.90–11.93 11.94–11.97 11.98–12.01 12.02–12.05 12.06–12.09 12.10–12.13

11.875 11.915 11.955 11.995 12.035 12.075 12.115

3 5 8 7 1 0 1

¦ f = 25

Class width =

Relative frequency 0.12 0.20 0.32 0.28 0.04 0.00 0.04 f =1 n

Max − Min 166 − 79 = = 14.5 15 Number of classes 6

Class

Midpoint

Frequency, f

79௘–௘93 94௘–௘108 109௘–௘123 124௘–௘138 139௘–௘153 154௘–௘168

86 101 116 131 146 161

9 12 5 4 2 1

¦ f = 33

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

7. Because the data entries go from a low of 20 to a high of 63, use stem values from 2 to 6. List the stems to the left of a vertical line. For each data entry, list a leaf to the right of its stem. Pollution Indices of U.S. Cities 2 0 3 5 7 9 Key: 2 | 0 = 20 3 0 0 1 1 2 6 7 9 9 4 2 3 4 5 5 5 7 5 0 2 6 3 Sample answer: Most U.S. cities have a pollution index from 30 to 47. 8.

Sample answer: Most U.S. cities have a pollution index from 30 to 47. 9.

Location

Sleeping Leisure and Sports Working Educational Activities Other

Frequency

8.8 4.0 2.3 3.5 5.4

¦ f = 24

Relative frequency 0.3667 0.1667 0.0958 0.1458 0.2250 f =1 n

Degrees

132° 60° 34° 52° 81°

Sample answer: Full-time university and college students spend the least amount of time working.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

10.

Sample answer: Full-time university and college students spend the most amount of time sleeping. 11.

Sample answer: The number of stories appears to increase with height. 12.

Sample answer: The real unemployment rate decreases steadily from 2010 to 2020, where it dramatically increases from 2020 to 2021. ¦ x 298.5 = = 29.85 n 10 24.0 28.0 29.0 29.0 29.5 30.0 30.0 30.0 31.5 37.5

13. x =

29.5 + 30.0 = 29.75 2 Mode = 30 (occurs three times)

median =

14. x is not possible Median is not possible Mode = “$250-999” The mean and median cannot be found because the data are at the nominal level of measurement.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

15.

Source Test 1 Test 2 Test 3 Test 4 Test 5 Test 6

16.

17.

¦ ( x ⋅ w)

¦w

Source Test 1 Test 2 Test 3 Test 4

Score, x 78 72 86 91 87 80

¦w

Weight, w 0.2 0.2 0.2 0.4 w =1

n = 20

18.

¦(x ⋅ f ) n

x 0 1 2 3 4 5 6

f 13 9 19 8 5 2 4

¦(x ⋅ f ) n

x·f

142.5 138 243 139.5 105 ( x ⋅ f ) = 768

768 = 38.4 20

n = 60 x=

x·w 19.2 17 18.2 34.4 ( x ⋅ w) = 88.8

88.8 = 88.8 1

Midpoint, Frequency, x f 28.5 5 34.5 4 40.5 6 46.5 3 52.5 2

x·w 11.7 10.8 12.9 13.65 13.05 20 ( x ⋅ w) = 82.1

82.1 = 82.1 1

Score, x 96 85 91 86

¦ ( x ⋅ w)

Weight, w 0.15 0.15 0.15 0.15 0.15 0.25 w =1

125 ≈ 2.1 60

x·f 0 9 38 24 20 10 24 ( x ⋅ f ) = 125

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

21. Skewed right 22. Skewed left 23. Mean, because the mean is to the right of the median in a skewed right distribution. 24. Median, because the mean is to the left of the median in a skewed left distribution. 25. Range = Max – Min = 15 – 1 = 14 ¦ x 96 μ= = ≈ 6.9 N 14 x−μ x

4 2 9 12 15 3 6 8 1 4 14 12 3 3

–2.9 –4.9 2.1 5.1 8.1 –3.9 –0.9 1.1 –5.9 –2.9 7.1 5.1 –3.9 –3.9

( x − μ )2 8.41 24.01 4.41 26.01 65.61 15.21 0.81 1.21 34.81 8.41 50.41 26.01 15.21 15.21

¦ x = 96 ¦ ( x − μ ) ≈ 0 ¦ ( x − μ ) = 295.74 2

σ = σ=

¦( x − μ )

N ¦(x − μ ) N

295.74 ≈ 21.1 14

295.74 ≈ 4.6 14

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

26. Range = Max – Min = 82 – 49 = 33 ¦ x 575 μ= = ≈ 63.89 N 9 x

x−μ

66 56 72 49 82 71 66 60 53

2.11 í7.89 8.11 í14.89 18.11 7.11 2.11 í3.89 í10.89

¦ x = 575 ¦ ( x − μ ) ≈ 0 2

σ = σ=

¦( x − μ )

N ¦(x − μ ) N

( x − μ )2 4.452 62.252 65.772 221.712 327.972 50.552 4.452 15.132 118.592

¦ ( x − μ ) = 870.888 2

870.888 ≈ 96.77 9

870.888 ≈ 9.84 9

27. Range = Max – Min = $7439 – $5395 = $2044 ¦ x 100, 269 x= = ≈ $6266.81 16 n x

x−x

( x − x)

5816 7220 6045 7439 5612 5395 6341 6908 6106 5561 7361 5710 6320 5538 6265 6632

−450.81 953.19 −221.81 1172.19 −654.81 −871.81 74.19 641.19 −160.81 −705.81 1094.19 −556.81 53.19 −728.81 −1.81 365.19

203,230 908,571 49,200 1,374,029 428,776 760,053 5504 411,125 25,860 498,168 1,197,252 310,037 2829 531,164 3 133,364

¦ x = 100, 269 ¦ ( x − x ) ≈ 0 ¦ ( x − x ) = 6,839,165 2

( s = 2

¦ x−x

n −1

) = 6,839,165 ≈ 455,944.3 2

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

(

¦ x−x

n −1

) = 6,839,165 ≈ $675.24 2

28. Range = Max – Min = 71,534 – 45,120 = $26,414 ¦ x 433,020 x= = ≈ $54,127.50 n 8 x

x−x

( x − x)

62,222 56,719 50,259 45,120 47,692 45,985 53,489 71,534

8094.5 2591.5 −3868.5 −9007.5 −6435.5 −8142.5 −638.5 17406.5

65,520,930.25 6,715,872.25 14,965,292.25 81,135,056.25 41,415,660.25 66,300,306.25 407,682.25 302,986,242.25

¦ x = 433,020 ¦ ( x − x ) ≈ 0 ¦ ( x − x ) = 579, 447,042 2

( s =

) = 579, 447,042 ≈ 82,778,148.86

(

) = 579, 447,042 ≈ $9098.25

¦ x−x

n −1

¦ x−x

n −1

29. 95% of the distribution lies within two standard deviations of the mean. x − 2 s = 110 − ( 2 )(17.50 ) = 75 x + 2 s = 110 + ( 2 )(17.50 ) = 145

95% of the distribution lies between $75 and $145.

(

30. ( 73.00, 102.00 ) → ( 87.50 − 1(14.50 ) , 87.50 + 1(14.50 ) ) → x − s , x + s

)

68% of the satellite television charges lie between $73.00 and $102.00.

(

)

31. x − 2 s, x + 2 s → ( 24, 40 ) are two standard deviations from the mean.

1−

1 1 1 =1− = 1 − = 0.75 2 2 4 k ( 2)

At least (40)(0.75) = 30 customers have a mean sale between $24 and $40.

(

)

1−

1 1 1 =1− = 1 − = 0.75 2 2 4 k ( 2)

32. x − 2 s , x + 2 s → ( 2.3, 17.5 ) are two standard deviations from the mean.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

33. x =

¦ xf 99 = ≈ 2.5 n 40

x−x

( x − x)

0 1 2 3 4 5

1 8 13 10 5 3

0 8 26 30 20 15

–2.5 –1.5 –0.5 0.5 1.5 2.5

6.25 2.25 0.25 0.25 2.25 6.25

n = 40

¦ xf = 99

(

) f = 60 ≈ 1.2

34. x =

¦ x−x

( x − x) f 2

6.25 18.00 3.25 2.50 11.25 18.75

¦ ( x − x ) f = 60 2

n −1

¦ xf 61 = ≈ 2.4 25 n

x−x

( x − x)

0 1 2 3 4 5 6

4 5 2 9 1 3 1

0 5 4 27 4 15 6

–2.4 –1.4 –0.4 0.6 1.6 2.6 3.6

5.76 1.96 0.16 0.36 2.56 6.76 12.96

n = 25

¦ xf = 61

(

) f = 72.2 ≈ 1.7

¦ x−x

n −1

( x − x) f 2

23.04 9.80 0.32 3.24 2.56 20.28 12.96

¦ ( x − x ) f = 72.2 2

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

35. Freshmen: x =

¦ x 23.1 = ≈ 2.567 n 9

x−x

( x − x)

2.8 1.8 4.0 3.8 2.4 2.0 0.9 3.6 1.8

0.233 -0.767 1.433 1.233 -0.167 -0.567 -1.667 1.033 -0.767

0.0543 0.5833 2.0535 1.5203 0.0279 0.3215 2.7789 1.0671 0.5883

¦ ( x − x ) = 9.0000 2

(

¦ x−x

n −1

) = 9.0000 ≈ 1.061 2

8 s 1.061 CV = ⋅ 100% = ⋅ 100% ≈ 41.3% x 2.567 ¦ x 26.6 Seniors: x = = ≈ 2.956 n 9

x−x

( x − x)

2.3 3.3 1.8 4.0 3.1 2.7 3.9 2.6 2.9

−0.656 0.344 −1.156 1.044 0.144 −0.256 0.944 −0.356 −0.056

0.4303 0.1183 1.3363 1.0899 0.0207 0.0655 0.8911 0.1267 0.0031

¦ ( x − x ) = 4.0822 2

(

¦ x−x

n −1

) = 4.0822 ≈ 0.714 2

s 0.714 ⋅ 100% = ⋅ 100% ≈ 24.2% x 2.956 Grade point averages are more variable for freshmen than seniors. CV =

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

¦ x 406 = = 50.75 N 8 x−μ x

36. Ages: μ =

66 54 47 61 36 59 50 33

15.25 3.25 −3.75 10.25 −14.75 8.25 −0.75 −17.75 ¦(x − μ) = 0

¦ x = 406

σ=

(

¦ x−x

( x − μ )2 232.5625 10.5625 14.0625 105.0625 217.5625 68.0625 0.5625 315.0625

¦ ( x − μ ) = 963.5 2

) = 963.5 ≈ 10.97 2

8 σ 10.97 CV = ⋅ 100% = ⋅ 100% ≈ 21.6% μ 50.75 ¦ x 185 Years of Experience: μ = = ≈ 23.1 8 N

x 37 20 23 32 14 29 22 8 ¦ x = 185

σ=

(

¦ x−x

x−μ

( x − μ )2

13.9 −3.1 −0.1 8.9 −9.1 5.9 −1.1 −15.1 ¦(x − μ) ≈ 0

193.21 9.61 0.01 79.21 82.81 34.81 1.21 228.01

¦ ( x − μ ) = 628.88 2

) = 628.88 ≈ 8.866 2

N 8 σ 8.866 CV = ⋅ 100% = ⋅ 100% ≈ 38.3% μ 23.13 Years of experience are more variable than ages for all lawyers at a firm. 37. 22, 22, 22, 23, 23, 26, 27, 30, 30, 31, 33, 33, 33, 35, 35, 36, 41, 45, 48, 50, 52, 56, 58, 105, 113, 113, 117, 118, 121, 141

Min = 22, Q1 = 30 , Q2 = 35.5 , Q3 = 58 , Max = 141 38. IQR = Q3 − Q1 = 58 − 30 = 28 miles per gallon

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

39.

40. Count the number of entries that are 58 or below: 23 vehicles 41. 24.0 28.0 29.0 29.0 29.5 30.0 30.0 30.0 31.5 37.5

median = 29.75 Q1 = 29.0 ; Q3 = 30.0 IQR = Q3 − Q1 = 30.0 − 29.0 = 1 inch 42. 145, 156, 167, 172, 173, 184, 185, 190, 190, 192, 195, 197, 205, 208, 212, 227, 228, 240 Min = 145, Q1 = 173 , Q2 = 191 , Q3 = 208 , Max = 240

43. The 65th percentile means that 65% had a test grade of 75 or less. So, 35% scored higher than 75. 44. If there are 115 stations with a larger daily audience, then this station has the 682 – 115 = 567th 567 largest audience. The percentile of 567 = ⋅ 100 ≈ 83rd percentile 682 16,500 − 11,830 ≈ 1.97 2370 Not unusual; The z-score is 1.97, so a towing capacity of 16,500 pounds is about 1.97 standard deviations above the mean.

45. z =

5500 − 11,830 ≈ −2.67 2370 Unusual; The z-score is −2.67, so a towing capacity of 5500 pounds is about 2.67 standard deviations below the mean.

46. z =

18,000 − 11,830 = 2.60 2370 Unusual; The z-score is 2.60, so a towing capacity of 18,000 pounds is about 2.60 standard deviations above the mean.

47. z =

48. z =

11,300 − 11,830 = −0.22 2370

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

Not unusual; The z-score is −0.22, so a towing capacity of 11,300 pounds is about 0.22 standard deviation below the mean.

CHAPTER 2 QUIZ SOLUTIONS 1. (a) Class width =

Max − Min 157 − 101 = = 11.2 12 Number of classes 5

Class

Midpoint

101–112

106.5

113–124

118.5

125–136

130.5

137–148

142.5

149–160

154.5

(b)

(c)

(d) Skewed right

Class boundaries 100.5– 112.5 112.5– 124.5 124.5– 136.5 136.5– 148.5 148.5– 160.5

Frequency, f

Relative frequency 0.11

Cumulative frequency 3

0.41

0.30

0.11

0.07

¦ f = 27

¦ n =1

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

(e)

(f) Weekly Exercise (in minutes) 10 11 12 13 14 15

1 8 Key: 10 | 8 = 108 1 4 6 7 8 9 9 0 0 3 3 4 7 7 8 0 1 1 2 5 9 9 2 0 7

(g) 101, 108, 111, 114, 116, 117, 118, 119, 119, 120, 120, 123, 123, 124, 127, 127, 128, 130, 131, 131, 132, 135, 139, 139, 142, 150, 157 Min = 101, Q1 = 118 , Q2 = 124 , Q3 = 132 , Max = 157

¦ xf 3403.5 = ≈ 126.1 n 27

Midpoint, Frequency, x f 106.5 3 118.5 11 130.5 8 142.5 3 154.5 2

(

x−x

( x − x)

( x − x) f

319.5 1303.5 1044 427.5 309.0

−19.6 −7.6 4.4 16.4 28.4

384.16 57.76 19.36 268.96 806.56

1152.48 635.36 154.88 806.88 1613.12

¦ xf = 3403.5

n = 27 ¦ x−x

) f = 4362.72 ≈ 13.0

n −1

¦ ( x − x ) f = 4362.72 2

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

3. (a)

Category Metals Metalloids Halogens Noble gases Rare earth elements Other nonmetals

Frequency 55 6 5 6 30 7

n = 109

Relative frequency 0.5046 0.0550 0.0459 0.0550 0.2752 0.0642 f =1 n

Degrees 182° 20° 17° 20° 99° 23°

(b)

¦ x 16, 262 = ≈ 1016.4 n 16 718, 720, 749, 790, 860, 891, 969, 976, 1062, 1100, 1100, 1124, 1248, 1255, 1316, 1384 976 + 1062 median = = 1019 2 Mode = 1100 (occurs twice) The mean or median best describes a typical salary because there are no outliers.

4. (a) x =

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

(b) Range = Max – Min = 1384 – 718 = 666 x

x−x

( x − x)

1100 749 720 1062 1384 1248 1124 891 1255 969 976 790 718 860 1316 1100

83.6 −267.4 −296.4 45.6 367.6 231.6 107.6 −125.4 238.6 −47.4 −40.4 −226.4 −298.4 −156.4 299.6 83.6

6989 71,503 87,853 2079 135,130 53,639 11,578 15,725 56,930 2247 1632 51,257 89,043 24,461 89,760 6989

¦ ( x − x ) = 706,815 2

( s =

) = 706,815 ≈ 47,120.9

(

) = 706,815 ≈ 217.1

¦ x−x

n −1

¦ x−x

n −1

s 217.1 ⋅ 100% = ⋅ 100% ≈ 21.4% x 1016.4

x − 2s = 180,000 − 2 ⋅ 15,000 = $150,000 x + 2s = 180,000 + 2 ⋅15,000 = $210,000 95% of the new home prices fall between $150,000 and $210,000. x − x 225,000 − 180,000 = = 3.0 15,000 s Unusual; The z-score is 3, so a new home price of $225,000 is about three standard deviations above the mean.

6. (a) x = 225,000 : z =

x − x 80,000 − 180,000 = ≈ −6.67 15,000 s Unusual; The z-score is −6.67, so a new home price of $80,000 is about 6.67 standard deviations below the mean.

(b) x = 80,000 :

x − x 200,000 − 180,000 = ≈ 1.33 15,000 s

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

Not unusual; The z-score is 1.33, so a new home price of $200,000 is about 1.33 standard deviations above the mean. x − x 147,000 − 180,000 = = −2.2 15,000 s Unusual; The z-score is −2.2, so a new home price of $147,000 is about 2.2 standard deviations below the mean.

(d) x = 147,000 : z =

7. 19 23 24 25 25 26 26 26 26 26 27 27 28 29 29 29 30 31 31 32 33 34 35 35 35 36 36 37 40 43

Q1 = 26 Q2 = 29 Min = 19, Q1 = 26 , Q2 = 29 , Q3 = 35 , Max = 43

Q3 = 35

CUMULATIVE REVIEW FOR CHAPTERS 1 AND 2 1. Systematic sampling is used because every 40th toothbrush from each assembly line is tested. It is possible for bias to enter into the sample if, for some reason, an assembly line makes a consistent error. 2. Simple random sampling is used because each telephone number has an equal chance of being dialed, and all samples of 1090 phone numbers have an equal chance of being selected. The sample may be biased because telephone sampling only samples those individuals who have telephones, who are available, and who are willing to respond. 3.

4. $58,700 is a parameter. The median salary is based on all first-year chemists. 5. 64% is a statistic. The percent is based on a subset of the population.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

6. (a) x = 86,500, s = 1500 (83,500, 89,500 ) = 86,500 ± 2 (1500 ) 2 standard deviations away from the mean. Approximately 95% of the electrical engineers will have salaries between $83,500 and $89,500. x − x 93,500 − 86,500 = ≈ 4.67 s 1500 x − x 85,600 − 86,500 x = $85,600 : z = = = −0.6 1500 s x − x 82,750 − 86,500 x = $82,750 : z = = = −2.5 s 1500 The salaries of $93,500 and $82,750 are unusual.

(b) x = $93,500 : z =

7. Population: Collection of opinions of all college students in bachelor’s degree programs Sample: Collection of opinions of the 3941 college students in bachelor’s degree programs surveyed 8. Population: Collections of responses of all undergraduate and graduate students Sample: Collection of responses of the 182 undergraduate and graduate students surveyed 9. Experiment. The study applies a treatment (digital device) to the subjects. 10. Observational study. The study does not attempt to influence the responses of the subjects. 11. Quantitative: The data are at the ratio level. 12. Qualitative: The data are at the nominal level. 13. (a)

Q1 Q2 0 0 0 0 0 0 0 0 1 2 2 2 2 2 3 3 4 5 6 6 6 7 7 10 15 18 19 21 21 22 22 23 27 27 28 31 34 35 38 41 54 55 57 65 69 71 75 78 102 127 Q3 Min = 0, Q1 = 2 , Q2 = 16.5 , Q3 = 35 , Max = 127

(b) The distribution of the number of tornadoes is skewed right.

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

14.

Source Test 1 Test 2 Test 3 Test 4 Test 5

Score, x 85 92 84 89 91

¦ ( x ⋅ w)

¦w

15. (a) x =

Weight, w 0.15 0.15 0.15 0.15 0.40 w =1

x·w 12.75 13.80 12.60 13.35 36.40 ( x ⋅ w) = 88.9

88.9 = 88.9 1

49.4 ≈ 5.49 9

Mode = none Both the mean and median accurately describe a typical American alligator tail length. (Answers will vary.) (b) Range = Max – Min = 7.5 – 3.4 = 4.1 x

x−x

( x − x)

3.4 3.9 4.2 4.6 5.4 6.5 6.8 7.1 7.5

–2.09 –1.59 –1.29 –0.89 –0.09 1.01 1.31 1.61 2.01

4.3681 2.5281 1.6641 0.7921 0.0081 1.0201 1.7161 2.5921 4.0401

¦ ( x − x ) = 18.7289 2

( s =

) = 18.7289 ≈ 2.34

(

) = 18.7289 ≈ 1.53

¦ x−x

n −1

¦ x−x

n −1

16. (a) An inference drawn from the study is that the life expectancies for Americans will continue to increase or remain stable.

(b) This inference may incorrectly imply that women will have less of a chance of dying of heart disease in the future. The study was only conducted over the past 5 years and deaths may not decrease in the next year.

100

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

17. Class width =

Max − Min 73 − 0 = = 9.125 10 Number of classes 8

Class

Midpoint

0–9 10–19 20–29 30–39 40–49 50–59 60–69 70–79

4.5 14.5 24.5 34.5 44.5 54.5 64.5 74.5

Class boundaries –0.5–9.5 9.5–19.5 19.5–29.5 29.5–39.5 39.5–49.5 49.5–59.5 59.5–69.5 69.5–79.5

Frequency, f

18 5 2 1 1 1 2 2

¦ f = 32

Relative frequency 0.56 0.16 0.06 0.03 0.03 0.03 0.06 0.06 f =1 n

18. The distribution is skewed right. 19.

Class with greatest frequency: 0 – 9 Classes with least frequency: 30 – 39, 40 – 49, and 50 – 59

CHAPTER 2 TEST SOLUTIONS 1. (a) x =

¦ x 964 = ≈ 80.3 n 12

63 67 68 72 73 81 85 87 87 88 94 99 N Median = 83 Mode = 87 (occurs twice) The median best represents the center of the data.

Cumulative frequency 18 23 25 26 27 28 30 32

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

(b) Range = Max – Min = 99 – 63 = 36 x

x−x

( x − x)

67 72 88 73 99 85 81 87 63 94 68 87

−13.3 −8.3 7.7 −7.3 18.7 4.7 0.7 6.7 −17.3 13.7 −12.3 6.7

176.89 68.89 59.29 53.29 349.69 22.09 0.49 44.89 299.29 187.69 151.29 44.89

¦ ( x − x) ≈ 0

( s =

) = 1458.68 ≈ 132.6

(

) = 1458.68 ≈ 11.5

¦ x−x

n −1

¦ x−x

n −1

¦ ( x − x ) = 1458.68 2

s 11.5 ⋅ 100% = ⋅ 100% ≈ 14.3% x 80.3

(d) Points scored

Key: 6 3 = 63

63 7 8 72 3 81 5 7 7 8 94 9

2. (a) Class width =

Range 221 − 18 = ≈ 33.8 34 Number of classes 6

Class

Midpoint

18–51 52–85 86–119 120–153 154–187 188–221

34.5 68.5 102.5 136.5 170.5 204.5

Class boundaries 17.5–51.5 51.5–85.5 85.5–119.5 119.5–153.5 153.5–187.5 187.5–221.5

Frequency, f

2 4 5 5 3 1

¦ f = 20

Relative frequency 0.10 0.20 0.25 0.25 0.15 0.05 f =1 n

Cumulative frequency 2 6 11 16 19 20

101

102

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

(b)

(c)

(d) The shape of the distribution is symmetric. (e)

¦ xf 2254 = = 112.7 n 20 Midpoint, Frequency, x f 34.5 2 68.5 4 102.5 5 136.5 5 170.5 3 204.5 1

(

x−x

( x − x)

( x − x) f

69.0 274.0 512.5 682.5 511.5 204.5

−78.2 −44.2 −10.2 23.8 57.8 91.8

6115.24 1953.64 104.04 566.44 3340.84 8427.24

12,230.5 7814.6 520.2 2832.2 10,022.5 8427.2

¦ xf = 2254

n = 20 ¦ x−x

¦ ( x − x ) f ≈ 41,847.2

) f = 41,847.2 ≈ 46.9

n −1

4. 149 is the 16th observation when the data are ordered. The percentile for 149 number of data entries less than 149 15 = ⋅ 100 = ⋅ 100 = 75th percentile total number of data entries 20

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

Certification Diamond Multi-Platinum Platinum Gold

f 6 26 42 48

¦ f = 122

Relative Frequency 0.049 0.213 0.344 0.393 f ≈1 n

Angle 18° 77° 124° 141° = 360°

(a)

(b)

Sample answer: It appears that there is no relation between minutes to complete the final exam and average score before the final exam. 7. (a) Find the five-number summary. 28 30 37 40 N 42 46 51 55 N 56 58 59 60 N 62 65 68

Q1 = 40 Q2 = 55 Q3 = 60 Min = 28, Q1 = 40 , Q2 = 55 , Q3 = 60 , Max = 68

103

104

CHAPTER 2 Ň DESCRIPTIVE STATISTICS

(

8. (a) ( 333.3, 354.1) → ( 343.7 − 1(10.4 ) , 343.7 + 1(10.4 ) ) → x − s , x + s

)

About 68% of the gestational lengths are between 333.3 and 354.1 days. Thus, about 0.68(208) ≈ 141 gestational lengths are between 333.3 and 354.1 days. x − x 318.4 − 343.7 = ≈ −2.43 ; Since the z-score is about –2.43 (between 2 and 3 standard s 10.4 deviations below the mean), the gestational length is unusual.

(b) z =

CHAPTER

Probability

3.1 BASIC CONCEPTS OF PROBABILITY AND COUNTING 3.1 TRY IT YOURSELF SOLUTIONS 1. (1)

6 outcomes Let Y = Yes, N = No, NS = Not sure, M = Male, and F = Female Sample space = {YM, YF, NM, NF, NSM, NSF} (2)

9 outcomes Let Y = Yes, N = No, NS = Not sure 50+ = 50 and older Sample space = {Y18–34, Y35–49, Y50+, N18–34, N35–49, N50+, NS18–34, NS35–49, NS50+} (3)

12 outcomes Let Y = Yes, N = No, NS = Not sure, NE = Northeast, S = South, MW = Midwest, and W = West Sample space = {YNE, YS, YMW, YW, NNE, NS, NMW, NW, NSNE, NSS, NSMW, NSW} 2. (1) Event C has six outcomes: choosing the ages 18, 19, 20, 21, 22, and 23. The event is not a simple event, because it consists of more than a single outcome. (2) Event D has one outcome: choosing the age 20. The event is a simple event, because it consists of a single outcome.

106

CHAPTER 3 Ň PROBABILITY

3. Manufacturer: 4 Size: 2 Color: 5

(4)(2)(5) = 40 ways

Tree Diagram for Car Selections

4. (1) Each letter is an event (26 choices for each). 26 ⋅ 26 ⋅ 26 ⋅ 26 ⋅ 26 ⋅ 26 = 308,915,776 (2) Each letter is an event (26, 25, 24, 23, 22, and 21 choices). 26 ⋅ 25 ⋅ 24 ⋅ 23 ⋅ 22 ⋅ 21 = 165,765,600 (3) Each letter is an event (22, 26, 26, 26, 26, and 26 choices). 22 ⋅ 26 ⋅ 26 ⋅ 26 ⋅ 26 ⋅ 26 = 261,390, 272 (4) Each digit is an event and each letter is an event (9, 26, 26, 26, 26, and 26 choices). 9 ⋅ 26 ⋅ 26 ⋅ 26 ⋅ 26 ⋅ 26 = 106,932,384 5

(1) P(9 of clubs) = (2) P(heart) =

1 ≈ 0.019 52

13 = 0.25 52

(3) P(diamond, heart, club, or spade) =

52 =1 52

CHAPTER 3 Ň PROBABILITY

107

7. Frequency = 546 Total of the frequencies = 3000 546 = 0.182 P(age 35 to 44) = 3000 8. The event is “salmon successfully passing through a dam on the Columbia River.” The probability is estimated from the results of an experiment. Empirical probability

459 = 0.153 3000 459 2541 = = 0.847 P(age is not 18 to 24) = 1 − 3000 3000 2541 847 = or 0.847 3000 1000

9. P(age 18 to 24) =

10. There are five outcomes in the event: {T1, T2, T3, T4, T5}. 5 ≈ 0.313 P(tail and less than 6) = 16 11. 10 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 = 10,000,000 1 10,000,000

3.1 EXERCISE SOLUTIONS 1. An outcome is the result of a single trial in a probability experiment, whereas an event is a set of one or more outcomes. 2. (a) Could represent the probability of an event. The probability of an event occurring must be contained in the interval [0, 1] or [0%, 100%]. (b) Could not represent the probability of an event. The probability of an event occurring cannot be greater than 100%. (c) Could not represent the probability of an event. The probability of an event occurring cannot be greater than 1. (d) Could not represent the probability of an event. The probability of an event occurring cannot be less than 0. (e) Could represent the probability of an event. The probability of an event occurring must be contained in the interval [0, 1] or [0%, 100%]. (f) Could not represent the probability of an event. The probability of an event occurring cannot be greater than 1.

108

CHAPTER 3 Ň PROBABILITY

3. It is impossible to have more than a 100% chance of rain. 4. The Fundamental Counting Principle counts the number of ways that two or more events can occur in sequence. 5. The law of large numbers states that as an experiment is repeated over and over, the probabilities found in the experiment will approach the actual probabilities of the event. Examples will vary. 6.

P ( E ) + P ( E ') = 1 , P ( E ) = 1 − P ( E ') , P ( E ') = 1 − P ( E )

7. False. The event “choosing false on a true or false question and choosing A or B on a multiple-choice question” is not simple, because it consists of two possible outcomes and can be represented as A = {FA, FB}. 8. False. You flip a fair coin nine times and it lands tails up each time. The probability it will land heads up on the 10th flip is 0.5. 9. False. A probability of less than

1 = 0.05 indicates an unusual event. 20

10. True 11. d

12. f

17. P( E ') = 1 − P( E ) = 1 −

13. b

14. c

15. a

16. e

5 6 5 1 = − = 6 6 6 6

18. P ( E ') = 1 − P ( E ) = 1 − 0.55 = 0.45 19. P ( E ') = 1 − P ( E ) = 1 − 0.03 = 0.97 20. P( E ') = 1 − P( E ) = 1 −

2 7 2 5 = − = 7 7 7 7

21. P ( E ) = 1 − P ( E ') = 1 − 0.95 = 0.05 22. P ( E ) = 1 − P ( E ') = 1 − 0.13 = 0.87 23. P( E ) = 1 − P( E ') = 1 −

3 4 3 1 = − = 4 4 4 4

24. P( E ) = 1 − P( E ') = 1 −

21 61 21 40 = − = 61 61 61 61

25. {A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z}; 26 26. {A, B, C, D, F}; 5

CHAPTER 3 Ň PROBABILITY

109

27. {AƆ, KƆ, QƆ, JƆ, 10Ɔ, 9Ɔ, 8Ɔ, 7Ɔ, 6Ɔ, 5Ɔ, 4Ɔ, 3Ɔ, 2Ɔ, AƇ, KƇ, QƇ, JƇ, 10Ƈ, 9Ƈ, 8Ƈ, 7Ƈ, 6Ƈ, 5Ƈ, 4Ƈ, 3Ƈ, 2Ƈ, AƄ, KƄ, QƄ, JƄ, 10Ƅ, 9Ƅ, 8Ƅ, 7Ƅ, 6Ƅ, 5Ƅ, 4Ƅ, 3Ƅ, 2Ƅ, Aƅ, Kƅ, Qƅ, Jƅ, 10ƅ, 9ƅ, 8ƅ, 7ƅ, 6ƅ, 5ƅ, 4ƅ, 3ƅ, 2ƅ}; 52 28. {(Blue, Blonde), (Blue, Black), (Blue, Brown), (Blue, Red), (Blue, Other), (Brown, Blonde), (Brown, Black), (Brown, Brown), (Brown, Red), (Brown, Other), (Green, Blonde), (Green, Black), (Green, Brown), (Green, Red), (Green, Other), (Hazel, Blonde), (Hazel, Black), (Hazel, Brown), (Hazel, Red), (Hazel, Other), (Gray, Blonde), (Gray, Black), (Gray, Brown), (Gray, Red), (Gray, Other), (Other, Blonde), (Other, Black), (Other, Brown), (Other, Red), (Other, Other),}; 30 29.

{HH, HT, TH, TT }; 4 30. {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}; 8

31. {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2,), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}; 36

32. {1HH1, 1HH2, 1HH3, 1HT1, 1HT2, 1HT3, 1TH1, 1TH2, 1TH3, 1TT1, 1TT2, 1TT3, 2HH1, 2HH2, 2HH3, 2HT1, 2HT2, 2HT3, 2TH1, 2TH2, 2TH3, 2TT1, 2TT2, 2TT3, 3HH1, 3HH2, 3HH3, 3HT1, 3HT2, 3HT3, 3TH1, 3TH2, 3TH3, 3TT1, 3TT2, 3TT3, 4HH1, 4HH2, 4HH3, 4HT1, 4HT2, 4HT3, 4TH1, 4TH2, 4TH3, 4TT1, 4TT2, 4TT3, 5HH1, 5HH2, 5HH3, 5HT1, 5HT2, 5HT3, 5TH1, 5TH2, 5TH3, 5TT1, 5TT2, 5TT3, 6HH1, 6HH2, 6HH3, 6HT1, 6HT2, 6HT3, 6TH1, 6TH2, 6TH3, 6TT1, 6TT2, 6TT3}; 72 33. One outcome; simple event, because it is an event that consists of a single outcome. 34. Number less than 500 = {1, 2, 3 . . . 499}; 499 outcomes Not a simple event, because it is an event that consists of more than a single outcome. 35. There are 13 diamonds or 13 outcomes. Not a simple, event because it is an event that consists of more than a single outcome.

110

CHAPTER 3 Ň PROBABILITY

36. One outcome; simple event, because it is an event that consists of a single outcome. 37. (6)(12)(8) = 576 38. (4)(3)(4)(6)(3) = 864 39. (9)(10)(10)(5) = 4500 40. (2)(2)(2)(2)(2)(2) = 64 41. P( A) =

1 ≈ 0.083 12

42. P( B) =

1 ≈ 0.083 12

43. P(C ) =

8 2 = ≈ 0.667 12 3

44. P( D) =

7 ≈ 0.583 12

45. P( E ) =

4 ≈ 0.333 12

46. P( F ) =

2 1 = ≈ 0.167 12 6

47. P ( does not have a tattoo ) = 48. P ( has two tattoos ) =

704 ≈ 0.700 1005

57 ≈ 0.057 1005

49. P(18 to 24) =

30.2 ≈ 0.092 328.2

50. P(25 to 44) =

87.6 ≈ 0.267 328.2

51. P(45 to 64) =

83.3 ≈ 0.254 328.2

52. P(65 years old and older ) =

54.1 ≈ 0.165 328.2

CHAPTER 3 Ň PROBABILITY

111

53. Empirical probability, because company records were used to calculate the frequency of a car needing repairs. 54. Classical probability, because each outcome in the sample space is equally likely to occur. 55. Subjective probability, because it is most likely based on an educated guess. 56. Empirical probability, because survey results were used to calculate the frequency of a high school counselor favoring significant changes. 57. Classical probability, because each outcome in the sample space is equally likely to occur. 58. Subjective probability, because it is most likely based on an educated guess. 59. P ( A) = 1 − P ( A ') = 1 −

16,598 ≈ 1 − 0.543 = 0.457 30,569

60. P ( B ) = 1 − P ( B ') = 1 −

5293 ≈ 1 − 0.173 = 0.827 30,569

61. P (C ) = 1 − P (C ') = 1 −

2416 ≈ 1 − 0.079 = 0.921 30,569

62. P ( D ) = 1 − P ( D ') = 1 −

1396 ≈ 1 − 0.046 = 0.954 30,569

63. P ( Republican ) =

497,944 ≈ 0.648 768,804

64. P (did not vote) =

5,640, 463 ≈ 0.333 16,955,519

65. (a)

1 = 0.001 1000

(b) 1 −

1 999 = = 0.999 1000 1000

66. Number of possible codes: 35 ⋅ 36 ⋅ 36 ⋅ 36 ⋅ 5 ⋅ 5 = 40,824,000 (a)

1 ≈ 0.0000000245 40,824,000

(b) 1 −

67. P( A) =

1 ≈ 1 − 0.00000002 = 0.99999998 40,824,000

1 ≈ 0.042 ; unusual 24

112

CHAPTER 3 Ň PROBABILITY

68. P( B) =

3 = 0.125 ; not unusual 24

69. P(C ) =

5 ≈ 0.208 ; not unusual 24

70. P( D) =

1 ≈ 0.042 ; unusual 24

71. There are eight outcomes in the sample space: {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG} 1 P ( GGG ) = = 0.125 8 72. There are eight outcomes in the sample space: {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG} 1 P ( BBB ) = = 0.125 8 73. There are eight outcomes in the sample space: {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG} 3 P ( BBG, BGB, GBB ) = = 0.375 8 74. There are eight outcomes in the sample space: {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG} 7 P ( BBB, BBG, BGB, BGG, GBB, GBG, GGB ) = = 0.875 8 1 OR: 1 − P ( GGG ) = 1 − = 0.875 8 75. P(doctorate) =

3 ≈ 0.033 91

76. P(associate's degree) =

77. P(master's degree) =

23 ≈ 0.253 91

25 ≈ 0.275 91

78. P(high school diploma) =

4 ≈ 0.044 91

79. No. None of the events have a probability of 0.05 or less. 80. Yes. The events in Exercises 75 and 78 can be considered unusual, because their probabilities are less than or equal to 0.05. 81. (a) P(pink) =

2 = 0.5 4

(b) P(red) =

1 = 0.25 4

CHAPTER 3 Ň PROBABILITY

82. P(same coloring as one of its parents) =

113

8 = 0.5 16

83. Total workers (in millions) = 18, 200,000 + 63,600,000 + 22,900,000 + 29,700,000 + 13, 400,000 = 147,800,000 13, 400,000 P ( natural resources, construction, or maintenance occupation ) = ≈ 0.091 147,800,000 84. Total workers (in millions) = 18, 200,000 + 63,600,000 + 22,900,000 + 29,700,000 + 13, 400,000 = 147,800,000 P ( not in a service occupation ) = 1 − P ( service occupation ) = 1 −

22,900,000 ≈ 1 − 0.155 = 0.845 147,800,000

85. Total workers (in millions) = 18, 200,000 + 63,600,000 + 22,900,000 + 29,700,000 + 13, 400,000 = 147,800,000 29,700,000 P ( sales or office occupation ) = ≈ 0.201 147,800,000 86. Total workers (in millions) = 18, 200,000 + 63,600,000 + 22,900,000 + 29,700,000 + 13, 400,000 = 147,800,000 P ( not in a production, transportation, or material moving occupation ) = 1 − P ( production, transportation, or material moving occupation ) = 1 −

87. (a) P ( at least 51) =

18, 200,000 ≈ 0.877 147,800,000

6 ≈ 0.047 ; This event is unusual because its probability is less than or equal to 127

0.05. (b) P ( between 21 and 31 inclusive ) = (c) P ( fewer than 14 ) =

47 ≈ 0.370 127

5 ≈ 0.039 ; This event is unusual because its probability is less than or 127

equal to 0.05. 88. (a) P (less than $21) = 0.25 ; Q1 = 21 (b) P (between $21 and $50) = 0.5 ; Q1 = 21 and Q3 = 50 (c) P ($30 or more) = 0.5 ; median = 30 89. The probability of randomly choosing a smoker whose mother did not smoke 90. The probability of randomly choosing a car with only one reason for showing its “CHECK ENGINE” light out of a population of vehicles showing “CHECK ENGINE” light.

114

CHAPTER 3 Ň PROBABILITY

91. No, the odds of winning a prize are 1:6 (one winning cap and six losing caps). So, the statement should read, “one in seven game pieces win a prize.” 92. The first game. The probability of winning the second game is 93. (a) P ( event will occur ) =

1 1 ≈ 0.091 , which is less than . 11 10

4 ≈ 0.444 9

(b) P ( event will not occur ) =

5 ≈ 0.556 9

94. 13: 39 = 1: 3 95. 39 :13 = 3:1 96. p = number of successful outcomes q = number of unsuccessful outcomes number of successful outcomes p P ( A) = = total number of outcomes p+q 97. (a)

Sum 2 3 4 5 6 7 8 9 10 11 12

Outcomes (1, 1) (1, 2), (2, 1) (1, 3), (2, 2,), (3, 1) (1, 4), (2, 3), (3, 2), (4, 1) (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) (3, 6), (4, 5), (5, 4), (6, 3) (4, 6), (5, 5), (6, 4) (5, 6), (6, 5) (6, 6)

(b) Answers will vary. (c) Answers will vary.

P(sum) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

Probability 0.028 0.056 0.083 0.111 0.139 0.167 0.139 0.111 0.083 0.056 0.028

CHAPTER 3 Ň PROBABILITY

115

3.2 CONDITIONAL PROBABILITY AND THE MULTIPLICATION RULE 3.2 TRY IT YOURSELF SOLUTIONS 1.

P ( 65 years old or olderȴnot ridden as a passenger in a self-driving vehicle ) =

2. (1) Dependent

196 ≈ 0.263 745

(2) Independent

3. (1) Let A = {first salmon swims successfully through the dam} B = {second salmon swims successfully through the dam} P ( A and B ) = P( A) ⋅ P( B) = (0.85) ⋅ (0.85) ≈ 0.723 (2) Let A = {selecting a heart} B = {selecting a second heart}

§ 13 · § 12 · P ( A and B ) = P( A) ⋅ P( B A) = ¨ ¸ ⋅ ¨ ¸ ≈ 0.059 © 52 ¹ © 51 ¹ 4. (1) P ( 3 surgeries are successful ) = (0.90) ⋅ (0.90) ⋅ (0.90) = 0.729 (2) P ( none are successful ) = (0.10) ⋅ (0.10) ⋅ (0.10) = 0.001 (3) P ( at least one rotator cuff surgery is successful ) = 1 − P(none are successful) = 1 − 0.001 = 0.999

5. A = {is female}; B = {works in health field} (1) P ( A and B ) = P( A) ⋅ P( B A) = (0.65) ⋅ (0.25) =0.163 (2) P ( A and B ') = P( A) ⋅ P( B ' A) = P( A) ⋅ (1 − P( B A)) = (0.65) ⋅ (0.75) =0.488 The events are not unusual because their probabilities are not less than or equal to 0.05.

3.2 EXERCISE SOLUTIONS 1. Two events are independent if the occurrence of one of the events does not affect the probability of the occurrence of the other event, whereas two events are dependent if the occurrence of one of the events does affect the probability of the occurrence of the other event. 2. (a) Sample answer: Roll a die twice. The outcome of the second roll is independent of the outcome of the first roll. (b) Sample answer: Draw two cards (without replacement) from a standard deck of 52 playing cards. The outcome of the second draw is dependent on the outcome of the first draw. 3. The notation P( B A) means the probability of event B occurring, given that event A has occurred.

116

CHAPTER 3 Ň PROBABILITY

4. The complement of “at least one” is “none.” So, the probability of getting at least one item is equal to 1 − P (none of the items) . 5. False. If two events are independent, then P( A B) = P( A) . 6. False. If events A and B are independent, then P ( A and B ) = P( A) ⋅ P( B) . 7. M = {student is male}; B = {student received business degree}; F = {student is female} 204,839 181,362 (a) P ( M | B ) = (b) P ( B | F ) = ≈ 0.530 ≈ 0.160 386, 201 1,135,684 8.

M = {worker is male}; R = {worker contributes to retirement savings}; F = {worker is female} 116 143 = 0.464 ≈ 0.552 (a) P ( R | M ) = (b) P ( F | R ) = 250 259

9. Independent. The outcome of the first card drawn does not affect the outcome of the second card drawn. 10. Dependent. The outcome of a father having hazel eyes affects the outcome of a daughter having hazel eyes. 11. Dependent. The outcome of returning a movie after its due date affects the outcome of receiving a late fee. 12. Dependent. The outcome of not putting money in a parking meter affects the outcome of getting a parking ticket. 13. Dependent. The sum of the rolls depends on which numbers came up on the first and second rolls. 14. Independent. The outcome of the first ball drawn does not affect the outcome of the second ball drawn. 15. Events: Achievements in math and achievements in music; Dependent. There is a strong relationship between achievements in math and music. 16. Events: Use of talc powder and incidence of ovarian cancer in women; Independent. There is no significant association between the use of talc powder and the incidence of ovarian cancer in women. 17. Events: Playing violent video games, aggressive or bullying behavior; Independent. Playing violent video games does not cause aggressive or bullying behavior in teens. 18. Events: High levels of self-leadership traits and attributing successes to their own efforts; Dependent. Business executives with high levels of self-leadership traits are more likely to attribute successes to their own efforts.

CHAPTER 3 Ň PROBABILITY

117

19. Let A = {card is a heart} and B = {card is a club | a heart card has been removed from the deck}. Since the card was not replaced, there are only 51 cards remaining for the second selection. 13 1 13 § 1 · § 13 · P( A) = = and P( B) = . Thus, P ( A and B ) = ¨ ¸ ⋅ ¨ ¸ ≈ 0.064 52 4 51 © 4 ¹ © 51 ¹ 20. Let A = {coin is tail} and B = {number on die is greater than 2}. 1 4 2 §1· §2· P( A) = and P( B) = = . Thus, P ( A and B ) = ¨ ¸ ⋅ ¨ ¸ ≈ 0.333 2 6 3 ©2¹ ©3¹ 21. Let A = {woman carries mutation of BRCA1 gene} and B = {woman develops breast cancer}. 1 64 § 1 · § 64 · P( A) = and P( B | A) = . Thus, P ( A and B ) = P( A) ⋅ P( B | A) = ¨ ¸⋅¨ ¸ = 0.0016 400 100 © 400 ¹ © 100 ¹ 22. Let A = {adult drives a Ford} and B = {adult drives a pickup truck}. 3 9 § 3 · § 9 · P( A) = and P( B | A) = . Thus, P ( A and B ) = P( A) ⋅ P( B | A) = ¨ ¸ ⋅ ¨ ¸ = 0.0675 20 20 © 20 ¹ © 20 ¹ 23. Let A = {first probable voter would like entertainers to address social and political issues} and B = {second probable voter would like entertainers to address social and political issues}. 828 827 P( A) = and P( B | A) = 1103 1102

§ 828 · § 827 · (a) P ( A and B ) = P( A) ⋅ P( B | A) = ¨ ¸⋅¨ ¸ ≈ 0.563 © 1103 ¹ © 1102 ¹ § 275 · § 274 · (b) P ( A ' and B ') = P( A ') ⋅ P( B ' | A ') = ¨ ¸⋅¨ ¸ ≈ 0.062 © 1103 ¹ © 1102 ¹ (c) P(at least one of the two probable voters would like entertainers to address social and political issues) = 1 − P ( A ' and B ') ≈ 1 − 0.062 = 0.938 24. Let A = {first child has lost a friend or relative to murder}, B = {second child has lost a friend or relative to murder}, C = {third child has lost a friend or relative to murder}, and D = {fourth child has lost a friend or relative to murder} 942 941 940 939 , P ( B | A) = , P (C | A and B ) = , and P ( D | A and B and C ) = P ( A) = 11,771 11,770 11,769 11,768 (a) P ( A and B and C and D ) = P( A) ⋅ P( B | A) ⋅ P(C | A and B) ⋅ P( D | A and B and C )

§ 942 · § 941 · § 940 · § 939 · =¨ ¸⋅¨ ¸⋅¨ ¸⋅¨ ¸ ≈ 0.00004 © 11,771 ¹ © 11,770 ¹ © 11,769 ¹ © 11,768 ¹

118

CHAPTER 3 Ň PROBABILITY

(b) P ( A ' and B ' and C ' and D ') = P( A ') ⋅ P( B ' | A ') ⋅ P(C ' | A ' and B ') ⋅ P( D ' | A ' and B ' and C ')

§ 10,829 · § 10,828 · § 10,827 · § 10,826 · =¨ ¸⋅¨ ¸⋅¨ ¸⋅¨ ¸ ≈ 0.716 © 11,771 ¹ © 11,770 ¹ © 11,769 ¹ © 11,768 ¹ (c) P(at least one of the four has lost a friend or relative to murder) = 1 − P ( A ' and B ' and C ' and D ') ≈ 1 − 0.716 = 0.284 25. Let A = {first adult said Barack Obama was the best president} and B = {second adult said Barack Obama was the best president}. 270 269 P( A) = and P( B | A) = . 1500 1499

§ 270 · § 269 · (a) P ( A and B ) = P( A) ⋅ P( B | A) = ¨ ¸⋅¨ ¸ ≈ 0.032 © 1500 ¹ © 1499 ¹ § 1230 · § 1229 · (b) P ( A ' and B ') = P( A ') ⋅ P( B ' | A ') = ¨ ¸⋅¨ ¸ ≈ 0.672 © 1500 ¹ © 1499 ¹ (c) P ( at least one of two adults says Barack Obama was the best president )

= 1 − P( A ' and B ') ≈ 1 − 0.672 = 0.328 (d) The event in part (a) is unusual because its probability is less than or equal to 0.05. 26. Let A = {first adult said Donald Trump was the worst president}, B = {second adult said Donald Trump was the worst president}, and C = {third adult said Donald Trump was the worst president}. 690 689 688 P( A) = , P( B | A) = , and P(C | A and B) = 1500 1499 1498

§ 690 · § 689 · § 688 · (a) P ( A and B and C ) = P( A) ⋅ P( B | A) ⋅ P(C | A and B) = ¨ ¸⋅¨ ¸⋅¨ ¸ ≈ 0.097 © 1500 ¹ © 1499 ¹ © 1498 ¹ § 810 · § 809 · § 808 · (b) P ( A ' and B 'and C ') = P( A ') ⋅ P( B ' | A ') ⋅ P(C ' | A ' and B ') = ¨ ¸⋅¨ ¸⋅¨ ¸ ≈ 0.157 © 1500 ¹ © 1499 ¹ © 1498 ¹ (c) P ( at most two adults say Donald Trump was the worst president )

= 1 − P( A and B and C ) ≈ 1 − 0.097 = 0.903 (d) None of the events can be considered unusual because they are all over 0.05. 27. (a) P ( all six have O+ ) = (0.39) ⋅ (0.39) ⋅ (0.39) ⋅ (0.39) ⋅ (0.39) ⋅ (0.39) ≈ 0.004 (b) P ( none have O+ ) = (0.61) ⋅ (0.61) ⋅ (0.61) ⋅ (0.61) ⋅ (0.61) ⋅ (0.61) ≈ 0.052

CHAPTER 3 Ň PROBABILITY

119

(c) P ( at least one has O+ ) = 1 − P(none have O+) ≈ 1 − 0.052 = 0.948 (d) The event in part (a) is unusual because the probability is less than or equal to 0.05. 28. (a) P ( all four have A+ ) = (0.29) ⋅ (0.29) ⋅ (0.29) ⋅ (0.29) ≈ 0.007 (b) P ( none have A+ ) = (0.71) ⋅ (0.71) ⋅ (0.71) ⋅ (0.71) ≈ 0.254 (c) P ( at least one has A+ ) = 1 − P(none have A+) ≈ 1 − 0.254 = 0.746 (d) The event in part (a) is unusual because the probability is less than or equal to 0.05. 29. Let A = {ART procedure resulted in pregnancy} and B = {ART pregnancy resulted in multiple birth}. P ( A) = 0.019 and P ( B | A) = 0.264 (a) P ( A and B ) = P( A) ⋅ P( B | A) = ( 0.019 ) ⋅ ( 0.264 ) ≈ 0.005 (b) P ( B ' | A) = 1 − P ( B | A) = 1 − 0.264 = 0.736 (c) Yes, this is unusual because the probability is less than or equal to 0.05. 30. Let A = {have taken a standardized test} and B = {do not plan to submit test score with application}. P ( A) = 0.578 and P ( B | A) = 0.356 (a) P ( A and B ) = P( A) ⋅ P( B | A) = ( 0.578) ⋅ ( 0.356 ) ≈ 0.206 (b) P ( A and B ') = P( A) ⋅ P( B ' A) = P( A) ⋅ (1 − P( B A)) = (0.578) ⋅ (0.644) ≈ 0.372 (c) No, neither is unusual because neither has a probability of less than or equal to 0.05. 31. Let A = {use social media to look up college} and B = {use Instagram}. P ( A) = 0.688 and P ( B | A) = 0.89

P ( A and B ) = P( A) ⋅ P( B | A) = ( 0.688) ⋅ ( 0.89 ) ≈ 0.612

32. Let A = {survives bypass surgery} and B = {make full recovery}. P ( A) = 0.60 and P ( B | A) = 0.70

P ( A and B ') = P( A) ⋅ P( B ' A) = P( A) ⋅ (1 − P( B A)) = (0.60) ⋅ (0.30) = 0.18

33. P( A B) =

P( A) ⋅ P( B A) P( A) ⋅ P( B A) + P( A ') ⋅ P( B A ')

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CHAPTER 3 Ň PROBABILITY

34. P( A B) =

P( A) ⋅ P( B A) P( A) ⋅ P( B A) + P( A ') ⋅ P( B A ')

§3· § 2· 1 ¨ 8 ¸⋅¨ 3 ¸ 2 © ¹ © ¹ = = 4 = = 0.4 5 5 §3· § 2· §5· §3· ¨ ¸⋅¨ ¸ + ¨ ¸⋅¨ ¸ 8 8 3 8 5 © ¹ © ¹ © ¹ © ¹ 35. P( A B) =

36. P( A B) =

37. P( A B) =

38. P( A B) =

39. P( A) =

P ( A) ⋅ P( B A) P ( A) ⋅ P ( B A) + P ( A ') ⋅ P ( B A ') (0.25) ⋅ (0.3) 0.075 = ≈ 0.167 (0.25) ⋅ (0.3) + (0.75) ⋅ (0.5) 0.45 P ( A) ⋅ P( B A) P ( A) ⋅ P ( B A) + P( A ') ⋅ P( B A ') (0.62) ⋅ (0.41) 0.2542 = ≈ 0.797 (0.62) ⋅ (0.41) + (0.38) ⋅ (0.17) 0.3188 P( A) ⋅ P ( B A) P ( A) ⋅ P ( B A) + P ( A ') ⋅ P ( B A ') (0.73) ⋅ (0.46) 0.3358 = ≈ 0.792 (0.73) ⋅ (0.46) + (0.17) ⋅ (0.52) 0.4242 P ( A) ⋅ P( B A) P ( A) ⋅ P ( B A) + P( A ') ⋅ P( B A ') (0.12) ⋅ (0.66) 0.0792 = ≈ 0.321 (0.12) ⋅ (0.66) + (0.88) ⋅ (0.19) 0.2464

1 = 0.005 , P( B A) = 0.8 , and P( B A ') = 0.05 200

(a) P( A B) = =

P( A) ⋅ P ( B A) P ( A) ⋅ P( B A) + P ( A ') ⋅ P ( B A ') (0.005) ⋅ (0.8) 0.004 = ≈ 0.074 (0.005) ⋅ (0.8) + (0.995) ⋅ (0.05) 0.05375

(b) P( A ' B ') = =

P ( A ') ⋅ P ( B ' A ') P ( A ') ⋅ P ( B ' A ') + P ( A) ⋅ P ( B ' A) (0.995) ⋅ (0.95) 0.94525 = ≈ 0.999 (0.995) ⋅ (0.95) + (0.005) ⋅ (0.2) 0.94625

CHAPTER 3 Ň PROBABILITY

40. (a) P ( different birthdays ) =

365 364 363 362 343 342 ⋅ ⋅ ⋅ ⋅ ... ⋅ ⋅ ≈ 0.462 365 365 365 365 365 365

(b) P(at least two have same birthday) = 1 – P(different birthdays) ≈ 1 − 0.462 = 0.538 (c) Answers will vary. 41. Let A = {flight departs on time} and B = {flight arrives on time}. P( A and B) 0.83 P( A B) = = ≈ 0.954 P( B) 0.87 42. Let A = {flight departs on time} and B = {flight arrives on time}. P( A and B) 0.83 P( B A) = = ≈ 0.933 P( A) 0.89

3.3 THE ADDITION RULE 3.3 TRY IT YOURSELF SOLUTIONS 1. (1) Not mutually exclusive; The events can occur at the same time.

(2) Mutually exclusive; The events cannot occur at the same time. 2. (1) Mutually exclusive Let A = {6} and B = {odd}. 1 3 1 P( A) = and P( B) = = 6 6 2 1 1 P( A or B) = P( A) + P( B) = + ≈ 0.667 6 2

(2) Not mutually exclusive Let A = {face card} and B = {heart}. 12 13 3 P( A) = , P( B) = , and P( A and B) = 52 52 52 12 13 3 P( A or B) = P( A) + P( B) − P( A and B) = + − ≈ 0.423 52 52 52 3. Let A = {sales between $0 and $24,999} Let B = {sales between $25,000 and $49,999}. A and B cannot occur at the same time. So, A and B are mutually exclusive. 3 5 P( A) = and P( B) = 36 36 3 5 P( A or B) = P( A) + P( B) = + ≈ 0.222 36 36

121

122

CHAPTER 3 Ň PROBABILITY

4. (1) Let A = {type B} and B = {type AB}. Let A = {type B} and B = {type AB}. A and B cannot occur at the same time. So, A and B are mutually exclusive. 45 16 P( A) = and P( B) = 409 409 45 16 P( A or B) = P( A) + P( B) = + ≈ 0.149 409 409

(2) If a donor does not have type O or type A blood, then the donor must have type B or type AB blood. See part (1). The probability is about 0.149. (3) Let A = {type O} and B = {Rh-positive}. A and B can occur at the same time. So, A and B are not mutually exclusive. 184 344 156 P( A) = , P( B) = , and P( A and B) = 409 409 409 184 344 156 P( A or B) = P( A) + P( B) − P( A and B) = + − ≈ 0.910 409 409 409 (4) Let A = {type A} and B = {Rh-negative}. A and B can occur at the same time. So, A and B are not mutually exclusive. 164 65 25 P( A) = , P( B) = , and P( A and B) = 409 409 409 164 65 25 P( A or B) = P( A) + P( B) − P( A and B) = + − ≈ 0.499 409 409 409 5. Let A = {linebacker} and B = {quarterback}. 28 13 P( A or B) = P( A) + P( B) = + ≈ 0.161 255 255 P(not a linebacker or quarterback) = 1 – P(A or B) ≈ 1 − 0.161 = 0.839

3.3 EXERCISE SOLUTIONS 1. P(A and B) = 0 because A and B cannot occur at the same time. 2. (a) Sample answer: Toss coin once: A = {head} and B = {tail}

(b) Sample answer: Draw one card: A = {ace} and B = {spade} 3. True 4. False. Two events being independent does not imply they are mutually exclusive. Example: Toss a 1 coin then roll a six-sided die. Let A = {head} and B = {6 on die}. P( B A) = = P( B) implies A and B 6 1 implies A and B are not mutually exclusive. are independent events. However, P(A and B) = 12 5. False. The probability that event A or event B will occur is P ( A or B ) = P ( A) + P ( B ) − P ( A and B ) . Copyright © 2023 Pearson Education, Inc.

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123

6. True 7. Not mutually exclusive. People can wear a polo shirt while wearing Bermuda shorts. 8. Mutually exclusive. A student cannot get “A”s for the semester in four out of their five classes and have a 0.0 GPA for the semester. 9. Not mutually exclusive. A freshman music major can be 20 years old. 10. Mutually exclusive. A student cannot have a birthday in both months. 11. Mutually exclusive. A voter cannot be both a Republican and a Democrat. 12. Not mutually exclusive. A member of Congress can be a male senator. 13. Let A = {physics major} and B = {minoring in math}. 12 16 3 P( A) = , P( B) = , and P ( B | A) = 40 40 12 12 3 3 ⋅ = Thus, P( A and B) = P( A) ⋅ P( B | A) = 40 12 40 12 16 3 25 P( A or B) = P( A) + P( B) − P( A and B) = + − = = 0.625 40 40 40 40 14. Let A = {college professor} and B = {male}. 3120 3595 1505 P( A) = , P( B) = , and P( B | A) = 6855 6855 3120 3120 1505 1505 ⋅ = Thus, P( A and B) = P( A) ⋅ P( B | A) = 6855 3120 6855 3120 3595 1505 P( A or B) = P( A) + P( B) − P( A and B) = + − ≈ 0.760 6855 6855 6855 15. Let A = {puncture} and B = {smashed corner}. P ( A) = 0.05 , P ( B ) = 0.08 , and P ( A and B ) = 0.004 . P ( A or B ) = P ( A) + P ( B ) − P ( A and B ) = 0.05 + 0.08 − 0.004 = 0.126 16. Let A = {does not have puncture} and B = {does not have smashed corner}. P ( A) = 0.96 , P ( B ) = 0.93 , and P ( A and B ) = 0.893 . P ( A or B ) = P ( A) + P ( B ) − P ( A and B ) = 0.96 + 0.93 − 0.893 = 0.997 17. (a) P ( club or 3) = P ( club ) + P ( 3) – P ( club and 3) =

13 4 1 + − ≈ 0.308 52 52 52

(b) P ( red or king ) = P ( red ) + P ( king ) – P ( red and king ) =

26 4 2 + − ≈ 0.538 52 52 52

124

CHAPTER 3 Ň PROBABILITY

4 12 + − 0 ≈ 0.308 52 52

18. (a) P ( 5 or greater than 3) = P ( 5) + P ( greater than 3) – P ( 5 and greater than 3)

1 3 1 + − = 0.5 6 6 6

(b) P ( 2 or odd ) = P ( 2 ) + P ( odd ) – P ( 2 and odd ) =

1 3 + − 0 ≈ 0.667 6 6

3 3 1 + − ≈ 0.833 6 6 6

19. (a) P ( under 10 ) = 0.121

(b) P ( 50 + ) = P ( 50 − 59 ) + P ( 60 − 69 ) + P ( 70 − 79 ) + P ( 80 + ) = 0.116 + 0.119 + 0.086 + 0.045 = 0.366 (c) P ( not 60 + ) = 1 − P ( 60 + ) = 1 − P(60 − 69) − P(70 − 79) − P(80+) = 1 − 0.119 − 0.086 − 0.045 = 0.750 (d) P ( not between 20 and 39 ) = 1 − P ( between 20 and 39 ) = 1 − P(20 − 29) − P(30 − 39) = 1 − 0.130 − 0.138 = 0.732 20. (a) P ( never used ) = 0.587

(b) P ( used ) = 1 − P ( never used ) = 1 − 0.587 = 0.413 (c) P ( used within last 30 days ) = P ( 2 weeks to 30 days ago ) + P ( less than 2 weeks ago ) = 0.032 + 0.127 = 0.159 (d) P ( not used within last 12 months )

= P ( more than 1 year ago ) + P ( never used ) = 0.138 + 0.587 = 0.725

21. (a) P ( not liberal or very liberal ) = 1 – P ( liberal or very liberal ) = 1 −

405 1095 = = 0.730 1500 1500

(b) P ( conservative or very conservative ) = P ( conservative ) + P ( very conservative ) =

285 195 480 + = = 0.320 1500 1500 1500

(c) P ( not very liberal or very conservative ) = 1 – P ( very liberal or very conservative ) = 1 − (d) P ( unsure or moderate ) = P ( unsure ) + P ( moderate ) =

180 435 615 + = = 0.410 1500 1500 1500

360 1140 = = 0.760 1500 1500

CHAPTER 3 Ň PROBABILITY

22. (a) P ( no difference ) =

125

190 ≈ 0.170 1115

(b) P ( very negative impact ) =

256 ≈ 0.230 1115

134 981 = ≈ 0.880 1115 1115

(d) P ( fairly positive or very positive )

= P ( fairly positive ) + P ( very positive ) =

178 134 312 + = ≈ 0.280 1115 1115 1115

23. Let A = {male}; B = {mechanical engineering degree}

(a) P ( A or B ) = P ( A ) + P ( B ) – P ( A and B ) =

94,847 35,182 30,150 + − ≈ 0.819 121,956 121,956 121,956

(b) P ( A 'or B ) = P ( A ' ) + P ( B ) – P ( A 'and B ) =

27,109 35,182 5,032 + − ≈ 0.470 121,956 121,956 121,956

94,847 86,774 64,697 + − ≈ 0.959 121,956 121,956 121,956

24. Let A = {none}, B = {major}, and C = {father}.

(a) P ( A or C ) = P ( A) + P ( C ) – P ( A and C ) =

2428 1144 744 + − ≈ 0.589 4805 4805 4805

(b) P ( B or C ') = P ( B ) + P ( C ') – P ( B and C ') =

491 3661 415 + − ≈ 0.778 4805 4805 4805

1192 + 694 + 2428 3661 1048 + 514 + 1684 + − ≈ 0.984 4805 4805 4805

25. Let A = {frequently}, B = {occasionally}, C = {not at all}, and D = {male}

(a) P ( D or A) = P ( D ) + P ( A) – P ( D and A) =

1084 563 346 + − ≈ 0.718 1811 1811 1811

(b) P ( D' or C ) = P ( D ') + P ( C ) – P ( D' and C ) = (c) P ( A or B ) =

563 495 + ≈ 0.584 1811 1811

727 753 312 + − ≈ 0.645 1811 1811 1811

126

CHAPTER 3 Ň PROBABILITY

(d) P ( D ' or A ') = P ( D ') + P ( A ') – P ( D ' and A ') =

727 495 + 753 198 + 312 + − ≈ 0.809 1811 1811 1811

26. Let A = {only contacts}, B = {only glasses}, C = {both}, D = {neither}, and E = {male}

(a) P ( A or B ) = P ( A) + P ( B ) =

253 1268 + ≈ 0.475 3203 3203

(b) P ( E or C ) = P ( E ) + P ( C ) – P ( E and C ) =

1538 545 177 + − ≈ 0.595 3203 3203 3203

1665 1137 681 + − ≈ 0.662 3203 3203 3203

(d) P ( E or ( A or D ) ) = P ( E ) + P ( A or D ) – P ( E and A) − P ( E and D )

1538 253 + 1137 64 456 + − − ≈ 0.752 3203 3203 3203 3203

27. P ( A or B or C ) = P ( A) + P ( B ) + P ( C ) – P ( A and B ) – P ( A and C )

– P ( B and C ) + P ( A and B and C )

= 0.40 + 0.10 + 0.50 – 0.05 – 0.25 – 0.10 + 0.03 = 0.63

28. P ( A or B or C ) = P ( A) + P ( B ) + P ( C ) – P ( A and B ) – P ( A and C )

– P ( B and C ) + P ( A and B and C )

= 0.38 + 0.26 + 0.14 – 0.12 – 0.03 – 0.09 + 0.01 = 0.55

29. If events A, B, and C are not mutually exclusive, P(A and B and C) must be added because P(A) + P(B) + P(C) counts the intersection of all three events three times and –P(A and B) – P(A and C) – P(B and C) subtracts the intersection of all three events three times. So, if P(A and B and C) is not added at the end, then it will not be counted. 30. No. If two events A and B are independent, then P(A and B) = P ( A) ⋅ P ( B ) . If two events are mutually exclusive, P (A and B) = 0. The only scenario when two events can be independent and mutually exclusive is when P(A) = 0 or P(B) = 0.

3.4 ADDITIONAL TOPICS IN PROBABILITY AND COUNTING 3.4 TRY IT YOURSELF SOLUTIONS 1. n = 10 schools The number of permutations is 10! = 3,628,800.

CHAPTER 3 Ň PROBABILITY

127

2. n = 8, r = 3 8! 8! 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = = = 8 ⋅ 7 ⋅ 6 = 336 (8 − 3)! 5! 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 There are 336 possible ways that the subject can pick a first, second, and third activity. 3. n = 12, r = 4 12! 12! = = 12 ⋅ 11 ⋅ 10 ⋅ 9 = 11,880 12 P4 = (12 − 4)! 8! 4.

n! 20! = = 77,597,520 n1 !⋅ n2 !⋅ n3 ! 6!⋅ 9!⋅ 5!

20 C3 =

20 P2 =

15 C5 =

20! 20! 20 ⋅ 19 ⋅ 18 ⋅ 17! = = = 1140 (20 − 3)!3! 17!3! 17!⋅ 3!

20! 20! 20 ⋅ 19 ⋅ 18! = = = 20 ⋅ 19 = 380 (20 − 2)! 18! 18! 1 P ( selecting the two members ) = ≈ 0.003 380

15! = 3003 5!(15 − 5)! 54! = 3,162,510 54 C5 = 5!(54 − 5)! C5 3003 == ≈ 0.0009 3,162,510 54 C5

P ( five diamonds from deck with two jokers ) = 15

5 C3 ⋅7 C0 = 12 C3 =

5! 7! ⋅ = 10 ⋅1 = 10 3!2! 0!7!

12! = 220 3!9!

P ( all three are men ) =

10 ≈ 0.045 220

3.4 EXERCISE SOLUTIONS 1. The number of ordered arrangements of n objects taken r at a time. Sample answer: An example of a permutation is the number of seating arrangements of you and three friends. 2. The number of ways to select r of the n objects without regard to order. Sample answer: An example of a combination is the number of selections of different playoff teams from a volleyball tournament. Copyright © 2023 Pearson Education, Inc.

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CHAPTER 3 Ň PROBABILITY

3. False. A permutation is an ordered arrangement of objects. 4. True

5. True

9 P5 =

9! 9! = = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 = 15,120 (9 − 5)! 4!

14 P3 =

14! 14! = = 14 ⋅ 13 ⋅ 12 = 2184 (14 − 3)! 11!

8 C3 =

8! 8! 8 ⋅ 7 ⋅ 6 ⋅ 5! = = = 56 (8 − 3)!3! 5!3! 5!⋅ 3!

10. 21 C8 =

6. True

21! 21! 21 ⋅ 20 ⋅ 19 ⋅ 18 ⋅ 17 ⋅ 16 ⋅ 15 ⋅ 14 ⋅ 13! = = = 203, 490 (21 − 8)!8! 13!8! 13!⋅ 8!

8! 8! (8 − 4)!4! 4!4! 70 8 C4 11. = = = ≈ 0.076 12! 12! 924 12 C6 (12 − 6)!6! 6!6! 10! 10! (10 − 7)!7! 3!7! 120 10 C7 12. = = = ≈ 0.035 14! 14! 3432 14 C7 (14 − 7)!7! 7!7! 3! 3! P 3⋅ 2 (3 − 2)! 13. 3 2 = = 1! = ≈ 0.462 13! 13! 13 13 P1 (13 − 1)! 12! 7! 7! P 7⋅6⋅5 (7 − 3)! 14. 7 3 = = 4! = ≈ 0.018 12! 12! 12 ⋅ 11 ⋅ 10 ⋅ 9 12 P 4 (12 − 4)! 8!

15. Permutation. The order of the 16 floats in line matters. 16. Combination. The order of the committee members does not matter. 17. Combination. The order does not matter because the position of one captain is the same as the other. 18. Permutation. The order of the letters matters in the password.

CHAPTER 3 Ň PROBABILITY

19. 7! = 5040

20. 8! = 40,320

21. 6! = 720

22. 10! = 3,628,800

23. 72 P3 =

72! 72! = = 357,840 (72 − 3)! 69!

24. 16 P5 =

16! 16! = = 524,160 (16 − 5)! 11!

25. 21 P6 =

21! 21! = = 39,070,080 (21 − 6)! 15!

26. 38 P4 =

38! 38! = = 1,771,560 (38 − 4)! 34!

17! = 2,042,040 8!6!3!

28.

19! = 11,085,360 9!3!7!

29. 3 S’s, 3 T’s, 1 A, 2 I’s, 1 C

30.

8! = 56 5!3!

27.

10! = 50,400 3!3!1!2!1! 31. 20 C10 =

20! = 184,756 20 − ( 10 )!10!

32. 36 C12 =

36! = 1, 251,677,700 (36 − 12)!12!

33. 40 C3 =

40! = 9880 (40 − 3)!3!

34. 52 C6 =

52! = 20,358,520 (52 − 6)!6!

35. 10 C1 ⋅8 C2 ⋅13 C1 =

36. 8 C6 ⋅6 C3 ⋅7 C3 =

10! 8! 13! ⋅ ⋅ = 3640 (10 − 1)!1! (8 − 2)!2! (13 − 1)!1!

8! 6! 7! ⋅ ⋅ = 19,600 (8 − 6)!6! (6 − 3)!3! (7 − 3)!3!

37. 5 C1 ⋅75 C5 =

5! 75! ⋅ = 86, 296,950 1!4! 70!5!

38. 6 C3 ⋅18 C5 =

6! 18! ⋅ = 171,360 3!3! 13!5!

39.

1 1 = ≈ 0.024 42 7 P2

40.

1 1 = = 0.002 506 23 P2

41.

1 1 = ≈ 0.008 120 10 C3

42.

1 1 = ≈ 0.012 84 9 C3

43. (a)

15 C3 56 C3

455 ≈ 0.016 27,720

C3 10,660 = ≈ 0.385 27,720 56 C3

(b) 41

129

130

CHAPTER 3 Ň PROBABILITY

44. (a)

6 C4 14 C4

15 ≈ 0.015 1001

(b)

8 C4 14 C4

70 ≈ 0.070 1001

45. (3.7%)(513) = (0.037)(513) ≈ 19 of the 513 say their primary source for new music is someone’s playlist. C 171 P ( two say playlist ) = 19 2 = ≈ 0.001 131, 328 513 C2 46. (26.1%)(513) = (0.261)(513) ≈ 134 of the 513 say their primary source for new music is subscription music services. C P ( three say subscription music services ) = 134 3 ≈ 0.018 513 C3 47. (15.8%)(513) = (0.158)(513) ≈ 81 of the 513 say their primary source for new music is friends or social media. 513 − 81 = 432 do not say their primary source for new music is friends or social media. C P ( none of the nine say friends or social media ) = 432 9 ≈ 0.210 513 C9 48. (45.9%)(513) = (0.459)(513) ≈ 235 of the 513 say their primary source for new music is personal search. 513 − 235 = 278 do not say their primary source for new music is personal search. C P ( none of the five say personal search ) = 278 5 ≈ 0.046 513 C5 49. 40 C5 = 658, 008 50. (a)

P (win) =

1 ≈ 0.0000015 658,008

200 C5 = 2,535,650,040

(b) 56 C5 = 3,819,816

C5 ≈ 0.190 ; This would be a biased sample because the probability that no 200 C5 males are selected is small.

(d) They could choose two from the males and three from the females seeing as there are more females than males. 51.

( 24 C5 )( 30 C3 ) = (42,504)(4060) ≈ 0.166

52.

( 17 C3 )( 37 C5 )

53.

( 13 C4 ) ( 41 C4 ) = (715)(101, 270) ≈ 0.070

54 C8

1,040, 465,790

(680)(435,897) ≈ 0.285 1,040, 465,790

1,040, 465,790

CHAPTER 3 Ň PROBABILITY

54.

55.

56.

( 17 C2 )( 13 C2 ) ( 24 C4 ) 54 C8 8 C3 10 C3 95 C4 100 C4

(136)(78)(10,626) ≈ 0.108 1,040, 465,790

( 8 C2 ) ( 2 C1 ) = 56 + (28)(2) = 112 ≈ 0.933 10 C3

120

( 95 C3 )( 5 C1 ) = 3,183,545 + (138, 415)(5) ≈ 100 C4

3,921, 225

0.988

§ ( C )( C ) ( C )( C ) · (1)(1)(1)(1) 57. 4 C2 ¨¨ 2 2 2 2 2 0 2 0 ¸¸ = 6 ⋅ ≈ 0.086 70 8 C4 © ¹

58.

4! § ( 2 C2 )( 2 C1 )( 2 C1 ) ( 2 C0 ) · (1)(2)(2)(1) ≈ 0.686 ¨¨ ¸¸ = 12 ⋅ 2!1!1! © 70 8 C4 ¹

59.

( 13 C2 )( 13 C1 )( 13 C1 )( 13 C1 ) = (78)(13)(13)(13) ≈ 0.066 52 C5

2,598,960

§ ( C )( C ) · § (1)(48) · 60. ( 13 C1 ) ¨¨ 4 4 48 1 ¸¸ = 13 ⋅ ¨ ¸ ≈ 0.0002 © 2,598,960 ¹ 52 C5 © ¹ § ( C )( C ) · § (4)(6) · 61. ( 13 C2 ) ¨¨ 4 3 4 2 ¸¸ = 78 ¨ ¸ ≈ 0.001 © 2,598,960 ¹ 52 C5 © ¹

62.

( 13 C1 ) § ( 4 C3 )( 48 C1 ) ( 44 C1 ) · = 13 ⋅ § (4)(48)(44) · ≈ 0.021 2

¨¨ ©

52 C5

¸¸ ¹

¨ ¸ 2 © 2,598,960 ¹

131

132

CHAPTER 3 Ň PROBABILITY

CHAPTER 3 REVIEW EXERCISE SOLUTIONS 1. Sample space: {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}

Event: Getting three heads {HHHT, HHTH, HTHH, THHH} There are four outcomes. 2. Sample space: {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2,), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Event: Sum of four or five {(1, 3), (1, 4), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)} There are seven outcomes.

CHAPTER 3 Ň PROBABILITY

133

3. Sample space: {January, February, March, April, May, June, July, August, September, October, November, December} Event: {January, June, July} There are three outcomes. 4. Sample space: {GGG, GGB, GBG, GBB, BGG, BGB, BBG, BBB}

Event: The family has two boys. {BGB, BBG, GBB} There are three outcomes. 5.

( 7 )( 4 )( 3) = 84

6. ( 26 )( 26 )( 26 )(10 )(10 )(10 )(10 ) = 175,760,000

7. Empirical probability, because prior counts were used to calculate the frequency of a part being defective. 8. Classical probability, because each outcome in the sample space is equally likely to occur. 9. Subjective probability, because it is based on opinion. 10. Empirical probability, because it is based on the observations obtained from the probability experiments. 11. Classical probability, because all the outcomes in the event and the sample space can be counted. 12. Empirical probability, because it is based on the observations obtained from the probability experiments. 13. P ( more than 4 ) = P ( 5-9) + P (10-19 ) + P ( 20-99) + P (100+ ) = 0.115 + 0.049 + 0.04 + 0.011 = 0.215 14. P ( fewer than 10 ) = P(1 − 4) + P(5 − 9) = 0.785 + 0.115 = 0.900 15.

1 = 1.25 × 10−7 (8)(10)(10)(10)(10)(10)(10)

16. 1 −

1 = 0.999999875 (8)(10)(10)(10)(10)(10)(10)

17. P first failed

27, 522 x 0.315 87, 297

134

CHAPTER 3 Ň PROBABILITY

18. P ( passed | repeated ) =

44,983 ≈ 0.429 104,758

19. Independent. The outcomes of the first four coin tosses do not affect the outcome of the fifth coin toss. 20. Dependent. The outcome of selecting an ace affects the outcome of selecting a jack. 21. Dependent. The outcome of taking a driver’s education course affects the outcome of passing the driver’s license exam. 22. Dependent. The outcome of getting high grades affects the outcome of being awarded an academic scholarship. 23. P ( correct toothpaste and correct dental rinse ) = P ( correct toothpaste ) ⋅ P ( correct dental rinse ) 1 1 1 = P ( correct toothpaste ) P ( correct dental rinse ) = ⋅ = = 0.025 8 5 40 The event is unusual because its probability is less than or equal to 0.05.

6 12 ⋅ ≈ 0.235 18 17 The event is not unusual because its probability is not less than or equal to 0.05.

24. P ( black and ( blue or white ) ) = P ( black ) ·P ( blue or whiteūblack ) =

25. Mutually exclusive. A jelly bean cannot be both completely red and completely yellow. 26. Not mutually exclusive. A person who loves cats can also own a dog. 27. P ( home or work ) = P ( work ) + P ( home ) – P ( home and work ) = 0.74 + 0.88 – 0.72 = 0.9 28. P ( black or sedan ) = P ( black ) + P ( sedan ) – P ( black and sedan ) = 29. P ( 4 − 8 or club ) = P ( 4 − 8) + P ( club ) – P ( 4 − 8 and club ) =

1560 3120 1170 + – = 0.54 6500 6500 6500

20 13 5 + − ≈ 0.538 52 52 52

30. P ( red or queen ) = P ( red ) + P ( queen ) – P ( red and queen ) =

26 4 2 + − ≈ 0.538 52 52 52

31. P ( odd or less than 4 ) = P ( odd ) + P ( less than 4 ) – P ( odd and less than 4 ) = 32. P ( multiple of 3 or greater than 5 )

= P ( multiple of 3) + P ( greater than 5) – P ( multiple of 3 and greater than 5) =

2 3 1 4 + − = = 0.5 8 8 8 8

6 3 2 + − ≈ 0.583 12 12 12

CHAPTER 3 Ň PROBABILITY

33. P ( fewer than 500 ) = P ( 300 − 499 ) + P ( fewer than 300 ) = 0.281 + 0.298 = 0.579 34. P ( 300 or more ) = 1 − P ( fewer than 300 ) = 1 − 0.298 = 0.702 35. P(more than 20) =

36. P(less than 11) =

714 867 + ≈ 0.180 8806 8806

1654 2452 + ≈ 0.466 8806 8806

37. P(not 6 to 10 hours) = 1 − P (6 to 10 hours) = 1 −

2452 ≈ 0.722 8806

38. P (not more than 15 hours) = 1 − [ P (16 to 20 hours) + P (21 to 25 hours) + P (25 + hours)]

867 · § 1309 714 =1− ¨ + + ¸ ≈ 0.672 © 8806 8806 8806 ¹ 39. No. You do not know whether events A and B are mutually exclusive. 40. Yes. The Addition Rule is P(A or B) = P(A) + P(B) – P(A and B). Substitute the given values for P(A or B) and P(A) + P(B) and then solve for P(A and B). 41. 11 P2 =

11! 11! = = 11 ⋅ 10 = 110 (11 − 2)! 9!

42. 8 P6 =

8! 8! = = 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 = 20,160 (8 − 6)! 2!

43. 7 C4 =

7! 7! = = 35 (7 − 4)!4! 3!4!

5! 5! C (5 − 3)!3! 2!3! 10 44. 5 3 = = = ≈ 0.083 10! 10! 120 C 10 3 (10 − 3)!3! 7!3!

45. Order is important: 15 P3 =

15! 15! = = 2730 (15 − 3)! 12!

46. 5! = 120 47. Order is not important: 17 C4 =

17! = 2380 (17 − 4)!4!

135

136

CHAPTER 3 Ň PROBABILITY

48. Order is not important: 13 C2 =

13! = 78 (13 − 2)!2!

49. P ( three kings and two queens ) =

50.

( 4 C3 ) ⋅ ( 4 C2 ) = 52 C5

4⋅6 ≈ 0.000009 2,598,960

1 1 = ≈ 0.000006 (23)(26)(26)(10) 155, 480 C3 1, 254, 890 = ≈ 0.955 1,313, 400 200 C3

51. (a) P ( no defective ) = 197

(b) P ( all defective ) =

3 C3 200 C3

1 ≈ 0.0000008 1,313, 400

(c) P ( at least one defective ) = 1 – P ( no defective ) ≈ 1 − 0.955 = 0.045 (d) P ( at least one non-defective ) = 1 – P ( all defective ) ≈ 1 − 0.0000008 ≈ 0.9999992

C3 56 = ≈ 0.006 9880 40 C3

52. (a) P ( all three receive an A ) = 8

C3 4060 = ≈ 0.411 9880 40 C3

(b) P ( all three receive a C or better ) = 30

C3 120 = ≈ 0.012 9880 40 C3

C3 1540 = ≈ 0.156 9880 40 C3

(d) P ( all three receive a B or a C ) = 22

C4 15 = ≈ 0.071 210 10 C4

53. (a) P ( four men ) = 6

C4 1 = ≈ 0.005 210 10 C4

(b) P ( four women ) = 4

( 6 C2 ) ⋅ ( 4 C2 )

(d) P ( one man and three women ) =

10 C4

15 ⋅ 6 ≈ 0.429 210

( 6 C1 ) ⋅ ( 4 C3 ) = 6 ⋅ 4 ≈ 0.114 10 C4

210

CHAPTER 3 Ň PROBABILITY

137

CHAPTER 3 QUIZ SOLUTIONS 1.

( 9)(10)(10)(10)(10)( 5) = 450,000

2. (a) P ( bachelor’s degree ) =

395.8 ≈ 0.700 565.4

(b) P(bachelor's degree computer science/engineering) =

220.3 ≈ 0.650 338.7

175.5 ≈ 0.774 226.7

(d) P ( bachelor’s degree or master's degree ) = P ( bachelor’s degree ) + P ( master's degree )

395.8 140.2 + ≈ 0.948 565.4 565.4

(e) P(doctorate degree computer science/engineering) =

13.0 ≈ 0.038 338.7

(f) P ( master’s degree or degree in natural sciences/mathematics )

= P ( master’s degree) + P (degree in natural sciences/mathematics )

− P ( master’s degree and degree in natural sciences/mathematics )

140.2 226.7 34.8 332.1 + − = ≈ 0.587 565.4 565.4 565.4 565.4

(g) P ( bachelor’s degree and degree in natural sciences/mathematics ) = (h) P(degree in computer science/engineering bachelor's degree) =

175.5 ≈ 0.310 565.4

220.3 ≈ 0.557 395.8

3. The event in part (e) is unusual because its probability is less than or equal to 0.05. 4. Not mutually exclusive. A bowler can have the highest game in a 40-game tournament and still lose the tournament. Dependent. One event can affect the occurrence of the second event. 5.

40 P4 =

6. (a)

40! = 2,193,360 (40 − 4)!

247 C3 =

247! = 2, 481,115 (247 − 3)!3!

138

CHAPTER 3 Ň PROBABILITY

(b) 3 C3 =

3! =1 (3 − 3)!3!

7. (a)

(c)

247 C3 250 C3

2, 481,115 ≈ 0.964 2,573,000

250 C3 −3 C3 250 C3

250! − 1 = 2,573,000 − 1 = 2,572,999 (250 − 3)!3!

(b)

3 C3 250 C3

1 ≈ .0000004 2,573,000

2,572,999 ≈ 0.9999996 2,573,000

CHAPTER 3 TEST SOLUTIONS

15! C 455 (15 − 3)!3! 15 3 = = ≈ 0.010 65! 43,680 65 C3 (65 − 3)!3!

2. (a)

1 1 = ≈ 0.00000015 (26)(26)(10)(10)(10)(10) 6,760,000

(b) ≈ 1 − 0.00000015 = 0.99999985 (c)

1 1 = ≈ 0.000008 (1)(26)(10)(10)(10)(5) 130,000

(d) Classical probability, because each outcome in the sample space is equally likely to occur. 3. Mutually exclusive. The month of February does not have 30 days. 4. (a) P(ninth grade) =

489.5 ≈ 0.278 1758.6

489.5 + 451.6 941.1 = ≈ 0.535 1758.6 1758.6 46.7 ≈ 0.241 (c) P(11th | Oklahoma) = 193.5 (b) P(ninth or 10th) =

(d) P(Texas | 12th) =

353.3 ≈ 0.889 397.4

CHAPTER 3 Ň PROBABILITY

(e) P(11th or Oklahoma) = P(11th) + P(Oklahoma) – P(11th and Oklahoma) 420.1 193.5 46.7 566.9 = + − = ≈ 0.322 1758.6 1758.6 1758.6 1758.6 (f) P(12th and Oklahoma) =

44.1 ≈ 0.025 1758.6

5. The event in part (f) is unusual because its probability is less than or equal to 0.05. 6. The events are dependent, because P (ninth grade) ≠ P (ninth | Texas) . 489.5 P(ninth grade) = ≈ 0.278 1758.6 437.2 P(ninth | Texas) = ≈ 0.279 1565.1 7. (a) 16 P3 =

(b)

16! = (16)(15)(14) = 3360 (16 − 3)!

16! = 50, 450, 400 3!5!4!4!

139

CHAPTER

Discrete Probability Distributions

4.1 PROBABILITY DISTRIBUTIONS 4.1 TRY IT YOURSELF SOLUTIONS 1. (1) The random variable is continuous because x can be any speed up to the maximum speed of a rocket. (2) The random variable is discrete because the number of calves born on a farm in one year is countable. (3) The random variable is discrete because the number of days of rain for the next three days is countable. x 0 1 2 3 4 5 6 7

f 16 19 15 21 9 10 8 2 n = 100

P(x) 0.16 0.19 0.15 0.21 0.09 0.10 0.08 0.02 P( x) 1

3. Each P(x) is between 0 and 1. P ( x) 1 Because both conditions are met, the distribution is a probability distribution. 4. (1) Probability distribution; The probability of each outcome is between 0 and 1, and the sum of all the probabilities is 1. (2) Not a probability distribution; The sum of all the probabilities is not 1. 5.

x 0 1 2 3 4 5 6 7

P(x) 0.16 0.19 0.15 0.21 0.09 0.10 0.08 0.02 P ( x) 1

xP(x) (0)(0.16) = 0.00 (1)(0.19) = 0.19 (2)(0.15) = 0.30 (3)(0.21) = 0.63 (4)(0.09) = 0.36 (5)(0.10) = 0.50 (6)(0.08) = 0.48 (7)(0.02) = 0.14 xP( x) 2.60

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

141

N xP( x) 2.6 On average, a new employee makes 2.6 sales per day. 6. From 5, N 2.6, x P(x) 0 0.16 1 0.19 2 0.15 3 0.21 4 0.09 5 0.10 6 0.08 7 0.02 P ( x) 1

x−μ −2.6 −1.6 −0.6 0.4 1.4 2.4 3.4 4.4

(x − μ)2 6.76 2.56 0.36 0.16 1.96 5.76 11.56 19.36

P(x)(x − μ)2 (0.16)(6.76) = 1.0816 (0.19)(2.56) = 0.4864 (0.15)(0.36) = 0.0540 (0.21)(0.16) = 0.0336 (0.09)(1.96) = 0.1764 (0.10)(5.76) = 0.5760 (0.08)(11.56) = 0.9248 (0.02)(19.36) = 0.3872 P( x)( x N)2 3.72

T 2 x 3.7

T T 2 x 3.7 x 1.9 Most of the data values differ from the mean by no more than 1.9 sales per day. 7.

Gain, x $ 1995 $ 995 $ 495 $ 245 $ 95 $ −5

P(x) 1 2000 1 2000 1 2000 1 2000 1 2000 1995 2000 P ( x) 1

xP(x) 1995 2000 995 2000 495 2000 245 2000 95 2000 9975 2000 xP ( x ) x 3.08

E ( x) xP( x) x $3.08 Because the expected value is negative, you can expect to lose an average of $3.08 for each ticket you buy.

4.1 EXERCISE SOLUTIONS 1. A random variable represents a numerical value associated with each outcome of a probability experiment. Examples: Answers will vary. 2. A discrete probability distribution lists each possible value a random variable can assume, together with its probability. Copyright © 2023 Pearson Education, Inc.

142

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

Condition 1: 0 b P ( x ) b 1 Condition 2: P( x) 1 3. No; The expected value may not be a possible value of x for one trial, but it represents the average value of x over a large number of trials. 4. The mean of a probability distribution represents the “theoretical average” of a probability experiment. 5. False. In most applications, discrete random variables represent counted data, while continuous random variables represent measured data. 6. True 7. False. The mean of the random variable of a probability distribution describes a typical outcome. The variance and standard deviation of the random variable of a probability distribution describe how the outcomes vary. 8. False. The expected value of a discrete random variable can be positive, negative, or zero. 9. Discrete; Attendance is a random variable that is countable. 10. Continuous; The length of time is a random variable that has an infinite number of possible outcomes and cannot be counted. 11. Continuous; The distance a baseball travels after being hit is a random variable that must be measured. 12. Discrete; The number of arrests is a random variable that is countable. 13. Discrete; The number of cars in a university parking lot is a random variable that is countable. 14. Continuous; The length of time it takes to complete an exam is a random variable that cannot be counted. 15. Continuous; The volume of blood drawn for a blood test is a random variable that must be measured. 16. Discrete; The population of the 50 U.S. states is a random variable that can be counted. 17. Discrete; The fitted sizes of hats is a random variable that can be measured. 18. Continuous; The amount of snow that fell in Nome, Alaska last winter is a random variable that cannot be counted.

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

19. (a)

x 0 1 2 3 or more

f 56 511 679 754

P(x) 0.028 0.256 0.340 0.377

n 2000

P( x) x 1

(b)

Skewed left 20. (a)

x 0 1 2 3 4 5 6

f 6 12 29 57 42 30 16

P(x) 0.031 0.063 0.151 0.297 0.219 0.156 0.083 P( x) 1

n 192 (b)

Approximately symmetric 21. (a) P (1 or 2) 0.256

0.340 0.596

(b) P (2 or more) 0.340

0.377 0.717

0.340

0.377 0.973

(d) P ( x b 2) 1 P ( x 3) 1 0.377 0.623

143

144

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

22. (a) P (1 or 2) 0.063

0.151 0.214

(b) P (2 or less) 0.031

0.063

0.151 0.245

0.219

0.156

(d) P (1 b x b 3) 0.063

0.151

0.297 0.511

(e) P ( x b 4) 0.031

0.063

0.151

0.083 0.755

0.297

0.219 0.761

23. P (0) 0.028 . Yes, it would be unusual for a household to have no HD televisions because the probability is less than 0.05. 24. P (2) 0.151 . No, because the probability is greater than 0.05. 25.

¦ P( x) = 1 → P(3) = 1 − (0.06 + 0.12 + 0.18 + 0.30) = 0.34

26.

¦ P( x) = 1 → P(1) = 1 − (0.05 + 0.23 + 0.21 + 0.17 + 0.11 + 0.08) = 0.15

27. Yes, because each P(x) is between 0 and 1, and P( x) 1, the distribution is a probability distribution. 28. No, P( x) 0.75 29. (a)

0.10

P(x)

0 1 2 3 4 5

0.686 0.195 0.077 0.022 0.013 0.006 P ( x) x 1

0.05

0.04

0.02

xP(x) 0 0.195 0.154 0.066 0.052 0.030 xP( x) 0.497

0.01 0.97 . (x − μ) −0.497 0.503 1.503 2.503 3.503 4.503

(x − μ)2 0.247 0.253 2.259 6.265 12.271 20.277

(x − μ)2P(x) 0.169 0.049 0.174 0.138 0.160 0.122 ( x N)2 P( x) 0.812

N xP( x) 0.497 x 0.5

T 2 ( x N)2 P( x) 0.812 x 0.8 T T 2 0.812 x 0.9 (b) The mean is 0.5, so the average number of dogs per household is about 0 or 1 dog. The standard deviation is 0.9, so most of the households differ from the mean by no more than 1 dog.

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

30. (a)

x 4 5 6 7 8

P(x) 0.164 0.241 0.216 0.345 0.034

xP(x) 0.656 1.205 1.296 2.415 0.272 xP( x) 5.844

(x − μ) −1.844 −0.844 0.156 1.156 2.156

(x − μ)2 3.400 0.712 0.024 1.336 4.648

145

(x − μ)2P(x) 0.558 0.172 0.005 0.461 0.158 ( x N)2 P( x) 1.354

N xP( x) 5.844 x 5.84 T 2 ( x N)2 P( x) 1.354 x 1.35 T T 2 1.354 x 1.16 (b) The mean is 5.84, so the average number of games played in the World Series from 1903 through 2020 is between 5 and 6. The standard deviation is 1.16, so most World Series differ from the mean by no more than about 1 game. 31. (a)

x 0 1 2 3 4 5

P(x) 0.263 0.285 0.243 0.154 0.041 0.014

xP(x) 0 0.285 0.486 0.462 0.164 0.070 xP( x) 1.467

(x − μ) −1.467 −0.467 0.533 1.533 2.533 3.533

(x − μ)2 2.152 0.218 0.284 2.350 6.416 12.482

(x − μ)2P(x) 0.566 0.062 0.069 0.362 0.263 0.175 ( x N)2 P( x) 1.497

N xP( x) x 1.5

T 2 ( x N)2 P( x) 1.497 x 1.5 T T 2 1.497 x 1.2 (b) The mean is 1.5, so the average batch of 1000 machine parts has about 1 or 2 defects. The standard deviation is 1.2, so most of the batches of 1000 differ from the mean by no more than about 1 defect. 32. (a)

x 0 1 2 3 4 5 6 7

P(x) 0.059 0.122 0.163 0.178 0.213 0.128 0.084 0.053

xP(x) 0 0.122 0.326 0.534 0.852 0.640 0.504 0.371 xP( x) 3.349

N xP( x) 3.349 x 3.3

(x − μ) −3.349 −2.349 −1.349 −0.349 0.651 1.651 2.651 3.651

(x − μ)2 11.216 5.518 1.820 0.122 0.424 2.726 7.028 13.330

(x − μ)2P(x) 0.662 0.673 0.297 0.022 0.090 0.349 0.590 0.706 ( x N)2 P( x) 3.389

146

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

T 2 ( x N)2 P( x) 3.389 x 3.4 T T 2 3.389 x 1.8 (b) The mean is 3.3, so the average student is involved in about 3 extracurricular activities. The standard deviation is 1.8, so most of the students differ from the mean by no more than about 2 activities. 33. (a)

x 1 2 3 4 5

P(x) 0.409 0.280 0.216 0.081 0.014

xP(x) 0.409 0.560 0.648 0.324 0.070 xP( x) 2.011

(x − μ) −1.011 í0.011 0.989 1.989 2.989

(x − μ)2 1.022 0.000 0.978 3.956 8.934

(x − μ)2P(x) 0.418 0.000 0.211 0.320 0.125 ( x N)2 P( x) 1.074

N xP( x) 2.011 x 2.01

T 2 ( x N)2 P( x) 1.074 x 1.07 T T 2 1.074 x 1.04 (b) The mean is 2.01, so the average category of hurricane is about 2. The standard deviation is 1.04, so most of the hurricanes differ from the mean by no more than about one category. 34. (a)

x 1 2 3 4 5

P(x) 0.009 0.005 0.007 0.055 0.925

xP(x) 0.009 0.010 0.021 0.220 4.625 xP( x) 4.885

(x − μ) −3.885 −2.885 −1.885 −0.885 0.115

(x − μ)2 15.093 8.323 3.553 0.783 0.013

(x − μ)2P(x) 0.136 0.042 0.025 0.043 0.012 ( x N)2 P( x) 0.258

N xP( x) 4.885 x 4.89

T 2 ( x N)2 P( x) 0.258 x 0.26 T T 2 0.258 x 0.51 (b) The mean is 4.89, so the average reviewer rating is about 5. The standard deviation is 0.51, so most reviewer ratings differ from the mean by less than 1. 35. An expected value of 0 means that the money gained is equal to the money spent, representing the breakeven point. 36. A “fair bet” in a game of chance has an expected value of 0, which means that the chances of losing are equal to the chances of winning.

1¬ ($35) ¸ x $0.05 38 ®

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

147

38. E ( x) N xP( x)

1 ¬ ($5450) ¸ 3500 ®

1 ¬ ($486) ¸ 3500 ®

18 ¬ ($90) ¸ 3500 ®

3480 ¬ $10 ¸ x $7.78 3500 ®

39. Area of bar 1 = 0.028; area of bar 2 = 0.256; area of bar 3 = 0.340; area of bar 4 = 0.377; sum ≈1. The area of each of the bars is the probability of each outcome, because the width is 1. The sum is 1, because all of the probabilities should add up to 1, as they are between 0 and 1. 40.

Games Played 4 5 6 7 8

Number of World Series 0.164(116) ≈ 19 0.241(116) ≈ 28 0.216(116) ≈ 25 0.345(116) ≈ 40 0.034(116) ≈ 4

x 4

P(x) 19 ≈ 0.164 116 28 ≈ 0.241 116 25 ≈ 0.216 116 40 ≈ 0.345 116 4 ≈ 0.034 116

5 6 7 8

xP(x) 0.656 1.205 1.296 2.415 0.272

N xP( x) 5.844 x 5.84 T 2 ( x N)2 P( x) 1.354 x 1.35 T T 2 1.354 x 1.16 The answers are the same. 41. N y a

bN x 600

1.03(46, 000) $47, 980

42. T y b Tx (1.035) 18,000,000 x $4391.13 43. N x y N x

N y 531

516 1047

σ x2+ y = σ x2 + σ y2 = (121)2 + (112)2 = 27,185 σ x+ y = σ x2+ y ≈ 164.9 44. μ x − y = μ x − μ y = 531 − 516 = 15

xP( x) 5.844

148

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

4.2 BINOMIAL DISTRIBUTIONS 4.2 TRY IT YOURSELF SOLUTIONS 1. Trial: answering a question Success: the question answered correctly Yes, the experiment satisfies the four conditions of a binomial experiment. It is a binomial experiment. n = 10, p = 0.25, q = 0.75, x = 0,1, 2,3, 4,5, 6, 7,8,9,10 2. Trial: drawing a card with replacement Success: card drawn is a club Failure: card drawn is not a club n = 5, p = 0.25, q = 0.75, x = 3 5! P(3) = (0.25)3 (0.75) 2 ≈ 0.088 (5 − 3)!3! 3. Trial: Selecting an adult and asking a question Success: Selecting an adult who goes online almost constantly Failure: Selecting an adult who does not go online almost constantly n = 5, p = 0.31, q = 0.69, x = 0,1, 2,3, 4,5 P (0) = 5 C 0 (0.31) 0 (0.69) 5 ≈ 0.156 P (1) = 5 C1 (0.31)1 (0.69) 4 ≈ 0.351 P (2) = 5 C 2 (0.31) 2 (0.69) 3 ≈ 0.316 P (3) = 5 C 3 (0.31) 3 (0.69) 2 ≈ 0.142 P (4) = 5 C 4 (0.31) 4 (0.69)1 ≈ 0.032 P (5) = 5 C 5 (0.31) 5 (0.69) 0 ≈ 0.003

x 0 1 2 3 4 5

P(x) 0.156 0.351 0.316 0.142 0.032 0.003

¦ P( x) = 1

n = 150, p = 0.28, x = 30 P (30) ≈ 0.006

n 5, p 0.16, q 0.84, x 2

(1) P(2) 5 C2 (0.16)2 (0.84)3 x 0.152

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

149

(2) P(2) 5 C2 (0.16)2 (0.84)3 10(0.16)2 (0.84)3 ; P(3) 5 C3 (0.16)3 (0.84)2 10(0.16)3 (0.84) 2

P(4) 5 C4 (0.16)4 (0.84)1 5(0.16)4 (0.84)1 ; P(5) 5 C5 (0.16)5 (0.84)0 1(0.16)5 (0.84)0 P ( x p 2) P (2)

10 0.16

P (3) 2

0.84

P (4) 3

P (5)

10(0.16)3 (0.84) 2

5(0.16) 4 (0.84)1

1(0.16)5 (0.84)0

x 0.183 (3) P(0) 5 C0 (0.16)0 (0.84)5 1(0.16)0 (0.84)5 ; P(1) 5 C1 (0.16)1 (0.84)4 5(0.16)1 (0.84)4 ;

P( x 2) P(0)

P(1) (0.16)0 (0.84)5

5(0.16)1 (0.84) 4 x 0.817

n 6, p 0.85, x 4 From Table 2 in Appendix B, the probability is 0.176.

P(0) 4 C0 (0.27)0 (0.73)4 x 0.284 P(1) 4 C1 (0.27)1 (0.73)3 x 0.420 P(2) 4 C2 (0.27)2 (0.73)2 x 0.233 P(3) 4 C3 (0.27)3 (0.73)1 x 0.057 P(4) 4 C4 (0.27)4 (0.73)0 x 0.005 x 0 1 2 3 4

P(x) 0.284 0.420 0.233 0.057 0.005

8. n = 31, p = 0.44, q = 0.56 N np (31)(0.44) x 13.6

T 2 npq (31)(0.44)(0.56) x 7.6

T npq (31)(0.44)(0.56) x 2.8 On average, there are about 14 clear days during the month of May. The standard deviation is about 3 days. Values that are more than 2 standard deviations from the mean are considered unusual. Because 13.6 − 2(2.8) = 8 and 13.6 + 2(2.8) = 19.2, a May with fewer than 8 clear days, or more than 19 clear days would be unusual.

150

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

4.2 EXERCISE SOLUTIONS 1. Each trial is independent of the other trials when the outcome of one trial does not affect the outcome of any of the other trials. 2. The random variable measures the number of successes in n trials. 3. c; Because the probability is greater than 0.5, the distribution is skewed left. 4. b; Because the probability is 0.5, the distribution is symmetric. 5. a; Because the probability is less than 0.5, the distribution is skewed right. 6. c; The histogram shows probabilities for 12 trials. 7. a; The histogram shows probabilities for 4 trials. 8. b; The histogram shows probabilities for 8 trials. As n increases, the distribution becomes more symmetric. 9. Combinations Rule 10. Unusual values have probabilities that are less than 0.05. (6) x 0,1, 2,3, 4,11,12 (7) x 0 (8) x 0,1, 2,8 11. N np (50)(0.4) 20 2

12. N np (84)(0.65) 54.6

T npq (50)(0.4)(0.6) 12

T 2 npq (84)(0.65)(0.35) x 19.1

T npq (50)(0.4)(0.6) x 3.5

T npq (84)(0.65)(0.35) x 4.4

13. N np (124)(0.26) x 32.2 2

14. N np (316)(0.82) x 259.1

T npq (124)(0.26)(0.74) x 23.9

T 2 npq (316)(0.82)(0.18) x 46.6

T npq (124)(0.26)(0.74) x 4.9

T npq (316)(0.82)(0.18) x 6.8

15. It is a binomial experiment. Success: Frequent gamers who own a virtual reality device n 10 , p 0.29 , q 0.71 , x 0,1, 2,3, 4,5, 6, 7,8,9,10 16. Not a binomial experiment because the probability of a success is not the same for each trial. 17. Not a binomial experiment because the probability of a success is not the same for each trial. 18. It is a binomial experiment. Success: making a free throw n 8 , p 0.78 , q 0.22 , x 0,1, 2,3, 4,5,6,7,8

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

19. n 8 , p 0.39 (a) P (6) x 0.037 (b) P (3) x 0.281 20. n 7 , p 0.59 (a) P (1) x 0.020 (b) P (5) x 0.252 21. n 10 , p 0.52 (a) P (4) x 0.188 (b) P ( x p 7) P (7) P (8) P (9) P (10) x 0.1364 0.0554 0.0133 0.0014 x 0.207 (c) P ( x 5) 1 P ( x p 5) 1 [ P (5) P (6) P (7) P (8) P (9) P (10)] x 1 (0.2441 0.2204 0.1364 0.0554 0.0133 0.0014) 0.329 22. n 6 , p 0.78 (a) P (6) x 0.225 (b) P ( x b 3) 1 P ( x 3) 1 [ P (4) P (5) P (6)] x 1 (0.2687 0.3811 0.2252) 0.125 (c) P ( x 3) P (4) P (5) P (6) x 0.2687 0.3811 0.2252 0.875 23. n 11 , p 0.14 (a) P (1) x 0.341 (b) P ( x b 5) 1 P ( x 5) 1 < P (6) x 1 < 0.0016

0.0002

P (7)

P (8)

P (9)

P (10)

0.0000

P (11) >

0.0000 > x 0.998

0.056 24. n 16 , p 0.45 (a) P (8) x 0.181 (b) P ( x p 10) P (10) P (11) P (12) P (13) P (14) P (15) P (16) ≈ 0.0755 + 0.0337 + 0.0115 + 0.0029 + 0.0005 + 0.0001 + 0.0000

0.0000

151

152

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

= 1 − [ P (12) + P (13) + P (14) + P (15) + P (16) ]

≈ 1 − [ 0.0115 + 0.0029 + 0.0005 + 0.0001 + 0.0000] = 0.985

25. n = 14 , p = 0.05 (a) P (2) ≈ 0.123 (b) P ( x > 2) = 1 − P ( x ≤ 2) = 1 − [ P (0) + P (1) + P (2)] ≈ 1 − (0.4877 + 0.3593 + 0.1229) ≈ 0.030 (c) P (2 ≤ x ≤ 5) = P (2) + P (3) + P (4) + P (5) ≈ 0.1229 + 0.0259 + 0.0037 + 0.0004 ≈ 0.153 26. n = 15 , p = 0.66 (a) P (7) ≈ 0.063 (b) P ( x > 7) = P (8) + P (9) + P (10) + P (11) + P (12) + P (13) + P (14) + P (15) ≈ 0.1217 + 0.1837 + 0.2140 + 0.1888 + 0.1222 + 0.0547 + 0.0152 + 0.0020 ≈ 0.902 (c) P (7 ≤ x ≤ 11) = P (7) + P (8) + P (9) + P (10) + P (11) ≈ 0.0627 + 0.1217 + 0.1837 + 0.2140 + 0.1888 ≈ 0.771 27. (a)

P(x)

0 1 2 3 4 5 6 7

0.008974 0.060355 0.173965 0.278572 0.267647 0.154291 0.049413 0.006782

(b)

Bell shaped (c) The values 0, 6, and 7 are unusual, because their probabilities are less than or equal to 0.05.

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

28. (a)

P(x)

0 1 2 3 4 5 6 7 8

0.040607 0.160003 0.275826 0.271709 0.167283 0.065915 0.016233 0.002284 0.000141

(b)

Bell shaped (c) The values 0, 6, 7, and 8 are unusual, because their probabilities are less than 0.05. 29. (a)

P(x)

0 1 2 3 4 5

0.00064 0.01077 0.07214 0.24151 0.40426 0.27068

(b)

Skewed Left (c) The values 0 and 1 are unusual because their probabilities are less than 0.05.

153

154

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

30. (a)

P(x)

0 1 2 3 4 5 6 7 8 9 10

0.000216 0.002865 0.017089 0.060407 0.140129 0.222904 0.246231 0.186514 0.092715 0.027311 0.003620

(b)

Skewed Left (c) The values 0, 1, 2, 9, and 10 are unusual because their probabilities are less than 0.05. 31. N np (6)(0.33) 1.98

T 2 npq (6)(0.33)(0.67) x 1.33

T npq (6)(0.33)(0.67) x 1.15 On average, 1.98 out of every 6 penalty shots are converted. The standard deviation is 1.15, so most samples of 6 penalty shots would differ from the mean by at most 1.15. Unusual values would be any above 4.28. 32. N np (7)(0.25) 1.75

T 2 npq (7)(0.25)(0.75) x 1.31

T npq (7)(0.25)(0.75) x 1.15 On average, 1.75 out of every 7 questions will be correct. The standard deviation is 1.15, so most samples of 7 questions would differ from the mean by at most 1.15. Unusual values would be any above 4.05. 33. N np (8)(0.79) x 6.3

T 2 npq (8)(0.79)(0.21) x 1.3

T npq (8)(0.79)(0.21) x 1.2

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

155

On average, 6.3 out of every 8 adults believe that life on other planets is possible. The standard deviation is 1.2, so most samples of 8 adults would differ from the mean by at most 1.2 adults. Unusual values would be any under 3.9. 34. N np (10)(0.85) 8.5

T 2 npq (10)(0.85)(0.15) x 1.28

T npq (10)(0.85)(0.15) x 1.13 On average, 8.5 out of every 10 U.S. adults believe cloning humans is morally wrong. The standard deviation is 1.13, so most samples of 10 U.S. adults would differ from the mean by at most 1.13. Unusual values would be any value under 6.24. 35. N np (12)(0.31) 3.72

T 2 npq (12)(0.31)(0.69) x 2.57

T npq (12)(0.31)(0.69) x 1.60 On average, 3.72 out of every 12 U.S. employees who are late for work blame oversleeping. The standard deviation is 1.60, so most samples of 12 U.S. employees who are late for work would differ from the mean by at most 1.60. Unusual values would be any under 0.52 or over 6.92. 36. N np (10)(0.10) 1.0

T 2 npq (10)(0.10)(0.90) 0.9

T npq (10)(0.10)(0.90) x 0.95 On average, one out of every 10 college graduates think that Judge Judy serves on the Supreme Court. The standard deviation is 0.95, so most samples of 10 college graduates would differ from the mean by at most 0.95. Unusual values would be any above 2.9. 37. n 9 , p

1 3

(a) P (5) x 0.102 (b) P ( x 5) P (6) P (7) P (8) P (9) x 0.0341 0.0073 0.0009 0.0001 x 0.042 ; Unusual, because it is less than or equal to 0.05 (c) P ( x 2) P (0)

P (1) x 0.0260

0.1171 x 0.143

38. n 12 , p 0.53 (a) P (9) x 0.075 (b) P ( x 4) P (0) P (1) P (2) P (3) x 0.0001 0.0016 0.0098 0.0367 x 0.048 ; Unusual, because it is less than or equal to 0.05 (c) P (8 b x b 11) P (8)

P (9)

P (10)

P (11) x 0.15038

0.07537

0.02550

0.00523 x 0.256

156

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS 5

10! 9 ¬ 3 ¬ 3 ¬ 1 ¬ 39. P 5, 2, 2,1 x 0.033 5!2!2!1! 16 ® 16 ® 16 ® 16 ® 10! 5 ¬ 4 ¬ 1 ¬ 6 ¬ 40. P 5,2, 2,1 x 0.002 5!2!2!1! 16 ® 16 ® 16 ® 16 ® 41. (a) P(0 defective) = P(1st not defective) + P(2nd not defective given 1st not defective) + P(3rd not defective given first 2 not defective) +… + P(10th not defective given the first 9 not defective) 8000 7999 7998 7991 P (0) q q q ...q x 0.107 10, 000 9999 9998 9991 (b) n 10 , p 0.20 P (0) x 0.107 (c) The results are the same.

4.3 MORE DISCRETE PROBABILITY DISTRIBUTIONS 4.3 TRY IT YOURSELF SOLUTIONS 1.

p 0.20, q 0.80, x 6

P(6) (0.20)(0.80)6 1 x 0.066 2.

30 (2.71828) 3 31 (2.71828) 3 P (1) x x 0.050 x 0.149 0! 1! 32 (2.71828) 3 33 (2.71828) 3 P (2) x P(3) x x 0.224 x 0.224 2! 3! 34 (2.71828) 3 P (4) x x 0.168 4! P (0) P (1) P (2) P (3) P (4) x 0.050 0.149 0.224 0.224 0.168 0.815 P ( x 4) 1 P ( x b 4) 1 0.815 0.185 P (0) x

2000 0.10, x 3 20, 000 P (3) 0.0002 The probability of finding three brown trout in any given cubic meter of the lake is 0.0002. Because 0.0002 is less than 0.05, this can be considered an unusual event. N

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

157

4.3 EXERCISE SOLUTIONS 1.

P(3) (0.65)(0.35)3 1 x 0.080

P(1) (0.45)(0.55)1 1 0.45

P(5) (0.09)(0.91)5 1 x 0.062

P(8) (0.28)(0.72)8 1 x 0.028

P(4) x

(5) 4 (2.71828) 5 x 0.175 4!

P(3) x

(6)3 (2.71828) 6 x 0.089 3!

P(0) x

(1.5)0 (2.71828) 1.5 x 0.223 0!

P(5) x

(9.8)5 (2.71828) 9.8 x 0.042 5!

9. In a binomial distribution, the value of x represents the number of successes in n trials. In a geometric distribution the value of x represents the first trial that results in a success. 10. In a binomial distribution, the value of x represents the number of successes in n trials. In a Poisson distribution the value of x represents the number of occurrences in an interval. 11. p 0.19 (a) P(1) (0.19)(0.81)1 1 x 0.190 (b) P(2) (0.19)(0.81)2 1 x 0.154 (c) P(5) (0.19)(0.81)5 1 x 0.082 12. N 5.5 (a) P (0) x

(5.5)0 (2.71828) 5.5 x 0.004 ; unusual 0!

(b) P(5) x

(5.5)5 (2.71828) 5.5 x 0.171 5!

(5.5)8 (2.71828) 5.5 x 0.085 8!

158

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

13. N 4 (a) P(3) x

(4)3 (2.71828) 4 x 0.195 3!

(b) P ( x b 3) P (0)

P (1)

P (2)

P (3) x 0.018

0.073

0.147

0.195 x 0.433

1 0.01 100

(a) P(10) (0.01)(0.99)10 1 x 0.009 . This event is unusual because its probability is less than 0.05. (b) P(defective on 1st , 2nd ,or 3rd part sold) P(1) 1 1

2 1

P(2)

P(3) 3 1

(0.01)(0.99) (0.01)(0.99) (0.01)(0.99) x 0.01 0.010 This event is unusual because its probability is less than 0.05.

0.010 0.030

(c) P(none of 1st 10 parts are defective) 1 P (1) P (2) P (3) P (4) P (5) P (6) P (7) P (8) P (9) P(10) 1 (0.01)(.99)1 1 (0.01)(.99) 2 1 (0.01)(.99)3 1 (0.01)(.99) 4 1 (0.01)(.99)5 1 (0.01)(.99)6 1 (0.01)(.99)7 1 (0.01)(.99)8 1 (0.01)(.99)9 1 (0.01)(.99)10 1 x 1 0.010 0.010 0.010 0.010 0.010 0.010 0.009 0.009 0.009 0.009 0.904

15. p 0.651 (a) P(2) (0.651)(0.349)2 1 x 0.227 . (b) P(completes 1st or 2nd pass) P(1)

P(2)

(0.651)(0.349)1 1

(0.651)(0.349)2 1 x 0.651 0.227 0.878

(c) P(does not complete his 1st two passes) 1 P(completes 1st or 2nd pass) x 1 0.878 0.122 16. p 0.75 (a) P(1) (0.75)(0.25)1 1 0.75 (b) P(2) (0.75)(0.25)2 1 x 0.188 (c) P ( x 2) 1 P (1) 17. n 8 , p 0.60 (a) P (4) x 0.232

P (2) x 1 (0.75

0.1875) x 0.063

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

(b) P ( x 5) 1 P ( x p 5) 1 < P (5) x 1 <0.2787

0.2090

P (6)

P (8) >

P (7)

0.0168 > x 0.406

0.0896

159

P (1)

P (2) >

0.0413 > x 0.950

0.0079

18. N 16 (a) P (12) x

(16)12 (2.71828) 16 x 0.066 12!

(b) P ( x p 8) 1 P ( x b 7) 1 P (0)

x 1 0.0000

0.0000

P (1)

P (2)

P (3)

P (4)

P (5)

P (6)

0.0001 0.0003

0.0010

0.0026

P (7)

0.0060

x 1 0.010 0.990 (c) P ( x b 10) P (0) P (1) P (2) P (3) P (4) P (5) P (6) P (7) P (8) P (9) x 0.0000 0.0000 0.0000 0.0001 0.0003 0.0010 0.0026 0.0060 0.0120 0.0213 0.0341 x 0.077

P (10)

19. N 1.8 (a) P (1) x

(1.8)1 (2.71828) 1.8 x 0.298 1!

(b) P ( x b 1) P (0)

P (1) x 0.1653

0.2975 x 0.463

P (5)

P (6)

P (3)

P (7) x 0.0203

P (4) x 0.2523

0.0027

0.0923

0.0002

0.000 x 0.023 ; unusual

0.0203 x 0.365

21. n 10 , p 0.76 (a) P (10) x 0.064 (b) P ( x 7) P (8)

P (9)

P (1)

P (10) x 0.2885 P (2)

P (3)

0.2030

P (4) x 0.0000

0.0643 x 0.556 0.0000

0.0003

0.0024

0.0134 x 0.016 ;

160

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

22. p

1 0.25 4

(a) P(4) (0.25)(0.75)4 1 x 0.105 (b) P prize with 1st , 2nd , or 3rd purchase P(1)

(0.25)(0.75)1 1

(0.25)(0.75)2 1

P (2)

0.188

P (3)

0.141

P(2)

P(3)

(0.25)(0.75)3 1 x 0.250

0.1875

0.1406 x 0.578

P (4)

0.105 x 1 0.684 0.316

23. N 14 (a) P (17) x

(14)17 (2.71828) 14 x 0.071 17!

(b) P ( x b 17) P (0)

P (1)

P (17) x 0.827

5! (0.68)3 (0.32) 2 x 0.322 2!3!

(b) P ( x 4) P ( x b 3) P (0) (c) P ( x p 3) P (3) 25. p

P (4)

P (1)

P (3) x 0.0034

P (2)

P (5) x 0.322

0.0357

0.1515

0.3220 x 0.513

0.145 0.809

0.342

1 0.002 500

(a) P(10) (0.002)(0.998)10 1 x 0.002 This event is unusual because its probability is less than 0.05. (b) P 1st , 2nd , or 3rd part is warped P (1)

P(2)

P(3)

(0.002)(0.998)1 1 (0.002)(0.998)2 1 (0.002)(0.998)3 1 x 0.002 This event is unusual because its probability is less than 0.05. (c) P ( x 10) 1 P ( x b 10) 1 P (1)

x 1 0.002

0.002

x 1 0.020 0.980

0.002

P (10)

0.002

P (2)

0.002

0.002 0.006

0.002

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

161

26. N 8 (a) P(8) x

(8)8 (2.71828) 8 x 0.140 8!

(b) P ( x b 3) P (0) P (1) P (2) P (3) x 0.0003 0.0027 0.0107 This is an unusual event, because the probability is less than 0.05. (c) P ( x 8) 1 P ( x b 8) 1 P (0) x 1 0.0003

0.0027

0.0286

P (8)

0.0573

0.0916

0.1221

0.1396

0.1396 x 1 0.593 0.407

1 0.0004 2500

27. (a) n 6000 , p

P(4)

0.0107

P (1)

0.0286 x 0.042

6000! (0.0004) 4 (0.9996)5996 x 0.12542 5996!4!

6000 2.4 cars with defects per 6000. 2500 (2.4) 4 (2.71828) 2.4 x 0.12541 P (4) 4! The results are approximately the same.

(b) N

2 28. (a) P(0)

(b) P(1)

2 (c) P(2)

(1)(286) 0.629 455

(2)(78) 0.343 455

(1)(13) 0.029 455

1 0.001 1000 1 1 1000 (a) N p 0.001 q 0.999 T2 2 999,000 p (0.001) 2

29. p

T T 2 x 999.5

(b) 1000 times, because it is the mean. (c) You would expect to lose money, because, on average, you would win $500 every 1000 times you play the lottery and pay $1000 to play it. So, the net gain would be −$500.

162

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

30. p 0.005 1 1 200 (a) N p 0.005

q 0.995 39,800 2 (0.005) 2 p

T T 2 x 199.5

(b) 200 records, because it is the mean. 31. N 3.9 (a) T 2 N 3.9 T T 2 x 1.97

The standard deviation is 1.97, so most of the scores will differ from the mean by no more than 2 strokes. (b) More than 8 strokes would be unusual. 32. N 2.5 (a) T 2 N 2.5 T T 2 x 1.58

The standard deviation is about 1.58, so most of the bankruptcies filed per hour will not differ from the mean by more than 1.58. (b) More than 5.66 business filings in a given hour would be unusual.

CHAPTER 4 REVIEW EXERCISE SOLUTIONS 1. Discrete; The grade on an exam is a random variable that can be counted. 2. Continuous; The weight of a truck at a weigh station is a random variable that cannot be counted. 3. (a)

x 0 1 2 3 4 5

f 29 62 33 12 3 1 n 140

P(x) 0.207 0.443 0.236 0.086 0.021 0.007 P ( x) 1

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

(b)

Skewed right 4. (a)

x 4 5 6 7 8 9 10

f 1 6 13 23 14 4 2 n = 63

P(x) 0.016 0.095 0.206 0.365 0.222 0.063 0.032

¦ P( x) ≈ 1

(b)

Approximately symmetric 5. Yes. 6.

No, P (5) 1 and P( x) v 1.

7. (a)

x 0 1 2 3 4 5 6

P(x) 0.020 0.140 0.272 0.292 0.168 0.076 0.032

xP(x) 0 0.140 0.544 0.876 0.672 0.380 0.192

¦ xP( x) = 2.804

(x − μ) −2.804 −1.804 −0.804 0.196 1.196 2.196 3.196

(x − μ)2 7.862 3.254 0.646 0.038 1.430 4.822 10.214

¦( x − μ) P(x) = 1.734 2

μ = ¦ xP( x) = 2.804 ≈ 2.8

(x − μ)2P(x) 0.157 0.456 0.176 0.011 0.240 0.367 0.327

σ = σ 2 = 1.734 ≈ 1.3

163

164

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

(b) The mean is 2.8, so the average number of cell phones per household is about 3. The standard deviation is 1.3, so most of the households differ from the mean by no more than about 1 cell phone. 8. (a)

x 15 30 60 90 120

P(x) 0.134 0.786 0.053 0.006 0.021

xP(x) 2.01 23.58 3.18 0.54 2.52

¦ xP(x) = 31.83

(x − μ) −16.83 −1.83 28.17 58.17 88.17

(x − μ)2 282.249 3.349 793.549 3383.749 7773.949

(x − μ)2P(x) 37.9554 2.6322 42.0581 20.3025 163.2529

¦( x − μ) P( x) = 266.2011 2

μ = ¦ xP( x) = 31.83 ≈ 31.8

σ 2 = ¦( x − μ )2 P( x) = 266.2011 ≈ 266.2 σ = σ 2 = 266.2011 ≈ 16.3 (b) The mean is 31.8, so the average length of an advertising block is 31.8 seconds. The standard deviation is 16.3, so most of the commercials differ from the mean by no more than about 16 seconds. 9.

x 100 0 −25

P(x) 0.125 0.250 0.625

xP(x) 12.500 0.000 −15.625

¦ xP( x) = −3.125

μ = ¦ xP( x) = −3.125 ≈ −$3.13 10.

x 99,995 95 45 −5

P(x) 0.00001 0.01000 0.02000 0.96999

xP(x) 0.99995 0.95000 0.90000 −4.84995 xP ( x) 2.00000

N xP( x) $2.00 11. It is a binomial experiment. Success: A green candy is selected n 12 , p 0.125 , q 0.875 , x 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 12. Not a binomial experiment because the experiment is not repeated for a fixed number of trials.

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

13. n 8 , p 0.53 (a) P (3) x 0.191 (b) P ( x p 3) 1 P x 3 1 P (0) x 1 0.0024

0.0215

P ( 2)

0.0848 x 1 0.109 0.891

P (1)

0.0848

P (2)

P (3)

0.1912 x 1 0.300 0.700

14. n 12 , p 0.42 (a) P (2) x 0.050 (b) P ( x p 2 ) 1 P x 2 1 P (0) x 1 0.0014

0.0126 0.986

P (1)

0.0126

P (1)

P ( 2)

0.0502 x 0.936

15. n 9 , p 0.72 (a) P (6) x 0.257 (b) P ( x p 6) P (6) P (7) P (8) P (9) x 0.2569 0.2831 0.1820 0.0520 0.774 (c) P ( x 6) P (7) P (8) P (9) x 0.2831 0.1820 0.0520 x 0.517 16. n 5 , p 0.62 (a) P (2) x 0.211 (b) P ( x p 2 ) 1 P x 2 1 P (0) x 1 0.008

0.065 x 1 0.073 0.927

17. (a)

P(x)

0 1 2 3 4 5

0.002 0.022 0.114 0.293 0.376 0.193

P (1)

0.0646

P (1)

P ( 2)

0.2109 x 1 0.283 0.717

165

166

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

(b)

Skewed left (c) The values 0 and 1 are unusual, because their probabilities are less than 0.05. 18. (a)

P(x)

0 1 2 3 4 5 6

0.018 0.101 0.244 0.312 0.225 0.086 0.014

(b)

Bell shaped (c) The values 0 and 6 are unusual, because their probabilities are less than 0.05. 19. n = 8 , p = 0.13 μ = np = (8)(0.13) ≈ 1.0 σ 2 = npq = (8)(0.13)(0.87) ≈ 0.9

σ = npq = (8)(0.13)(0.87) ≈ 1.0 On average, 1 out of every 8 drivers are uninsured. The standard deviation is 1.0, so most samples of 8 drivers would differ from the mean by at most 1 driver. Unusual values would be any above 3.

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

167

20. n = 10 , p = 0.33 μ = np = (10)(0.33) = 3.3 σ 2 = npq = (10)(0.33)(0.67) = 2.211

σ = npq = (10)(0.33)(0.67) ≈ 1.487 On average, 3.3 out of every 10 NCAA student-athletes have a job waiting for them when they graduate college. The standard deviation is about 1.487, so most samples of 10 NCAA studentathletes would differ from the mean by at most 1.487. Unusual values would be under 0.326 or above 6.274. 21. p = 0.14 (a) P (3) = (0.14)(0.86) 3−1 ≈ 0.104 (b) P (4 th or 5 th ) = P (4) + P (5) = (0.14)(0.86)3 + (0.14)(0.86) 4 ≈ 0.166 (c) P (not one of the 1st through 6th) = 1 − [ P (1) + P (2) + P (3) + P (4) + P (5) + P (6) ]

≈ 1 − [ 0.1400 + 0.1204 + 0.1035 + 0.0890 + 0.0766 + 0.0659 ] ≈ 0.405

22. μ ≈ 0.26 (a) P (0) ≈

(0.26)0 (2.71828) −0.26 ≈ 0.771 0!

(b) P ( x ≤ 2 ) = P (0) + P (1) + P (2) ≈ 0.7711 + 0.2005 + 0.0261 ≈ 0.99 8 (c) P ( x > 1) = 1 − P ( x ≤ 1) = 1 − ( P (0) + P (1) ) ≈ 1 − ( 0.7711 + 0.2005 ) ≈ 0.028 This event is unusual, because its probability is less than 0.05. 23. n= 7 , p = 0.36 (a) P (4) ≈ 0.154

(b) P ( x < 2 ) = P (0) + P (1) ≈ 0.044 + 0.173 = 0.217 (c) P ( x ≥ 6 ) = P (6) + P (7) ≈ 0.010 + 0.001 = 0.0 11 This event would be unusual because its probability is less than 0.05. 24. p =

25 ≈ 0.455 55

(a) P (1) ≈ (0.455)(0.545) 0 ≈ 0.455 (b) P (2) ≈ (0.455)(0.545)1 ≈ 0.248 (c) P ( x ≤ 3) = P (1) + P (2) + P (3) ≈ (0.455)(0.545) 0 + (0.455)(0.545)1 + (0.455)(0.545) 2 ≈ 0.838 (d) 1 − P ( x ≤ 3) ≈ 1 − 0.838 = 0.162

168

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

25. μ = 6.4

(a) P (3) ≈

(6.4)3 (2.71828) −6.4 ≈ 0.073 3!

(b) P ( x > 6) = 1 − P ( x ≤ 6) = 1 − ( P (0) + P (1) + P (2) + P (3) + P (4) + P (5) + P (6) )

≈ 1 − ( 0.0017 + 0.0106 + 0.0340 + 0.0726 + 0.1162 + 0.1487 + 0.1586 ) ≈ 0.458

(c) P ( x ≤ 5) = P (0) + P (1) + P (2) + P (3) + P (4) + P (5) ≈ 0.0017 + 0.0106 + 0.0340 + 0.0726 + 0.1162 + 0.1487 ≈ 0.384 26. n = 10 , p = 0.69 (a) P (7) ≈ 0.266

(b) P ( x > 8) = P (9) + P (10) ≈ 0.1099 + 0.0245 ≈ 0.134 (c) P (2 ≤ x ≤ 4) = P (2) + P (3) + P (4) ≈ 0.0018 + 0.0108 + 0.0422 ≈ 0.055

CHAPTER 4 QUIZ SOLUTIONS 1. (a) Discrete; The number of lightning strikes that occur in Wyoming during the month of June is a random variable that is countable.

(b) Continuous; The fuel (in gallons) used by a jet during takeoff is a random variable that has an infinite number of possible outcomes and cannot be counted. (c) Discrete; The final score in a game of bowling is a random variable that is countable. 2. (a)

x 0 1 2 3 4 5

f 277 471 243 105 46 22

n = 1164 (b)

Skewed right

P(x) 0.238 0.405 0.209 0.090 0.040 0.019 P( x) ≈ 1

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

(c)

(x − μ) −1.348 −0.348 0.652 1.652 2.652 3.652

xP(x) 0.000 0.405 0.418 0.270 0.160 0.095

¦ xP( x) = 1.348

(x − μ)2 1.817 0.121 0.425 2.729 7.033 13.337

169

(x − μ)2P(x) 0.432 0.049 0.089 0.246 0.281 0.253

¦ ( x − μ ) P( x) = 1.35 2

μ = ¦ xP( x) = 1.348 ≈ 1.3

σ 2 = ¦ ( x − μ )2 P( x) = 1.35 ≈ 1.4 σ = σ 2 = 1.35 ≈ 1.2 The mean is 1.3, so the average number of wireless devices per household is 1.3. The standard deviation is 1.2, so most households will differ from the mean by no more than about 1.2 wireless devices. (d) P( x ≥ 4) = P(4) + P(5) = 3.

46 22 + ≈ 0.058 1164 1164

n = 9 , p = 0.33 (a) P (3) ≈ 0.273 (b) P ( x ≤ 4) = P (0) + P (1) + P (2) + P (3) + P (4) ≈ 0.0272 + 0.1206 + 0.2376 + 0.2731 + 0.2017 ≈ 0.860 (c) P ( x > 5) = P (6) + P (7) + P (8) + P (9) ≈ 0.0326 + 0.0069 + 0.0008 + 0.0000 ≈ 0.040

n = 6 , p = 0.75 (a)

P(x)

0 1 2 3 4 5 6

0.0002 0.004 0.033 0.132 0.297 0.356 0.178

170

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

(b)

Skewed left (c) μ = np = (6)(0.75) = 4.5

σ 2 = npq = (6)(0.75)(0.25) ≈ 1.13

σ = npq = (6)(0.75)(0.25) ≈ 1.06 On average, 4.5 out of every 6 patients survive to five years. The standard deviation is about 1.06, so most samples of 6 patients would differ from the mean by at most 1.06 survivors. 5.

μ =5 (a) P (5) ≈

(5)5 (2.71828) −5 ≈ 0.175 5!

(b) P ( x < 5) = P (0) + P (1) + P (2) + P (3) + P (4) ≈ 0.007 + 0.034 + 0.084 + 0.140 + 0.175 = 0.440 (c) P (0) ≈ 6.

(5)0 (2.71828) −5 ≈ 0.007 0!

p = 0.43

(a) P (3) ≈ (0.43)(0.57) 2 ≈ 0.140 (b) P (4th or 5th) = P (4) + P (5) ≈ (0.43)(0.57)3 + (0.43)(0.57) 4 ≈ 0.125 (c) P (not 1st through 6th) = 1 − [ P (1) + P (2) + P (3) + P (4) + P (5) + P (6) ]

(

= 1 − (0.43)(0.57)0 + (0.43)(0.57)1 + (0.43)(0.57) 2 + (0.43)(0.57)3 + (0.43)(0.57) 4 + (0.43)(0.57)5 ≈ 1 − 0.966 = 0.034 7. Event (c) in problem 5 and event (c) in problem 6 are both unusual, because the probabilities are less than 0.05.

CHAPTER 4 TEST SOLUTIONS 1.

1 ≈ 0.0238 42

)

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

(b) P (1st or 2 nd ) = P (1) + P (2) ≈ (0.0238)(0.9762) 0 + (0.0238)(0.9762)1 ≈ 0.047 This event is unusual, because its probability is less than 0.05. (c) P ( x > 5) = 1 − P ( x ≤ 5) = 1 − ( P (1) + P (2) + P (3) + P (4) + P (5) )

≈ 1 − ( 0.0238 + 0.0232 + 0.0227 + 0.0221 + 0.0216 ) ≈ 1 − 0.113 = 0.887

n = 6 , p = 0.53 (a) P (2) ≈ 0.206 (b) P ( x ≥ 3) = 1 − P ( x < 3) = 1 − ( P (0) + P (1) + P (2)) ≈ 1 − (0.0108 + 0.0729 + 0.2056) ≈ 0.711 (c) P ( x < 4) = P (0) + P (1) + P (2) + P (3) ≈ 0.0108 + 0.0729 + 0.2056 + 0.3091 ≈ 0.598

μ = 1.5 (a) P (3) ≈

(1.5)3 (2.71828) −1.5 ≈ 0.126 3!

(b) P ( x > 4) = 1 − P ( x ≤ 4) = 1 − ( P (0) + P (1) + P (2) + P (3) + P (4) ) ≈ 1 − ( 0.2231 + 0.3347 + 0.2510 + 0.1255 + 0.0471) ≈ 1 − 0.981 ≈ 0.019 This event is unusual, because its probability is less than 0.05.

(b) No, 5. (a)

¦ P( x) ≠ 1 . x 17 18 19 20 21 22

f 2 13 4 3 2 1

n = 25 (b) Skewed right

P(x) 0.08 0.52 0.16 0.12 0.08 0.04

¦ P( x) = 1

171

172

CHAPTER 4 Ň DISCRETE PROBABILITY DISTRIBUTIONS

(c)

(x − μ) −1.72 −0.72 0.28 1.28 2.28 3.28

xP(x) 1.36 9.36 3.04 2.40 1.68 0.88

¦ xP( x) = 18.72 μ = ¦ xP( x) = 18.72 ≈ 18.7 σ = ¦ ( x − μ ) P( x) = 1.5616 ≈ 1.6 2

(x − μ)2 2.9584 0.5184 0.0784 1.6384 5.1984 10.7584

(x − μ)2P(x) 0.2367 0.2696 0.0125 0.1966 0.4159 0.4303

¦( x − μ ) P( x) = 1.5616 2

σ = σ 2 = 1.5616 ≈ 1.2 The mean is 18.7, so the average student in the course is about 19 years old. The standard deviation is 1.210, so most of the ages will differ from the mean by no more than about 1 year. (d) P ( x < 20) = P (17) + P (18) + P (19) = 0.08 + 0.52 + 0.16 = 0.76 6.

n = 5 , p = 0.56 (a) x 0 1 2 3 4 5

P(x) 0.016 0.105 0.267 0.340 0.216 0.055

¦ P( x) ≈ 1

(b)

Bell shaped

σ 2 = npq = (5)(0.56)(0.44) = 1.232

σ = npq = (5)(0.56)(0.44) ≈ 1.110 On average, 2.8 out of every 5 student loans are in repayment. The standard deviation is about 1.110, so most samples of 5 student loans would differ from the mean by at most 1.110.

CHAPTER

Normal Probability Distributions

5.1 INTRODUCTION TO NORMAL DISTRIBUTIONS AND THE STANDARD NORMAL DISTRIBUTION 5.1 TRY IT YOURSELF SOLUTIONS 1. (1) A: x 45 , B: x 60 , C: x 45 (B has the greatest mean.) (2) Curve C is more spread out, so curve C has the greatest standard deviation. 2. Because a normal curve is symmetric about the mean, the estimated mean is: P 18.0 Inflection points are approximately 12.2 and 23.8 Because the inflection points are one standard deviation from the mean, V 5.8 3. (1) 0.0143

(2) 0.9850

Area = 0.9834

Area

1 0.0154

0.9846

Area 0.0885 0.0152 0.0733

5.1 EXERCISE SOLUTIONS 1. When it is a normal distribution, the mean is equal to the median. 2. 1 3. Inflection points are where the curve changes concavity. Inflection points occur one standard deviation away from the mean on both the right and left. The x-values would be P V and P V . 173 Copyright © 2023 Pearson Education, Inc.

174

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

4. Answers will vary. 5. Answers will vary. Similarities: The two curves will have the same line of symmetry. Differences: The curve with the larger standard deviation will be more spread out than the curve with the smaller standard deviation. 6. Answers will vary. Similarities: The two curves will have the same shape because they have equal standard deviations. Differences: The two curves will have different lines of symmetry. 7.

P 0, V 1

8. Transform each data value x into an z-score by subtracting the mean from x and dividing the result by x P . the standard deviation. In symbols, z

9. “The” standard normal distribution is used to describe one specific normal distribution ( P 0 , V 1 ). “A” normal distribution is used to describe a normal distribution with any mean and standard deviation. 10. (c) is true because a z-score equal to 0 indicates that the corresponding x-value is equal to the mean. (a) and (b) are not true because it is possible to have a z-score equal to 0 and the mean is not 0 or the corresponding x-value is not 0. 11. No, the graph is skewed left. 12. Yes, the graph fulfills the properties of the normal distribution. P | 18.5 , V | 2 13. No, the graph is symmetric but it is not bell-shaped. 14. No, the graph crosses the x-axis. 15. Yes, the graph fulfills the properties of the normal distribution. P | 11.5 , V | 1.5 16. No, the graph is skewed to the right. 17. (Area left of z 0.6) 0.7257 18. (Area left of z 1.5) 0.0668 19. (Area right of z 2) 1 (Area left of z 2) 1 0.9772 0.0228 20. (Area right of z 0.9) 1 (Area left of z 0.9) 1 0.1841 0.8159 21. (Area left of z 0) (Area left of z 2.25) 0.5000 0.0122 0.4878 22. (Area left of z 1.2) (Area left of z 0.7) 0.8849 0.2420 0.6429

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

23. 0.6293

24. 0.0008

25. 0.0470

26. 0.9139

27. 1 0.2578 0.7422

28. 1 0.9994 0.0006

29. 1 0.3613 0.6387

30. 1 0.9866 0.0134

31. 0.9979 0.5000 0.4979

32. 0.5000 0.0630 0.4370

33. 0.9394 0.0606 0.8788

34. 0.9901 0.0099 0.9802

35. 0.1003 0.1003 0.2006

36. 0.0749 (1 0.9864) 0.0749 0.0136 0.0885

175

37. (a)

It is reasonable to assume that the life spans are normally distributed because the histogram is symmetric and bell-shaped. (b) x

37,234.7; s 6259.2

(c) The sample mean of 37,234.7 hours is less than the claimed mean, 40,000 on average, the tires in the sample lasted for a shorter time. The sample standard deviation of 6259.2 is greater than the claimed standard deviation, so the tires in the sample had greater variation in life span than the manufacturer’s claim. 38. (a)

It is reasonable to assume that the weekly milk consumptions are normally distributed because the histogram is nearly symmetric and bell-shaped. (b) x

45.9; s 9.5

(c) The mean of your sample is 2.8 fluid ounces less than that of the previous study, so the average milk consumption from the sample is less than in the previous study. The standard deviation is about 0.9 fluid ounce greater than that of the previous study, so milk consumptions are slightly more spread out than in the previous study. Copyright © 2023 Pearson Education, Inc.

176

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x P

98 100 0.125 V 16 x P 65 100 | 2.188 x 65 z V 16 x P 106 100 0.375 x 106 z V 16 x P 124 100 1.5 x 124 z V 16

39. (a) x 98 z

(b) x 65 is unusual because the corresponding z-score ( 2.188) lies more than 2 standard deviations away from the mean.

x P

1050 1051 | 0.005 V 211 x P 960 1051 | 0.431 x 960 z V 211 x P 870 1051 | 0.858 x 870 z V 211 x P 1440 1051 | 1.844 x 1440 z V 211

40. (a) x 1050 z

(b) None of the values above are unusual because the corresponding z-scores lie within 2 standard deviations away from the mean. 43. 1 0.0168 0.9832

41. 0.2660

42. 0.9750

44. 1 0.7881 0.2119

45. 0.8413 0.1587 0.6826 (Tech: 0.6827)

46. 0.9535 0.5000 0.4535 47. P z 0.53

0.7019

48. P( z 1.11) 0.1335

49. Using technology: P z ! 2.175 50. P z ! 1.85

1 P z 2.175

1 0.9852 0.0148

1 P( z 1.85) 1 0.0322 0.9678

51. P( 0.89 z 0) 0.5 0.1867 0.3133 52. Using technology: P(0 z 0.835) | 0.7981– 0.5 0.2981 53. P( 1.78 z 1.78) 0.9625 0.0375 0.9250 (Tech: 0.9249) 54. P( 1.54 z 1.54) 0.9382 0.0618 0.8764

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

177

55. P( z 2.58 or z ! 2.58) 2(0.0049) 0.0098 (Tech: 0.0099) 56. P( z 1.22 or z ! 1.32) 0.111232 (1 0.906582) | 0.2047 57.

The normal distribution curve is centered at its mean (60) and has 2 points of inflection (48 and 72) representing P r V . 58.

The normal distribution curve is centered at its mean (450) and has 2 points of inflection (400 and 500) representing P r V . § 1 · b a 59. (1) The area under the curve is (b a) ¨ 1. ¸ ©b a¹ b a (Because a < b, you do not have to worry about division by 0.) § 1 · (2) All of the values of the probability density function are positive because ¨ ¸ is positive when ©b a¹ a b. § 1 · 60. (a) P(2 x 8) (8 2) ¨ ¸ 0.25 © 25 1 ¹ § 1 · (b) P(4 x 12) (12 4) ¨ ¸ 0.333 © 25 1 ¹ § 1 · (c) P(5 x 17) (17 5) ¨ ¸ 0.5 © 25 1 ¹

§ 1 · (d) P(8 x 14) (14 8) ¨ ¸ 0.25 © 25 1 ¹

178

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

5.2 NORMAL DISTRIBUTIONS: FINDING PROBABILITIES 5.2 TRY IT YOURSELF SOLUTIONS 1.

x P

70 67 | 0.857 3.5

P( z ! 0.857) 1 P( z 0.857) | 1 0.8043 | 0.1957 The probability that a randomly selected vehicle is violating the speed limit of 70 miles per hour is 0.1957. 2.

z z

x P

29 41 1 12 56 41 1.25 12

V P(29 x 56) P( 1 z 1.25) P( z 1.25) P( z 1) 0.8944 0.1587

0.7357

When 200 shoppers enter the store, you would expect 200(0.7357) 147.14 or about 147 shoppers to be in the store between 29 and 56 minutes. 3.

P 300 x 350 P(0.256 z 0.897) | 0.8151 0.6010 | 0.214 The probability that a randomly selected state has between 300 and 350 physicians per 100,000 resident population is about 0.214.

5.2 EXERCISE SOLUTIONS 1.

P( x 170) P( z 0.2) 0.4207

P( x 200) P( z 1.3) 0.9032

P( x ! 182) P( z ! 0.4) 1 0.6554 0.3446

P( x ! 155) P( z ! 0.95) 1 0.1711 0.8289

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

P(160 x 170) P( 0.7 z 0.2) 0.4207 0.2420 0.1787

P(172 x 192) P( 0.1 z 0.9) 0.8159 0.4602 0.3557

7. (a) Using technology: P( x 57) P( z 2.0909) | 0.0183 (b) Using technology: P(60 x 65) P( 1.1818 z 0.3333) | 0.6305 0.1186

0.5119

(c) Using technology: P( x ! 70.5) P( z ! 2) 1 P( z 2) 1 0.9772 0.0228 Both the events a and c are unusual because the probability is less than 0.05. 8. (a) Using technology: P( x 41) P( z 1.2667) | 0.1026 (b) Using technology: P(41 x 45) P( 1.2667 z 1.4) | 0.9192 0.1026 0.8166 (c) Using technology: P( x ! 45.8) P( z ! 1.9333) 1 P( z 1.9333) | 1 0.9734 0.0266 Event c is unusual because the probability is less than 0.05. 9. (a) Using technology: P( x 490) P( z 1.0283) | 0.1519 (b) Using technology: P(490 x 510) P( 1.0283 z 0.8585) | 0.8047 0.1519 0.6528 (c) Using technology: P( x ! 515) P( z ! 1.3302) 1 P( z 1.3302) | 1 0.9083 0.0917 None of the events are unusual because none of the probabilities are less than 0.05. 10. (a) Using technology: P( x 120) P( z 1.6552) | 0.0489 (b) Using technology: P(122 x 127) P( 0.96552 z 0.75862) | 0.77596 0.16714 | 0.6088

(c) Using technology: P( x ! 131) P( z ! 2.1379) 1 P( z 2.1379) | 1 0.9837 0.0163 Events a and c are unusual because the probabilities are less than 0.05. 11. (a) Using technology: P( x 70) P( z 2.5) 0.0062 (b) Using technology: P(90 x 120) P( 0.8333 z 1.6667) 0.95221 0.20233 0.7499

(c) Using technology: P( x ! 140) P( z ! 3.3333) 1 P( z 3.3333) 1 0.9996 0.0004 12. (a) Using technology: P( x 17) P( z 0.6) 0.2743

179

180

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

(b) Using technology: P(20 x 28) P(0 z 1.6) 0.9452 0.5 0.4452 (c) Using technology: P( x ! 30) P( z ! 2) 1 0.9772 0.0228 13. Using technology: P(750 x 1000) P( 1.4265 z 0.2417) | 0.4045 0.0769

0.3276

14. Using technology: P(25 x 30) P(0.72881 z 1.57627) | 0.94252 0.76694 | 0.1756 15. Using technology: P(285 x 294) P(1.8 z 2.7) 0.99653 0.96407 0.0325 16. Using technology: P(4.5 x 5.5) P( 2.25 z 0.25) 0.5987 0.0122 0.5865 17. (a) Using technology: P( x 1300) P( z 1.1801) | 0.8810 88.10% (b) Using technology: P( x ! 1100) P( z ! 0.2322) 1 P( z 0.2322) | 1 0.5918 0.4082 (1000)(0.4082) 408.2 408 scores 18. (a) Using technology: P( x 19) P( z 0.2281) | 0.4098 40.98% (b) Using technology: P( x ! 21) P( z ! 0.0508) 1 P( z 0.0508) | 1 0.5203 0.4797 (1500)(0.4797) 719.55 720 scores 19. (a) Using technology: P( x 290) P( z 2.3) 0.9893 98.93% (b) Using technology: P(260 x 300) P( 0.7 z 3.3) 0.99952 0.24196 0.7576 75.76% (c) Using technology: P( x ! 287) P( z ! 2) 1 P( z 2) 1 0.9772 0.0228 (250)(0.0228) 5.7 6 mothers 20. (a) Using technology: P( x 6) P( z 1.5) 0.9332 93.32% (b) Using technology: P(4.7 x 5.3) P( 1.75 z 0.25) 0.4013 0.0401 0.3612 36.12% (c) Using technology: P( x ! 4.8) P( z ! 1.5) 1 P( z 1.5) 1 0.0668 0.9332 (200)(0.9332) 186.64 187 adult males 21. Out of control, because the 10th observation is more than 3 standard deviations beyond the mean. 22. Out of control, because two out of three consecutive points lie more than 2 standard deviations from the mean. (8th and 10th observations) 23. Out of control, because there are nine consecutive points below the mean, and two out of three consecutive points more than 2 standard deviations from the mean. 24. In control, because none of the three warning signals detected a change.

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

181

5.3 NORMAL DISTRIBUTIONS: FINDING VALUES 5.3 TRY IT YOURSELF SOLUTIONS 1. (1)

(2)

1.77

z 1.96

2. (1)

(2)

use area = 0.1003 z

(3)

use area = 0.2005

1.28

0.84

use area = 0.9901 z

2.33

P 52, V 15 (1) z

2.33 x P zV

(2) z 3 x

P zV

52 2.33 15

52 3 15

17.05 pounds

97 pounds

(3) z 0.58 x P zV 52 0.58 15 60.70 pounds 17.05 pounds is to the left of the mean, 97 pounds is to the right of the mean, and 60.7 pounds is to the right of the mean. 4.

2.33

x P zV 132 2.33 5.18 | 119.93 So, the longest braking distance one of these cars could have and still be in the bottom 1% is about 120 feet.

182

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

1.28

x P zV 11.2 1.28 2.1 8.512 The maximum length of time an employee could have worked and still be laid off is about 8.5 years.

5.3 EXERCISE SOLUTIONS 1.

0.81

2. z

0.16

3. z

0.84

5. z

1.645

6. z 1.555

z 1.04

8. z

2.35

9. z

0.44

12. z

2.17

0.45

10. z

0.44

11. z 1.96

13. z

0.67

14. z

0.25

15. z 1.34

16. z

0.67

17. z

0.38

18. z

0.25

19. z 1.99

20. z

0.58

21. z

r1.645

22. z

r1.96

23. The z-score has an area of 0.0119 to its left. z

1.18

24. The z-score has an area of 0.785 to its left. z

0.79

25. The z-score has an area of 0.637 to its right and has an area of 1 0.637 0.363 to its left. z

0.35

26. The z-score has an area of 0.209 to its right and has an area of 1 0.209 0.791 to its left. z

0.81

27. The z-score has an area of 0.02275 to its left. z

2.00

28. The z-score has an area of 0.841345 to its right and has an area 1 0.841325 0.158655 to its left. z

1.00

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

183

29. Since the area between the z-scores is 0.80, then the combined area remaining (to the left of -z and to the right of z) is 1 0.80 0.20 . The positive z-score that has an area of 0.10 to its right has an area 1 0.10 0.90 to its left. z 1.28

30. Since the area between the z-scores is 0.12, then the combined area remaining (to the left of -z and to the right of z) is 1 0.12 0.88 . The positive z-score that has an area of 0.44 to its right has an area 1 0.44 0.56 to its left. z

0.15

31. (a) 95th percentile Area = 0.95 z | 1.6449 (using technology) x P zV 166.7 1.6449 49.3 | 247.8 pounds (b) 43rd percentile Area = 0.43 z | 0.1764 (using technology) x P zV 166.7 ( 0.1764) 49.3 | 158.0 pounds (c) 1st quartile Area = 0.25 z | 0.6745 (using technology) x P zV 166.7 0.6745 49.3 | 133.45 pounds 32. (a) 88th percentile Area = 0.88 z | 1.17499 (using technology) x P zV 50.6 1.17499 29.0 | 84.67 (b) 61st percentile Area = 0.61 z | 0.2793 (using technology) x P zV 50.6 0.2793 29.0 | 58.70 (c) 1st quartile Area = 0.25 z | 0.6745 (using technology) x P zV 50.6 0.6745 29.0 | 31.04 33. (a) 5th percentile Area = 0.05 z | 1.64485 (using technology) x P zV 187.9 1.64485 32.8 | 133.95 seconds (b) 17th percentile Area = 0.17 z | 0.954165 (using technology) x P zV 187.9 0.954165 32.8 | 156.60 seconds (c) 3rd quartile Area = 0.75 z | 0.6745 (using technology) x P zV 187.9 0.6745 32.8 | 210.02 seconds 34. (a) 80th percentile Area = 0.80 z | 0.8416 (using technology) x P zV 1.34 0.8416 0.8 | 2.01 cubic meters (b) 29th percentile Area = 0.29 z | 0.5534 (using technology) x P zV 1.34 0.5534 0.80 | 0.90 cubic meters (c) 3rd quartile Area = 0.75 z | 0.6745 (using technology) x P zV 1.34 0.6745 0.80 | 1.88 cubic meters Copyright © 2023 Pearson Education, Inc.

184

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

35. (a) Top 5% Area = 0.95 z | 1.6449 (using technology) x P zV 500 1.6449 100 | 664.5 z 0.6745 (using technology) (b) Middle 50% Area = 0.25 to 0.75 x P zV 500 r0.6745 100 | 432.6 to 567.5

36. (a) Bottom 20% Area = 0.20 z | 0.8416 (using technology) x P zV 3.58 0.8416 0.85 | 2.86 (b) Middle 80% Area = 0.10 to 0.90 z | r1.2816 (using technology) x P zV 3.58 r1.2816 0.85 | 2.49 to 4.67 37. (a) Upper 25% Area = 0.75 z | 0.6745 (using technology) x P zV 5.4 0.6745 0.4 | 5.67 millions of cells per microliter (b) Upper 5% Area = 0.95 z | 1.6449 (using technology) x P zV 5.4 1.6449 0.4 | 6.06 millions of cells per microliter Bottom 5% Area = 0.05 z | 1.6449 (using technology) x P zV 5.4 1.6449 0.4 | 4.74 millions of cells per microliter Red blood cell counts less than 4.74 million or greater than 6.06 million would be considered unusual. 38. (a) Upper 10% Area = 0.90 z | 1.2816 (using technology) x P zV 267 1.2816 10 | 279.8 days (b) Upper 5% Area = 0.95 z | 1.6449 (using technology) x P zV 267 1.6449 10 | 283.45 days Bottom 5% Area = 0.05 z | 1.6449 (using technology) x P zV 267 1.6449 10 | 250.55 days Pregnancy durations less than 250.55 days or greater than 283.45 days would be considered unusual. 39. Upper 4.5% Area = 0.955 z | 1.6954 (using technology) x P zV 32 1.6954 0.36 | 32.61 ounces 40. Bottom 10% Area = 0.10 z | 1.2816 (using technology) x P zV 30,000 1.2816 2500 | 26,796 Tires that wear out by 26,796 miles will be replaced free of charge.

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

185

41. Top 1% Area = 0.99 z | 2.3264 (using technology) x P zV 8 P 2.3264 0.03 P | 7.93 ounces 42. (a) Yes, because x P zV 67 0.675 15 | 77 x P zV 67 0.6745 15 | 77 So a 75th percentile score of 77 would make the grades normally distributed. (b) No, because x P zV 72 1.2816 9 | 83.5 A 90th percentile score would be 83.5, not 93, if the grades could be normally distributed. (c) No, because the distribution is not symmetric about the mean. (d) Yes, because the distribution appears to be symmetric about the mean. (e) No, because lowest grade for the extra credit assignment is less than one standard deviation from the mean. (f)

No, because the 80th percentile score would be about 184, not 200, with the given 35the percentile and 65th percentile in a normal distribution. 80th percentile Area = 0.80 z | 0.8416 (using technology) 35th percentile Area = 0.35 z | 0.3853 (using technology) 65th percentile Area = 0.65 z | 0.3853 (using technology)

160 140 (0.3853)V 20 (0.3853)V 52 | V

120 140 ( 0.3853)V 20 ( 0.3853)V 52 | V

80th percentile: x

P zV 140 0.8416 52 | 184

186

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

5.4 SAMPLING DISTRIBUTIONS AND THE CENTRAL LIMIT THEOREM 5.4 TRY IT YOURSELF SOLUTIONS 1.

Sample 1, 1, 1 1, 1, 3 1, 1, 5 1, 3, 1 1, 3, 3 1, 3, 5 1, 5, 1 1, 5, 3 1, 5, 5

Mean 1 1.67 2.33 1.67 2.33 3 2.33 3 3.67

Sample 3, 1, 1 3, 1, 3 3, 1, 5 3, 3, 1 3, 3, 3 3, 3, 5 3, 5, 1 3, 5, 3 3, 5, 5

x 1 1.67 2.33 3 3.67 4.33 5

Probability

1 3 6 7 6 3 1

0.03704 0.11111 0.22222 0.25926 0.22222 0.11111 0.03704 ¦ P( x) 1

n = 27

3, V x2 | 0.889, V x | 0.943

P 3,

2 x

Vx 2.

V2 n

V n

Mean 1.67 2.33 3 2.33 3 3.67 3 3.67 4.33

Sample 5, 1, 1 5, 1, 3 5, 1, 5 5, 3, 1 5, 3, 3 5, 3, 5 5, 5, 1 5, 5, 3 5, 5, 5

Mean 2.33 3 3.67 3 3.67 4.33 3.67 4.33 5

3 | 0.889, 3 8 3 | 0.943 3

6.9, V x

V n

1.5 64

0.1875

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

P 3.5, V x

P 37.0, V x

x 36.2 : z

x P

0.05

5.8 100

36.2 37.0 5.8 100

n x 37.8 : z

0.2 16

x P

37.8 37.0 5.8 100

187

0.58

0.8 | 1.38 0.58

P(36.2 x 37.8) P( 1.38 z 1.38) 0.9162 0.0838 0.8324 Of the samples of 100 drivers ages 30 to 49, about 83.24% will drive a mean distance each day that is between 36.2 and 37.8 miles. 5.

296,700, V x

275,000 : z

x P

V n

50,000 | 14, 433.76 12

275,000 296,700 21,700 | | 1.50 50,000 14, 433.76 12

P( x ! 275,000) P( z ! 1.50) 1 P( z 1.50) 1 0.0668 0.9332 About 93% of samples of 12 existing homes will have a mean sales price greater than $275,000. 6.

x P

200 208 | 0.07 V 107 x P 200 208 8 x 200 : z | | 0.24 V 107 33.84 n 10 P ( z 0.07) 0.4702 P ( z 0.24) 0.4065 x 200 : z

There is about a 47% chance that a monitor will cost less than $200. There is about a 41% chance that the mean of a sample of 10 computer monitors is less than $200.

188

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

5.4 EXERCISE SOLUTIONS 1.

P 150 V 25

Vx 3.

2. P x

P 790 V 48

| 3.571

4. P x

250

| 3.036

P 45 V 15 n

100

P 1275 V 6 n

1000

1.5

| 0.190

5. False. As the size of a sample increases, the mean of the distribution of sample means does not change. 6. False. As the size of a sample increases, the standard deviation of the distribution of sample means decreases. 7. False. A sampling distribution is normal if either n t 30 or the population is normal. 8. True 9. (c) Because P x

16.5, V x

10. (b) Because P x

5.8, V x

V n

11.9 100

2.3 | 0.23, and the graph approximates a normal curve. 100

1.19, and the graph approximates a normal curve.

11. (a) The mean and the standard deviation of the population (64, 48, 19, 79, 56) are:

P 53.2, V | 19.9

(b)

Sample

Mean

Sample

Mean

Sample

Mean

19, 19 19, 48 19, 56 19, 64 19, 79 48, 19 48, 48 48, 56 48, 64

19 33.5 37.5 41.5 49 33.5 48 52 56

48, 79 56, 19 56, 48 56, 56 56, 64 56, 79 64, 19 64, 48 64, 56

63.5 37.5 52 56 60 67.5 41.5 56 60

64, 64 64, 79 79, 19 79, 48 79, 56 79, 64 79, 79

64 71.5 49 63.5 67.5 71.5 79

19.9 | 14.1 2 The means are equal, but the standard deviation of the sampling distribution is smaller.

53.2, V x |

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

12. (a) The mean and the standard deviation of the population (1.000, 1.004, 1.001, 1.003) are:

P 1.002, V | 0.0016

(b)

Sample

Mean

1.000, 1.000 1.000, 1.001 1.000, 1.003 1.000, 1.004 1.001, 1.000 1.001, 1.001 1.001, 1.003 1.001, 1.004 1.003, 1.000 1.003, 1.001 1.003, 1.003 1.003, 1.004 1.004, 1.000 1.004, 1.001 1.004, 1.003 1.004, 1.004

1.000 1.0005 1.0015 1.002 1.0005 1.001 1.002 1.0025 1.0015 1.002 1.003 1.0035 1.002 1.0025 1.0035 1.004

0.0016 | 0.0011 2 The means are equal, but the standard deviation of the sampling distribution is smaller.

1.002, V x |

13. (a) The mean and the standard deviation of the population (350, 399, 418) are:

P 389, V | 28.65

(b)

Sample

Mean

Sample

Mean

Sample

Mean

350, 350, 350 350, 350, 399 350, 350, 418 350, 399, 350 350, 399, 399 350, 399, 418 350, 418, 350 350, 418, 399 350, 418, 418

350 366.33 372.67 366.33 382.67 389 372.67 389 395.33

399, 350, 350 399, 350, 399 399, 350, 418 399, 399, 350 399, 399, 399 399, 418, 350 399, 418, 399 399, 418, 418 418, 350, 350

366.33 382.67 389 382.67 399 389 405.33 411.67 372.67

418, 350, 399 418, 350, 418 418, 399, 350 418, 399, 399 418, 399, 418 418, 418, 350 418, 418, 399 418, 418, 418

389 395.33 389 405.33 411.67 395.33 411.67 418

28.65 | 16.54 3 The means are equal, but the standard deviation of the sampling distribution is smaller.

389, V x |

189

190

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

14. (a) The mean and the standard deviation of the population (70, 85, 81, 67) are:

P 75.75, V | 7.46 Sample

Mean

Sample

Mean

Sample

67, 67, 67 67, 67, 70 67, 67, 81 67, 67, 85 67, 70, 67 67, 70, 70 67, 70, 81 67, 70, 85 67, 81, 67 67, 81, 70 67, 81, 81 67, 81, 85 67, 85, 67 67, 85, 70 67, 85, 81 67, 85, 85 70, 67, 67 70, 67, 70 70, 67, 81 70, 67, 85 70, 70, 67 70, 70, 70

67 68 71.67 73 68 69 72.67 74 71.67 72.67 76.33 77.67 73 74 77.67 79 68 69 72.67 74 69 70

70, 70, 81 70, 70, 85 70, 81, 67 70, 81, 70 70, 81, 81 70, 81, 85 70, 85, 67 70, 85, 70 70, 85, 81 70, 85, 85 81, 67, 67 81, 67, 70 81, 67, 81 81, 67, 85 81, 70, 67 81, 70, 70 81, 70, 81 81, 70, 85 81, 81, 67 81, 81, 70 81, 81, 81 81, 81, 85

73.67 75 72.67 73.67 77.33 78.67 74 75 78.67 80 71.67 72.67 76.33 77.67 72.67 73.67 77.33 78.67 76.33 77.33 81 82.33

81, 85, 67 81, 85, 70 81, 85, 81 81, 85, 85 85, 67, 67 85, 67, 70 85, 67, 81 85, 67, 85 85, 70, 67 85, 70, 70 85, 70, 81 85, 70, 85 85, 81, 67 85, 81, 70 85, 81, 81 85, 81, 85 85, 85, 67 85, 85, 70 85, 85, 81 85, 85, 85

(b)

Mean 77.67 78.67 82.33 83.67 73 74 77.67 79 74 75 78.67 80 77.67 78.67 82.33 83.67 79 80 83.67 85

7.46 | 4.31 3 The means are equal, but the standard deviation of the sampling distribution is smaller.

15. z

75.75, V x |

x P

24.3 24 0.3 | | 1.92 1.25 0.156 64

P( x 24.3) P( z 1.92) 0.9726 The probability is not unusual because it is greater than 0.05. 16. z

x P

24.3 24 1.25 100

0.3 0.125

2.40

P( x ! 24.3) P( z ! 2.40) 1 P( z 2.40) 1 0.9918 0.0082 The probability is unusual because it is less than 0.05.

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

17. z

x P

551 550 1 | | 1.8130 3.7 0.552 45

P( x ! 551) P( z ! 1.8130) 1 P( z 1.8130) | 1 0.9651 0.0349 (using technology) The probability is unusual because it is less than 0.05. 18. z

x P

12,753 12,750 3 | | 10.60 1.7 0.283 36

P( x 12,750 or x ! 12,753) P( z 0 or z ! 10.60) | 0.5 0.000 0.5 (using technology) The probability is not unusual because it is greater than 0.05. 19. P x

21. P x

23. P x

20. P x

31.58

V n

12.293 | 1.419 75

22. P x

97.2

V n

0.83 | 0.138 36

$54,000

V n

$11,000 | $1,886.48 34

24. P x

158.51

V n

33.424 | 3.342 100

96.8

V n

0.61 | 0.111 30

$136,000

V n

$11,500 | $1,818.31 40

191

192

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

25. P x

26. P x

603

V n

111 16

27.75

110 25

628

V n

27. n 40 : P x n 60 : P x

603, V x 603, V x

V n

111 | 17.55 40 111 | 14.33 60

As the sample size increases, the standard deviation of the sample means decreases, while the mean of the sample means remains constant. 28. n 72 : P x n 108 : P x

628, V x 628, V x

V n

110 | 12.964 72 110 | 10.585 108

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

193

As the sample size increases, the standard deviation of the sample means decreases, while the mean of the sample means remains the same. 29. z

x P

n x P

400 652 | 0.9257 1540 32 700 652 | 0.1763 1540 32

P(400 x 700) P( 0.9257 z 0.1763) | 0.56997 0.17730 | 0.3927 (using technology) About 39% of samples of 32 years will have a mean gain between 400 and 700. 30. z

x P

n x P

0.10 0.1259 | 0.81542 0.1958 38 0.11 0.1259 | 0.50058 0.1958 38

P(0.10 x 0.11) P( 0.81542 z 0.50058) | 0.30833 0.20742 | 0.101 (using technology) About 10.1% of samples of 38 years will have a mean return of between 10% and 11%. 31. z

x P

0.10 0.0951 | 2.2939 0.0117 30

P( x ! 0.10) P( z ! 2.2939) | 1 0.9891 0.0109 (using technology) About 1.1% of samples of 30 U.S. states will have a mean asthma rate greater than 10%. 32. z

x P

58 59.0 | 1.1562 5.47 40

P( x 58) P( z 1.1562) | 0.1238 (using technology) About 12.4% of samples of 40 states will have a fertility rate less than 58. 33. z

x P

0.10 0.0951 | 0.4188 0.0117

V P( x 0.10) P( z 0.4188) | 0.6623 (using technology)

194

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x P

0.10 0.0951 | 1.3244 0.0117 10

P( x 0.10) P( z 1.3244) | 0.9073 (using technology) It is more likely to select a sample of 10 states with a mean asthma prevalence of less than 10%, because the sample of 10 states has a higher probability. 34. z

x P

65 59.0 | 1.0969 5.47

V P( x 65) P( z 1.0969) | 0.86366 (using technology) x P 65 59.0 z | 4.2482 V 5.47 n

P( x 65) P( z 4.2482) | 0.99999 (using technology) It is more likely to select 15 states in which the mean of the state fertility rate is less than 65 because the sample of 15 states has a higher probability. x P

127.9 128 0.1 | | 3.1623 V 0.20 0.0316 n 40 P( x 127.9) P( z 3.1623) | 0.0008 (using technology) Yes, it is very unlikely that you would have randomly sampled 40 cans with a mean less than or equal to 127.9 ounces because it is more than 3 standard deviations from the mean of the sample means.

35. z

36. z

x P

64.05 64 0.05 | | 2.8748 0.11 0.0174 40

P( x ! 64.05) P( z ! 2.8748) 1 P( z 2.8748) 1 0.9980 0.0020 (using technology) Yes, it is very unlikely that we would have randomly sampled 40 containers with a mean greater than or equal to 64.05 ounces because it is more than 2 standard deviations from the mean of the sample means. 37. (a) P

96; V x P

0.5 96.25 96 z 0.5 V 0.5 P( x ! 96.25) P( z ! 0.5) 1 P( z 0.5) | 1 0.6915 0.3085 (using technology) x P

96.25 96 0.25 | | 3.1623 V 0.5 0.079 n 40 P( x ! 96.25) P( z ! 3.1623) 1 P( z 3.1623) | 1 0.9992 0.0008 (using technology)

(b) z

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

195

38. (a) P 10; V

0.5 x P 10.21 10 z 0.42 V 0.5 P( x ! 10.21) P( z ! 0.42) 1 P( z 0.42) | 1 0.6628 0.3372 (using technology) x P

10.21 10 0.21 2.1 0.5 0.1 n 25 P( x ! 10.21) P( z ! 2.1) 1 P( z 2.1) | 1 0.9821 0.0179 (using technology)

(b) z

39. Yes, Use the finite correction factor, because n 60 ! 0.05(1000) 50. x P 40 49.83 z | 1.5052 V N n 52.15 1000 60 1000 1 n N 1 60 P ( x 40) P ( z 1.5052) | 0.06614 (using technology) 40. Yes, Use the finite correction factor since n 30 ! 0.05(100) 5. x P 120 129.58 z | 0.5749 V N n 108.54 100 30 100 1 n N 1 30 140 129.58 x P | 0.6253 z V N n 108.54 100 30 100 1 30 n N 1

P(120 x 140) P( 0.5749 z 0.6253) | 0.7341 0.2827 0.4514 (using technology)

(0.63)(0.37) pq | 0.0471 105 n 0.08 0.55 0.63 pˆ p 1.6979 z 0.0471 0.63(0.37) pq 105 n P( pˆ 0.55) P( z 1.6979) 0.0448 (using technology) The probability that less than 55% of a sample of 105 residents are in favor of building a new high school is about 4.5%. Because the probability is less than 0.05, this is an unusual event.

41. P

p 0.63; V

196

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

pq (0.74)(0.26) | 0.0418 n 110 pˆ p 0.80 0.74 0.06 z | 1.4346 0.0418 pq 0.74(0.26) n 110 P( pˆ ! 0.80) P( z ! 1.4346) 1 P( z 1.4346) | 1 0.9243 0.0757 (using technology) The probability that more than 80% of a sample of 110 residents are making an effort to conserve water or electricity is about 7.6%.

42. P

p 0.74; V

5.5 NORMAL APPROXIMATIONS TO BINOMIAL DISTRIBUTIONS 5.5 TRY IT YOURSELF SOLUTIONS 1.

n 100, p 0.35, q 0.65 np 100(0.35) 35, nq 100(0.65) 65 Because np t 5 and nq t 5 , the normal distribution can be used. P np 100(0.35) 35 V

100(0.35)(0.65) | 4.77

npq

2. (1) The discrete midpoint values are 57, 58, …, 83. The corresponding interval is 56.5 x 83.5 . The normal probability distribution is P(56.5 x 83.5). (2) The discrete midpoint values are …, 52, 53, 54. The corresponding interval is x 54.5 . The normal probability distribution is P( x 54.5). 3.

P np 100(0.35) 35 V npq 100(0.35)(0.65) | 4.77 P( x ! 30.5) x P 30.5 35 | | 0.94 z V 4.77 P( x ! 30.5) P( z ! 0.94) 1 P( z 0.94) 1 0.1736 0.8264 P x 90.5

x P

90.5 104

| 1.91

V 200(0.52)(0.48) P( x 90.5) P( z 1.91) 0.0281

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

n 100, p 0.081 np 100(0.081) 8.1 t 5 and nq 100(0.919) 91.9 t 5 The normal distribution can be used. P np 100(0.081) 8.1

npq

100(0.081)(0.919) | 2.73

P 3.5 x 4.5 x P

3.5 8.1 | 1.69 V 100(0.081)(0.919) x P 4.5 8.1 z2 | 1.32 V 100(0.081)(0.919) P( z 1.69) 0.0455 P( z 1.32) 0.0934 P( 1.69 z 1.32) 0.0934 0.0455 0.0479 z1

5.5 EXERCISE SOLUTIONS 1.

24 0.85

20.4 t 5

2. np

nq 24 0.15 3.6 5 Cannot use normal distribution. 3. np 18 0.90 16.2 t 5

6. d

10.5 t 5

nq 15 0.30 4.5 5 Cannot use normal distribution. 4. np 20 0.65 13 t 5

nq 18 0.10 1.8 5 Cannot use normal distribution. 5. a

15 0.70

nq 20 0.35 7 t 5 Can use normal distribution.

7. c

8. b

9. The probability of getting fewer than 25 successes; P( x 24.5) 10. The probability of getting at least 110 successes; P( x ! 109.5) 11. The probability of getting exactly 33 successes; P(32.5 x 33.5) 12. The probability of getting more than 65 successes; P( x ! 65.5) 13. The probability of getting at most 150 successes; P( x 150.5) 14. The probability of getting between 55 and 60 successes; P(55.5 x 59.5) 15. Binomial: P(5 d x d 7)

P( x 5) P( x 6) P( x 7) 16 C5

0.4

0.6

16 C6 0.4

0.6

16 C7 0.4

0.6

197

198

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

Normal: P

x P

np 16(0.4)

6.4, V

npq

16(0.4)(0.6) | 1.9596

16. Binomial: P(3 d x d 5)

P( x 3) P( x 4) P( x 5) 12 C3

Normal: P

x P

| 0.3679 np 12(0.5) 6, V

2.5 6 | 2.0207 V 1.7321 x P 5.5 6 | | 0.2887 z V 1.7321 P(3 d x d 5) | P(2.5 d x d 5.5) z

0.5

0.5 12 C4 0.5 npq

0.5 12 C5 0.5

12(0.5)(0.5) | 1.7321

P( 2.0207 d z d 0.2887) | 0.38641 0.02166 | 0.3648

(using technology) The results are about the same. 17. n 20, p 0.22, q 0.78 np 20(0.22) 4.4 5, nq 20(0.78) 15.6 t 5 Cannot use normal distribution because np < 5. 18. n 20, p 0.74, q 0.26 np 20(0.74) 14.8 t 5, nq

20(0.74) 14.8, V

20(0.26) 5.2 t 5 npq

20(0.74)(0.26) | 1.96

19. n 40, p 0.62, q 0.38 np 40(0.62) 24.8 t 5, nq 40(0.38) 15.2 t 5

P np 40(0.62) 24.8, V

npq

40(0.62)(0.38) | 3.0699

Can use normal distribution. (a) z

x P

24.5 24.8 | 0.0977 V 3.0699 x P 25.5 24.8 | | 0.2280 V 3.0699 |

P( x 25) | P(24.5 x 25.5)

0.5

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

(b) P( x t 30) | P( x ! 29.5) P ( z ! 1.5310) 1 P ( z 1.5310) | 1 0.9371 0.0629 (using technology)

No unusual events because all of the probabilities are greater than 0.05. 20. n 500, p 0.55, q 0.45 np 500(0.55) 275 t 5, nq 500(0.45) 225 t 5

P np 500(0.55) 275, V

npq

500(0.55)(0.45) | 11.1243

Can use normal distribution. (a) z

x P

249.5 275 | 2.2923 11.1243

V P( x 250) | P( x 249.5)

P ( z 2.2923) | 0.0109 (using technology)

(b) z

x P

299.5 275 | 2.2024 11.1243

V P( x t 300) | P( x ! 299.5)

199

200

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

P( z ! 2.2024) 1 P( z 2.2024) | 1 0.9862 0.0138 (using technology)

x P

239.5 275 | 3.1912 V 11.1243 x P 280.5 275 | | 0.4944 V 11.1243 |

P(239.5 x 280.5) P( 3.1912 z 0.4944) | 0.6895 0.0007 0.6888 (using technology)

The events in parts (a) and (b) are unusual, because the probabilities are less than 0.05. 21. n 100, p 0.43, q 0.57 np 100(0.43) 43 t 5, nq 100(0.57) 57 t 5

P np 100(0.43) 43, V

npq

100(0.43)(0.57) | 4.9508

Can use normal distribution. (a) z

x P

50.5 43 | 1.5149 4.9508

V P( x ! 50) | P( x ! 50.5) P( z ! 1.5149) 1 P( z 1.5149) | 1 0.9351 0.0649 (using

technology)

(b) z

x P

50.5 43

| 1.5149

V 4.9508 P( x d 50) | P( x 50.5)

P ( z 1.5149) | 0.9351 (using technology)

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x P

39.5 43 | 0.7070 V 4.9508 x P 45.5 43 | | 0.5050 V 4.9508 |

P(40 d x d 45) | P(39.5 x 45.5) P ( 0.7070 z 0.5050) | 0.6932 0.2398 0.4534 (using technology)

None of the events are unusual because their probabilities are greater than 0.05. 22. n 150, p 0.71, q 0.29 np 150(0.71) 106.5 t 5, nq 150(0.29) 43.5 t 5

P np 150(0.71) 106.5, V

npq

150(0.71)(0.29) | 5.5574

Can use normal distribution. (a) z

x P

100.5 106.5 | 1.0796 5.5574

P( x d 100) | P( x 100.5) P( z 1.0796) | 0.1402 (using technology)

(b) z

x P

120.5 106.5 | 2.5192 5.5574

V P( x ! 120) | P( x ! 120.5) P( z ! 2.5192) 1 P( z 2.5192) | 1 0.9941 0.0059 (using

technology)

201

202

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x P

89.5 106.5 | 3.0590 V 5.5574 x P 110.5 106.5 | | 0.7198 V 5.5574 |

P(90 d x d 110) | P(89.5 x 110.5) P( 3.0590 z 0.7198) | 0.7642 0.0011 0.7631 (using technology)

The event in part (b) is unusual because its probability is less than 0.05. 23. n 25, p 0.84, q 0.16 np 25(0.84) 21 t 5, nq 25(0.16) 4 5 Cannot use normal distribution because nq < 5. (a) P( x t 24) P(24) P(25)

| 0.0609 0.0128 0.0737 (b) P ( x 23) 1 P( x t 23) 1 > P (23) P(24) P(25) @ | 1 [0.1392 0.0609 0.0128] 1 0.2129 0.7871

| 0.7509 None of the events are unusual, because their probabilities are greater than 0.05. 24. n 20, p 0.36, q 0.64 np 20(0.36) 7.2 t 5, nq 20(0.64) 12.8 t 5

P np 20(0.36) 7.2, V

npq

20(0.36)(0.64) | 2.1466

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

Can use normal distribution. (a) z

x P

8.5 7.2 | 0.6056 2.1466

P( x d 8) | P( x 8.5) P ( z 0.6056) | 0.7276 (using technology)

(b) z

x P

6.5 7.2 | 0.3261 V 2.1466 x P 7.5 7.2 | | 0.1398 V 2.1466 |

P( x 7) | P(6.5 x 7.5) P( 0.3261 z 0.1398) | 0.5556 0.3722 0.1834 (using technology)

x P

5.5 7.2 | 0.7920 V 2.1466 x P 9.5 7.2 | | 1.0715 V 2.1466 |

P(6 d x d 9) | P(5.5 x 9.5) P( 0.7920 z 1.0715) | 0.8580 0.2142 0.6438 (using technology)

None of the events are unusual, because their probabilities are greater than 0.05. 25. n 200, p 0.19, q 0.81 np 200(0.19) 38 t 5, nq 200(0.81) 162 t 5

P np 200(0.19) 38, V

npq

200(0.19)(0.81) | 5.5480

203

204

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

Can use normal distribution.

x P

(a) z

29.5 38 | 1.5321 5.5480

P( x t 30) | P( x ! 29.5) P( z ! 1.5321) 1 P( z 1.5321) | 1 0.0627 0.9373 (using technology)

(b) z

x P

40.5 38 | 0.4506 5.5480

P( x ! 40) | P( x ! 40.5) P( z ! 0.4506) 1 P( z 0.4506) | 1 0.6739 0.3261 (using technology)

x P

34.5 38 | 0.6309 V 5.5480 x P 45.5 38 | | 1.3518 V 5.5480 |

P(35 d x d 45) | P(34.5 x 45.5) P( 0.6309 z 1.3518) | 0.9118 0.2641 0.6477 (using technology)

None of the events are unusual because their probabilities are greater than 0.05.

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

205

26. n 500, p 0.77, q 0.23 np 500(0.77) 385 t 5, nq 500(0.23) 115 t 5

P np 500(0.77) 385, V

npq

500(0.77)(0.23) | 9.4101

Can use normal distribution.

x P

(a) z

400.5 385 | 1.6472 9.4101

P( x d 400) | P( x 400.5) P ( z 1.6472) | 0.9502 (using technology)

(b) z

x P

360.5 385 | 2.6036 9.4101

V P( x ! 360) | P( x ! 360.5) P( z ! 2.6036) 1 P(z 2.6036) | 1 0.0046 0.9954 (using

technology)

x P

379.5 385 | 0.58448 V 9.4101 x P 390.5 385 | | 0.58448 V 9.4101 |

P(380 d x d 390) | P(379.5 x 390.5) P( 0.58448 z 0.58448) | 0.72055 0.27945 0.4411 (using technology)

None of the events are unusual because their probabilities are greater than 0.05.

206

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

27. (a) n 110, p 0.83, q 0.17 np 110(0.83) 91.3 t 5, nq 110(0.17) 18.7 t 5 Can use normal distribution. P np 110(0.83) 91.3, V = npq 110(0.83)(0.17) | 3.9397

P( x ! 100) | P( x ! 100.5) P( z ! 2.3352) 1 P ( z 2.3352) | 1 0.9902 0.0098 (using technology)

(b) n 125, p 0.83, q 0.17 np 125(0.83) 103.75 t 5, nq 125(0.17) 21.25 t 5 Can use normal distribution. P np 125(0.83) 103.75, V = npq 125(0.83)(0.17) | 4.1997

P( x ! 100) | P( x ! 100.5) P( z ! 0.7739) 1 P( z 0.7739) | 1 0.2195 0.7805 (using technology)

(c) n 150, p 0.83, q 0.17 np 150(0.83) 124.5 t 5, nq 150(0.17) 25.5 t 5 Can use normal distribution. P np 150(0.83) 124.5, V = npq 150(0.83)(0.17) | 4.6005

P( x ! 100) | P( x ! 100.5) P( z ! 5.2168) 1 P( z 5.2168) | 1 0.0000 1 (using technology)

28. (a) n 250, p 0.35, q 0.65 np 250(0.35) 87.5 t 5, nq 250(0.65) 162.5 t 5 Can use normal distribution. P np 250(0.35) 87.5, V = npq 250(0.35)(0.65) | 7.5416

P( x 100) | P( x 99.5) P ( z 1.5912) | 0.9442 (using technology)

(b) n 300, p 0.35, q 0.65 np 300(0.35) 105 t 5, nq 300(0.65) 195 t 5 Can use normal distribution. P np 300(0.35) 105, V = npq 300(0.35)(0.65) | 8.2614

P( x 100) | P( x 99.5) P( z 0.6657) | 0.2528 (using technology)

400(0.65)

260 t 5

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

Can use normal distribution. P np 400(0.35) 140, V = npq

207

400(0.35)(0.65) | 9.5394

P( x 100) | P( x 99.5) P( z 4.2456) | 0.0000 (using technology).

29. n 250, p 0.68, q 0.32, P

250(0.68) 170, V

250(0.68)(0.32) | 7.3756

npq

80% say no o n 250(0.80) 200 say no while 50 say yes. x P 50.5 170 | | 16.2021 z V 7.3756

P(less than or equal to 50 say yes) P( x d 50) | P( x 50.5) P( z 16.2021) | 0

It is highly unlikely that 80% responded no. Answers will vary. 30. n 72, p 0.13, q 0.87, P np 72(0.13) 9.36 V 50% of 72 = 36 people x P 36.5 9.36 | | 9.5108 z V 2.8536

npq

72(0.13)(0.87) | 2.8536

P( x d 36) | P( x 36.5) P( z 9.5108) 1

It is probable that 36 of the 72 people responded that they participate in volleyball. Answers will vary. 31. n 100, p 0.75, P

x P

np 100(0.75)

75, V

npq

100(0.75)(0.25) | 4.3301

69.5 75 | 1.2702 4.3301

V P(reject claim) P(x 70) | P(x 69.5) | P(z 1.2702) | 0.1020 32. n 100, p 0.65, P

np 100(0.65)

65, V

npq

100(0.65)(0.35) | 4.7697

69.5 69 5 65 0.9435 0 943 4.7697 4.7 4 7697 7 P(accept claim) P( x t 70) | P( x ! 69.5) | P( z ! 0.9435) 1 P( z 0.9435) 1 0.8273 0.1727 z

CHAPTER 5 REVIEW EXERCISE SOLUTIONS 1.

P 15, V 3

2. P 56, V

3. Curve B has the greatest mean because its line of symmetry occurs the farthest to the right. 4. Curve A has the greatest standard deviation because it is the most spread out. 5. 0.6772

6. 0.2119 – 0.0094 = 0.2025

7. 0.6293

8. 0.0256

9. 1 0.2843 0.7157

10. 1 0.9994 0.0006

208

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

11. 0.00236 (using technology)

12. 1 0.5060 0.4940

13. 0.5 0.0505 0.4995

14. 0.85083 0.06057 | 0.7903 (using technology)

15. 0.95637 0.51994 | 0.4364 (using technology) 16. 0.99632 0.00368 | 0.9926 (using technology) 17. 0.0668 0.0668 0.1336 18. 0.73891 (1 0.99968) | 0.7392 (using technology) x P 17 21.2 | 0.61 19. x 17 o z V 6.9 x P 29 21.2 x 29 o z | 1.13 V 6.9 x P 8 21.2 | 1.91 x 8o z V 6.9 x P 23 21.2 | 0.26 x 23 o z V 6.9 20. None of the events are unusual, because they all lie within two standard deviations of the mean. 21. P( z 1.28) 0.8997 22. P( z

0.74) 1 P(z

23. P( 2.15

x 1.55)

0.74) 1 .2296

0.93943 0.01578

0.7704

0.9237 (using technology)

24. P(0.42 z 3.15) 0.99918 0.66276 | 0.3364 (using technology) 25. P( z 2.50 or z ! 2.50) 2(0.0062) 0.0124 26. P( z 0 or z ! 1.68) 0.5 (1 .9535) 0.5465 27. z

x P

84 74 1.25 8

V P( x 84) P( z 1.25) 0.8944

28. z

x P

55 74 8

2.375

V P( x 55) P( z 2.375) 0.0088 (using technology)

29. z

x P

80 74

0.75

V 8 P( x ! 80) P( z ! 0.75) 1 P( z 0.75) 1 0.7734 0.2266

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

30. z

x P

71.6 74 8

0.3

V P( x ! 71.6) P( z ! 0.3) 1 P( z 0.3) 1 0.3821 0.6179

31. z

x P

60 74 8 70 74 8

x P

1.75 0.5

V P(60 x 70) P( 1.75 z 0.5) | 0.30854 0.04001 | 0.2685 (using technology) 32. z

x P

72 74 0.25 8 82 74 1.00 8

x P

V P(72 x 82) P( 0.25 z 1.00) | 0.84135 0.40129 | 0.4401 (using technology) 33. (a) z

x P

5500 5774 | 0.8228 333

V P( x 5500) P( z 0.8228) | 0.2053 (using technology)

(b) z

x P

6000 5774 | 0.6789 333 6500 5774 | 2.1802 333

V P(6000 x 6500) P(0.6789 z 2.1802) | 0.9854 0.7514 0.2340 (using technology) (c) z

x P

5900 5774 | 0.3784 333

V P( x ! 5900) P( z ! 0.3784) 1 P( z 0.3784) | 1 0.6474 0.3526 (using technology)

34. (a) z

x P

5.1 9.1 1.1429 3.5 15.7 9.1 1.8857 3.5

V P(5.1 x 15.7) P( 1.1429 z 1.8857) | 0.9703 0.1265 | 0.8438 (using technology) (b) z

x P

10.5 9.1 0.4 3.5 12.3 9.1 | 0.9143 3.5

V P(10.5 x 12.3) P(0.4 z 0.9143) | 0.8197 0.6554 | 0.1643 (using technology)

209

210

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x P

11.1 9.1 | 0.5714 3.5

V P( x ! 11.1) P( z ! 0.5714) 1 P( z 0.5714) 1 0.7161 0.2839

35. No unusual events because all of the probabilities are greater than 0.05. 36. No unusual events because all of the probabilities are greater than 0.05. 37. z

0.07

38. z

1.28

39. z

2.457 (using technology)

40. z

2.05

41. z 1.04

42. z

0.10

43. 30.5% to right 69.5% to left, z

0.51

44. 94% between 97% below, z 1.88 45. x

P zV 132 2.75 4.53 | 119.54 feet

46. x

P zV 132 1.6 4.53 | 139.25 feet

47. 90th percentile Area 0.90 z 1.2816 x P zV 132 1.2816 4.53 | 137.81 feet 48. 1st quartile Area 0.25 z | 0.6745 (using technology) x P zV 132 ( 0.6745) 4.53 | 128.94 feet 49. Top 15% Area 0.85 z 1.036 (using technology) x P zV 132 1.036 4.53 | 136.70 feet 50. Bottom 20% Area 0.20 z 0.8416 x P zV 132 0.8416 4.53 128.19 feet 51. (a) P 1.5, V | 1.118 (b)

Sample 0, 0 0, 1 0, 2 0, 3 1, 0 1, 1 1, 2 1, 3

Mean 0 0.5 1 1.5 0.5 1 1.5 2

Sample 2, 0 2, 1 2, 2 2, 3 3, 0 3, 1 3, 2 3, 3

Mean 1 1.5 2 2.5 1.5 2 2.5 3

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

1.118 | 0.791 2 The means are equal, but the standard deviation of the sampling distribution is smaller.

1.5, V x |

52. (a) P 140, V | 50.990 (b)

Sample 90,90,90 90,90,120 90,90,210 90,120,90 90,120,120 90,120,210 90,210,90 90,210,120 90,210,210

Mean 90 100 130 100 110 140 130 140 170

Sample 120,90,90 120,90,120 120,90,210 120,120,90 120,120,120 120,120,210 120,210,90 120,210,120 120,210,210

Mean 100 110 140 110 120 150 140 150 180

Sample Mean 210,90,90 130 210,90,120 140 210,90,210 170 210,120,90 140 210,120,120 150 210,120,210 180 210,210,90 170 210,210,120 180 210,210,210 210

50.990 | 29.439 3 The means are equal, but the standard deviation of the sampling distribution is smaller.

140, V x |

53. P x

199.6, V x

54. P x

151.88, V x

55. (a) z

x P

265.4 | 44.86 35

V n

9.95 | 1.57 40

5500 5774 | 1.1636 333 2

P( x 5500) P( z 1.1636) | 0.1223 (using technology) (b) z

x P

6000 5774 | 0.9598 333 2

211

212

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x P

6500 5774 | 3.0832 333 2

P(6000 x 6500) P(0.9598 z 3.0832) | 0.9990 0.8314 0.1676 (using technology) (c) z

x P

5900 5774 | 0.5351 333 2

P( x ! 5900) P( z ! 0.5351) 1 P( z 0.5351) | 1 0.7037 0.2963 (using technology) The probabilities are smaller. 56. (a) z

x P

n x P

5.1 9.1 | 2.7994 3.5 6 15.7 9.1 | 4.6190 3.5 6

P(5.1 x 15.7) | P( 2.7994 z 4.6190) | 0.999999 0.002560 | 0.9974 (using technology) (b) z

x P

n x P

10.5 9.1 | 0.9798 3.5 6 12.3 9.1 | 2.2395 3.5 6

P(10.5 x 12.3) | P(0.9798 z 2.2395) | 0.9874 0.8364 0.1510 (using technology) (c) z

x P

11.1 9.1 | 1.3997 3.5 6

P( x ! 11.1) | P( z ! 1.3997) | 1 0.9192 | 0.0808 (using technology) The probabilities in parts (b) and (c) are smaller, and the probability in part (a) is larger. 57. (a) z

x P

22 20.7 | 1.3220 5.9 36

P( x 22) P( z 1.3220) 0.9069 (b) z

x P

23 20.7 | 2.33898 5.9 36

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x P

n x P

20 20.7 | 0.7119 5.9 36 21.5 20.7 | 0.8136 5.9 36

P(20 x 21.5) P( 0.7119 z 0.8136) | 0.7921 0.2383 0.5538 (using technology) 58. (a) z

x P

503 500.9 | 1.1207 10.6 32

P( x 503) P( z 1.1207) | 0.8688 (using technology) (b) z

x P

502 500.9 | 0.5870 10.6 32

P( x ! 502) P( z ! 0.5870) | 1 0.7214 0.2786 (using technology) (c) z

x P

n x P

498 500.9 | 1.5476 10.6 32 501 500.9 | 0.0534 10.6 32

P(498 x 501) P( 1.5476 z 0.0534) | 0.5213 0.0609 0.4604 (using technology) 59. (a) z

x P

75,000 72,000 | 1.8295 11,000 45

P( x 75,000) P( z 1.8295) | 0.9663 (using technology) (b) z

x P

68,000 72,000 | 2.4393 11,000 45

P( x ! 68,000) P( z ! 2.4393) | 1 0.0074 0.9926 (using technology) 60. (a) z

x P

84,000 87,000 | 2.0203 10,500 50

P( x 84,000) P( z 2.0203) | 0.0217 (using technology)

213

214

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

(b) z

x P

85,000 87,000 | 1.3469 10,500 50

P( x ! 85,000) P( z ! 1.3469) 1 P( z 1.3469) | 1 0.0890 0.9110 (using technology) 61. n 20, p 0.94, q 0.06 np 20(0.94) 18.8 t 5 , nq 20(0.06) 1.2 5 Cannot use the normal distribution because nq 5 62. n 45, p 0.69, q 0.31

np 45(0.69) 31.05 t 5 , nq 45(0.31) 13.95 t 5 Can use the normal distribution. P np 45(0.69) 31.05, V npq

45 0.69 0.31 | 3.10

63. The probability of getting at least 25 successes; P( x t 25) | P( x ! 24.5) 64. The probability of getting at most 36 successes; P( x d 36) | P( x 36.5) 65. The probability of getting exactly 45 successes; P( x 45) | P(44.5 x 45.5) 66. The probability of getting more than 14 successes; P( x ! 14) | P( x ! 14.5) 67. The probability of getting less than 60 successes; P( x 60) | P( x 59.5) 68. The probability of getting between 54 and 64 successes; P(54 x 64) | P(54.5 x 63.5) 69. n 70, p 0.72, q 0.28 np 70(0.72) 50.4 t 5 , nq 70(0.28) 19.6 t 5 Can use the normal distribution.

P np 70(0.72) 50.4 V (a) z

x P

npq

40.5 50.4 | 2.6354 3.7566

49.5 50.4 | 0.2396 3.7566

70 0.72 0.28 | 3.7566

V P( x d 40) | P( x 40.5) P( z 2.6354) | 0.0042 (using technology)

(b) z1

x P

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x P

50.5 50.4 | 0.0266 3.7566

V P( x 50) | P(49.5 x 50.5) P( 0.2396 z 0.0266) | 0.5106 0.4053 0.1053 (using technology)

x P

60.5 50.4 | 2.6886 3.7566

V P( x ! 60) | P( x ! 60.5) P( z ! 2.6886) 1 P( z 2.6886) | 1 0.9964 0.0036 (using

technology)

70. n 20, p 0.34, q 0.66 np 20(0.34) 6.8 t 5 , nq 20(0.66) 13.2 t 5 Can use normal distribution.

P np 20(0.34) 6.8, V (a) z1

x P

npq

20 0.34 0.66 | 2.1185

6.5 6.8 | 0.1416 V 2.1185 x P 7.5 6.8 | | 0.3304 V 2.1185 |

P( x 7) | P(6.5 x 7.5) P( 0.1416 z 0.3304) | 0.6295 0.4437 0.1858 (using technology)

(b) z

x P

6.5 6.8 | 0.1416 2.1185

V P( x d 6) | P( x 6.5) P( z 0.1416) | 0.4437 (using technology)

215

216

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x P

4.5 6.8 | 1.0857 V 2.1185 x P 10.5 6.8 | | 1.7465 V 2.1185 |

P(5 d x d 10) | P(4.5 x 10.5) P( 1.0857 z 1.7465) | 0.9596 0.1388 0.8208 (using technology)

No unusual event because all of the probabilities are greater than 0.05.

CHAPTER 5 QUIZ SOLUTIONS 1. (a) P( z ! 1.68) 1 0.0465 0.9535 (b) P( z 2.23) 0.9871 (c) P( 0.47 z 0.47) | 0.6808 0.3192 | 0.3616 (using technology) (d) P( z 1.992 or z ! 0.665) | 0.0232 (1 0.2530) 0.7702 (using technology) 2. (a) z

x P

5.97 9.2 | 1.9938 1.62

V P( x 5.97) | P( z 1.9938) | 0.0231 (using technology)

(b) z

x P

40.5 87 | 2.4474 19

V P( x ! 40.5) | P( z ! 2.4474) | 1 0.0072 | 0.9928 (using technology)

x P

5.36 5.5 1.75 0.08 5.64 5.5 1.75 0.08

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

217

x P

125 100 | 1.6667 15

V P( x ! 125) | P( z ! 1.6667) | 1 0.9522 | 0.0478 (using technology) Yes, the event is unusual because its probability is less than 0.05.

z z

x P

95 100 | 0.3333 15 105 100 | 0.3333 15

V P(95 x 105) | P( 0.3333 z 0.3333) | 0.63056 0.36944 | 0.2611 (using technology) No, the event is not unusual because its probability is greater than 0.05. 5.

x P

112 100 15

0.80

V P( x ! 112) P( z ! 0.80) 1 .7881 0.2119 o 21.19% (using technology) z

x P

90 100 | 0.6667 15

V P( x 90) P( z 0.6667) 0.2525 (using technology) (2000)(0.2525) 505 people

7. Top 5% o z | 1.6449

P zV 100 (1.6449)(15) | 124.7 o125

8. Bottom 10% o z | 1.2816

P zV 100 ( 1.2816)(15) | 80.8 o 80

(Because you are finding the highest score that would still place you in the bottom 10%, round down, because if you rounded up you would be outside the bottom 10%.) 9.

x P

105 100 5 | | 2.5820 15 1.9365 60

P( x ! 105) | P( z ! 2.5820) | 1 0.9951 | 0.0049 (using technology) About 0.5% of samples of 60 students will have a mean IQ score greater than 105. This is a very unusual event.

218

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

10. z

x P

105 100 | 0.3333 15

V P( x ! 105) | P( z ! 0.3333) | 1 0.6306 | 0.3694 (using technology) x P 105 100 5 z | | 1.2910 V 15 3.873 n

P( x ! 105) | P( z ! 1.2910) | 1 0.9016 | 0.0984 (using technology) You are more likely to select one student with a test score greater than 105 because the standard error of the mean is less than the standard deviation. 11. n 250, p 0.81, q 0.19

np 250(0.81) 202.5 t 5 , nq 250(0.19) 47.5 t 5 Can use normal distribution. P np 250(0.81) 202.5, V 12. (a) z

x P

npq

200.5 202.5 | 0.3224 6.2028

209.5 202.5 | 1.1285 6.2028

250(0.81)(0.19) | 6.203

V P( x d 200) | P( x 200.5) P( z 0.3224) | 0.3736 (using technology)

(b) z

x P

V P( x 210) | P( x 209.5) P( z 1.1285) | 0.8704 (using technology)

x P

201.5 202.5 | 0.1612 V 6.2028 x P 202.5 202.5 | 0 V 6.2028 |

P( x 202) | P(201.5 x 202.5) P( 0.1612 z 0) | 0.5 0.4360 0.064

(using technology) No unusual events because all of the probabilities are greater than 0.05.

CUMULATIVE REVIEW, CHAPTERS 3-5 1. (a) np 30 0.61 18.3 t 5, nq 30 0.39 Can use normal distribution. (b) P

np 30 0.61

x P

18.3, V

npq

14.5 18.3 | 1.4224 2.6715

11.7 t 5

30 0.61 0.39 | 2.6715

V P( x d 14) | P( x 14.5) | P( z 1.4224) | 0.0775 (using technology)

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x P

219

13.5 18.3 | 1.7967 2.6715

V P( x 14) | P(13.5 x 14.5) | P( 1.7967 z 1.4224) | 0.07746 0.03619 | 0.0413

(using technology) Yes, because the probability is less than 0.05. 2.

x 1 2 3 4 5 6 7 8 9

P(x) 0.282 0.348 0.151 0.127 0.058 0.023 0.007 0.003 0.002

xP(x) 0.282 0.696 0.453 0.508 0.290 0.138 0.049 0.024 0.018 ¦ xP( x) 2.458

(a) P

¦ xP( x) 2.458

(b) V 2

¦ x P P( x) | 2.00

V 2 | 1.42

(d) E( x)

¦ xP(x) | 2.458

x P

–1.458 −0.458 0.542 1.542 2.542 3.542 4.542 5.542 6.542

x P P(x) 0.5995 0.0730 0.0444 0.3020 0.3748 0.2886 0.1444 0.0921 0.0856

x P 2.1258 0.2098 0.2938 2.3778 6.4618 12.5458 20.6298 30.7138 42.7978

¦ x P P( x) | 2.0044 2

The size of an average household is about 2.5 persons. The standard deviation is 1.42, so most households differ from the mean by no more than about one person. 3.

x 0 1 2 3 4 5 6

P(x) 0.044 0.162 0.221 0.368 0.103 0.088 0.015

xP(x) 0.000 0.162 0.442 1.104 0.412 0.440 0.090 xP ¦ ( x) 2.65

(a) P

¦ xP(x) 2.65

(b) V 2

¦ x P P( x) | 1.73

V 2 | 1.32

x P −2.65 −1.65 −0.65 0.35 1.35 2.35 3.35

x P 7.0225 2.7225 0.4225 0.1225 1.8225 5.5225 11.2225

x P P(x) 0.3090 0.4410 0.0934 0.0451 0.1877 0.4860 0.1683

¦ x P P( x) | 1.731 2

220

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

(d) E( x) P 2.65 The number of fouls per game for DeMar DeRozan is about 2.65 fouls. The standard deviation is 1.32, so most of DeRozan’s games differ from the mean by no more than about 1 or 2 fouls. 4. (a) P( x 4) 0.044 0.162 0.221 0.368 0.795 (b) P( x t 3) 0.368 0.103 0.088 0.015 0.574 (c) P(2 d x d 4) 0.221 0.368 0.103 0.692 5. (a) (16)(15)(14)(13) 43,680 (b)

7 6 5 4 16 15 14 13

840 | 0.019 43,680

6. 0.7642

7. 0.0010

8. 1 0.2005 0.7995

9. 0.9984 0.500 0.4984

10. 0.3974 0.1112 0.2862

11. 0.5478 1 0.9573

0.5905

12. n 25, p 0.49 (a) P( x 12) | 0.1573 (using technology) (b) P(8 d x d 11) | 0.0384 0.0697 0.1071 0.1404 0.3556 (using technology) (c) P( x 2) 0.00000005 0.000001 | 0.000001 (using technology) (d) The event in part (c) is unusual because the probability is less than 0.05. 13. p

1 200

0.005

(a) P( x 5) (0.005)(0.995) 4 | 0.0049 (b) P( x d 3) | (0.005)(0.995)0 (0.005)(0.995)1 (0.005)(0.995)2

| 0.005 0.004975 0.004950 | 0.0149 (c) P( x ! 20) 1 P( x d 20) | 1 0.0954 0.9046 14. (a)

167, 400 | 0.3288 509,168

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

(b)

1,003,014 | 0.8685 1,154,879

(c) Dependent. P(being a public school teacher │ having 20 years or more of full-time teaching experience) ≠ P(being a public school teacher) (d)

1,154,879 509,168 151,865 | 0.3730 4,054, 454

15. (a) P x

V n

(b) P( x d 69)

16. (a) P( x 36)

1.2 | 0.190 40

§ · ¨ 69 70 ¸ ¸ | P( z 3.2275) | 0.0006 (using technology) P¨ z d 1.2 ¸ ¨ ¨ ¸ 15 ¹ © 36 44 · § P¨ z P( z 1.6) 0.0548 (using technology) 5 ¸¹ ©

(b) P(42 x 60)

60 44 · § 42 44 P¨ z 5 ¸¹ © 5

P( 0.4 z 3.2) | 0.99931 0.34458 | 0.6547 (using technology)

x P zV 44 (1.6449)(5) | 52.2 months

17. (a) 12 C4 (b)

495

(1)(1)(1)(1) 12 C4

0.0020

221

222

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

18. (a)

x 0 1 2 3 4 5 6 7 8 9 10

P(x) 0.00002 0.00031 0.00284 0.01538 0.05465 0.13315 0.22528 0.26136 0.19899 0.08978 0.01823

(b)

Skewed left (c) The values 0, 1, 2, 3, and 10 are unusual because their probabilities are less than 0.05.

CHAPTER 5 TEST SOLUTIONS

1. (a) P x

(b) z

7.46%, V x

x P

V n

53.47 | 7.56% 50

25 7.46 | 2.3196 53.47 50

P( x ! 25) P( z ! 2.3196) 1 P( z 2.3196) | 1 0.9898 0.0102 (using technology) (c) z

x P

n x P

10 7.46 | 2.30897 53.47 50 30 7.46 | 2.98077 53.47 50

P(10 x 30) P( 2.30897 z 2.98077) | 0.99856 0.01047 | 0.9881 (using technology)

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

2. (a) z

x P

20 18 | 0.2632 7.6

V P( x ! 20) | P( z ! 0.2632) | 1 0.6038 | 0.3962 (using technology)

(b) z

x P

0 18 | 2.3684 V 7.6 x P 5 18 | | 1.7105 V 7.6 |

P(0 x 5) | P( 2.3684 z 1.7105) | 0.04358 0.00893 | 0.0347 (using technology) (c) z

x P

9 18 | 1.1842 V 7.6 x P 27 18 | | 1.1842 V 7.6 |

P( x 9 or x ! 27) | P( z 1.1842 or z ! 1.1842) | 0.11817 (1 0.88184) | 0.2363 (using technology) 3.

88.3% z | 1.1901 (using technology)

64.8% toright 35.2% toleft z | 0.3799 (using technology) x P zV 18 ( 0.3799)(7.6) | 15.113

n 18, p 0.37, q 0.63 np 18(0.37) 6.66 t 5 , and nq 18(0.63) 11.34 t 5

x P zV 18 (1.1901)(7.6) | 27.045

Can use the normal distribution.

P np 18(0.37) 6.66, V (a) z

x P

npq

18 0.37 0.63 | 2.048

6.5 6.66 | 0.0781 V 2.048 x P 7.5 6.66 | | 0.4102 V 2.048 |

P( x 7) | P(6.5 x 7.5) P( 0.0781 z 0.4102) | 0.6592 0.4689 0.1903 (using technology)

(b) z

x P

4.5 6.66 | 1.0547 2.048

V P( x 5) | P( x 4.5) P( z 1.0547) | 0.1458 (using technology)

223

224

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x P

9.5 6.66 | 1.3867 2.048

V P( x t 10) | P( x ! 9.5) P( z ! 1.3867) 1 P( z 1.3867) | 1 0.9172 0.0828 (using

technology)

None of the events are unusual because the probability is more than 0.05. 6.

n 12, p 0.59, q 0.41 np 12(0.59) 7.08 t 5 , and nq 12(0.41) 4.92 5 Cannot use the normal distribution because nq 5 . (a) P(7) 12C7 0.59

0.41 | 0.2284

(b) P( x ! 8) P( x t 9)

12C9 0.59

12C12 0.59 220 0.59

0.41 12C10 0.59 3

0.41 66 0.59 0.59

0.41

0.41 10

0.41 12C11 0.59

0.41

0.41 12 0.59

0.41

| 0.2047 (c) P( x 4) P(0) P(1) P(2) P(3)

12C0 0.59

0.59

0.41

12C1 0.59

12C3 0.59

0.41

12 0.59

220 0.59

0.41

12C2 0.59

66 0.59

0.41

| 0.0183 The event in part (c) is unusual because its probability is less than 0.05.

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x P

225

45,000 44,000 | 0.4082 2450

V P( x ! 45,000) P( z ! 0.4082) 0.3416 No, the event is not unusual because its probability is greater than 0.05.

z z

x P

40,000 44,000 | 1.6327 V 2450 x P 42,000 44,000 | | 0.8163 V 2450 |

P(40,000 x 42,000) | P( 1.6327 z 0.8163) | 0.20716 0.05127 | 0.1559 (using technology) (800)(0.1559) | 124 residents 9.

Middle 60% 20% toleft and 20% to right z | r 0.8416 (using technology) x P zV 44,000 ( 0.8416)(2450) | 41,938.08 x P zV 44,000 (0.8416)(2450) | 46,061.92 The middle 60% of disposable incomes lie between $41,938.08 and $46,061.92.

10. Yes; By the Central Limit Theorem, when a population is normally distributed, the sampling distribution of sample means is normally distributed for any sample size.

Confidence Intervals

CHAPTER

6.1 CONFIDENCE INTERVALS FOR THE MEAN (V KNOWN) 6.1 TRY IT YOURSELF SOLUTIONS

¦ x 114 2.85

n 40 Another point estimate for the population mean is 2.85.

zc 1.96, n 40, V 1.4 V 1.4 E zc | 0.43 1.96 n 40 You are 95% confident that the margin of error for the population mean is about 0.43 hour.

x 2.85, E | 0.43 x E | 2.85 0.43 2.42 x E | 2.85 0.43 3.28 (2.42, 3.28)

4. 75% CI: (19.49, 20.36) 85% CI: (19.38, 20.47) 90% CI: (19.30, 20.55) As the confidence level increases, so does the width of the interval. 5.

n 30, x 22.9, V 1.5, zc 1.645 V 1.5 E zc | 0.5 1.645 n 30 x E | 22.9 0.5 22.4 x E | 22.9 0.5 23.4 (22.4, 23.4) With 90% confidence, you can say that the mean age of the students is between 22.4 and 23.4 years. Because of the larger sample size, the confidence interval is slightly narrower.

zc 1.96, E 0.75, V | 2.4

§ z V · § 1.96 2.4 · n ¨ c ¸ ¨ ¸ | 39.34 o 40 © E ¹ © 0.75 ¹ Because of the larger margin of error, the sample size needed is smaller.

CHAPTER 6 │ CONFIDENCE INTERVALS

227

6.1 EXERCISE SOLUTIONS 1. You are more likely to be correct using an interval estimate because it is unlikely that a point estimate will exactly equal the population mean. 2. b 3. d; As the level of confidence increases, zc increases, causing wider intervals. 4. No, the 95% confidence interval means that with 95% confidence you can say that the population mean is in this interval. If a large number of samples is collected and a confidence interval created for each, approximately 95% of these intervals will contain the population mean. 5. 1.28 9.

6. 1.44

x P 3.8 4.27 0.47

11. x P 13. E

15. E

26.43 24.67 1.76 V n

V n

7. 1.15

8. 2.17

10. x P 9.5 8.76 0.74 12. x P

1.96

5.2 | 1.861 30

14. E

1.28

1.3 | 0.192 75

16. E

46.56 48.12 1.56 V n

V n

1.645

2.9 | 0.675 50

2.24

4.6 | 1.030 100

17. Because c = 0.88 is the lowest level of confidence, the interval associated with it will be the narrowest. Thus, this matches (c). 18. Because c = 0.90 is the second lowest level of confidence, the interval associated with it will be the second narrowest. Thus, this matches (d). 19. Because c = 0.95 is the third lowest level of confidence, the interval associated with it will be the third narrowest. Thus, this matches (b). 20. Because c = 0.98 is the highest level of confidence, the interval associated with it will be the widest. Thus, this matches (a). 21. x r zc

22. x r zc

23. x r zc

V n

12.3 r 1.645

1.5 | 12.3 r 0.349 | (12.0, 12.6) 50

31.39 r 1.96

0.8 | 31.3 r 0.173 | (31.22, 31.56) 82

10.5 r 2.575

2.14 | 10.5 r 0.821 | (9.7, 11.3) 45

228

CHAPTER 6 │ CONFIDENCE INTERVALS

24. x r zc

20.6 r 1.28

4.7 | 20.6 r 0.602 | (20.0, 21.2) 100

14.8 12.0 13.4, E 14.8 13.4 1.4 2

25. (12.0, 14.8) x

30.15 21.61 25.88, E 30.15 25.88 4.27 2

26. (21.61, 30.15) x

27. (1.71, 2.05) x

2.05 1.71 1.88, E 2

28. (3.144, 3.176) x

3.176 3.144 2

2.05 1.88 0.17

3.16, E 3.176 3.16 0.016

29. c 0.90 zc 1.645

§zV· n ¨ c ¸ © E ¹

§ (1.645)(6.8) · ¨ ¸ | 125.13 126 1 © ¹

30. c 0.95 zc 1.96

§zV· n ¨ c ¸ © E ¹

§ (1.96)(2.5) · ¨ ¸ | 24.01 25 1 © ¹

31. c 0.80 zc 1.28

§zV· n ¨ c ¸ © E ¹

§ (1.28)(4.1) · ¨ ¸ | 6.89 7 2 © ¹

32. c 0.98 zc

§zV· n ¨ c ¸ © E ¹

2.33 2

§ (2.33)(10.1) · ¨ ¸ | 138.45 139 2 © ¹

33. (26.2, 30.1) 2E 30.1 26.2 3.9 E 1.95 and x

26.2 E

26.2 1.95 28.15

34. (244.07, 280.97) 2E 280.97 244.07 36.9 E 18.45 and x 244.07 E 244.07 18.45 262.52

232.09 | 1404.09 r 55.106 | (1348.98, 1459.20) n 48 V 232.09 | 1404.09 r 65.659 | (1338.43, 1469.75) 1404.09 r 1.96 95% CI: x r zc n 48 With 90% confidence, you can say that the population mean price is between $1348.98 and $1459.20. With 95% confidence, you can say that the population mean price is between $1338.43 and $1469.75. The 95% CI is wider.

35. 90% CI: x r zc

1404.09 r 1.645

CHAPTER 6 │ CONFIDENCE INTERVALS

229

21.77 | 97.17 r 5.969 | (91.20, 103.14) 36 n V 21.77 | 97.17 r 7.112 | (90.06, 104.28) 97.17 r 1.96 95% CI: x r zc n 36 With 90% confidence, you can say that the population mean price is between $91.20 and $103.14. With 95% confidence, you can say that the population mean price is between $90.06 and $104.28. The 95% CI is wider.

36. 90% CI: x r zc

97.17 r 1.645

48.43 | 54.97 r 11.267 | (43.70, 66.24) n 50 V 48.43 | 54.97 r 13.424 | (41.55, 68.39) 54.97 r 1.96 95% CI: x r zc n 50 With 90% confidence, you can say that the population mean price is between $43.70 and $66.24. With 95% confidence, you can say that the population mean price is between $41.55 and $68.39. The 95% CI is wider.

37. 90% CI: x r zc

54.97 r 1.645

111 | 100 r 30.864 | (69.14, 130.86) n 35 V 111 | 100 r 36.774 | (63.23, 136.77) 100 r 1.96 95% CI: x r zc n 35 With 90% confidence, you can say that the population mean tornadoes per month is between 69.14 and 130.86. With 95% confidence, you can say that the population mean tornadoes per month is between 63.23 and 136.77. The 95% CI is wider.

38. 90% CI: x r zc

100 r 1.645

39. Yes; The right endpoint of the 95% confidence interval is 1469.75. 40. No; The left endpoint of the 95% confidence interval is 90.06. 41. No; The right endpoint of the 95% confidence interval is 68.39. 42. Yes; The midpoint of the confidence intervals is 100. Therefore, the population mean is as likely to be less than 100 as it is to be greater than 100. 43. (a) An increase in the level of confidence will widen the confidence interval and the less certain you can be about a point estimate. (b) An increase in the sample size will narrow the confidence interval because it decreases the standard error. (c) An increase in the population standard deviation will widen the confidence interval because small standard deviations produce more precise intervals, which are smaller. 44. Answers will vary.

230

45. x

CHAPTER 6 │ CONFIDENCE INTERVALS

¦ x 1167 | 36.47 n

14.9 | 36.47 r 4.333 | (32.1, 40.8) n 32 V 14.9 | 36.47 r 6.782 | (29.7, 43.3) 36.47 r 2.575 99% CI: x r zc n 32 With 90% confidence, you can say that the population mean commute time is between 32.1 and 40.8 minutes. With 99% confidence, you can say that the population mean commute time is between 29.7 and 43.3 minutes. The 99% CI is wider.

90% CI: x r zc

36.47 r 1.645

¦x

795.71 | 22.10 n 36 V 7.61 | 22.10 r 2.09 | (20.01, 24.19) 22.10 r 1.645 90% CI: x r zc n 36 V 7.61 | 22.10 r 3.27 | (18.83, 25.37) 22.10 r 2.575 99% CI: x r zc n 36 With 90% confidence, you can say that the population mean concentration is between 20.01 and 24.19 grams per liter. With 99% confidence, you can say that the population mean concentration is between 18.83 and 25.37 grams per liter. The 99% CI is wider.

46. x

§zV· 47. n ¨ c ¸ © E ¹

§ 1.96 4.8 · ¨ ¸ | 88.510 o 89 1 © ¹

§zV· 48. n ¨ c ¸ © E ¹

§ 2.575 1.4 · ¨ ¸ | 3.249 o 4 2 © ¹

§zV· 49. (a) n ¨ c ¸ © E ¹

§ 1.96 3.10 · ¨ 0.75 ¸ | 65.6 o 66 servings © ¹

3.10 | 29 r 0.748 | (28.252, 29.748) n 66 No. The 95% CI is (28.252, 29.748). If the population mean is within 3% of the sample mean, then it falls outside the CI. Yes. If the population mean is within 0.3% of the sample mean, then it falls within the CI.

(b) 95% CI: x r zc

§zV· 50. (a) n ¨ c ¸ © E ¹

29 r 1.96

§ 1.645 1.6 · ¨ 1.5 ¸ | 3.1 o 4 students © ¹

CHAPTER 6 │ CONFIDENCE INTERVALS

231

1.6 | 20 r 1.316 | (18.684, 21.316) 4 n No; No; The 90% CI is (18.684, 21.316). If the population mean is within 7% or 8% of the sample mean, then it falls outside the CI.

(b) 90% CI: x r zc

§zV· 51. (a) n ¨ c ¸ © E ¹

20 r 1.645

§ 1.645 0.75 · ¨ ¸ | 6.1 o 7 cans 0.5 © ¹

0.75 | 127.75 r 0.44 | (127.3, 128.2) n 8 Yes; The 90% CI is (127.3, 128.2) and 128 ounces falls within that interval.

(b) 90% CI: x r zc

§zV· 52. (a) n ¨ c ¸ © E ¹

127.75 r 1.645

§ 1.96 1 · ¨ 0.25 ¸ | 61.5 o 62 bottles © ¹

1 | 64 r 0.238 | (63.762, 64.238) n 68 Yes; The 95% CI is (63.762, 64.238) and there are amounts greater than 63.85 fluid ounces that fall within that interval.

(b) 95% CI: x r zc

§zV· 53. (a) n ¨ c ¸ © E ¹

64 r 1.96

§ 2.575 0.5 · ¨ 0.15 ¸ | 73.7 o 74 balls © ¹

0.5 | 27.5 r 0.140 | (27.360, 27.640) n 84 Yes; The 99% CI is (27.360, 27.640) and there are amounts less than 27.6 inches that fall within that interval.

(b) 99% CI: x r zc

§zV· 54. (a) n ¨ c ¸ © E ¹

27.5 r 2.575

§ 2.575 0.10 · ¨ ¸ | 26.5 o 27 balls 0.05 © ¹

0.10 | 8.3 r 0.0442 | (8.2558, 8.3442) n 34 Yes; The 99% CI is (8.2558, 8.3442) and 8.258 inches falls within that interval.

(b) 99% CI: x r zc

8.3 r 2.575

55. Sample answer: A 99% CI may not be practical to use in all situations. It may produce a CI so wide that is has no practical application. 56. (a) An increase in the level of confidence will increase the minimum sample size required because the more data you have, the narrower the interval. (b) An increase in the error tolerance will decrease the minimum sample size required because the less data you have, the less accurate your results will be. (c) An increase in the population standard deviation will increase the minimum sample size required because the data is more spread out. Copyright © 2023 Pearson Education, Inc.

232

CHAPTER 6 │ CONFIDENCE INTERVALS

57. (a)

N n N 1

1000 500 | 0.707 1000 1

(b)

N n N 1

1000 100 | 0.949 1000 1

(c)

N n N 1

1000 75 | 0.962 1000 1

(d)

N n N 1

1000 50 | 0.975 1000 1

(e)

N n N 1

100 50 | 0.711 100 1

(f)

N n N 1

400 50 | 0.937 400 1

(g)

N n 700 50 N n 1200 50 (h) | 0.964 | 0.979 N 1 700 1 N 1 1200 1 The finite population correction factor approaches 1 as the population size increases and the sample size remains the same. The finite population correction factor approaches 1 as the population size increases and the sample size remains the same.

58. (a) 99% CI: x r zc

(b) 90% CI: x r zc

(d) 80% CI: x r zc

V n

N n N 1

8.6 r 2.575

4.9 25

N n N 1

10.9 r 1.645

2.8 500 50 | 10.9 r 0.619 | (10.3, 11.5) 50 500 1

N n N 1

40.3 r 1.96

0.5 300 68 | 40.3 r 0.105 | (40.2, 40.4) 68 300 1

N n N 1

56.7 r 1.28

9.8 36

59. Sample answer: zcV Write original equation. E n E n

zcV

Multiply each side by n .

zV n= c E §zV· n ¨ c ¸ © E ¹

Divide each side by E. 2

Square each side.

200 25 | 8.6 r 2.366 | (6.2, 11.0) 200 1

400 36 | 56.7 r 1.997 | (54.7, 58.7) 400 1

CHAPTER 6 │ CONFIDENCE INTERVALS

233

6.2 CONFIDENCE INTERVALS FOR THE MEAN (V UNKNOWN) 6.2 TRY IT YOURSELF SOLUTIONS 1.

d.f. n 1 22 1 21

90% confidence level

tc 1.721 For a t-distribution curve with 21 degrees of freedom, 90% of the area under the curve lies between t r1.721 .

2. d.f. n 1 16 1 15 90% CI: tc 1.753 s 10 E tc 1.753 | 4.4 n 16 90% CI: x r E | 162 r 4.4 (157.6, 166.4) With 90% confidence, you can say that the population mean temperature of coffee sold is between 157.6qF and 166.4qF. 99% CI: tc 2.947 s 10 E tc 2.947 | 7.4 n 16 99% CI: x r E | 162 r 7.4 (154.6, 169.4) With 99% confidence, you can say that the population mean temperature of coffee sold is between 154.6qF and 169.4qF. 3.

d.f. n 1 36 1 35 90% CI: tc 1.690 s 2.39 E tc 1.690 | 0.67 n 36 90% CI: x r E | 9.75 r 0.67 (9.08, 10.42) With 90% confidence, you can say that the population mean number of days the car model sits on the lot is between 9.08 and 10.42 days. 95% CI: tc 2.030 s 2.39 E tc 2.030 | 0.81 n 36 95% CI: x r E | 9.75 r 0.81 (8.94, 10.56) With 95% confidence, you can say that the population mean number of days the car model sits on the lot is between 8.94 and 10.56 days. The 90% confidence interval is slightly narrower.

234

CHAPTER 6 │ CONFIDENCE INTERVALS

4. Is V known? No Is n t 30? No Is the population normally distributed? Yes Use the t-distribution because V is not known and the population is normally distributed.

6.2 EXERCISE SOLUTIONS 1. The mean, median, and mode are all equal to 0. The distribution is bell-shaped and symmetric about the mean. The total area under the curve is equal to 1. 2. As the sample size increases, the t-distribution approaches the standard normal distribution. 3.

tc 1.833

2.201

2.947

2.426

s n

2.131

x r tc

s n

12.5 r 2.015

2.0 | 12.5 r 1.645 | (10.9, 14.1) 6

10. x r tc

s n

13.4 r 2.365

0.85 | 13.4 r 0.711 | (12.7, 14.1) 8

11. x r tc

s n

4.3 r 2.650

12. x r tc

s n

24.7 r 2.678

5 | 2.664 16

s n

0.34 | 4.3 r 0.241 | (4.1, 4.5) 14 4.6 | 24.7 r 1.742 | (23.0, 26.4) 50

13. (14.7, 22.1) x

14.7 22.1 18.4 E 2

14. (6.17, 8.53) x

6.17 8.53 2

15. t

x P s n

70.3 64.8 | 3.10 7.1 16

16. t

x P s n

22.9 24.5 | 2.35 4.3 40

22.1 18.4 3.7

7.35 E 8.53 7.35 1.18

4.032

3 | 4.938 6

CHAPTER 6 │ CONFIDENCE INTERVALS

17. E

s n

2.365

235

7.2 | 6.0 8

x r E | 35.5 r 6.0 (29.5, 41.5) With 95% confidence, you can say that the population mean commute time to work is between 29.5 and 41.5 minutes. 18. E

s n

2.776

5.8 | 7.2 5

x r E | 22.2 r 7.2 (15.0, 29.4) With 95% confidence, you can say that the population mean driving distance to work is between 15.0 and 29.4 miles. 19. E

s n

2.365

184.00 | 153.85 8

x r E | 526.50 r 153.85 (372.65,680.35) With 95% confidence, you can say that the population mean cell phone price is between $372.65 and $680.35. 20. E

s n

2.201

33.61 | 21.35 12

x r E | 90.42 r 21.35 (69.07, 111.77) With 95% confidence, you can say that the population mean repair cost is between $69.07 and $111.77. 21. E

V n

1.96

9.3 | 6.4 8

x r E | 35.5 r 6.4 (29.1, 41.9) With 95% confidence, you can say that the population mean commute time to work is between 29.1 and 41.9 minutes. This confidence interval is slightly wider than the one found in Exercise 17. 22. E

V n

1.96

5.2 | 4.6 5

x r E | 22.2 r 4.6 (17.6, 26.8) With 95% confidence, you can say that the population mean driving distance to work is between 17.6 and 26.8 miles. This confidence interval is narrower than the one found in Exercise 18. 23. Yes, $431.61 falls between $372.65 and $680.35. 24. Yes, $89.56 falls between $69.07 and $111.77. 25. (a) x 1185 (b) s | 168.1 (c) x r tc

s 168.1 | 1185 r 3.106 | 1185 r 150.72 | (1034.3, 1335.7) n 12

236

CHAPTER 6 │ CONFIDENCE INTERVALS

26. (a) x | 2.51 (b) s | 0.87 (c) x r tc

s 0.87 | 2.51 r 3.012 | 2.51 r 0.70 (1.81, 3.21) n 14

27. (a) x | 7.49 (b) s | 1.64 (c) x r tc

s 1.64 | 7.49 r 2.947 | 7.49 r 1.21 (6.28, 8.70) n 16

28. (a) x | 12.19 (b) s | 1.75 (c) x r tc

s 1.755 | 12.19 r 2.898 | 12.19 r 1.20 (10.99, 13.39) n 18

29. Yes, 1051 falls between 1034.3 and 1335.7. 30. No, 7.8 does not fall between 10.99 and 13.39. 31. (a) x | 68,757.94 (b) s | 15,834.18 (c) Use a t-distribution because V unknown and n t 30 . s 15,834.18 x r tc | 68,757.94 r 2.453 | 68,757.94 r 6866.227 | (61,892, 75,624) n 32 32. (a) x

95, 443

(b) s | 17,673.10 (c) Use a t-distribution because V unknown and n t 30 . s 17,673.10 x r tc | 95, 443 r 2.441 | 95, 443 r 7292 (88,151.00, 102,735.00) n 35 33. Yes, $67,319 falls between $61,892.21 and $75,623.67. 34. Yes, $93,867 falls between $88,151.00 and $102,735.00.

CHAPTER 6 │ CONFIDENCE INTERVALS

237

35. Use a t-distribution because V unknown and n t 30 . s 6.12 x r tc 27.7 r 2.009 | 27.7 r 1.74 | (26.0, 29.4) n 50 With 95% confidence, you can say that the population mean BMI is between 26.0 and 29.4. 36. Use a t-distribution because V is unknown and the interest rates are normally distributed. s 0.49 x r tc 3.95 r 2.110 | 3.95 r 0.24 (3.71, 4.19) n 18 With 95% confidence, you can say that the population mean interest rate is between 3.71% and 4.19%. 37. Use the standard normal distribution because V is known and the weights are known to be normally distributed. V 2.49 x r zc | 13.68 r 1.543 | (12.14, 15.22) 13.68 r 1.96 n 10 With 95% confidence, you can say that the population mean weight of two-year-old males is between 12.14 pounds and 15.22 pounds. 38. Neither distribution can be used because n 30 and the data used are not normally distributed. 39. Neither distribution can be used because n < 30 and the mileages are not normally distributed. 40. Use the standard normal distribution because V is known and the yards per carry are normally distributed. V 1.76 x r zc | 4.392 r 0.957 | (3.44, 5.35) 4.392 r 1.96 n 13 With 95% confidence, you can say that the population mean yards per carry is between 3.44 and 5.35. 41. n 25, x 56.0, s 0.25 x P 56.0 55.5 t 10 s 0.25 n 25 t0.99 2.797, t0.99 2.797 No; They are not making good tennis balls because for this sample the t-value is t 10 , which is not between t0.99 2.797 and t0.99 2.797 . 42. n 16, x 1015, s 25 x P 1015 1000 t 2.4 s 25 n 16 t0.99 2.947, t0.99 2.947 Yes; They are making good light bulbs because for this sample the t-value is t between t0.99 2.947 and t0.99 2.947 .

2.4 , which is

238

CHAPTER 6 │ CONFIDENCE INTERVALS

6.3 CONFIDENCE INTERVALS FOR POPULATION PROPORTIONS 6.3 TRY IT YOURSELF SOLUTIONS 1.

x 147, n 147 420 378 105 1050 x 147 pˆ 0.14 14% n 1050

pˆ 0.14, qˆ 1 0.14 0.86 npˆ (1050)(0.14) 147 ! 5 nqˆ (1050)(0.86) 903 ! 5 Distribution of p̂ is approximately normal. zc 1.645 ˆˆ pq (0.14)(0.86) | 0.018 1.645 n 1050 pˆ r E | 0.14 r 0.018 (0.122, 0.158)

n 800, pˆ 0.52 qˆ 1 pˆ 1 0.52 0.48 npˆ 800 0.52 416 ! 5 nqˆ 800 0.48 384 ! 5 Distribution of p̂ is approximately normal. ˆˆ pq (0.52)(0.48) | 0.045 2.575 n 800 pˆ r E | 0.52 r 0.045 (0.475, 0.565)

2.575, E

4. (1) pˆ 0.5, qˆ 0.5, zc 1.645, E 0.02 2

§z · § 1.645 · ˆ ˆ ¨ c ¸ (0.5)(0.5) ¨ pq ¸ | 1691.266 o 1692 © 0.02 ¹ ©E¹ At least 1692 people should be included in the sample. n

(2) pˆ 0.622, qˆ 0.378, zc 1.645, E 0.02 2

§z · § 1.645 · ˆ ˆ ¨ c ¸ (0.622)(0.378) ¨ n pq ¸ | 1590.574 o 1591 © 0.02 ¹ ©E¹ At least 1591 people should be included in the sample.

CHAPTER 6 │ CONFIDENCE INTERVALS

239

6.3 EXERCISE SOLUTIONS 1. True 2. False. To find the minimum sample size required for a confidence interval, solve for n in the formula for margin of error and round n up to the next whole number. 3.

pˆ

x n

62 | 0.060, qˆ 1 pˆ | 0.940 1040

pˆ

x n

200 | 0.570, qˆ 1 pˆ | 0.430 351

pˆ

x n

218 | 0.330, qˆ 1 pˆ | 0.670 661

pˆ

x n

1025 | 0.240, qˆ 1 pˆ | 0.760 4272

0.905, 0.933 o pˆ

0.905 0.933 2

0.919 E

0.933 0.919 0.014

0.245, 0.475 o pˆ

0.245 0.475 2

0.360 E

0.475 0.360 0.115

0.512, 0.596 o pˆ

0.512 0.596 2

0.554 E 0.596 0.554 0.042

10. 0.087, 0.263 o pˆ

0.087 0.263 2

0.175 E

0.263 0.175 0.088

x n

1325 | 0.740, qˆ 1 pˆ | 0.260 1790 ˆˆ pq (0.740)(0.260) 90% CI: pˆ r zc | 0.740 r 1.645 | 0.740 r 0.017 (0.723, 0.757) n 1790 ˆˆ pq (0.740)(0.260) 95% CI: pˆ r zc | 0.740 r 1.96 | 0.740 r 0.020 (0.720, 0.760) n 1790 With 90% confidence, you can say that the population proportion of the U.S. adults who say they have made a New Year’s resolution is between 72.3% and 75.7%. With 95% confidence, you can say that it is between 72.0% and 76.0%. The 95% confidence interval is slightly wider.

11. pˆ

x n

816 | 0.456, qˆ 1 pˆ | 0.544 1790 ˆˆ pq (0.456)(0.544) 90% CI: pˆ r zc | 0.456 r 1.645 | 0.456 r 0.019 (0.437, 0.475) n 1790

12. pˆ

ˆˆ pq (0.456)(0.544) | 0.456 r 1.96 | 0.456 r 0.023 (0.433, 0.479) n 1790 With 90% confidence, you can say that the population proportion of the U.S. adults who say they made a New Year’s resolution related to their health is between 43.7% and 47.5%. With 95% confidence, you can say that it is between 43.3% and 47.9%. The 95% confidence interval is slightly wider.

95% CI: pˆ r zc

240

CHAPTER 6 │ CONFIDENCE INTERVALS

364 | 0.360, qˆ 1 pˆ | 0.640 1010 ˆˆ pq (0.360)(0.640) pˆ r zc | 0.360 r 2.575 | 0.360 r 0.039 (0.321, 0.399) n 1010 With 99% confidence, we can say that the population proportion of the U.S. adults who say they worry a great deal about the possibility of a future terrorist attack is between 32.1% and 39.9%.

13. pˆ

x n

14. pˆ

x n

15. pˆ

x n

556 | 0.550, qˆ 1 pˆ | 0.450 1010 ˆˆ pq (0.550)(0.450) pˆ r zc | 0.550 r 2.575 | 0.550 r 0.040 (0.510, 0.590) n 1010 With 99% confidence, we can say that the population proportion of the U.S. adults who say they worry a great deal about hunger and homelessness is between 51.0% and 59.0%.

pˆ r zc

16. pˆ

x n

pˆ r zc

860 | 0.056, qˆ 1 pˆ | 0.944 15,349

ˆˆ pq (0.056)(0.944) | 0.056 r 1.96 | 0.056 r 0.004 (0.052, 0.060) n 15,349

378 | 0.546, qˆ 1 pˆ | 0.454 692 ˆˆ pq (0.546)(0.454) | 0.546 r 1.645 | 0.546 r 0.031 (0.515, 0.577) n 692 §z · ˆ ˆ¨ c ¸ pq ©E¹

17. (a) n

§ 1.96 · (0.5) (0.5) ¨ ¸ | 600.25 o 601 voters © 0.04 ¹

(b) n

§z · ˆ ˆ¨ c ¸ pq ©E¹

§ 1.96 · (0.21) (0.79) ¨ ¸ | 398.33 o 399 voters © 0.04 ¹

§z · ˆ ˆ¨ c ¸ pq ©E¹

18. (a) n

§ 2.575 · (0.5) (0.5) ¨ ¸ | 4144.14 o 4145 adults © 0.02 ¹

(b) n

§z · ˆ ˆ¨ c ¸ pq ©E¹

§ 2.575 · (0.34)(0.66) ¨ ¸ | 3719.78 o 3720 adults © 0.02 ¹

CHAPTER 6 │ CONFIDENCE INTERVALS

§z · ˆ ˆ¨ c ¸ pq ©E¹

19. (a) n

§ 1.645 · (0.5)(0.5) ¨ ¸ | 751.67 o 752 families © 0.03 ¹

§z · ˆ ˆ¨ c ¸ pq ©E¹

(b) n

§ 1.645 · (0.83)(0.17) ¨ ¸ | 424.24 o 425 families © 0.03 ¹

§z · ˆ ˆ¨ c ¸ pq ©E¹

20. (a) n

§ 1.96 · (0.5) (0.5) ¨ ¸ | 384.16 o 385 adults © 0.05 ¹

§z · ˆ ˆ¨ c ¸ pq ©E¹

(b) n

§ 1.96 · (0.28) (0.72) ¨ ¸ | 309.79 o 310 adults © 0.05 ¹

(c) Having an estimate of the population proportion reduces the minimum sample size needed. 21. 90% CI (0.723, 0.757) 95% CI = (0.720, 0.760) No; 0.725 falls within both confidence intervals. 22. 90% CI (0.437, 0.475) 95% CI (0.433, 0.479) Yes; 0.48 falls outside the confidence intervals. 23. Yes; The minimum sample size needed is 399 adults. 24. Yes; The minimum sample size needed is 3720 adults. 25. France: pˆ 0.32, qˆ 0.68, n 1051 pˆ r zc

ˆˆ pq n

0.32 r 2.575

(0.32)(0.68) | 0.32 r 0.037 (0.283, 0.357) 1051

Germany: pˆ 0.25, qˆ 0.75, n 1042 ˆˆ pq (0.25)(0.75) | 0.25 r 0.035 (0.215, 0.285) 0.25 r 2.575 n 1042 United Kingdom: pˆ 0.31, qˆ 0.69, n 1003 pˆ r zc

ˆˆ pq (0.31)(0.69) | 0.31 r 0.038 (0.272, 0.348) 0.31 r 2.575 n 1003 United States: pˆ 0.35, qˆ 0.65, n 1000 pˆ r zc

pˆ r zc

ˆˆ pq n

0.35 r 2.575

(0.35)(0.65) | 0.35 r 0.039 (0.311, 0.389) 1000

241

242

CHAPTER 6 │ CONFIDENCE INTERVALS

26. Germany and the United States. Their confidence intervals do not overlap. 27. Cyberterrorism: pˆ 0.82, qˆ 1 pˆ 0.18 ˆˆ (0.82)(0.18) pq 0.82 r 1.96 | 0.82 r 0.024 (0.796, 0.844) 1021 n Development of nuclear weapons by Iran: pˆ 0.75, qˆ 1 pˆ 0.25 pˆ r zc

ˆˆ pq (0.75)(0.25) | 0.75 r 0.027 (0.723, 0.777) 0.75 r 1.96 n 1021 The spread of infectious diseases throughout the world: pˆ 0.72, qˆ 1 pˆ 0.28 pˆ r zc

ˆˆ pq (0.72)(0.28) | 0.72 r 0.028 (0.692, 0.748) 0.72 r 1.96 n 1021 Global warming or climate change: pˆ 0.58, qˆ 1 pˆ 0.42 pˆ r zc

ˆˆ pq (0.58)(0.42) | 0.58 r 0.030 (0.550, 0.610) 0.58 r 1.96 n 1021 The military power of Russia: pˆ 0.44, qˆ 1 pˆ 0.56 pˆ r zc

ˆˆ pq (0.44)(0.56) | 0.44 r 0.030 (0.410, 0.470) 0.44 r 1.96 n 1021 The conflict between the Israelis and Palestinians: pˆ 0.32, qˆ 1 pˆ 0.68 pˆ r zc

pˆ r zc

ˆˆ pq n

0.32 r 1.96

(0.32)(0.68) | 0.32 r 0.029 (0.291, 0.349) 1021

28. “Development of nuclear weapons by Iran” and “the spread of infectious diseases throughout the world.” Their confidence intervals overlap. 29. 31% r 2.9% o (28.1%, 33.9%) o (0.281, 0.339) E

ˆˆ pq o zc n

n ˆˆ pq

0.029

1502 |o zc (0.31)(0.69)

2.43

P( 2.43 z 2.43) 0.9925 0.0075 0.985 (0.281, 0.339) is approximately a 98.5% CI. 30. 65% r 7.9% o (57.1%, 72.9%) o (0.571, 0.729) E

ˆˆ pq o zc n

n ˆˆ pq

0.079

220 |o zc (0.65)(0.35)

P( 2.46 z 2.46) 0.9931 0.0069 0.9862 (0.571, 0.729) is approximately a 98.6% CI.

2.46

CHAPTER 6 │ CONFIDENCE INTERVALS

243

31. 71% r 3% o (68%, 74%) o (0.68, 0.74) E

ˆˆ pq o zc n

n ˆpqˆ

0.03

1000 |o zc (0.71)(0.29)

2.09

P( 2.09 z 2.09) 0.9817 0.0183 0.9634 (0.68, 0.74) is approximately a 96.3% CI. 32. 73% r 4.8% o (68.2%, 77.8%) o (0.682, 0.778) E

ˆˆ pq o zc n

n ˆˆ pq

0.048

880 |o zc (0.73)(0.27)

3.21

P( 3.21 z 3.21) 0.9993 0.0007 0.9986 (0.682, 0.778) is approximately a 99.9% CI. 33. 57% r 3.6% o (53.4%, 60.6%) o (0.534, 0.606) E

ˆˆ pq o zc n

n ˆˆ pq

0.036

2094 |o zc (0.57)(0.43)

3.33

P( 3.33 z 3.33) 0.9996 0.0004 0.9992 (0.534, 0.606) is approximately a 99.9% CI. 34. 68% r 3% o (65%, 71%) o (0.65, 0.71) E

ˆˆ pq o zc n

n ˆˆ pq

0.03

1052 |o zc (0.68)(0.32)

2.09

P( 2.09 z 2.09) 0.9817 0.0183 0.9634 (0.65, 0.71) is approximately a 96.3% CI. 42% r 3% o (39%, 45%) o (0.39, 0.45) E

ˆˆ pq o zc n

n ˆˆ pq

0.03

1052 |o zc (0.42)(0.58)

1.97

P( 1.97 z 1.97) 0.9756 0.0244 0.9512 (0.39, 0.45) is approximately a 95.1% CI. 35. If npˆ 5 or nqˆ 5, the sampling distribution of p̂ may not be normally distributed, so zc cannot be used to calculate the confidence interval. 36. Sample answer: ˆˆ pq E zc n E n

ˆˆ zc pq

ˆˆ pq

§z · ˆ ˆ¨ c ¸ pq ©E¹

Write original equation. Multiply each side by n .

zc E

Divide each side by E. 2

Square each side.

244

37.

CHAPTER 6 │ CONFIDENCE INTERVALS

ˆˆ pq p̂ qˆ 1 pˆ p̂ 0.0 1.0 0.00 0.9 0.1 0.9 0.09 1.0 0.2 0.8 0.16 0.45 0.3 0.7 0.21 0.46 0.4 0.6 0.24 0.47 0.5 0.5 0.25 0.48 0.6 0.4 0.24 0.49 0.7 0.3 0.21 0.50 0.8 0.2 0.16 0.51 ˆ ˆ. pˆ 0.5 gives the maximum value of pq

qˆ 1 pˆ 0.1 0.0 0.55 0.54 0.53 0.52 0.51 0.50 0.49

ˆˆ pq 0.09 0.00 0.2475 0.2484 0.2491 0.2496 0.2499 0.2500 0.2499

p̂ 0.52 0.53 0.54 0.55

qˆ 1 pˆ 0.48 0.47 0.46 0.45

ˆˆ pq 0.2496 0.2491 0.2484 0.2475

6.4 CONFIDENCE INTERVALS FOR VARIANCE AND STANDARD DEVIATION 6.4 TRY IT YOURSELF SOLUTIONS 1.

d.f. n 1 29 level of confidence = 0.90 Area to the right of F R2 is 0.05. Area to the right of F L2 is 0.95.

F R2 42.557 , F L2 17.708 For a chi-square distribution curve with 29 degrees of freedom, 90% of the area under the curve lies between 17.708 and 42.557. 2. 90% CI: F R2 95% CI: F R2

42.557 , F L2 17.708 45.722 , F L2

16.047

§ (n 1) s 2 (n 1) s 2 · , 90% CI for V 2 : ¨ ¸ 2 F L2 ¹ © FR § (n 1) s 2 (n 1) s 2 · , 95% CI for V 2 : ¨ ¸ 2 F L2 ¹ © FR 90% CI for V :

0.981,

§ 29 (1.2) 2 29 (1.2) 2 · , ¨ ¸ | (0.98, 2.36) 17.708 ¹ © 42.557 § 29 (1.2) 2 29 (1.2) 2 · , ¨ ¸ | (0.91, 2.60) 16.047 ¹ © 45.722

2.358 | 0.99, 1.54

95% CI for V : 0.913, 2.602 | 0.96, 1.61 With 90% confidence, you can say that the population variance is between 0.98 and 2.36 and that the population standard deviation is between 0.99 and 1.54. With 95% confidence, you can say that the population variance is between 0.91 and 2.60, and that the population standard deviation is between 0.96 and 1.61.

CHAPTER 6 │ CONFIDENCE INTERVALS

245

6.4 EXERCISE SOLUTIONS 1. Yes. 2. It approaches the shape of the normal curve. 3.

F R2 14.067 , F L2

2.167

F R2

31.319 , F L2

4.075

F R2

32.852 , F L2

8.907

F R2

44.314 , F L2 11.524

F R2

52.336 , F L2

13.121

F R2

63.167 , F L2

37.689

§ (n 1) s 2 (n 1) s 2 · § 29 (11.56) 29 (11.56) · , , 9. (a) ¨ ¸ ¨ ¸ | (7.33, 20.89) 2 16.047 ¹ F L2 ¹ © 45.722 © FR (b)

7.3321,

20.8911 | (2.71, 4.57)

§ (n 1) s 2 (n 1) s 2 · § 6 (0.64) 6 (0.64) · , , 10. (a) ¨ ¸ ¨ ¸ | (0.21, 5.68) 2 0.676 ¹ F L2 ¹ © 18.548 © FR (b)

0.2070,

5.6805 | (0.46, 2.38)

§ (n 1) s 2 (n 1) s 2 · § 17 (35) 2 17 (35) 2 · , , 11. (a) ¨ ¸ ¨ ¸ | (755, 2401) 2 2 27.587 8.672 F F R L © ¹ © ¹ (b)

754.885,

2401.407 | (27, 49)

§ (n 1) s 2 (n 1) s 2 · § 40 (278.1) 2 40 (278.1) 2 · , , 12. (a) ¨ ¸ ¨ ¸ | (48,571.8, 139,577.0) 2 22.164 ¹ F L2 ¹ © 63.691 © FR (b)

48,571.77, 139,576.99 | (220.4, 373.6)

13. (a) s 2 | 0.0756 § (n 1) s 2 (n 1) s 2 · § 17 (0.0756) 17 (0.0756) · , , ¨ ¸|¨ ¸ | (0.0426, 0.1699) 2 7.564 ¹ F L2 ¹ © 30.191 © FR (b)

0.04257, 0.16991 | (0.2063, 0.4122) With 95% confidence, you can say that the population variance is between 0.0426 and 0.1699, and the population standard deviation is between 0.2063 and 0.4122 inch.

246

CHAPTER 6 │ CONFIDENCE INTERVALS

14. (a) s2 | 18.116 § (n 1) s 2 (n 1) s 2 · § 14 (18.116) 14 (18.116) · , , ¨ ¸|¨ ¸ | (10.71, 38.60) 2 6.571 ¹ F L2 ¹ © 23.685 © FR (b)

10.71, 38.60 | (3.27, 6.21) With 90% confidence, you can say that the population variance is between 10.71 and 38.60, and the population standard deviation is between 3.27 and 6.21 minutes.

15. (a) s 2 104.811 § (n 1) s 2 (n 1) s 2 · § 20 (104.811) 20 (104.811) · , , ¨ ¸|¨ ¸ | (52.41, 281.98) 2 7.434 F L2 ¹ © 39.997 ¹ © FR (b)

52.41, 281.98 | (7.24, 16.79) With 99% confidence, you can say that the population variance is between 52.41 and 281.98, and the population standard deviation is between 7.24 and 16.79 thousand dollars.

16. (a) s2 | 253.913 § (n 1) s 2 (n 1) s 2 · § 23 (253.913) 23 (253.913) · , , ¨ ¸|¨ ¸ | (153.4, 499.6) 2 11.689 F L2 ¹ © 38.076 ¹ © FR (b)

153.4, 499.6 | (12.4, 22.4) With 95% confidence, you can say that the population variance is between 153.4 and 499.6, and the population standard deviation is between 12.4 and 22.4.

§ (n 1) s 2 (n 1) s 2 · § 13 (3.54) 2 13 (3.54) 2 · , , 17. (a) ¨ ¸|¨ ¸ | (5.46, 45.70) 2 3.565 ¹ F L2 ¹ © 29.819 © FR (b)

5.46, 45.70 | (2.34, 6.76) With 99% confidence, you can say that the population variance is between 5.46 and 45.70, and the population standard deviation is between 2.34 and 6.76 days.

§ (n 1) s 2 (n 1) s 2 · § 14 (26.1) 2 14 (26.1) 2 · , , 18. (a) ¨ ¸ ¨ ¸ | (452.76, 1224.25) 2 7.790 ¹ F L2 ¹ © 21.064 © FR (b)

452.76, 1224.25 | (21.28, 34.99) With 80% confidence, you can say that the population variance is between 452.76 and 1224.25, and the population standard deviation is between 21.28 and 34.99 service aces.

§ (n 1) s 2 (n 1) s 2 · § 18 (15) 2 18 (15) 2 · , , 19. (a) ¨ ¸ ¨ ¸ | (128, 492) 2 8.231 ¹ F L2 ¹ © 31.526 © FR

CHAPTER 6 │ CONFIDENCE INTERVALS

(b)

247

128.465, 492.042 | (11, 22) With 95% confidence, you can say that the population variance is between 128 and 492, and the population standard deviation is between 11 and 22 grains per gallon.

§ (n 1) s 2 (n 1) s 2 · § 29(3600) 2 29(3600) 2 · , , 20. (a) ¨ ¸ ¨ ¸ | (8,831, 450, 21,224,305) 2 17.708 ¹ F L2 ¹ © 42.557 © FR (b)

8,831,450, 21,224,305 | (2972, 4607) With 90% confidence, you can say that the population variance is between 8,831,450 and 21,224,305, and the population standard deviation is between $2972 and $4607.

§ (n 1) s 2 (n 1) s 2 · § 17 (0.25) 2 17 (0.25) 2 · , , 21. (a) ¨ ¸ ¨ ¸ | (0.04, 0.11) 2 10.085 ¹ F L2 ¹ © 24.769 © FR (b)

0.0429, 0.1054 | (0.21, 0.32) With 80% confidence, you can say that the population variance is between 0.04 and 0.11, and the population standard deviation is between 0.21 and 0.32 day.

§ (n 1) s 2 (n 1) s 2 · § 60 (6.46)2 60 (6.46)2 · , , 22. (a) ¨ ¸ ¨ ¸ | (28.33, 66.80) 2 37.485 ¹ F L2 ¹ © 88.379 © FR (b)

28.33,

66.80 | (5.32, 8.17)

With 98% confidence, you can say that the population variance is between 28.33 and 66.80, and the population standard deviation is between 5.32 and 8.17 inches.

§ (n 1) s 2 (n 1) s 2 · § 27 (56.1) 2 27 (56.1) 2 · , , 23. (a) ¨ ¸ ¨ ¸ | (1809.40, 6597.92) 2 12.879 ¹ F L2 ¹ © 46.963 © FR (b)

1809.40, 6597.92 | (42.54, 81.23) With 98% confidence, you can say that the population variance is between 1809.40 and 6597.92, and the population standard deviation is between 42.54 and 81.23 seconds.

§ (n 1) s 2 ( n 1) s 2 · § 19 (0.91) 2 19 (0.91) 2 · , , 24. (a) ¨ ¸ ¨ ¸ | (0.52, 1.56) 2 10.117 ¹ F L2 ¹ © 30.144 © FR (b)

0.52, 1.56 | (0.72, 1.25) With 90% confidence, you can say that the population variance is between 0.52 and 1.56, and the population standard deviation is between 0.72 and 1.25 seconds.

25. 95% CI for V : (0.2063, 0.4122) Yes, because all of the values in the confidence interval are less than 0.5.

248

CHAPTER 6 │ CONFIDENCE INTERVALS

26. 90% CI for V : (3.27, 6.21) Yes, because all the values in the confidence interval are less than 10. 27. 80% CI for V : (0.21, 0.32) No, because 0.25 is contained in the confidence interval. 28. 98% CI for V : (42.54, 81.23) No, because 1 minute is contained in the confidence interval. 29. Answers will vary. Sample answer: Unlike a confidence interval for a population mean or proportion, a confidence interval for a population variance does not have a margin of error. The left and right endpoints must be calculated separately.

CHAPTER 6 REVIEW EXERCISE SOLUTIONS 1. (a) x | 103.5 (b) E

V n

1.645

45 | 11.7 40

1.96

8 | 2.9 30

2. (a) x | 9.5 (b) E

V n

45 | 103.5 r 11.7 (91.8, 115.2) n 40 With 90% confidence, you can say that the population mean waking time is between 91.8 and 115.2 minutes past 5:00 A.M.

3. (a) x r zc

103.5 r 1.645

(b) x r 0.10x 103.5 r 0.10(103.5) 103.5 r 10.35 (93.15, 113.85) Yes; If the population mean is within 10% of the sample mean, then it falls inside the confidence interval.

8 | 9.5 r 2.9 (6.6, 12.4) n 30 With 95% confidence, you can say that the population mean driving distance is between 6.6 and 12.4 miles.

4. (a) x r zc

9.5 r 1.96

(b) x r 0.10x 9.5 r 0.125(9.5) 9.5 r 1.19 (8.31, 10.69) No; 12.5 is outside the confidence interval.

CHAPTER 6 │ CONFIDENCE INTERVALS

20.75, 24.10 o x

20.75 24.10 2

22.425 o E

24.10 22.425 1.675

7.428, 7.562 o x

7.428 7.562 2

7.495 o E

7.562 7.495 0.067

§zV· n ¨ c ¸ © E ¹

§ 1.96 45 · ¨ 10 ¸ | 77.79 78 people © ¹

§zV· n ¨ c ¸ © E ¹

§ 2.575 8 · ¨ ¸ | 106.09 107 people 2 © ¹

tc 1.383

13. (a) E

(b) x r tc

14. (a) E

(b) x r tc

15. (a) E tc

(b) x r tc

16. (a) E tc

(b) x r tc

s n

10. tc

2.069

1.753

25.6 | 11.2 16

11. tc

2.624

12. tc

2.756

s | 72.1 r 11.2 (60.9, 83.3) n s n

2.064

1.1 | 0.5 25

s | 3.5 r 0.5 (3.0, 4.0) n

s n

§ 0.9 · 2.718 ¨ ¸ | 0.7 © 12 ¹

s | 6.8 r 0.7 n

s n

(6.1, 7.5)

§ 16.5 · 2.861¨ ¸ | 10.6 © 20 ¹

s | 25.2 r 10.6 (14.6, 35.8) n

s 67 165 r 1.690 | 165 r 18.87 | (146, 184) n 36 With 90% confidence, you can say that the population mean height is between 146 and 184 feet.

17. x r tc

18. Yes; 160 feet falls within the confidence interval, (146, 184).

249

250

CHAPTER 6 │ CONFIDENCE INTERVALS

19. (a) pˆ

x n

383 | 0.420, qˆ 1 pˆ | 0.580 912

(b) 90% CI: pˆ r zc 95% CI: pˆ r zc

ˆˆ 0.420 0.580 pq | 0.420 r 1.645 | 0.420 r 0.027 (0.393, 0.447) 912 n ˆˆ 0.420 0.580 pq | 0.420 r 1.96 | 0.420 r 0.032 (0.388, 0.452) 912 n

(c) With 90% confidence, you can say that the population proportion of the U.S. adults in Generation Z who say they are at least somewhat likely to consider an electric vehicle for their next vehicle is between 39.3% and 44.7%. With 95% confidence, you can say that it is between 38.8% and 45.2%. The 95% confidence interval is slightly wider. 20. (a) pˆ

x n

5362 | 0.390, qˆ 1 pˆ | 0.610 13,749

(b) 90% CI: pˆ r zc

ˆˆ pq 0.390 0.610 | 0.390 r 1.645 | 0.390 r 0.007 (0.383, 0.397) n 13,749

95% CI: pˆ r zc

ˆˆ pq 0.390 0.610 | 0.390 r 1.96 | 0.390 r 0.008 (0.382, 0.398) n 13,749

(c) With 90% confidence, you can say that the population proportion of the U.S. adults who say that they are at least somewhat likely to consider an electric vehicle for their next vehicle purchase is between 38.3% and 39.7%. With 95% confidence, you can say that it is between 38.2% and 39.8%. The 95% confidence interval is slightly wider. 21. (a) pˆ

x n

23,991 | 0.325, qˆ 1 pˆ | 0.675 73,901

(b) 90% CI: pˆ r zc

ˆˆ pq 0.325 0.675 | 0.325 r 1.645 | 0.325 r 0.003 (0.322, 0.328) n 73,901

95% CI: pˆ r zc

ˆˆ pq 0.325 0.675 | 0.325 r 1.96 | 0.325 r 0.003 (0.322, 0.328) n 73,901

(c) With 90% confidence, you can say that the population proportion of college graduates who earn a postgraduate degree is between 32.2% and 32.8%. With 95% confidence, you can say that it is between 32.2% and 32.8%. The confidence intervals are the same. 22. (a) pˆ

x n

1334 | 0.600, qˆ 1 pˆ | 0.400 2223

(b) 90% CI: pˆ r zc 95% CI: pˆ r zc

ˆˆ 0.600 0.400 pq | 0.600 r 1.645 | 0.600 r 0.017 (0.583, 0.617) 2223 n ˆˆ 0.600 0.400 pq | 0.600 r 1.96 | 0.600 r 0.020 (0.580, 0.620) 2223 n

CHAPTER 6 │ CONFIDENCE INTERVALS

251

(c) With 90% confidence, you can say that the population proportion of the U.S. adults who say an occupation as an athlete is prestigious is between 58.3% and 61.7%. With 95% confidence, you can say it is between 58.0% and 62.0%. The 95% confidence interval is slightly wider. 23. Yes; it falls outside both confidence intervals. 24. No; it is in the 95% confidence interval.

§z · ˆ ˆ¨ c ¸ pq ©E¹

25. (a) n

§ 1.96 · 0.50 0.50 ¨ ¸ | 384.16 o 385 adults © 0.05 ¹

(b) n

§z · ˆ ˆ¨ c ¸ pq ©E¹

§ 1.96 · 0.32 0.68 ¨ ¸ | 334.4 o 335 adults © 0.05 ¹

(c) Having an estimate of the population proportion reduces the minimum sample size needed. 26. Yes, because the minimum sample size needed is 335 adults. 27. F R2

23.337 , F L2

28. F R2

42.980 , F L2 10.856

29. F R2

24.996 , F L2

7.261

30. F R2

23.589 , F L2

1.735

4.404

31. s2 | 25.833

§ (n 1) s 2 (n 1) s 2 · § 12 (25.833) 12 (25.833) · , , (a) 95% CI for V 2 : ¨ ¸|¨ ¸ | (13.28, 70.39) 2 4.404 ¹ F L2 ¹ © 23.337 © FR (b) 95% CI for V :

13.28,

70.39 | (3.64, 8.39)

With 95% confidence, we can say that the population variance is between 13.28 and 70.39, and the population standard deviation is between 3.64 and 8.39. 32. s2 | 0.011

§ (n 1) s 2

(a) 98% CI for V 2 : ¨

F R2

(n 1) s 2 · § (30)(0.011) (30)(0.011) · , ¸|¨ ¸ | (0.006, 0.022) 14.954 ¹ F L2 ¹ © 50.892

§ (30)(0.011) (30)(0.011) · (b) 98% CI for V : ¨ , ¸ | (0.081, 0.149) ¨ 50.892 14.954 ¸¹ © With 98% confidence, you can say that the population variance is between 0.006 and 0.022, and the population standard deviation is between 0.081 and 0.149. Copyright © 2023 Pearson Education, Inc.

252

CHAPTER 6 │ CONFIDENCE INTERVALS

CHAPTER 6 QUIZ SOLUTIONS

1. (a) x (b) E

¦x

49.17 | 2.46 20

1.96

0.068 | 0.030 20

| 2.46 r 0.030 (2.43, 2.49) n With 95% confidence, you can say that the population mean winning time is between 2.43 and 2.49 hours.

(d) No; It falls outside the confidence interval. 2

§zV· n ¨ c ¸ © E ¹

3. (a) x

§ · ¨ 2.575 0.068 ¸ ¨ ¸ | 27.59 o 28 champions 2 ¨¨ ¸¸ 60 © ¹

6.61, s | 3.376

s 3.376 | 6.61 r 1.833 | 6.61 r 1.957 | (4.65, 8.57) n 10 With 90% confidence you can say that the population mean amount of time is between 4.65 and 8.57 minutes.

(b) x r tc

3.5 | 6.61 r 1.82 (4.79, 8.43) n 10 With 90% confidence you can say that the population mean amount of time is between 4.79 and 8.43 minutes. This confidence interval is narrower than the one found in part (b).

| 6.61 r 1.645

s 36,729 133,326 r 2.201 | 133,326 r 23,336 (109,990, 156,662) n 12 With 95% confidence you can say that the population mean annual earnings is between $109,990 and $156,662. x r tc

5. Yes. $131,935 falls within the confidence interval, (109,990, 156,662). 6. (a) pˆ

x n

838 | 0.830 1010

ˆˆ 0.830 0.170 pq | 0.830 r 1.645 | 0.830 r 0.019 (0.811, 0.849) 1010 n With 90% confidence, you can say that the population proportion of the U.S. adults who say that the energy situation in the United States is very or fairly serious is between 81.1% and 84.9%.

(b) pˆ r zc

CHAPTER 6 │ CONFIDENCE INTERVALS

253

(d) n

§z · § 2.575 · ˆ ˆ ¨ c ¸ | 0.830 0.170 ¨ pq ¸ | 584.74 o 585 adults © 0.04 ¹ ©E¹

§ (n 1) s 2 (n 1) s 2 · § 9 (3.38) 2 9 (3.38) 2 · , , 7. (a) ¨ ¸|¨ ¸ | (5.41, 38.08) 2 2.700 ¹ F L2 ¹ © 19.023 © FR (b)

38.0813 | (2.32, 6.17)

5.4050,

With 95% confidence you can say that the population standard deviation is between 2.32 and 6.17 minutes.

CHAPTER 6 TEST SOLUTIONS

1. (a) pˆ

x n

1740 | 0.830 2096

ˆˆ 0.830 0.170 pq | 0.830 r 1.96 | 0.830 r 0.016 (0.814, 0.846) 2096 n With 95% confidence, you can say that the population proportion of the U.S. adults who think football teams of all levels should require players who suffer a head injury to take a set amount of time off from playing to recover is between 81.4% and 84.6%.

(b) pˆ r zc

(d) n 2. (a) x

§z · § 2.575 · ˆ ˆ ¨ c ¸ | 0.83 0.17 ¨ pq ¸ | 1039.5 o 1040 adults © 0.03 ¹ ©E¹

214.1, s | 51.1

s 51.07 | 214.1 r 2.262 | 214.1 r 36.53 | (177.6, 250.6) n 10 With 95% confidence you can say that the population mean weight is between 177.6 and 250.6 pounds.

(b) x r tc

§ (n 1) s 2 (n 1) s 2 · § (9)(51.1) 2 (9)(51.1) 2 · ¸ ¨ ¸ | (31.6, 116.4) | , , (c) 99% CI for V : ¨ 2 2 ¨ ¸ ¨ 23.589 1.735 ¸ F F R L © ¹ © ¹ With 99% confidence, you can say that the population standard deviation is between 31.6 and 116.4 pounds.

254

CHAPTER 6 │ CONFIDENCE INTERVALS

3. (a) x

605

108 | 605 r 51.29 (553.71, 656.29) n 12 With 90% confidence, you can say that the population mean test score is between 553.71 and 656.29.

(b) x r zc

605 r 1.645

§zV· (d) n ¨ c ¸ © E ¹

§ 1.96 108 · ¨ 10 ¸ | 448.08 o 449 students © ¹

4. (a) Use the t-distribution because V is unknown and n t 30 . s 7.5 x r tc 20 r 2.023 | 20 r 2.4 (17.6, 22.4) n 40 With 95% confidence, you can say that the population mean waiting time is between 17.6 and 22.4 minutes. (b) Use the standard normal distribution because V is known and the weights are normally distributed. V 0.05 x r zc | 11.89 r 0.021 | (11.87, 11.91) 11.89 r 1.645 n 15 With 90% confidence, you can say that the population mean weight is between 11.87 and 11.91 ounces.

CHAPTER

Hypothesis Testing with One Sample

7.1 INTRODUCTION TO HYPOTHESIS TESTING 7.1 TRY IT YOURSELF SOLUTIONS 1. (1) The mean is not 74 months; μ ≠ 74 . H 0 : μ = 74; H a : μ ≠ 74 (claim) (2) The variance is less than or equal to 2.7; σ 2 ≤ 2.7 .

σ 2 > 2.7 H 0 : σ 2 ≤ 2.7 (claim); H a : σ 2 > 2.7 (3) The proportion is more than 24%; p > 0.24 . H0 : p ≤ 0.24; H a : p > 0.24 (claim) 2.

H0 : p ≤ 0.01; H a : p > 0.01 A type I error will occur if the actual proportion is less than or equal to 0.01, but you reject H 0 . A type II error will occur if the actual proportion is greater than 0.01, but you fail to reject H 0 . A type II error is more serious because you would be misleading the consumer, possibly causing serious injury or death.

3. (1) H 0 : The mean life of a certain type of automobile battery is 74 months. H a : The mean life of a certain type of automobile battery is not 74 months.

H 0 : μ = 74; H a : μ ≠ 74 Two-tailed

(2) H0: The variance of the life of a manufacturer’s home theater systems is less than or equal to 2.7. Ha: The variance of the life of a manufacturer’s home theater systems is greater than 2.7. H0 : σ2 ≤ 2.7 ; H a : σ2 > 2.7 Right-tailed

(3) H 0 : The proportion of homeowners who feel their house is too small for their family is less than or equal to 24%.

256

CHAPTER 7 Ň HYPOTHESIS TESTING WITH ONE SAMPLE

H a : The proportion of homeowners who feel their house is too small for their family is greater than 24%. H0 : p ≤ 0.24; H a : p > 0.24 Right-tailed

4. (1) There is enough evidence to support the claim that the mean life of a certain type of automobile battery is not 74 months. There is not enough evidence to support the claim that the mean life of a certain type of automobile battery is not 74 months. (2) There is enough evidence to support the realtor’s claim that the proportion of homeowners who feel their house is too small for their family is more than 24%. There is not enough evidence to support the realtor’s claim that the proportion of homeowners who feel their house is too small for their family is more than 24%. 5. (1) H 0 : μ ≥ 650; H a : μ < 650 (claim) If you reject H0, then you will support the claim that the mean repair cost per automobile is less than $650. (2) H 0 : μ = 98.6 (claim); H a : μ ≠ 98.6 If you reject H0, then you will reject the claim that the mean temperature is about 98.6°F.

7.1 EXERCISE SOLUTIONS 1. The two types of hypotheses used in a hypothesis test are the null hypothesis and the alternative hypothesis. The alternative hypothesis is the complement of the null hypothesis. 2. Type I Error: The null hypothesis is rejected when it is true. Type II Error: The null hypothesis is not rejected when it is false. 3. You can reject the null hypothesis, or you can fail to reject the null hypothesis. 4. No; Failing to reject the null hypothesis means that there is not enough evidence to reject it. 5. False. In a hypothesis test, you assume the null hypothesis is true. 6. False. A statistical hypothesis is a statement about a population. 7. True 8. True 9. False. A small P-value in a test will favor a rejection of the null hypothesis. Copyright © 2023 Pearson Education, Inc.

CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

257

10. False. To support a claim, state it so that it becomes the alternative hypothesis. 11. H0 : P d 645 (claim); H a : P ! 645

12. H0 : P t 128 ; Ha : P 128 (claim)

13. H0 : V

14. H 0 : V 2 t 1.2 (claim); H a : V 2 1.2

5 ; H a : V z 5 (claim)

15. H0 : p t 0.45 ; H a : p 0.45 (claim)

16. H0 : p 0.21 (claim); H a : p z 0.21

17. c; H 0 : P d 3

18. d; H 0 : P t 3

19. b; H 0 : P

20. a; H0 : P d 2

21. Right-tailed

22. Left-tailed

23. Two-tailed

24. Two-tailed

25. P ! 8 H 0 : P d 8 ; H a : P ! 8 (claim)

26. V 3 H0 : V t 3 ; H a : V 3 (claim)

27. V d 320 H0 : V d 320 (claim); H a : V ! 320

28. P t 20,000 H0 : P t 20,000 (claim); Ha : P 20,000

29. p 0.54 H0 : p 0.54 (claim); H a : p z 0.54

30. p 0.14 H0 : p 0.14 (claim); H a : p z 0.14

31. A type I error will occur if the actual proportion of new customers who return to buy their next textbook is at least 0.60, but you reject H0 : p t 0.60 . A type II error will occur if the actual proportion of new customers who return to buy their next textbook is less than 0.60, but you fail to reject H0 : p t 0.60 . 32. A type I error will occur when the actual noontime mean traffic flow rate is 35 cars per minute, but you reject H0: μ = 35. A type II error will occur when the actual noontime mean traffic flow rate is not 35 cars per minute, but you fail to reject H0: μ = 35. 33. A type I error will occur if the actual standard deviation of the length of time to play a game is less than or equal to 12 minutes, but you reject H 0 : V d 12 . A type II error will occur if the actual standard deviation of the length of time to play a game is greater than 12 minutes, but you fail to reject H 0 : V d 12 . 34. A type I error will occur when the actual proportion of the U.S. gamers who are women is 0.5, but you reject H0 : p 0.5 . A type II error will occur when the actual proportion of the U.S. gamers who are women is not 0.5, but you fail to reject H0 : p 0.5 .

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35. A type I error will occur when the actual proportion of applicants who become campus security officers is at most 0.25, but you reject H0: p 0.25. A type II error will occur when the actual proportion of applicants who become campus security officers is greater than 0.25, but you fail to reject H0: p 0.25. 36. A type I error will occur when the actual mean cost of repairing a phone screen is greater than or equal to $120, but you reject H0: ȝ 120. A type II error will occur when the actual mean cost of repairing a phone screen is less than $120, but you fail to reject H0: ȝ 120. 37. H0: The proportion of homeowners who have a home security alarm is greater than or equal to 14%. Ha: The proportion of homeowners who have a home security alarm is less than 14%. H0: p 0.14; Ha: p < 0.14 Left-tailed because the alternative hypothesis contains <.

38. H0: The mean time that the manufacturer’s clocks lose is less than or equal to 0.02 second per day. Ha: The mean time that the manufacturer’s clocks lose is greater than 0.02 second per day. H0: ȝ 0.02; Ha: ȝ > 0.02 Right-tailed because the alternative hypothesis contains >.

39. H0: The standard deviation of the 18-hole scores for a golfer is greater than or equal to 2.1 strokes. Ha: The standard deviation of the 18-hole scores for a golfer is less than 2.1 strokes. H 0 :σ ≥ 2.1; H a : σ < 2.1 Left-tailed because the alternative hypothesis contains <.

40. H0: The percentage of cancer diagnoses attributable to lung cancer is 25%. Ha: The percentage of cancer diagnoses attributable to lung cancer is not 25%. H0: p = 0.25; Ha: p ≠ 0.25 Two-tailed because the alternative hypothesis contains ≠.

41. H0: The percentage of responses to a survey is 100%. Ha: The percentage of responses to a survey is not 100%. H0: p = 1; Ha: p ≠ 1 Two-tailed because the alternative hypothesis contains ≠.

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42. H0: The mean graduation rate at a high school is less than or equal to 97%. Ha: The mean graduation rate at a high school is greater than 97%. H 0 : μ ≤ 97; H a : μ > 97 Right-tailed because the alternative hypothesis contains >.

43. Alternative hypothesis (a) There is enough evidence to support the scientist’s claim that the mean incubation period for swan eggs is less than 40 days. (b) There is not enough evidence to support the scientist’s claim that the mean incubation period for swan eggs is less than 40 days. 44. Alternative hypothesis (a) There is enough evidence to support the report’s claim that more than 30% of households in a Wisconsin county struggle to afford basic necessities. (b) There is not enough evidence to support the report’s claim that more than 30% of households in a Wisconsin county struggle to afford basic necessities. 45. Null hypothesis (a) There is enough evidence to reject the researcher’s claim that the standard deviation of the life of the lawn mower is at most 2.8 years. (b) There is not enough evidence to reject the researcher’s claim that the standard deviation of the life of the lawn mower is at most 2.8 years. 46. Null hypothesis (a) There is enough evidence to reject the automotive manufacturer’s claim that the standard deviation for the gas mileage of its vehicles is 3.9 miles per gallon. (b) There is not enough evidence to reject the automotive manufacturer’s claim that the standard deviation for the gas mileage of its vehicles is 3.9 miles per gallon. 47. Alternative hypothesis (a) There is enough evidence to support the fitness equipment company’s claim that its competitor’s home gym does not have a customer satisfaction rate of 99%. (b) There is not enough evidence to support the fitness equipment company’s claim that its competitor’s home gym does not have a customer satisfaction rate of 99%. 48. Null hypothesis (a) There is enough evidence to reject the study’s claim that at least 20% of renters are behind on rent payments in New Jersey. (b) There is not enough evidence to reject the study’s claim that at least 20% of renters are behind on rent payments in New Jersey.

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49. H 0 : μ ≥ 60; H a : μ < 60

50. H 0 : μ = 16; H a : μ ≠ 16

51. (a) H0: ȝ 5; Ha: ȝ < 5

52. (a) H 0 : μ ≤ 28; H a : μ > 28

(b) H0: ȝ 5; Ha: ȝ > 5

(b) H 0 : μ ≥ 28; H a : μ < 28

53. If you decrease α , then you are decreasing the probability that you reject H0 . Therefore, you are increasing the probability of failing to reject H0 . This could increase β , the probability of failing to reject H 0 when H 0 is false. 54. If α = 0, the null hypothesis cannot be rejected and the hypothesis test is useless. 55. Yes; If the P-value is less than α = 0.05, it is also less than α = 0.10. 56. Not necessarily; A P-value less than α = 0.10 may or may not also be less than α = 0.05. 57. (a) Fail to reject H 0 because the CI includes 70. (b) Reject H 0 because the CI does not include 70. (c) Fail to reject H 0 because the CI includes 70. 58. (a) Fail to reject H 0 because the CI includes values less than 54. (b) Fail to reject H 0 because the CI includes values less than 54. (c) Reject H 0 because the CI is located to the right of 54. 59. (a) Reject H 0 because the CI is located entirely to the right of 0.20. (b) Fail to reject H 0 because the CI includes values less than 0.20. (c) Fail to reject H 0 because the CI includes values less than 0.20. 60. (a) Fail to reject H 0 because the CI includes values greater than 0.73. (b) Reject H 0 because the CI is located entirely to the left of 0.73. (c) Fail to reject H 0 because the CI includes values greater than 0.73.

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7.2 HYPOTHESIS TESTING FOR THE MEAN (σ KNOWN) 7.2 TRY IT YOURSELF SOLUTIONS 1. (1) Fail to reject H 0 because 0.0745 > 0.05 . (2) Reject H 0 because 0.0745 < 0.10 . 2.

P = 0.0436 Reject H 0 because P = 0.0436 < 0.05 .

3. Area to the left of z = 1.64 is 0.9495 . P = 2(area) = 2(0.0505) = 0.1010

Fail to reject H 0 because P = 0.1010 > 0.10 . 4. The claim is “the mean speed is greater than 35 miles per hour.” H 0 : μ ≤ 35; Hα : μ > 35 (claim) α = 0.05 x − μ 36 − 35 = = 2.5 z= σ 4 100 n P-value = {Area right of z = 2.50 } = 0.0062 Reject H 0 because P-value = 0.0062 < 0.05 . There is enough evidence at the 5% level of significance to support the claim that the average speed is greater than 35 miles per hour.

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5. The claim is “the mean is different from 3.6.” H 0 : μ = 3.6 ; Hα : μ ≠ 3.6 (claim) α = 0.01 x − μ 4 − 3.6 z= = ≈ 1.33 σ 1.5 n 25 P-value = 2{Area to the right of z = 1.33 } ≈ 2(0.0918) = 0.1836 Fail to reject H 0 because P-value = 0.1836 > 0.01 . There is not enough evidence at the 1% level of significance to support the claim that the mean number of workdays missed due to illness or injury in the past 12 months is different from 3.6 days. 6.

P = 0.0440 > 0.01 = α . Fail to reject H 0 .

Area = 0.1003 z0 = −1.28 Rejection region: z < −1.28

0.0401 and 0.9599 − z0 = −1.75 and z0 = 1.75 Rejection region: z < −1.75, z > 1.75

9. The claim is “the mean work day of the company’s mechanical engineers is less than 8.5 hours.” H 0 : μ ≥ 8.5; H a : μ < 8.5 (claim) α = 0.01 z0 = −2.33; Rejection region: z < −2.33 x − μ 8.2 − 8.5 z= = = −3.00 σ 0.5 n 25

Because −3.00 < −2.33, reject H 0 . There is enough evidence at the 1% level of significance to support the claim that the mean work day is less than 8.5 hours. 10. α = 0.01 − z0 = −2.575, z0 = 2.575; Rejection regions: z < −2.575, z > 2.575 Because z ≈ −1.89 is not in the rejection region, we fail to reject H 0 . There is not enough evidence at the 1% level of significance to reject the claim that the mean cost of raising a child between the ages of 1 and 3 in the United States is $13,350. Copyright © 2023 Pearson Education, Inc.

CHAPTER 7 Ň HYPOTHESIS TESTING WITH ONE SAMPLE

7.2 EXERCISE SOLUTIONS 1. In the z-test using rejection region(s), the test statistic is compared with critical values. The z-test using a P-value compares the P-value with the level of significance α . 2. An z-test can be used if the population is normally distributed. 3. (a) Fail to reject H0 because P = 0.0461 > 0.01 = α . (b) Reject H0 because P = 0.0461 < 0.05 = α . (c) Reject H0 because P = 0.0461 < 0.10 = α . 4. (a) Fail to reject H0 because P = 0.0691 > 0.01 = α . (b) Fail to reject H0 because P = 0.0691 > 0.05 = α . (c) Reject H0 because P = 0.0691 < 0.10 = α . 5. (a) Fail to reject H0 because P = 0.1271 > 0.01 = α . (b) Fail to reject H0 because P = 0.1271 > 0.05 = α . (c) Fail to reject H0 because P = 0.1271 > 0.10 = α . 6. (a) Fail to reject H0 because P = 0.0107 > 0.01 = α . (b) Reject H0 because P = 0.0107 < 0.05 = α . (c) Reject H0 because P = 0.0107 < 0.10 = α . 7. (a) Fail to reject H0 because P = 0.0838 > 0.01 = α . (b) Fail to reject H0 because P = 0.0838 > 0.05 = α . (c) Reject H0 because P = 0.0838 < 0.10 = α . 8. (a) Reject H0 because P = 0.0062 < 0.01 = α . (b) Reject H0 because P = 0.0062 < 0.05 = α . (c) Reject H0 because P = 0.0062 < 0.10 = α . 9. (b). The smaller P-value corresponds to the smaller area. 10. (a). The larger P-value corresponds to the larger area. 11. (a). The smaller z-statistic corresponds to the smaller area. Copyright © 2023 Pearson Education, Inc.

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12. (b). The larger z-statistic corresponds to the larger area. 13.

P = 0.0934 ; Reject H 0 because P = 0.0934 < 0.10 .

14.

P = 0.0606 ; Fail to reject H 0 because P = 0.0606 > 0.05 .

15.

P = 0.0069 ; Reject H 0 because P = 0.0069 < 0.01 .

16.

P = 0.1093 ; Fail to reject H 0 because P = 0.1093 > 0.10 .

17.

P = 2(Area) = 2(0.0465) = 0.0930 ; Fail to reject H 0 because P = 0.0930 > 0.05 .

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18.

P = 2(Area) = 2(0.0256) = 0.0512 ; Reject H 0 because P = 0.0512 < 0.08 19. Fail to reject H 0 , ( P = 0.0628 > 0.05). 20. Reject H 0 , ( P = 0.0065 < 0.01). 21. (a) Fail to reject H 0 because z < 1.285 . (b) Fail to reject H 0 because z < 1.285 . (c) Fail to reject H 0 because z < 1.285 . (d) Reject H 0 because z > 1.285 . 22. (a) Reject H 0 because z > 1.96 . (b) Fail to reject H 0 because −1.96 < z < 1.96 . (c) Fail to reject H 0 because −1.96 < z < 1.96 . (d) Reject H 0 because z < −1.96 . 23. Critical value: z0 = −1.88 ; Rejection region: z < −1.88

24. Critical value: z0 = −1.34 ; Rejection region: z < −1.34

25. Critical value: z0 = 1.645 ; Rejection region: z > 1.645

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26. Critical value: z0 = 1.41 ; Rejection region: z > 1.41

27. Critical values: − z0 = −2.33, z0 = 2.33 ; Rejection regions: z < −2.33, z > 2.33

28. Critical values: −z0 = −1.555, z0 = 1.555 Rejection regions: z < −1.555, z > 1.555

29. H 0 : μ = 40 (claim); H a : μ ≠ 40 α = 0.05 → z0 = ±1.96 x − μ 39.2 − 40 z= = ≈ −2.030 σ 1.97 n 25 Because z = −2.03 < −1.96 , reject H 0 . There is enough evidence at the 5% level of significance to reject the claim. 30. H0 : μ ≥ 1475 (claim); H a : μ < 1475 α = 0.07 → z0 = −1.48 x − μ 1468 − 1475 z= = ≈ −1.231 σ 29 n 26 Because z = −1.231 is not less than −1.48 , fail to reject H 0 . There is not enough evidence at the 7% level of significance to support the claim. 31. H0 : μ = 5880; H a : μ ≠ 5880 (claim) α = 0.03 → z0 = ±2.17

x −μ

5771 − 5880 ≈ −2.16 σ 413 n 67 Because −2.17 < −2.16 < 2.17 , fail to reject H 0 . There is not enough evidence at the 3% level of significance to support the claim. z=

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32. H 0 : μ ≤ 22,500 (claim); H a : μ > 22,500 α = 0.01 → z0 = 2.33 x − μ 23,500 − 22,500 z= = ≈ 5.590 σ 1200 n 45 Because z = 5.590 > 2.33 , reject H 0 . There is enough evidence at the 1% level of significance to reject the claim. 33. (a) The claim is “the mean total score for the school’s applicants is more than 503.” H 0 : μ ≤ 503 ; H a : μ > 503 (claim) (b) z =

x −μ

505 − 503 ≈ 1.89 10.6 100

(c) P-value = {Area to the right of z = 1.89 } ≈ 0.0294 (d) Because P = 0.0294 > 0.01 = α , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to support the report’s claim that the mean total score for the school’s applicants is more than 503. 34. (a) The claim is “the average activating temperature is at least 135oF.” H 0 : μ ≥ 135 (claim); H a : μ < 135 (b) z =

x −μ

133 − 135 ≈ −3.43 3.3 32

(c) P-value = {Area to left of z = −3.43 } = 0.0003 (d) Because P = 0.0003 < 0.10 = α , reject H 0 . (e) There is enough evidence at the 10% level of significance to reject the manufacturer’s claim that the average activating temperature is at least 135° F. 35. (a) The claim is “the mean winning times for Boston Marathon women’s open division champions is at least 2.6 hours.” H 0 : μ ≥ 2.6 (claim); H a : μ < 2.6 (b) z =

x −μ

2.52 − 2.6 ≈ −2.43 0.18 30

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(d) Because P = 0.0075 < 0.05 = α , reject H 0 . (e) There is enough evidence at the 5% level of significance to reject the statistician’s claim that the mean winning times for Boston Marathon women’s open division champions is at least 2.6 hours. 36. (a) The claim is “the mean acceleration time from 0 to 60 miles per hour for all 3-cylinder vehicles is 9.4 seconds.” H 0 : μ = 9.4 (claim); H a : μ ≠ 9.4 (b) z =

x −μ

10.2 − 9.4 ≈ 2.00 2.3 33

(c) P-value = 2{Area to right of z = 2.00 } ≈ 0.0456 (d) Because P = 0.0456 < 0.05 = α , reject H 0 . (e) There is enough evidence at the 5% level of significance to reject the claim that the mean acceleration time from 0 to 60 miles per hour for all 3-cylinder vehicles is 9.4 seconds. 37. (a) The claim is “the mean vertical drop of top-rated roller coasters is 163 feet.” H 0 : μ = 163 (claim); H a : μ ≠ 163 (b) z =

x −μ

144.97 − 163 ≈ −1.66 65.1 36

(c) P-value = 2{Area to left of z = −1.66 } ≈ 0.097 (d) Because P = 0.097 > 0.05 = α , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to reject the claim that the mean vertical drop of top-rated roller coasters is 163 feet. 38. (a) The claim is “the mean annual salary for intermediate level electrical engineers in Baltimore, Maryland, is more than the national mean, $86,700.” H 0 : μ ≤ 86,700 ; H a : μ >86,700 (claim) (b) z =

x −μ

89,827.29 − 86,700 ≈ 1.05 13,600 21

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(e) There is not enough evidence at the 9% level of significance to support the analyst’s claim that the mean annual salary for intermediate level electrical engineers in Baltimore, Maryland, is more than the national mean, $86,700.” 39. (a) The claim is “the mean caffeine content per 12-ounce bottle of a population of caffeinated soft drinks is 37.7 milligrams.” H0 : μ = 37.7 (claim); H a : μ ≠ 37.7 (b) − z0 = −2.575, z0 = 2.575; Rejection regions: z < −2.575, z > 2.575 (c) z =

x −μ

36.4 − 37.7 ≈ −0.72 10.8 36

(d) Because −2.575 < z < 2.575 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to reject the consumer research organization’s claim that the mean caffeine content per 12-ounce bottle of a population of caffeinated soft drinks is 37.7 milligrams. 40. (a) The claim is “the mean high school 4-year graduation rate per state in the United States is 83%.” H 0 : p = 0.83 (claim); H a : p ≠ 0.83 (b) − z0 = −1.96, z0 = 1.96; Rejection regions: z < −1.96 , z > 1.96 (c) z =

x −μ

0.85 − 0.83 ≈ 2.81 0.039 30

(d) Because z > 1.96 , reject H 0 . (e) There is enough evidence at the 5% level of significance to reject the claim that the mean high school 4-year graduation rate per state in the United States is 83%. 41. (a) The claim is “the mean sodium content in one of its breakfast sandwiches is less the 920 milligrams.” H 0 : N p 9 2 0 ; H a : N 9 2 0 (claim )

(b) z0 = −1.645; Rejection region: z < −1.645 (c) z =

x −μ

≈

925 − 920 ≈ 1.76 18 40

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(d) Because z > −1.645 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to support the claim that the mean sodium content in one of its breakfast sandwiches is less than 920 milligrams. 42. (a) The claim is “the mean life of a certain type of light bulb is at least 750 hours.” H 0 : μ ≥ 750 (claim); H a : μ < 750 (b) z0 = −2.05; Rejection region: z < −2.05 (c) z =

x −μ

745 − 750 = −0.42 60 25

(d) Because z > −2.05 , fail to reject H 0 . (e) There is not enough evidence at the 2% level of significance to reject the claim that the mean life of the bulb is at least 750 hours. 43. (a) The claim is “the mean life of a CFL bulb is at least 10,000 hours.” H 0 : μ ≥ 10,000 (claim); H a : μ < 10,000 (b) z0 = −1.23; Rejection region: z < −1.23 (c) x ≈ 9580.9 x − μ 9580.9 − 10,000 z= ≈ ≈ −1.28 σ 1850 n 32 (d) Because z < −1.23 , reject H 0 . (e) There is enough evidence at the 11% level of significance to reject the lamp manufacturer’s claim that the mean life of CFL bulb is at least 10,000 hours. 44. (a) The claim is “the mean GDP per country in a recent year is greater than $400 billion.” H 0 : μ ≤ 400 ; H a : μ > 400 (claim) (b) z0 = 1.555; Rejection region: z > 1.555 (c) x ≈ 231.35 x − μ 231.35 − 400 z= ≈ ≈ −0.521 σ 2099 n 42 (d) Because z < 1.555 , fail to reject H 0 .

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(e) There is not enough evidence at the 6% level of significance to support the politician’s claim that the mean GDP per country in a recent year is greater than $400 billion. 45. Outside; When the standardized test statistic is inside the rejection region, P < α . 46. To the right; When P < α , the standardized test statistic is inside the rejection region.

7.3 HYPOTHESIS TESTING FOR THE MEAN (σ UNKNOWN) 7.3 TRY IT YOURSELF SOLUTIONS 1.

d.f = n − 1 = 14 − 1 = 13 and α = 0.01 , One tail (left-tailed) t0 = −2.650

2. d.f = n − 1 = 9 − 1 = 8 and α = 0.10 , One tail (right-tailed) t0 = 1.397 3. d.f = n − 1 = 16 − 1 = 15 and α = 0.05 , Two tail −t0 = −2.131, t0 = 2.131 4. The claim is “the mean transaction price of all new vehicles sold in the past 12 months is less than $43,500.” H 0 : μ ≥ 43,500; H a : μ < 43,500 (claim) α = 0.10 and d.f. = n − 1 = 25 − 1 = 24 t0 = −1.318; Rejection region: t < −1.318 x − μ 40,573 − 43,500 t= = ≈ −2.34 s 6250 n 25 Because t < −1.318 , reject H 0 . There is enough evidence at the 10% level of significance to support the analyst’s claim that the mean transaction price of new vehicles sold in the past 12 months is less than $43,500. 5. The claim is “the mean conductivity of the river is 1890 milligrams per liter.” H0 : μ = 1890 (claim); H a : μ ≠ 1890 α = 0.01 and d.f. = n − 1 = 38 −t0 = −2.712, t0 = 2.712; Rejection regions: t < −2.712, t > 2.712 x − μ 2350 − 1890 t= = ≈ 3.192 s 900 n 39 Because t > 2.712 , reject H 0 .

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There is enough evidence at the 1% level of significance to reject the company’s claim that the mean conductivity of the river is 1890 milligrams per liter. 6. The claim is “the mean wait time is at most 18 minutes.” H 0 : μ ≤ 18 minutes (claim); H a : μ > 18 minutes P − value = 0.9997 P − value = 0.9997 > 0.05 = α Fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the office’s claim that the mean wait time is at most 18 minutes.

7.3 EXERCISE SOLUTIONS 1. Specify the level of significance α and the degrees of freedom, d.f. = n − 1 . Find the critical value(s) using the t-distribution table in the row with n − 1 d.f. When the hypothesis test is: (1) left-tailed, use the “One Tail, α ” column with a negative sign. (2) right-tailed, use the “One Tail, α ” column with a positive sign. (3) two-tailed, use the “Two Tail, α ,” column with a negative and a positive sign. 2. Identify the claim. State H 0 and H a . Specify the level of significance. Identify the degrees of freedom. Determine the critical value(s) and rejection region(s). Find the standardized test statistic. Make a decision and interpret it in the context of the original claim. The sample must be a random sample, and either the population must be normally distributed or n must be greater than or equal to 30. 3. Critical value: t0 = −1.328 ; Rejection region: t < −1.328 4. Critical value: t0 = −2.441 ; Rejection region: t < −2.441 5. Critical value: t0 = 1.717 ; Rejection region: t > 1.717 6. Critical value: t0 = 2.457 ; Rejection region: t > 2.457 7. Critical values: − t0 = −2.056, t0 = 2.056 ; Rejection regions: t < −2.056, t > 2.056 8. Critical values: − t0 = −1.687, t0 = 1.687 ; Rejection region: t < −1.687, t > 1.687 9. (a) Fail to reject H 0 because t > −2.086 . (b) Fail to reject H 0 because t > −2.086 . (c) Reject H 0 because t < −2.086 .

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10. (a) Fail to reject H0 because t < 1.402. (b) Reject H0 because t > 1.402. (c) Fail to reject H0 because t < 1.402. 11. (a) Reject H 0 because t < −1.725 . (b) Fail to reject H 0 because −1.725 < t < 1.725 . (c) Reject H 0 because t > 1.725 . 12. (a) Reject H0 because t < −1.071. (b) Fail to reject H0 because −1.071 < t < 1.071. (c) Reject H0 because t > 1.071. 13. H 0 : μ = 15 (claim); H a : μ ≠ 15 α = 0.01 and d.f. = n − 1 = 35 t0 = ±2.724 x − μ 13.9 − 15 t= = ≈ −2.043 s 3.23 n 36 Because −2.724 < t < 2.724 , fail to reject H 0 . There is not enough evidence at the 1% level of significance to reject the claim. 14. H 0 : μ ≤ 25; H a : μ > 25 (claim) α = 0.05 and d.f. = n − 1 = 16 t0 = 1.746 x − μ 26.2 − 25 t= = ≈ 2.133 s 2.32 n 17 Because t > 1.746 , reject H 0 . There is enough evidence at the 5% level of significance to support the claim. 15. H0 : μ ≥ 8000 (claim); H a : μ < 8000 α = 0.01 and d.f. = n − 1 = 24 t0 = −2.492 x − μ 7700 − 8000 t= = ≈ −3.333 s 450 n 25 Because t < −2.492 , reject H 0 . There is enough evidence at the 1% level of significance to reject the claim. Copyright © 2023 Pearson Education, Inc.

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16. H0 : μ ≤ 1600 (claim); H a : μ > 1600 α = 0.02 and d.f. = n − 1 = 45 t0 = 2.115 x − μ 1550 − 1600 t= = ≈ −2.06 s 165 n 46 Because t < 2.115 , fail to reject H 0 . There is not enough evidence at the 2% level of significance to reject the claim. 17. H 0 : μ ≥ 4915; H a : μ < 4915 (claim) α = 0.02 and d.f. = n − 1 = 50 t0 = −2.109 x − μ 5017 − 4915 t= = ≈ 0.13 s 5613 n 51 Because t > −2.109 , fail to reject H 0 . There is not enough evidence at the 2% level of significance to support the claim. 18. H 0 : μ = 52,200; H a : μ ≠ 52,200 (claim) α = 0.05 and d.f. = n − 1 = 33 t0 = ±2.035 x − μ 53,220 − 52, 200 t= = ≈ 2.203 s 2700 n 34 Because t > 2.035 , reject H 0 . There is enough evidence at the 5% level of significance to support the claim. 19. (a) The claim is “Apple AirPods have a mean listening time of at least 5.0 hours on a single charge.” H 0 : μ ≥ 5.0 (claim); H a : μ < 5.0 (b) t0 = −2.567 Rejection region: t < −2.567 (c) t =

x − μ 4.69 − 5.0 = ≈ −2.31 s 0.57 n 18

(d) Because t > −2.567 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to reject the claim that Apple AirPods have a mean listening time of at least 5.0 hours on a single charge. 20. (a) The claim is “the mean wait time for various services at different locations is at most 6 minutes.” H 0 : μ ≤ 6 (claim); H a : μ > 6 Copyright © 2023 Pearson Education, Inc.

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(b) t0 = 2.445; Rejection region: t > 2.445 (c) t =

x − μ 10.3 − 6 = ≈ 3.13 s 8.0 n 34

(d) Because t > 2.445 , reject H 0 . (e) There is enough evidence at the 1% level of significance to reject the state Department of Transportation’s claim that the mean wait time for various services at different locations is at most 6 minutes. 21. (a) The claim is “the mean credit card debt in Colorado is greater than $5540 per borrower.” H 0 : μ ≤ 5540 ; H a : μ > 5540 (claim) (b) t0 = 1.699; Rejection region: t > 1.699 (c) t =

x − μ 5594 − 5540 = ≈ 0.495 s 597 n 30

(d) Because t < 1.699 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to support the claim that the mean credit card debt in Colorado is greater than $5540 per borrower. 22. (a) The claim is “the mean price of a three-year-old sports utility vehicle (in good condition) is $20,000.” H 0 : μ = 20,000 (claim); H a : μ ≠ 20,000 (b) −t0 = −2.080, t0 = 2.080; Rejection region: t < −2.080, t > 2.080 (c) t =

x − μ 20,640 − 20,000 = ≈ 1.508 s 1990 n 22

(d) Because −2.080 < t < 2.080 , fail to reject. (e) There is not enough evidence at the 5% level of significance to reject the claim that the mean price of a three-year-old sports utility vehicle (in good condition) is $20,000. 23. (a) The claim is “the mean amount of carbon monoxide in the air in the U.S. cities is less than 1.80 parts per million.” H 0 : N p 1.80; H a : N 1.80 (claim)

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CHAPTER 7 Ň HYPOTHESIS TESTING WITH ONE SAMPLE

(b) t0 = −1.295; Rejection region: t < −1.295 (c) t =

x − μ 1.92 − 1.80 = ≈ 0.691 s 1.39 n 64

(d) Because t > −1.295 , fail to reject H 0 . (e) There is not enough evidence at the 10% level of significance to support the claim that the mean amount of carbon monoxide in the air in the U.S. cities is less than 1.80 parts per million. 24. (a) The claim is “the mean amount of lead in the air in the U.S. cities is less than 0.032 microgram per cubic meter.” H0 : μ ≥ 0.032; H a : μ < 0.032 (claim) (b) t0 = −2.396; Rejection region: t < −2.396 (c) t =

x − μ 0.021 − 0.032 = ≈ −2.42 s 0.034 n 56

(d) Because t < −2.396 , reject H 0 . (e) There is enough evidence at the 1% significance level to support the claim that the mean amount of lead in the air in the U.S. cities is less than 0.032 microgram per cubic meter. 25. (a) The claim is “the mean annual salary for mid-level product engineers is $86,000.” H 0 : μ = 86,000 (claim); H a : μ ≠ 86,000 (b) −t0 = −2.131, t0 = 2.131; Rejection region: t < −2.131, t > 2.131 (c) x ≈ 79,314.56 s ≈ 12, 426.52 x − μ 79,314.56 − 86,000 t= ≈ ≈ −2.152 s 12, 426.52 n 16 (d) Because t < −2.131 , reject H 0 . (e) There is enough evidence at the 5% level of significance to reject the claim that the mean annual salaries for mid-level product engineers is $86,000. 26. (a) The claim is “the mean annual salary for senior level statisticians is more than $124,000.” H 0 : μ ≤ 124,000; H a : μ > 124,000 (claim) (b) t0 = 1.363; Rejection region: t > 1.363

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277

(c) x ≈ 129,366.67 s ≈ 10,587.76 x − μ 129,366.67 − 124,000 t= ≈ ≈ 1.756 s 10,587.76 n 12 (d) Because t > 1.363 , reject H 0 . (e) There is enough evidence at the 10% level of significance to support the claim that the mean annual salary for senior level statisticians is more than $124,000. 27. (a) The claim is “the mean minimum time it takes for a sedan to travel a quarter mile is greater than 15.3 seconds.” H 0 : μ ≤ 15.3; H a : μ > 15.3 (claim)

x − μ 15.8 − 15.3 = ≈ 0.994 s 2.36 n 22 P-value = {Area to right of t = 0.994 } ≈ 0.166

(b) t =

(c) Because P > 0.10 = α , fail to reject H 0 . (d) There is not enough evidence at the 10% level of significance to support the consumer group’s claim that the mean minimum time it takes for a sedan to travel a quarter mile is greater than 15.3 seconds. 28. (a) The claim is “the mean dive duration of a North Atlantic right whale is 11.5 minutes.” H0 : μ = 11.5 (claim); H a : μ ≠ 11.5

x − μ 12.2 − 11.5 = ≈ 1.855 s 2.2 n 34 P-value = 2{Area to right of t = 1.855 } ≈ 2(0.03627) ≈ 0.0725

(b) t =

(c) Because P < 0.10 = α , reject H 0 . (d) There is enough evidence at the 10% level of significance to reject the oceanographer’s claim that the mean dive duration of a North Atlantic right whale is 11.5 minutes. 29. (a) The claim is “the mean number of classroom hours per week for full-time faculty is 11.0.” H0 : μ = 11.0 (claim); H a : μ ≠ 11.0 (b) x = 10.05 s ≈ 2.485 x − μ 10.05 − 11.0 t= ≈ ≈ −1.081 s 2.485 n 8 P-value = 2{Area to left of t = −1.081 } ≈ 2(0.15776) ≈ 0.3155

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CHAPTER 7 Ň HYPOTHESIS TESTING WITH ONE SAMPLE

(c) Because P > 0.01 = α , fail to reject H 0 . (d) There is not enough evidence at the 1% level of significance to reject the dean’s claim that the mean number of classroom hours per week for full-time faculty is 11.0. 30. (a) The claim is “the mean class size for full-time faculty is fewer than 32 students.” H 0 : μ ≥ 32; H a : μ < 32 (claim) (b) x ≈ 30.167 s ≈ 4.004 x − μ 30.167 − 32 t= = ≈ −1.942 s 4.004 n 18 P-value = {Area to left of t = −1.942 } ≈ 0.0344 (c) Because P < 0.05 = α , reject H 0 . (d) There is enough evidence at the 5% level of significance to support the brochure’s claim that the mean class size for full-time faculty is fewer than 32 students. 31. Because σ is unknown, n < 30, the sample is random, and the gas mileage is normally distributed, use the t-distribution. H 0 : μ ≥ 23 (claim); H a : μ < 23 x − μ 22 − 23 t= ≈ ≈ −0.559 s 4 n 5 P-value = {Area left of t = −0.559 } ≈ 0.3030 Because P > 0.05 = α , fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the car company’s claim that the mean gas mileage for the luxury sedan is at least 23 miles per gallon. 32. Because σ is known, the sample is random, and n ≥ 30, use the standard normal distribution. H 0 : μ ≤ 10,500; H a : μ > 10,500 (claim) x − μ 10,931 − 10,500 z= ≈ ≈ 0.992 σ 2380 n 30 P-value = {Area right of z = 0.992 } ≈ 0.1606 Because P > 0.01 = α , fail to reject H 0 . There is not enough evidence at the 1% level of significance to support the education publication’s claim that the mean in-state tuition and fees at public four-year institutions by state is more than $10,500. 33. More likely; The tails of the t-distribution curve are thicker than those of a standard normal distribution curve. So, if you incorrectly use a standard normal sampling distribution instead of a t-sampling distribution, then the area under the curve at the tails will be smaller than what it would be for the t-test, meaning the critical value(s) will lie closer to the mean. This makes it more likely for the test statistic to be in the rejection region(s). This result is the same regardless of whether the test is left-tailed, righttailed, or two-tailed; in each case, the tail thickness affects the location of the critical value(s).

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7.4 HYPOTHESIS TESTING FOR PROPORTIONS 7.4 TRY IT YOURSELF SOLUTIONS 1.

np = (150)( 0.36 ) = 54 > 5, nq = (150 )( 0.64) = 96 > 5 The claim is “more than 36% of the U.S. adults would consider buying a product or service directly from an advertisement that is run during streaming programming.” H 0 : p ≤ 0.36; H a : p > 0.36 (claim) α = 0.01 z0 = 2.33; Rejection region: z > 2.33 lp − p 0.33 − 0.36 = ≈ −0.77 z= pq ( 0.36 )( 0.64 ) n 150 Fail to reject H 0 . There is not enough evidence at the 1% level of significance to support the researcher’s claim that more than 36% of the U.S. adults would consider buying a product or service directly from an advertisement that is run during streaming programming.

np = ( 7400 )( 0.70) = 5180 > 5, nq = ( 7400 )( 0.30) ≈ 2220 > 5 The claim is “70% of the U.S. adults ages 22 to 59 who have at least one parent with a bachelor’s degree or beyond have completed a bachelor’s degree themselves.” H 0 : p = 0.70 (claim); H a : p ≠ 0.70 α = 0.10 − z0 = −1.645, z0 = 1.645; Rejection region: z < −1.645, z > 1.645 § 5110 · ¨ 7400 ¸ − 0.70 ¹ z= =© ≈ −1.78 pq ( 0.70 ) (0.30) n 7400 Reject H 0 . There is enough evidence at the 10% level of significance to reject the researcher’s claim that 70% of the U.S. adults ages 22 to 59 who have at least one parent with a bachelor’s degree or more education have completed a bachelor’s degree themselves. lp − p

7.4 EXERCISE SOLUTIONS 1. If np ≥ 5 and nq ≥ 5, the normal distribution can be used. 2. Verify that np ≥ 5 and nq ≥ 5. State H 0 and H a . Specify the level of significance α . Determine the critical value(s) and rejection region(s). Find the standardized test statistic. Make a decision and interpret in the context of the original claim. Copyright © 2023 Pearson Education, Inc.

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CHAPTER 7 Ň HYPOTHESIS TESTING WITH ONE SAMPLE

np = ( 40 )( 0.12 ) = 4.8 < 5 nq = (40)(0.88) = 35.2 > 5 Cannot use normal distribution because np < 5 .

np = ( 90 )( 0.48 ) = 43.2 > 5 nq = (90)(0.52) = 46.8 > 5 → use normal distribution H 0 : p ≥ 0.48 (claim); H a : p < 0.48 z0 = −1.405; Rejection region: z < −1.405 lp − p 0.40 − 0.48 z= = ≈ −1.52 pq ( 0.48)( 0.52 ) n 90 Reject H 0 . There is enough evidence at the 8% level of significance to reject the claim.

np = ( 500 )( 0.15 ) = 75 > 5 nq = (500)(0.85) = 425 > 5 → use normal distribution

H 0 : p = 0.15; H a : p ≠ 0.15 (claim) − z0 = −1.96, z0 = 1.96; Rejection region: z < −1.96, z > 1.96 lp − p 0.12 − 0.15 z= = ≈ −1.88 pq ( 0.15)( 0.85) n 500 H Fail to reject 0 . There is not enough evidence at the 5% level of significance to support the claim. 6.

np = ( 225)( 0.70 ) = 157.5 > 5 nq = (225)(0.30) = 67.5 > 5 → use normal distribution

H 0 : p ≤ 0.70; H a : p > 0.70 (claim) z0 = 1.75; Rejection region: z > 1.75 lp − p 0.64 − 0.70 z= = ≈ −1.96 pq ( 0.70 )( 0.30 ) n 225 Fail to reject H 0 . There is not enough evidence at the 4% level of significance to support the claim. 7. (a) The claim is “more than 80% of the U.S. adults feel children should be vaccinated to attend school.” H 0 : p ≤ 0.80; H a : p > 0.80 (claim)

(b) z0 = 2.33; Rejection region: z > 2.33

CHAPTER 7 Ň HYPOTHESIS TESTING WITH ONE SAMPLE

lp − p

pq n

0.84 − 0.80

( 0.80 )( 0.20 )

281

≈ 1.414

200 (d) Because z < 2.33 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to support the claim that more than 80% of the U.S. adults feel children should be vaccinated to attend school. 8. (a) The claim is “at least 55% of the U.S. adults feel that COVID-19 vaccinations should be required for high school students to attend school in the fall.” H 0 : p ≥ 0.55 (claim); H a : p < 0.55

(b) z0 = −1.28; Rejection region: z < −1.28 (c) z =

lp − p

pq n

0.56 − 0.55

( 0.55)( 0.45)

≈ 0.284

200

(d) Because z > −1.28 , fail to reject H 0 . (e) There is not enough evidence at the 10% level of significance to reject the reporter’s claim that at least 55% of the U.S. adults feel that COVID-19 vaccinations should be required for high school students to attend school in the fall. 9. (a) The claim is “at least 60% of tax filers are expecting a tax refund.” H 0 : p ≥ 0.60 (claim); H a : p < 0.60

(b) z0 = −1.645; Rejection region: z < −1.645 (c) z =

lp − p

pq n

0.56 − 0.60

( 0.60 )( 0.40 )

≈ −4.078

2494

(d) Because z < −1.645 , reject H 0 . (e) There is enough evidence at the 5% level of significance to reject the analyst’s claim that at least 60% of tax filers are expecting a tax refund. 10. (a) The claim is “65% of full-time college students work year-round.” H 0 : p = 0.65 (claim); H a : p ≠ 0.65

(b) − z0 = −1.645, z0 = 1.645; Rejection regions: z < −1.645, z > 1.645

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CHAPTER 7 Ň HYPOTHESIS TESTING WITH ONE SAMPLE

§ 66 · ¨ 105 ¸ − 0.65 ¹ = © ≈ −0.460 pq ( 0.65)( 0.35) n 105

lp − p

(d) Because −1.645 < z < 1.645 , fail to reject H 0 . (e) At the 10% level of significance, there is not enough evidence to reject the researcher’s claim that 65% of full-time college students work year-round. 11. (a) The claim is “more than half of all nurses feel they became better professionals during the coronavirus pandemic.” H 0 : p ≤ 0.5 ; H a : p > 0.5 (claim)

(b) z0 = 2.33; Rejection region: z > 2.33

§ 174 · ¨ ¸ − 0.5 300 ¹ © = ≈ 2.771 pq ( 0.5)( 0.5 ) n 300

lp − p

(d) Because z > 2.33 , reject H 0 . (e) At the 1% level of significance, there is enough evidence to support the manager’s claim that more than half of all nurses feel they became better professionals during the coronavirus pandemic. 12. (a) The claim is “40% of U.S. adults would consider changing jobs.” H 0 : p = 0.4(claim); H a : p ≠ 0.4

(b) − z0 = −1.645, z0 = 1.645; Rejection regions: z < −1.645, z > 1.645

§ 25 · ¨ 50 ¸ − 0.4 = © ¹ ≈ 1.443 pq ( 0.4 )( 0.6 ) n 50

lp − p

(d) Because −1.645 < z < 1.645 , fail to reject H 0 . (e) At the 10% level of significance, there is not enough evidence to reject the researcher’s claim that 40% of the U.S. adults would consider changing jobs. 13. (a) The claim is “27% of U.S. adults would travel into space on a commercial flight if they could afford it.” H0 : p = 0.27 (claim); H a : p ≠ 0.27 Copyright © 2023 Pearson Education, Inc.

CHAPTER 7 Ň HYPOTHESIS TESTING WITH ONE SAMPLE

(b) z =

lp − p

pq n

0.30 − 0.27

( 0.27 )( 0.73)

283

≈ 2.14

1000 P-value = 2{Area to right of z = 2.14 } = 2(0.0162) ≈ 0.03 (c) Because P < 0.05 = α , reject H 0 . (d) There is enough evidence at the 5% level of significance to reject the research center’s claim that 27% of U.S. adults would travel into space on a commercial flight if they could afford it. 14. (a) The claim is “at most 35% of the U.S. adults seek to buy naturally raised and fed meat and poultry.” H 0 : p ≤ 0.35 (claim); H a : p > 0.35

(b) z =

lp − p

0.38 − 0.35

≈ 1.989 pq ( 0.35)( 0.65) n 1000 P-value = Area to right of z = 1.989 ≈ 0.023

(c) Because P > 0.01 = α , fail to reject H 0 . (d) At the 1% level of significance, there is not enough evidence to reject the claim that at most 35% of the U.S. adults seek to buy naturally raised and fed meat and poultry. 15. (a) The claim is “more than 25% of the U.S. adults believe moral values in the country are getting better.” H 0 : p ≤ 0.25; H a : p > 0.25 (claim)

§ 17 · ¨ 50 ¸ − 0.25 (b) z = = © ¹ ≈ 1.470 pq ( 0.25 )( 0.75) n 50 P-value = Area to right of z = 1.470 ≈ 0.071

lp − p

(c) Because P < 0.10 = α , reject H 0 . (d) At the 10% level of significance, there is enough evidence to support the politician’s claim that more than 25% of the U.S. adults believe moral values in the country are getting better. 16. (a) The claim is “25% of the U.S. households have taken in a stray cat.” H 0 : p = 0.25 (claim); H a : p ≠ 0.25

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CHAPTER 7 Ň HYPOTHESIS TESTING WITH ONE SAMPLE

§ 105 · ¨ 500 ¸ − 0.25 ¹ (b) z = = © ≈ −2.066 pq ( 0.25)( 0.75 ) n 500 P-value = 2{Area to left of z = −2.066 } = 2(0.0197) ≈ 0.039

lp − p

(c) Because P < 0.05 = α , reject H 0 . (d) At the 5% level of significance, there is enough evidence to reject the advocate’s claim that 25% of the U.S. households have taken in a stray cat. 17. H 0 : p ≥ 0.64 (claim); H a : p < 0.64 z0 = −1.645: Rejection region: z < −1.645 lp − p 0.58 − 0.64 z= = = −1.25 pq ( 0.64 )( 0.36 ) n 100

Because z > −1.645 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the claim that at least 64% of adults make an effort to live in ways that help protect the environment some of the time. 18. Answers will vary. 19. (a) The claim is “more than 80% of the U.S. adults feel children should be vaccinated to attend school.” H 0 : p ≤ 0.80; H a : p > 0.80 (claim)

(b) z0 = 2.33; Rejection region: z > 2.33 (c) z =

x − np npq

168 − 200(0.80) 200(0.80)(0.20)

≈ 1.414

(d) Because z < 2.33 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to support the claim that more than 80% of the U.S. adults feel children should be vaccinated to attend school. The results are the same.

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285

20. Answers will vary. Sample answer: lp − p z= Write the original equation pq n §x· ¨n¸− p x Substitute for pˆ =© ¹ n pq n x − np n Multiply by = n pq n n x − np Use the fact that n = n 2 and combine radicals = 2 n pq n x − np Simplify = npq

7.5 HYPOTHESIS TESTING FOR VARIANCE AND STANDARD DEVIATION 7.5 TRY IT YOURSELF SOLUTIONS 1.

d.f. = 17, α = 0.01

χ 02 = 33.409 2.

d.f. = 29, α = 0.05

χ02 = 17.708 3.

d.f. = 50, α = 0.01

χ R2 = 79.490 χ L2 = 27.991 4. The claim is “the variance of the amount of sports drink in a 12-ounce bottle is no more than 0.40.” H 0 : σ 2 ≤ 0.40 (claim); H a : σ 2 > 0.40 α = 0.01 and d.f. = n − 1 = 30

χ02 = 50.892; Rejection region: χ 2 > 50.892 2

χ =

( n − 1) s 2 ( 30)( 0.75) σ2

0.40

= 56.250

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CHAPTER 7 Ň HYPOTHESIS TESTING WITH ONE SAMPLE

Because χ 2 > 50.892 , reject H 0 . There is enough evidence at the 1% level of significance to reject the bottling company’s claim that the variance of the amount of sports drink in a 12-ounce bottle is no more than 0.40. 5. The claim is “the standard deviation of the lengths of response times is less than 3.7 minutes.” H 0 : σ ≥ 3.7; H a : σ < 3.7 (claim) α = 0.05 and d.f. = n − 1 = 8

χ02 = 2.733; Rejection region: χ 2 < 2.733

( n − 1) s 2 (8)( 3.0)2 = ≈ 5.259 χ = σ2 ( 3.7 )2 2

Because χ 2 > 2.733 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to support the police chief’s claim that the standard deviation of the lengths of response times is less than 3.7 minutes. 6. The claim is “the variance of the weight losses is 25.5.” H 0 : σ 2 = 25.5 (claim); H a : σ 2 ≠ 25.5 α = 0.10 and d.f. = n − 1 = 12

χ L2 = 5.226 and χ R2 = 21.026; Rejection region: χ 2 > 21.026, χ 2 < 5.226 2

χ =

( n − 1) s 2 (12 )(10.8) =

≈ 5.082

25.5 σ2 2 Because χ < 5.226 , reject H 0 .

There is enough evidence at the 10% level of significance to reject the company’s claim that the variance of the weight losses of users is 25.5.

7.5 EXERCISE SOLUTIONS 1. Specify the level of significance α . Determine the degrees of freedom. Determine the critical values

using the χ 2 -distribution . For a right-tailed test, use the value that corresponds to d.f. and α . For a left-tailed test, use the value that corresponds to d.f. and 1 − α . For a two-tailed test, use the value that 1 1 corresponds to d.f. and α , and d.f. and 1 − α . 2 2 2. No; In a χ 2 -distribution , all are greater than or equal to 0 because anything squared is greater than or equal to 0. 3. As decreases, the right critical value for a two-tailed test increases and the left critical value for a twotailed test decreases. 4. To test for standard deviation, the variance is needed, so the standard deviation is squared.

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287

5. The requirement of a normal distribution is more important when testing a standard deviation than when testing a mean. When the population is not normal, the results of the chi-square test can be misleading because the chi-square test is not as robust as the tests for the population mean. 6. Verify that the sample is random and the population is normally distributed. State H 0 and H a and identify the claim. Specify the level of significance. Determine the degrees of freedom. Determine the critical value(s) and rejection region(s). Find the standardized test statistic. Make a decision and interpret it in the context of the original claim. 7. Critical value: χ 02 = 38.885 ; Rejection region: χ 2 > 38.885 8. Critical value: χ02 = 14.684 ; Rejection region: χ 2 > 14.684 9. Critical value: χ02 = 0.872 ; Rejection region: χ 2 < 0.872 10. Critical value: χ02 = 13.091 ; Rejection region: χ 2 < 13.091 11. Critical values: χ L2 = 60.391, χ R2 = 101.879 ; Rejection regions: χ 2 < 60.391, χ 2 > 101.879 12. Critical values: χ L2 = 35.534, χ R2 = 91.952 ; Rejection regions: χ 2 < 35.534, χ 2 > 91.952 13. (a) Fail to reject H 0 because χ 2 < 6.251 .

(b) Fail to reject H 0 because χ 2 < 6.251 . (c) Reject H 0 because χ 2 > 6.251 . 14. (a) Fail to reject H 0 because 8.547 < χ 2 < 22.307 .

(b) Reject H 0 because χ 2 > 22.307 . (c) Reject H 0 because χ 2 < 8.547 . 15. H 0 : σ 2 = 0.52 (claim); H a : σ 2 ≠ 0.52

χ L2 = 7.564, χ R2 = 30.191; Rejection regions: χ 2 < 7.564, χ 2 > 30.191 χ2 =

( n − 1) s 2 (17 )( 0.508 ) = ≈ 16.608 σ2 ( 0.52 )

Because 7.564 < χ 2 < 30.191 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the claim. 16. H 0 : σ 2 = 63 (claim); H a : σ 2 ≠ 63

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CHAPTER 7 Ň HYPOTHESIS TESTING WITH ONE SAMPLE

χ2 =

( n − 1) s 2 ( 28)( 58) = ≈ 25.78 σ2 ( 63)

Because 12.461 < χ 2 < 50.993 , Fail to reject H 0 . There is not enough evidence at the 1% level of significance to reject the claim. 17. H 0 : σ 2 ≥ 8.5 (claim); H a : σ 2 < 8.5

χ02 = 12.338; Rejection region: χ 2 < 12.338 χ2 =

( n − 1) s 2 = ( 22)( 7.45) ≈ 19.28

8.5 σ2 2 Because χ > 12.338 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the claim. 18. H 0 : σ ≤ 0.92 (claim); H a : σ > 0.92

χ02 = 63.691; Rejection region: χ 2 > 63.691

( n − 1) s2 ( 40)( 0.67 )2 = ≈ 21.215 χ = σ2 ( 0.92)2 2

Because χ 2 < 63.691 , fail to reject H 0 . There is not enough evidence at the 1% level of significance to reject the claim. 19. H 0 : σ ≥ 40; H a : σ < 40 (claim)

χ02 = 3.053; Rejection region: χ 2 < 3.053

( n − 1) s 2 (11)( 40.8)2 = ≈ 11.444 χ = σ2 ( 40)2 2

Because χ 2 > 3.053 , fail to reject H 0 . There is not enough evidence at the 1% level of significance to support the claim. 20. H 0 : σ 2 ≤ 19; H a : σ 2 > 19 (claim)

χ02 = 23.542; Rejection region: χ 2 > 23.542 χ2 =

( n − 1) s 2 = (16)( 28) ≈ 23.58 σ2 (19)

Because χ 2 > 23.542 , reject H 0 . There is enough evidence at the 10% level of significance to support the claim. 21. H 0 : σ 2 = 32.8; H a : σ 2 ≠ 32.8 (claim)

χ L2 = 77.929, χ R2 = 124.342; Rejection regions: χ 2 < 77.929, χ 2 > 124.342 χ2 =

( n − 1) s 2 (100)( 40.9) = ≈ 124.7 σ2 ( 32.8)

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289

Because χ 2 > 124.342 , reject H 0 . There is enough evidence at the 10% level of significance to support the claim. 22. H 0 : T 24.9 ; H a : T v 24.9 (claim)

DL2 34.764, DR2 67.505; Rejection regions: D2 34.764, D2 67.505 2

n 1 s 2 T2

50 29.1 24.9

x 68.290

Because D 67.505 , reject H 0 . There is enough evidence at the 10% level of significance to reject the claim. 23. (a) The claim is “the variance of the diameters in a certain tire model is 8.6.” H 0 : σ 2 = 8.6 (claim); H a : σ 2 ≠ 8.6

(b) χ L2 = 1.735, χ R2 = 23.589; Rejection regions: χ 2 < 1.735, χ 2 > 23.589 (c) χ 2 =

( n − 1) s 2 = ( 9)( 4.3) = 4.5 σ2

8.6

(d) Because 1.735 < χ 2 < 23.589 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to reject the tire manufacturer’s claim that the variance of the diameters in a certain tire model is 8.6. 24. (a) The claim is “the variance of the gas mileages in a model of hybrid vehicle is 0.16.” H 0 : σ 2 = 0.16 (claim); H a : σ 2 ≠ 0.16

(b) χ L2 = 16.047, χ R2 = 45.722; Rejection regions: χ 2 < 16.047, χ 2 > 45.722 (c) χ 2 =

( n − 1) s 2 ( 29)( 0.26) σ2

0.16

= 47.125

(d) Because χ 2 > 45.722 , reject H 0 . (e) There is enough evidence at the 5% level of significance to reject the auto manufacture’s claim that the variance of the gas mileages in a model of hybrid vehicle is 0.16. 25. (a) The claim is “the standard deviation for grade 12 students on a mathematics assessment test is less than 37 points.” H 0 : σ ≥ 37; H a : σ < 37 (claim)

(b) χ02 = 18.114; Rejection region: χ 2 < 18.114

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( n − 1) s 2 ( 27 )( 34)2 = ≈ 22.799 (c) χ = σ2 ( 37 )2 2

(d) Because χ 2 > 18.114 , fail to reject H 0 . (e) There is not enough evidence at the 10% level of significance to support the school administrator’s claim that the standard deviation for grade 12 students on a mathematics assessment test is less than 37 points. 26. (a) The claim is “the standard deviation for grade 12 students on a reading assessment test is greater than 41 points.” H 0 : σ ≤ 41; H a : σ > 41 (claim)

(b) χ02 = 42.980; Rejection region: χ 2 > 42.980 2 (c) χ =

( n − 1) s 2 ( 24)( 46)2 = ≈ 30.211 σ2 ( 41)2

(d) Because χ 2 < 42.980 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to support the school administrator’s claim that the standard deviation for grade 12 students on a reading assessment test is greater than 41 points. 27. (a) The claim is “the standard deviation of the waiting times experienced by patients is no more than 0.5 minute.” H 0 : σ ≤ 0.5 (claim); H a : σ > 0.5

(b) χ02 = 33.196; Rejection region: χ 2 > 33.196 2 n − 1) s 2 ( 24 )( 0.7 ) ( = ≈ 47.04 (c) χ = σ2 ( 0.5)2 2

(d) Because χ 2 > 33.196 , reject H 0 . (e) There is enough evidence at the 10% level of significance to reject the hospital’s claim that the standard deviation of waiting times experienced by patients is no more than 0.5 minute. 28. (a) The claim is “the standard deviation of the room rates for two adults at three-star hotels in Denver is at least $68.” H 0 : σ ≥ 68 (claim); H a : σ < 68

CHAPTER 7 Ň HYPOTHESIS TESTING WITH ONE SAMPLE

2 (c) χ =

291

( n − 1) s 2 (17 )( 40)2 = ≈ 5.88 σ2 ( 68)2

(d) Because χ 2 < 6.408 , reject H 0 . (e) There is enough evidence at the 1% level of significance to reject the travel analyst’s claim that the standard deviation of the room rates for two adults at three-star hotels in Denver is at least $68. 29. (a) The claim is “the standard deviation of the annual salaries of senior level graphic design specialists is different from $13,056.” H 0 : σ = 13,056; H a : σ ≠ 13,056 (claim)

(b) χ L2 = 5.629, χ R2 = 26.119 ; Rejection regions: χ 2 < 5.629, χ 2 > 26.119 (c) s ≈ 18, 286.32 2 n − 1) s 2 (14 )(18,286.32) ( = ≈ 27.464 χ = σ2 (13,056)2 2

(d) Because χ 2 > 26.119 , reject H 0 . (e) There is enough evidence at the 5% level of significance to support the claim that the standard deviation of the annual salaries of senior level graphic design specialists is different from $13,056. 30. (a) The claim is “the standard deviation of the annual salaries of nursing supervisors is $18,630.” H 0 : σ = 18,630 (claim); H a : σ ≠ 18,630

(b) χ L2 = 4.575, χ R2 = 19.675 ; Rejection regions: χ 2 < 4.575, χ 2 > 19.675 (c) s ≈ 14,516.60

( n − 1) s 2 (11)(14,516.60)2 = ≈ 6.679 χ = σ2 (18,630)2 2

(d) Because 4.575 < χ 2 < 19.675 , fail to reject H 0 . (e) There is not enough evidence at the 10% level of significance to reject the claim that the standard deviation of the annual salaries of nursing supervisors is $18,630. 31. χ 2 ≈ 22.799 , d.f. = n − 1 = 27

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32. χ 2 ≈ 30.211 , d.f. = n − 1 = 24

P-value = {Area right of χ 2 = 30.211 } ≈ 0.178 Fail to reject H 0 because P-value ≈ 0.178 > 0.01 . 33. χ 2 ≈ 47.04 , d.f. = n − 1 = 24

P-value = {Area right of χ 2 = 47.04 } ≈ 0.0033 Reject H 0 because P-value ≈ 0.0033 < 0.10 . 34. χ 2 ≈ 5.88 , d.f. = n − 1 = 17

P-value = {Area left of χ 2 = 5.88 } ≈ 0.0060 Reject H 0 because P-value ≈ 0.0060 < 0.01 .

CHAPTER 7 REVIEW EXERCISE SOLUTIONS 1.

H 0 : μ ≤ 375 (claim); H a : μ > 375

2. H 0 : μ = 82 (claim); H a : μ ≠ 82

H 0 : p ≥ 0.205; H a : p < 0.205 (claim)

4. H 0 : μ = 150,020; H a : μ ≠ 150,020 (claim)

H 0 : σ ≤ 1.9; H a : σ > 1.9 (claim)

6. H 0 : p ≥ 0.64 (claim); H a : p < 0.64

7. (a) H0: p = 0.81 (claim); Ha: p ≠ 0.81

(b) A type I error will occur when the actual proportion of the U.S. adults who believe that Earth’s temperature has been increasing over the past 100 years is 81%, but you reject H0: p = 0.81. A type II error will occur when the actual proportion is not 81%, but you fail to reject H0: p = 0.81. (c) Two-tailed, because the alternative hypothesis contains ≠. (d) There is enough evidence to reject the polling organization’s claim that the proportion of the U.S. adults who believe that Earth’s temperature has been increasing over the past 100 years is 81%. (e) There is not enough evidence to reject the polling organization’s claim that the proportion of the U.S. adults who believe that Earth’s temperature has been increasing over the past 100 years is 81%. 8. (a) H 0 : μ ≥ 400 (claim); H a : μ < 400

(b) A type I error will occur if the actual mean shelf life of the dried fruit is at least 400 days, but you reject H0 : μ ≥ 400 . A type II error will occur if the actual mean shelf life of the dried fruit is less than 400 days, but you fail to reject H0 : μ ≥ 400 . (c) Left-tailed because the alternative hypothesis contains < Copyright © 2023 Pearson Education, Inc.

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293

(d) There is enough evidence to reject the agricultural cooperative’s claim that the mean shelf life of the dried fruit is at least 400 days. (e) There is not enough evidence to reject the agricultural cooperative’s claim that the mean shelf life of the dried fruit is at least 400 days. 9. (a) H0: ı 2900 (claim); Ha: ı > 2900

(b) A type I error will occur when the actual standard deviation of the starting prices of its top-rated vehicles for a recent year is no more than $2900, but you reject H0: ı 2900. A type II error will occur when the actual standard deviation of the starting prices of its top-rated vehicles for a recent year is more than $2900, but you fail to reject H0: ı 2900. (c) Right-tailed because the alternative hypothesis contains >. (d) There is enough evidence to reject the nonprofit consumer organization’s claim that the standard deviation of the starting prices of its top-rated vehicles for a recent year is no more than $2900. (e) There is not enough evidence to reject the nonprofit consumer organization’s claim that the standard deviation of the starting prices of its top-rated vehicles for a recent year is no more than $2900. 10. (a) H 0 : μ ≥ 25; H a : μ < 25 (claim)

(b) A type I error will occur if the actual mean number of grams of carbohydrates in one bar is at least 25, but you reject H 0 : μ ≥ 25 . A type II error will occur if the actual mean number of grams of carbohydrates in one bar is less than 25, but you fail to reject H0 : μ ≥ 25 . (c) Left-tailed because the alternative hypothesis contains <. (d) There is enough evidence to support the energy bar maker’s claim that the mean number of grams of carbohydrates in one bar is less than 25. (e) There is not enough evidence to support the energy bar maker’s claim that the mean number of grams of carbohydrates in one bar is less than 25. 11. P-value = {Area to left of z = −0.94 } = 0.1736 Fail to reject H 0 . 12. P-value = 2{Area to right of z = 2.57 } = 2(0.0051) = 0.0102 Reject H 0 . 13. Critical value: z0 ≈ −2.05 ; Rejection region: z < −2.05

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14. Critical values: − z0 = −2.81, z0 = 2.81 ; Rejection regions: z < −2.81, z > 2.81

15. Critical value: z0 = 1.96 ; Rejection region: z > 1.96

16. Critical values: − z0 = −2.17, z0 = 2.17 ; Rejection regions: z < −2.17, z > 2.17

17. Fail to reject H 0 because −1.645 < z < 1.645 . 18. Reject H 0 because z > 1.645 . 19. Fail to reject H 0 because −1.645 < z < 1.645 . 20. Reject H 0 because z < −1.645 . 21. H 0 : μ ≤ 45 (claim); H a : μ > 45 z0 = 1.645; Rejection region: z > 1.645 x − μ 47.2 − 45 z= = ≈ 1.54 6.7 σ n 22 Because z < 1.645 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the claim. 22. H0 : μ = 8.45; H a : μ ≠ 8.45 (claim) − z0 = −2.17, z0 = 2.17; Rejection regions: z < −2.17, z > 2.17 x − μ 7.88 − 8.45 z= = ≈ −2.52 σ 1.75 n 60 Because z < −2.17 , reject H 0 . There is enough evidence at the 3% level of significance to support the claim.

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23. H 0 : μ ≥ 5.500; H a : μ < 5.500 (claim) z0 = −2.33; Rejection region: z < −2.33 x − μ 5.497 − 5.500 z= = ≈ −1.636 0.011 σ n 36 Because z > −2.33 , fail to reject H 0 . There is not enough evidence at the 1% level of significance to support the claim. 24. H 0 : μ = 7450 (claim); H a : μ ≠ 7450 − z0 = −1.645, z0 = 1.645; Rejection regions: z < −1.645, z > 1.645 x − μ 7495 − 7450 z= = ≈ 0.962 243 σ n 27 Because −1.645 < z < 1.645 , fail to reject H 0 . There is not enough evidence at the 10% level of significance to reject the claim. 25. (a) The claim is “the mean air concentration of nitrogen dioxide in the U.S. cities is 9 parts per billion.” H 0 : μ = 9 (claim); H a : μ ≠ 9

(b) z =

x −μ

8.1 − 9 ≈ −1.15 5.65 52

(c) P-value = 2{Area to left of z = −1.15 } = 2(0.1251) = 0.2502 (d) Because P-value > 0.05 = α , fail to reject H0. (e) There is not enough evidence at the 5% level of significance to reject the researcher’s claim that the mean air concentration of nitrogen dioxide in the U.S. cities is 9 parts per billion. 26. (a) The claim is “the mean air concentration of fine particulate matter in the U.S. cities is greater than 9 micrograms per cubic meter.” H 0 : μ ≤ 9; H a : μ > 9(claim)

(b) z =

x −μ

8.25 − 9 ≈ −2.72 2.53 84

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(e) There is not enough evidence at the 1% level of significance to support the researcher’s claim that mean air concentration of fine particulate matter in the U.S. cities is greater than 9 micrograms per cubic meter. 27. (a) The claim is “the mean annual drug overdose death rate for the 50 states is at least 25 deaths per 100,000 people.” H 0 : μ ≥ 25 (claim); H a : μ < 25

(b) z0 = −2.33; Rejection region: z < −2.33 (c) z =

x −μ

22.48 − 25 ≈ −1.29 10.69 30

(d) Because z > −2.33 , fail to reject H 0 . (e) At the 1% level of significance, there is not enough evidence to reject the claim that the mean annual drug overdose death rate for the 50 states is at least 25 deaths per 100,000 people. 28. (a) The claim is “the mean price of a round-trip flight from New York City to Los Angeles is less than $725.” H 0 : μ ≥ 725; H a : μ < 725(claim)

(b) z0 = −1.645; Rejection region: z < −1.645 (c) z =

x −μ

712 − 725 ≈ −0.67 133 47

(d) Because z > −1.645 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to support the claim that the mean price of a round-trip flight from New York City to Los Angeles is less than $725. 29. Critical values: −t0 = −2.093, t0 = 2.093 ; Rejection region: t < −2.093, t > 2.093 30. Critical value: t0 = 2.449 ; Rejection region: t > 2.449 31. Critical value: t0 = 2.098 ; Rejection region: t > 2.098 32. Critical value: t0 = −1.678 ; Rejection region: t < −1.678 33. Critical value: t0 = −2.977 ; Rejection region: t < −2.977 34. Critical values: −t0 = −2.718, t0 = 2.718 ; Rejection region: t < −2.718, t > 2.718

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35. H 0 : μ ≤ 12,700; H a : μ > 12,700 (claim) t0 = 2.845; Rejection region: t > 2.845 x − μ 12,855 − 12,700 t= = ≈ 2.864 s 248 n 21 Because t > 2.845 , reject H 0 . There is enough evidence at the 0.5% level of significance to support the claim. 36. H 0 : μ ≥ 0 (claim); H a : μ < 0 t0 = −1.310; Rejection region: t < −1.310 x − μ −0.45 − 0 t= = ≈ −1.053 s 2.38 n 31 Because t > −1.310 , fail to reject H 0 . There is not enough evidence at the 10% level of significance to reject the claim. 37. H 0 : μ ≤ 51 (claim); H a : μ > 51 t0 = 2.426; Rejection region: t > 2.426 x − μ 52 − 51 t= = ≈ 2.530 s 2.5 n 40 Because t > 2.426 , reject H 0 . There is enough evidence at the 1% level of significance to reject the claim. 38. H 0 : μ ≥ 850; H a : μ < 850 (claim) t0 = −2.160; Rejection region: t < −2.160 x − μ 875 − 850 t= = ≈ 3.742 s 25 n 14 Because t > −2.160 , fail to reject H 0 . There is not enough evidence at the 2.5% level of significance to support the claim. 39. H 0 : μ = 195 (claim); H a : μ ≠ 195 −t0 = −1.660; t0 = 1.660 Rejection regions: t < −1.660; t > 1.660 x − μ 190 − 195 t= = ≈ −1.396 s 36 n 101 Because −1.660 < t < 1.660 , fail to reject H 0 . There is not enough evidence at the 10% level of significance to reject the claim.

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40. H 0 : μ = 333; H a : μ ≠ 333 (claim) −t0 = −2.032; t0 = 2.032 ; Rejection regions: t < −2.032; t > 2.032 x − μ 328 − 333 t= = ≈ −2.28 s 13 n 35 Because t < −2.032 , reject H 0 . There is enough evidence at the 5% level of significance to support the claim. 41. (a) The claim is “the mean monthly cost of joining a health club is $25.” H 0 : μ = 25 (claim); H a : μ ≠ 25

(b) −t0 = −1.740, t0 = 1.740; Rejection regions: t < −1.740, t > 1.740 (c) t =

x − μ 26.25 − 25 = ≈ 1.642 s 3.23 n 18

(d) Because −1.740 < t < 1.740 , fail to reject H 0 . (e) There is not enough evidence at the 10% level of significance to reject the claim that the mean monthly cost of joining a health club is $25. 42. (a) The claim is “the mean cost of a yoga session is no more than $14.” H 0 : μ ≤ 14 (claim); H a : μ > 14

(b) t0 = 2.040; Rejection region: t > 2.040 (c) t =

x − μ 15.59 − 14 = ≈ 3.46 s 2.60 n 32

(d) Because t > 2.040 , reject H 0 . (e) There is enough evidence at the 2.5% level of significance to reject the claim that the mean cost of a yoga session is no more than $14. 43. (a) The claim is “the mean score for grade 12 students on a science achievement test is more than 145.” H 0 : μ ≤ 145; H a : μ > 145 (claim) s ≈ 43.93 (b) x ≈ 155.39 x − μ 155.39 − 145 t= = ≈ 1.419 s 43.93 n 36 P-value = {Area to right of t = 1.419 } ≈ 0.0824

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(c) Because P-value < 0.10 = α , reject H 0 . (d) There is enough evidence at the 10% level of significance to support the education publication’s claim that the mean score for grade 12 students on a science achievement test is more than 145. 44. (a) The claim is “the overall average score of 15-year-old students on an international mathematics literacy test is 489.” H 0 : μ = 489 (claim); H a : μ ≠ 489 s ≈ 28.0 (b) x ≈ 491.9 x − μ 491.9 − 489 t= = ≈ 0.567 s 28 n 30 P-value = 2{Area to right of t = 0.567 } ≈ 2(0.2875) = 0.575

(c) Because P-value > 0.05 = α , fail to reject H 0 . (d) There is not enough evidence at the 5% level of significance to reject the claim that the overall average score of 15-year-old students on an international mathematics literacy test is 489. 45. np = ( 40 )( 0.15 ) = 6 > 5

nq = ( 40 )( 0.85 ) = 34 > 5 → can use normal distribution

H 0 : p = 0.15 (claim); H a : p ≠ 0.15 − z0 = −1.96, z0 = 1.96; Rejection regions: z < −1.96, z > 1.96 lp − p 0.09 − 0.15 z= = ≈ −1.063 pq ( 0.15)( 0.85) n 40 Because −1.96 < z < 1.96 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the claim. 46. np = (116 )( 0.65 ) = 75.4 > 5

nq = (116 )( 0.35 ) = 40.6 > 5 → can use normal distribution

H 0 : p = 0.65 (claim); H a : p ≠ 0.65 − z0 = −2.17, z0 = 2.17; Rejection regions: z < −2.17, z > 2.17 lp − p 0.76 − 0.65 z= = ≈ 2.48 pq ( 0.65)( 0.35) n 116 Because z > 2.17 , reject H 0 . There is not enough evidence at the 3% level of significance to support the claim.

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47. np = ( 68 )( 0.70 ) = 47.6 > 5

nq = ( 68 )( 0.30 ) = 20.4 > 5 → can use normal distribution

H 0 : p ≥ 0.70; H a : p < 0.70 (claim) z0 = −2.33; Rejection region: z < −2.33 lp − p 0.50 − 0.70 z= = ≈ −3.599 pq ( 0.70 )( 0.30 ) n 68 Because z < −2.33 , reject H 0 . There is enough evidence at the 1% level of significance to support the claim. 48. np = ( 30 )( 0.04 ) = 1.2 < 5

nq = ( 30 )( 0.96 ) = 28.8 > 5

Because np = 1.2 < 5 , the normal distribution cannot be used to approximate the binomial distribution. 49. (a) The claim is “over 56% of the U.S. adults think it is likely that robots and computers will do most jobs 25 years from now.” H 0 : p ≤ 0.56; H a : p > 0.56 (claim)

(b) z0 = 2.33; Rejection region: z > 2.33 (c) z =

lp − p

pq n

0.59 − 0.56

( 0.56 )( 0.44 )

≈ 1.911

1000

(d) Because z < 2.33 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to support the claim that over 56% of the U.S. adults think it is likely that robots and computers will do most jobs 25 years from now. 50. (a) The claim is “40% of the U.S. adults have a positive view of the sports industry.” H 0 : p = 0.40 (claim); H a : p ≠ 0.40

(b) − z0 = −1.96, z0 = 1.96; Rejection regions: z < −1.96, z > 1.96 x 165 = 0.3 1 (c) lp = = n 550 lp − p 0.3 − 0.40 z= = ≈ −4.787 pq ( 0.40 )( 0.60 ) n 550

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(e) At the 5% level of significance, there is enough evidence to reject the sport’s analyst’s claim that 40% of the U.S. adults have a positive view of the sports industry. 51. Critical value: χ 02 = 30.144 ; Rejection region: χ 2 > 30.144 52. Critical values: χ L2 = 3.565, χ R2 = 29.819 ; Rejection regions: χ 2 < 3.565, χ 2 > 29.819 53. Critical values: χ L2 = 26.509, χ R2 = 55.758 ; Rejection regions: χ 2 < 26.509, χ 2 > 55.758 54. Critical value: χ02 = 1.145 ; Rejection region: χ 2 < 1.145 55. H 0 : σ 2 ≤ 2; H a : σ 2 > 2 (claim)

χ02 = 24.769; Rejection region: χ 2 > 24.769 χ2 =

( n − 1) s 2 = (17 )( 2.95) = 25.075 σ2 ( 2)

Because χ 2 > 24.769 , reject H 0 . There is enough evidence at the 10% level of significance to support the claim. 56. H 0 : σ 2 ≤ 60 (claim); H a : σ 2 > 60

χ02 = 26.119; Rejection region: χ 2 > 26.119 χ2 =

( n − 1) s 2 = (14 )( 72.7 ) ≈ 16.963 σ2 ( 60 )

Because χ 2 < 26.119 , fail to reject H 0 . There is not enough evidence at the 2.5% level of significance to reject the claim. 57. H 0 : σ = 1.25 (claim); H a : σ ≠ 1.25

χ L2 = 0.831, χ R2 = 12.833; Rejection regions: χ 2 < 0.831, χ 2 > 12.833 2 n − 1) s 2 ( 5)(1.03) ( = ≈ 3.395 χ = σ2 (1.25)2 2

Because 0.831 < χ 2 < 12.833 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the claim. 58. H 0 : σ = 0.035; H a : σ ≠ 0.035 (claim)

χ L2 = 4.601, χ R2 = 32.801; Rejection regions: χ 2 < 4.601, χ 2 > 32.801

( n − 1) s 2 (15)( 0.026)2 χ = = ≈ 8.278 σ2 ( 0.035)2 2

Because 4.601 < χ 2 < 32.801 , fail to reject H 0 . There is not enough evidence at the 1% level of significance to support the claim.

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59. (a) The claim is “the variance of the bolt widths is at most 0.01.” H0 : σ 2 ≤ 0.01 (claim); H a : σ 2 > 0.01

(b) χ02 = 49.645; Rejection region: χ 2 > 49.645 n − 1) s 2 ( 27 )( 0.064 ) ( (c) χ = = = 172.8 σ2 ( 0.01) 2

(d) Because χ 2 > 49.645 , reject H 0 . (e) There is enough evidence at the 0.5% level of significance to reject the claim that the variance is at most 0.01. 60. (a) The claim is “the standard deviation of the lengths of serving times is 3 minutes.” H0 : σ = 3 (claim); H a : σ ≠ 3

(b) χ L2 = 11.160, χ R2 = 48.290; Rejection regions: χ 2 < 11.160, χ 2 > 48.290

( n − 1) s 2 ( 26)( 3.9)2 = = 43.94 (c) χ = σ2 ( 3) 2 2

(d) Because 11.160 < χ 2 < 48.290 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to reject the claim that the standard deviation of the lengths of serving times is 3 minutes. 61. χ02 = 46.963; Rejection region: χ 2 > 46.963

From Exercise 59, χ 2 = 172.8 . You can reject H 0 at the 1% level of significance because χ 2 = 172.8 > 46.963 . 62. χ L2 = 13.844, χ R2 = 41.923; Rejection regions: χ 2 < 13.844, χ 2 > 41.923

From Exercise 60, χ 2 = 43.94 . You can reject H 0 at the 5% level of significance because χ 2 = 43.94 > 41.923 .

CHAPTER 7 QUIZ SOLUTIONS 1. (a) The claim is “the mean hat size for a male is at least 7.25.” H0 : μ ≥ 7.25 (claim); H a : μ < 7.25

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(c) Sample answer: z0 = −2.33; Rejection region: z < −2.33 x − μ 7.15 − 7.25 z= = ≈ −1.283 0.27 σ n 12 (d) Because z > −2.33 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to reject the company’s claim that the mean hat size for a male is at least 7.25. 2. (a) The claim is “the mean daily base price for renting a full-size or less expensive vehicle in Vancouver, British Columbia, is more than $86.” H0: μ 86; Ha: μ > 86 (claim)

(b) Right-tailed, because the alternative hypothesis contains >; z-test, because ı is known and n 30. (c) Sample answer: z0 = 1.28; Rejection region: z > 1.28 x − μ 93.23 − 86 z= = ≈ 1.582 28.90 σ n 40 (d) Because z > 1.28 , reject H 0 . (e) There is enough evidence at the 10% level of significance to support the travel analyst’s claim that the mean daily base price for renting a full-size or less expensive vehicle in Vancouver, British Columbia, is more than $86. 3. (a) The claim is “the mean amount of earnings for full-time workers ages 18 to 24 with a bachelor’s degree in a recent year is $52,133.” H0: μ = 52,133 (claim); Ha: μ ≠ 52,133

(b) Two-tailed, because the alternative hypothesis contains ≠; t-test, because ı is unknown and the population is normally distributed. (c) Sample answer: −t0 = −2.145; t0 = 2.145 ; Rejection regions: t < −2.145; t > 2.145 x − μ 48,400 − 52,133 t= = ≈ −2.165 s 6679 n 15 (d) Because t < −2.145 , reject H 0 . (e) There is enough evidence at the 5% level of significance to reject the government agency’s claim that the mean amount of earnings for full-time workers ages 18 to 24 with a bachelor’s degree in a recent year is $52,133.

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4. (a) The claim is “program participants have a mean weight loss of at least 10.5 pounds after 1 month.” H0: μ ≥ 10.5 (claim); Ha: μ < 10.5

(b) Left-tailed because the alternative hypothesis contains <; t-test because ı is unknown and n 30. (c) Sample answer: t0 = −2.426; Rejection region: t < −2.426; x − μ 9.233 − 10.5 t= = ≈ −3.09 s 2.593 n 40 (d) Because reject H 0 . (e) There is enough evidence at the 1% level of significance to reject the weight loss program’s claim that program participants have a mean weight loss of at least 10.5 pounds after 1 month. 5. (a) The claim is “less than 25% of the televisions the organization rated in a recent year have an overall score of 70 or more.” H0: p 0.25; Ha: p < 0.25 (claim)

(b) Left-tailed, because the alternative hypothesis contains <; z-test, because np 5 and nq 5. (c) Sample answer: z0 = −1.645; Rejection region: z < −1.645 lp − p 0.23 − 0.25 z= = ≈ −0.273 pq ( 0.25)( 0.75) n 35 (d) Because z > −1.645 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to support the claim that less than 25% of the televisions the organizations rated in a recent year have an overall score of 70 or more. 6. (a) The claim is “the standard deviation of the television rating scores is 10.1.” H0: ı = 10.1 (claim); Ha: ı ≠ 10.1.

(b) Two-tailed, because the alternative hypothesis contains ≠; chi-square test, because the test is for a standard deviation and the population is normally distributed. (c) Sample answer: χ L2 = 18.493, χ R2 = 43.773; Rejection regions: χ 2 < 18.493, χ 2 > 43.773

χ2 =

( n − 1) s 2 = ( 34)(10.9)2 ≈ 39.599 σ2 (10.1)2

(d) Because 18.493 < χ 2 < 43.773 , fail to reject H 0 .

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(e) There is not enough evidence at the 10% level of significance to reject the claim that the standard deviation of the TV rating scores is 10.1.

CHAPTER 7 TEST SOLUTIONS 1. (a) The claim is “at least 25% have purchased a meal kit in a recent year.” H0: p 0.25 (claim); Ha: p < 0.25

(b) Left-tailed, because the alternative hypothesis contains <; z-test, because np ≥ 5 and nq ≥ 5. (c) Sample answer: z0 = −1.28; Rejection region: z < −1.28 lp − p 0.19 − 0.25 z= = ≈ −0.831 pq ( 0.25)( 0.75) n 36 (d) Because z > −1.28 , fail to reject H 0 . (e) At the 10% level of significance, there is not enough evidence to reject the retail grocery chain owner’s claim that at least 25% of adults have purchased a meal kit in a recent year. 2. (a) The claim is “the mean of the room rates for two adults at three-star hotels in Salt Lake City is $134.” H0: ȝ = 134 (claim); Ha: ȝ ≠ 134

(b) Two-tailed because the alternative hypothesis contains ≠; z-test because s is known and n ≥ 30. (c) Sample answer: − z0 = −1.645, z0 = 1.645; Rejection regions: z < −1.645, z > 1.645 x − μ 143 − 134 z= = ≈ 1.82 30 σ n 37 (d) Because z > 1.645 , reject H 0 . (e) There is enough evidence at the 10% level of significance to reject the travel analyst’s claim that the mean of the room rates for two adults at three-star hotels in Salt Lake City is $134. 3. (a) The claim is “the mean price of a meal for a family of 4 in a resort restaurant is at most $100.” H0: ȝ ≤ 100 (claim); Ha: ȝ > 100

(b) Right-tailed because the alternative hypothesis contains >; t-test because s is unknown and n ≥30. (c) Sample answer: t0 = 2.449; Rejection region: t > 2.449 x − μ 110 − 100 t= = ≈ 3.02 s 19 n 33 Copyright © 2023 Pearson Education, Inc.

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CHAPTER 7 Ň HYPOTHESIS TESTING WITH ONE SAMPLE

(d) Because t > 2.449 , reject H 0 . (e) There is enough evidence at the 1% level of significance to reject the travel analyst’s claim that the mean price of a meal for a family of 4 in a resort restaurant is at most $100.” 4. (a) The claim is “more than 80% of U.S. adults think that mothers should have paid maternity leave.” H0: p ≤ 0.80 ; Ha: p > 0.80 (claim)

(b) Right-tailed because the alternative hypothesis contains >; z-test because np ≥ 5 and nq ≥ 5. (c) Sample answer: z0 = 1.645; Rejection region: z > 1.645 lp − p 0.82 − 0.80 z= = ≈ 0.35 pq ( 0.80 )( 0.20 ) n 50 (d) Because z < 1.645 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to support the researcher’s center’s claim that more than 80% of U.S. adults think that mothers should have paid maternity leave. 5. (a) The claim is “the standard deviation of the number of grams of carbohydrates in a bar is 1.11 grams.” H0 : σ = 1.11 (claim); H a : σ ≠ 1.11

(b) Two-tailed because the alternative hypothesis contains ≠; chi-square test because the test is for a standard deviation and the population is normally distributed. (c) Sample answer: χ L2 = 13.120, χ R2 = 40.646; Rejection regions: χ 2 < 13.120, χ 2 > 40.646

( n − 1) s 2 ( 25)(1.19)2 = ≈ 28.733 χ = σ2 (1.11)2 2

(d) Because 13.120 < χ 2 < 40.646 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to reject the claim that the standard deviation of the number of grams of carbohydrates in a bar is 1.11 grams. 6. (a) The claim is “the mean rating of gas grills the organization rated in a recent year is less than 64.” H0: μ ≥ 64 ; Ha: μ < 64 (claim)

(b) Left-tailed, because the alternative hypothesis contains <; t-test, because σ is unknown and n ≥ 30.

CHAPTER 7 Ň HYPOTHESIS TESTING WITH ONE SAMPLE

307

(c) Sample answer: t0 = −2.403; Rejection region: t < −2.403 x − μ 62.9 − 64 t= = ≈ −0.960 s 8.1 n 50 (d) Because t > −2.403 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to support the claim that the mean rating of gas grills the organization rated in a recent year is less than 64. 7. (a) The claim is “the mean age of the residents of a small town is more than 38 years.” H0: μ ≤ 38 (claim); Ha: μ > 38

(b) Right-tailed because the alternative hypothesis contains >; z-test because σ is known and n ≥ 30. (c) Sample answer: z0 = 1.28; Rejection region: z > 1.28 x − μ 42.0667 − 38 z= = ≈ 2.47 9 σ n 30 (d) Because z > 1.28 , reject H 0 . (e) There is enough evidence at the 10% level of significance to support the researcher’s claim that the mean age of the residents of a small town is more than 38 years.

Hypothesis Testing with Two Samples

CHAPTER

8.1 TESTING THE DIFFERENCE BETWEEN MEANS (INDEPENDENT SAMPLES, V1 and V2 KNOWN) 8.1 TRY IT YOURSELF SOLUTIONS Note: Answers may differ due to rounding. 1. (1) Independent; Because each sample represents blood pressures of different individuals, and it is not possible to form a pairing between the members of the samples. (2) Dependent; Because the samples represent exam scores of the same students, the samples can be paired with respect to each student. 2. The claim is “there is a difference in the mean annual wages for forensic science technicians working for local and state governments.”

H0 : P1 P2 ; Ha : P1 z P2 (claim) D = 0.10 z0 1.645, z0 1.645; Rejection regions: z 1.645, z ! 1.645 ( x1 x2 ) ( P1 P2 ) (63,560 62,070) (0) z

V 12 n1

V 22

(6200) 2 (5575) 2 100 100

| 1.787

Since z ! 1.645 , reject H 0 . There is enough evidence at the 10% level of significance to support the claim that there is a difference in mean annual wages. 3.

( x1 x2 ) ( P1 P2 )

2 1

(1610 1603) (0)

2 2

(210) 2 (190) 2 200 300

| 0.379

P-value = {area right of z 0.379 } | 0.352 Because P-value ! 0.05 D , fail to reject H 0 . There is not enough evidence at the 5% level of significance to support the claim that the average monthly rent for a one-bedroom apartment in Atlanta, GA, is greater than the average monthly rent for a one-bedroom apartment in Nashville, TN.

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

309

8.1 EXERCISE SOLUTIONS 1. Two samples are dependent when each member of one sample corresponds to a member of the other sample. Example: The weights of 22 people before starting an exercise program and the weights of the same 22 people 6 weeks after starting the exercise program. Two samples are independent when the sample selected from one population is not related to the sample selected from the other population. Example: The weights of 25 cats and the weights of 20 dogs. 2. State the hypotheses and identify the claim. Specify the level of significance. Find the critical values(s) and identify the rejection region(s). Find the standardized test statistic. Make a decision and interpret it in the context of the claim. 3. Use P-values. 4. (1) The population standard deviations are known. (2) The samples are randomly selected. (3) The samples are independent. (4) The populations are normally distributed or each sample size is at least 30. 5. Dependent because the same football players were sampled. 6. Independent because different individuals were sampled. 7. Independent because different boats were sampled. 8. Dependent because the same workers were sampled. 9. Because z | 2.96 ! 1.96 and P | 0.0031 0.05, reject H0 . 10. Because z | 1.94 2.33 and P | 0.0261 ! 0.01, fail to reject H0 . 11. H0 : P1 P2 (claim); Ha : P1 z P2 Rejection region: z 2.575, z ! 2.575 ( x1 x2 ) ( P1 P2 ) (16 14) (0) z | 2.89 2 2 V1 V 2 (3.4)2 (1.5) 2 29 28 n1 n2 Because z ! 2.575 , reject H0 . There is enough evidence at the 1% level of significance to reject the claim.

310

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

12. H0 : P1 d P2 ; Ha : P1 ! P2 (claim) Rejection region: z ! 1.28 ( x1 x2 ) ( P1 P2 ) (500 495) (0) z | 1.15 V 12 V 22 (40) 2 (15) 2 100 75 n1 n2 Because z 1.28 , fail to reject H0 . There is not enough evidence at the 10% level of significance to support the claim. 13. H0 : P1 t P2 ; Ha : P1 P2 (claim) Rejection region: z 1.645 ( x1 x2 ) ( P1 P2 ) (2435 2432) (0) z | 0.18 V 12 V 22 (75)2 (105)2 35 90 n1 n2 Because z ! 1.645 , fail to reject H0 . There is not enough evidence at the 5% level of significance to support the claim. 14. H0 : P1 d P2 (claim); Ha : P1 ! P2 Rejection region: z ! 1.88 ( x1 x2 ) ( P1 P2 ) (5004 4895) (0) z | 5.29 V 12 V 22 (136)2 (215)2 144 156 n1 n2 Because z ! 1.88 , reject H0 . There is enough evidence at the 3% level of significance to reject the claim. 15. (a) The claim is “the mean braking distances are different for the two categories of SUVs.”

H0 : P1 P2 ; Ha : P1 z P2 (claim)

(b) z0 (c) z

1.645, z0 1.645; Rejection regions: z 1.645, z ! 1.645 ( x1 x2 ) ( P1 P2 )

V 12 n1

V 22 n2

(131.8 132.8) (0) (5.5) 2 (6.7) 2 16 11

| 0.409

(d) Because 1.645 z 1.645 , fail to reject H0 . (e) There is not enough evidence at the 10% level of significance to support the claim that the mean braking distances are different for the two categories of SUVs. 16. (a) The claim is “the mean rating of traditional mattresses is greater than the mean rating of mattresses in a box.”

H0 : P1 d P2 ; Ha : P1 ! P2 (claim)

(b) z0

2.33; Rejection region: z ! 2.33

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

( x1 x2 ) ( P1 P2 )

V 12 n1

V 22

(70.9 68.7) (0) (5.6)2 (6.6)2 30 30

311

| 1.39

(d) Because z 2.33 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to support the claim that the mean rating of traditional mattresses is greater than the mean rating of mattresses in the box. 17. (a) The claim is “the wind speed in Region A is less than the wind speed in Region B.”

H0 : P1 t P2 ; Ha : P1 P2 (claim)

(b) z0

1.645; Rejection region: z 1.645

( x1 x2 ) ( P1 P2 )

2 1

2 2

(14 15.1) (0) (2.9)2 (3.3)2 60 60

| 1.94

(d) Because, z 1.645 reject H0 . (e) There is enough evidence at the 5% level of significance to support the claim that the wind speed in Region A is less than the wind speed in Region B. 18. (a) The claim is “the mean repair costs for Model A and Model B are equal.”

H0 : P1 P2 (claim); Ha : P1 z P2

(b) z0 (c) z

2.575, z0

2.575; Rejection regions: z 2.575, z ! 2.575

( x1 x2 ) ( P1 P2 )

V 12 n1

V 22

(208 221) (0) (18)2 (22) 2 24 26

| 2.29

(d) Because 2.575 z 2.575 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to reject the claim that the mean repair costs for Model A and Model B are equal. 19. (a) The claim is “ACT mathematics and science scores are equal.”

H0 : P1 P2 (claim); Ha : P1 z P2

(b) z0

2.575, z0

2.575; Rejection regions: z 2.575, z ! 2.575

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

( x1 x2 ) ( P1 P2 )

V12 n1

V 22

(20.2 20.6) (0) (5.7) 2 (5.9) 2 60 75

| 0.399

(d) Because 2.575 z 2.575 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to reject the claim that ACT mathematics and science scores are equal. 20. (a) The claim is “ACT reading scores are higher than ACT English scores.” Note that P1 represents English and P2 represents reading.

H0 : P1 t P2 ; Ha : P1 P2 (claim)

(b) z0

1.28; Rejection region: z 1.28

( x1 x2 ) ( P1 P2 )

V 12 n1

V 22

(19.9 21.2) (0) (7.2)2 (7.1) 2 120 150

| 1.483

(d) Because z 1.28 , reject H0 . (e) There is enough evidence at the 10% level of significance to support the claim that ACT reading scores are higher than ACT English scores. 21. (a) The claim is “the mean home sales price in Casper, Wyoming, is the same as in Cheyenne, Wyoming.”

H0 : P1 P2 (claim); Ha : P1 z P2

(b) z0 (c) z

2.575, z0

2.575; Rejection regions: z 2.575, z ! 2.575

( x1 x2 ) ( P1 P2 )

2 1

2 2

(349, 237 435, 244) (0) (158,005)2 (154,716) 2 35 41

| 2.388

(d) Because 2.575 z 2.575 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to reject the claim that the mean home sales price in Casper, Wyoming, is the same as in Cheyenne, Wyoming. 22. (a) The claim is “the mean home sales price in Casper, Wyoming, is the same as in Cheyenne, Wyoming.”

H0 : P1 P2 (claim); Ha : P1 z P2

(b) z0

2.575, z0

2.575; Rejection regions: z 2.575, z ! 2.575

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

( x1 x2 ) ( P1 P2 )

V 12 n1

V 22

(343, 202.64 436,666.76) (0)

(154,716)2 (158,005)2 42 36

313

| 2.629

(d) Because z 2.575 , reject H0 . (e) There is enough evidence at the 1% level of significance to reject the claim that the mean home sales price in Casper, Wyoming, is the same as in Cheyenne, Wyoming. Yes, the new samples lead to a different conclusion. 23. (a) The claim is “the precipitation in Seattle, Washington, was greater than in Birmingham, Alabama.”

H0 : P1 d P2 ; Ha : P1 ! P2 (claim)

(b) z0 1.645; Rejection region: z ! 1.645 (c) x1 | 0.2336666667, n1 30

x2 | 0.1263333333, n2 30 ( x1 x2 ) ( P1 P2 ) (0.2336666667 0.1263333333) (0) z | 1.019 V12 V 22 (0.25)2 (0.52) 2 30 30 n1 n2 (d) Because z 1.645 , fail to reject H0 . (e) There is not enough evidence at the 5% level of significance to support the claim that the precipitation in Seattle, Washington, was greater than in Birmingham, Alabama. 24. (a) The claim is “the temperature in Seattle, Washington, was lower than in Birmingham, Alabama.” H0 : P1 t P2 ; Ha : P1 P2 (claim) (b) z0

2.33; Rejection region: z 2.33

x2 | 74.36666667, n2 30 ( x1 x2 ) ( P1 P2 ) (62.06666667 74.36666667) (0) z | 3.776 V 12 V 22 (12)2 (13.2)2 30 30 n1 n2 (d) Because z 2.33 , reject H0 . (e) There is enough evidence at the 1% level of significance to support the climatologist’s claim that the temperature in Seattle, Washington, was lower than in Birmingham, Alabama.

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

25. They are equivalent through algebraic manipulation of the equation.

P1 P2 P1 P2 0

26. They are equivalent through algebraic manipulation of the inequality. P1 t P2 P1 P2 t 0 27. H0 : P1 P2 d 4000; Ha : P1 P2 ! 4000 (claim) z0 1.645; Rejection region: z ! 1.645 ( x1 x2 ) ( P1 P2 ) (88,900 81,600) (4000) z | 1.089 V 12 V 22 (14,060) 2 (13,050) 2 42 38 n1 n2 Because z 1.645 , fail to reject H0 . There is not enough evidence at the 5% level of significance to support the claim that the difference between the mean annual salaries of entry-level software engineers in Santa Clara, California, and Greenwich, Connecticut, is more than $4000. 28. H0 : P1 P2

9000 (claim); Ha : P1 P2 z 9000 z0 2.575, z0 2.575; Rejection regions: z 2.575, z ! 2.575 ( x1 x2 ) ( P1 P2 ) (58,300 54, 240) (9000) z | 3.072 V 12 V 22 (6560)2 (6100)2 32 30 n1 n2

Because z 2.575 , reject H0 . There is enough evidence at the 1% level of significance to reject the claim that the difference between the mean annual salaries of entry-level architects in Denver, Colorado, and Lincoln, Nebraska, is equal to $9000. ( x1 x2 ) zc

29. (88,900 81,600) 1.96

V 12

V 22 n2

P1 P2 ( x1 x2 ) zc

V 12 n1

V 22 n2

(14,060)2 (13,050)2 (14,060)2 (13,050)2 P1 P2 (88,900 81,600) 1.96 42 38 42 38

7300 1.96 9,188,397.118 P1 P2 7300 1.96 9,188,397.118 $1359 P1 P2 $13, 241

30.

( x1 x2 ) zc (58,300 54, 240) 2.575

V 12 n1

V 22 n2

P1 P2 ( x1 x2 ) zc

V 12 n1

V 22 n2

(6560)2 (6100)2 (6560)2 (6100)2 P1 P2 (58,300 54, 240) 2.575 32 30 32 30

4060 2.575 2,585,133.333 P1 P2 4060 2.575 2,585,133.333 $80 P1 P2 $8200

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

315

8.2 TESTING THE DIFFERENCE BETWEEN MEANS (INDEPENDENT SAMPLES, V1 and V2 UNKNOWN) 8.2 TRY IT YOURSELF SOLUTIONS 1. The claim is “there is a difference in the mean annual earnings based on the level of education.”

H0 : P1 P2 ; Ha : P1 z P2 (claim) D 0.05; d.f. min{n1 1, n2 1} min ^25 1, 16 1` 15 t0 2.131, t0 2.131; Rejection regions: t 2.131, t ! 2.131 ( x1 x2 ) ( P1 P2 ) (41,775 51,650) (0) t

s12 s22 n1 n2

(6050) 2 (9580) 2 25 16

| 3.680

Because t 2.131 , reject H0 . There is enough evidence at the 5% level of significance to support the claim that there is a difference in the mean annual earnings based on the level of education. 2. The claim is “the mean driving cost per mile of the manufacturer’s minivans is less than that of its leading competitor.”

H0 : P1 t P2 ; Ha : P1 P2 (claim) D 0.10; d.f. n1 n2 2 34 38 2 70 t0 1.294; Rejection region: t 1.294 ( x1 x2 ) ( P1 P2 ) t

(n1 1) s12 (n2 1) s22 n1 n2 2

1 1 n1 n2

(0.57 0.59) (0)

(34 1)(0.09) 2 (38 1)(0.08) 2 34 38 2

1 1 34 38

| 0.998

Because t ! 1.294 , fail to reject H0 . There is not enough evidence at the 10% level of significance to support the manufacturer’s claim that the mean driving cost per mile of its minivans is less than that of its leading competitor.

8.2 EXERCISE SOLUTIONS 1. (1) The population standard deviations are unknown. (2) The samples are randomly selected. (3) The samples are independent (4) The populations are normally distributed or each sample size is at least 30. 2. State hypotheses and identify the claim. Specify the level of significance. Determine the degrees of freedom. Find the critical value(s) and identify the rejection region(s). Find the standardized test statistic. Make a decision and interpret it in the context of the original claim.

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

3. (a) d.f. n1 n2 2 23

4. (a) d.f. n1 n2 2 25

1.714, t0 1.714

(b) d.f. min{n1 1, n2 1} 10

1.812, t0 1.812

1.746

1.943

2.708, t0

2.708

(b) d.f. min{n1 1, n2 1} 18

7. (a) d.f. n1 n2 2 19

2.878, t0

2.878

8. (a) d.f. n1 n2 2 60

t0 1.729

(b) d.f. min{n1 1, n2 1} 7

t0 1.895 9.

2.718

6. (a) d.f. n1 n2 2 39

(b) d.f. min{n1 1, n2 1} 6

(b) d.f. min{n1 1, n2 1} 11

5. (a) d.f. n1 n2 2 16

2.485

1.296

(b) d.f. min{n1 1, n2 1} 29

1.311

H0 : P1 P2 (claim); Ha : P1 z P2 d.f. n1 n2 2 27 Rejection regions: t 2.771, t ! 2.771 ( x1 x2 ) ( P1 P2 ) (33.7 35.5) (0) t | 1.70 (n1 1) s12 (n2 1) s22 1 1 (12 1)(3.5) 2 (17 1)(2.2) 2 1 1 12 17 2 12 17 n1 n2 2 n1 n2 Because 2.771 t 2.771 , fail to reject H0 . There is not enough evidence at the 1% level of significance to reject the claim.

10. H0 : P1 t P2 ; Ha : P1 P2 (claim)

d.f. n1 n2 2 18 Rejection region: t 1.330 ( x1 x2 ) ( P1 P2 )

(0.345 0.515) (0)

(n1 1) s12 (n2 1) s22 n1 n2 2

(11 1)(0.305) 2 (9 1)(0.215) 2 1 1 11 9 2 11 9

1 1 n1 n2

| 1.41

Because t 1.330 , reject H0 . There is enough evidence at the 10% level of significance to support the claim. 11. H0 : P1 d P2 (claim); Ha : P1 ! P2

d.f. min{n1 1, n2 1} 9 Rejection region: t ! 1.833

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

( x1 x2 ) ( P1 P2 )

(2410 2305) (0)

s12 s22 n1 n2

(175) 2 (52) 2 13 10

317

| 2.05

Because t ! 1.833 , reject H0 . There is enough evidence at the 5% level of significance to reject the claim. 12. H0 : P1 d P2 ; Ha : P1 ! P2 (claim)

d.f. min{n1 1, n2 1} 31 Rejection region: t ! 2.453 ( x1 x2 ) ( P1 P2 ) (52 50) (0) t

s12 s22 n1 n2

(4.8)2 (1.2) 2 32 40

| 2.30

Because t 2.453 , fail to reject H0 . There is not enough evidence at the 1% level of significance to support the claim. 13. (a) The claim is “the mean annual costs of food for dogs and cats are the same.”

H0 : P1 P2 (claim); Ha : P1 z P2

(b) d.f. n1 n2 2 32

t0 | 1.694, t0 | 1.694; Rejection regions: t 1.694, t ! 1.694 (c) t

( x1 x2 ) ( P1 P2 ) (n1 1) s12 (n2 1) s22 n1 n2 2

1 1 n1 n2

(255 231) (0) (16 1)(30) 2 (18 1)(28) 2 1 1 16 18 2 16 18

| 2.412

(d) Because t ! 1.694 , reject H0 . (e) There is enough evidence at the 10% level of significance to reject the pet association’s claim that the mean annual costs of food for dogs and cats are the same. 14. (a) The claim is “the mean amount spent by a customer at Burger Stop is greater than the mean amount spent by a customer at Fry World.” H0 : P1 d P2 ; Ha : P1 ! P2 (claim) (b) d.f. n1 n2 2 50

t0 1.676; Rejection region: t ! 1.676 (c) t

( x1 x2 ) ( P1 P2 ) (n1 1) s12 (n2 1) s22 n1 n2 2

1 1 n1 n2

(d) Because t 1.676 , fail to reject H0 .

(5.46 5.12) (0) (22 1)(0.89) 2 (30 1)(0.79) 2 22 30 2

1 1 22 30

| 1.45

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

(e) There is not enough evidence at the 5% level of significance to support the claim that the mean amount spent by a customer at Burger Stop is greater than the mean amount spent by a customer at Fry World. 15. (a) The claim is “the stomachs of blue crabs from one location contain more fish than the stomachs of blue crabs from another location.”

H0 : P1 d P2 ; Ha : P1 ! P2 (claim)

(b) d.f. n1 n2 2 38 t0 2.429; Rejection region: t ! 2.429 (c) t

( x1 x2 ) ( P1 P2 ) (n1 1) s12 (n2 1) s22 n1 n2 2

1 1 n1 n2

(320 280) (0) (25 1)(60) 2 (15 1)(80) 2 25 15 2

1 1 25 15

| 1.80

(d) Because t 2.429 , fail to reject H0 . (e) There is not enough evidence to support the claim at the 1% level of significance that the stomachs of blue crabs from one location contain more fish than the stomachs of blue crabs from another location. 16. (a) The claim is “the mean fork length of yellowfin tuna is different in two zones in the eastern tropical Pacific Ocean.”

H0 : P1 P2 ; Ha : P1 z P2 (claim)

(b) d.f. n1 n2 2 55

t0 | 2.678, t0 | 2.678; Rejection regions: t 2.678, t ! 2.678 (c) t

( x1 x2 ) ( P1 P2 ) (n1 1) s12 (n2 1) s22 n1 n2 2

1 1 n1 n2

(76.2 80.8) (0) (26 1)(16.5) 2 (31 1)(23.4) 2 26 31 2

1 1 26 31

| 0.84

(d) Because 2.678 t 2.678 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to support the marine biologist’s claim that the mean fork length of yellowfin tuna is different in two zones in the eastern tropical Pacific Ocean. 17. (a) The claim is “the mean household income in a recent year is greater in York County, South Carolina, than it is in Elmore County, Alabama.”

H0 : P1 d P2 ; Ha : P1 ! P2 (claim)

(b) d.f. min{n1 1, n2 1} 18

t0 1.734; Rejection region: t ! 1.734

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

( x1 x2 ) (0)

(64,900 59,500) (0)

s12 s22 n1 n2

(16,000) 2 (23,600) 2 23 19

319

| 0.849

(d) Because t 1.734 , fail to reject H0 . (e) There is not enough evidence at the 5% level of significance to support the claim that the mean household income in a recent year is greater in York County, South Carolina, than it is in Elmore County, Alabama. 18. (a) The claim is “the mean household income in a recent year is the same in Eagle County, Colorado, and Boulder County, Colorado.”

H0 : P1 P2 (claim); Ha : P1 z P2

(b) d.f. min{n1 1, n2 1} 14 t0 1.761; t0 1.761 Rejection regions: t 1.761, t ! 1.761 (c) t

( x1 x2 ) (0)

(85, 400 83,300) (0)

s12 s22 n1 n2

(26,100)2 (18, 200)2 18 15

| 0.271

(d) Because 1.761 t 1.761 , fail to reject H0 . (e) There is not enough evidence at the 10% level of significance to reject the claim that the mean household income in a recent year is the same in Eagle County, Colorado, and Boulder County, Colorado. 19. (a) The claim is “an experimental method makes a difference in the tensile strength of steel bars.”

H0 : P1 P2 ; Ha : P1 z P2 (claim)

(b) d.f. n1 n2 2 22

2.819, t0

2.819; . Rejection regions: . t 2.819, t ! 2.819 .

x2 | 388.214, s2 | 14.797, n2 14 ( x1 x2 ) ( P1 P2 ) t | (n1 1) s12 (n2 1) s22 1 1 n1 n2 2 n1 n2

(368.3 388.214) (0) (10 1)(22.301) 2 (14 1)(14.797) 2 1 1 10 14 2 10 14

| 2.64

(d) Because 2.831 t 2.831 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to support the claim that an experimental method makes a difference in the tensile strength of steel bars.

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

20. (a) The claim is “the experimental method produces steel with greater mean tensile strength.” H0 : P1 d P2 ; Ha : P1 ! P2 (claim) (b) d.f. min{n1 1, n2 1} 13 t0 1.350; Rejection region: t ! 1.350 (c) x1 | 403.167, s1 | 11.137, n1 18

x2 384, s2 | 17.698, n2 14 ( x1 x2 ) ( P1 P2 ) (403.167 384) (0)

s12 s22 n1 n2

(11.137)2 (17.698)2 18 14

| 3.543

(d) Because t ! 1.350 , reject H0 . (e) There is enough evidence at the 10% level of significance to support the claim that the experimental method produces steel with greater mean tensile strength. 21. (a) The claim is “the new method of teaching reading produces higher reading test scores than the old method”

H0 : P1 t P2 ; Ha : P1 P2 (claim)

(b) d.f. n1 n2 2 42 t0 | 1.303; Rejection region: t 1.303 (c) x1 | 56.684, s1 | 6.961, n1 19 x2 | 67.4, s2 | 9.014, n2 25 ( x1 x2 ) ( P1 P2 ) t | (n1 1) s12 (n2 1) s22 1 1 n1 n2 2 n1 n2

(56.684 67.4) (0) (19 1)(6.961) 2 (25 1)(9.014) 2 1 1 19 25 2 19 25

| 4.295

(d) Because t 1.303 , reject H0 . (e) There is enough evidence at the 10% level of significance to support the claim that the new method of teaching reading produces higher reading test scores than the old method. 22. (a) The claim is “the mean science test score is lower for students taught using the traditional lab method than it is for students taught using the interactive simulation software.”

H0 : P1 t P2 ; Ha : P1 P2 (claim)

(b) d.f. n1 n2 2 39

2.426; Rejection region: t 2.426

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

x2 83, s2 | 7.645, n2 19 ( x1 x2 ) ( P1 P2 )

321

(n1 1) s12 (n2 1) s22

1 1 n1 n2

n1 n2 2

(79.091 83) (0) (22 1)(6.900) 2 (19 1)(7.645) 2 22 19 2

1 1 22 19

| 1.72

(d) Because t ! 2.426 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to support the claim that the mean science test score is lower for students taught using the traditional lab method than it is for students taught using the interactive simulation software. 23. ( x1 x2 ) r tc

s12 s22 (34.1)2 (17.7) 2 o (65.8 65.3) r 1.333 20 18 n1 n2 o 0.5 r 11.586 o 11.086 P1 P2 12.086 o 11.1 P1 P2 12.1

24. ( x1 x2 ) r tc

s12 s22 (6)2 (12)2 o (267 244) r 2.132 9 5 n1 n2 o 23 r 12.21 o 10.78 P1 P2 35.21 o 10.8 P1 P2 35.2

25. Vˆ

(n1 1)s12 (n2 1) s22 n1 n2 2

( x1 x2 ) r tcVˆ

(20 1)(39.7)2 (17 1)(42.4)2 | 40.956 20 17 2

1 1 1 1 o (28 26) r 1.69 40.956 n1 n2 20 17

o 2 22.833 o P1 P2 2 22.833 o 20.8 P1 P2 24.8

26. Vˆ

(n1 1)s12 (n2 1) s22 n1 n2 2

( x1 x2 ) r tcVˆ

(20 1)(12.3)2 (18 1)(14.5)2 | 13.384 20 18 2

1 1 1 1 o (40 39) r 2.028 13.384 n1 n2 20 18

o 1 r 8.8 o 7.8 P1 P2 9.8

322

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

8.3 TESTING THE DIFFERENCE BETWEEN MEANS (DEPENDENT SAMPLES) 8.3 TRY IT YOURSELF SOLUTIONS 1. The claim is “people who play golf can improve (decrease) their average golf scores.”

H0 : Pd d 0; Ha : Pd ! 0 (claim) D 0.10 , d.f. n 1 8 1 7 t0 1.415; Rejection region: t ! 1.415 Before 89 84 96 82 74 92 85 91

After 83 83 92 84 76 91 80 91

d 6 1 4 −2 −2 1 5 0

¦ d 13

¦d n

13 1.625 8

¦d sd

d2 36 1 16 4 4 1 25 0 d 2 87

2 ª d º » « « n » ¬« ¼» n 1

13 8 7

| 3.0677

d Pd 1.625 0 | | 1.498 3.0677 sd 8 n

Because t ! 1.415 , reject H0 . There is enough evidence at the 10% level of significance to support the claim that people who play golf can improve (decrease) their average golf scores by using the company’s newly designed golf clubs. 2. The claim is “the drug changes the body’s temperature.” H0 : Pd 0; Ha : Pd z 0 (claim) D 0.05 , d.f. n 1 6 t0 2.447, t0 2.447; Rejection regions: t 2.447, t ! 2.447

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

Before 101.8 98.5 98.1 99.4 98.9 100.2 97.9

¦d n

¦ sd

After 99.2 98.4 98.2 99 98.6 99.7 97.8

d 2.6 0.1 0.1 0.4 0.3 0.5 0.1

¦ d 3.9 ¦

323

d2 6.76 0.01 0.01 0.16 0.09 0.25 0.01 d 2 7.29

3.9 | 0.5771 7 2 ª d º « » d « n » ¬« ¼» n 1 2

3.9 7.29 7 6

| 0.9235

d Pd 0.5571 0 | | 1.596 0.9235 sd 7 n

Because 2.447 t 2.447 , fail to reject H0 . There is not enough evidence at the 5% level of significance to conclude that the drug changes the body’s temperature.

8.3 EXERCISE SOLUTIONS 1. (1) The samples are randomly selected. (2) The samples are dependent. (3) The populations are normally distributed or the number n of pairs of data is at least 30. 2. The symbol d represents the mean of the differences between the paired data entries in dependent samples. The symbol sd represents the standard deviation of the differences between the paired data entries in the dependent samples. 3.

H0 : Pd t 0; Ha : Pd 0 (claim) D

0.05 and d.f. n 1 13

1.771 ; Rejection region: t 1.771

324

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

d Pd sd n

1.5 0 | 1.754 3.2 14

Because t ! 1.771 , fail to reject H0 . There is not enough evidence at the 5% level of significance to support the claim. 4.

H0 : Pd

0 (claim); Ha : Pd z 0

0.01 and d.f. n 1 7

3.499, t0 3.499; Rejection regions: t 3.499, t ! 3.499

d Pd sd n

3.2 0 | 1.071 8.45 8

Because 3.499 t 3.499 , fail to reject H0 . There is not enough evidence at the 1% level of significance to reject the claim. 5.

H0 : Pd d 0 (claim); Ha : Pd ! 0 D

0.10 and d.f. n 1 15

t0 1.341; Rejection region: t ! 1.341 t

d Pd sd n

6.5 0 | 2.725 9.54 16

Because t ! 1.341 , reject H0 . There is enough evidence at the 10% level of significance to reject the claim. 6.

H0 : Pd d 0; Ha : Pd ! 0 (claim) D

0.05 and d.f. n 1 27

t0 1.703; Rejection region: t ! 1.703 t

d Pd sd n

0.55 0 | 2.940 0.99 28

Because t ! 1.703 , reject H0 . There is enough evidence at the 5% level of significance to support the claim. 7.

H0 : Pd t 0 (claim); Ha : Pd 0 D

0.01 and d.f. n 1 14

2.624; Rejection region: t 2.624

d Pd sd n

2.3 0 | 7.423 1.2 15

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

325

Because t 2.624 , reject H0 . There is enough evidence at the 1% level of significance to reject the claim. 8.

H0 : Pd

0; Ha : Pd z 0 (claim)

0.10 and d.f. n 1 19

1.729, t0 1.729; Rejection regions: t 1.729, t ! 1.729

d Pd sd n

1 0 | 1.626 2.75 20

Because 1.729 t 1.729 , fail to reject H0 . There is not enough evidence at the 10% level of significance to support the claim. 9. (a) The claim is “injections of onabotulinumtoxinA reduce the number of days per month that chronic migraine sufferers have headaches."

H0 : Pd d 0; Ha : Pd ! 0 (claim)

(b) t0

2.567; Rejection region: t ! 2.567

d Pd sd n

13.33 0 | 7.914 7.146 18

(e) Because t ! 2.567 , reject H0 . (f) There is enough evidence at the 1% level of significance to support the claim that injections of onabotulinumtoxinA reduce the number of days per month that chronic migraine sufferers have headaches. 10. (a) The claim is “the Dow Jones Industrial Average lost value from the closing price on the previous day to the current prices.”

H0 : Pd d 0; Ha : Pd ! 0 (claim)

(b) t0

3.143; Rejection region: t ! 3.143

d Pd sd n

1.659 0 | 2.205 1.991 7

326

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

(f) There is not enough evidence at the 1% level of significance to support the claim that the Dow Jones Industrial Average lost value from the closing price on the previous day to the current prices. 11. (a) The claim is “an SAT preparation course will improve the test scores of students.”

H0 : Pd t 0; Ha : Pd 0 (claim)

(b) t0

2.821; Rejection region: t 2.821

60 and sd | 27.889

(d) t

d Pd sd n

60 0 | 6.803 27.889 10

(e) Because t 2.821 , reject H0 . (f) There is enough evidence at the 1% level of significance to support the SAT preparation course’s claim that it improve the test scores of students. 12. (a) The claim is “sprint interval training improves running performance in trained athletes.” H0 : Pd t 0; Ha : Pd 0 (claim) (b) t0

1.415; Rejection region: t 1.415

0.4375 and sd | 0.623

(d) t

d Pd sd n

0.4375 0 | 1.986 0.623 8

(e) Because t 1.415 , reject H0 . (f) There is enough evidence at the 10% level of significance to support the claim that sprint interval training improves running performance in trained athletes. 13. (a) The claim is “a baseball clinic will help players raise their batting average.”

H0 : Pd t 0; Ha : Pd 0 (claim)

(b) t0

1.771; Rejection region: t 1.771

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

(d) t

d Pd sd n

327

0.002 0 | 0.499 0.015 14

(e) Because t ! 1.771 , fail to reject H0 . (f) There is not enough evidence at the 5% level of significance to support the claim that the baseball clinic will help players raise their batting averages. 14. (a) The claim is “the use of a specific type of therapeutic tape reduces pain in patients with chronic tennis elbow.”

H0 : Pd d 0; Ha : Pd ! 0 (claim)

(b) t0 1.761; Rejection region: t ! 1.761 (c) d | 2.133 and sd | 1.885 (d) t

d Pd sd n

2.133 0 | 4.383 1.885 15

(e) Because t ! 1.761 , reject H0 . (f) There is enough evidence at the 5% level of significance to support the physical therapist’s claim that the use of a specific type of therapeutic tape reduces pain in patients with chronic tennis elbow. 15. (a) The claim is “U.S. gasoline prices increased from May to June of 2021.”

H0 : Pd t 0; Ha : Pd 0 (claim)

(b) t0

3.106; Rejection region: t 3.106

d Pd sd n

0.029 0 | 3.315 0.0303 12

(e) Because t 3.106 , reject H0 . (f) There is enough evidence at the 0.5% level of significance to support the claim that the U.S. gasoline prices increased from May to June of 2021.

328

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

16. (a) The claim is “entry-level receptionists earn more than entry-level customer support representatives in Dallas, Texas.”

H0 : Pd t 0; Ha : Pd 0 (claim)

(b) t0

3.707; Rejection region: t 3.707

d Pd sd n

2060.71 0 | 1.420 3839.37 7

(e) Because t ! 3.707 , fail to reject H0 . (f) There is not enough evidence at the 0.5% level of significance to support the claim that entrylevel receptionists earn more than entry-level customer support representatives in Dallas, Texas. 17. (a) The claim is “the product ratings have changed from last year to this year.”

H0 : Pd

0; Ha : Pd z 0 (claim)

(b) t0

2.365, t0

1 and sd | 1.309

(d) t

d Pd sd n

2.365; Rejection regions: t 2.365, t ! 2.365

1 0 | 2.161 1.309 8

(e) Because 2.365 t 2.365 , fail to reject H0 . (f) There is not enough evidence at the 5% level of significance to support the claim that the product ratings have changed from last year to this year. 18. (a) The claim is “passing percentage is different for home and away games.”

H0 : Pd

0; Ha : Pd z 0 (claim)

(b) t0

1.383, t0 1.383; Rejection regions: t 1.383, t ! 1.383

0.83 and sd | 9.202

(d) t

d Pd sd n

0.83 0 | 0.285 9.202 10

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

329

(e) Because 1.383 t 1.383 , fail to reject H0 . (f) There is not enough evidence at the 20% level of significance to support the claim that passing percentage is different for home and away games. 19. (a) The claim is “eating new cereal as part of a daily diet lowers total blood cholesterol levels.”

H0 : Pd d 0; Ha : Pd ! 0 (claim)

(b) t0 1.943; Rejection region: t ! 1.943 (c) d | 2.857 and sd | 4.451 (d) t

d Pd sd n

2.857 0 | 1.698 4.451 7

(e) Because t 1.943 , fail to reject H0 . (f) There is not enough evidence at the 5% level of significance to support the claim that eating new cereal as part of a daily diet lowers total blood cholesterol levels. 20. (a) The claim is “contestants’ times have changed.”

H0 : Pd

0; Ha : Pd z 0 (claim)

(b) t0

3.499, t0 3.499; Rejection regions: t 3.499, t ! 3.499

8.2375 and sd | 5.222

(d) t

x Pd sd n

8.2375 0 | 4.462 5.222 8

(e) Because t ! 3.499 , reject H0 . (f) There is enough evidence at the 1% level of significance to support the claim that the contestants’ times have changed. 21. Yes; P | 0.0033 0.005, so you reject H0 . 22. Yes; P | 0.7819 ! 0.20, so you fail to reject H0 .

330

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

23. d

1.525 and sd | 0.542 s s d tD /2 d Pd d tD /2 d n n

§ 0.542 · § 0.542 · 1.525 1.753 ¨ ¸ Pd 1.525 1.753 ¨ ¸ © 16 ¹ © 16 ¹ 1.525 0.238 Pd 1.525 0.238 1.76 Pd 1.29 24. d | 0.436 and sd | 0.677 s s d tD /2 d Pd d tD /2 d n n

§ 0.677 · § 0.677 · 0.436 2.160 ¨ ¸ Pd 0.436 2.160 ¨ ¸ © 14 ¹ © 14 ¹ 0.436 0.391 Pd 0.436 0.391 0.83 Pd 0.05

8.4 TESTING THE DIFFERENCE BETWEEN PROPORTIONS 8.4 TRY IT YOURSELF SOLUTIONS 1.

x1 x2 367 6290 | 0.2111, q 1 p | 0.7889 n1 n2 1593 29948 n1 p 1593(0.2111) | 336.3 ! 5, n1q 1593(0.7889) | 1256.7 ! 5 n2 p 29,948(0.2111) | 6322.0 ! 5, n2 q 29,948(0.7889) | 23,626.0 ! 5 The claim is “there is a difference between the proportion of people who use yoga who are 40- to 49year-olds and the proportion of people who do not use yoga who are 40- to 49-year-olds.” H0 : p1 p2 ; H a : p1 z p2 (claim) D 0.05 z0 1.96, z0 1.96; Rejection regions: z 1.96, z ! 1.96 . ( pˆ1 pˆ 2 ) ( p1 p2 ) (0.230 0.210) (0) z | 1.91 1 · §1 1 · § 1 0.211(0.789) ¨ pq ¨ ¸ ¸ © 1593 29,948 ¹ © n1 n2 ¹ p

********* Because 1.96 z 1.96 , fail to reject H0 . There is not enough evidence at the 5% level of significance to support the claim that there is a difference between the proportion of people who use yoga who are 40- to 49-year-olds and the proportion of people who do not use yoga who are 40- to 49-year-olds.

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

331

x1 x2 415 5691 | 0.1936, q 1 p | 0.8064 n1 n2 1593 29,948 n1 p 1593(0.1936) | 308.4 ! 5, n1q 1593(0.8064) | 1284.6 ! 5 n2 p 29,948(0.1936) | 5797.9 ! 5, n2 q 29,948(0.8064) | 24,150.1 ! 5 The claim is “the proportion of people who use yoga who are from the western region of the United States is greater than or equal to the proportion of people who do not use yoga who are from the western region of the United States.” H0 : p1 t p2 (claim); Ha : p1 p2 D 0.05 z0 1.645; Rejection region: z 1.645 ( pˆ1 pˆ 2 ) ( p1 p2 ) (0.26 0.19) (0) z | 6.89 1 · §1 1 · § 1 0.1936 0.8064 ¨ pq ¨ ¸ ¸ © 1593 29,948 ¹ © n1 n2 ¹ p

Because z ! 1.645 , fail to reject H0 . There is not enough evidence at the 5% level of significance to reject the claim that the proportion of people who use yoga who are from the western region of the United States is greater than or equal to the proportion of people who do not use yoga who are from the western region of the United States.

8.4 EXERCISE SOLUTIONS 1. (1) The samples are randomly selected. (2) The samples are independent. (3) n1 p t 5, n1q t 5, n2 p t 5, n2q t 5 2. State the hypotheses and identify the claim. Specify the level of significance. Find the critical value(s) and rejection region(s). Find p and q . Find the standardized test statistic. Make a decision and interpret it in the context of the claim. The test can also be performed by calculating the P-value and comparing it to α. 3.

x1 x2 35 36 | 0.5462; q 1 p | 0.4538 n1 n2 70 60 n1 p 70(0.5462) | 38.2 ! 5, n1q 70(0.4538) | 31.8 ! 5 n2 p 60(0.5462) | 32.8 ! 5, n2 q 60(0.4538) | 27.2 ! 5 Because all conditions are met above, the normal sampling distribution can be used. H0 : p1 p2 ; H a : p1 z p2 (claim) p

2.575, z0

2.575; Rejection regions: z 2.575, z ! 2.575

332

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

( pˆ1 pˆ 2 ) ( p1 p2 )

(0.5 0.6) (0)

§1 1 · pq ¨ ¸ © n1 n2 ¹

1 · § 1 0.5462 0.4538 ¨ ¸ 70 60 © ¹

| 1.142

Because 2.575 z 2.575 , fail to reject H0 . There is not enough evidence at the 1% level of significance to support the claim. 4.

x1 x2 471 372 | 0.6744; q 1 p | 0.3256 n1 n2 785 465 n1 p | 785(0.6744) | 529.404 ! 5, n1q | 785(0.3256) | 255.596 ! 5 n2 p | 465(0.6744) | 313.596 ! 5, and n 2 q | 465(0.3256) | 151.404 ! 5 p

Because all conditions are met above, the normal sampling distribution can be used.

H0 : p1 t p2 ; Ha : p1 p2 (claim)

1.645; Rejection region: z 1.645

( pˆ1 pˆ 2 ) ( p1 p2 )

(0.600 0.800) (0)

§1 1 · pq ¨ ¸ © n1 n2 ¹

1 · § 1 0.6744 0.3256 ¨ ¸ © 785 465 ¹

| 7.293

Because z 1.645 , reject H0 . There is enough evidence at the 5% level of significance to support the claim. 5.

x1 x2 n1 n2

42 76 | 0.337, q 1 p | 0.663 150 200

n1 p | 150(0.337) | 50.57 ! 5, n1q | 150(0.663) | 99.44 ! 5 . n2 p | 200(0.337) | 67.42 ! 5, and n 2q | 200(0.663) | 132.58 ! 5 Because all conditions are met above, the normal sampling distribution can be used.

H0 : p1 z0 z

p2 (claim); Ha : p1 z p2

1.645, z0 1.645; Rejection regions: z 1.645, z ! 1.645 ( pˆ1 pˆ 2 ) ( p1 p2 )

(0.28 0.38) (0)

§1 1 · pq ¨ ¸ © n1 n2 ¹

1 · § 1 (0.337)(0.663) ¨ ¸ © 150 200 ¹

| 1.96

Because z 1.645 , reject H0 . There is enough evidence at the 10% level of significance to reject the claim. 6.

x1 x2 n1 n2

6 4 20 30

0.2, q 1 p 0.8

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

333

n2 p 30(0.2) 6 ! 5, and n 2q 30(0.8) 24 ! 5 Because n1 p 4 5, cannot use normal sampling distribution. 7. (a) The claim is “there is a difference in the proportion of subjects who had no 12-week confirmed disability progression.”

H0 : p1

p2 ; H a : p1 z p2 (claim)

(b) z0

2.575, z0

x1 x2 n1 n2

2.575; Rejection regions: z 2.575, z ! 2.575

327 148 | 0.649, q 1 p | 0.351 488 244

( pˆ1 pˆ 2 ) ( p1 p2 )

(0.670 0.607) (0)

§1 1 · pq ¨ ¸ © n1 n2 ¹

1 · § 1 (0.649)(0.341) ¨ ¸ © 488 244 ¹

| 1.7

(d) Because 2.575 z 2.575 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to support the claim that there is a difference in the proportion of subjects who had no 12-week confirmed disability progression. 8. (a) The claim is “the proportion of 12-year survivors is greater for subjects who were given chemotherapy than for subjects who were given the placebo.”

H0 : p1 d p2 ; Ha : p1 ! p2 (claim) (b) z0 1.28; Rejection region: z ! 1.28 (c) p

x1 x2 n1 n2

293 268 | 0.738 q 1 p | 0.262 384 376

( pˆ1 pˆ 2 ) ( p1 p2 ) §1 1 · pq ¨ ¸ © n1 n2 ¹

§ 293 268 · ¨ ¸ (0) © 384 376 ¹ | 1.58 1 · § 1 (0.738)(0.262) ¨ ¸ © 384 376 ¹

(d) Because z ! 1.28 , reject H0 . (e) There is enough evidence at the 10% level of significance to support the claim that the proportion of 12-year survivors is greater for subjects who were given chemotherapy than for subjects who were given the placebo. 9. (a) The claim is “there is a difference in the proportion of those employed between the two groups.”

H0 : p1 (b) z0

p2 ; H a : p1 z p2 (claim)

2.575, z0

2.575; Rejection regions: z 2.575, z ! 2.575

334

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

x1 x2 2807 1728 | 0.825, q 1 p | 0.175 n1 n2 3500 2000 ( pˆ1 pˆ 2 ) ( p1 p2 ) (0.802 0.864) (0) §1 1 · pq ¨ ¸ © n1 n2 ¹

1 · § 1 (0.825)(0.175) ¨ ¸ © 3500 2000 ¹

| 5.821

(d) Because z 2.575 , reject H0 . (e) There is enough evidence at the 1% level of significance to support the claim that there is a difference in the proportion of those employed between the two groups. 10. (a) The claim is “the proportion of males ages 20 to 24 who were neither in school nor working is greater than the proportion of females ages 20 to 24 who were neither in school nor working.”

H0 : p1 d p2 ; Ha : p1 ! p2 (claim) (b) z0 1.28; Rejection region: z ! 1.28 (c) p z

x1 x2 133 124 n1 n2 1000 1000 ( pˆ1 pˆ 2 ) ( p1 p2 ) §1 1 · pq ¨ ¸ © n1 n2 ¹

0.1285, q 1 p 0.8715 (0.133 0.124) (0) 1 · § 1 (0.1285)(0.8715) ¨ ¸ © 1000 1000 ¹

| 0.601

(d) Because z 1.28 , fail to reject H0 . (e) There is not enough evidence at the 10% level of significance to support the claim that the proportion of males ages 20 to 24 who were neither in school nor working is greater than the proportion of females ages 20 to 24 who were neither in school nor working. 11. (a) The claim is “the proportion of drivers who wear seat belts is greater in the West than in the Northeast.”

H0 : p1 d p2 ; Ha : p1 ! p2 (claim) (b) z0 1.645; Rejection region: z ! 1.645 (c) p z

x1 x2 934 909 n1 n2 1000 1000 ( pˆ1 pˆ 2 ) ( p1 p2 ) §1 1 · pq ¨ ¸ © n1 n2 ¹

0.9215, q 1 p 0.0785 (0.934 0.909) (0) 1 · § 1 (0.9215)(0.0785) ¨ ¸ 1000 1000 © ¹

| 2.08

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

335

(e) There is enough evidence at the 5% level of significance to support the claim that the proportion of drivers who wear seat belts is greater in the West than in the Northeast. 12. (a) The claim is “the proportion of drivers who wear seat belts in the Midwest is less than the proportion of drivers who wear seat belts in the South.”

H0 : p1 t p2 ; Ha : p1 p2 (claim) (b) z0 (c) p z

1.28; Rejection region: z 1.28 x1 x2 855 909 n1 n2 1000 1000 ( pˆ1 pˆ 2 ) ( p1 p2 ) §1 1 · pq ¨ ¸ © n1 n2 ¹

0.882, q 1 p 0.118 (0.855 0.909) (0) 1 · § 1 (0.882)(0.118) ¨ ¸ © 1000 1000 ¹

| 3.74

(d) Because z 1.28 , reject H0 . (e) There is enough evidence at the 10% level of significance to support the claim that the proportion of drivers who wear seat belts in the Midwest is less than the proportion of drivers who wear seat belts in the South.” 13. The claim is “the proportion of 18- to 24-year-olds who benefit mentally from exercising in parks is the same as the proportion who benefit mentally from socializing in parks.” H0 : p1 p2 (claim); Ha : p1 z p2

z0 p z

1.645, z0 1.645; Rejection regions: z 1.645, z ! 1.645 x1 x2 114 102 0.54, q 1 p 0.46 n1 n2 200 200 ( pˆ1 pˆ 2 ) ( p1 p2 )

(0.57 0.51) (0)

§1 1 · pq ¨ ¸ © n1 n2 ¹

1 · § 1 (0.54)(0.46) ¨ ¸ © 200 200 ¹

| 1.204

Because 1.645 z 1.645 , fail to reject H0 . No, there is not enough evidence at the 10% level of significance to reject the claim that the proportion of 18- to 24-year-olds who benefit mentally from exercising in parks is the same as the proportion who benefit mentally from socializing in parks. 14. The claim is “the proportion of 18- to 24-year-olds who benefit mentally from taking classes less than the proportion who benefit mentally from enjoying nature in parks.”

H0 : p1 t p2 ; Ha : p1 p2 (claim) z0 1.645; Rejection region: z 1.645 x x 78 88 p 1 2 0.415, q 1 p 0.585 n1 n2 200 200

336

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

( pˆ1 pˆ 2 ) ( p1 p2 )

(0.39 0.44) (0)

| 1.015 1 · §1 1 · § 1 (0.415)(0.585) ¨ pq ¨ ¸ 200 200 ¸¹ © n n 2 ¹ © 1 Because z ! 1.645 , fail to reject H0 . No, there is not enough evidence at the 5% level of significance to support the claim that the proportion of 18- to 24-year-olds who benefit mentally from taking classes is less than the proportion who benefit mentally from enjoying nature in parks. z

15. The claim is “the proportion of 18- to 24-year-olds who benefit mentally from socializing in parks is greater than the proportion who benefit mentally from enjoying nature in parks.” H0 : p1 d p2 ; Ha : p1 ! p2 (claim) z0 1.28; Rejection region: z ! 1.28 x x 102 88 p 1 2 0.475, q 1 p 0.525 n1 n2 200 200 ( pˆ1 pˆ 2 ) ( p1 p2 ) (0.51 0.44) (0) | 1.402 z 1 · §1 1 · § 1 (0.475)(0.525) ¨ pq ¨ ¸ ¸ © 200 200 ¹ © n1 n2 ¹ Because z ! 1.28 , reject H0 . Yes, there is enough evidence at the 10% level of significance to support the claim that the proportion of 18- to 24-year-olds who benefit mentally from socializing in parks is greater than the proportion who benefit mentally from enjoying nature in parks. 16. The claim is “the proportion of 18- to 24-year-olds who benefit mentally from socializing in parks is different from the proportion who benefit mentally from taking classes in parks.”

H0 : p1 p2 ; Ha : p1 z p2 (claim) z0 1.96, z0 1.96; Rejection regions: z 1.96, z ! 1.96 x x 102 78 p 1 2 0.45, q 1 p 0.55 n1 n2 200 200 ( pˆ1 pˆ 2 ) ( p1 p2 )

(0.51 0.39) (0)

| 2.412 1 · §1 1 · § 1 (0.45)(0.55) ¨ pq ¨ ¸ ¸ © 200 200 ¹ © n1 n2 ¹ Because z ! 1.96 , reject H0 . Yes, there is enough evidence at the 5% level of significance to support the claim that the proportion of 18- to 24-year-olds who benefit mentally from socializing in parks is different from the proportion who benefit mentally from taking classes in parks. z

17. The claim is “the proportion of 18- to 24-year-olds who benefit mentally from enjoying nature in parks is less than the proportion who benefit mentally from exercising in parks.”

H0 : p1 t p2 ; Ha : p1 p2 (claim) z0 2.33; Rejection region: z 2.33 x x 88 114 p 1 2 0.505, q 1 p 0.495 n1 n2 200 200

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

( pˆ1 pˆ 2 ) ( p1 p2 )

337

(0.44 0.57) (0)

| 2.600 1 · §1 1 · § 1 (0.505)(0.495) ¨ pq ¨ ¸ 200 200 ¸¹ © n n 2 ¹ © 1 Because z 2.33 , reject H0 . Yes, there is enough evidence at the 1% level of significance to support the claim that the proportion of 18- to 24-year-olds who benefit mentally from enjoying nature in parks is less than the proportion who benefit mentally from exercising in parks. z

18. The claim is “the proportion of 18- to 24-year-olds who benefit mentally from exercising in parks is greater than or equal to the proportion who benefit mentally from taking classes in parks.” H0 : p1 t p2 (claim); H a : p1 p2 z0 2.33; Rejection region: z ! 2.33 x x 114 78 p 1 2 0.48, q 1 p 0.52 n1 n2 200 200 ( pˆ1 pˆ 2 ) ( p1 p2 ) (0.57 0.39) (0) | 3.603 z 1 · §1 1 · § 1 (0.48)(0.52) ¨ pq ¨ ¸ ¸ © 200 200 ¹ © n1 n2 ¹ Because z ! 2.33 , reject H0 . Yes, there is enough evidence at the 1% level of significance to reject the claim that the proportion of 18- to 24-year-olds who benefit mentally from exercising in parks is greater than or equal to the proportion who benefit mentally from taking classes in parks. 19. The claim is “the proportion of men who work is greater than the proportion of men who socialize and communicate on an average day.”

H0 : p1 d p2 ; Ha : p1 ! p2 (claim) z0 2.33 Rejection region: z ! 2.33 x x 98 66 p 1 2 0.41, q 1 p 0.59 n1 n2 200 200 ( pˆ1 pˆ 2 ) ( p1 p2 )

(0.49 0.33) (0)

| 3.253 1 · §1 1 · § 1 (0.41)(0.59) ¨ pq ¨ ¸ ¸ © 200 200 ¹ © n1 n2 ¹ Because z ! 2.33 , reject H0 . Yes, there is enough evidence at the 1% level of significance to support the claim that the proportion of men who work is greater than the proportion of men who socialize and communicate on an average day. z

20. The claim is “the proportion of women who work is the same as the proportion of women who socialize and communicate on an average day.”

H0 : p1 p2 (claim); H a : p1 z p2 z0 2.575, z0 2.575 Rejection regions: z 2.575, z ! 2.575 x x 111 111 p 1 2 0.37, q 1 p 0.63 n1 n2 300 300

338

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

( pˆ1 pˆ 2 ) ( p1 p2 )

(0.37 0.37) (0)

0 1 · §1 1 · § 1 (0.37)(0.63) ¨ pq ¨ ¸ 300 300 ¸¹ © n n 2 ¹ © 1 Because 2.575 z 2.575 , fail to reject H0 . No, there is not enough evidence at the 1% level of significance to reject the claim that the proportion of women who work is the same as the proportion of women who socialize and communicate on an average day.

21. The claim is “the proportion of women who work is less than the proportion of men who work on an average day.” H0 : p1 t p2 ; Ha : p1 p2 (claim) z0 2.33; Rejection region: z 2.33 x x 98 111 p 1 2 | 0.418, q 1 p | 0.582 n1 n2 200 300 ( pˆ1 pˆ 2 ) ( p1 p2 ) (0.37 0.49) (0) | 2.665 z 1 · §1 1 · § 1 (0.418)(0.582) ¨ pq ¨ ¸ ¸ © 200 300 ¹ © n1 n2 ¹ Because z 2.33 , reject H0 . Yes, there is enough evidence at the 1% level of significance to support the claim that the proportion of women who work is less than the proportion of men who work on an average day. 22. The claim is “the proportion of women who socialize and communicate is greater than the proportion of men who socialize and communicate on an average day.” H0 : p1 d p2 ; Ha : p1 ! p2 (claim)

z0 1.28; Rejection region: z ! 1.28 x x 111 66 p 1 2 0.354, q 1 p 0.646 n1 n2 300 200 ( pˆ1 pˆ 2 ) ( p1 p2 )

(0.37 0.33) (0)

| 0.916 1 · §1 1 · § 1 (0.354)(0.646) ¨ pq ¨ ¸ ¸ © 300 200 ¹ © n1 n2 ¹ Because z 1.28 , fail to reject H0 . No, there is not enough evidence at the 10% level of significance to support the claim that the proportion of women who socialize and communicate is greater than the proportion of men who socialize and communicate on an average day. z

23.

( pˆ1 pˆ 2 ) zc (0.07 0.08) 1.96

pˆ1qˆ1 pˆ 2 qˆ2 p1 p2 ( pˆ1 pˆ 2 ) zc n1 n2

pˆ1qˆ1 pˆ 2 qˆ2 n1 n2

(0.07)(0.93) (0.08)(0.92) (0.07)(0.93) (0.08)(0.92) p1 p2 (0.07 0.08) 1.96 10, 000 8000 10, 000 8000 0.01 0.00777 p1 p2 0.01 0.00777 0.0178 p1 p2 0.0022

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

24.

pˆ1qˆ1 pˆ 2 qˆ2 p1 p2 ( pˆ1 pˆ 2 ) zc n1 n2

( pˆ1 pˆ 2 ) zc (0.09 0.07) 1.645

339

pˆ1qˆ1 pˆ 2 qˆ2 n1 n2

(0.09)(0.91) (0.07)(0.93) (0.09)(0.91) (0.07)(0.93) p1 p2 (0.09 0.07) 1.645 10, 000 8000 10, 000 8000 0.02 0.006647 p1 p2 0.02 0.006647 0.0134 p1 p2 0.0266

25.

( pˆ1 pˆ 2 ) zc

pˆ1qˆ1 pˆ 2 qˆ2 p1 p2 ( pˆ1 pˆ 2 ) zc n1 n2

pˆ1qˆ1 pˆ 2 qˆ2 n1 n2

(0.82)(0.18) (0.72)(0.28) (0.82)(0.18) (0.72)(0.28) p1 p2 (0.82 0.72) 1.96 1021 1021 1021 1021 0.10 0.036248 p1 p2 0.10 0.036248

(0.82 0.72) 1.96

0.0638 p1 p2 0.1362

Answers will vary. 26.

( pˆ1 pˆ 2 ) zc (0.82 0.72) 2.575

pˆ1qˆ1 pˆ 2 qˆ2 p1 p2 ( pˆ1 pˆ 2 ) zc n1 n2

pˆ1qˆ1 pˆ 2 qˆ2 n1 n2

(0.82)(0.18) (0.72)(0.28) (0.82)(0.18) (0.72)(0.28) p1 p2 (0.82 0.72) 2.575 1021 1021 1021 1021 0.1 0.047621 p1 p2 0.1 0.047621 0.0524 p1 p2 0.1476

Answers will vary.

CHAPTER 8 REVIEW EXERCISE SOLUTIONS 1. Dependent, because the same children were sampled. 2. Independent, because different fruits were sampled. 3. Independent, because different vehicles were sampled. 4. Dependent because the same cars were sampled. 5.

H0 : P1 t P2 (claim); Ha : P1 P2 z0

1.645; Rejection region: z 1.645

( x1 x2 ) ( P1 P2 )

2 1

2 2

(1.28 1.34) (0) (0.30)2 (0.23) 2 96 85

| 1.519

Because z ! 1.645 , fail to reject H0 . There is not enough evidence at the 5% level of significance to reject the claim.

340

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

H0 : P1 P2 (claim); Ha : P1 z P2

z0 z

2.575, z0

2.575; Rejection regions: z 2.575, z ! 2.575

( x1 x2 ) ( P1 P2 )

2 1

(5595 5575) (0)

2 2

(52) 2 (68) 2 156 216

| 3.213

Because z ! 2.575 , reject H0 . There is enough evidence at the 1% level of significance to reject the claim. 7.

H0 : P1 t P2 ; Ha : P1 P2 (claim)

1.28; Rejection regions: z 1.28

( x1 x2 ) ( P1 P2 )

2 1

(0.28 0.33) (0)

2 2

(0.11)2 (0.10) 2 41 34

| 2.060

Because z 1.28 , reject H0 . There is enough evidence at the 10% level of significance to support the claim. 8.

H0 : P1 P2 ; Ha : P1 z P2 (claim)

z0 z

1.96, z0 1.96; Rejection regions: z 1.96, z ! 1.96 ( x1 x2 ) ( P1 P2 )

V 12 n1

(87 85) (0)

V 22

(14)2 (15)2 410 340

| 1.87

Because 1.96 z 1.96 , fail to reject H0 . There is not enough evidence at the 5% level of significance to support the claim. 9. (a) The claim is “the mean sodium content of sandwiches at Restaurant A is less than the mean sodium content of sandwiches at Restaurant B.”

H0 : P1 t P2 ; Ha : P1 P2 (claim)

(b) z0

1.645; Rejection region: z 1.645

( x1 x2 ) ( P1 P2 )

2 1

2 2

(670 690) (0) (20) 2 (30) 2 22 28

(d) Because z 1.645 , reject H0 .

| 2.82

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

341

(e) There is enough evidence at the 5% level of significance to support the claim that the mean sodium content of sandwiches at Restaurant A is less than the mean sodium content of sandwiches at Restaurant B. 10. (a) The claim is “the mean annual salaries of entry-level paralegals in Dayton, Ohio, and Coventry, Rhode Island, are the same.”

H0 : P1 P2 (claim); Ha : P1 z P2

(b) z0

1.645, z0 1.645; Rejection regions: z 1.645, z ! 1.645 ( x1 x2 ) ( P1 P2 )

2 1

(58,180 61,120) (0)

2 2

(10,990) 2 (11,850) 2 40 35

| 1.109

(d) Because 1.645 z 1.645 , fail to reject H0 . (e) There is not enough evidence at the 10% level of significance to reject the claim that the mean annual salary of entry-level paralegals in Dayton, Ohio, and Coventry, Rhode Island, are the same. 11. H0 : P1

P2 (claim); Ha : P1 z P2

d.f. n1 + n2 2 31 t0 t

2.040, t0

2.040; Rejection regions: t 2.040, t ! 2.040

( x1 x2 ) ( P1 P2 ) (n1 1) s12 (n2 1) s22

1 1 n1 n2

n1 n2 2

(228 207) (0) (20 1)(27) 2 (13 1)(25) 2 20 13 2

1 1 20 13

| 2.25

Because t ! 2.040 , reject H0 . There is enough evidence at the 5% level of significance to reject the claim. 12. H0 : P1 t P2 ; Ha : P1 P2 (claim)

d.f. min{n1 1, n2 1} 5

1.476; Rejection region: t 1.476

( x1 x2 ) ( P1 P2 ) s12 n1

s22 n2

(0.015 0.019) (0) (0.011) 2 (0.004) 2 8 6

| 0.948

Because t ! 1.476 , fail to reject H0 . There is not enough evidence at the 10% level of significance to support the claim.

342

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

13. H0 : P1 d P2 (claim); Ha : P1 ! P2

d.f. min{n1 1, n2 1} 39 t0 1.304; Rejection region: t ! 1.304 t

( x1 x2 ) ( P1 P2 )

(664.5 665.5) (0)

s12 s22 n1 n2

(2.4) 2 (4.1) 2 40 40

| 1.33

Because t 1.304 , fail to reject H0 . There is not enough evidence at the 10% level of significance to reject the claim. 14. H0 : P1 t P2 (claim); Ha : P1 P2

d.f. n1 n2 2 33

t0 t

2.445; Rejection region: t 2.445 ( x1 x2 ) ( P1 P2 ) (n1 1) s12 (n2 1) s22

1 1 n1 n2

n1 n2 2

(44.5 49.1) (0) (17 1)(5.85) 2 (18 1)(5.25) 2 1 1 17 18 2 17 18

| 2.45

Because t 2.445 , reject H0 . There is enough evidence at the 1% level of significance to reject the claim. 15. H0 : P1

P2 ; Ha : P1 z P2 (claim)

d.f. n1 n2 2 10 t0 t

3.169, t0 3.169; Rejection regions: t 3.169, t ! 3.169 ( x1 x2 ) ( P1 P2 )

(61 55) (0)

(n1 1) s12 (n2 1) s22 1 1 n1 n2 2 n1 n2

(5 1)(3.3) 2 (7 1)(1.2) 2 1 1 5 7 2 5 7

| 4.484

Because t ! 3.169 , reject H0 . There is enough evidence at the 1% level of significance to support the claim. 16. H0 : P1 d P2 ; Ha : P1 ! P2 (claim)

d.f. min{n1 1, n2 1} 5

t0 1.476; Rejection region: t ! 1.476 t

( x1 x2 ) ( P1 P2 ) s12 n1

(520 500) (0)

s22 n2

(25)2 (55)2 7 6

| 0.821

343

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

Because t 1.476 , fail to reject H0 . There is not enough evidence at the 10% level of significance to support the claim. 17. (a) The claim is “the new method of teaching mathematics produces higher mathematics test scores than the old method does.” Note: P1 represents the new curriculum and P2 represents the old curriculum.

H0 : P1 d P2 ; Ha : P1 ! P2 (claim)

(b) d.f. n1 n2 2 70

t0 1.667; Rejection region: t ! 1.667 (c) x1 | 62.8056, s1 | 22.8446, n1 36

x2 | 48.6944, s2 | 28.5915, n2 36 ( x1 x2 ) ( P1 P2 ) t | (n1 1) s12 (n2 1) s22 1 1 n1 n2 2 n1 n2

(62.8056 48.6944) (0) (36 1)(22.8446) 2 (36 1)(28.5915) 2 36 36 2

1 1 36 36

| 2.313

(d) Because t ! 1.667 , reject H0 . (e) There is enough evidence at the 5% level of significance to support the claim the new method of teaching mathematics produces higher mathematics test scores than the old method does. 18. (a) The claim is “there is no difference between the mean household incomes of two neighborhoods.”

H0 : P1 P2 (claim); Ha : P1 z P2

(b) d.f. n1 n2 2 20

t0 (c) t

2.845, t0

2.845; Rejection regions: t 2.845, t ! 2.845

( x1 x2 ) ( P1 P2 ) (n1 1) s12 (n2 1) s22 n1 n2 2

1 1 n1 n2

(52,750 51, 200) (0) (12 1)(2900) 2 (10 1)(2225) 2 1 1 12 10 2 12 10

| 1.383

(d) Because 2.845 t 2.845 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to reject the agent’s claim that there is no difference between the mean household incomes of the two neighborhoods. 19. H0 : Pd 0 (claim); H a : Pd z 0 D 0.01 and d.f. n 1 15 t0 2.947, t0 2.947; Rejection regions: t 2.947, t ! 2.947 t

d Pd sd n

8.5 0 | 3.178 10.7 16

344

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

Because t ! 2.947 , reject H0 . There is enough evidence at the 1% level of significance to reject the claim. 20. H0 : Pd t 0; Ha : Pd 0 (claim) D 0.10 and d.f. n 1 24 t0 1.318 ; Rejection region: t 1.318 d Pd 3.2 0 | 2.817 5.68 sd 25 n Because t ! 1.318 , fail to reject H0 . There is not enough evidence at the 10% level of significance to support the claim. t

21. H0 : Pd d 0 (claim); Ha : Pd ! 0 D 0.10 and d.f. n 1 33 t0 1.308 ; Rejection region: t ! 1.308 d Pd 10.3 0 | 3.253 18.19 sd 33 n Because t ! 1.308 , reject H0 . There is enough evidence at the 10% level of significance to reject the claim. t

22. H0 : Pd 0; Ha : Pd z 0 (claim) D 0.05 and d.f. n 1 36 t0 2.028, t0 2.028; Rejection regions: t 2.028, t ! 2.028 d Pd 17.5 0 | 26.284 4.05 sd 37 n Because t ! 2.028 , reject H0 . There is enough evidence at the 5% level of significance to support the claim. t

23. (a) The claim is “the numbers of runs scored in a season by NCAA Division 1 baseball teams changed from 2019 to 2021.”

H0 : Pd

0; Ha : Pd z 0 (claim)

2.262; ; Rejection regions: t 2.262, t ! 2.262

(b) t0

2.262, t0

38.1 and sd | 66.10

(d) t

d Pd sd n

38.1 0 | 1.823 66.10 10

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

345

(f) There is not enough evidence at the 5% level of significance to support the claim that the numbers of runs scored in a season by NCAA Division 1 baseball teams changed from 2019 to 2021. 24. (a) The claim is “the weight loss supplement will help users lose weight after two weeks.”

H0 : Pd d 0; Ha : Pd ! 0 (claim)

(b) t0 1.397 ; Rejection region: t ! 1.397 (c) d | 1.444 and sd | 2.744 d Pd sd n

(d) t

1.444 0 | 1.579 2.744 9

(e) Because t ! 1.397 , reject H0 . (f) There is enough evidence at the 10% level of significance to support the physical fitness instructor’s claim that the weight loss supplement will help users lose weight after two weeks. 25. p

x1 x2 n1 n2

425 410 | 0.522, q 1 p | 0.478 840 760

n1 p | 840(0.522) | 438.48 ! 5, n1q | 840(0.478) | 401.52 ! 5, n2 p | 760(0.522) | 396.72 ! 5, and n 2q | 760(0.478) | 363.28 ! 5 . Can use normal sampling distribution.

H0 : p1 z0 z

p2 (claim); Ha : p1 z p2

1.96, z0 1.96; Rejection regions: z 1.96, z ! 1.96 ( pˆ1 pˆ 2 ) ( p1 p2 )

(0.506 0.539) (0)

§1 1 · pq ¨ ¸ © n1 n2 ¹

1 · § 1 0.522(0.478) ¨ ¸ © 840 760 ¹

| 1.320

Because 1.96 z 1.96 , fail to reject H0 . There is not enough evidence at the 5% level of significance to reject the claim. 26. p

x1 x2 n1 n2

36 46 | 0.273, q 1 p | 0.727 100 200

n1 p | 100(0.273) | 27.3 ! 5, n1q | 100(0.727) | 72.7 ! 5, n2 p | 200(0.273) | 54.6 ! 5, and n 2 q | 200(0.727) | 145.4 ! 5 . Can use normal sampling distribution. H0 : p1 d p2 (claim) : H a : p1 ! p2 Copyright © 2023 Pearson Education, Inc.

346

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

2.33; Rejection region: z ! 2.33

( pˆ1 pˆ 2 ) ( p1 p2 )

(0.36 0.23) (0)

§1 1 · pq ¨ ¸ © n1 n2 ¹

1 · § 1 0.273(0.727) ¨ ¸ © 100 200 ¹

| 2.383

Because z ! 2.33 , reject H0 . There is enough evidence at the 1% level of significance to reject the claim. 27. p

x1 x2 n1 n2

261 207 | 0.450, q 1 p | 0.550 556 483

n1 p | 556(0.450) | 250.2 ! 5, n1q | 556(0.550) | 305.8 ! 5, n2 p | 483(0.450) | 217.35 ! 5, and n 2q | 483(0.550) | 265.65 ! 5 . Can use normal sampling distribution.

H0 : p1 d p2 ; Ha : p1 ! p2 (claim) z0 1.28; Rejection region: z ! 1.28 z

( pˆ1 pˆ 2 ) ( p1 p2 )

(0.469 0.429) (0)

§1 1 · pq ¨ ¸ © n1 n2 ¹

1 · § 1 0.450(0.550) ¨ ¸ © 556 483 ¹

| 1.293

Because z ! 1.28 , reject H0 . There is enough evidence at the 10% level of significance to support the claim. 28. p

x1 x2 n1 n2

86 107 | 0.092, q 1 p | 0.908 900 1200

n1 p | 900(0.092) | 82.8 ! 5, n1q | 900(0.908) | 817.2 ! 5, n2 p | 1200(0.092) | 110.4 ! 5, and n 2 q | 1200(0.908) | 1089.6 ! 5 . Can use normal sampling distribution.

H0 : p1 t p2 ; Ha : p1 p2 (claim) z0

1.645; Rejection region: z 1.645

( pˆ1 pˆ 2 ) ( p1 p2 )

(0.096 0.089) (0)

§1 1 · pq ¨ ¸ © n1 n2 ¹

1 · § 1 0.092(0.908) ¨ ¸ 900 1200 © ¹

| 0.549

Because z ! 1.645 , fail to reject H0 . There is not enough evidence at the 5% level of significance to support the claim. 29. (a) The claim is “the proportion of subjects who had at least 24 weeks of accrued remission is the same for the two groups.”

H0 : p1

p2 (claim); Ha : p1 z p2

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

(b) z0

347

1.96, z0 1.96; Rejection regions: z 1.96, z ! 1.96 x1 x2 19 2 | 0.1544, q 1 p | 0.8456 n1 n2 68 68 ( pˆ1 pˆ 2 ) ( p1 p2 ) (0.2794 0.0294) (0) §1 1 · pq ¨ ¸ © n1 n2 ¹

1 · § 1 0.1544(0.8456) ¨ ¸ 68 68 © ¹

| 4.03

(d) Because z ! 1.96 , reject H0 . (e) There is enough evidence at the 5% level of significance to reject the medical research team’s claim that the proportion of subjects who had at least 24 weeks of accrued remission is the same for the two groups. 30. (a) The claim is “the proportion of motorcyclists, in states where helmet use is required, who use helmets is decreased from 2019 to 2020.”

H0 : p1 d p2 ; Ha : p1 ! p2 (claim) (b) z0 2.33 Rejection region: z ! 2.33 x1 x2 346.988 283.92 (c) p | 0.86782, q 1 p | 0.13218 n1 n2 389 338 z

( pˆ1 pˆ 2 ) ( p1 p2 )

(0.892 0.840) (0)

§1 1 · pq ¨ ¸ © n1 n2 ¹

1 · § 1 0.86782(0.13218) ¨ ¸ © 389 338 ¹

| 2.065

(d) Because z 2.33 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to support the claim that the proportion of motorcyclists, in states where helmet use is required, who use helmets is decreased from 2019 to 2020.

CHAPTER 8 QUIZ SOLUTIONS 1. (a) The claim is “the mean score on the reading assessment test for male high school students is less than the mean score for female high school students.”

H0 : P1 t P2 ; Ha : P1 P2 (claim)

(b) Left-tailed, because H a contains <; z-test, because V1 and V 2 are known, the samples are random samples, the samples are independent, and n1 t 30 and n2 t 30 . (c) z0

1.645; Rejection region: z 1.645

348

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

(d) z

( x1 x2 ) ( P1 P2 )

V 12 n1

V 22 n2

(279 292) (0) (41) 2 (39) 2 49 50

| 1.616

(e) Because z ! 1.645 , fail to reject H0 . (f) There is not enough evidence at the 5% level of significance to support the claim that the mean score on the reading assessment test for male high school students is less than for the female high school students. 2. (a) The claim is “the mean scores on a music assessment test for eighth grade students in public and private schools are equal.”

H0 : P1 P2 (claim); Ha : P1 z P2

(b) Two-tailed, because H a contains z; t-test, because V1 and V 2 are unknown, the samples are random samples, the samples are independent, and the populations are normally distributed. (c) d.f. n1 n2 2 26

t0 (d) t

1.706, t0 1.706; Rejection regions: t 1.706, t ! 1.706 ( x1 x2 ) ( P1 P2 ) (n1 1) s12 (n2 1) s22 n1 n2 2

1 1 n1 n2

(146 160) (0) (13 1)(49) 2 (15 1)(42) 2 1 1 13 15 2 13 15

| 0.814

(e) Because 1.706 t 1.706 , fail to reject H0 . (f) There is not enough evidence at the 10% level of significance to reject the claim that the mean scores on a music assessment test for eighth grade students in public and private schools are equal. 3. (a) The claim is “the seminar helps adults increase their credit scores.”

H0 : Pd t 0; Ha : Pd 0 (claim)

(b) Left-tailed because H a contains ; t-test because both populations are normally distributed and the samples are dependent. (c) d.f. n 1 11 t0 2.718; Rejection region: t 2.718 (d) t

d Pd sd n

51.167 0 | 5.073 34.938 12

(e) Because t 2.718 , reject H0 .

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

349

(f) There is enough evidence at the 1% level of significance to support the claim that the seminar helps adults increase their credit scores. 4. (a) The claim is “the proportion of the U.S. adults who approve of the job the Supreme Court is doing is greater than it was 3 years prior.”

H0 : p1 d p2 ; Ha : p1 ! p2 (claim) (b) Right-tailed, because H a contains >; z-test, because you are testing proportions, the samples are random samples, the samples are independent, and the quantities n1 p, n1q , n2 p, and n2 q are at least 5. (c) z0 1.645; Rejection region: z ! 1.645 (d) p z

x1 x2 584 501 | 0.534746, q 1 p | 0.465254 n1 n2 1007 1022 ( pˆ1 pˆ 2 ) ( p1 p2 ) (0.5799404171 0.4902152642) (0) §1 1 · pq ¨ ¸ © n1 n2 ¹

1 · § 1 0.534746(0.465254) ¨ ¸ © 1007 1022 ¹

| 4.051

(e) Because z ! 1.645 , reject H0 . (f) There is enough evidence at the 5% level of significance to support the claim that the proportion of the U.S. adults who approve of the job the Supreme Court is doing is greater than it was 3 years prior.

CUMULATIVE REVIEW, CHAPTERS 6 8 1. (a) pˆ 0.77, qˆ 0.23

pˆ r zc

ˆˆ pq n

0.77 r 1.96

(0.77)(0.23) | 0.77 r 0.0118 | (0.758, 0.782) 4860

(b) H0 : p d 0.75; Ha : p ! 0.75 (claim) z0 1.645; Rejection region: z ! 1.645 pˆ p 0.77 0.75 z | 3.220 pq (0.75)(0.25) n 4860 Because z ! 1.645 , reject H0 . There is enough evidence at the 5% level of significance to support the claim that more than 75% of the U.S. adults say they would date or have already dated someone whose religion was different from theirs.

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

H0 : Pd t 0; Ha : Pd 0 (claim) D 0.10 d.f. n 1 7 t0 1.415; Rejection region: t 1.415

1.575 and sd | 0.803

d Pd 1.575 0 | | 5.548 sd 0.803 8 n Because t 1.415 , reject H0 . t

There is enough evidence at the 10% level of significance to support the claim that the fuel additive improved gas mileage.

x r zc

x r tc

s n

3.46 r 2.131

1.63 | 3.46 r 0.87 16

x r tc

s n

12.1 r 2.787

2.64 | 12.1 r 1.4 (10.7, 13.5); t-distribution 26

x r zc

8.21 r 1.645

0.62 | 8.21 r 0.36 (7.85, 8.57); z-distribution 8

H0 : P t 33 H a : P 33 (claim)

H0 : p t 0.19 (claim) H a : p 0.19

V n

26.97 r 1.96

3.4 | 26.97 r 1.03 (25.94, 28.00); z-distribution 42

(2.59, 4.33); t-distribution

H0 : V 0.63 (claim) H a : V z 0.63

10. H0 : P

2.28 Ha : P z 2.28 (claim)

11. H0 : P1 d P2 ; Ha : P1 ! P2 (claim) z0 1.28; Rejection region: z ! 1.28 ( x1 x2 ) 0 (3086 2263) (0) z | 8.464 V 12 V 22 (563)2 (624) 2 85 68 n1 n2 Because z ! 1.28 , reject H0 . There is enough evidence at the 10% level of significance to support the pediatrician’s claim that the mean birth weight of a single-birth baby is greater than the mean birth weight of a baby that has a twin.

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

351

§ (n 1) s 2 (n 1) s 2 · § (26 1)(31) 2 (26 1)(31) 2 · , , 12. (a) ¨ ¸ ¨ ¸ | (511.95, 2283.75) 2 10.520 ¹ F L2 ¹ © 46.928 © FR

(b)

511.95,

2283.74 | (22.63, 47.79)

F02 F2

44.314; Rejection region: F 2 ! 44.314

n 1 s2

(25)(31) 2 | 26.7 (30) 2

Because F 2 44.314 , fail to reject H0 . There is not enough evidence at the 1% level of significance to reject the travel analyst’s claim that the standard deviation of the mean room rate for two adults at three-star hotels in Cincinnati is at most $30. 13. H0 : P1

P2 ; Ha : P1 z P2 (claim)

d.f. n1 n2 2 26 18 2 42 t0

2.018 t0

2.018; Rejection regions: t 2.018, t ! 2.018

( x1 x2 ) (0)

(n1 1) s12 (n2 1) s22 n1 n2 2

1189 1376 1 1 n1 n2

(26 1)(218) (18 1)(186) 2 26 18 2

1 1 26 18

| 2.97

Because t 2.018 , reject H0 . There is enough evidence at the 5% level of significance to support the organization’s claim that the mean SAT scores for male athletes and male non-athletes at a college are different. 14. (a) x | 58,168.03 , s | 9429.707 s 9429.707 x r tc | 58,168.03 r 2.045 n 30 | (54,647.31, 61,688.75) (b) H0 : P 55,000 (claim); Ha : P z 55,000

t0 t

2.045, t0 x P s n

2.045; Rejection regions: t 2.045, t ! 2.045

58,168.03 55,000 | 1.840 9429.71 30

Because 2.045 t 2.045 , fail to reject H0 . There is not enough evidence at the 5% level of significance to reject the claim that the mean annual earnings for locksmiths is $55,000.

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

15. H0 : p 1 d p2 ; H a : p1 ! p2 (claim)

x1 x2 n1 n2

6 1 | 0.065, q 1 p | 0.935 52 56

z0 1.28; Rejection regions: z ! 1.28; z0

( pˆ1 pˆ 2 ) 0

(0.1154 0.0179)

§1 1 · pq ¨ ¸ © n1 n2 ¹

1 · § 1 (0.065)(0.935) ¨ ¸ 52 56 © ¹

| 2.05

Because z ! 1.28 , reject H0 . There is enough evidence at the 10% level of significance to support the medical research team’s claim that the proportion of monthly convulsive seizure reduction is greater for the group that received the extract than for the group that received the placebo. 16. (a) x r z

V n

42 r 1.96

1.6 | 42 r 0.496 | (41.5, 42.5) 40

(b) H0 : P t 45 (claim); H a : P 45

z0 z

1.645; Rejection region: z 1.645; x P s n

42 45 | 11.86 1.6 40

Because z 1.645; , reject H0 . There is enough evidence at the 5% level of significance to reject the zoologist’s claim that the mean incubation period for ostriches is at least 45 days. 17. A type I error will occur when the actual proportion of people who purchase their eyeglasses online is 0.05, but you reject H0 . A type II error will occur when the actual proportion of people who purchase their eyeglasses online is different from 0.05, but you fail to reject H0 .

CHAPTER 8 TEST SOLUTIONS 1. (a) The claim is “the proportion of students taking the SAT who are undecided on an intended college major has not changed.” H0 : p1 p2 (claim); Ha : p1 z p2 (b) Two-tailed because H a contains z ; z-test because you are testing proportions, the samples are random samples, the samples are independent and the quantities n1 p, n1q , n2 p, and n2 q are at least 5. (c) z0

1.645, z0 1.645; Rejection regions: z 1.645, z ! 1.645

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

(d) p z

x1 x2 n1 n2

353

350 360 | 0.04176, q 1 p | 0.95824 5000 12,000

( pˆ1 pˆ 2 ) ( p1 p2 )

(0.07 0.03) (0)

§1 1 · pq ¨ ¸ © n1 n2 ¹

1 · § 1 0.04176(0.95824) ¨ ¸ © 5000 12,000 ¹

| 11.88

(e) Because z ! 1.645 , reject H0 . (f) There is enough evidence at the 10% level of significance to reject the claim that the proportion of students taking the SAT who are undecided on an intended college major has not changed. 2. (a) The claim is “the mean home sales price in Olathe, Kansas, is greater than in Rolla, Missouri.” H0 : P1 d P2 ; Ha : P1 ! P2 (claim) (b) Right-tailed, because H a contains >; z-test, because V1 and V 2 are known, the samples are random samples, the samples are independent, and n1 t 30 and n2 t 30 . (c) z0 1.645; Rejection region: z ! 1.645 (d) z

( x1 x2 ) ( P1 P2 )

V 12 n1

V 22 n2

(392, 453 285,787) (0) (224,902)2 (330,578)2 39 38

| 1.651

(e) Because z ! 1.645 , reject H0 . (f) There is enough evidence at the 5% level of significance to support the claim that the mean home sales price in Olathe, Kansas, is greater than in Rolla, Missouri. 3. (a) The claim is “soft tissue massage therapy helps to reduce the lengths of time patients suffer from headaches.”

H0 : Pd d 0; Ha : Pd ! 0 (claim)

(b) Right-tailed because H a contains !; t-test because both populations are normally distributed and the samples are dependent. (c) d.f. n 1 17 t0 1.740; Rejection region: t ! 1.740 (d) t

d Pd sd n

1.233 0 | 4.33 1.207 18

(e) Because t ! 1.740 , reject H0 .

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

(f) There is enough evidence at the 5% level of significance to support the physical therapist’s claim that soft tissue massage therapy helps to reduce the lengths of time patients suffer from headaches. 4. (a) The claim is “the mean household income in a recent year is different for native-born households and foreign-born households.”

H0 : P1 P2 ; Ha : P1 z P2 (claim)

t0 (d) t

2.898, t0

2.898; Rejection regions: t 2.898, t ! 2.898

( x1 x2 ) ( P1 P2 ) s12 n1

s22 n2

(69, 474 64,900) (0) (21, 249)2 (17,896) 2 18 21

| 0.720

(e) Because 2.898 t 2.898 , fail to reject H0 . (f) There is not enough evidence at the 1% level of significance to support the claim that the mean household income in a recent year is different for native-born households and foreign-born households.

CHAPTER

Correlation and Regression

9.1 CORRELATION 9.1 TRY IT YOURSELF SOLUTIONS 1.

It appears that there is a negative linear correlation. As the number of years out of school increases, the annual contribution tends to decreases. 2.

It appears that there is no linear correlation between height and pulse rate. 3.

It appears that there is a positive linear correlation. As the team salary increases, the average home game attendances tends to increase.

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CHAPTER 9 Ň CORRELATION AND REGRESSION

n=7

¦ r= = =

x 1 10 5 15 3 24 30 x = 88

y 12.5 8.7 14.6 5.2 9.9 3.1 2.7 y = 56.7

xy 12.5 87.0 73.0 78.0 29.7 74.4 81.0 xy = 435.6

x2 1 100 25 225 9 576 900 2 x = 1836

y2 156.25 75.69 213.16 27.04 98.01 9.61 7.29 2 y = 587.05

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

n ¦ y2 − (¦ y )

( 7 )( 435.6 ) − (88 )( 56.7 ) ( 7 )(1836 ) − (88)2 ( 7 )( 587.05 ) − ( 56.7 )2 −1940.4 5108 894.46

≈ −0.908

Because r is close to −1 , this suggests a strong negative linear correlation. As the number of years out of school increases, the annual contribution tends to decrease. 5. 0.792; Because r is close to 1, this suggest a strong positive linear correlation. As the team salaries increase, the average home game attendances tends to increase. 6.

n = 7 , α = 0.01 , 0.875, r ≈ −0.908 > 0.875; The correlation is significant. There is enough evidence at the 1% level of significance to conclude that there is a significant linear correlation between the number of years out of school and the annual contribution.

7. There is enough evidence at the 1% level of significance to conclude that there is a significant linear correlation between the salaries and average home game attendances for the teams in Major League Baseball.

9.1 EXERCISE SOLUTIONS 1. Increase; Decrease 2. The range of values for the correlation coefficient is −1 to 1, inclusive. 3. The sample correlation coefficient r measures the strength and direction of a linear relationship between two variables; r = −0.932 indicates a stronger correlation because −0.932 = 0.932 is closer

to 1 than 0.918 = 0.918 .

CHAPTER 9 Ň CORRELATION AND REGRESSION

357

4. Answers will vary. Sample answer: Perfect positive linear correlation: price per gallon of gasoline and total cost of gasoline Perfect negative linear correlation: distance from door and height of wheelchair ramp 5. A table can be used to compare r with a critical value, or a hypothesis test can be performed using a t-test. 6. r is the sample correlation coefficient, while ρ is the population correlation coefficient. 7. Since the t-test is non-directional, the null hypothesis, or hypothesis of no difference, will include =, instead of > or <, and may be interpreted as saying there is no significant linear correlation between the variables. The alternate hypothesis is the opposite of the null hypothesis and states that there is a significant linear correlation between the variables. The two hypotheses are written as: H 0 : ρ = 0 (no significant linear correlation)

H a : ρ ≠ 0 (significant linear correlation) The null hypothesis is rejected when the calculated t is greater than the critical t-value. In this case, t is in the rejection area of the standard normal curve. 8. If two variables show a correlation, a researcher cannot interpret this result as one variable caused the other. For example, there may be a linear correlation between final exam grades in U.S. History and final exam grades in Algebra I, but it does not mean that final exam grades in one subject caused the final exam grades in the other subject. 9. Explanatory variable: Amount of water consumed Response variable: Weight loss 10. Explanatory variable: Hours of safety classes Response variable: Number of driving accidents 11. Strong negative linear correlation 12. No linear correlation 13. No linear correlation 14. Perfect positive linear correlation 15. (c), You would expect a positive linear correlation between age and income. 16. (d), You would not expect age and height to be correlated. 17. (b), You would expect a negative linear correlation between age and balance on student loans. 18. (a), You would expect the relationship between age and body temperature to be fairly constant. 19. Answers will vary. Sample answer: People who can afford an expensive home likely have enough money to seek medical attention and purchase necessary medication.

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CHAPTER 9 Ň CORRELATION AND REGRESSION

20. Answers will vary. Sample answer: People who use alcohol are more likely to use tobacco because they are less concerned about their health than the general population. 21. Answers will vary. Sample answer: Ice cream sales increase in the warm months of the year, when more people spend time outdoors. Homicide rates increase when more people are outside. 22. Answers will vary. This correlation appears to be the result of a coincidence. 23. (a)

(b)

x 1 2 3 4 5 6 3 5 2 4 6

y 3 220 540 1,100 2,100 2,600 730 2,200 260 1,200 2,500

xy 3 440 1,620 4,400 10,500 15,600 2,190 11,000 520 4,800 15,000

x2 1 4 9 16 25 36 9 25 4 16 36

y2 9 48,400 291,600 1,210,000 4,410,000 6,760,000 532,900 4,840,000 67,600 1,440,000 6,250,000

¦ x = 41 ¦ y = 13, 453 ¦ xy = 66,073 ¦ x = 181 ¦ y = 25,850,509 r=

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

n ¦ y2 − ( ¦ y )

11(66,073) − (41)(13, 453) 11(181) − (41)2 11(25,850,509) − (13, 453)2

175, 230 ≈ 0.979 310 103,372,390

(c) Strong positive linear correlation. As age increases, the number of words in the children’s vocabulary increases, as well. (d) There is enough evidence at the 1% level of significance to conclude that there is a significant linear correlation between children’s ages and number of words in their vocabulary.

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359

24. (a)

(b)

x 62 58 65 67 59 64 65 57

y 109 102 107 114 96 110 116 128

xy 6,758 5,916 6,955 7,638 5,664 7,040 7,540 7,296

¦ x = 497

¦ y =882

¦ xy = 54,807

r= =

x2 3,844 3,364 4,225 4,489 3,481 4,096 4,225 3,249 2 ¦ x = 30,973

y2 11,881 10,404 11,449 12,996 9,216 12,100 13,456 16,384 2 ¦ y = 97,886

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

n ¦ y2 − ( ¦ y )

8(54,807) − (497)(882) 8(30,973) − (497) 2 8(97,886) − (882) 2 102 ≈ 0.051 775 5,164

(c) No linear correlation. The heights of high school girls do not appear to be related to their IQ scores. (d) There is not enough evidence at the 1% level of significance to conclude that there is a significant linear correlation between high school girls’ heights and their IQ scores. 25. (a)

360

CHAPTER 9 Ň CORRELATION AND REGRESSION

(b)

¦ r= = =

x 190 185 155 180 175 170 150 160 160 180 190 210 x =2105

y 60 57 54 60 56 64 52 51 49 57 59 64 y =683

xy 11,400 10,545 8370 10,800 9800 10,880 7800 8160 7840 10,260 11,210 13,440 xy =120,505

x2 36,100 34,225 24,025 32,400 30,625 28,900 22,500 25,600 25,600 32,400 36,100 44,100 x 2 =372,575

y2 3600 3249 2916 3600 3136 4096 2704 2601 2401 3249 3481 4096 2 y =39,129

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

n ¦ y2 − ( ¦ y )

12 (120,505 ) − ( 2105 )( 683) 12 ( 372,575 ) − ( 2105 )

12 ( 39,129 ) − ( 683)

8345 ≈ 0.756 39,875 3059

(c) Strong positive linear correlation; As the maximum weight for one repetition of a half squat increases, the jump height tends to increase. (d) There is enough evidence at the 1% level of significance to conclude that there is a significant linear correlation between maximum weight for one repetition of a half squat and jump height. 26. (a)

CHAPTER 9 Ň CORRELATION AND REGRESSION

(b)

¦ r= = =

x 175 180 155 210 150 190 185 160 190 180 160 170 x =2105

y 1.80 1.77 2.05 1.42 2.04 1.61 1.70 1.91 1.60 1.63 1.98 1.90 y = 21.41

xy 315 318.6 317.75 298.2 306 305.9 314.5 305.6 304 293.4 316.8 323 xy =3718.75

x2 30,625 32,400 24,025 44,100 22,500 36,100 34,225 25,600 36,100 32,400 25,600 28,900 x 2 =372,575

361

y2 3.24 3.1329 4.2025 2.0164 4.1616 2.5921 2.89 3.6481 2.56 2.6569 3.9204 3.61 2 y =38.6309

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

n ¦ y2 − ( ¦ y )

12 ( 3718.75 ) − ( 2105 )( 21.41) 12 ( 372,575 ) − ( 2105 )

12 ( 38.6309 ) − ( 21.41)

−443.05 ≈ −0.975 39,875 5.1827

(c) Strong negative linear correlation; As maximum weight for one repetition of a half squat increases, the time to run 10-meter sprint tends to decrease. (d) There is enough evidence at the 1% level of significance to conclude that there is a significant linear correlation between maximum weight for one repetition of a half squat and time to run a 10-meter sprint. 27. (a)

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CHAPTER 9 Ň CORRELATION AND REGRESSION

(b)

x 3.53 1.01 3.16 5.55 2.61 8.64

y 1.18 0.88 2.64 2.16 0.90 2.68

xy 4.1654 0.8888 8.3424 11.988 2.349 23.1552

¦ x = 24.5

¦ y =10.44

¦ xy = 50.8888

r= r= =

x2 12.4609 1.0201 9.9856 30.8025 6.8121 74.6496 2 ¦ x =135.7308

y2 1.3924 0.7744 6.9696 4.6656 0.81 7.1824 2 ¦ y = 21.7944

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

n ¦ y2 − ( ¦ y )

6(50.8888) − (24.5)(10.44) 6(135.7308) − (24.5) 2 6(21.7944) − (10.44)2 49.5528 214.1348 21.7728

≈ 0.7257

(c) Strong positive linear correlation. As the earnings per share increase, the dividends per share tend to increase. (d) There is not enough evidence at the 1% level of significance to conclude that there is a significant linear correlation between earnings per share for the companies and their dividends per share. 28. (a)

(b)

x 0 5 10 15 20 25 30 35 40 45 50 x =275

y 1116.3 1096.9 1077.3 1057.2 1036.8 1015.8 994.5 969.0 967.7 967.7 967.7 y =11, 266.9

xy 0.000 5484.5 10,773 15,858 20,736 25,395 29,835 33,915 38,708 43,547 48,385 xy =272,636

x2 0 25 100 225 400 625 900 1225 1600 2025 2500 x 2 =9625

y2 1,246,125.69 1,203,189.61 1,160,575.29 1,117,671.84 1,074,984.24 1,031,849.64 989,030.25 938,961.00 936,443.29 936,443.29 936,443.29 2 y =11,571,687.43

CHAPTER 9 Ň CORRELATION AND REGRESSION

r= = =

363

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

n ¦ y2 − ( ¦ y )

11( 272,636 ) − ( 275 )(11, 266.9 ) 11( 9625 ) − ( 275 )

11(11,571,687.43) − (11, 266.9 )

−99, 401.5 ≈ −0.972 30, 250 345,526.12

(c) Strong negative linear correlation; As the altitude increases, the speed of sound tends to decrease. (d) There is enough evidence at the 1% level of significance to conclude that there is a significant linear correlation between altitude and speed of sound. 29. The new correlation coefficient is approximately 0.863, showing a decrease in r by approximately 0.116. Thus, the strength of the linear relationship becomes weaker with the addition of the one data point. 30. The new correlation coefficient is approximately 0.845, showing an increase in r by approximately 0.794. Thus, the strength of the linear relationship becomes much stronger with the deletion of the one data point. 31. The correlation coefficient gets stronger, going from r ≈ 0.756 to r ≈ 0.908 . 32. The correlation coefficient gets weaker, going from r ≈ −0.975 to r ≈ −0.655 . 33. r ≈ 0.623 n = 8 and α = 0.01

critical value = 0.834 r ≈ 0.623 < 0.834 The correlation is not significant. OR H 0 : ρ = 0; H a : ρ ≠ 0

α = 0.01 d.f. = n − 2 = 6 −t0 = −3.707, t0 = 3.707; Rejection regions: t < −3.707 or t > 3.707. t=

0.623

≈ 1.951

1 − ( 0.623) 8−2 Because −3.707 < t < 3.707 , fail to reject H 0 . There is not enough evidence at the 1% level of significance to conclude that there is a significant linear correlation between vehicle weight and the variability in braking distance on a dry surface. 2

1− r n−2

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CHAPTER 9 Ň CORRELATION AND REGRESSION

34. r ≈ 0.955 n = 8 and α = 0.05

critical value = 0.707 r = 0.955 > 0.707 The correlation is significant. OR H 0 : ρ = 0; H a : ρ ≠ 0

α = 0.05 d.f. = n − 2 = 6 −t0 = −2.447, t0 = 2.447; Rejection regions: t < −2.447 or t > 2.447. t=

0.955

≈ 7.887

1 − ( 0.955 ) 8−2 Because t > 2.447 , reject H 0 . There is enough evidence at the 5% level of significance to conclude that there is a significant linear correlation between vehicle weight and the variability in braking distance on a wet surface. 2

1− r n−2

35. r ≈ 0.756 n = 12 and α = 0.05

critical value = 0.576 r =| 0.756 |> 0.576 The correlation is significant. OR H 0 : ρ = 0; H a : ρ ≠ 0

α = 0.05 d.f. = n − 2 = 10 −t0 = −2.228, t0 = 2.228; Rejection regions: t < −2.228 or t > 2.228. t=

0.756

≈ 3.652 2 1 − ( 0.756 ) 12 − 2 Because t > 2.228 , reject H 0 . There is enough evidence at the 5% level of significance to conclude that there is a significant linear correlation between the maximum weight for one repetition of a half squat and the jump height. 1− r2 n−2

36. r ≈ −0.975 n = 12 and α = 0.01

critical value = 0.708 r =| −0.975 |> 0.708 The correlation is significant. OR H 0 : ρ = 0; H a : ρ ≠ 0

CHAPTER 9 Ň CORRELATION AND REGRESSION

365

−t0 = −3.169, t0 = 3.169; Rejection regions: t < −3.169 or t > 3.169. r

−0.975

≈ −13.876 2 1 − ( −0.975 ) 12 − 2 Because t < −3.169 , reject H 0 . There is enough evidence at the 1% level of significance to conclude that there is a significant linear correlation between the maximum weight for one repetition of a half squat and the time to run a 10-meter sprint. 1− r2 n−2

37.

¦ r= =

x 1.80 1.77 2.05 1.42 2.04 1.61 1.70 1.91 1.60 1.63 1.98 1.90 x =21.41

y 175 180 155 210 150 190 185 160 190 180 160 170 y =2105

xy 315 318.6 317.75 298.2 306 305.9 314.5 305.6 304 293.4 316.8 323 xy =3718.75

x2 3.24 3.1329 4.2025 2.0164 4.1616 2.5921 2.89 3.6481 2.56 2.6569 3.9204 3.61 2 x =38.6309

y2 30,625 32,400 24,025 44,100 22,500 36,100 34,225 25,600 36,100 32,400 25,600 28,900 y 2 =372,575

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

n ¦ y2 − ( ¦ y )

12 ( 3718.75 ) − ( 21.41)( 2105 ) 12 ( 38.6309 ) − ( 21.41)

12 ( 372,575 ) − ( 2105 )

−443.05 ≈ −0.975 5.1827 39,875 The correlation coefficient remains unchanged when the x-values and the y-values are switched. =

38. Answers will vary.

9.2 LINEAR REGRESSION 9.2 TRY IT YOURSELF SOLUTIONS 1.

n = 7, m=

¦ x = 88, ¦ y = 56.7, ¦ xy = 435.6, ¦ x = 1836 2

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

( 7 )( 435.6 ) − (88)( 56.7 ) −1940.4 = ≈ −0.379875 5108 ( 7 )(1836 ) − (88 )2

366

CHAPTER 9 Ň CORRELATION AND REGRESSION

(88) ¦y ¦ x ( 56.7 ) −m ≈ − ( −0.379875 ) ≈ 12.8756 n n 7 7 y = −0.380 x + 12.876

b = y − mx =

yˆ = 152.932x + 7811.244

3. (1) y = 12.481( 2 ) + 33.683 y = 58.645

(2)

58.645 minutes

y = 12.481( 3.32 ) + 33.683 y = 75.120 75.120 minutes

9.2 EXERCISE SOLUTIONS 1. A residual is the difference between the observed y-value of a data point and the predicted y-value on the regression line for the x-coordinate of the data point. A residual is positive when the data point is above the line, negative when the point is below the line, and zero when the observed y-value equals the predicted y-value. 2. Positive 3. Substitute a value of x into the equation of a regression line and solve for y . 4. Prediction values are meaningful only for x-values in (or close to) the range of the original data. 5. The correlation between variables must be significant. 6. Answers will vary. Sample answer: Because the regression line models the trend of the given data, and it is not known if the trend continues beyond the range of those data. 7. b

8. a

12. d 13. b 14. c 15. d 16. a

9. e

10. c

11. f

CHAPTER 9 Ň CORRELATION AND REGRESSION

17.

x 1002 992 901 780 762 756 752 741 735

y 75 71 64 56 53 55 48 47 47

xy 75,150 70,432 57,664 43,680 40,386 41,580 36,096 34,827 34,545

¦ x = 7,421

¦ y = 516

¦ xy = 434,360

367

x2 1,004,004 984,064 811,801 608,400 580,644 571,536 565,504 549,081 540,225 2 ¦ x = 6, 215, 259

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

9(434,360) − (7421)(516) 80,004 = ≈ 0.092 9(6, 215, 259) − (7421)2 866,090 ¦y ¦x b = y − mx = −m n n 516 § 80,004 · § 7421 · b= −¨ ≈ −18.834 ¸¨ 9 © 866,090 ¹ © 9 ¸¹ m=

yˆ = 0.092x − 18.834

(a) yˆ = 0.092(950) − 18.834 ≈ 69 stories (b) yˆ = 0.092(850) − 18.834 ≈ 59 stories (c) yˆ = 0.092(800) − 18.834 ≈ 55 stories (d) It is not meaningful to predict the value of y for x = 650 feet because x = 650 is outside the range of the original data.

368

CHAPTER 9 Ň CORRELATION AND REGRESSION

18.

x 1438 1280 1680 5627 1040 1180 876 1552 2180

y 120.0 109.5 289.0 915.0 49.9 79.9 59.9 94.9 199.9

xy 172,560 140,160 485,520 5,148,705 51,896 94,282 52,472.4 147,284.8 435,782

¦ x =16,853

¦ y =1918

¦ xy = 6,728,662.2

x2 2,067,844 1,638,400 2,822,400 31,663,129 1,081,600 1,392,400 767,376 2,408,704 4,752,400 2 ¦ x = 48,594, 253

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

9(6,728,662.2) − (16,853)(1918) 28, 233,905.8 = ≈ 0.184 153,324,668 9(48,594, 253) − (16,853)2 ¦y ¦x b = y − mx = −m n n 1918 § 28, 233,905.8 · § 16,853 · b= −¨ ¸¨ ¸ ≈ −131.71 9 © 153,324,668 ¹ © 9 ¹ m=

yˆ = 0.184x − 131.71

(a) y = 0.184(1100) − 131.71 = $70,690 (b) It is not meaningful to predict the value of y for x = 725 because x = 725 is outside the

range of the original data.

CHAPTER 9 Ň CORRELATION AND REGRESSION

19.

¦ m=

x 0 2 4 5 5 5 6 7 8 x =42

y 40 51 64 69 73 75 93 90 95 y =650

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x − (¦ x) 2

xy 0 102 256 345 365 375 558 630 760 xy =3391

369

x2 0 4 16 25 25 25 36 49 64 x 2 =244

( 9 )( 3391) − ( 42 )( 650 ) 3219 = ≈ 7.451389 432 ( 9 )( 244 ) − ( 42 )2

§ 650 · § 42 · b = y − mx ≈ ¨ − ( 7.451389 ) ¨ ¸ ≈ 37.4491 ¸ © 9 ¹ © 9 ¹ y = 7.451x + 37.449

(a) y = 7.451( 3) + 37.449 ≈ 60 (b) y = 7.451( 6.5 ) + 37.449 ≈ 86 (c) It is not meaningful to predict the value of y for x = 13 because x = 13 is outside the range of the original data. (d) y = 7.451( 4.5 ) + 37.449 ≈ 71

370

CHAPTER 9 Ň CORRELATION AND REGRESSION

20.

x 51 29 26 22 18 15 23 16 8 í1

y 27 21 20 19 20 19 18 18 18 17

xy 1377 609 520 418 360 285 414 288 144 í17

¦ x = 207

¦ y =197

¦ xy = 4398

x2 2601 841 676 484 324 225 529 256 64 1 2 ¦ x = 6001

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

10(4398) − (207)(197) 3201 = ≈ 0.187 ; 17,161 10(6001) − (207) 2 ¦y ¦x b = y − mx = −m n n 197 § 3201 · § 207 · b= −¨ ≈ 15.839 ¸¨ 10 © 17,161 ¹ © 10 ¸¹ y = 0.187 x + 15.839 m=

(a) y = 0.187(0) + 15.839 ≈ 16 wins (b) y = 0.187(20) + 15.839 ≈ 20 wins (c) y = 0.187(40) + 15.839 ≈ 23 wins (d) It is not meaningful to predict the value of y for x = 60 because x = 60 is outside the range of the original data.

CHAPTER 9 Ň CORRELATION AND REGRESSION

21.

¦ m=

x 60 75 62 68 84 97 66 65 86 78 93 75 88 x =997

y 403 363 381 367 341 317 401 384 342 377 329 377 349 y =4731

xy 24,180 27,225 23,622 24,956 28,644 30,749 26,466 24,960 29,412 29,406 30,597 28,275 30,712 xy =359, 204

371

x2 3600 5625 3844 4624 7056 9409 4356 4225 7396 6084 8649 5625 7744 2 x =78, 237

x ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

(13)( 359, 204 ) − ( 997 )( 4731) (13)( 78, 237 ) − ( 997 )2

−47,155 ≈ −2.0438 23,072

§ 4731 · § 997 · b = y − mx ≈ ¨ − ( −2.04382 ) ¨ ¸ ¸ ≈ 520.6683 © 13 ¹ © 13 ¹ y = −2.044 x + 520.668

(a) It is not meaningful to predict the value of y for x = 120 because x =120 is outside the range of the original data. (b) y = −2.044 ( 67 ) + 520.668 = 384 milliseconds (c) y = −2.044 ( 90 ) + 520.668 = 337 milliseconds (d) y = −2.044 (83) + 520.668 = 351 milliseconds

372

CHAPTER 9 Ň CORRELATION AND REGRESSION

22.

¦ m=

x 137 168 152 145 159 159 124 137 155 148 147 146 x =1777

y 106 130 116 106 125 119 103 104 120 110 107 109 y =1355

x ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

xy 14,522 21,840 17,632 15,370 19,875 18,921 12,772 14,248 18,600 16,280 15,729 15,914 xy =201,703

(12 )( 201,703) − (1777 )(1355 ) 12,601 = ≈ 0.682352 18, 467 (12 )( 264,683) − (1777 )2

§ 1355 · § 1777 · b = y − mx ≈ ¨ ¸ − ( 0.682352 ) ¨ ¸ ≈ 11.87166 12 © ¹ © 12 ¹ y = 0.682 x + 11.872

(a) y = 0.682 (140 ) + 11.872 = 107 centimeters (b) y = 0.682 (172 ) + 11.872 = 129 centimeters (c) y = 0.682 (164 ) + 11.872 = 124 centimeters (d) y = 0.682 (158) + 11.872 = 120 centimeters

x2 18,769 28,224 23,104 21,025 25,281 25,281 15,376 18,769 24,025 21,904 21,609 21,316 x 2 =264,683

CHAPTER 9 Ň CORRELATION AND REGRESSION

23.

¦ m=

x 180 220 230 90 160 190 150 110 110 160 140 150 x =1,890

y 510 740 740 280 530 580 490 480 330 640 480 460 y = 6, 260

xy 91,800 162,800 170,200 25,200 84,800 110,200 73,500 52,800 36,300 102,400 67,200 69,000 xy =1,046, 200

373

x2 32,400 48,400 52,900 8,100 25,600 36,100 22,500 12,100 12,100 25,600 19,600 22,500 x 2 = 317,900

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

12(1,046, 200) − (1,890)(6, 260) 723,000 = ≈ 2.979 ; 242,700 12(317,900) − (1,890)2 ¦y ¦x b = y − mx = −m n n 6, 260 § 723,000 · § 1,890 · b= −¨ ¸¨ ¸ ≈ 52.476 12 © 242,700 ¹ © 12 ¹ y = 2.979 x + 52.476 m=

(a) y = 2.979(170) + 52.476 ≈ 559 milligrams (b) y = 2.979(100) + 52.476 ≈ 350 milligrams (c) It is not meaningful to predict the value of y for x = 260 because x = 260 is outside the

range of the observed data.

374

CHAPTER 9 Ň CORRELATION AND REGRESSION

24.

¦ m=

x 9000 42,000 18,000 19,700 7600 7300 30,200 14,300 7000 46,000 800 2600 141,000 15,500 x = 361,000

y 1300 5162 2886 1728 1182 1964 2096 2179 1054 3612 316 615 10,571 2160 y = 36,825

xy 11,700,000 216,804,000 51,948,000 34,041,600 8,983,200 14,337,200 63,299,200 31,159,700 7,378,000 166,152,000 252,800 1,599,000 1,490,511,000 33,480,000 xy = 2,131,645,700

x2 81,000,000 1,764,000,000 324,000,000 388,090,000 57,760,000 53,290,000 912,040,000 204,490,000 49,000,000 2,116,000,000 640,000 6,760,000 19,881,000,000 240,250,000 2 x = 26,078,320,000

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

14(2,131,645,700) − (361,000)(36,825) 16,549, 214,800 = ≈ 0.07 234,775, 480,000 14(26,078,320,000) − (361,000)2

¦y ¦x −m n n 36,825 § 16,549, 214,800 · § 361,000 · b= −¨ ¸¨ ¸ ≈ 812.73 14 © 234,775, 480,000 ¹ © 14 ¹ y = 0.07 x + 812.73 b = y − mx =

(a) y = 0.07(32,500) + 812.73 ≈ 3087.73 million dollars (b) y = 0.07(6000) + 812.73 ≈ 1232.73 million dollars (c) y = 0.07(1350) + 812.73 ≈ 907.23 million dollars (d) y = 0.07(100,000) + 812.73 ≈ 7812.73 million dollars Copyright © 2023 Pearson Education, Inc.

CHAPTER 9 Ň CORRELATION AND REGRESSION

25.

¦ m=

x 8.5 9.0 9.0 9.5 10.0 10.0 10.5 10.5 11.0 11.0 11.0 12.0 12.0 12.5 x =146.5

y 66.0 68.5 67.5 70.0 70.0 72.0 71.5 69.5 71.5 72.0 73.0 73.5 74.0 74.0 y =993

x ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

xy 561.00 616.50 607.50 665.00 700.00 720.00 750.75 729.75 786.50 792.00 803.00 882.00 888.00 925.00 xy =10, 427

375

x2 72.25 81.00 81.00 90.25 100.00 100.00 110.25 110.25 121.00 121.00 121.00 144.00 144.00 156.25 x 2 =1552.25

(14 )(10, 427 ) − (146.5)( 993) 503.5 = ≈ 1.870009 269.25 (14 )(1552.25 ) − (146.5 )2

§ 993.0 · § 146.5 · b = y − mx ≈ ¨ − (1.870 ) ¨ ¸ ¸ ≈ 51.3603 © 14 ¹ © 14 ¹ y = 1.870 x + 51.360

(a) y = 1.870 (11.5 ) + 51.360 = 72.865 inches (b) y = 1.870 ( 8.0 ) + 51.360 = 66.32 inches (c) It is not meaningful to predict the value of y for x = 15.5 because x = 15.5 is outside the range of the original data. (d) y = 1.870 (10.0 ) + 51.360 = 70.06 inches

376

CHAPTER 9 Ň CORRELATION AND REGRESSION

26.

¦ m=

x 0.1 0.2 0.4 0.7 0.6 0.9 0.1 0.2 0.4 0.9 x = 4.5

y 14.5 14.3 14.1 13.9 13.9 13.7 14.3 14.2 14 13.8 y =140.7

xy 1.45 2.86 5.64 9.73 8.34 12.33 1.43 2.84 5.6 12.42 xy = 62.64

x2 0.01 0.04 0.16 0.49 0.36 0.81 0.01 0.04 0.16 0.81 x 2 = 2.89

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

10(62.64) − (4.5)(140.7) −6.75 = ≈ −0.780 ; 8.65 10(2.89) − (4.5) 2 ¦y ¦x b = y − mx = −m n n 140.7 § −6.75 ·§ 4.5 · b= −¨ ¸¨ ¸ ≈ 14.421 10 © 8.65 ¹© 10 ¹ y = −0.780 x + 14.421 m=

(a) y = −0.780(0.3) + 14.421 = 14.187 hours (b) It is not meaningful to predict the value of y for x = 3.9 because the data is examining infants, and a child almost 4 years of age is not an infant. (c) y = −0.780(0.6) + 14.421 = 13.953 hours (d) y = −0.780(0.8) + 14.421 = 13.797 hours 27. Strong positive linear correlation; As the years of experience of the registered nurses increase, their salaries tend to increase.

CHAPTER 9 Ň CORRELATION AND REGRESSION

28.

x 0.5 2 4 5 7 9 10 12.5 13 16 18 20 22 25 x =164

¦ m=

y 48.3 53.4 58.5 63.4 65.7 67.3 70.6 71.8 69.8 73.2 75.5 74.3 78.9 76.6 y = 947.3

xy 24.15 106.8 234 317 459.9 605.7 706 897.5 907.4 1171.2 1359 1486 1735.8 1915 xy =11,925.45

377

x2 0.25 4 16 25 49 81 100 156.25 169 256 324 400 484 625 x 2 = 2,689.5

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

14(11,925.45) − (164)(947.3) 11,599.1 = ≈ 1.078 10,757 14(2,689.5) − (164) 2 ¦y ¦x b = y − mx = −m n n 947.3 § 11,599.1 · § 164 · b= −¨ ¸¨ ¸ ≈ 55.033 14 © 10,757 ¹ © 14 ¹ y = 1.078 x + 55.033 m=

29. No, it is not meaningful to predict a salary for a registered nurse with 28 years of experience because x = 28 is outside the range of the original data. 30. critical value = 0.661 r ≈ 0.920 > 0.661 There is enough evidence at the 1% level of significance to conclude that there is a significant linear correlation between years of experience and annual salary.

378

CHAPTER 9 Ň CORRELATION AND REGRESSION

31. (a)

(b)

y = −4.297 x + 94.200

y = −0.141x + 14.763

(b)

y = 0.453 x − 26.448

(b)

b ≈ 21.024 y = 0.139 x + 21.024

(c)

x 38 34 40 46 43 48 60 55 52

Y 24 22 27 32 30 31 27 26 28

y = 0.139 x + 21.024 26.306 22.750 26.584 27.418 27.001 27.696 29.364 28.669 28.252

y − y –2.306 –3.750 0.416 4.582 2.999 3.304 –2.364 –2.669 –0.252

(d) The residual plot shows a pattern because the residuals do not fluctuate about 0. This implies that the regression line is not a good representation of the relationship between the variables.

CHAPTER 9 Ň CORRELATION AND REGRESSION

34. (a) m ≈ 1.711

379

(b)

b ≈ 3.912 y = 1.711x + 3.912

(c)

x 8 4 15 7 6 3 12 10 5

y 18 11 29 18 14 8 25 20 12

y = 1.711x + 3.912 17.600 10.756 29.577 15.889 14.178 9.045 24.444 21.022 12.467

y − y 0.400 0.244 –0.577 2.111 –0.178 –1.045 0.556 –1.022 –0.467

(d) The residual plot shows no pattern in the residuals because the residuals fluctuate about 0. This suggests that the regression line is a good representation of the data. 35. (a)

(b) The point (44, 8) may be an outlier. (c) Excluding the point (44, 8) y = −0.711x + 35.263 . The point (44, 8) is not influential because using all 8 points y = −0.607 x + 34.160 . The slopes and y-intercepts of the regression lines with the point included and without the point are not significantly different. 36. (a)

(b) The point (14, 3) may be an outlier.

380

CHAPTER 9 Ň CORRELATION AND REGRESSION

(c) The point (14, 3) is influential because using all 6 points y = 0.212 x + 6.445 . However, excluding the point (14, 3) y = 0.905 x + 3.568 The slopes and y-intercepts of the regression lines with the point included and without the point included are significantly different. 37. m ≈ 654.536

b ≈ −1214.857 y = 654.536 x − 1214.857

38.

x 1 2 3 4 5 6 7

y log y 165 2.218 280 2.447 468 2.670 780 2.892 1310 3.117 1920 3.283 4900 3.690 log y = 0.233x + 1.969

39. Using a technology tool y = 93.028 (1.712 ) . x

40. The exponential equation is a much better model for the data. The graph of the exponential equation fits the data better than the regression line. 41. m = −78.929

b = 576.179 y = −78.929 x + 576.179

CHAPTER 9 Ň CORRELATION AND REGRESSION

42.

381

x y log x log y 1 695 0 2.842 2 410 0.301 2.613 3 256 0.477 2.408 4 110 0.602 2.041 5 80 0.699 1.903 6 75 0.778 1.875 7 68 0.845 1.833 8 74 0.903 1.869 log y = m ( log x ) + b log y = −1.251 log x + 2.893 A linear model is more appropriate for the transformed data.

43. Using a technology tool y = 782.300 x −1.251 .

44. The power equation is a much better model for the data. The graph of the power equation fits the data much better than the regression line. 45. y = a + b ln x = 25.035 + 19.599ln x

46.

y = 13.758 − 0.291ln x

47. The logarithmic equation is a better model for the data. The graph of the logarithmic equation fits the data better than the regression line. 48. The logarithmic equation is a better model for the data. The graph of the logarithmic equation fits the data better than the regression line.

382

CHAPTER 9 Ň CORRELATION AND REGRESSION

9.3 MEASURES OF REGRESSION AND PREDICTION INTERVALS 9.3 TRY IT YOURSELF SOLUTIONS

r ≈ 0.979 r 2 ≈ ( 0.979 ) ≈ 0.958 About 95.8% of the variation in the times is explained. About 4.2% of the variation is unexplained. 2

y i

yi − y i

( y − y )

15 20 20 30 40 45 50 60

26 32 38 56 54 78 80 88

28.386 35.411 35.411 49.461 63.511 70.536 77.561 91.611

–2.386 –3.411 2.589 6.539 –9.511 7.464 2.439 –3.611

5.692996 11.634921 6.702921 42.758521 90.459121 55.711296 5.948721 13.039321 ¦ = 231.947818

n =8 se =

(

¦ yi − l yi

) = 231.947818 ≈ 6.218 2

6 n−2 The standard error of estimate of the weekly sales for a specific radio ad time is about $621.80. 3.

n = 10, d.f. = 8, tc = 2.306, se ≈ 150.669 y = 187.660(4) + 44.663 = 795.303 2 n x0 − x 1 E = tc se 1 + + n n ¦ x 2 − ( ¦ x )2

(

)

E = (2.306)(150.669) 1 +

)

1 10(4 − 2.57) 2 + ≈ 384.078 10 10(82.81) − (25.7) 2

y − E = 795.303 − 384.078 = 411.225 y + E = 795.303 + 384.078 = 1179.381

411.225 < y < 1179.381 You can be 95% confident that when the gross domestic product is $4 trillion, the carbon dioxide emissions will be between 411.225 and 1179.381 million metric tons.

CHAPTER 9 Ň CORRELATION AND REGRESSION

383

9.3 EXERCISE SOLUTIONS 1. The total variation is the sum of the squares of the differences between the y-values of each ordered

(

)

pair and the mean of the y-values of the ordered pairs or ¦ yi − y . 2. The explained variation is the sum of the squares of the differences between the predicted y-values 2 and the mean of the y-values of the ordered pairs or ¦ y − y .

(

)

3. The unexplained variation is the sum of the squares of the differences between the observed y-values 2 and the predicted y-values or ¦ y − y .

(

)

( 4. The coefficient of determination r = 2

¦ y i − y

(

¦ yi − y

) is the ratio of the explained variation to the total 2

)

variation and is the percent of variation of y that is explained by the relationship between x and y; 1− r 2 is the percent of variation that is unexplained. 5. Two variables that have perfect positive or perfect negative linear correlation have a correlation coefficient of 1 or −1 , respectively. In either case, the coefficient of determination is 1, which means that 100% of the variation in the response variable is explained by the variation in the explanatory variable. 6. Two variables have a bivariate normal distribution when, for any fixed values of either variable, the corresponding values of the other variable are normally distributed. 7.

r 2 = ( 0.465 ) ≈ 0.216; About 21.6% of the variation is explained. About 78.4% of the variation is unexplained.

r 2 = ( −0.328 ) ≈ 0.108; About 10.8% of the variation is explained. About 89.2% of the variation is unexplained.

r 2 = ( −0.957 ) ≈ 0.916; About 91.6% of the variation is explained. About 8.4% of the variation is unexplained.

10. r 2 = ( 0.881) ≈ 0.776; About 77.6% of the variation is explained. About 22.4% of the variation is unexplained. 2

( 11. (a) r = 2

¦ y i − y

(

¦ yi − y

) ≈ 0.592 2

)

384

CHAPTER 9 Ň CORRELATION AND REGRESSION

(b) se =

(

¦ yi − l yi

) = 859, 474,635 ≈ 9270.785 2

10 n−2 The standard error of estimate of the proceeds for a specific number of offerings is about $9,270,785,000.

( 12. (a) r 2 =

¦ yi − y

(

¦ yi − y

) ≈ 0.940

)

About 94% of the variation is explained by the relationship between the median and mean hourly wages and about 6% of the variation is unexplained.

(b) se =

(

¦ yi − l yi

) = 2.539471299 ≈ 0.563

n−2 8 The standard error of estimate of the mean for a specific median is about $0.56.

( 13. (a) r 2 =

¦ yi − y

(

¦ yi − y

) ≈ 0.764

)

About 76.4% of the variation is explained by the relationship between the goals allowed and points and about 23.6% of the variation is unexplained.

(b) se =

(

¦ yi − l yi

) = 565.084608 ≈ 6.593

n−2 13 The standard error of estimate of the points for a specific number of goals allowed is about 6.593.

( 14. (a) r = 2

¦ y i − y

(

¦ yi − y

) ≈ 0.671 2

)

About 67.1% of the variation in the trunk diameters can be explained by the relationship between tree heights and the trunk diameters, and about 32.9% of the variation is unexplained.

(b) se =

(

¦ yi − l yi

) = 19.0193 ≈ 1.780 2

6 n−2 The standard error of estimate of the trunk diameters for a specific tree height is about 1.780 inches.

CHAPTER 9 Ň CORRELATION AND REGRESSION

( 15. (a) r 2 =

¦ yi − y

(

¦ yi − y

385

) ≈ 0.845

)

About 84.5% of the variation is explained by the relationship between the percentages of employment in STEM occupations and about 15.5% of the variation is unexplained.

(b) se =

(

¦ yi − l yi

) = 235.1604316 ≈ 4.427 .

n−2 12 The standard error of estimate of the mean annual wage for a specific percentage of employment in STEM occupations is about $4427.

( 16. (a) r = 2

¦ y i − y

(

¦ yi − y

) ≈ 0.715 2

)

About 71.5% of the variation is explained by the relationship between the U.S. voting age populations and the number of ballots cast and about 28.5% of the variation is unexplained.

(b) se =

(

¦ yi − l yi

) = 593.8212733 ≈ 8.616 . 2

8 n−2 The standard error of estimate of the ballots cast in federal elections for a specific voting age population is about 8,616,000 people.

( 17. (a) r = 2

¦ y i − y

(

¦ yi − y

) ≈ 0.902 2

)

About 90.2% of the variation is explained by the relationship between the marketed productions of natural gas in the U.S. and the exports of natural gas from the U.S. and about 9.8% of the variation is unexplained.

(b) se =

(

¦ yi − l yi

) = 1,193,708.172 ≈ 488.612 . 2

5 n−2 The standard error of estimate of the exports of natural gas for a specific marketed production of natural gas is about 488,612,000,000.

( 18. (a) r = 2

¦ y i − y

(

¦ yi − y

) ≈ 0.920 2

)

About 92% of the variation is explained by the relationship between IRA assets and federal DB plan assets and about 8% of the variation is unexplained.

386

CHAPTER 9 Ň CORRELATION AND REGRESSION

(b) se =

(

¦ yi − l yi

) = 49,597.72678 ≈ 78.738 . 2

8 n−2 The standard error of estimate of assets in federal DB plans for a specific total of IRA assets is about $78,738,000,000.

( 19. (a) r = 2

¦ y i − y

(

¦ yi − y

) ≈ 0.700 2

)

About 70% of the variation is explained by the relationship between the new vehicle sales of Ford and General Motors and about 30% of the variance is unexplained.

(b) se =

(

¦ yi − l yi

) = 356,933.1042 ≈ 199.146 . 2

9 n−2 The standard error of estimate of the new vehicle sales of General Motors for a specific amount of new vehicle sales of Ford is about 199,146 new vehicles.

( 20. (a) r = 2

¦ y i − y

(

¦ yi − y

) ≈ 0.951 2

)

About 95.1% of the variation is explained by the relationship between new vehicle sales of Toyota and Honda and about 4.9% of the variation is unexplained.

(b) se =

(

¦ yi − l yi

) = 16,793.63511 ≈ 43.197 . 2

9 n−2 The standard error of estimate of new vehicle sales of Honda for a specific amount of new vehicle sales of Toyota is about 43,197 new vehicles.

21. n = 12, d.f. = 10, tc = 2.228, se ≈ 9270.785 y = 222.346 x + 5001.340 = 222.346 ( 200 ) + 5001.340 = 49, 470.54

(

)

n x0 − x 1 E = tc se 1 + + n n ¦ x 2 − ( ¦ x )2

(

)

≈ ( 2.228 )( 9270.785 ) 1 +

12 ( 200 − 1461 / 12 ) 1 + 12 12 ( 203,063) − (1461)2 2

≈ 23,789.004 y ± E → 49, 470.54 ± 23,789.004 → ( 25,681.536, 73,259.544 ) You can be 95% confident that the proceeds will be between $25,681,536,000 and $73,259,544,000 when the number of initial offerings is 200 issues.

CHAPTER 9 Ň CORRELATION AND REGRESSION

387

22. n = 10, d.f. = 8, tc = 2.306, se ≈ 0.563 y = 1.208(21.50) + 1.495 = 27.467 2 n x0 − x 1 E = tc se 1 + + n n ¦ x 2 − ( ¦ x )2

(

E = (2.306)(0.563) 1 +

)

1 10(21.50 − 19.506)2 + ≈ 1.450 10 10(3831.918) − (195.06) 2

y − E = 27.467 − 1.450 ≈ 26.02 ; y + E = 27.467 + 1.450 ≈ 28.92

26.02 < y < 28.92 You can be 95% confident that the mean hourly wage will be between $26.02 and $28.92 when the median wage is $21.50. 23. n = 15, d.f. = 13, tc = 1.771, se ≈ 6.593 y = −0.552(140) + 152.584 = 75.304 E = tc se 1 +

(

n x0 − x

)

1 + n n ¦ x 2 − ( ¦ x )2

(

E = (1.771)(6.593) 1 +

)

1 15(140 − 2474 / 15) 2 + ≈ 12.635 15 15(414,008) − (2474) 2

y − E = 75.304 − 12.635 ≈ 62.67 ; y + E = 75.304 + 12.635 ≈ 87.94

62.67 < y < 87.94 You can be 90% confident that the total points earned will be between 63 and 88 when the number of goals allowed is 140. 24. n = 8, d.f. = 6, tc = 1.943, se ≈ 1.780 y = 0.479 x − 24.086 = 0.479 ( 80 ) − 24.086 ≈ 14.234

E = tc se 1 +

(

n x−x

)

1 + n n ¦ x 2 − ( ¦ x )2

(

)

8 ( 80 − 615 / 8 ) 1 ≈ (1.943)(1.780 ) 1 + + 8 8 ( 47447 ) − ( 615 )2 2

≈ 3.761 y ± E → 14.234 ± 3.761 → (10.473, 17.995 ) You can be 90% confident that the trunk diameter will be between 10.473 and 17.995 inches when the tree height is 80 feet.

388

CHAPTER 9 Ň CORRELATION AND REGRESSION

25. n = 14, d.f. = 12, tc = 3.055, se ≈ 4.427 y = 1.224(13) + 81.643 = 97.555

(

)

n x0 − x 1 E = tc se 1 + + n n ¦ x 2 − ( ¦ x )2

(

E = (3.055)(4.427) 1 +

)

1 14(13 − 104.7 / 14) 2 + ≈ 14.231 14 14(1636.05) − (104.7) 2

y − E = 97.555 − 14.231 ≈ 83.324 ; y + E = 97.555 + 14.231 = 111.786

83.324 < y < 111.786 You can be 99% confident that the mean annual wage will be between $83,324 and $111,786 when the percentage of employment in STEM occupations is 13% in the industry. 26. n = 10, d.f. = 8, tc = 3.355, se ≈ 8.616 y = 0.429(210) − 10.763 = 79.327

(

)

n x0 − x 1 E = tc se 1 + + n n ¦ x 2 − ( ¦ x )2

(

E = (3.355)(8.616) 1 +

)

1 10(210 − 2109.2 / 10) 2 + ≈ 30.319 10 10(452,953.7) − (2109.2) 2

y − E = 79.327 − 30.319 = 49.008 ; y + E = 79.327 + 30.319 = 109.646

49.008 < y < 109.646 You can be 99% confident that the number of ballots cast will be between 49,008,000 and 109,646,000 when the voting age population is 210 million. 27. n = 7, d.f. = 5, tc = 2.571, se ≈ 488.612 y = 0.355(31,000) − 7960.686 = 3044.14

(

)

n x0 − x 1 E = tc se 1 + + n n ¦ x 2 − ( ¦ x )2

(

)

E = (2.571)(488.612) 1 +

1 7(31,000 − 219,605 / 7) 2 + ≈ 1343.889 7 7(6,976,813,867) − (219,605) 2

y − E = 3044.314 − 1343.889 = 1700.425 ; y + E = 3044.314 + 1343.889 = 4388.203

1700.425 < y < 4388.203 You can be 95% confident that the export of natural gas from the United States will be between 1,700,425,000,000 and 4,388,203,000,000 cubic feet when the production of natural gas in the United States is 31 trillion cubic feet.

CHAPTER 9 Ň CORRELATION AND REGRESSION

389

28. n = 10, d.f. = 8, tc = 1.86, se ≈ 78.738 y = 0.122(6400) + 558.880 = 1339.68

(

)

n x0 − x 1 E = tc se 1 + + n n ¦ x 2 − ( ¦ x )2

(

)

E = (1.860)(78.738) 1 +

1 10(6400 − 83, 299 / 10) 2 + ≈ 160.219 10 10(732,336,375) − (83, 299) 2

y − E = 1339.68 − 160.219 = 1179.461 ; y + E = 1339.68 + 160.219 = 1499.899

1179.461 < y < 1499.899 You can be 90% confident that the total assets of defined benefit plans will be between $1,179,461,000,000 and $1,499,899,000,000 when the total assets in IRAs are $6400 billion. 29. n = 11, d.f. = 9, tc = 2.262, se ≈ 199.146 y = 0.937(2028) + 637.674 = 2537.91

(

)

n x0 − x 1 E = tc se 1 + + n n ¦ x 2 − ( ¦ x )2

(

)

E = (2.262)(199.146) 1 +

1 11(2028 − 24, 245 / 11) 2 + ≈ 477.481 11 11(54,388,789) − (24, 245) 2

y − E = 2537.91 − 477.481 = 2060.429 ; y + E = 2537.91 + 477.481 = 3015.391

2060.429 < y < 3015.391 You can be 95% confident that the new vehicle sales of General Motors will be between 2,060,429 and 3,015,391 when the new vehicle sales of Ford are 2,028,000. 30. n = 11, d.f. = 9, tc = 3.250, se ≈ 43.197 y = 0.584(2159) + 169.227 = 1430.083 E = (3.250)(43.197) 1 +

1 11(2159 − 23,637 / 11) 2 + ≈ 146.640 11 11(51,733,601) − (23,637) 2

y − E = 1430.083 − 146.640 = 1283.443 ; y + E = 1430.083 + 146.640 = 1576.723

1283.443 < y < 1576.723 You can be 99% confident that the new vehicle sales of Honda will be between 1,283,443 and 1,576,723 when the new vehicle sales of Toyota are 2159 thousand. 31.

390

CHAPTER 9 Ň CORRELATION AND REGRESSION

32. y = 0.089 x − 168.421

33. r 2 ≈ 0.960; About 96.0% of the variation in the average age of vehicles can be explained by the relationship between the year and the average age of vehicles and about 4% of the variation is unexplained.

34. se =

(

¦ yi − l yi

) = 2.66814 ≈ 0.667 . The standard error of estimate of the average age of vehicles 2

6 n−2 for a specific year is about 0.667 year.

35. critical value = ±3.707 m ≈ −0.205

se ≈ 0.554

( ¦ x ) ≈ −0.205 838.55 − ( 79.5) ≈ −2.578 m t= ¦ x2 − se n 0.554 8 2

Because −3.707 < t < 3.707 , fail to reject H0 : M = 0 . There is not enough evidence at the 1% level of significance to support the claim that there is a linear relationship between weight and number of hours slept. 36. critical value = ±2.306 m ≈ 1.328

se ≈ 5.086

( ¦ x ) ≈ 1.328 14,660 − ( 374 ) ≈ 6.771 m t= ¦ x2 − se n 5.086 10 2

Because t > 2.306 , reject H0 : M = 0 . There is enough evidence at the 5% level of significance to support the claim that there is a linear relationship between age and salary. 2

1 x 1 (2.57) 2 + ≈ 2.306 (150.669) + ≈ 244.216 37. E = tc se ( ) n ¦ x 2 − ª( ¦ x ) 2 / n º 10 (25.7) 2 82.81 − ¬ ¼ 10 b + E = 44.663 + 244.216 = 288.879 ; b − E = 44.663 − 244.216 = −199.553 tc se (2.306)(150.669) ≈ ≈ 84.866 E= 2 2 (25.7) ¦ x ( ) 82.81 − ¦ x2 − 10 n Copyright © 2023 Pearson Education, Inc.

CHAPTER 9 Ň CORRELATION AND REGRESSION

391

m + E = 187.660 + 84.866 = 272.526 ; m − E = 187.660 − 84.866 = 102.794 Confidence intervals for B is −199.553 < B < 288.879 ; Confidence intervals for M is 102.794 < M < 272.526 2

1 x 1 (2.57) 2 + ≈ 3.355 (150.669) + ≈ 355.310 ( ) n ¦ x 2 − ª( ¦ x ) 2 / n º 10 (25.7) 2 82.81 − ¬ ¼ 10 b + E = 44.663 + 355.310 = 399.973 ; b − E = 44.663 − 355.310 = −310.647 tc se (3.355)(150.669) = = 123.471 E= 2 2 (25.7) (¦ x) 82.81 − ¦ x2 − 10 n m + E = 187.660 + 123.471 = 311.131 ; m − E = 187.660 − 123.471 = 64.189 . Confidence intervals for B is −310.647 < B < 399.973 ; Confidence intervals for M is 64.189 < M < 311.131

38. E = tc se

9.4 MULTIPLE REGRESSION 9.4 TRY IT YOURSELF SOLUTIONS 1. Enter the data.

y = 46.385 + 0.540 x − 4.897 x 1 2

2. (1) y = 46.385 + 0.540 ( 89 ) − 4.897 (1) y = 89.548 = 90

(2) y = 46.385 + 0.540 ( 78 ) − 4.897 ( 3) y = 73.814 = 74 (3) y = 46.385 + 0.540 ( 83) − 4.897 ( 2 ) y = 81.411 = 81

9.4 EXERCISE SOLUTIONS 1. The predicted y value, y , will change by í8.2 when the variable x2 is changed by one unit and x1

and x3 are held constant. 2. The number of dependent variables is less than the number of independent variables in a multiple regression equation and equal to the number of independent variables in a single regression equation. Copyright © 2023 Pearson Education, Inc.

392

CHAPTER 9 Ň CORRELATION AND REGRESSION

y = 24,791 + 4.508 x − 4.723 x 1 2

(a) y = 24,791 + 4.508 ( 36,500 ) − 4.723 ( 36,100 ) = 18,832.7 pounds per acre (b) y = 24,791 + 4.508 ( 38,100 ) − 4.723 ( 37,800 ) = 18,016.4 pounds per acre (c) y = 24,791 + 4.508 ( 39,000 ) − 4.723 ( 38,800 ) = 17,350.6 pounds per acre (d) y = 24,791 + 4.508 ( 42, 200 ) − 4.723 ( 42,100 ) = 16,190.3 pounds per acre 4.

y = 80.1 − 20.2 x + 21.2 x 1 2

(a) y = 80.1 − 20.2 ( 5.5 ) + 21.2 ( 3.9 ) = 51.68 bushels per acre (b) y = 80.1 − 20.2 ( 8.3) + 21.2 ( 7.3) = 67.2 bushels per acre (c) y = 80.1 − 20.2 ( 6.5 ) + 21.2 ( 5.7 ) = 69.64 bushels per acre (d) y = 80.1 − 20.2 ( 9.4 ) + 21.2 ( 7.8) = 55.58 bushels per acre 5.

y = −52.2 + 0.3 x + 4.5 x 1 2

(a) y = −52.2 + 0.3 ( 70 ) + 4.5 (8.6 ) = 7.5 cubic feet (b) y = −52.2 + 0.3 ( 65 ) + 4.5 (11.0 ) = 16.8 cubic feet (c) y = −52.2 + 0.3 ( 83) + 4.5 (17.6 ) = 51.9 cubic feet (d) y = −52.2 + 0.3 ( 87 ) + 4.5 (19.6 ) = 62.1 cubic feet 6.

y = −4016 + 11.5 x + 7.55 x + 12.5 x 1 2 3

(a) y = −4016 + 11.5 ( 421) + 7.55 ( 224 ) + 12.5 (144 ) = 4316.7 kilograms (b) y = −4016 + 11.5 ( 311) + 7.55 (171) + 12.5 (102 ) = 2126.55 kilograms (c) y = −4016 + 11.5 ( 376 ) + 7.55 ( 226 ) + 12.5 (124 ) = 3564.3 kilograms

CHAPTER 9 Ň CORRELATION AND REGRESSION

393

(d) y = −4016 + 11.5 ( 231) + 7.55 (135) + 12.5 ( 86 ) = 734.75 kilograms 7. (a) y = 25,581.174 − 1314.611x1 − 72.451x2

(b) 2002.030 (c) 0.845 The standard error of estimate of the predicted price given the specific age and mileage of pre-owned Honda Civic Sedans is about $2002.03. The multiple regression model explains about 84.5% of the variation. 8. (a) y = −5.199 + 8.102 x1 + 4.271x2 + 5.275 x3

(b) 7.535 (c) 0.996 The standard error of estimate of the predicted calories given specific amounts of fat, carbs, and protein is about 7.535. The multiple regression model explains about 99.6% of the variation. 9.

n = 9 k = 2, r 2 = 0.845 ª (1 − 0.845)(9 − 1) º r 2 adj = 1 − « » ≈ 0.793 9 − 2 −1 ¬ ¼ About 79.3% of the variation in y can be explained by the relationship between variables; r 2 adj < r 2 .

10. n = 8 k = 3, r 2 = 0.996 ª (1 − 0.996)(8 − 1) º r 2 adj = 1 − « » = 0.993 8 − 3 −1 ¬ ¼

99.3% of the variation in y can be explained by the relationship between variables; r 2 adj < r 2 .

CHAPTER 9 REVIEW EXERCISE SOLUTIONS 1. (a)

394

CHAPTER 9 Ň CORRELATION AND REGRESSION

(b)

¦ r=

x 626 595 608 492 390 443 368 x = 3522

y 4581 4336 3803 3733 2942 2933 2657 y = 24,985

xy 2,867,706 2,579,920 2,312,224 1,836,636 1,147,380 1,299,319 977,776 xy =13,020,961

x2 391,876 354,025 369,664 242,064 152,100 196,249 135,424 2 x =1,841, 402

y2 20,985,561 18,800,896 14,462,809 13,935,289 8,655,364 8,602,489 7,059,649 2 y = 92,502,057

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

n ¦ y2 − ( ¦ y )

7(13,020,961) − (3522)(24,985) 7(1,841, 402) − (3522)2 7(92,502,057) − (24,985)2

3,149,557 ≈ 0.937 485,330 23, 264,174

(c) Strong positive linear correlation; As the number of pass attempts increase, the number of passing yards tends to increase. 2. (a)

(b)

x 47.6 63.3 68.2 67.7 71.5 58.1 50.5 59.0 x = 485.9

y 4.3 3.6 10.1 5.5 10.0 8.8 4.7 10.1 y = 57.1

xy 204.68 227.88 688.82 372.35 715 511.28 237.35 595.9 xy = 3,553.26

x2 2265.76 4006.89 4651.24 4583.29 5112.25 3375.61 2550.25 3481 2 x = 30,026.29

y2 18.49 12.96 102.01 30.25 100 77.44 22.09 102.01 y 2 = 465.25

CHAPTER 9 Ň CORRELATION AND REGRESSION

r= = =

395

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

n ¦ y2 − ( ¦ y )

8(3553.26) − (485.9)(57.1) 8(30,026.29) − (485.9) 2 8(465.25) − (57.1) 2 681.19 ≈ 0.494 4111.51 461.59

(c) Weak positive linear correlation; The number of wildland fires does not appear to be related to the number of wildland acres burned. 3. (a)

(b)

¦ r= = =

x 89 95 107 120 73 108 92 80 127 x =891

y 1400 1485 1570 1550 1446 1210 1620 1710 1570 y =13,561

xy 124,600 141,075 167,990 186,000 105,558 130,680 149,040 136,800 199,390 xy =1,341,133

x2 7921 9025 11,449 14,400 5329 11,664 8464 6400 16,129 x 2 =90,781

y2 1,960,000 2,205,225 2,464,900 2,402,500 2,090,916 1,464,100 2,624,400 2,924,100 2,464,900 y 2 =20,601,041

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

n ¦ y2 − ( ¦ y )

9 (1,341,133) − ( 891)(13,561) 9 ( 90,781) − ( 891)

9 ( 20,601,041) − (13,561)

−12,654 ≈ −0.068 23,148 1,508,648

396

CHAPTER 9 Ň CORRELATION AND REGRESSION

4. (a)

(b)

¦ r= = =

x 2.1 5.0 6.3 6.5 7.7 8.7 11.6 x =47.9

y 0.59 1.51 1.55 1.70 2.18 2.10 2.73 y =12.36

xy 1.239 7.55 9.765 11.05 16.786 18.27 31.668 xy =96.328

x2 4.41 25.00 39.69 42.25 59.29 75.69 134.56 x 2 =380.89

y2 0.3481 2.2801 2.4025 2.8900 4.7524 4.4100 7.4529 y 2 =24.536

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x2 − ( ¦ x )

n ¦ y2 − (¦ y )

7 ( 96.328 ) − ( 47.9 )(12.36 ) 7 ( 380.89 ) − ( 47.9 )

7 ( 24.536 ) − (12.36 )

82.252 ≈ 0.979 371.82 18.9824

H0 : ρ = 0; Ha : ρ ≠ 0 α = 0.05, d.f. = n − 2 = 5

−t0 = −2.571, t0 = 2.571; Rejection regions: t < −2.571 or t > 2.571 t=

0.937

≈ 5.998

1 − ( 0.937 ) 7−2 Because t > 2.571 , reject H 0 .There is enough evidence at the 5% level of significance to conclude that there is a significant linear correlation between a quarterback’s pass attempts and passing yards. 6.

1− r n−2

H0 : ρ = 0; Ha : ρ ≠ 0 α = 0.05, d.f. = n − 2 = 6

−t0 = −2.447, t0 = 2.447; Rejection regions: t < −2.447 or t > 2.447

CHAPTER 9 Ň CORRELATION AND REGRESSION

0.494

≈ 1.392 2 1 − ( 0.494 ) 8−2 Because −2.447 < t < 2.447 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to conclude that there is a significant linear correlation between the number of wildland fires and the number of acres burned. 7.

1− r2 n−2

397

H0 : ρ = 0; Ha : ρ ≠ 0

α = 0.01, d.f. = n − 2 = 7

−t0 = −3.499, t0 = 3.499; Rejection regions: t < −3.499 or t > 3.499 t=

1− r n−2

−0.068

≈ −0.180 2 1 − ( −0.068 ) 9−2 Because −3.499 < t < 3.499 , fail to reject H 0 . There is not enough evidence at the 1% level of significance to conclude that a significant linear correlation between IQ scores and brain sizes. 2

H0 : ρ = 0; Ha : ρ ≠ 0

α = 0.01, d.f. = n − 2 = 5

−t0 = −4.032, t0 = 4.032; Rejection regions: t < −4.032 or t > 4.032 t=

0.979

≈ 10.738 2 1 − ( 0.979 ) 7−2 Because t > 4.032 , reject H 0 . There is enough evidence at the 1% level of significance to conclude that there is a linear correlation between sugar consumption and number of cavities. 9.

1− r2 n−2

≈

y = 0.097 x − 695.161

(a) It is not meaningful to predict the value of y for x = 9080 because x = 9080 is outside the range of the original data. (b) y = 0.097(9230) − 695.161 = 200.149 billions of pounds (c) y = 0.097(9340) − 695.161 = 210.819 billions of pounds (d) y = 0.097(9400) − 695.161 = 216.639 billions of pounds

398

CHAPTER 9 Ň CORRELATION AND REGRESSION

10. y = 0.669 x + 0.755

(a) y = 0.669(1.97) + 0.755 ≈ 2.073 hours (b) y = 0.669(2.21) + 0.755 ≈ 2.233 hours (c) y = 0.669(2.27) + 0.755 ≈ 2.274 hours (d) It is not meaningful to predict the value of y when x = 2.43 because x = 2.43 is outside the range of the original data. 11. y = −0.086 x + 10.450

(a) It is not meaningful to predict the value of y for x = 16 because x = 16 is outside the range of the original data. (b) y = −0.086(25) + 10.450 = 8.3 hours (c) It is not meaningful to predict the value of y for x = 85 because x = 85 is outside the range of the original data. (d) y = −0.086(50) + 10.450 = 6.15 hours

CHAPTER 9 Ň CORRELATION AND REGRESSION

399

12. y = −0.090 x + 44.675

(a) It is not meaningful to predict the value of y for x = 86 because x = 86 is outside the range of the original data. (b) y = −0.090(198) + 44.675 = 26.855 miles per gallon (c) y = −0.090(289) + 44.675 = 18.665 miles per gallon (d) It is not meaningful to predict the value of y for x = 407 because x = 407 is outside the range of the original data. 13. r 2 = ( −0.450 ) ≈ 0.203 About 20.3% of the variation in y is explained. About 79.7% of the variation in y is unexplained. 2

14. r 2 = ( −0.937 ) ≈ 0.878 About 87.8% of the variation in y is explained. About 12.2% of the variation in y is unexplained. 2

15. r 2 = ( 0.642 ) ≈ 0.412 About 41.2% of the variation in y is explained. About 58.8% of the variation in y is unexplained. 2

16. r 2 = ( 0.795 ) ≈ 0.632 About 63.2% of the variation in y is explained. About 36.8% of the variation in y is unexplained. 2

( 17. (a) r = 2

¦ y i − y

(

¦ yi − y

) ≈ 0.876 2

)

About 87.6% of the variation is explained by the relationship between the time to accelerate from 0 to 60 miles per hour and top speed for electric cars and about 12.4% of the variation is unexplained.

400

CHAPTER 9 Ň CORRELATION AND REGRESSION

(b) se =

(

¦ yi − l yi

) = 531.4304 ≈ 9.411 2

6 n−2 The standard error of estimate of the top speed for a specific time to accelerate from 0 to 60 miles per hour is about 9.411 miles per hour.

( 18. (a) r = 2

¦ y i − y

(

¦ yi − y

) ≈ 0.62 2

)

About 62% of the variation is explained by the relationship between cooking areas of gas grills and their prices and about 38% of the variation is unexplained.

(b) se =

(

¦ yi − l yi

) = 586,797.9814 ≈ 191.51. 2

16 n−2 The standard error of estimate of the price for a specific cooking area of a gas grill is about $191.51.

19. y = 0.097(9275) − 695.161 = 204.514

(

)

n x−x 1 1 8(9275 − 9333.5) 2 = + + (1.943)(3.754) 1 E = tc se 1 + + 8 8(696,943,606) − (74,668) 2 n n ¦ x 2 − ( ¦ x )2

(

)

≈ 8.121 y + E = 204.514 + 8.121 = 212.635 ; y − E = 204.514 − 8.121 = 196.393 196.393 < y < 212.635 You can be 90% confident that the amount of milk produced will be between 196.393 billion pounds and 212.635 billion pounds when the average number of cows is 9275 thousand.

20. y = 0.669(2.25) + 0.755 = 2.26025

(

)

n x−x 1 1 10(2.25 − 2.092) 2 = (1.860)(0.053) 1 + + ≈ 0.114 E == tc se 1 + + 10 10(43.8722) − (20.92)2 n n ¦ x 2 − ( ¦ x )2

(

)

y − E = 2.26025 − 0.114 ≈ 2.146 ; y + E = 2.26025 + 0.114 ≈ 2.374

2.146 < y < 2.374 You can be 90% confident that the average time per day spent watching television for adults ages 35 to 44 will be between 2.146 hours and 2.347 hours when the average time per day spent watching television for adults ages 25 to 34 is 2.25 hours.

CHAPTER 9 Ň CORRELATION AND REGRESSION

401

21. y = −0.086 ( 45 ) + 10.450 = 6.580

(

)

2 n x−x 7 ( 45 − 337 / 7 ) 1 1 ≈ ( 2.571)( 0.622 ) 1 + + E = tc se 1 + + 7 7 (18,563) − ( 337 )2 n n ¦ x 2 − ( ¦ x )2

(

)

≈ 1.714 y ± E → 6.580 ± 1.714 → ( 4.866, 8.294 ) You can be 95% confident that the hours slept will be between 4.866 and 8.294 hours for a person who is 45 years old.

22. y = −0.090 ( 265 ) + 44.675 = 20.825

(

)

2 n x−x 7 ( 265 − 1516 / 7 ) 1 1 ≈ ( 2.571)(1.476 ) 1 + + E = tc se 1 + + 7 7 ( 369,382 ) − (1516 )2 n n ¦ x 2 − ( ¦ x )2

(

)

≈ 4.157

y ± E → 20.825 ± 4.157 → (16.668, 24.982 ) You can be 95% confident that the fuel efficiency will be between 16.668 miles per gallon and 24.982 miles per gallon when the engine displacement is 265 cubic inches. 23. y = −14.399(5.9) + 196.996 = 112.0419 E = (3.707)(9.411) 1 +

1 8(5.9 − 4.6875) 2 + ≈ 38.315 8 8(193.89) − (37.5) 2

y − E = 112.0419 − 38.315 ≈ 73.727 y + E = 112.0419 + 38.315 ≈ 150.357

73.727 < y < 150.357 You can be 99% confident that the top speed of an electric car will be between 73.727 miles per hour and 150.357 miles per hour when it has a time of 5.9 seconds to accelerate from 0 to 60 miles per hour. 24. y = 1.501(900) − 341.501 = 1009.399

(

)

n x−x 1 1 18(900 − 604.333) 2 ≈ + + (2.921)(191.51) 1 E = tc se 1 + + 18 18(6,999,174) − (10,878) 2 n n ¦ x 2 − ( ¦ x )2

(

)

≈ 628.208

y ± E → 1009.399 ± 628.208 → ( 381.191,1637.607 ) 381.191 < y < 1637.607 You can be 99% confident that the price of a gas grill will be between $381.19 and $1637.61 when the cooking area is 900 square inches.

402

CHAPTER 9 Ň CORRELATION AND REGRESSION

25. (a) y = 1.64 + 3.47 x1 − 32.3 x2

(b) se ≈ 0.838; The standard error of estimate of the carbon monoxide content given specific contents of tar and nicotine is about 0.838 milligram. (c) r 2 ≈ 0.943; The multiple regression model explains about 94.3% of the variation. 26. (a) y = 59,837 + 0.0123 x1 − 0.0282 x2

(b) se ≈ 360.394; The standard error of estimate of the annual yield of potatoes given specific number of acres planted and number of acres harvested is about 360.394 pounds. (c) r 2 ≈ 0.935; The multiple regression model explains about 93.5% of the variation. 27. (a) 21.705 miles per gallon

(b) 25.21 miles per gallon

(d) 25.86 miles per gallon

28. y = 1.64 + 3.47 x1 − 32.3 x2 : (a) 10.26 milligrams; y = 1.64 + 3.47(9.0) − 32.3(0.70) = 10.26

(b) 3.975 milligrams; y = 1.64 + 3.47(3.0) − 32.3(0.25) = 3.975 (c) 10.02 milligrams; y = 1.64 + 3.47(8.0) − 32.3(0.60) = 10.02 (d) 4.826 milligrams; y = 1.64 + 3.47(5.2) − 32.3(0.46) = 4.826

CHAPTER 9 QUIZ SOLUTIONS 1.

The data appear to have a positive linear correlation. As x increases, y tends to increase. 2.

r ≈ 0.995; Strong positive linear correlation; As the average annual salaries of secondary school teachers increase, the average annual salaries of elementary school teachers tend to increase.

CHAPTER 9 Ň CORRELATION AND REGRESSION

403

H0 : ρ = 0; Ha : ρ ≠ 0

α = 0.05, d.f. = n − 2 = 9

−t0 = −2.262, t0 = 2.262; Rejection regions: t < −2.262 or t > 2.262 r

0.995

≈ 29.89 2 1 − ( 0.995 ) 11 − 2 Because t > 2.262 , reject H 0 . There is enough evidence at the 5% level of significance to conclude that there is a significant linear correlation between the average annual salaries of secondary school teachers and the average annual salaries of elementary school teachers. 1− r2 n−2

≈

x 56.0 56.8 57.8 58.3 59.3 60.4 61.4 62.9 64.3 65.9 67.3

y 54.3 55.3 56.1 56.3 56.8 57.7 59.0 60.8 62.2 63.9 65.4

xy 3040.8 3141.04 3242.58 3282.29 3368.24 3485.08 3622.6 3824.32 3999.46 4211.01 4401.42

¦ x = 670.4

¦ y = 647.8

¦ xy = 39,618.84

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x − (¦ x) 2

b = y − mx =

11(39,618.84) − (670.4)(647.8) ≈ 0.976 11(40,999.58) − (670.4) 2

¦y § ¦ x · 647.8 § 1522.12 ·§ 670.4 · − m¨ −¨ ¸≈ ¸¨ ¸ ≈ −0.604 n 11 © n ¹ © 1559.22 ¹© 11 ¹

y = 0.976 x − 0.604

x2 3136 3226.24 3340.84 3398.89 3516.49 3648.16 3769.96 3956.41 4134.49 4342.81 4529.29 2 ¦ x = 40,999.58

y = 0.976(63.5) − 0.604 = 61.372 $ 61,372

404

CHAPTER 9 Ň CORRELATION AND REGRESSION

r 2 ≈ ( 0.9954 ) ≈ 0.991 About 99.1% of the variation in the average annual salaries of elementary school teachers can be explained by the relationship between the average annual salaries of secondary school teachers and elementary school teachers, and about 0.9% of the variation is unexplained.

se ≈ 0.372;

The standard error of estimate of the average annual salaries of elementary school teachers for a specific average annual salary of secondary school teachers is about $372. 8.

y = 0.976(63.5) − 0.604 = 61.372

se =

1.24865088 ≈ 0.372 9

E = (2.262)(0.372) 1 +

1 11(63.5 − 60.94545455) 2 + ≈ 0.897 11 11(40,999.58) − (670.4) 2

y − E = 61.372 − 0.897 = 60.475 y + E = 61.372 + 0.897 = 62.269

60.475 < y < 62.269 You can be 95% confident that the average annual salary of elementary school teachers is between $60,475 and $62,269 when the average annual salary of secondary school teachers is $63,500. 9. (a) y = −86 + 7.46 ( 27.6 ) − 1.61(15.3) = $95.26

(b) y = −86 + 7.46 ( 24.1) − 1.61(14.6 ) = $70.28 (c) y = −86 + 7.46 ( 23.5 ) − 1.61(13.4 ) = $67.74 (d) y = −86 + 7.46 ( 22.8 ) − 1.61(15.3) = $59.46

CHAPTER 9 TEST SOLUTIONS 1. (a) y = 23,769 + 9.18 (1057 ) − 8.41( 3698 ) = $2372.08 million

(b) y = 23,769 + 9.18 (1012 ) − 8.41( 3659 ) = $2286.97 million (c) y = 23,769 + 9.18 ( 952 ) − 8.41( 3601) = $2223.95 million (d) y = 23,769 + 9.18 ( 914 ) − 8.41( 3594 ) = $1933.98 million

CHAPTER 9 Ň CORRELATION AND REGRESSION

405

The data appear to have a positive linear correlation. As x increases, y tends to increase. 3.

r ≈ 0.900; Strong positive linear correlation; As the average annual salaries of librarians increase, the average annual salaries of library science teachers tend to increase.

H0 : ρ = 0; Ha : ρ ≠ 0

α = 0.01, d.f. = n − 2 = 10

−t0 = −3.169, t0 = 3.169; Rejection regions: t < −3.169 or t > 3.169 t=

¦ m=

1− r n−2

≈

0.900

≈ 6.53 2 1 − ( 0.900 ) 12 − 2 Because t > 3.169 , reject H 0 . There is enough evidence at the 1% level of significance to conclude that there is a significant linear correlation between the average annual salaries of librarians and the average annual salaries of library science teachers. 2

x 55.7 56.4 57.0 57.2 57.6 58.1 58.9 59.9 60.8 61.5 61.9 63.6 x = 708.6

y 64.3 67.0 70.0 70.8 73.3 72.4 73.0 72.3 73.7 75.5 77.2 77.6 y = 867.1

n ¦ xy − ( ¦ x )( ¦ y ) n ¦ x − (¦ x) 2

xy 3581.51 3778.8 3990 4049.76 4222.08 4206.44 4299.7 4330.77 4480.96 4643.25 4778.68 4935.36 xy = 51, 297.31

x2 3102.49 3180.96 3249 3271.84 3317.76 3375.61 3469.21 3588.01 3696.64 3782.25 3831.61 4044.96 2 x = 41,910.34

12(51, 297.31) − (708.6)(867.1) 1140.66 = ≈ 1.408 810.12 12(41,910.34) − (708.6) 2

406

CHAPTER 9 Ň CORRELATION AND REGRESSION

b = y − mx =

¦y § ¦ x · 867.1 § 1140.66 ·§ 708.6 · − m¨ −¨ ¸≈ ¸¨ ¸ ≈ −10.885 n 12 © n ¹ © 810.12 ¹© 12 ¹

y = 1.408 x − 10.885

y = 1.408(61) − 10.885 = 75.003 $75,003

r 2 ≈ ( 0.8995 ) ≈ 0.809 About 80.9% of the variation in the average annual salaries of library science teachers can be explained by the relationship between the average annual salaries of librarians and library science teachers, and about 19.1% of the variation is unexplained.

se ≈ 1.777; The standard error of estimate of the average annual salary of library science teachers for

a specific average annual salary of librarians is about $1777.

9. y = 1.408(61) − 10.885 = 75.003 E = (3.169)(1.777) 1 +

1 12(61 − 59.05) 2 + ≈ 6.012 12 12(41,910.34) − (708.6) 2

y − E = 75.003 − 6.012 = 68.991 ; y + E = 75.003 + 6.012 = 81.015 68.991 < y < 81.015 ; You can be 99% confident that the average annual salary of library science teachers is between $68,991 and $81,015 when the average annual salary of librarians is $61,000.

CHAPTER

Chi-Square Tests and the F-Distribution

10.1 GOODNESS-OF-FIT TEST 10.1 TRY IT YOURSELF SOLUTIONS 1.

Tax Preparation Method Accountant By hand Computer software Friend/family Tax preparation service

% of people 24% 20% 35% 6% 15%

Expected frequency 500(0.24) = 120 500(0.20) = 100 500(0.35) = 175 500(0.06) = 30 500(0.15) = 75

2. The expected frequencies are 64, 80, 32, 56, 60, 48, 40, and 20, all of which are at least 5. Claimed distribution: Ages 0−9 10−19 20−29 30−39 40−49 50−59 60−69 70+

Distribution 16% 20% 8% 14% 15% 12% 10% 5%

H 0 : The distribution of the ages are 16% ages 0−9, 20% ages 10−19, 8% ages 20−29, 14% ages

30−39, 15% ages 40−49, 12% ages 50−59, 10% ages 60-69 and 5% ages 70+ (claim). H a : The distribution of ages differs from the claimed distribution.

α = 0.05 d.f. = n − 1 = 7 2 χ0 = 14.067 ; Rejection region: χ 2 > 14.067 Ages 0−9 10−19 20−29 30−39 40−49 50−59 60−69 70+

Distribution

Observed

Expected

(O E)2 E

16% 20% 8% 14% 15% 12% 10% 5%

76 84 30 60 54 40 42 14

64 80 32 56 60 48 40 20

2.250 0.200 0.125 0.286 0.600 1.333 0.100 1.800 407

408

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

χ 2 ≈ 6.694

Because χ < 14.067 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to support the sociologist’s claim that the distribution of ages differs from the age distribution 10 years ago. 3. The expected frequency for each category is 30 which is at least 5. Claimed distribution: Color Brown Yellow Red Blue Orange Green

Distribution 16.6% 16.6% 16.6% 16.6% 16.6% 16.6%

H 0 : The distribution of colors is uniform. (claim) H a : The distribution of colors is not uniform. α = 0.05 d.f. = n − 1 = 5

χ02 = 11.071 ; Rejection region: χ 2 > 11.071 Color

Distribution

Observed

Expected

Brown Yellow Red Blue Orange Green

16.6% 16.6% 16.6% 16.6% 16.6% 16.6%

22 27 22 41 41 27

30 30 30 30 30 30

χ 2 ≈ 12.933

(O E)2 E 2.133

0.300 2.133 4.033 4.033

0.300 12.933

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

409

Because χ > 11.071 , reject H 0 . There is enough evidence at the 5% level of significance to reject the claim that the distribution of different colored candies in bags of peanut M&M’s is uniform.

10.1 EXERCISE SOLUTIONS 1. A multinomial experiment is a probability experiment consisting of a fixed number of independent trials in which there are more than two possible outcomes for each trial. The probability of each outcome is fixed, and each outcome is classified into categories. 2. The observed frequencies must be obtained using a random sample, and each expected frequency must be greater than or equal to 5. 3.

Ei = npi = (150)(0.3) = 45

Ei = npi = (500)(0.9) = 450

Ei = npi = (230)(0.25) = 57.5

Ei = npi = (415)(0.08) = 33.2

7. (a) Claimed Distribution: Age 2−11 12−17 18−24 25−39 40–49 50+

Distribution 15% 11% 11% 25% 13% 25%

H 0 : The distribution of the ages of moviegoers is 15% ages 2−11, 11% ages 12−17, 11% ages

18−24, 25% ages 25−39, 13% ages 40–49, and 25% ages 50+. (claim) H a : The distribution of the ages differs from the claimed or expected distribution.

(b) χ 02 = 9.236 , Rejection region: χ 2 > 9.236

410

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(c)

Age

Distribution

Observed

Expected

2−11 12−17 18−24 25−39 40–49 50+

15% 11% 11% 25% 13% 25%

146 116 131 232 117 258

150 110 110 250 130 250

(O E )2 E 0.1067 0.3273 4.0091 1.296 1.3 0.256 7.295

χ 2 ≈ 7 .2 9 5

(d) Because χ 2 < 9 .2 3 6 , fail to reject H 0 . (e) There is not enough evidence at the 10% level of significance to reject the claim that the distribution of the ages of moviegoers and the expected distribution are the same. 8. (a) Claimed Distribution: Response 0 cups 1 cup 2 cups 3 cups 4 or more cups

Distribution 36% 26% 19% 8% 11%

H 0 : The distribution of the number of cups of coffee U.S. adults drink per day is 36% 0 cups,

26% 1 cup, 19% 2 cups, 8% 3 cups, and 11% 4 or more cups. (claim) H a : The distribution of amounts differs from the expected distribution.

(b) χ 02 = 9 .4 8 8 , Rejection region: χ 2 > 9 .4 8 8 (c)

Response

Distribution

Observed

Expected

0 cups 1 cup 2 cups 3 cups 4 or more cups

36% 26% 19% 8% 11%

570 432 282 152 164

576 416 304 128 176

(O E )2 E

0.0625 0.6154 1.5921 4.5 0.8182 7.5882

χ 2 ≈ 7 .5 8 8 2

(d) Because χ 2 < 9 .4 8 8 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to reject the claim that the distribution of the number of cups of coffee U.S. adults drink per day and the expected distribution are the same. Copyright © 2023 Pearson Education, Inc.

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

411

9. (a) Claimed distribution: Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday

Distribution 14% 10% 11% 12% 13% 21% 19%

H 0 : The distribution of the days people order pizza is 14% Sunday, 10% Monday, 11%

Tuesday, 12% Wednesday, 13% Thursday, 21% Friday, and 19% Saturday. H a : The distribution of days differs from the claimed or expected distribution. (claim)

(b) χ 02 = 1 6 .8 1 2 , Rejection region: χ 2 > 1 6 .8 1 2 (c)

Day

Distribution

Observed

Expected

Sunday Monday Tuesday Wednesday Thursday Friday Saturday

14% 10% 11% 12% 13% 21% 19%

54 32 44 75 59 114 122

70 50 55 60 65 105 95

(O E )2 E

3.6571 6.48 2.2 3.75 0.5538 0.7714 7.6737 25.086

χ 2 ≈ 25.086

(d) Because χ 2 > 16.812 , reject H 0 . (e) There is enough evidence at the 1% level of significance to conclude that the distribution of days differs from the expected distribution. 10. (a) Claimed Distribution: Response Cash Debit or credit Check Digital wallet/other

Distribution 29% 59% 5% 7%

H 0 : The distribution of people’s preferences on how to pay for goods is 29% cash, 59% debit or

credit, 5% check, and 7% digital wallet/other. H a : The distribution of preferred ways to pay differs from the expected distribution. (claim)

412

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(b) χ 02 = 1 1 .3 4 5 , Rejection region: χ 2 > 1 1 .3 4 5 (c)

Response

Distribution

Observed

Expected

Cash Debit or credit Check Digital wallet/other

29% 59% 5% 7%

194 338 21 47

174 354 30 42

(O E )2 E

2.2989 0.7232 2.7 0.5952 6.317

χ 2 ≈ 6.317

(d) Because χ 2 < 1 1 .3 4 5 , fail to reject H 0 . (e) There is not enough evidence at the 1% level to conclude that the distribution of preferred ways to pay for goods differs from the expected distribution. 11. (a) Claimed Distribution: County Alameda Contra Costa Fresno Kern Los Angeles Monterey Orange Riverside Sacramento San Bernardino San Diego San Francisco San Joaquin Santa Clara Stanislaus Tulare

Distribution 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25%

H 0 : The distribution of the number of homicide crimes in California by county is uniform.

(claim) The distribution of homicides by county is not uniform. Ha : (b) χ 02 = 30.57 8 , Rejection region: χ 2 > 30.57 8

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(c)

County

Distribution

Observed

Expected

Alameda Contra Costa Fresno Kern Los Angeles Monterey Orange Riverside Sacramento San Bernardino San Diego San Francisco San Joaquin Santa Clara Stanislaus Tulare

6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25% 6.25%

58 44 53 79 151 34 40 86 62 98 69 38 55 49 45 39

62.5 62.5 62.5 62.5 62.5 62.5 62.5 62.5 62.5 62.5 62.5 62.5 62.5 62.5 62.5 62.5

413

(O E )2 E

0.324 5.476 1.444 4.356 125.316 12.996 8.1 8.836 0.004 20.164 0.676 9.604 0.9 2.916 4.9 8.836 214.848

χ 2 ≈ 2 1 4 .8 4 8

(d) Because χ 2 > 30.578 , reject H 0 . (e) There is enough evidence at the 1% level of significance to reject the claim that the distribution of the number of homicide crimes in California by county is uniform. 12. (a) Claimed Distribution: Month January February March April May June July August September October November December

Distribution 8.33% 8.33% 8.33% 8.33% 8.33% 8.33% 8.33% 8.33% 8.33% 8.33% 8.33% 8.33%

H 0 : The distribution of the number of homicide crimes in California by month is uniform.

(claim) H a : The distribution of homicides by month is not uniform. 2 (b) χ 02 = 1 7 .2 7 5 , Rejection region: χ > 17.275

414

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(c)

Month

Distribution

Observed

Expected

January February March April May June July August September October November December

8.33% 8.33% 8.33% 8.33% 8.33% 8.33% 8.33% 8.33% 8.33% 8.33% 8.33% 8.33%

134 111 133 134 158 161 184 181 205 215 184 200

166.67 166.67 166.67 166.67 166.67 166.67 166.67 166.67 166.67 166.67 166.67 166.67

(O E )2 E

6.403845323 18.59452151 6.801877362 6.403845323 0.4510043799 0.1928895422 1.801937361 1.232068759 8.814957101 14.01445311 1.801937361 6.665200096 73.18

χ 2 ≈ 73.18

(d) Because χ 2 > 1 7 .2 75 , reject H 0 . (e) There is enough evidence at the 10% level of significance to reject the claim that the distribution of the number of homicide crimes in California by month is uniform. 13. (a) Claimed Distribution: Response Four-year college Two-year college Non-college training program All other pathways

Distribution 46% 7% 19% 28%

H 0 : The distribution of the opinions of Democrat U.S. parents on what path they would prefer

their child take after high school is 46% four-year college, 7% two-year college, 19% noncollege training program, and 28% all other pathways. H a : The distribution of opinions differs from the expected distribution. (claim) (b) χ 02 = 7.815 , Rejection region: χ 2 > 7.815

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(c)

Response Four-year college Two-year college Non-college training program All other pathways

Distribution

Observed

Expected

(O E )2 E

46% 7% 19%

140 22 14

92 14 38

25.0435 4.5714 15.1579

28%

18.2857 63.059

415

χ 2 ≈ 63.059

(d) Because χ 2 > 7.815 , reject H 0 . (e) There is enough evidence at the 5% level of significance to support the claim that the distribution of opinions of Democrats and Republicans is different. 14. (a) Claimed distribution: Response Completely trust Trust with certain aspects Do not trust Not sure

Distribution 65.6% 27.8% 5.7% 0.9%

H 0 : The distribution of how much married U.S. female adults trust their spouses to manage their

finances is 65.6% completely trust, 27.8% trust with certain aspects, 5.7% do not trust, and 0.9% not sure. (claim) H a : The distribution of how much married U.S. female adults trust their spouses to manage their finances differs from the distribution of how much married U.S. male adults trust their spouses to manage their finances. (b) χ 02 = 6 .251 , Rejection region: χ 2 > 6.2 51 (c)

Response Completely trust Trust with certain aspects Do not trust Not sure

χ 2 ≈ 29.057 2 (d) Because χ > 6.251 , reject H 0 .

Distribution

Observed

Expected

65.6% 27.8% 5.7% 0.9%

243 108 36 13

262 111 23 4

(O E )2 E 1.3779 0.0811 7.3478 20.2500 29.0568

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CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(e) There is enough evidence at the 10% level of significance to reject the claim that the distribution of how much married U.S. female adults trust their spouses to manage their finances is the same as the distribution of how much married U.S. male adults trust their spouses to manage their finances. 15. (a) Claimed distribution: Response Larger Same size Smaller

Distribution 33. 3% 33. 3% 33. 3%

H 0 : The distribution of prospective home buyers by the size they want their next house to be is

uniform. H a : The distribution of prospective home buyers by the size they want their next house to be is

not uniform. (claim) 2

(b) χ0 = 5.991, Rejection region: χ > 5.991 (c)

Response

Distribution

Observed

Expected

Larger Same size Smaller

33. 3% 33. 3% 33. 3%

285 224 291

266.66 266.66 266.66

(O E )2 E

1.2604 6.8267 2.2204 10.3075

χ 2 ≈ 10.308 2 (d) Because χ > 5.991 , reject H 0 .

(e) There is enough evidence at the 5% level of significance to conclude that the distribution of prospective home buyers by the size they want their next house to be is not uniform. 16. (a) Claimed Distribution: Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday

Distribution 14.29% 14.29% 14.29% 14.29% 14.29% 14.29% 14.29%

H 0 : The distribution of the number of births by day of the week is uniform. (claim) H a : The distribution of the number of births by day of the week is not uniform.

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

417

2 2 (b) χ0 = 10.645 , Rejection region: χ > 10.645

Distribution

Observed

Expected

Sunday Monday Tuesday Wednesday Thursday Friday Saturday

14.29% 14.29% 14.29% 14.29% 14.29% 14.29% 14.29%

69 106 113 111 112 111 78

100 100 100 100 100 100 100

(O E ) 2 E 9.61 0.36 1.69 1.21 1.44 1.21 4.84 20.36

χ 2 ≈ 20.36 2 (d) Because χ > 10.645 , reject H 0 .

(e) There is enough evidence at the 10% level of significance to reject the claim that the distribution of the number of births by day of the week is uniform. 17. (a) Frequency distribution: μ = 69.435; σ ≈ 8.337 Lower Boundary 49.5 58.5 67.5 76.5 85.5 Class Boundaries 49.5−58.5 58.5−67.5 67.5−76.5 76.5−85.5 85.5−94.5

Upper Boundary 58.5 67.5 76.5 85.5 94.5

Distribution 8.67% 31.39% 39.33% 17.09% 2.55%

Lower z-score −2.39 −1.31 −0.23 0.85 1.93

Frequency 19 61 82 34 4 200

H 0 : Test scores have a normal distribution. (claim) H a : Test scores do not have a normal distribution. 2 2 (b) χ0 = 13.277 ; Rejection region. χ > 13.277

Upper z-score −1.31 −0.23 0.85 1.93 3.01

Expected 17 63 79 34 5

Area 0.0867 0.3139 0.3933 0.1709 0.0255

(O − E ) 2 E 0.2353 0.0635 0.1139 0 0.2000 0.6127

418

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

2 (c) χ = 0.613 2 (d) Because χ < 13.277 , fail to reject H 0 .

(e) There is not enough evidence at the 1% level of significance to reject the claim that the test scores are normally distributed. 18. (a) Frequency distribution: μ = 74.325; σ ≈ 10.434 Lower Upper Lower Boundary Boundary z-score 50.5 60.5 −2.28 60.5 70.5 −1.32 70.5 80.5 −0.37 80.5 90.5 0.59 90.5 100.5 1.55 Class Boundaries 50.5−60.5 60.5−70.5 70.5−80.5 80.5−90.5 90.5−100.5

Distribution 8.21% 26.23% 36.67% 21.70% 5.45%

Frequency 46 88 151 97 18 400

Upper z-score −1.32 −0.37 0.59 1.55 2.51

Expected 33 105 147 87 22

Area 0.0821 0.2623 0.3667 0.2170 0.0545

(O − E ) 2 E 5.1212 2.7524 0.1088 1.1494 0.7273 9.859

H 0 : Test scores have a normal distribution. (claim) H a : Test scores do not have a normal distribution. 2 2 (b) χ0 = 9.488 ; Rejection region. χ > 9.488 2 (c) χ = 9.859 2 (d) Because χ > 9.488 , reject H 0 .

(e) There is enough evidence at the 5% level of significance to reject the claim that the test scores are normally distributed.

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

10.2 INDEPENDENCE 10.2 TRY IT YOURSELF SOLUTIONS 1. Borrowed Did not borrow Total

With parents, rent-free 364 484

With parents, pay rent 92 58

On campus

Off campus, alone 42 28

Total

303 236

Off campus, w/others 203 186

848

150

539

389

1996

With parents, rent-free 426.549 421.451

With parents, pay rent 75.451 74.549

On campus

1004 992

n = 1996

Borrowed Did not borrow

271.120 267.880

Off campus, w/others 195.669 193.331

2. The claim is that “the student’s living arrangement depends on borrowing status.” H 0 : The student’s living arrangement is independent of borrowing status. H a : The student’s living arrangement is dependent on borrowing status. (claim) α = 0.01 (r − 1)(c − 1) = 4

χ02 = 13.277; Rejection region: χ 2 > 13.277 O

O−E

( O E )2

364 92 303 203 42 484 58 236 186 28

426.549 75.451 271.120 195.669 35.210 421.451 74.549 267.880 193.331 34.790

í62.549 16.549 31.88 7.331 6.79 62.549 í16.549 í31.88 í7.331 í6.79

3912.377401 273.869401 1016.3344 53.743561 46.1041 3912.377401 273.869401 1016.3344 53.743561 46.1041

χ 2 ≈ 36.489

(O E ) 2 E 9.172164044 3.629765026 3.74865152 0.2746656905 1.309403579 9.283113342 3.673683094 3.793991339 0.2779872912 1.325211268 36.489

Off campus, alone 35.210 34.790

419

420

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

Because χ > 13.277 , reject H 0 . There is enough evidence at the 1% level of significance to conclude that student’s living arrangement depends on borrowing status. 3.

H 0 : Whether or not a tax credit would influence an adult to make a charitable donation is

independent of age. H a : Whether or not a tax credit would influence an adult to make a charitable donation is dependent on age. (claim) Enter the data.

χ02 = 9.210; Rejection region: χ 2 > 9.210

χ 2 ≈ 15.306 2

Because χ > 9.210 , reject H 0 . There is enough evidence at the 1% level of significance to conclude that whether or not a tax credit would influence an adult to make a charitable donation is dependent on age.

10.2 EXERCISE SOLUTIONS 1. Find the sum of the row and the sum of the column in which the cell is located. Find the product of these sums. Divide the product by the sample size. 2. In a contingency table, a marginal frequency is the frequency that an entire category of a variable occurs, whereas a joint frequency is a frequency from a cell in the interior of a contingency table. 3. Answer will vary. Sample answer: For both the chi-square test for independence and the chi-square goodness-of-fit test, you are testing a claim about data that are in categories. However, the chi-square goodness-of-fit test has only one data value per category, while the chi-square test for independence has multiple data values per category. Both tests compare observed and expected frequencies. However, the chi-square goodness-of-fit test simply compares the distributions, whereas the chi-square test for independence compares them and then draws a conclusion about the dependence or independence of the variables.

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

421

4. A chi-square independence test is always a right-tailed test because if the variables are dependent, then the chi-square test statistic will be large, which is evidence for rejecting the null hypothesis. 5. False. If the two variables of a chi-square test for independence are dependent, then you can expect a large difference between the observed frequencies and the expected frequencies. 6. True. 7. (a) Result Inquiry No Inquiry

Athlete has Stretched Not stretched 18 22 211 189 229 211

(b) Result Inquiry No Inquiry

Athlete has Stretched 20.82 208.18

Not stretched 19.18 191.82

8. (a) Result Nausea No Nausea Total

Drug 36 254 290

(b)

Athlete has Placebo 13 262 275

Nausea No Nausea 9. (a)

New procedure 92 76 168

Preference Old Procedure No preference 351 50 42 8 393 58

(b) Bank employee Teller Customer service 10. (a) Size Seats 100 or fewer Seats over 100

Total 49 516 565

Athlete has Drug Placebo 25.15 23.85 264.85 251.15

Result

Bank employee Teller Customer service Total

Total 40 400 440

New procedure 133.80 34.20

Preference Old Procedure 313.00 80.00

Rating Excellent 182 180 362

Fair 203 311 514

Total 493 126 619

No preference 46.19 11.81 Poor 165 159 324

Total 550 650 1200

422

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(b)

Rating Excellent 165.92 196.08

Size Seats 100 or fewer Seats over 100 11. (a)

Fair 235.58 278.42

Poor 148.50 175.50

Type of car SUV 21 20 41

Truck/Van 22 14 36 Truck/Van 19.8 16.2

Gender Male Female

Compact 28 24 52

Full-size 39 32 71

Gender Male Female

Compact 28.6 23.4

Type of car Full-size SUV 39.05 22.55 31.95 18.45

(b)

12. (a)

Total 110 90 200

Age Type of movie rented Comedy Action Drama Total

18−24 38 15 12 65

25−34 30 17 11 58

(b)

35−44 24 16 19 59

45−64 10 9 25 44

65 and older 8 5 13 26

Total 110 62 80 252

Age Type of movie rented Comedy Action Drama

18−24 28.37 15.99 20.63

25−34 25.32 14.27 18.41

35−44 25.75 14.52 18.73

45−64 19.21 10.83 13.97

65 and older 11.35 6.40 8.25

13. (a) The claim is “an athlete’s injury result is not related to whether or not the athlete has stretched.” H 0 : An athlete’s injury result is independent of whether or not the athlete has stretched. (claim) H a : An athlete’s injury result is dependent on whether or not the athlete has stretched. (b) d.f. = (r − 1)(c − 1) = 1

χ02 = 6.635 ; Rejection region: χ 2 > 6.635 (c)

O−E

(O E ) 2

18 22 211 189

20.818 19.182 208.182 191.818

–2.818 2.818 2.818 –2.818

7.941124 7.941124 7.941124 7.941124

(O E )2 E 0.3814547027 0.4139883224 0.0381451038 0.0413992639 0.875

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

423

2 (d) Because χ < 6.635 , fail to reject H 0 .

(e) There is not enough evidence at the 1% level of significance to reject the claim that an athlete’s injury result is independent of whether or not the athlete has stretched. 14. (a) The claim is “nausea result is related to treatment.” H 0 : Nausea result is independent of treatment. H a : Nausea result is dependent on treatment. (claim) (b) d.f. = (r − 1)(c − 1) = 1

χ02 = 3.841 ; Rejection region: χ 2 > 3.841 (c)

O−E

(O E ) 2

36 13 254 262

25.15 23.85 264.85 251.15

10.85 –10.85 –10.85 10.85

117.7225 117.7225 117.7225 117.7225

(O E )2 E 4.6808 4.9360 0.4445 0.4687 10.53

χ 2 ≈ 10.53 2 (d) Because χ > 3.841 , reject H 0 .

(e) There is enough evidence at the 5% level of significance to conclude that nausea result is dependent on treatment. 15. (a) The claim is “result is related to type of treatment.” H 0 : The result is independent of the type of treatment. H a : The result is dependent on the type of treatment. (claim) (b) d.f. = (r − 1)(c − 1) = 1

χ02 = 2.706 ; Rejection region: χ 2 > 2.706 (c)

OíE

(O − E )2

58 81 42 19

69.5 69.5 30.5 30.5

–11.5 11.5 11.5 –11.5

132.25 132.25 132.25 132.25

χ 2 ≈ 12.478 2 (d) Because χ > 2.706 , reject H 0 .

(O − E ) 2 E 1.9029 1.9029 4.3361 4.3361 12.478

424

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(e) There is enough evidence at the 10% level of significance to conclude that the result is dependent on the type of treatment. 16. (a) The claim is “attitudes about the safety steps taken by the school staff are related to the type of school.” H 0 : Attitudes about safety are independent of the type of school. H a : Attitudes about safety are dependent on the type of school. (claim) (b) d.f. = ( r − 1)(c − 1) = 1

χ02 = 6.635 ; Rejection region: χ 2 > 6.635 (c)

OíE

(O − E )2

40 51 64 34

50.074 40.926 53.926 44.074

−10.074 10.074 10.074 −10.074

101.4855 101.4855 101.4855 101.4855

(O − E ) 2 E 2.0267 2.4797 1.8819 2.3026 8.6909

χ 2 ≈ 8.691 2 (d) Because χ > 6.635 , reject H 0 .

(e) There is enough evidence at the 1% level of significance to conclude that attitudes about the safety steps taken by the school staff are dependent on the type of school. 17. (a) The claim is “the number of times former smokers tried to quit before they were habit-free is related to gender.” H 0 : The number of times former smokers tried to quit is independent of gender. H a : The number of times former smokers tried to quit is dependent of gender. (claim) (b) d.f. = (r − 1)(c − 1) = 2

χ02 = 5.991 ; Rejection region: χ 2 > 5.991 (c)

O−E

(O − E )2

271 257 149 146 139 80

270.930 257.286 148.784 146.070 138.714 80.216

0.070 −0.286 0.216 −0.070 0.286 −0.216

0.004900 0.081796 0.046656 0.004900 0.081796 0.046656

χ 2 ≈ 0.002

(O − E ) 2 E 0 0.0003 0.0003 0 0.0006 0.0006 0.0018

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

425

2 (d) Because χ < 5.991 , fail to reject H 0 .

(e) There is not enough evidence at the 5% level of significance to conclude that the number of times former smokers tried to quit is dependent on gender. 18. (a) The claim is “achieving a basic skill level is related to the location of the school.” H 0 : Skill level in a subject is independent of location. (claim) H a : Skill level in a subject is dependent on location. (b) d.f. = (r − 1)(c − 1) = 2

χ02 = 9.210 ; Rejection region: χ 2 > 9.210 (c)

O−E

(O E ) 2

43 42 38 63 66 65

41.129 41.905 39.965 64.871 66.095 63.035

1.871 0.095 −1.965 −1.871 −0.095 1.965

3.500641 0.009025 3.861225 3.500641 0.009025 3.861225

(O E )2 E 0.0851 0.0002 0.0966 0.0540 0.0001 0.0613 0.2973

χ 2 ≈ 0.297 2 (d) Because χ < 9.210 , fail to reject H 0 .

(e) There is not enough evidence at the 1% level of significance to reject the claim that skill level in a subject is independent of location. 19. (a) The claim is “the reason and the type of worker are dependent.” H 0 : Reasons are independent of the type of worker. H a : Reasons are dependent on the type of worker. (claim) (b) d.f. = (r − 1)(c − 1) = 2

χ02 = 9.210 ; Rejection region: χ 2 > 9.210 (c)

O−E

(O E ) 2

30 36 41 47 25 30

39.421 31.230 36.349 37.579 29.770 34.651

−9.421 4.770 4.651 9.421 −4.770 −4.651

88.7552 22.7529 21.6318 88.7552 22.7529 21.6318

(O E )2 E 2.2515 0.7286 0.5951 2.3618 0.7643 0.6243 7.3256

426

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

2 (d) Because χ < 9.210 , fail to reject H 0 .

(e) There is not enough evidence at the 1% level of significance to conclude that reasons for continuing education are dependent on the type or worker. 20. (a) The claim is “age is related to which aspect of career development is considered to be most important.” H 0 : The aspect of career development that is considered to be the most important is independent of age. H a : The aspect of career development that is considered to be the most important is dependent on age. (claim) (b) d.f. = (r − 1)(c − 1) = 4

χ02 = 7.779 ; Rejection region: χ 2 > 7.779 (c)

O−E

(O E ) 2

31 22 21 27 31 33 19 14 8

27.6602 24.0680 22.2718 34.0146 29.5971 27.3883 15.3252 13.3350 12.3398

3.3398 −2.0680 −1.2718 −7.0146 1.4029 5.6117 3.6748 0.6650 −4.3398

11.1543 4.2765 1.6176 49.2041 1.9682 31.4906 13.5038 0.4423 18.8339

(O E )2 E 0.4033 0.1777 0.0726 1.4466 0.0665 1.1498 0.8812 0.0332 1.5263 5.7572

χ 2 ≈ 5.757 2 (d) Because χ < 7.779 , fail to reject H 0 .

(e) There is not enough evidence at the 10% level of significance to conclude that the aspect of career development that is considered to be most important is dependent on age. 21. (a) The claim is “a family’s choice to visit colleges before enrolling is related to income.” H 0 : A family’s choice to visit colleges before enrolling is independent of their income. H a : A family’s choice to visit colleges before enrolling is dependent on their income. (claim) (b) d.f. = (r − 1)(c − 1) = 6

χ02 = 16.812 ; Rejection region: χ 2 > 16.812

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(c)

O−E

(O E )2

155 60 88 53 461 136 171 85 415 52 130 52

197.544 47.518 74.534 36.405 473.328 113.856 178.588 87.228 360.129 86.626 135.878 66.367

í42.544 12.482 13.466 16.595 í12.328 22.144 í7.588 í2.228 54.871 í34.626 í5.878 í14.367

1809.991936 155.800324 181.333156 275.394025 151.979584 490.356736 57.577744 4.963984 3010.826641 1198.959876 34.550884 206.410689

(O E )2 E 9.162474871 3.278764342 2.432891781 7.564730806 0.3210872461 4.306815065 0.3224054472 0.0569081488 8.360411522 13.84064687 0.252787206 3.110140416 53.012

χ 2 ≈ 53.012 2 (d) Because χ > 16.812 , reject H 0 .

(e) There is enough evidence at the 1% level of significance to conclude that a family’s choice to visit colleges before enrolling is dependent on their income. 22. (a) The claim is “the deciding factor in choosing a college is related to the income of the family.” H 0 : The deciding factor in choosing a college is independent of the family’s income. H a : The deciding factor in choosing a college is dependent on the family’s income. (claim) (b) d.f. = (r − 1)(c − 1) = 6

χ02 = 16.812 ; Rejection region: χ 2 > 16.812 (c)

O−E

(O E )2

138 121 78 14 290 298 239 26 168 298 175 13

112.592 135.450 92.945 10.012 273.621 329.172 225.875 24.332 209.787 252.378 173.180 18.656

25.408 í14.45 í14.945 3.988 16.379 í31.172 13.125 1.668 í41.787 45.622 1.82 í5.656

645.566464 208.8025 223.353025 15.904144 268.271641 971.693584 172.265625 2.782224 1746.153369 2081.366884 3.3124 31.990336

χ 2 ≈ 34.381

(O E )2 E 5.733679693 1.541546696 2.403066599 1.58850819 0.9804497498 2.95193268 0.7626591035 0.114344238 8.323458408 8.247021864 0.01912692 1.714747856 34.381

427

428

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

2 (d) Because χ > 16.812 , reject H 0 .

(e) There is enough evidence at the 1% level of significance to conclude that the deciding factor in choosing a college is dependent on the family’s income. 23. (a) The claim is “the point of initial impact of crashes with occupant fatalities depends on the type of vehicle.” H 0 : Point of initial impact of crashes with occupant fatalities is independent from type of vehicle. H a : Point of initial impact of crashes with occupant fatalities is dependent on type of vehicle. (claim) (b) d.f. = ( r − 1)(c − 1) = 2

χ02 = 5.991 ; Rejection region: χ 2 > 5.991 (c)

O−E

(O E )2

1435 697 161 523 150 34

1496.565 647.390 149.045 461.435 199.610 45.955

í61.565 49.61 11.955 61.565 í49.61 í11.955

3790.249225 2461.1521 142.922025 3790.249225 2461.1521 142.922025

(O E ) 2 . E 2.532632545 3.801652945 0.9589186152 8.21404797 12.32980362 3.110042977 30.947 .

χ 2 ≈ 30.947 2 (d) Because χ > 5.991 , reject H 0 .

(e) There is enough evidence at the 5% level of significance to conclude that point of initial impact of crashes with occupant fatalities is dependent on the type of vehicles. 24. (a) The claim is “age and gender are related.” H 0 : Age and gender are independent. H a : Age and gender are dependent. (claim) (b) d.f. = (r − 1)(c − 1) = 5

χ02 = 11.071 ; Rejection region: χ 2 > 11.071

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(c)

O−E

(O E ) 2

27 148 107 71 66 54 9 45 26 17 16 14

28.38 152.148 104.848 69.373 64.643 53.607 7.62 40.852 28.152 18.627 17.357 14.393

í1.38 í4.148 2.152 1.627 1.357 0.393 1.38 4.148 í2.152 í1.627 í1.357 í0.393

1.9044 17.205904 4.631104 2.647129 1.841449 0.154449 1.9044 17.205904 4.631104 2.647129 1.841449 0.154449

429

(O E )2 E 0.0671035941 0.1130866262 0.0441696933 0.0381579145 0.0284864409 0.0028811349 0.2499212598 0.4211765397 0.1645035521 0.1421124711 0.1060925851 0.0107308414 1.388

χ 2 ≈ 1.388 2 (d) Because χ < 11.071 , fail to reject H 0 .

(e) There is not enough evidence at the 5% level of significance to conclude that age and gender are dependent. 25. (a) The claim is “procedure preference is related to bank employee.” H 0 : Procedure preference is independent of bank employee. H a : Procedure preference is dependent on bank employee. (claim) (b) d.f. = ( r − 1)(c − 1) = 2

χ02 = 5.991 ; Rejection region: χ 2 > 5.991 (c)

O−E

(O E ) 2

92 351 50 76 42 8

133.8029 313.0032 46.19386 34.19709 79.99677 11.80614

í41.802908 37.996769 3.80613893 41.8029079 í37.996769 í3.8061389

1747.483 1443.754 14.487 1747.483 1443.754 14.487

(O E ) 2 E 13.060 4.613 0.314 51.100 18.047 1.227 88.361

χ 2 ≈ 88.361 2 (d) Because χ > 5.991 , reject H 0 .

(e) There is enough evidence at the 5% level of significance to conclude that procedure preference is dependent on bank employee.

430

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

26. (a) The claim is “Restaurant rating is related to size of restaurant.” H 0 : Restaurant rating is independent of size of restaurant. H a : Restaurant rating is dependent on size of restaurant. (claim) (b) d.f. = ( r − 1)(c − 1) = 2

χ02 = 9.210 ; Rejection region: χ 2 > 9.210 (c)

O−E

(O E )2

182 203 165 180 311 159

165.9167 235.5833 148.5 196.0833 278.4167 175.5

16.0833333 í32.583333 16.5 í16.083333 32.5833333 í16.5

258.674 1061.674 272.250 258.674 1061.674 272.250

(O E ) 2 E 1.559 4.507 1.834 1.319 3.813 1.551 14.583

χ 2 ≈ 14.583 2 (d) Because χ > 9.210 , reject H 0 .

(e) There is enough evidence at the 1% level of significance to conclude that restaurant rating is dependent on size of restaurant. 27. (a) The claim is “Type of car is not related to gender.” H 0 : Type of car is independent of gender (claim) H a : Type of car is dependent on gender. (b) d.f. = (r − 1)(c − 1) = 3

χ02 = 6.251 ; Rejection region: χ 2 > 6.251 (c)

O−E

(O − E )2

28 39 21 22 24 32 20 14

28.60 39.05 22.55 19.80 23.40 31.95 18.45 16.20

–0.60 –0.05 –1.55 2.2 0.6 0.05 1.55 –2.2

0.36 0.0025 2.4025 4.84 0.36 0.0025 2.4025 4.84

χ 2 ≈ 0.808 2 (d) Because χ < 6.251 , fail to reject H 0 .

(O − E ) 2 E 0.0126 0.0001 0.1065 0.2444 0.0154 0.0001 0.1302 0.2988 0.808

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

431

(e) There is not enough evidence at the 10% level of significance to conclude that type of car is dependent on gender. 28. (a) The claim is “type of movie rented and age are related.” H 0 : Type of movie rented is independent of age. H a : Type of movie rented is dependent on age. (claim) (b) d.f. = (r − 1)(c − 1) = 8

χ02 = 13.362 ; Rejection region: χ 2 > 13.362 (c)

O−E

(O E ) 2

38 30 24 10 8 15 17 16 9 5 12 11 19 25 13

28.37 25.32 25.75 19.21 11.35 15.99 14.27 14.52 10.83 6.40 20.63 18.41 18.73 13.97 8.25

9.63 4.68 –1.75 –9.21 –3.35 –0.99 2.73 1.48 –1.83 –1.4 –8.63 –7.41 0.27 11.03 4.75

92.7369 21.9024 3.0625 84.8241 11.2225 0.9801 7.4529 2.1904 3.3489 1.96 74.4769 54.9081 0.0729 121.6609 22.5625

(O E )2 E 3.2688 0.8650 0.1189 4.4156 0.9888 0.0613 0.5223 0.1509 0.3092 0.3063 3.6101 2.9825 0.0039 8.7087 2.7348 29.0471

χ 2 ≈ 29.0471 2

(d) Because χ > 13.362 , reject H 0 . (e) There is enough evidence at the 10% level of significance to conclude that type of movie rented is dependent on age. 29. The claim is “the proportions of motor vehicle crash deaths involving males and females are the same for each group.” H 0 : The proportions are equal. (claim) H a : At least one of proportion is different from the others. d.f. = (r − 1)(c − 1) = 7

χ02 = 14.067 ; Rejection region: χ 2 > 14.067

432

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

O−E

(O − E )2

80 97 76 73 81 51 30 12 33 35 26 27 26 23 19 9

80.946 94.556 73.066 71.633 76.648 53.009 35.100 15.043 32.054 37.444 28.934 28.367 30.352 20.991 13.900 5.957

í0.946 2.444 2.934 1.367 4.352 í2.009 í5.1 í3.043 0.946 í2.444 í2.934 í1.367 í4.352 2.009 5.1 3.043

0.894916 5.973136 8.608356 1.868689 18.939904 4.036081 26.01 9.259849 0.894916 5.973136 8.608356 1.868689 18.939904 4.036081 26.01 9.259849

(O − E ) 2 E 0.0110557162 0.0631703541 0.1178161662 0.026086985 0.2471023901 0.0761395423 0.741025641 0.6155586652 0.0279190117 0.159521846 0.2975169697 0.0658754539 0.6240084344 0.1922767376 1.871223022 1.55444838 6.691

χ 2 ≈ 6.691 2

Because χ < 14.067 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the claim that the proportions of motor vehicle crash deaths involving males and females are the same for each group. 30. The claim is “the proportions of the results for drug and placebo treatments or the same.” H 0 : The proportions are equal. (claim) H a : At least one of proportion is different from the others. d.f. = ( r − 1)(c − 1) = 1

χ02 = 2.706 ; Rejection region: χ 2 > 2.706 O

O−E

(O − E )2

39 25 54 70

31.66 32.34 61.34 62.66

7.34 −7.34 −7.34 7.34

53.8756 53.8756 53.8756 53.8756

(O − E ) 2 E 1.7017 1.6659 0.8783 0.8598 5.1057

χ 2 ≈ 5.106 2

Because χ > 2.706 , reject H 0 . There is enough evidence at the 10% level of significance to reject the claim that the proportions of the results for drug and placebo treatments or the same. 31. Right-tailed 32. Answers will vary. Sample answer: Both tests are very similar, but the chi-square test for independence tests whether the occurrence of one variable affects the probability of the occurrence of another variable, while the chi-square homogeneity of proportions test determines whether the proportions for categories from a population follow the same distribution as another population. Copyright © 2023 Pearson Education, Inc.

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

33.

Educational Attainment

Status Employed

Not a High School Graduate 0.037

High School Graduate 0.148

Some College, No Degree 0.088

Associate’s, Bachelor’s, or Advanced Degree 0.335

Unemployed

0.004

0.009

0.005

0.011

Not in the labor force

0.050

0.119

0.059

0.136

34. Several of the expected frequencies are less than 5. 35. (a) 1.1% (b) 5.9% 36. (a) 14.8% (b) 36.41% (c) 9.06% Educational Attainment

37.

0.243

Some College, No Degree 0.145

Associate’s, Bachelor’s, or Advanced Degree 0.551

0.127

0.317

0.175

0.381

0.137

0.328

0.163

0.373

High School Graduate

Status Employed

Not a High School Graduate 0.061

Unemployed Not in the labor force 38. 55.1% 39. 16.3% 40.

Educational Attainment

0.535

Some College, No Degree 0.579

Associate’s, Bachelor’s, or Advanced Degree 0.695

0.040

0.033

0.032

0.022

0.550

0.433

0.388

0.282

High School Graduate

Status Employed

Not a High School Graduate 0.411

433

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CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

10.3 COMPARING TWO VARIANCES 10.3 TRY IT YOURSELF SOLUTIONS 1. α = 0.05

F0 = 2.45

2. α = 0.01

F0 = 18.31

3. The claim is “the variance of the time required for nutrients to enter the bloodstream is less with the specially treated intravenous solution than the variance of the time without the solution.”

H0 : σ12 ≤ σ 22 ; Ha : σ12 > σ 22 (claim) α = 0.01 d.f.N = n1 − 1 = 24 d.f.D = n2 − 1 = 19

F0 = 2.92 ; Rejection region: F > 2.92

s 2 180 F = 12 = ≈ 3.21 56 s2 Because F > 2.92 , reject H 0 . There is enough evidence at the 1% level of significance to support the researcher’s claim that a specially treated intravenous solution decreases the variance of the time required for nutrients to enter the bloodstream. 4. The claim is “the pH levels of the soil in two geographic regions have equal standard deviations.” H 0 : σ 1 = σ 2 (claim); H a : σ 1 ≠ σ 2

α = .01

d.f.N = n1 − 1 = 15 d.f.D = n2 − 1 = 21

F0 = 3.43; Rejection region: F > 3.43

s12 (0.95)2 = ≈ 1.48 s22 (0.78)2 Because F < 3.43 . fail to reject H 0 . There is not enough evidence at the 1% level of significance to reject the biologist’s claim that the pH levels of the soil in the two geographic regions have equal standard deviations. F=

10.3 EXERCISE SOLUTIONS 1. Specify the level of significance α . Determine the degrees of freedom for the numerator and denominator. Use Table 7 in Appendix B to find the critical value F. 2. (1) The F-distribution is a family of curves determined by two types of degrees of freedom, d.f.N and d.f.D . Copyright © 2023 Pearson Education, Inc.

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(2) F-distributions are positively skewed and therefore the distribution is not symmetrical. (3) The area under the F-distribution curve is equal to 1. (4) All values of F are greater than or equal to 0. (5) For all F-distributions, the mean value of F is approximately equal to 1. 3. (1) The samples must be randomly selected, (2) the samples must be independent, and (3) each population must have a normal distribution. 4. Determine the sample whose variance is greater. Use the size of this sample n1 to find d.f.N = n1 − 1. Use the size of the other sample n 2 to find d.f.D = n2 − 1 . 5.

F0 = 2.54

F0 = 7.21

F0 = 14.62

F0 = 9.16

10. F0 = 1.91

11. F0 = 1.80 2

F0 = 2.06

12. F0 = 2.42 2

13. H0 : σ1 ≤ σ 2 ; Ha : σ1 > σ 2 (claim) d.f.N = 4 , d.f.D = 5 F0 = 3.52; Rejection region: F > 3.52

s 2 773 F = 12 = ≈ 1.010 s2 765 Because F < 3.52 , fail to reject H 0 . There is not enough evidence at the 10% level of significance to support the claim. 2

14. H 0 : σ1 = σ 2 (claim); H a : σ1 ≠ σ 2 d.f.N = 6 , d.f.D = 7 F0 = 5.12; Rejection region: F > 5.12

s 2 310 F = 12 = ≈ 1.044 s2 297 Because F < 5.12 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the claim. 2

15. H 0 : σ1 ≤ σ 2 (claim); H a : σ1 > σ 2 d.f.N = 10 , d.f.D = 9 F0 = 5.26; Rejection region: F > 5.26

s 2 842 F = 12 = ≈ 1.007 s2 836 Because F < 5.26 , fail to reject H 0 . There is not enough evidence at the 1% level of significance to reject the claim.

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CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION 2

16. H0 : σ1 = σ 2 ; H a : σ1 ≠ σ 2 (claim) d.f.N = 30 , d.f.D = 27 F0 = 2.13; Rejection region: F > 2.13

s 2 245 F = 12 = ≈ 2.188 s2 112 Because F > 2.13 , reject H 0 . There is enough evidence at the 5% level of significance to support the claim. 2

17. H 0 : σ1 = σ 2 (claim); H a : σ1 ≠ σ 2 d.f.N = 12 , d.f.D = 19 F0 = 3.76; Rejection region: F > 3.76

s 2 9.8 F = 12 = ≈ 3.920 s2 2.5 Because F > 3.76 , reject H 0 . There is enough evidence at the 1% level of significance to reject the claim. 2

18. H0 : σ1 ≤ σ 2 ; Ha : σ1 > σ 2 (claim) d.f.N = 15 , d.f.D = 11 F0 = 2.72; Rejection region: F > 2.72

s 2 44.6 F = 12 = ≈ 1.135 s2 39.3 Because F < 2.72 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to support the claim. 19. Population 1: Company B; Population 2: Company A

(a) The claim is “the variance of the life of Company A’s appliances is less than the variance of the life of Company B’s appliances.”

H0 : σ12 ≤ σ 22 ; Ha : σ12 > σ 22 (claim) (b) d.f.N = 24 , d.f.D = 19 F0 = 2.45; Rejection region: F > 2.45 s 2 3.9 (c) F = 12 = ≈ 2.17 s2 1.8 (d) Because F < 2.45 , fail to reject H 0 . (e) There is not enough evidence at the 2.5% level of significance to support Company A’s claim that the variance of life of its appliances is less than the variance of the life of Company B’s appliances.

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20. Population 1: Competitor; Population 2: Auto Manufacturer

(a) The claim is “the variance of the carbon monoxide emissions for the make and model of one of an auto manufacturer’s vehicles is less than that of a top competitor’s equivalent vehicles.”

H0 : σ12 ≤ σ 22 ; Ha : σ12 > σ 22 (claim) (b) d.f.N = 20 , d.f.D = 18 F0 = 1.84 Rejection region: F > 1.84 s 2 0.045 (c) F = 12 = = 5.625 s2 0.008 (d) Because F > 1.84 , reject H 0 . (e) There is enough evidence at the 10% level of significance to conclude that the variance of carbon monoxide emissions for the company’s make and model of one of its vehicles is less than that of the competitor’s equivalent vehicles. 21. Population 1: Golfer 1; Population 2: Golfer 2

(a) The claim is “the variances of the driving distances differ between the two golfers.”

H0 : σ12 = σ 22 ; H a : σ12 ≠ σ 22 (claim) (b) d.f.N = 9 , d.f.D = 9 F0 = 3.18; Rejection region: F > 3.18 2 2 (c) s1 = 169.7333, s2 = 23.2111

s 2 169.7333 F = 12 = ≈ 7.31 23.2111 s2 (d) Because F > 3.18 , reject H 0 . (e) There is enough evidence at the 10% level of significance to conclude that the variances of the driving distances differ between the two golfers. 22. Population 1: Age group 18–34; Population 2: Age group 35–49

(a) The claim is “the variances of the waiting times differ between the two age groups.”

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(d) Because F < 4.82 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to conclude that the variances of the waiting times differ between the two age groups. 23. Population 1: District 1; Population 2: District 2

(a) The claim is “the standard deviations of science achievement test scores for eighth grade students are the same in Districts 1 and 2.”

H0 : σ12 = σ 22(claim); H a : σ12 ≠ σ 22 (b) d.f.N = 11 , d.f.D = 13 F0 = 2.635; Rejection region: F > 2 .6 3 5

s12 (36.8) 2 F = = ≈ 1.282 (c) s22 (32.5) 2 (d) Because F < 2 .6 3 5 , fail to reject H 0 . (e) There is not enough evidence at the 10% level of significance to reject the administrator’s claim that the standard deviations of science assessment test scores for eighth grade students are the same in Districts 1 and 2. 24. Population 1: District 1; Population 2: District 2

(a) The claim is “the standard deviations of U.S. history assessment scores for eighth grade students are the same in Districts 1 and 2.”

H0 : σ12 = σ 22(claim); H a : σ12 ≠ σ 22 (b) d.f.N = 9 , d.f.D = 12 F0 = 5.20; Rejection region: F > 5 .2 0

s 2 (30.9)2 ≈ 1.29 (c) F = 12 = s2 (27.2)2 (d) Because F < 5.2 0 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to reject the administrator’s claim that the standard deviations of U.S. history assessment test scores for eighth grade students are the same in Districts 1 and 2. 25. Population 1: New York; Population 2: California

(a) The claim is “the standard deviation of the annual salaries for actuaries is less in California than in New York.” H0 : σ12 ≤ σ 22 ; Ha : σ12 > σ 22 (claim)

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(b) d.f.N = 40 , d.f.D = 60 F0 = 1.59 Rejection region: F > 1 .5 9

s12

(42,800)2 ≈ 2.240 (c) F = 2 = s2 (28, 600)2 (d) Because F > 1 .5 9 , reject H 0 . (e) There is enough evidence at the 5% level of significance to conclude that the standard deviation of the annual salaries for actuaries is less in California than in New York. 26. Population 1: Florida; Population 2: Louisiana

(a) The claim is “the standard deviation of the annual salaries for public relations managers is less in Louisiana than in Florida.”

H0 : σ12 ≤ σ 22 ; Ha : σ12 > σ 22 (claim) (b) d.f.N = 27 , d.f.D = 23 F0 = 1.985 Rejection region: F > 1.985

s12

(39,100)2 ≈ 2.066 (c) F = 2 = s2 (27, 200) 2 (d) Because F > 1 .9 8 5 , reject H 0 . (e) There is enough evidence at the 5% level of significance to conclude that the standard deviation of the annual salaries for public relations managers is less in Louisiana than in Florida. 27. Right-tailed: FR = 14.73 Left-tailed: d.f.N = 3 and d.f.D = 6 F = 6.60

Critical value is

1 1 = ≈ 0.15 . F 6.60

28. Right-tailed: FR = 2.33 Left-tailed: d.f.N = 15 and d.f.D = 20 F = 2 .2 0

Critical value is

29.

1 1 = ≈ 0.45 . F 2.20

2 s12 1 σ12 s12 1 σ12 § 10.89 ·§ 1 · σ1 § 10.89 · < < → < < → < < 3.422 (3.02) 0.340 ¨ ¸¨ ¸ ¨ ¸ 2 s22 FR σ 22 s22 FL σ 22 © 9.61 ¹© 3.33 ¹ σ 2 © 9.61 ¹

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30.

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION 2 s12 1 σ12 s12 1 σ12 § 5.29 ·§ 1 · σ1 § 5.29 · < < → < < → < < 4.425 (3.02) 0.440 ¨ ¸¨ ¸ ¨ ¸ 2 s22 FR σ 22 s22 FL σ 22 © 3.61 ¹© 3.33 ¹ σ 2 © 3.61 ¹

10.4 ANALYSIS OF VARIANCE 10.4 TRY IT YOURSELF SOLUTIONS 1. The claim is “there is a difference in the mean a monthly sales among the sales regions.” H 0 : μ1 = μ 2 = μ 3 = μ 4 H a : At least one mean is different from the others. (claim) α = 0.05

d.f.N = 3 ; d.f.D = 14 F0 = 3.34 ; Rejection region: F > 3 .3 4

Variation Between Within

Sum of Squares 549.8 608.0

Degrees of Freedom 3 14

Mean Squares 183.3 43.4

F 4.22

F ≈ 4 .2 2

Because F > 3 .3 4 , reject H 0 .

There is enough evidence at the 5% level of significance for the analyst to conclude that there is a difference in the mean monthly sales among the sales regions. 2. The claim is “there is a difference in the means of the GPAs.” H 0 : μ1 = μ 2 = μ 3 = μ 4 H a : At least one mean is different from the others. (claim) Enter the data. Variation Between Within

Sum of Squares 0.584 4.360

F = 1 .3 4 → P -valu e = 0 .2 8 0

Degrees of Freedom 3 30

Mean Squares 0.195 0.145

F 1.34

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0.280 > 0.05

Because P -value > 0.05 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to conclude that there is a difference in the means of the GPAs.

10.4 EXERCISE SOLUTIONS 1.

H 0 : μ1 = μ 2 = ! = μ k H a : At least one mean is different from the others.

2. Each sample must be randomly selected from a normal, or approximately normal, population. The samples must be independent of each other. Each population must have the same variance. 3. The MS B measures the differences related to the treatment given to each sample. The MSW measures the differences related to entries within the same sample. 4. There are three null hypotheses for a two-way ANOVA test. There is one for each main effect and one for the interaction effect. H 0 A : There is no difference among the treatment means of Factor A. H aA : There is at least one difference among the treatment means of Factor A. H 0 B : There is no difference among the treatment means of Factor B. H aB : There is at least one difference among the treatment means of Factor A. H 0 AB : There is no interaction between Factor A and Factor B. H aAB : There is interaction between Factor A and Factor B. 5. (a) The claim is that “at least one mean cost per ounce is different from the others.” H 0 : μ1 = μ 2 = μ 3 H a : At least one mean is different from the others. (claim)

(b) d.f.N = k − 1 = 2 ; d .f.D = N − k = 1 2 F0 = 3.89 ; Rejection region: F > 3.89 (c) Variation Between Within

Sum of Squares 3.35334 15.1813

Degrees of Freedom 2 12

Mean Squares 1.67667 1.26511

F 1.325

F ≈ 1.325 (d) Because F < 3.89 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to conclude that at least one mean cost per ounce is different from the others.

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6. (a) The claim is that “at least one mean lifespan is different from the others.” H 0 : μ1 = μ 2 = μ 3 H a : At least one mean is different from the others. (claim)

(b) d.f.N = k − 1 = 2 ; d.f.D = N − k = 15 F0 = 3.68 ; Rejection region: F > 3.68 (c) Variation Between Within

Sum of Squares 4.17333 37.94667

Degrees of Freedom 2 15

Mean Squares 2.08667 2.52978

F 0.825

F ≈ 0.825 (d) Because F < 3.68 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to conclude that at least one of the dog breeds has a mean lifespan that is different from the others. 7. (a) The claim is that “at least one mean vacuum cleaner weight is different from the others.” H 0 : μ1 = μ 2 = μ 3 H a : At least one mean is different from the others. (claim)

(b) d.f.N = k − 1 = 2 ; d.f.D = N − k = 15 F0 = 6.36 ; Rejection region: F > 6.36 (c)

Variation

Between Within

Sum of Squares 122.1111 56.8333

Degrees of Freedom 2 15

Mean Squares 61.0556 3.7889

F 16.11

F ≈ 16.11 (d) Because F > 6.36 , reject H 0 . (e) There is enough evidence at the 1% level of significance to conclude that at least one mean vacuum cleaner weight is different from the others 8. (a) The claim is that “at least one mean salary is different from the others.” H 0 : μ1 = μ 2 = μ 3 H a : At least one mean is different from the others. (claim)

(b) d.f.N = k − 1 = 2 ; d .f.D = N − k = 2 7 F0 = 5.49 ; Rejection region: F > 5.49 (c) Variation Between Within

Sum of Squares 5062.6287 6486.19

Degrees of Freedom 2 27

Mean Squares 2531.3143 240.22926

F 10.537

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(d) Because F > 5.49 , reject H 0 . (e) There is enough evidence at the 1% level of significance to conclude that at least one mean salary is different from the others. 9. (a) The claim is “at least one mean age is different from the others.” H 0 : μ1 = μ 2 = μ 3 = μ 4 H a : At least one mean is different from the others. (claim)

(b) d.f.N = k − 1 = 3 ; d .f.D = N − k = 4 8 F0 = 2.84 ; Rejection region: F > 2.84 (c) Variation Between Within

Sum of Squares 10.07692 262.1538

Degrees of Freedom 3 48

Mean Squares 3.358974 5.461538

F 0.62

F ≈ 0.62 (d) Because F < 2.84 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to conclude that at least one mean age is different from the others. 10. (a) The claim is that “at least one mean cost per mile is different from the others.” H 0 : μ1 = μ 2 = μ 3 = μ 4 = μ 5 H a : At least one mean is different from the others. (claim)

(b) d.f.N = k − 1 = 4 ; d.f.D = N − k = 21 F0 = 4.37 ; Rejection region: F > 4.37 (c) Variation Between Within

Sum of Squares 1699.5551 668.48333

Degrees of Freedom 4 21

Mean Squares 424.88878 31.83254

F 13.348

F ≈ 13.348 (d) Because F > 4.37 , reject H 0 . (e) There is enough evidence at the 1% level of significance to conclude that at least one mean cost per mile is different from the others. 11. (a) The claim is that “the mean scores are the same for all regions.” H 0 : μ1 = μ 2 = μ 3 = μ 4 (claim) H a : At least one mean is different from the others.

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Sum of Squares 7.777 21.2124

Degrees of Freedom 3 30

Mean Squares 2.5923 0.7071

F 3.67

F ≈ 3.67 (d) Because F > 2.28 , reject H 0 . (e) There is enough evidence at the 10% level of significance to reject the claim that the mean scores are the same for all regions. 12. (a) The claim is “the mean number of days patients spend at the hospital is the same for all four regions.” H 0 : μ1 = μ 2 = μ 3 = μ 4 (claim) H a : At least one mean is different from the others.

(b) d.f.N = k − 1 = 3 ; d .f.D = N − k = 2 9 F0 = 4.54 ; Rejection region: F > 4.54 (c) Variation Between Within

Sum of Squares 5.607504 97.30159

Degrees of Freedom 3 29

Mean Squares 1.869168 3.355227

F 0.56

F ≈ 0.56 (d) Because F < 4.54 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to reject the claim that the mean number of days patients spend at the hospital is the same for all four regions. 13. (a) The claim is that “the mean salary is different in at least one of the areas.” H 0 : μ1 = μ 2 = μ 3 = μ 4 = μ 5 = μ 6 H a : At least one mean is different from the others. (claim)

(b) d.f.N = k − 1 = 5 ; d .f.D = N − k = 3 0 F0 = 2.53 ; Rejection region: F > 2.53 (c) Variation Between Within

Sum of Squares 836335706 1722674771

F ≈ 2.913 (d) Because F > 2.53 , reject H 0 .

Degrees of Freedom 5 30

Mean Squares 167267141 57422492

F 2.913

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(e) There is enough evidence at the 5% level of significance to conclude that at least one mean salary is different from the others. 14. (a) The claim is that “at least one mean sale price is different from the others.” H 0 : μ1 = μ 2 = μ 3 H a : At least one mean is different from the others. (claim)

(b) d.f.N = k − 1 = 2 ; d .f.D = N − k = 2 9 F0 = 2.50 ; Rejection region: F > 2.50 (c) Variation Between Within

Sum of Squares 39330.8076 709707.252

Degrees of Freedom 2 29

Mean Squares 19665.4038 24472.6639

F 0.804

F ≈ 0.804 (d) Because F < 2.50 , fail to reject H 0 . (e) There is not enough evidence at the 10% level of significance to conclude that at least one mean sale price is different from the others. 15. H 0 : Advertising medium has to effect on mean ratings. H a : Advertising medium has an effect on mean ratings. H 0 : Length of ad has no effect on mean ratings. H a : Length of ad has an effect on mean ratings. H 0 : There is no interaction effect between advertising medium and length of ad on mean ratings. H a : There is an interaction effect between advertising medium and length of ad on mean ratings. Source Ad medium Length of ad Interaction Error Total

d.f. 1 1 1 16 19

SS 1.25 0.45 0.45 34.80 36.95

MS 1.25 0.45 0.45 2.175

F 0.57 0.21 0.21

P 0.459 0.655 0.655

Fail to reject all null hypotheses. The interaction between the advertising medium and length of the ad has no effect on the rating and therefore there is no significant difference in the means of the ratings. 16. H 0 : Type of vehicle has no effect on the mean number of vehicles sold. H a : Type of vehicle has an effect on the mean number of vehicles sold. H 0 : Gender has no effect on the mean number of vehicles sold. H a : Gender has an effect on the mean number of vehicles sold. H 0 : There is no interaction effect between type of vehicle and gender on the mean number of vehicles sold. H a : There is an interaction effect between type of vehicle and gender on the mean number of vehicles sold. Copyright © 2023 Pearson Education, Inc.

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Source Type of vehicle Gender Interaction Error Total

d.f. 2 1 2 18 23

SS 84.083 0.375 12.250 22.250 118.958

MS 42.042 0.375 6.125 1.236

F 34.01 0.30 4.96

P 0.000 0.589 0.019

Type of vehicle has an effect on the mean number of vehicles sold in a month. There is an interaction effect between type of vehicle and gender on the number of vehicles sold in a month. 17. H 0 : Age has no effect on mean GPA. H a : Age has an effect on mean GPA. H 0 : Gender has no effect on mean GPA. H a : Gender has an effect on mean GPA. H 0 : There is no interaction effect between age and gender on mean GPA. H a : There is no interaction effect between age and gender on mean GPA. Source Age Gender Interaction Error Total

d.f. 3 1 3 16 23

SS 0.4146 0.1838 0.2912 18.6600 19.5496

MS 0.138 0.184 0.097 1.166

F 0.12 0.16 0.08

P 0.948 0.697 0.968

Fail to reject all null hypotheses. The interaction between age and gender has no effect GPA and therefore there is no significant difference in the means of the GPAs. 18. H 0 : Technicians have no effect on mean repair time. H a : Technicians have an effect on mean repair time. H 0 : Brand has no effect on mean repair time. H a : Brand has an effect on mean repair time. H 0 : There is no interaction effect between technicians and brand on mean repair time. H a : There is interaction effect between technicians and brand on mean repair time. Source Technicians Brand Interaction Error Total

d.f. 3 2 6 24 35

SS 714.31 382.17 2006.94 3277.33 6380.75

MS 238.102 191.083 334.491 136.556

F 1.74 1.40 2.45

P 0.185 0.266 0.054

There is an of interaction effect between technicians and brand on the mean repair time it takes to repair a laptop at the 10% level of significance.

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

19.

Pop 1 Pop2 Pop3

Mean 20.5 18 24.33

Size 6 6 6

SSW = 56.83

¦ (n − 1) = N − k = 15 i

F0 = 6.36 → CVScheffe′ = 6.36(3 − 1) = 12.72

( x1 − x2 ) 2 ≈ 4.95 → No difference ª1 1 º SSW « + » (ni − 1) ¬ n1 n2 ¼

( x1 − x3 ) 2 ≈ 11.62 → No difference ª1 1º SSW « + » ( ni − 1) ¬ n1 n3 ¼

( x2 − x3 ) 2 ≈ 31.73 → Significant difference ª1 SSW 1º « + » (ni − 1) ¬ n2 n3 ¼

¦ 20.

Pop 1 Pop 2 Pop 3

Mean 85.41 74.67 54.10

Size 10 10 10

SSW = 6486.19

¦ (n − 1) = N − k = 27 i

F0 = 5.49 → CVScheffe′ = 5.49(3 − 1) = 10.98

( x1 − x2 ) 2 ≈ 2.40 → No difference ª1 1 º SSW « + » (ni − 1) ¬ n1 n2 ¼

( x1 − x3 ) 2 ≈ 20.40 → Significant difference ª1 1º SSW « + » (ni − 1) ¬ n1 n3 ¼

( x2 − x3 ) 2 ≈ 8.81 → No difference ª1 SSW 1º « + » (ni − 1) ¬ n2 n3 ¼

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CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

21.

Pop 1 Pop 2 Pop 3 Pop 4 Pop 5

Mean 41.00 51.33 63.40 57.75 59.40

Size 6 6 5 4 5

SSW = 668.48333

¦ (n − 1) = N − k = 21 i

F0 = 4.37 → CVScheffe′ = 4.37(5 − 1) = 17.48

( x1 − x2 ) 2 ≈ 10.06 → No difference ª1 1 º SSW « + » ( ni − 1) ¬ n1 n2 ¼

( x1 − x3 ) 2 ≈ 42.99 → Significant difference ª1 1º SSW « + » (ni − 1) ¬ n1 n3 ¼

( x1 − x4 ) 2 ≈ 21.15 → Significant difference ª1 1 º SSW « + » (ni − 1) ¬ n1 n4 ¼

( x1 − x5 ) 2 ≈ 29.01 → Significant difference ª1 1º SSW « + » ( ni − 1) ¬ n1 n5 ¼

( x2 − x3 ) 2 ≈ 12.48 → No difference ª1 SSW 1º « + » (ni − 1) ¬ n2 n3 ¼

( x2 − x4 ) 2 ≈ 3.11 → No difference ª1 SSW 1º « + » (ni − 1) ¬ n2 n4 ¼

( x2 − x5 ) 2 ≈ 5.58 → No difference ª1 SSW 1º « + » (ni − 1) ¬ n2 n5 ¼

( x3 − x4 ) 2 ≈ 2.23 → No difference ª1 SSW 1º « + » (ni − 1) ¬ n3 n4 ¼

( x3 − x5 ) 2 ≈ 1.26 → No difference ª1 SSW 1º « + » (ni − 1) ¬ n3 n5 ¼

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

( x4 − x5 ) 2 ≈ 0.19 → No difference ª1 SSW 1º « + » ( ni − 1) ¬ n4 n5 ¼

¦ 22.

Pop 1 Pop 2 Pop 3 Pop 4

Mean 62.55 62.4 61.67 62.88

Size 6 8 11 9

SSW = 21.2124

¦ (n − 1) = N − k = 30 i

F0 = 2.28 → CVScheffe′ = 2.28(4 − 1) = 6.84

( x1 − x2 ) 2 ≈ 0.11 → No difference ª1 1 º SSW « + » (ni − 1) ¬ n1 n2 ¼

( x1 − x3 ) 2 ≈ 4.23 → No difference ª1 1º SSW « + » (ni − 1) ¬ n1 n3 ¼

( x1 − x4 ) 2 ≈ 0.55 → No difference ª1 1 º SSW « + » (ni − 1) ¬ n1 n4 ¼

( x2 − x3 ) 2 ≈ 3.46 → No difference ª1 SSW 1º « + » (ni − 1) ¬ n2 n3 ¼

( x2 − x4 ) 2 ≈ 1.37 → No difference ª1 SSW 1º « + » (ni − 1) ¬ n2 n4 ¼

( x3 − x4 ) 2 ≈ 10.17 → Significant difference ª1 SSW 1º « + » (ni − 1) ¬ n3 n4 ¼

CHAPTER 10 REVIEW EXERCISE SOLUTIONS 1. (a)

Age Less than 40 40–49 50–59 60–69 70 and older

Distribution 20% 26% 28% 22% 4%

449

450

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

H 0 : The age distribution of U.S. physicians is 20% are less than 40, 26% are 40–49, 28% are

50–59, 22% are 60–69, and 4% are 70 and older. H a : The age distribution of U.S. physicians differs from the expected distribution. (claim) 2 2 (b) χ0 = 13.277 ; Rejection region: χ > 13.277

(c)

Age

Distribution

Observed

Expected

20% 26% 28% 22% 4%

100 85 125 70 20

80 104 112 88 16

Less than 40 40–49 50–59 60–69 70 and older

(O E )2 E 5 3.4712 1.5089 3.6818 1 14.662

χ 2 ≈ 14.662 2 (d) Because χ > 13.277 , reject H 0 .

(e) There is not enough evidence at the 1% level of significance to conclude that the distribution of the ages differs from the expected distribution. 2. (a)

Response $5 or less $6–$10 $11–$20 $21–$50 $51 or more

Distribution 14% 22% 35% 22% 7%

H 0 : The distribution of the allowance amounts is 14% $5 or less, 22% $6–$10, 35% $11–$20,

22% $21–$50, and 7% $51 or more. H a : The distribution of allowance amounts differs from the expected distribution. (claim) 2 2 (b) χ0 = 7.779 ; Rejection region: χ > 7.779

(c)

Response

Distribution

Observed

Expected

$5 or less $6–$10 $11–$20 $21–$50 $51 or more

14% 22% 35% 22% 7%

220 312 539 337 101

211.26 331.98 528.15 331.98 105.63

χ 2 ≈ 2.066 2 (d) Because χ < 7.779 , fail to reject H 0 .

(O E ) 2 E 0.3616 1.2025 0.2229 0.0759 0.2029 2.066

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

451

(e) There is not enough evidence at the 10% level of significance to conclude that the distribution of allowance amounts differs from the expected distribution. 3. (a)

Response Slow play Poor course conditions Poor etiquette High green fees

Distribution 65.1% 18.5% 11.1% 5.3%

H 0 : The distribution of opinions of golfers about what irritates them the most is 65.1% slow

play, 18.5% poor course conditions, 11.1% poor etiquette, and 5.3% high green fees. (claim) H a : The distribution of opinions of golfers about what irritates them the most differs from the expected distribution. 2 2 (b) χ0 = 7.815 ; Rejection region: χ > 7.815

(c)

Response

Distribution

Observed

Expected

Slow play Poor course conditions Poor etiquette High green fees

65.1% 18.5% 11.1% 5.3%

646 201 126 45

662.718 188.33 112.998 53.954

(O E ) 2 E 0.4217 0.8524 1.4961 1.4860 4.256

χ 2 ≈ 4.256 2 (d) Because χ < 7.815 , fail to reject H 0 .

(e) There is not enough evidence at the 5% level of significance to conclude that the distribution of opinions of golfers about what irritates them the most differs from the expected distribution. 4. (a)

Response 15,000–17,499 17,500–19,999 20,000–22,499 22,500–24,499 25,000+

Distribution 20% 20% 20% 20% 20%

H 0 : The distribution of charges for tuition, fees, room, and board at 4-year degree granting

postsecondary institutions is uniform. (claim) H a : The distribution of charges for tuition, fees, room, and board is not uniform. 2 2 (b) χ0 = 9.488 , Rejection region: χ > 9.488

452

(c)

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

Response

Distribution

Observed

Expected

Alameda Contra Costa Fresno Kern Los Angeles

20% 20% 20% 20% 20%

138 154 246 169 93

160 160 160 160 160

(O E )2 E

3.025 0.225 46.225 0.506 28.056 78.037

χ 2 ≈ 78.037 2 (d) Because χ > 9.488 , reject H 0 .

(e) There is enough evidence at the 5% level of significance to reject the claim that the distribution of charges for tuition, fees, room, and board at 4-year degree-granting postsecondary institutions is uniform. 5. (a) Expected frequencies: Level of instruction Elementary Secondary Total

Years of full-time teaching experience Less than 3 20 years or years more 3−9 years 10−20 years 93.089 293.379 423.703 240.829 216.911 683.621 987.297 561.171 310 977 1411 802

Total 1051 2449 3500

(b) The claim is “level of instruction is related to the years of full-time teaching experience.” H 0 : The years of full-time teaching experience is independent of the level of instruction. H a : The years of full-time teaching experience is dependent on the level of instruction. (claim) 2 2 (c) d.f. = (r − 1)(c − 1) = 3 ; χ0 = 11.345 ; Rejection region: χ > 11.345

(d)

O−E

(O E )2

99 307 410 235 211 670 1001 567

93.089 293.379 423.703 240.829 216.911 683.621 987.297 561.171

5.911 13.621 í13.703 í5.829 í5.911 í13.621 13.703 5.829

34.939921 185.531641 187.772209 33.977241 34.939921 185.531641 187.772209 33.977241

χ 2 ≈ 2.275 2 (e) Because χ < 11.345 , fail to reject H 0 .

(O E )2 E 0.37533888 0.6323957782 0.4431694111 0.1410845081 0.1610795257 0.2713954677 0.1901881693 0.0605470365 2.275

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

453

(f) There is not enough evidence at the 1% level of significance to conclude that the years of fulltime teaching experience is dependent on the level of instruction. 6. (a) Expected frequencies: Generation Gen X (ages 41– 56) Millennial (ages 25 – 40) Total

Car 94.25

Type of Vehicle Owned Truck SUV Van 81.2 50.75 5.8

Total 232

100.75

86.8

54.25

6.2

248

195

168

105

480

(b) The claim is “type of vehicle owned is related to generation.” H 0 : The type of vehicle owned is independent of generation. H a : The type of vehicle owned is dependent on generation. (claim) 2 2 (c) d.f. = ( r − 1)(c − 1) = 3 ; χ0 = 11.345 ; Rejection region: χ > 11.345

(d)

O−E

(O E ) 2

85 95 44 8 110 73 61 4

94.25 81.2 50.75 5.8 100.75 86.8 54.25 6.2

í9.25 13.8 í6.75 2.2 9.25 í13.8 6.75 í2.2

85.5625 190.44 45.5625 4.84 85.5625 190.44 45.5625 4.84

(O E ) 2 E 0.9078249337 2.345320197 0.8977832512 0.8344827586 0.8492555831 2.194009217 0.8398617512 0.7806451613 9.649

χ 2 ≈ 9.649 2 (e) Because χ < 11.345 , fail to reject H 0 .

(f) There is not enough evidence at the 1% level of significance to conclude that the type of vehicle owned is dependent on generation. 7. (a) Expected frequencies: Status Endangered Threatened Total

Mammals

Vertebrate Group Birds Reptiles Amphibians

Fish

Total

132.431 38.569 171

117.717 34.283 152

72.024 20.976 93

388 113 501

48.790 14.210 63

17.038 4.962 22

(b) The claim is “a species’ status is not related to vertebrate group.” H 0 : A species’ status is independent of vertebrate group. (claim) H a : A species’ status is dependent on vertebrate group. Copyright © 2023 Pearson Education, Inc.

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CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

2 2 (c) d.f. = ( r − 1)(c − 1) = 4 ; χ0 = 9.488 ; Rejection region: χ > 9.488

(d)

O−E

(O E ) 2

148 132 39 14 55 23 20 24 8 38

132.431 117.717 48.790 17.038 72.024 38.569 34.283 14.210 4.962 20.976

15.569 14.283 í9.79 í3.038 í17.024 í15.569 í14.283 9.79 3.038 17.024

242.393761 204.004089 95.8441 9.229444 289.816576 242.393761 204.004089 95.8441 9.229444 289.816576

(O E ) 2 E 1.830340034 1.733004485 1.964420988 0.5416976171 4.023888926 6.284678395 5.950590351 6.744834624 1.86002499 13.81657971 44.750

χ 2 ≈ 44.750 2 (e) Because χ > 9.488 , reject H 0 .

(f) There is enough evidence at the 5% level of significance to reject the claim that a species status (endangered or threatened) is independent of vertebrate group. 8. (a) Expected frequencies Victim

12.–5:59 A.M.

Pedestrian Bicyclist Total

879.344 116.656 996

6.–11:59 A.M. 622.427 82.573 705

Time of Day 12.–5:59 P.M. 672.751 89.249 762

6.–11:59 P.M. 2001.478 265.522 2267

Total

4176 554 4730

(b) The claim is “the time of day of a collision is related to type of crash victim.” H 0 : The type of crash victim and time of day are independent. H a : The type of crash victim is dependent on the time of day. (claim) 2 2 (c) d.f. = (r − 1)(c − 1) = 3 ; χ0 = 6.251 ; Rejection region: χ > 6.251

(d)

O−E

(O E ) 2

924 581 617 2054 72 124 145 213

879.344 622.427 672.751 2001.478 116.656 82.573 89.249 265.522

44.656 í41.427 í55.751 52.522 í44.656 41.427 55.751 í52.522

1994.158336 1716.196329 3108.174001 2758.560484 1994.158336 1716.196329 3108.174001 2758.560484

(O E ) 2 E 2.267779545 2.757265236 4.620095698 1.378261707 17.09434865 20.78398906 34.82586921 10.38919745 94.117

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

455

2 (e) Because χ > 6.251 , reject H 0 .

(f) There is enough evidence at the 10% level of significance to conclude that the type of crash victim is dependent on the time of day. 9.

F0 ≈ 2.295

10. F0 = 4.71

11. F0 = 2.39

12. F0 = 2.01

13. F0 = 2.06

14. F0 = 4.36

15. F0 = 2.08

16. F0 = 4.73

17. (a) The claim is “the standard deviations of hotel room rates for Sacramento, CA and San Francisco, CA are the same.”

H0 : σ1 = σ 2 (claim); Ha : σ1 ≠ σ 2 (b) d.f.N = 35 ; d.f.D = 30 F0 = 1.81; Rejection region: F > 1.81 s 2 (51) 2 (c) F = 12 = ≈ 1.900 s2 (37) 2 (d) Because F > 1.81 , reject H 0 . (e) There is enough evidence at the 10% level of significance to reject the claim that the standard deviations of hotel room rates in Sacramento, CA, and San Francisco, CA, are the same. 18. (a) The claim is “the variation in wheat production is greater in Garfield County than in Kay County.”

H0 : σ12 ≤ σ 22; H a : σ12 > σ 22 (claim) (b) d.f.N = 20 ; d.f.D = 15 ; F0 = 1.92; Rejection region: F > 1.92 s 2 (0.76) 2 (c) F = 12 = ≈ 1.72 s2 (0.58) 2 (d) Because F < 1.92 , fail to reject H 0 . (e) There is not enough evidence at the 10% level of significance to support the claim that the variation in wheat production is greater in Garfield County than in Kay County. 19. (a) The claim is “the variance of SAT evidence-based reading and writing scores is different than the variance of SAT math scores.”

H0 : σ12 = σ 22; H a : σ12 ≠ σ 22 (claim) (b) d.f.N = 11 ; d.f.D = 11 F0 = 5.32 Rejection region: F > 5.32

456

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

s 2 20790.17934 (c) F = 12 = ≈ 1.37 s2 15169.61723 (d) Because F < 5.32 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to support the claim that the test score variance for reading and writing is different from that for math. 20. (a) The claim is “the new mold produces inserts that are less variable in diameter than the inserts the current mold produces.”

H0 : σ12 ≤ σ 22; H a : σ12 > σ 22 (claim) (b) d.f.N = 11 ; d.f.D = 11 F0 = 2.82; Rejection region: F > 2.82 s 2 0.00146 (c) F = 12 = ≈ 2.91 s2 0.00050 (d) Because F > 2.82 , reject H 0 . (e) There is enough evidence at the 5% level of significance to support the claim that the new mold produces inserts that are less variable in diameter than those produced with the current mold. 21. (a) The claim is “the mean monthly electric bill is different in at least one of the regions.” H 0 : μ1 = μ 2 = μ 3 = μ 4 H a : At least one mean is different from the others. (claim)

(b) d.f.N = k − 1 = 3 ; d .f.D = N − k = 3 2 F0 = 2.26 ; Rejection region: F > 2.26 (c) Variation Between Within F ≈ 2.18

Sum of Squares 2488.0945 12155.0036

Degrees of Freedom 3 32

Mean Squares 829.364832 379.843863

F 2.18

(d) Because F < 2.26 , fail to reject H 0 . (e) There is not enough evidence at the 10% level of significance to conclude that at least one monthly electric bill is different from the others. 22. (a) The claim is “at least one of the mean incomes is different from the others.” H 0 : μ1 = μ 2 = μ 3 = μ 4 H a : At least one mean is different from the others. (claim)

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

457

(b) d.f.N = k − 1 = 3 ; d .f.D = N − k = 2 0 F0 = 3.10 ; Rejection region: F > 3.10 (c) Variation Between Within

Sum of Squares 842026719 207849599

Degrees of Freedom 3 20

Mean Squares 280675578 10392479.9

F 27.01

F ≈ 27.01 (d) Because F > 3.10 , reject H 0 . (e) There is enough evidence at the 5% level of significance to conclude that at least one of the mean incomes is different from the others.

CHAPTER 10 QUIZ SOLUTIONS 1. (a) H 0 : The distribution of educational attainment for people in the United States ages 30−34 is 3.5% none–8th grade, 5.5% 9th–11th grade, 27.6% high school graduates, 15.2% some college, no degree, 10.6% associate’s degree, 23.4% bachelor’s degree, 10.6% master’s degree, and 3.6% professional/doctoral degree. H a : The distribution of educational attainment for people in the United States ages 30−34 differs from the distribution for people ages 25 and older. (claim) 2 2 (b) χ0 = 14.067 ; Rejection region: χ > 14.067

(c) None–8th grade 9th–11th grade High school graduate Some college, no degree Associate’s degree Bachelor’s degree Master’s degree Professional/Doctoral degree

Distribution

Observed

Expected

3.5% 5.5% 27.6% 15.2% 10.6% 23.4% 10.6% 3.6%

10 23 80 56 34 75 31 11

11.2 17.6 88.32 48.64 33.92 74.88 33.92 11.52

320

(O E ) 2 E 0.1285714286 1.656818182 0.7837681159 1.113684211 0.0001886792453 0.0001923076923 0.2513679245 0.0234722222 3.958

χ 2 ≈ 3.958 2 (d) Because χ < 14.067 , fail to reject H 0 .

(e) There is not enough evidence at the 5% level of significance to conclude that the distribution for people in the United States ages 30−34 differs from the distribution for people ages 25 and older.

458

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

2. (a) H 0 : Age and educational attainment are independent. H a : Age and educational attainment are dependent. (claim) 2 2 (b) d.f. = ( r − 1)(c − 1) = 7 ; χ0 = 18.475 ; Rejection region: χ > 18.475

(c)

10 23 80 56 34 75 31 11 23 26 136 77 41 78 41 18

O−E

13.89474 20.63158 90.94737 56 31.57895 64.42105 30.31579 12.21053 19.10526 28.36842 125.0526 77 43.42105 88.57895 41.68421 16.78947

í3.8947368 2.36842105 í10.947368 0 2.42105263 10.5789474 0.68421053 í1.2105263 3.89473684 í2.3684211 10.9473684 0 í2.4210526 í10.578947 í0.6842105 1.21052632

(O E ) 2 15.169 5.609 119.845 0 5.861 111.914 0.468 1.465 15.169 5.609 119.845 0 5.861 111.914 0.468 1.465

(O E )2 E 1.092 0.272 1.318 0 0.186 1.737 0.015 0.121 0.794 0.198 0.958 0 0.135 1.263 0.011 0.087 8.187

χ 2 ≈ 8.187 2 (d) Because χ < 18.475 , fail to reject H 0 .

(e) There is not enough evidence at the 1% level of significance to conclude that educational attainment is dependent on age. 2

3. Population 1: Ithaca → s1 = 72.59 2

Population 2: Little Rock → s2 = 52.48 2 2 2 2 (a) H 0 : σ1 = σ 2 ; H a : σ1 ≠ σ 2 (claim)

(b) d.f.N = 12 , d.f.D = 14 ; F0 = 4.43 ; Rejection region: F > 4.43 s 2 72.59 (c) F = 12 = ≈ 1.38 s2 52.48 (d) Because F < 4.43 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to conclude the variances in annual wages for Ithaca, NY and Little Rock, AR are different. Copyright © 2023 Pearson Education, Inc.

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

459

4. (a) H 0 : μ1 = μ 2 = μ 3 (claim) H a : At least one mean is different from the others.

(b) d.f.N = 2 , d.f.D = 40 F0 = 2.44 ; Rejection region: F > 2.44 (c) Variation Between Within

Sum of Squares 1700.22829 2264.2159

Degrees of Freedom 2 40

Mean Squares 850.114144 56.6053974

F 15.02

F ≈ 15.02 (d) Because F > 2.44 , reject H o . (e) There is enough evidence at the 10% level of significance to reject the claim that the mean annual wages are equal for all three cities.

CUMULATIVE REVIEW, CHAPTERS 9–10 1. (a) r ≈ 0.774 ; strong positive linear correlation

(b) There is enough evidence at the 5% level of significance to conclude that there is a significant linear correlation between the men’s and women’s winning 100-meter times. (c) y = 1.228 x − 1.225

460

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

2. The claim is that “the mean expenditures are the same for all four regions.” H 0 : μ1 = μ 2 = μ 3 = μ 4 (claim) H a : At least one mean is different from the others d.f.N = k − 1 = 3 , d .f.D = N − k = 2 8 F0 = 2.29 ; Rejection region: F > 2.29 Variation Between Within

Sum of Squares 451179.25 1253020.25

Degrees of Freedom 3 28

Mean Squares 150393.08 44750.7232

F 3.36

F ≈ 3.36 Because F > 2.29 , reject H o . There is enough evidence at the 10% level of significance to reject the claim that the mean expenditures are the same for all four regions. 3. (a) y = 17,106 − 0.318(110,000) + 0.303(100,000) = 12, 426 12,426 pounds per acre

(b) y = 17,106 − 0.318(125,000) + 0.303(115,000) = 12, 201 12,201 pounds per acre 4.

H0 : σ12 = σ 22 (claim); H a : σ12 ≠ σ 22 d.f.N = 15 , d.f.D = 14 F0 = 2.46; Rejection region: F > 2.46

s 2 33.12 F = 12 = ≈ 1.083 s2 31.82 Because F < 2.46 , fail to reject H 0 . There is not enough evidence at the 10% level of significance to reject the administrator’s claim that the standard deviations of reading test scores for eighth-grade students are the same in Illinois and Georgia. 5.

H 0 : The distribution of how families pay for college is distributed as 7.7% student income and savings,

8.4% parent borrowing, 1.3% relatives and friends, 43.8% parent income and savings, 25.4% scholarships and grants, and 13.4% student borrowing. (claim) H a : The distribution of how families pay for college differs from the claimed or expected distribution.

χ02 = 11.071 ; Rejection region: χ 2 > 11.071

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

Distribution

Observed

Expected

0.438

393

394.2

(O − E ) 2 E 0.003652968

0.254

222

228.6

0.1905511811

0.134

120

120.6

0.0029850746

0.077

69.3

3.556854257

0.084 0.013

66 14

75.6 11.7

1.219047619 0.4521367521

Parent income and savings Scholarships and grants Student borrowing Student income and savings Parent borrowing Relatives and friends

461

5.43

χ 2 ≈ 5.43 2

Because χ < 11.071 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the claim that the distributions are the same. 6.

H 0 : Rating and generation are independent. H a : Rating and generation are dependent. (claim) 2 2 d.f. = (r − 1)(c − 1) = 3 ; χ0 = 7.815 ; Rejection region: χ > 7.815

O−E

(O E ) 2

97 42 26 5 101 33 25 11

99 37.5 25.5 8 99 37.5 25.5 8

í2 4.5 0.5 í3 2 í4.5 í0.5 3

4 20.25 0.25 9 4 20.25 0.25 9

(O E )2 E 0.0404040404 0.54 0.0098039216 1.125 0.0404040404 0.54 0.0098039216 1.125 3.43

χ 2 ≈ 3.43 2

Because χ < 7.815 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to conclude that the adults’ ratings of the movies are dependent on generation.

( 7. (a) r = 2

¦ y i − y

(

¦ yi − y

) ≈ 0.743 2

)

About 74.3% of the variation in height can be explained by the relationship between metacarpal bone length and height, and about 25.7% of the variation is unexplained.

462

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION 2

yi l yi

154.170592 x 3.93 n 2 10 The standard error of estimate of the height for a specific metacarpal bone length is about 3.93 centimeters.

(b) se

x 50 y 1.746x

92.536 1.746 50

E t c se 1

1 n

n x0 x

n x2 x

x 2.228 3.93

1 12

92.536 179.836

12 50 44.66667

12 24088 536

x 9.896

y o E l 179.836 o 9.896 l 169.94, 189.73 You can be 95% confident that the height will be between 169.94 centimeters and 189.73 centimeters when the metacarpal bone length is 50 millimeters.

CHAPTER 10 TEST SOLUTIONS 2 2 2 2 1. (a) H0 : σ1 = σ 2 (claim); H a : σ1 ≠ σ 2

(b) d.f.N = 10 , d.f.D = 11 ; F0 = 3.53 ; Rejection region: F > 3.53 s 2 ( 3.726 ) (c) F = 12 ≈ ≈ 1.01 s2 ( 3.704 )2 2

(d) Because F < 3.53 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to reject the claim that the variances of the hourly wages are the same. 2 2 2 2 2. (a) H0 : σ1 ≤ σ 2 ; H a : σ1 > σ 2 (claim)

(b) d.f.N = 9 , d.f.D = 11 ; F0 = 4.63 ; Rejection region: F > 4.63

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

463

s 2 (3.937) 2 (c) F = 12 = ≈ 1.130 s2 (3.704) 2 (d) Because F < 4.63 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to support the claim that the variance of the hourly wages in Oklahoma is greater than the variance of the hourly wages in Massachusetts. 3. (a) H 0 : μ1 = μ 2 = μ 3 (claim) H a : At least one mean is different from the others.

(b) d.f.N = 2 , d.f.D = 30 F0 = 5.39 ; Rejection region: F > 5.39 (c) Variation Between Within F ≈ 11.12

Sum of Squares 318.095755 429.221596

Degrees of Freedom 2 30

Mean Squares 159.047878 14.3073865

F 11.12

(d) Because F > 5.39 , reject H o . (e) There is enough evidence at the 1% level of significance to reject the claim that the mean hourly wages are the same for all three states. 4. (a) H 0 : The distribution of ages of workers in Oklahoma is 7.1% ages 16–19, 11.6% ages 20–24, 43.1% ages 25–44, 19.6% ages 45–54, 7.8% ages 55–59, and 10.8% ages 60+. H a : The distribution of the ages of workers in Oklahoma differs from the distribution of ages of workers in Maine. (claim) 2 2 (b) χ0 = 9.236 , Rejection region: χ > 9.236

(c)

Ages

Distribution

Observed

Expected

16–19 20–24 25–44 45–54 55–59 60+

7.1% 11.6% 43.1% 19.6% 7.8% 10.8%

12 25 95 36 15 17

14.2 23.2 86.2 39.2 15.6 21.6

χ 2 ≈ 2.64 2 (d) Because χ < 9.236 , fail to reject H 0 .

(O E )2 E 0.3408450704 0.1396551724 0.8983758701 0.2612244898 0.0230769231 0.9796296296 2.64

464

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(e) There is not enough evidence at the 10% level of significance to support the claim that the distribution of ages of workers in Oklahoma differs from the distribution of ages of workers in Maine. 5. (a) H 0 : The distribution of ages of workers in Massachusetts is 7.1% ages 16–19, 11.6% ages 20–24, 43.1% ages 25–44, 19.6% ages 45–54, 7.8% ages 55–59, and 10.8% ages 60+. H a : The distribution of the ages of workers in Massachusetts differs from the distribution of ages of workers in Maine. (claim) 2 2 (b) χ0 = 15.086 , Rejection region: χ > 15.086

(c)

Ages

Distribution

Observed

Expected

16–19 20–24 25–44 45–54 55–59 60+

7.1% 11.6% 43.1% 19.6% 7.8% 10.8%

16 21 86 37 15 25

14.2 23.2 86.2 39.2 15.6 21.6

(O E ) 2 E 0.2281690141 0.2086206897 0.000464037123 0.1234693878 0.0230769231 0.5351851852 1.12

χ 2 ≈ 1.12 2 (d) Because χ < 15.086 , fail to reject H 0 .

(e) There is not enough evidence at the 1% level of significance to reject the claim that the distribution of ages of workers in Massachusetts is the same as the distribution of ages in workers in Maine. 6. (a) The claim is “state and age are not related.” H 0 : State and age are independent. (claim) H a : State and age are dependent. 2 2 (b) d.f. = (r − 1)(c − 1) = 5 ; χ0 = 11.071 ; Rejection region: χ > 11.071

(c)

O−E

(O E ) 2

12 25 95 36 15 17 16 21 86 37 15 25

14 23 90.5 36.5 15 21 14 23 90.5 36.5 15 21

í2 2 4.5 í0.5 0 í4 2 í2 í4.5 0.5 0 4

4 4 20.25 0.25 0 16 4 4 20.25 0.25 0 16

(O E ) 2 E 0.2857142857 0.1739130435 0.2237569061 0.0068493151 0 0.7619047619 0.2857142857 0.1739130435 0.2237569061 0.0068493151 0 0.7619047619 2.90

CHAPTER 10 Ň CHI-SQUARE TESTS AND THE F-DISTRIBUTION

465

2 (d) Because χ < 11.071 , fail to reject H 0 .

(e) There is not enough evidence at the 5% level of significance to reject the claim that state and age are independent.

CHAPTER

Nonparametric Tests

11 11.1 THE SIGN TEST

11.1 TRY IT YOURSELF SOLUTIONS 1. The claim is “the median number of days a home is on the market in its city is no more than 120.” H 0 : median d 120 (claim); H a : median > 120 D 0.025 n 23 The critical value is 6. x 6 Because x d 6 , reject H0 . There is enough evidence at the 2.5% level of significance to reject the agency’s claim that the median number of days a home is on the market in its city is no more than 120. 2. The claim is “the median age of museum educator employees in the United States is 44 years old.” H 0 : median 44 (claim) H a : median z 44 D 0.10 n 93 The critical value is –1.645. x 45 ( x 0.5) 0.5n (45 0.5) 0.5(93) 1 z | | 0.21 4.82 93 n 2 2 Because z ! 1.645 , fail to reject H0 . There is not enough evidence at the 10% level of significance to reject the claim that the median age of museum educator employees in the United States is 44 years old. 3. The claim is “a new vaccine will decrease the number of colds in adults.” H 0 : The number of colds will not decrease. H a : The number of colds will decrease. (claim) D 0.05 n 11 The critical value is 2. x 2 Because x d 2 , reject H0 . There is enough evidence at the 5% level of significance to support the researcher’s claim that the new vaccine will decrease the number of colds in adults.

CHAPTER 11 │ NONPARAMETRIC TESTS

467

11.1 EXERCISE SOLUTIONS 1. A nonparametric test is a hypothesis test that does not require any specific conditions concerning the shapes of populations or the values of any population parameters. A nonparametric test is usually easier to perform than its corresponding parametric test, but the nonparametric test is usually less efficient. 2. Median 3. When n is less than or equal to 25, the test statistic is equal to x (the smaller number of + or signs). ( x 0.5) 0.5n . When n is greater than 25, the test statistic is equal to z n 2 4. Answers will vary. Sample answer: It is called the sign test because each entry in a sample is compared to the hypothesized median and assigned a + or sign, based on whether the sample entry is greater than (+) the hypothesized median or less than ( ) the hypothesized median. The numbers of + signs and signs are used to determine whether the null hypothesis should be rejected. 5. Verify that the sample is random. Identify the claim and state H 0 and H a . Identify the level of significance and sample size. Find the critical value using Table 8 (if n d 25 ) or Table 4 ( n ! 25 ). Calculate the test statistic. Make a decision and interpret it in the context of the problem. 6. A sample must be randomly selected from each population, and the samples must be dependent. 7. (a) The claim is “the median credit card balance of college students is more than $500.” H 0 : median d $500; H a : median > $500 (claim) (b) The critical value is 1. (c) x

(d) Because x ! 1 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance for the accountant to conclude that the median credit card balance of college students is more than $500. 8. (a) The claim is “the median daily high temperature for the month of July in Pittsburgh is 83q Fahrenheit.” H 0 : median = 83 (claim); H a : median z 83 (b) The critical value is 1. (c) x

(d) Because x ! 1 , fail to reject H0 .

468

CHAPTER 11 │ NONPARAMETRIC TESTS

(e) There is not enough evidence at the 1% level of significance to reject the meteorologist’s claim that the median daily high temperature for the month of July in Pittsburgh is 83q Fahrenheit. 9. (a) The claim is “the median sales price of new privately owned one-family homes sold in a recent month is $500,000 or less.” H 0 : median d $500,000 (claim); H a : median ! $500,000 (b) The critical value is 1. (c) x 3 (d) Because x ! 1 , fail to reject H0 . (e) There is not enough evidence at the 5% level of significance to reject the agent’s claim that the median sales price of new privately owned one-family homes sold in a recent month is $500,000 or less. 10. (a) The claim is “the median daily high temperature for the month of January in San Diego is 66° Fahrenheit.” H 0 : median = 66 (claim); H a : median z 66 (b) The critical value is 2. (c) x

(d) Because x d 2 , reject H0 . (e) There is enough evidence at the 1% level of significance to reject the meteorologist’s claim that the median daily high temperature for the month of January in San Diego is 66° Fahrenheit. 11. (a) The claim is “the median annual amount households with employer health insurance coverage in Hawaii spend on health insurance premiums is at least $600.” H 0 : median t $600 (claim); H a : median $600 (b) The critical value is z0 (c) x 47 ( x 0.5) 0.5n z n 2

2.05 .

(47 0.5) 0.5(112) 8.5 | | 1.61 5.29 112 2

(d) Because z ! 2.05 , fail to reject H0 . (e) There is not enough evidence at the 2% level of significance to reject the agent’s claim that the median annual amount households with employer health insurance coverage in Hawaii spend on health insurance premiums is at least $600.

CHAPTER 11 │ NONPARAMETRIC TESTS

469

12. (a) The claim is “the median annual income of a family of four in California is less than $100.000.” H 0 : median t $100,000 ; H a : median $100,000 (claim) (b) The critical value is z0

1.96 .

(24 0.5) 0.5(66) 8.5 | | 2.09 4.06 66 2

(d) Because z 1.96 , reject H0 . (e) There is enough evidence at the 2.5% level of significance to support the accountant’s claim that the median annual income of a family of four in California is less than $100,000. 13. (a) The claim is “the median age of the users of a social media website is greater than 30 years old.” H 0 : median d 30; H a : median > 30 (claim) (b) The critical value is 4. (c) x 10 (d) Because x ! 4 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to support the research group’s claim that the median age of social media website users is greater than 30 years old. 14. (a) The claim is “the median age of the users of a social networking website is less than 32 years old.” H 0 : median t 32; H a : median < 32 (claim) (b) The critical value is 5. (c) x 5 (d) Because x d 5 , reject H0 . (e) There is enough evidence at the 5% level of significance to support the research group’s claim that the median age of the users of a social networking website is less than 32 years old. 15. (a) The claim is “the median number of rooms in renter-occupied units is four.” H 0 : median = 4; (claim) H a : median z 4 (b) The critical value is z0

1.96.

470

CHAPTER 11 │ NONPARAMETRIC TESTS

(29 0.5) 0.5(82) 11.5 | | 2.54 4.53 82 2

(d) Because z 1.96 , reject H0 . (e) There is enough evidence at the 5% level of significance to reject the organization’s claim that the median number of rooms in renter-occupied units is four. 16. (a) The claim is “the median square footage of renter-occupied units is 1000 square feet.” H 0 : median = 1000 (claim); H a : median z 1000 (b) The critical value is 5 (c) x

(d) Because x ! 5 , fail to reject H0 . (e) There is not enough evidence at the 10% level of significance to reject the organization’s claim that the median square footage of renter-occupied units is 1000 square feet. 17. (a) The claim is “the median hourly wage of computer systems analysts is $41.93.” H 0 : median $41.93 (claim); H a : median z $41.93 (b) The critical value is –2.575 (c) x 18 ( x 0.5) 0.5(n) z n 2

(18 0.5) 0.5(43) 3 | | 0.91 3.28 43 2

(d) Because z ! 2.575 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to reject the labor organization’s claim that the median hourly wage of computer systems analysts is $41.93. 18. (a) The claim is “the median hourly wage of podiatrists is at least $70.00.” H 0 : median t $70.00 (claim); H a : median < $70.00 (b) The critical value is 6. (c) x 5 (d) Because x d 6 , reject H0 . (e) There is enough evidence at the 5% level of significance to reject the organization’s claim that the median hourly wage of podiatrists is at least $70.00. Copyright © 2023 Pearson Education, Inc.

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471

19. (a) The claim is “the lower back pain intensity scores will decrease after acupuncture treatment.” H 0 : The lower back pain intensity scores will not decrease. H a : The lower back pain intensity scores will decrease. (claim) (b) The critical value is 1. (c) x

(d) Because x d 1 , reject H0 . (e) There is enough evidence at the 5% level of significance to support the physician’s claim that lower back pain intensity scores will decrease. 20. (a) The claim is “the lower back pain intensity scores will decrease after taking the anti-inflammatory drugs.” H 0 : The lower back pain intensity scores will not decrease. H a : The lower back pain intensity scores will decrease. (claim) (b) The critical value is 2. (c) x

(d) Because x ! 2 , fail to reject H0 . (e) There is not enough evidence at the 5% level of significance to support the physician’s claim that lower back pain intensity scores will decrease after taking the anti-inflammatory drugs. 21. (a) The claim is “the student’s math SAT scores will improve.” H 0 : The SAT scores will not improve. H a : The SAT scores will improve. (claim) (b) The critical value is 1. (c) x 1 (d) Because x d 1 , reject H0 . (e) There is enough evidence at the 5% level of significance to support the agency’s claim that the SAT scores will improve. 22. (a) The claim is “the student’s SAT scores will improve the second time they take the SAT.” H 0 : The SAT scores will not improve. H a : The SAT scores will improve. (claim) (b) The critical value is 0. (c) x 3 Copyright © 2023 Pearson Education, Inc.

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CHAPTER 11 │ NONPARAMETRIC TESTS

(d) Because x ! 0 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to support the guidance counselor’s claim that the SAT scores will improve. 23. (a) The claim is “the proportion of adults who feel older than their real age is equal to the proportion of adults who feel younger than their real age.” H 0 : The proportion of adults who feel older is equal to the proportion of adults who feel younger. (claim) H a : The proportion of adults who feel older is different from the proportion of adults who feel younger. The critical value is 3. x 3 Because x d 3 , reject H0 . (b) There is enough evidence at the 5% level of significance to reject the claim that the proportion of adults who feel older is equal to the proportion of adults who feel younger than their real age. 24. (a) The claim is “the proportion of adults who contact their parents by phone weekly is equal to the proportion of adults who contact their parents by phone daily.” H 0 : The proportion of adults who contact their parents by phone weekly is equal to the proportion of adults who contact their parents by phone daily. (claim) H a : The proportion of adults who contact their parents by phone weekly is different from the proportion of adults who contact their parents by phone daily. The critical value is 5. x 8 Because x ! 5 , fail to reject H0 . (b) There is not enough evidence at the 5% level of significance to reject the claim that the proportion of adults who contact their parents by phone weekly is equal to the proportion of adults who contact their parents by phone daily. 25. (a) The claim is “the median weekly earnings of 16- to 24-year-old workers is less than or equal to $600.” H 0 : median d $600 (claim); H a : median ! $600 (b) The critical value is z0

2.33.

(29 0.5) 0.5(47) 5 | | 1.46 3.43 47 2

(d) Because z 2.33 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to reject the organization’s claim that the median weekly earnings of 16- to 24-year-old workers is less than or equal to $600. Copyright © 2023 Pearson Education, Inc.

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473

26. (a) The claim is “the median weekly earnings of 25- to 34-year-old workers is greater than $900.” H 0 : median d $900 ; H a : median ! $900 (claim) (b) The critical value is z0

2.33.

(48 0.5) 0.5(73) 11 | | 2.57 4.272002 73 2

(d) Because z ! 2.33 , reject H0 . (e) There is enough evidence at the 1% level of significance to support the organization’s claim that the median weekly earnings of 25- to 24-year-old workers is greater than $900. 27. (a) The claim is “the median age of women at the time of their first marriage is less than or equal to 27 years old.” H0 : median d 27 (claim); Ha : median > 27 (b) The critical value is z0 1.645. (c) x 35 ( x 0.5) 0.5(n) z n 2

(35 0.5) 0.5(59) 5 | | 1.30 3.841 59 2

(d) Because z 1.645 , fail to reject H0 . (e) There is not enough evidence at the 5% level of significance to reject the counselor’s claim that the median age of women at the time of their first marriage is less than or equal to 27 years old. 28. (a) The claim is “the median age of men at the time of their first marriage is greater than 28 years.” H0 : median d 28; Ha : median > 28 (claim) (b) The critical value is z0 1.645. (c) x 33 ( x 0.5) 0.5(n) z n 2

(33 0.5) 0.5(56) 4.5 | | 1.20 3.742 56 2

(d) Because z 1.645 , fail to reject H0 . (e) There is not enough evidence at the 5% level of significance to support the counselor’s claim that the median age of men at the time of their first marriage is greater than 28 years.

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CHAPTER 11 │ NONPARAMETRIC TESTS

11.2 THE WILCOXON TESTS 11.2 TRY IT YOURSELF SOLUTIONS 1. The claim is “a spray-on water repellant is effective.” H0 : The water repellent does not increase the water repelled. Ha : The water repellent increases the water repelled. (claim) D 0.01 n 11 The critical value is 7. No repellent 8 7 7 4 6 10 9 5 9 11 8 4

Repellent applied 15 12 11 6 6 8 8 6 12 8 14 8

Difference 7 5 4 2 0 2 1 1 3 3 6 4

Absolute value 7 5 4 2 0 2 1 1 3 3 6 4

Rank 11 9 7.5 3.5 — 3.5 1.5 1.5 5.5 5.5 10 7.5

Signed rank 11 9 7.5 3.5 — 3.5 1.5 1.5 5.5 5.5 10 7.5

Sum of negative ranks = 55.5 Sum of positive ranks = 10.5 ws 10.5 Because ws ! 7 , fail to reject H0 . There is not enough evidence at the 1% level of significance to support the claim that the spray-on water repellent is effective. 2. The claim is “there is a difference in the claims paid by the companies.” H0 : There is no difference in the claims paid by the companies. Ha : There is a difference in the claims paid by the companies. (claim) D 0.05 The critical values are z0 r1.96. n1 12 and n2 12

CHAPTER 11 │ NONPARAMETRIC TESTS

Ordered data 1.7 1.8 2.2 2.5 3.0 3.0 3.4 3.9 4.1 4.4 4.5 4.7

Sample B B B A A B B A B B A B

Rank 1 2 3 4 5.5 5.5 7 8 9 10 11 12

Ordered data 5.3 5.6 5.8 6.0 6.2 6.3 6.5 7.3 7.4 9.9 10.6 10.8

Sample B B A A A A A B A A A B

475

Rank 13 14 15 16 17 18 19 20 21 22 23 24

R = sum ranks of company B = 120.5 (or R 179.5) n1 (n1 n2 1) 12(12 12 1) 150 PR 2 2 n1n2 (n1 n2 1) (12)(12)(12 12 1) | 17.321 VR 12 12 R PR 120.5 150 | | 1.703 (or z 1.703) z 17.321 VR

Because 1.96 z 1.96 , fail to reject H0 . There is not enough evidence at the 5% level of significance to conclude that there is a difference in the claims paid by the companies.

11.2 EXERCISE SOLUTIONS 1. When the samples are dependent, use the Wilcoxon signed-rank test. When the samples are independent, use the Wilcoxon rank sum test. 2. The sample size for both samples must be at least 10 to use the Wilcoxon rank sum test. 3. (a) The claim is “there was no reduction in diastolic blood pressure.” H0 : There is no reduction in diastolic blood pressure. (claim) Ha : There is a reduction in diastolic blood pressure. (b) Wilcoxon signed-rank test (c) The critical value is 10.

476

CHAPTER 11 │ NONPARAMETRIC TESTS

(d)

Before treatment

After treatment

Difference

108 109 120 129 112 111 117 135 124 118 130 115

99 115 105 116 115 117 108 122 120 126 128 106

9 6 15 13 3 6 9 13 4 8 2 9

Absolute difference 9 6 15 13 3 6 9 13 4 8 2 9

Rank

Signed rank

8 4.5 12 10.5 2 4.5 8 10.5 3 6 1 8

Sum of negative ranks = 17 Sum of positive ranks = 61 ws 17 (e) Because ws ! 10 , fail to reject H0 . (f) There is not enough evidence at the 1% level of significance to reject the claim that there was no reduction in diastolic blood pressure. 4. (a) The claim is “there is no difference in the salaries earned by workers in the wholesale trade and manufacturing industries.” H0 : There is no difference in salaries. (claim) Ha : There is a difference in salaries. (b) Wilcoxon rank sum test (c) The critical value is z0 (d)

r1.645.

Ordered data 62

Sample M

Rank 1

Ordered data 88

Sample WT

Rank 11

4.5

100

CHAPTER 11 │ NONPARAMETRIC TESTS

477

R = sum ranks of Wholesale Trade = 111.5 (or R 98.5) n1 (n1 n2 1) 10(10 10 1) PR 105 2 2 n1n2 (n1 n2 1) (10)(10)(10 10 1) | 13.229 VR 12 12 R PR 111.5 105 | | 0.491 (or z | 0.491 ) z 13.229 VR (e) Because 1.645 z 1.645 , fail to reject H0 . (f) There is not enough evidence at the 10% level of significance to reject the analyst’s claim that there is no difference in the salaries earned by workers in the wholesale trade and manufacturing industries. 5. (a) The claim is “there is a difference in the earnings of people with bachelor’s degrees and those with advanced degrees.” H0 : There is no difference in the earnings. Ha : There is a difference in the earnings. (claim) (b) Wilcoxon rank sum test (c) The critical values are z0 (d)

Ordered data 50 50 63 67 67 69 71 74 82 84 88 93 96 99 99 102 104 113 114 116 138

r2.575.

Sample B B B B B B B B B A A B A B A A A A A A A

R = sum ranks of advanced degree = 159.5

Rank 1.5 1.5 3 4.5 4.5 6 7 8 9 10 11 12 13 14.5 14.5 16 17 18 19 20 21

478

CHAPTER 11 │ NONPARAMETRIC TESTS

n1 (n1 n2 1) 10(10 11 1) 110 2 2 n1n2 (n1 n2 1) (10)(11)(10 11 1) | 14.201 VR 12 12 R PR 159.5 110 | | 3.486 z 14.201 VR

(e) Because z ! 2.575 , reject H 0 . (f) There is enough evidence at the 1% level of significance to support the administrator’s claim that there is a difference in the earnings of people with bachelor’s degrees and those with advanced degrees. 6. (a) The claim is “the new drug affects the number of headache hours.” H0 : The new drug does not affect the number of headache hours. Ha : The new drug affects the number of headache hours. (claim) (b) Wilcoxon signed-rank test (c) The critical value is 2. (d)

Headache hours before 0.8 2.4 2.8 2.6 2.7 0.9 1.2

Headache hours after 1.6 1.3 1.6 1.4 1.5 1.6 1.7

Difference 0.8 1.1 1.2 1.2 1.2 0.7 0.5

Absolute difference 0.8 1.1 1.2 1.2 1.2 0.7 0.5

Rank

Signed rank

3 4 6 6 6 2 1

Sum of negative ranks = 6 Sum of positive ranks = 22 ws 6 (e) Because ws ! 2 , fail to reject H0 . (f) There is not enough evidence at the 5% level of significance to conclude that the new drug affects the number of headache hours. 7. (a) The claim is “there is a difference in the salaries earned by teachers in Virginia and Missouri.” H0 : There is no difference in salaries. Ha : There is a difference in salaries. (claim) (b) Wilcoxon rank sum test (c) The critical values are z0

r1.96.

CHAPTER 11 │ NONPARAMETRIC TESTS

(d)

Ordered data 37 40 46 46 46 49 50 51 53 53 56 56 57 59 60 60 61 61 63 64 66 66 66

Sample M M M M M M V M V M V M M M V V V V V V V V M

479

Rank 1 2 4 4 4 6 7 8 9.5 9.5 11.5 11.5 13 14 15.5 15.5 17.5 17.5 19 20 22 22 22

R = sum ranks of Wisconsin = 177 n1 (n1 n2 1) 11(11 12 1) 132 PR 2 2 n1n2 (n1 n2 1) (11)(12)(11 12 1) | 16.248 VR 12 12 R PR 177 132 | | 2.77 z 16.248 VR (e) Because z ! 2.575 , reject H 0 . (f) There is enough evidence at the 5% level of significance to support the representative’s claim that there is a difference in the salaries earned by teachers in Virginia and Missouri. 8. (a) The claim is “the experimental medication affects an individual’s heart rate.” H0 : The experimental medication does not affect an individual’s heart rate. Ha : The experimental medication affects an individual’s heart rate. (claim) (b) Wilcoxon signed-rank test Copyright © 2023 Pearson Education, Inc.

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CHAPTER 11 │ NONPARAMETRIC TESTS

Before 72 81 75 76 79 74 65 67 76 83 66 75 76 78 68

After 73 80 75 79 74 76 73 67 74 77 70 77 76 75 74

Difference 1 1 0 3 5 2 8 0 2 6 4 2 0 3 6

Absolute value 1 1 0 3 5 2 8 0 2 6 4 2 0 3 6

Rank 1.5 1.5 — 6.5 9 4 12 — 4 10.5 8 4 — 6.5 10.5

Signed rank 1.5 1.5 — 6.5 9 4 12 — 4 10.5 8 4 — 6.5 10.5

Sum of negative ranks = 46.5 Sum of positive ranks = 31.5 ws 31.5 (e) Because ws ! 14 , fail to reject H0 . (f) There is not enough evidence at the 5% level of significance for the physician to conclude that the experimental medication affects an individual’s heart rate. 9. The claim is “a certain fuel additive improves a car’s gas mileage.” H0 : The fuel additive does not improve gas mileage. Ha : The fuel additive does improve gas mileage. (claim) The critical value is z0 1.28.

CHAPTER 11 │ NONPARAMETRIC TESTS

Before 36.4 36.4 36.6 36.6 36.8 36.9 37.0 37.1 37.2 37.2 36.7 37.5 37.6 37.8 37.9 37.9 38.1 38.4 40.2 40.5 40.9 35.0 32.7 33.6 34.2 35.1 35.2 35.3 35.5 35.9 36.0 36.1 37.2

After 36.7 36.9 37.0 37.5 38.0 38.1 38.4 38.7 38.8 38.9 36.3 38.9 39.0 39.1 39.4 39.4 39.5 39.8 40.0 40.0 40.1 36.3 32.8 34.2 34.7 34.9 34.9 35.3 35.9 36.4 36.6 36.6 38.3

Difference 0.3 0.5 0.4 0.9 1.2 1.2 1.4 1.6 1.6 1.7 0.4 1.4 1.4 1.3 1.5 1.5 1.4 1.4 0.2 0.5 0.8 1.3 0.1 0.6 0.5 0.2 0.3 0 0.4 0.5 0.6 0.5 1.1

Absolute value 0.3 0.5 0.4 0.9 1.2 1.2 1.4 1.6 1.6 1.7 0.4 1.4 1.4 1.3 1.5 1.5 1.4 1.4 0.2 0.5 0.8 1.3 0.1 0.6 0.5 0.2 0.3 0 0.4 0.5 0.6 0.5 1.1

Rank 4.5 11 7 17 19.5 19.5 25 30.5 30.5 32 7 25 25 21.5 28.5 28.5 25 25 2.5 11 16 21.5 1 14.5 11 2.5 4.5 — 7 11 14.5 11 18

Signed rank 4.5 11 7 17 19.5 19.5 25 30.5 30.5 32 7 25 25 21.5 28.5 28.5 25 25 2.5 11 16 21.5 1 14.5 11 2.5 4.5 — 7 11 14.5 11 18

Sum of negative ranks = 484.5 Sum of positive ranks = 43.5 ws 43.5

n(n 1) 4 n(n 1)(2n 1) 24 ws

32(32 1) 4 32(32 1) > (2)32 1@ 24 43.5

220.5 2860

4.123

Note: n = 32 because one of the differences is zero and should be discarded. Because z 1.28 , reject H0 . There is enough evidence at the 10% level of significance for the engineer to conclude that the gas mileage is improved.

481

482

CHAPTER 11 │ NONPARAMETRIC TESTS

10. The claim is “a fuel additive improves gas mileage.” H0 : The fuel additive does not improve gas mileage. Ha : The fuel additive does improve gas mileage. (claim) The critical value is z0 1.645. Before 34.0 34.2 34.4 34.4 34.6 34.8 35.6 35.7 30.2 31.6 32.3 33.0 33.1 33.7 33.7 33.8 35.7 36.1 36.1 36.6 36.6 36.8 37.1 37.1 37.2 37.9 37.9 38.0 38.0 38.4 38.8 42.1

After 36.6 36.7 37.2 37.2 37.3 37.4 37.6 37.7 34.2 34.9 34.9 34.9 35.7 36.0 36.2 36.5 37.8 38.1 38.2 38.3 38.3 38.7 38.8 38.9 39.1 39.1 39.2 39.4 39.8 40.3 40.8 43.2

Difference 2.6 2.5 2.8 2.8 2.7 2.6 2 2 4 3.3 2.6 1.9 2.6 2.3 2.5 2.7 2.1 2 2.1 1.7 1.7 1.9 1.7 1.8 1.9 1.2 1.3 1.4 1.8 1.9 2 1.1

Absolute value 2.6 2.5 2.8 2.8 2.7 2.6 2 2 4 3.3 2.6 1.9 2.6 2.3 2.5 2.7 2.1 2 2.1 1.7 1.7 1.9 1.7 1.8 1.9 1.2 1.3 1.4 1.8 1.9 2 1.1

Rank 24.5 21.5 29.5 29.5 27.5 24.5 15.5 15.5 32 31 24.5 11.5 24.5 20 21.5 27.5 18.5 15.5 18.5 6 6 11.5 6 8.5 11.5 2 3 4 8.5 11.5 15.5 1

Signed rank 24.5 21.5 29.5 29.5 27.5 24.5 15.5 15.5 32 31 24.5 11/5 24.5 20 21.5 27.5 18.5 15.5 18.5 6 6 11.5 6 8.5 11.5 2 3 4 8.5 11.5 15.5 1

Sum of negative ranks = 528 Sum of positive ranks = 0 ws 0 n(n 1) 32(32 1) ws 0 0 264 4 4 z 4.937 n(n 1)(2n 1) 32(32 1)[2(32) 1] 2860 24 24 Because z 1.645 , reject H0 . There is enough evidence at the 5% level of significance to support the engineer’s claim that the fuel additive improves gas mileage. Copyright © 2023 Pearson Education, Inc.

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483

11.3 THE KRUSKAL-WALLIS TEST 11.3 TRY IT YOURSELF SOLUTIONS 1. The claim is “the distribution of the veterinarians’ salaries in at least one state is different from the others.” H0 : The distribution of the salaries is the same in all three states. Ha : The distribution of the salaries in at least one state is different from the others. (claim) D 0.05 d.f. k 1 2

F 02 5.991 ; Rejection region: F 2 ! 5.991 Ordered data 55.9 77.4 80.9 85.5 90.1 91.5 101.1 104.3 107.9 109.6 109.8 111.3 112 113.5 119.7 123 123.6 124 127.1 129.4 141.7 144.6 145.2 149.4 152.4 155.4 180.6 183.3 Copyright © 2023 Pearson Education, Inc.

State CA TX CA FL FL TX FL TX CA TX CA TX FL FL TX CA TX FL CA FL CA FL TX CA FL CA TX TX

Rank 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

484

CHAPTER 11 │ NONPARAMETRIC TESTS

R1 148 , R2 128 , R3 130

| 0.024 Because H 5.991 , fail to reject H0 . There is not enough evidence at the 5% level of significance to support the claim that the distribution of the veterinarians’ salaries in at least one state is different from the others.

11.3 EXERCISE SOLUTIONS 1. The conditions for using a Kruskal-Wallis test are that the samples must be random and independent, and the size of each sample must be at least 5. 2. The Kruskal-Wallis test is always a right-tailed test because you only reject the null hypothesis when H is significantly large. 3. (a) The claim is “the distributions of the SAT scores for at least one school is different from the others.” H0 : The distribution of SAT scores is the same in all three schools. Ha : The distribution of SAT scores for at least one school is different from the others. (claim) (b) F 02 (c)

5.991 ; Rejection region: F 2 ! 5.991 Ordered data 1190 1220 1250 1250 1260 1280 1280 1290 1290 1300 1340

41.5, R2 125, R3

Sample PS PS PS P P PS P PS PS P PS 64.5

Rank 1 2 3.5 3.5 5 6.5 6.5 8.5 8.5 10 11.5

Ordered data 1340 1430 1460 1470 1490 1490 1500 1520 1540 1570

Sample P P CM P CM CM CM CM CM CM

Rank 11.5 13 14 15 16.5 16.5 18 19 20 21

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485

§ R12 R2 2 R32 · 12 ¨ ¸ 3( N 1) N ( N 1) ¨© n1 n2 n3 ¸¹ § (41.5) 2 (125) 2 (64.5) 2 · 12 ¨ ¸ 3(21 1) 21(21 1) © 7 7 7 ¹ | 13.805

(d) Because H ! 5.991 , reject H0 . (e) There is enough evidence at the 5% level of significance to conclude that the distribution of SAT scores for at least one school is different from the others. 4. (a) The claim is “the distribution of the hourly pay rates of registered nurses in at least one state is different from the others.” H0 : The distribution of the hourly pay rates is the same in all three states. Ha : The distribution of the hourly pay rates in at least one state is different from the others. (claim) (b) F 02

4.605 ; Rejection region: F 2 ! 4.605

(c)

Ordered data 26.66 27.46 27.77 28.54 29.34 29.53 29.74 30 30.3 31.52 31.85 33.12 34.78 35.88 36.26 37.35 38.43 39.7 39.72 43.96 R1

65, R2

81, R3

Sample OH IN IN KY OH IN IN KY OH KY KY KY IN OH IN KY OH OH IN KY 64

Rank 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

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(d) Because H 4.605 , fail to reject H0 . (e) There is not enough evidence at the 10% level of significance to conclude that the distribution of the hourly pay rates in at least one state is different from the others. 5. (a) The claim is “the distribution of construction costs in at least one city is different from the others.” H0 : The distribution of construction costs is the same in all four cities. Ha : The distribution of construction costs in at least one city is different from the others. (claim) (b) F 02 (c)

6.251 ; Rejection region: F 2 ! 6.251 Ordered data 174 228 243 255 259 260 271 288 291 294 299 320 328 337

R1 134, R2 102, R3

Sample A A A A A A A C NY C SF C C SF

Rank 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Ordered data 338 343 350 357 370 378 391 395 401 408 412 418 441 452

28, R4 142

Sample SF C NY NY C SF NY NY SF C SF SF NY NY

Rank 15 16 17 18 19 20 21 22 23 24 25 26 27 28

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487

(e) There is enough evidence at the 10% level of significance to conclude that the distribution of construction costs in at least one city is different from the others. 6. (a) The claim is “the distribution of the amounts of caffeine in at least one beverage is different from the others.” H0 : The distribution of the amounts of caffeine is the same for all four beverages. Ha : The distribution of the amounts of caffeine in at least one beverage is different from the others. (claim) (b) F 02 11.345 ; Rejection region: F 2 ! 11.345 (c)

Ordered data 10 15 32 42 47 51 56 71 72 95 96 100 106 141 150 152 154 160 166 200 206 266 300 320 R1 105, R2 H

Sample Teas Teas Teas Teas Soft drinks Soft drinks Soft drinks Soft drinks Soft drinks Soft drinks Soft drinks Teas Teas Energy drinks Coffees Energy drinks Energy drinks Energy drinks Energy drinks Energy drinks Coffees Coffees Coffees Coffees

56, R3 104, R4

Rank 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

§ R12 R22 R32 R42 · 12 ¨ ¸ 3( N 1) N ( N 1) ¨© n1 n2 n3 n4 ¸¹ § (105) 2 (56) 2 (104) 2 (35) 2 · 12 ¨ ¸ 3(24 1) 24(24 1) © 5 7 6 6 ¹ | 18.197

(d) Because H ! 11.345 , reject H0 . (e) There is enough evidence at the 1% level of significance to conclude that the distribution of the amounts of caffeine in at least one beverage is different from the others. Copyright © 2023 Pearson Education, Inc.

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7. (a) The claim is “the number of days patients spend in the hospital is different in at least one region of the United States.” H0 : The number of days spent in the hospital is the same for all four regions. Ha : The number of days spent in the hospital is different in at least one region. (claim)

F 02 11.345 ; Rejection region: F 2 ! 11.345 Ordered data 1 1 1 1 2 3 3 3 3 3 3 4 4 4 4 5 5 5 5

Sample Rank NE 2.5 MW 2.5 S 2.5 S 2.5 W 5 NE 8.5 NE 8.5 MW 8.5 MW 8.5 W 8.5 W 8.5 MW 13.5 MW 13.5 MW 13.5 W 13.5 NE 19 MW 19 S 19 S 19

Ordered data 5 5 5 6 6 6 6 6 6 6 7 7 8 8 8 8 9 11

220.5, R2 171.5, R3 159.5, R4 151.5

Sample S W W NE NE NE MW W W W MW S NE NE S S MW NE

Rank 19 19 19 26 26 26 26 26 26 26 30.5 30.5 33.5 33.5 33.5 33.5 36 37

§ (220.5) 2 (171.5) 2 (159.5)2 (151.5)2 · 12 ¨ ¸ 3(37 1) 37(37 1) © 10 10 8 9 ¹ | 1.507 Because H 11.345 , fail to reject H0 . There is not enough evidence at the 1% level of significance to support the underwriter’s claim that the number of days patients spend in the hospital is different in at least one region of the United States.

(b) Variation Between Within

Sum of squares 9.17 194.72

Degrees of freedom 3 33

Mean squares 3.06 5.90

F 0.52

p 0.673

Because D 0.01 , the critical value is about 4.51. Because F | 0.52 is less than the critical value, the decision is to fail to reject H0 . There is not enough evidence at the 1% level of significance to support the underwriter’s claim that the number of days patients spend in the hospital is different in at least one region of the United States.

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489

(c) Both tests come to the same decision, which is that there is not enough evidence to support the claim that the number of days patients spend in the hospital is different in at least one region of the United States. 8. (a) The claim is “the energy consumed is different in at least one region.” H0 : The energy consumed is the same in all four regions. Ha : The energy consumed is different in at least one region. (claim)

F 02 11.345 ; Rejection region: F 2 ! 11.345 . Ordered data 35 35 37 39 46 59 61 62 63 65 67 69 70 72 77 81 84 85 86 86

Sample S W MW W W MW NE S W S S S W MW W W NE W S S

Rank 1.5 1.5 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19.5 19.5

Ordered data 88 89 93 95 95 97 104 113 123 123 125 127 140 140 142 158 163 169 187

251.5, R2

Sample MW NE NE NE MW NE MW W NE MW W NE NE MW S MW NE MW MW

Rank 21 22 23 24.5 24.5 26 27 28 29.5 29.5 21 32 33.5 33.5 35 36 37 38 39

271.5, R3 116.5, R4 140.5

§ (251.5) 2 (271.5) 2 (116.5) 2 (140.5) 2 · 12 ¨ ¸ 3(39 1) 39(39 1) © 10 11 8 10 ¹ | 8.438 Because H 11.345 , fail to reject H0 . There is enough evidence at the 1% level of significance to support the claim that the energy consumed is different in at least one region.

(b) Variation Between Within

Sum of squares 11,993 45,794

Degrees of freedom 3 35

Mean squares 3998 1308

F 3.06

p 0.041

Because D 0.01 , the critical value is about 4.41. Because F | 3.06 is not greater than the critical value, the decision is to fail to reject H0 . There is not enough evidence at the 1% level of significance to support the claim that the energy consumed is different in at least one region. Copyright © 2023 Pearson Education, Inc.

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(c) Both tests come to the same decision, which is that there is not enough evidence to support the claim that the energy consumed is different in at least one region.

11.4 RANK CORRELATION 11.4 TRY IT YOURSELF SOLUTIONS 1. The claim is “there is a significant correlation between the oat and wheat prices.” H 0 : Us 0 ; Ha : Us z 0 (claim) The critical value is 0.714. Oat 2.97 2.69 3.00 3.11 3.07 3.41 3.42

Rank 2 1 3 5 4 6 7

¦d 1 6

Wheat 5.24 5.46 5.48 5.83 5.86 6.04 6.46

Rank 1 2 3 4 5 6 7

d 1 −1 0 1 −1 0 0

d2 1 1 0 1 1 0 0

¦d

6(4) | 0.929 n( n 1) 7(7 2 1) Because rs ! 0.714 , reject H0 . There is enough evidence at the 10% level of significance to conclude that there is a significant correlation between the oat and wheat prices. rs

11.4 EXERCISE SOLUTIONS 1. The Spearman rank correlation coefficient can be used to describe the relationship between linear and nonlinear data. Also, it can be used for data at the ordinal level and it is easier to calculate by hand than the Pearson correlation coefficient. 2. Both the Spearman rank correlation coefficient and the Pearson correlation coefficient range from 1 to 1, inclusive. 3. The ranks of corresponding data pairs are exactly identical when rs is equal to 1. The ranks are in “reverse” order when rs is equal to 1. The ranks of have no relationship when rs is equal to 0.

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491

4. The value of rs represents the correlation coefficient of the sample data, whereas Us represents the correlation coefficient of the entire population. 5. (a) The claim is “there is a significant correlation between purchased seed expenses and fertilizer and lime expenses in the farming business.” H0 : Us 0 ; Ha : Us z 0 (claim) (b) The critical value is 0.738. (c)

Seed expenses(in millions of dollars) 1760 510 960 234 1570 400 748 760

¦d 2

Rank 8 3 6 1 7 2 4 5

n(n 1)

Fertilizer/lime expense (in millions of dollars) 2280 430 950 248 1330 430 783 680

Rank

8 2.5 6 1 7 2.5 5 4

0 0.5 0 0 0 −0.5 −1 1

0 0.25 0 0 0 0.25 1 1

¦d

2.5

6(2.5) | 0.970 8(82 1)

(d) Because rs ! 0.738 , reject H 0 . (e) There is enough evidence at the 5% level of significance to conclude that there is a significant correlation between purchased seed expenses and fertilizer and lime expenses. 6. (a) The claim is “there is a correlation between the overall score and the price.” H 0 : Us 0 ; H a : Us z 0 (claim) (b) The critical value is 0.700. (c)

Score 77 75 73 71 66 66 64 62 58

Rank 9 8 7 6 4.5 4.5 3 2 1

Price 3700 1700 1300 900 1000 1400 1800 1000 700

Rank 9 7 5 2 3.5 6 8 3.5 1

d 0 1 2 4 1 1.5 5 1.5 0

d2 0 1 4 16 1 2.25 25 2.25 0 2 d 51.5

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¦d

n(n 1)

6(51.5) | 0.571 9(92 1)

(d) Because rs 0.700 , fail to reject H0 . (e) There is not enough evidence at the 5% level of significance to conclude that there is a significant correlation between the overall score and the price. 7. (a) The claim is “there is a significant correlation between per capita spending for housing and utilities and for health care.” H0 : Us 0 ; Ha : Us z 0 (claim) (b) The critical value is 0.833. (c)

Housing, Utilities 7.0 6.7 7.2 6.8 6.8 9.1 5.8 7.4 11.1

¦d 1 6

n(n 1)

Rank 5 2 6 3.5 3.5 8 1 7 9

Health care 5.3 6.6 7.2 7.6 7.0 11.3 6.3 8.2 7.7

Rank 1 3 5 6 4 9 2 8 7

d 4 −1 1 −2.5 −0.5 −1 −1 −1 2

d2 16 1 1 6.25 0.25 1 1 1 4

¦d

31.5

6(31.5) | 0.7375 9(92 1)

(d) Because rs 0.833 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to conclude that there is a significant correlation between per capita spending for housing and utilities and for health care. 8. (a) The claim is “there is a significant correlation between the overall score and the price.” H0 : Us 0 ; Ha : Us z 0 (claim) (b) The critical value is 0.497

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(c)

Score 65 71 69 47 55 38 47 47 47 57 34 65

Rank 9.5 12 11 4.5 7 2 4.5 4.5 4.5 8 1 9.5

¦d 1 6

n(n 1)

Price 150 200 550 350 470 90 80 130 210 190 300 260

Rank 4 6 12 10 11 2 1 3 7 5 9 8

d2 30.25 36 1 30.25 16 0 12.25 2.25 6.25 9 64 2.25

d 5.5 6 1 5.5 4 0 3.5 1.5 2.5 3 8 1.5

¦d

209.5

6(209.5) | 0.267 12(122 1)

(d) Because rs 0.497 , fail to reject H0 . (e) There is not enough evidence at the 10% level of significance to conclude that there is a significant correlation between the overall score and the price. 9. The claim is “there is a significant correlation between science achievement scores and GNI.” H0 : Us 0 ; Ha : Us z 0 (claim) Science average

Rank

GNI

Rank

503 404 518 503 396 468 419

7.5 2 9 7.5 1 4 3

53,730 7,850 43,440 46,980 3870 32,200 8480

7 2 5 6 1 4 3

0.5 0 4 1.5 0 0 0

0.25 0 16 2.25 0 0 0

499

53,800

−2

495

87,950

−4

¦d

16 2

38.5

6(38.5) | 0.679 n(n 1) 9(92 1) Because rs 0.700 , fail to reject H0 . There is not enough evidence at the 5% level of significance to conclude that there is a significant correlation between science achievement scores and GNI. rs

493

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10. The claim is “there is a significant correlation between mathematics achievement scores and GNI.” H0 : Us 0 ; Ha : Us z 0 (claim) The critical value is 0.700 Math average

Rank

GNI

Rank

491 384 512 500 379 487 409 502 515

5 2 8 6 1 4 3 7 9

53,730 7850 43,440 46,980 3870 32,200 8480 53,800 87,950

7 2 5 6 1 4 3 8 9

−2 0 3 0 0 0 0 −1 0

4 0 9 0 0 0 0 1 0

¦d 1 6

¦d

6(14) | 0.883 n(n 1) 9(92 1) Because rs ! 0.700 , reject H 0 . There is enough evidence at the 5% level of significance to conclude that there is a significant correlation between mathematics achievement scores and GNI. rs

11. The claim is “there is a significant correlation between science and mathematics achievement scores.” H0 : Us 0 ; Ha : Us z 0 (claim) The critical value is 0.700. Science average

Rank

Math average

Rank

503 404 518 503 396 468 419 499 495

7.5 2 9 7.5 1 4 3 6 5

491 384 512 500 379 487 409 502 515

5 2 8 6 1 4 3 7 9

2.5 0 1 1.5 0 0 0 −1 −4

6.25 0 1 2.25 0 0 0 1 16

¦d r |1 6

¦d

26.5

6(26.5) | 0.779 n(n 1) 9(92 1) Because rs ! 0.700 , reject H 0 . There is enough evidence at the 5% level of significance to conclude that there is a significant correlation between science and mathematics achievement scores. s

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495

12. No. Sample answer: Although there is a strong correlation between math and science scores, and math scores and per capita GNI, there is not a significant correlation between science scores and per capita GNI. 13. The claim is “there is a significant correlation between average hours worked and the number of onthe-job injuries.” H 0 : Us 0 ; H a : Us z 0 (claim) rz r1.645 | r0.291. The critical values are n 1 33 1 Hours worked 46 43 41 40 41 42 45 45 42 45 44 44 45 46 47 47 46 46 49 50 50 42 41 42 41 41 41 41 40 39 38 39 39

¦d 1 6

Rank 26.5 18 10 5.5 10 15.5 22.5 22.5 15.5 22.5 19.5 19.5 22.5 26.5 29.5 29.5 26.5 26.5 31 32.5 32.5 15.5 10 15.5 10 10 10 10 5.5 3 1 3 3

n(n 1)

Injuries 22 25 18 17 20 22 28 29 24 26 26 25 27 29 29 30 29 29 30 30 30 23 22 23 21 19 18 18 17 16 16 16 16

6(246) | 0.959 33(332 1)

Rank 14 19.5 8 5.5 11 14 24 27 18 21.5 21.5 19.5 23 27 27 31.5 27 27 31.5 31.5 31.5 16.5 14 16.5 12 10 8 8 5.5 2.5 2.5 2.5 2.5

D 12.5 1.5 2 0 1 1.5 1.5 4.5 2.5 1 2 0 0.5 0.5 2.5 2 0.5 0.5 0.5 1 1 1 4 1 2 0 2 2 0 0.5 1.5 0.5 0.5

D2 156.25 2.25 4 0 1 2.25 2.25 20.25 6.25 1 4 0 0.25 0.25 6.25 4 0.25 0.25 0.25 1 1 1 16 1 4 0 4 4 0 0.25 2.25 0.25 0.25

¦d

246

496

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Because rs ! 0.291 , reject H0 . There is enough evidence at the 10% level of significance to conclude that there is a significant correlation between average hours worked and the number of on-the-job injuries. 14. The claim is “there is a significant correlation between average hours worked and the number of onthe-job injuries.” H 0 : Us 0 ; H a : Us z 0 (claim) rz r1.96 | r0.336. The critical values are n 1 35 1 Hours worked

Rank

Injuries

Rank

38 38 37 38 38 40 39 39 39 40 39 41 41 42 41 41 41 42 42 42 42 41 41 39 38 38 39 39 36 37 36 37 37 37 37

11.5 11.5 5.5 11.5 11.5 22.5 18 18 18 22.5 18 27 27 33 27 27 27 33 33 33 33 27 27 18 11.5 11.5 18 18 1.5 5.5 1.5 5.5 5.5 5.5 5.5

11 11 9 10 10 17 15 14 14 16 15 17 17 21 18 18 18 22 21 19 21 18 17 12 12 11 13 12 6 6 6 6 7 8 7

12 12 8 9.5 9.5 24.5 20.5 18.5 18.5 22 20.5 24.5 24.5 33 28.5 28.5 28.5 35 33 31 33 28.5 24.5 15 15 12 17 15 2.5 2.5 2.5 2.5 5.5 7 5.5

0.5 0.5 2.5 2 2 2 2.5 0.5 0.5 0.5 2.5 2.5 2.5 0 1.5 1.5 1.5 2 0 2 0 1.5 2.5 3 3.5 0.5 1 3 1 3 1 3 0 1.5 0

0.25 0.25 6.25 4 4 4 6.25 0.25 0.25 0.25 6.25 6.25 6.25 0 2.25 2.25 2.25 4 0 4 0 2.25 6.25 9 12.25 0.25 1 9 1 9 1 9 0 2.25 0

¦d

121.5

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¦d

497

6(121.5) | 0.983 n(n 1) 35(352 1) Because rs ! 0.336 , reject H0 . There is enough evidence at the 5% level of significance to conclude that there is a correlation between average hours worked and the number of on-the-job injuries. rs

11.5 THE RUNS TEST 11.5 TRY IT YOURSELF SOLUTIONS 1. PPP F P F PPPP FF P F PP FFF PPP F PPP 13 groups 13 runs 3, 1, 1, 1, 4, 2, 1, 1, 2, 3, 3, 1, 3 2. The claim is “the sequence of seat locations is not random.” H0 : The sequence of seat locations is random. Ha : The sequence of seat locations is not random. (claim) D 0.05 M BBB MM BB M B MM BBB n1 = number of M’s = 6 n2 = number of B’s = 9 G = number of runs = 8 Lower critical value = 4 Upper critical value = 13 G=8 Because 4 G 13 , fail to reject H0 . There is not enough evidence at the 5% level of significance to support the claim that the sequence of seat locations is not random. 3. The claim is “the sequence of weather conditions is not random.” H0 : The sequence of weather conditions is random. Ha : The sequence of weather conditions is not random. (claim) D 0.05 n1 = number of N’s = 21 n2 = number of S’s = 10 G = number of runs = 17 The critical values are z0 r1.96. 2n1n2 2(21)(10) PG 1 1 14.55 21 10 n1 n2

2n1n2 (2n1n2 n1 n2 ) (n1 n2 ) 2 (n1 n2 1)

2(21)(10)(2(21)(10) 21 10) | 2.38 (21 10) 2 (21 10 1)

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G PG

17 14.55 | 1.03 VG 2.38 Because z 1.96 , fail to reject H0 . There is not enough evidence at the 5%level of significance to support the claim that the sequence of weather conditions is not random. z

11.5 EXERCISE SOLUTIONS 1. Answers will vary. Sample answer: It is called the runs test because it considers the number of runs of data in a sample to determine whether the sequence of data was randomly selected. 2. When both n1 and n2 are less than or equal to 20, the test statistic is equal to the number of runs G. G PG When either n1 or n2 is greater than 20, the test statistic is equal to .

3. Number of runs = 8 Run lengths = 1, 1, 1, 1, 3, 3, 1, 1

4. Number of runs = 9 Run lengths = 2, 2, 1, 1, 2, 2, 1, 1, 2

5. Number of runs = 9 Run lengths = 1, 1, 1, 1, 1, 6, 3, 2, 4

6. Number of runs = 10 Run lengths = 3, 3, 1, 2, 6, 1, 2, 1, 1, 2

n1 = number of T’s = 6 n2 = number of F’s = 6

n1 = number of U’s = 8 n2 = number of D’s = 6

n1 = number of M’s = 10 n2 = number of F’s = 10

10. n1 = number of A’s = 13 n2 = number of B’s = 9

11. n1 = number of T’s = 6 n1 = number of F’s = 6 too high: 11; too low: 3

12. n1 = number of M’s = 9 n2 = number of F’s = 3 too high: 8; too low: 2

13. n1 = number of N’s = 11 n1 = number of S’s = 7 too high: 14; too low: 5

14. n1 = number of X’s = 7 n2 = number of Y’s = 14 too high: 15; too low: 5

15. (a) The claim is “the tosses were not random.” H0 : The coin tosses were random. Ha : The coin tosses were not random. (claim) (b) n1 = number of H’s = 7 n2 = number of T’s = 9 lower critical value = 4 upper critical value = 14

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(c) G = 9 runs (d) Because 4 G 14 , fail to reject H0 . (e) There is not enough evidence at the 5% level of significance to support the claim that the coin tosses were not random. 16. (a) The claim is “the sequence of majority parties is not random.” H0 : The sequence of majority parties is random. Ha : The sequence of majority parties is not random. (Claim) (b) n1 = number of R’s = 21 n2 = number of D’s = 35 The critical values are z0 r1.96 . (c) G = 16 runs 2n1n2 PG 1 n1 n2 2(21)(35) PG 1 27.25 21 35

VG Z Z

2n1n2 2n1n2 n1 n2 n1 n2

n1 n2 1

2(21)(35) 2(21)(35) 21 35 21 35

21 35 1

| 3.47

G PG

16 27.25 | 3.24 3.47

(d) Because z 1.96 , reject H0 . (e) There is enough evidence at the 5% level of significance to conclude that the sequence of majority parties is not random. 17. (a) The claim is “the sequence of leagues of World Series winning teams is not random.” H0 : The sequence of leagues of winning teams is random. Ha : The sequence of leagues of winning teams is not random. (claim) (b) n1 = number of N’s = 23 n2 = number of A’s = 27 The critical values are z0 r1.96

499

500

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VG Z Z

n1 n2

n1 n2 1

2(23)(27) 2(23)(27) 23 27 23 27

23 27 1

| 3.48

G PG

33 25.84 | 2.06 3.48

(d) Because z ! 1.96 , reject H0 . (e) There is enough evidence at the 5% level of significance to conclude that the sequence of leagues of World Series winning teams is not random. 18. (a) The claim is “the digits were not randomly generated.” H0 : The sequence of digits was randomly generated. Ha : The sequence of digits was not randomly generated. (claim) (b) n1 = number of O’s = 16 n2 = number of E’s = 16 lower critical value = 11; upper critical value = 23 (c) G = 9 runs (d) Because G 9 , reject H0 . (e) There is enough evidence at the 5% level of significance to support the claim that the sequence of digits was not randomly generated. 19. (a) The claim is “the microchips are random by gender.” H0 : The microchips are random by gender. (claim) Ha : The microchips are not random by gender. (b) n1 = number of M’s = 9 n2 = number of F’s = 20 lower critical value = 8; upper critical value = 18 (c) G = 12 runs

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501

(d) Because 8 G 18 , fail to reject H0 . (e) There is not enough evidence at the 5% level of significance to reject the claim that the microchips are random by gender. 20. (a) The claim is “the sequence of past winners is not random.” H0 : The sequence of past winners is random. Ha : The sequence of past winners is not random. (claim) (b) n1 = number of F’s = 42 n2 = number of A’s = 16 The critical values are z0 r1.96 . (c) G = 32 runs 2n1n2 2(42)(16) PG 1 1 | 24.17 42 16 n1 n2

VG z

2n1n2 (2n1n2 n1 n2 ) (n1 n2 )2 (n1 n2 1) G PG

2(42)(16)(2(42)(16) 42 16) | 3.0 (42 16) 2 (42 16 1)

32 24.17 | 2.61 3.00

(d) Because z ! 1.96 , reject H0 . (e) There is enough evidence at the 5% level of significance to support the claim that the sequence of past winners is not random. 21. The claim is “the daily high temperatures do not occur randomly.” H0 : Daily high temperatures occur randomly. Ha : Daily high temperatures do not occur randomly. (claim) median = 87 n1 = number of above = 15 n2 = number of below = 13 lower critical value = 9 upper critical value = 21 G = 11 runs Because 9 G 21 , fail to reject H0 , There is not enough evidence at the 5% level of significance to support the claim that the daily high temperatures do not occur randomly. 22. The claim is “the scores occur randomly.” H0 : The sequence of exam scores is random. (claim) Ha : The sequence of exam scores is not random. median = 87 n1 = number above median = 14 n2 = number below median = 15 Copyright © 2023 Pearson Education, Inc.

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lower critical value = 9 upper critical value = 22 G = 15 runs Because 9 G 22 , fail to reject H0 , There is not enough evidence at the 5% level of significance to reject the claim that the scores occur randomly. 23. Answers will vary.

CHAPTER 11 REVIEW EXERCISE SOLUTIONS 1. (a) The claim is “the median number of customers per day is no more than 650.” H0: median d 650 (claim) Ha: median ! 650 (b) The critical value is 2. (c) x

(d) Because x ! 2 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to reject the store manager’s claim that the median number of customers per day is no more than 650. 2. (a) The claim is “median credit score for U.S. adults is at least 710.” H0 : median t 710 (claim); Ha : median < 710 (b) The critical value is 2. (c) x 5 (d) Because x ! 2 , fail to reject H0 . (e) There is not enough evidence at the 5% level of significance to reject the company’s claim that the median credit score for U.S. adults is at least 710. 3. (a) The claim is “median sentence length for all federal prisoners is 2 years.” H0 : median = 2 (claim); Ha : median z 2 (b) The critical value is z0

1.645.

(65 0.5) 0.5(174) 21.5 | | 3.26 6.595 174 2

CHAPTER 11 │ NONPARAMETRIC TESTS

503

(d) Because z0 1.645 , reject H0 . (e) There is enough evidence at the 10% level of significance to reject the agency’s claim that the median sentence length for all federal prisoners is 2 years. 4. (a) The claim is “there was no reduction in diastolic blood pressure.” H0 : There was no reduction in diastolic blood pressure. (claim) Ha : There was a reduction in diastolic blood pressure. (b) The critical value is 1. (c) x

(d) Because x ! 1 , fail to reject H0 . (e) There is not enough evidence at the 5% level of significance to reject the claim that there was no reduction in diastolic blood pressure. 5. (a) The claim is “there was no reduction in diastolic blood pressure.” H0 : There was no reduction in diastolic blood pressure. (claim) Ha : There was a reduction in diastolic blood pressure. (b) The critical value is 2. (c) x 3 (d) Because x ! 2 , fail to reject H0 . (e) There is not enough evidence at the 5% level of significance to reject the claim that there was no reduction in diastolic blood pressure. 6. (a) The claim is “the median annual salary of lawyers is $126,930.” H0: median $126,930 (claim) Ha: median z $126,930 (b) The critical value is z0

1.96

504

CHAPTER 11 │ NONPARAMETRIC TESTS

(e) There is enough evidence at the 5% level of significance to reject the claim that the median annual salary is $126,930. 7. (a) The claim is “there is a difference in the total times required to earn a doctorate degree by female and male graduate students.” H0: There is no difference in the total times required to earn a doctorate degree by female and male graduate students. Ha: There is a difference in the total times required to earn a doctorate degree by female and male graduate students. (claim) (b) Wilcoxon rank sum test (c) The critical values are z0 (d)

Ordered data 6 6 7 7 7 7 8 8 8 8 8 9 9 9 9 9 9 10 10 10 11 11 12 13

r2.575.

Sample F F M M M M F F M M M F F F M M M F M M F F F F

Rank 1.5 1.5 4.5 4.5 4.5 4.5 9 9 9 9 9 14.5 14.5 14.5 14.5 14.5 14.5 19 19 19 21.5 21.5 23 24

n1 (n1 n2 1) 12(12 12 1) 150 2 2 n1n2 (n1 n2 1) (12)(12)(12 12 1) | 17.32 VR 12 12 R PR 173.5 150 | | 1.357 z 17.32 VR z = –1.357 or 1.357.

CHAPTER 11 │ NONPARAMETRIC TESTS

505

(e) Because z ! 2.575 , fail to reject H0 . (f) There is not enough evidence at the 1% level of significance to support the claim that there is a difference in the total times required to earn a doctorate degree by female and male graduate students. 8. (a) The claim is “a new drug affects the number of headache hours experienced by headache sufferers.” H0 : The new drug does not affect the number of headache hours experienced. Ha : The new drug does affect the number of headache hours experienced. (claim) (b) Wilcoxon signed-rank test (c) The critical value is 4. (d)

Before 0.9 2.3 2.7 2.4 2.9 1.9 1.2 3.1

After 1.4 1.5 1.4 1.8 1.3 0.6 0.7 1.9

Difference 0.5 0.8 1.3 0.6 1.6 1.3 0.5 1.2

Absolute value 0.5 0.8 1.3 0.6 1.6 1.3 0.5 1.2

Rank 1.5 4 6.5 3 8 6.5 1.5 5

Signed rank 1.5 4 6.5 3 8 6.5 1.5 5

Sum of negative ranks = 1.5 Sum of positive ranks = 34.5 ws 1.5 (e) Because ws 4 , reject H0 . (f) There is enough evidence at the 5% level of significance to support the claim that the new drug affect the number of headache hours experienced. 9. (a) The claim is “the distributions of the ages of the doctorate recipients in at least one field of study is different from the others.” H0 : The distribution of the ages of doctorate recipients is the same in all three fields of study. Ha : The distribution of the ages of doctorate recipients in at least one field of study is different from the others. (claim) (b) F 02

9.210 ; Rejection region: F 2 ! 9.210 .

506

CHAPTER 11 │ NONPARAMETRIC TESTS

(c)

Ordered data 29 29 30 30 30 30 30 30 31 31 31 31 31 31 31 31 32

Sample L P L P P P P S L L L P P P S S L

R1 191.5, R2 126, R3

Rank 1.5 1.5 5.5 5.5 5.5 5.5 5.5 5.5 12.5 12.5 12.5 12.5 12.5 12.5 12.5 12.5 20

Ordered data 32 32 32 32 32 32 33 33 33 34 34 34 35 35 35 36

Sample L L P P S S P S S L L S L S S S

Rank 20 20 20 20 20 20 25 25 25 28 28 28 31 31 31 33

243.5

§ R12 R22 R32 · 12 ¨ ¸ 3( N 1) N ( N 1) ¨© n1 n2 n3 ¸¹ § (191.5) 2 (126) 2 (243.5) 2 · 12 ¨ ¸ 3(33 1) 33(33 1) © 11 11 11 ¹ | 6.741

(d) Because H 9.210 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to conclude that the distribution of ages of the doctorate recipients in at least one field of study is different from the others. 10. (a) The claim is “the distribution of the starting salaries in at least one field of engineering is different from the others.” H0: The distribution of the starting salaries is the same in all four fields of engineering. Ha: The distribution of the starting salaries in at least one field of engineering is different from the others. (claim) (b) F 02

7.815 ; Rejection region: F 2 ! 7.815

CHAPTER 11 │ NONPARAMETRIC TESTS

(c)

Ordered data 67.1 67.8 68 68 68.1 68.5 68.6 68.8 69 69.2 69.4 69.5 69.6 69.7 69.8 69.9 70.4 70.5 70.6 70.7 70.9 71 71.2 71.2 71.3 71.4 71.6 71.7 71.7 71.7 72 72.1 72.1 72.2 72.3 72.5 72.8 73 73.7 74.2

Sample Comp Mech Comp Mech Mech Mech Comp Elec Mech Mech Elec Mech Mech Mech Elec Comp Mech Chem Comp Comp Elec Elec Elec Chem Elec Chem Comp Chem Chem Elec Chem Chem Chem Chem Comp Chem Elec Elec Comp Comp

Rank 1 2 3.5 3.5 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23.5 23.5 25 26 27 29 29 29 31 32.5 32.5 34 35 36 37 38 39 40

507

508

CHAPTER 11 │ NONPARAMETRIC TESTS

291.5, R2

207.5, R3

229.5, R4

91.5

(d) Because H ! 7.815 , reject H0 . (e) There is enough evidence at the 5% level of significance to conclude that the distribution of the starting salaries in at least one field of engineering is different from the others. 11. (a) The claim is “there is a correlation between the engine size and the price.” H 0 : Us 0 ; H a : Us z 0 (claim) (b) The critical value is 0.829. (c)

Size 1700 1000 999 645 471 399

6¦d2 2

n n 1

Rank 6 5 4 3 2 1

6 2 6 62 1

Price 16,799 8,999 11,599 8799 6699 5999

Rank 6 4 5 3 2 1

d 0 1 −1 0 0 0

¦d

d2 0 1 1 0 0 0 2

0.943

(d) Because rs ! 0.829 , reject H0 . (e) There is enough evidence at the 10% level of significance to conclude that there is a significant correlation between the engine size and the price. 12. (a) The claim is “there is a correlation between the number of car problems in the first and third years.” H 0 : Us 0 ; H a : Us z 0 (claim) (b) The critical value is 0.643.

CHAPTER 11 │ NONPARAMETRIC TESTS

(c)

First Year 148 147 151 182 146 240 128 144

6¦d 2

n n 1

Rank 5 4 6 7 3 8 1 2

6 56.5 8 82 1

Third Year 102 97 125 125 128 127 123 116

Rank 2 1 5.5 5.5 8 7 4 3

d 3 3 0.5 1.5 −5 1 −3 −1

509

d2 9 9 0.25 2.25 25 1 9 1 2 d 56.5

| 0.327

(d) Because rs 0.643 , fail to reject H0 . (e) There is not enough evidence at the 10% level of significance to conclude that there is a significant correlation between the number of car problems in the first and third years. 13. (a) The claim is “the stops were not random by gender.” H0 : The traffic stops were random by gender. Ha : The traffic stops were not random by gender. (claim) (b) n1 = number of F’s = 12 n2 = number of M’s = 13 lower critical value = 8 upper critical value = 19 (c) G = 14 runs (d) Because 8 G 19 , fail to reject H0 . (e) There is not enough evidence at the 5% level of significance to support the claim that the stops were not random by gender. 14. (a) The claim is “the departure status of the buses is not random.” H0 : The departure status of buses is random. Ha : The departure status of buses is not random. (claim) (b) n1 = number of T’s = 11 n2 = number of L’s = 7 lower critical value = 5 upper critical value = 14 (c) G = 5 runs (d) Because G d 5 , reject H0 . Copyright © 2023 Pearson Education, Inc.

510

CHAPTER 11 │ NONPARAMETRIC TESTS

(e) There is enough evidence at the 5% level of significance support the claim that the departure status of the buses is not random.

CHAPTER 11 QUIZ SOLUTIONS 1. (a) The claim is “the median number of annual volunteer hours is 50.” H0: median 50 (claim) Ha: median z 50 (b) Sign test (c) The critical values are z0 (d) x 23 x 0.5 0.5n z n 2

r1.96

23 0.5 0.5 70 70 2

2.75

(e) Because z 1.96 , reject H0. (f) There is enough evidence at the 5% level of significance to reject the organization’s claim that the median number of annual volunteer hours is 50. 2. (a) The claim is “there is a difference in the hourly earnings of union employees and nonunion employees in state and local governments.” H0 : There is no difference in the hourly earnings. Ha : There is a difference in the hourly earnings. (claim) (b) Wilcoxon rank sum test (c) The critical values are z0 (d)

Ordered data 23.45 24.20 24.40 25.05 25.25 25.50 26.10 27.75 29.15 29.75

r1.645 .

Sample N N N N N N N N N U

Rank 1 2 3 4 5 6 7 8 9 10

Ordered data 29.95 30.35 30.60 30.85 31.15 32.05 32.75 35.30 35.88 38.52

Sample N U U U U U U U U U

Rank 11 12 13 14 15 16 17 18 19 20

CHAPTER 11 │ NONPARAMETRIC TESTS

511

n1 (n1 n2 1) 10(10 10 1) 105 2 2 n1n2 (n1 n2 1) (10)(10)(10 10 1) | 13.229 VR 12 12 R PR 56 105 | | 3.70 (or 3.70) z 13.229 VR

(e) Because z 1.645 , reject H0 . (f) There is enough evidence at the 10% level of significance to support the claim that there is a difference in the hourly earnings of union employees and nonunion employees in state and local governments. 3. (a) The claim is “the distribution of the sales prices in at least one region is different from the others.” H0: The distribution of the sales prices is the same in all four regions. Ha: The distribution of the sales prices in at least one region is different from the others. (claim) (b) Kruskal-Wallis test (c) The critical value is F 2 11.345 (d)

Ordered data

Sample

Rank

Ordered data

Sample

Rank

174.6 180

MW MW

1 2

299.8 300.5

NE NE

17 18

189.4

307.4

191.2 193 193.9 195 196 205 205.1 208.7 209.2 213.6 216.7 219.6 234.6

MW MW S S MW MW S S MW S S S S

4 5 6 7 8 9 10 11 12 13 14 15 16

311.9 314.4 321.3 327.8 332.1 360.5 372.5 376.9 389.1 390.6 396.3 400.6 426.3

NE NE NE NE NE W W W W W W W W

20 21 22 23 24 25 26 27 28 29 30 31 32

92, R4

228

R1 164, R2

44, R3

512

CHAPTER 11 │ NONPARAMETRIC TESTS

§ R12 R22 R32 R42 · 12 ¨ ¸ 3( N 1) N ( N 1) ¨© n1 n2 n3 n4 ¸¹ § (164) 2 (44) 2 (92) 2 (228) 2 · 12 ¨ ¸ 3(32 1) 32(32 1) © 8 8 8 8 ¹ | 27.818

(e) Because H ! 11.345 , reject H0. (f) There is enough evidence at the 1% level of significance to conclude that the distribution of the sales prices in at least one region is different from the others. 4. (a) The claim is “there is a correlation between the number of emails sent and the number of emails received.” H 0 : Us 0 ; H a : Us z 0 (claim) (b) Spearman rank correlation coefficient. (c) The critical value is 0.833. (d)

Emails sent 30 30 25 26 24 18 18 25 28

Rank

Emails received 32 36 21 22 20 20 22 23 23

8.5 8.5 4.5 6 3 1.5 1.5 4.5 7

¦d 1 6

n(n 1)

Rank

8 9 3 4.5 1.5 1.5 4.5 6.5 6.5

0.5 0.5 1.5 1.5 1.5 0 3 2 0.5

0.25 0.25 2.25 2.25 2.25 0 9 4 0.25 d 2 20.5

6(20.5) | 0.829 9(92 1)

(e) Because rs 0.833 , fail to reject H0 . (f) There is not enough evidence at the 1% level of significance to conclude that there is a significant correlation between the number of emails sent and the number of emails received. 5. (a) The claim is “days with rain are not random.” H0 : The days with rain are random. Ha : The days with rain are not random. (claim) (b) Runs test

CHAPTER 11 │ NONPARAMETRIC TESTS

513

(c) n1 = number of N’s = 15 n2 = number of R’s = 15 lower critical value = 10 upper critical value = 22 (d) G = 16 runs (e) Because 10 G 22 , fail to reject H0 . (f) There is not enough evidence at the 5% level of significance for the meteorologist to conclude that days with rain are not random.

CHAPTER 11 TEST SOLUTIONS 1. (a) The claim is “the selection of members was not random.” H0 : The selection of members was random. Ha : The selection of members was not random. (claim) (b) Runs test (c) n1 = number of R’s = 18 n2 = number of D’s = 16 lower critical value = 11 upper critical value = 25 (d) G = 19 runs (e) Because 11 G 25 , fail to reject H0 . (f) There is not enough evidence at the 5% level of significance to conclude the selection of members was not random. 2. (a) The claim is “the distribution of the annual household incomes in at least one region is different from the others.” H0: The distribution of annual household incomes is the same in all four regions. Ha: The distribution of annual household incomes in at least one region is different from the others. (claim) (b) Kruskal–Wallis test (c) The critical value is F 2 11.345 .

514

CHAPTER 11 │ NONPARAMETRIC TESTS

(d)

Ordered data

Sample

Rank

Ordered data

Sample

Rank

57.1 60 61.2 62 62.6 63.2 64.8 64.8 66.5 67.3 68.7 69.8 70 71

S S S S S MW S S MW MW MW MW MW NE

1 2 3 4 5 6 7.5 7.5 9 10 11 12 13 14

72.4 72.9 73.9 73.9 75 75.5 75.5 76.2 76.4 77.4 78.2 78.3 78.7 79.6

MW W NE W W NE W W NE W NE W NE NE

15 16 17.5 17.5 19 20.5 20.5 22 23 24 25 26 27 28

R1 155, R2 H

(e) Because H ! 11.345 , reject H0. (f) There is enough evidence at the 1% level of significance to conclude that the distribution of annual household incomes in at least one region is different from the others. 3. (a) The claim is “the median age of individuals heading mutual fund–owning households is 50 years.” H0 : median = 50 (claim); Ha : median z 50 (b) Sign test (c) The critical value is 3. (d) x

(e) Because x ! 3 , fail to reject H0 . (f) There is not enough evidence at the 1% level of significance to reject the claim that the median age of individuals heading mutual fund–owning households is 50 years.

CHAPTER 11 │ NONPARAMETRIC TESTS

515

4. (a) The claim is “there is a difference in the weekly earnings of employees who are union members and employees who are not union members.” H0: There is no difference in the weekly earnings. Ha: There is a difference in the weekly earnings. (claim) (b) Wilcoxon rank sum test (c) The critical values are z0 (d)

r1.96

Ordered data

Sample

Rank

805 843 886 903 928 939 946 1006 1030 1036 1051 1059 1062 1091 1110 1120 1140 1166 1227 1230 1276

N N N N M N N N M M N M N M N M M M M M M

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

R = sum ranks of nonmember workers = 70 n1 (n1 n2 1) 10(10 11 1) 110 PR 2 2 n1n2 (n1 n2 1) (10)(11)(10 11 1) | 14.201 VR 12 12 R PR 70 110 | | 2.82 z 14.201 VR (e) Because z 1.96 , reject H0. (f) There is enough evidence at the 5% level of significance to support the claim that there is a difference in the weekly earnings of employees who are union members and employees who are not union members.

516

CHAPTER 11 │ NONPARAMETRIC TESTS

5. (a) The claim is “there is a correlation between the expert review rating and the price.” H 0 : Us 0 ; H a : Us z 0 (claim) (b) Spearman rank correlation coefficient. (c) The critical value is 0.738. (d)

Rating 4.8 4.7 4.6 4.5 4.4 4.3 4.2 4.1

Rank 8 7 6 5 4 3 2 1

¦d 1 6

n(n 1)

Price 20.9 22.0 26.3 25.7 21.1 17.0 27.5 26.2

Rank 2 4 7 5 3 1 8 6

d 6 3 −1 0 1 2 −6 −5

¦d

d2 36 9 1 0 1 4 36 25 2

112

6(112) | 0.333 8(82 1)

(e) Because rs 0.738 , fail to reject H0 . (f) There is not enough evidence at the 5% level of significance to conclude that there is a significant correlation between the expert review rating and the price.

CHAPTER

Confidence Intervals

6.1 CONFIDENCE INTERVALS FOR THE MEAN (σ KNOWN) 6.1 TRY IT YOURSELF SOLUTIONS

¦ x = 114 = 2.85

n 40 Another point estimate for the population mean is 2.85.

zc = 1.96, n = 40, σ = 1.4 σ 1.4 E = zc = 1.96 ≈ 0.43 n 40 You are 95% confident that the margin of error for the population mean is about 0.43 hour.

x = 2.85, E ≈ 0.43 x − E ≈ 2.85 − 0.43 = 2.42 x + E ≈ 2.85 + 0.43 = 3.28 (2.42, 3.28)

4. 75% CI: (19.49, 20.36) 85% CI: (19.38, 20.47) 90% CI: (19.30, 20.55) As the confidence level increases, so does the width of the interval. 5.

n = 30, x = 22.9, σ = 1.5, zc = 1.645 σ 1.5 E = zc = 1.645 ≈ 0.5 30 n

x − E ≈ 22.9 − 0.5 = 22.4 x + E ≈ 22.9 + 0.5 = 23.4 (22.4, 23.4) With 90% confidence, you can say that the mean age of the students is between 22.4 and 23.4 years. Because of the larger sample size, the confidence interval is slightly narrower. 6.

zc = 1.96, E = 0.75, σ ≈ 2.4 2

§ z σ · § 1.96 ⋅ 2.4 · n=¨ c ¸ =¨ ¸ ≈ 39.34 → 40 © E ¹ © 0.75 ¹ Because of the larger margin of error, the sample size needed is smaller.

518

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

6.1 EXERCISE SOLUTIONS 1. You are more likely to be correct using an interval estimate because it is unlikely that a point estimate will exactly equal the population mean. 2. b 3. d; As the level of confidence increases, zα 2 increases, causing wider intervals. 4. No, the 95% confidence interval means that with 95% confidence you can say that the population mean is in this interval. If a large number of samples is collected and a confidence interval created for each, approximately 95% of these intervals will contain the population mean. 5. 1.28 9.

6. 1.44

7. 1.15

x − μ = 3.8 − 4.27 = −0.47

10. x − μ = 9.5 − 8.76 = 0.74

11. x − μ = 26.43 − 24.67 = 1.76 13. E = zc

15. E = zc

σ n

= 1.96

= 1.28

5.2 30 1.3 75

8. 2.17

12. x − μ = 46.56 − 48.12 = −1.56

≈ 1.861

14. E = zc

≈ 0.192

16. E = zc

σ n

= 1.645

= 2.24

2.9 50 4.6 100

≈ 0.675

≈ 1.030

22. x ± zc

23. x ± zc

σ n

= 12.3 ± 1.645

= 31.39 ± 1.96

= 10.5 ± 2.575

1.5 50 0.8 82 2.14 45

≈ 12.3 ± 0.349 ≈ (12.0, 12.6)

≈ 31.3 ± 0.173 ≈ (31.22, 31.56)

≈ 10.5 ± 0.821 ≈ (9.7, 11.3)

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

24. x ± zc

σ n

= 20.6 ± 1.28

25. (12.0, 14.8) x =

≈ 20.6 ± 0.602 ≈ (20.0, 21.2)

100

14.8 + 12.0 = 13.4, E = 14.8 − 13.4 = 1.4 2

26. (21.61, 30.15) x = 27. (1.71, 2.05) x =

4.7

519

30.15 + 21.61 = 25.88, E = 30.15 − 25.88 = 4.27 2

2.05 + 1.71 = 1.88, E = 2.05 − 1.88 = 0.17 2

28. (3.144, 3.176) x =

3.176 + 3.144 = 3.16, E = 3.176 − 3.16 = 0.016 2

29. c = 0.90 zc = 1.645 2

§ z σ · § (2.33)(10.1) · n=¨ c ¸ =¨ ¸ ≈ 138.45 139 2 ¹ © E ¹ © 33. (26.2, 30.1) 2 E = 30.1 − 26.2 = 3.9 E = 1.95 and x = 26.2 + E = 26.2 + 1.95 = 28.15 34. (244.07, 280.97) 2 E = 280.97 − 244.07 = 36.9 E = 18.45 and x = 244.07 + E = 244.07 + 18.45 = 262.52 35. 90% CI: x ± zc

95% CI: x ± zc

σ n

= 1404.09 ± 1.645

232.09

48 232.09

≈ 1404.09 ± 55.106 ≈ (1348.98, 1459.20)

= 1404.09 ± 1.96 ≈ 1404.09 ± 65.659 ≈ (1338.43, 1469.75) 48 n With 90% confidence, you can say that the population mean price is between $1348.98 and $1459.20. With 95% confidence, you can say that the population mean price is between $1338.43 and $1469.75. The 95% CI is wider.

520

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

36. 90% CI: x ± zc

95% CI: x ± zc

σ n

= 97.17 ± 1.645

21.77

36 21.77

≈ 97.17 ± 5.969 ≈ (91.20, 103.14)

= 97.17 ± 1.96 ≈ 97.17 ± 7.112 ≈ (90.06, 104.28) 36 n With 90% confidence, you can say that the population mean price is between $91.20 and $103.14. With 95% confidence, you can say that the population mean price is between $90.06 and $104.28. The 95% CI is wider.

37. 90% CI: x ± zc

95% CI: x ± zc

σ n

= 54.97 ± 1.645

48.43

50 48.43

≈ 54.97 ± 11.267 ≈ (43.70, 66.24)

= 54.97 ± 1.96 ≈ 54.97 ± 13.424 ≈ (41.55, 68.39) 50 n With 90% confidence, you can say that the population mean price is between $43.70 and $66.24. With 95% confidence, you can say that the population mean price is between $41.55 and $68.39. The 95% CI is wider.

38. 90% CI: x ± zc

111

≈ 100 ± 30.864 ≈ (69.14, 130.86) 35 σ 111 95% CI: x ± zc = 100 ± 1.96 ≈ 100 ± 36.774 ≈ (63.23, 136.77) 35 n With 90% confidence, you can say that the population mean tornadoes per month is between 69.14 and 130.86. With 95% confidence, you can say that the population mean tornadoes per month is between 63.23 and 136.77. The 95% CI is wider. n

= 100 ± 1.645

(b) An increase in the sample size will narrow the confidence interval because it decreases the standard error. (c) An increase in the population standard deviation will widen the confidence interval because small standard deviations produce more precise intervals, which are smaller. 44. Answers will vary. 45. x =

¦ x = 1167 ≈ 36.47 n

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

90% CI: x ± zc 99% CI: x ± zc

σ n

= 36.47 ± 1.645

14.9

32 14.9

521

≈ 36.47 ± 4.333 ≈ (32.1, 40.8)

= 36.47 ± 2.575 ≈ 36.47 ± 6.782 ≈ (29.7, 43.3) 32 n With 90% confidence, you can say that the population mean commute time is between 32.1 and 40.8 minutes. With 99% confidence, you can say that the population mean commute time is between 29.7 and 43.3 minutes. The 99% CI is wider.

46. x =

¦ x = 795.71 ≈ 22.10 n

36 σ 7.61 90% CI: x ± zc = 22.10 ± 1.645 ≈ 22.10 ± 2.09 ≈ (20.01, 24.19) 36 n σ 7.61 99% CI: x ± zc = 22.10 ± 2.575 ≈ 22.10 ± 3.27 ≈ (18.83, 25.37) 36 n With 90% confidence, you can say that the population mean concentration is between 20.01 and 24.19 grams per liter. With 99% confidence, you can say that the population mean concentration is between 18.83 and 25.37 grams per liter. The 99% CI is wider. 2

= 29 ± 1.96

3.10

≈ 29 ± 0.748 ≈ (28.252, 29.748) 66 n No. The 95% CI is (28.252, 29.748). If the population mean is within 3% of the sample mean, then it falls outside the CI. Yes. If the population mean is within 0.3% of the sample mean, then it falls within the CI. 2

= 20 ± 1.645

1.6

≈ 20 ± 1.316 ≈ (18.684, 21.316) 4 n No; No; The 90% CI is (18.684, 21.316). If the population mean is within 7% or 8% of the sample mean, then it falls outside the CI.

522

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

= 127.75 ± 1.645

0.75

≈ 127.75 ± 0.44 ≈ (127.3, 128.2) 8 n Yes; The 90% CI is (127.3, 128.2) and 128 ounces falls within that interval. 2

= 64 ± 1.96

≈ 64 ± 0.238 ≈ (63.762, 64.238) 68 n Yes; The 95% CI is (63.762, 64.238) and there are amounts greater than 63.85 fluid ounces that fall within that interval. 2

= 27.5 ± 2.575

0.5

≈ 27.5 ± 0.140 ≈ (27.360, 27.640) 84 n Yes; The 99% CI is (27.360, 27.640) and there are amounts less than 27.6 inches that fall within that interval. 2

= 8.3 ± 2.575

0.10

≈ 8.3 ± 0.0442 ≈ (8.2558, 8.3442) 34 n Yes; The 99% CI is (8.2558, 8.3442) and 8.258 inches falls within that interval.

(b) An increase in the error tolerance will decrease the minimum sample size required because the less data you have, the less accurate your results will be. (c) An increase in the population standard deviation will increase the minimum sample size required because the data is more spread out. 57. (a)

N −n 1000 − 500 = ≈ 0.707 N −1 1000 − 1

(b)

N −n 1000 − 100 = ≈ 0.949 N −1 1000 − 1

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

(c)

N −n 1000 − 75 = ≈ 0.962 N −1 1000 − 1

(d)

N −n 1000 − 50 = ≈ 0.975 N −1 1000 − 1

(e)

N −n 100 − 50 = ≈ 0.711 N −1 100 − 1

(f)

N −n = N −1

(g)

400 − 50 ≈ 0.937 400 − 1

N −n 700 − 50 N −n 1200 − 50 (h) = ≈ 0.964 = ≈ 0.979 N −1 700 − 1 N −1 1200 − 1 The finite population correction factor approaches 1 as the sample size decreases and the population size remains the same. The finite population correction factor approaches 1 as the population size increases and the sample size remains the same.

58. (a) 99% CI: x ± zc

(b) 90% CI: x ± zc

(d) 80% CI: x ± zc 59. Sample answer: zα 2σ E= n

σ n

N −n 0.5 300 − 68 = 40.3 ± 1.96 ≈ 40.3 ± 0.105 ≈ (40.2, 40.4) N −1 68 300 − 1 N −n 9.8 = 56.7 ± 1.28 N −1 36

Divide each side by E.

200 − 25 ≈ 8.6 ± 2.366 ≈ (6.2, 11.0) 200 − 1

N −n 2.8 500 − 50 = 10.9 ± 1.645 ≈ 10.9 ± 0.619 ≈ (10.3, 11.5) N −1 50 500 − 1

Multiply each side by n .

zα 2σ

N −n 4.9 = 8.6 ± 2.575 N −1 25

Write the original equation.

E n = zα 2σ n=

Square each side.

400 − 36 ≈ 56.7 ± 1.997 ≈ (54.7, 58.7) 400 − 1

523

524

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

6.2 CONFIDENCE INTERVALS FOR THE MEAN (σ UNKNOWN) 6.2 TRY IT YOURSELF SOLUTIONS 1.

d.f. = n − 1 = 22 − 1 = 21 90% confidence level tc = 1.721 For a t-distribution curve with 21 degrees of freedom, 90% of the area under the curve lies between t = ±1.721 .

2. d.f. = n − 1 = 16 − 1 = 15 90% CI: tc = 1.753 s 10 = 1.753 ≈ 4.4 n 16 90% CI: x ± E ≈ 162 ± 4.4 = (157.6, 166.4) With 90% confidence, you can say that the population mean temperature of coffee sold is between 157.6°F and 166.4°F. 99% CI: tc = 2.947 s 10 E = tc = 2.947 ≈ 7.4 n 16 99% CI: x ± E ≈ 162 ± 7.4 = (154.6, 169.4) With 99% confidence, you can say that the population mean temperature of coffee sold is between 154.6°F and 169.4°F. E = tc

d.f. = n − 1 = 36 − 1 = 35 90% CI: tc = 1.690 s 2.39 E = tc = 1.690 ≈ 0.67 n 36 90% CI: x ± E ≈ 9.75 ± 0.67 = (9.08, 10.42) With 90% confidence, you can say that the population mean number of days the car model sits on the lot is between 9.08 and 10.42 days. 95% CI: tc = 2.030 s 2.39 E = tc = 2.030 ≈ 0.81 n 36 95% CI: x ± E ≈ 9.75 ± 0.81 = (8.94, 10.56) With 95% confidence, you can say that the population mean number of days the car model sits on the lot is between 8.94 and 10.56 days. The 90% confidence interval is slightly narrower.

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

525

4. Is σ known? No Is n ≥ 30? No Is the population normally distributed? Yes Use the t-distribution because σ is not known and the population is normally distributed.

tc = 1.833

tc = 2.201

tc = 2.947

tc = 2.426

E = tc

s 5 = 2.131 ≈ 2.664 n 16

x ± tc

s 2.0 = 12.5 ± 2.015 ≈ 12.5 ± 1.645 ≈ (10.9, 14.1) n 6

10. x ± tc

s 0.85 = 13.4 ± 2.365 ≈ 13.4 ± 0.711 ≈ (12.7, 14.1) n 8

11. x ± tc

12. x ± tc

s n

= 4.3 ± 2.650

0.34 14

E = tc

s 3 = 4.032 ≈ 4.938 n 6

≈ 4.3 ± 0.241 ≈ (4.1, 4.5)

s 4.6 = 24.7 ± 2.678 ≈ 24.7 ± 1.742 ≈ (23.0, 26.4) n 50

13. (14.7, 22.1) x =

14.7 + 22.1 = 18.4 E = 22.1 − 18.4 = 3.7 2

14. (6.17, 8.53) x =

6.17 + 8.53 = 7.35 E = 8.53 − 7.35 = 1.18 2

15. t =

16. t =

x−μ s

70.3 − 64.8 7.1 16 22.9 − 24.5 4.3

≈ 3.10

≈ −2.35

526

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

s 7.2 = 2.365 ≈ 6.0 n 8 x ± E ≈ 35.5 ± 6.0 = (29.5, 41.5) With 95% confidence, you can say that the population mean commute time to work is between 29.5 and 41.5 minutes.

17. E = tc

s 5.8 = 2.776 ≈ 7.2 n 5 x ± E ≈ 22.2 ± 7.2 = (15.0, 29.4) With 95% confidence, you can say that the population mean driving distance to work is between 15.0 and 29.4 miles.

18. E = tc

19. E = tc

184.00

≈ 153.85 n 8 x ± E ≈ 526.50 ± 153.85 = (372.65,680.35) With 95% confidence, you can say that the population mean cell phone price is between $372.65 and $680.35.

20. E = tc

= 2.365

33.61

≈ 21.35 n 12 x ± E ≈ 90.42 ± 21.35 = (69.07, 111.77) With 95% confidence, you can say that the population mean repair cost is between $69.07 and $111.77.

= 2.201

9.3 ≈ 6.4 8 n x ± E ≈ 35.5 ± 6.4 = (29.1, 41.9) With 95% confidence, you can say that the population mean commute time to work is between 29.1 and 41.9 minutes. This confidence interval is slightly wider than the one found in Exercise 17.

21. E = zc

= 1.96

5.2 ≈ 4.6 5 n x ± E ≈ 22.2 ± 4.6 = (17.6, 26.8) With 95% confidence, you can say that the population mean driving distance to work is between 17.6 and 26.8 miles. This confidence interval is narrower than the one found in Exercise 18.

22. E = zc

= 1.96

23. Yes, $431.61 falls between $372.65 and $680.35. 24. Yes, $89.56 falls between $69.07 and $111.77. 25. (a) x = 1185

(b) s ≈ 168.1 (c) x ± tc

s 168.1 ≈ 1185 ± 3.106 ≈ 1185 ± 150.72 ≈ (1034.3, 1335.7) n 12

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

26. (a) x ≈ 2.51

(b) s ≈ 0.87 (c) x ± tc

s 0.87 ≈ 2.51 ± 3.012 ≈ 2.51 ± 0.70 = (1.81, 3.21) n 14

27. (a) x ≈ 7.49

(b) s ≈ 1.64 (c) x ± tc

s 1.64 ≈ 7.49 ± 2.947 ≈ 7.49 ± 1.21 = (6.28, 8.70) n 16

28. (a) x ≈ 12.19

(b) s ≈ 1.75 (c) x ± tc

s n

≈ 12.19 ± 2.898

1.755 18

≈ 12.19 ± 1.20 = (10.99, 13.39)

29. Yes, 1051 falls between 1034.3 and 1335.7. 30. No, 7.8 does not fall between 10.99 and 13.39. 31. (a) x ≈ 68,757.94

(b) s ≈ 15,834.18 (c) Use a t-distribution because σ unknown and n ≥ 30 . s 15,834.18 x ± tc ≈ 68,757.94 ± 2.453 ≈ 68,757.94 ± 6866.227 ≈ (61,892, 75,624) n 32 32. (a) x = 95,443

(b) s ≈ 17,673.10 (c) Use a t-distribution because σ unknown and n ≥ 30 . s 17,673.10 x ± tc ≈ 95, 443 ± 2.441 ≈ 95, 443 ± 7292 = (88,151.00, 102,735.00) n 35 33. Yes, $67,319 falls between $61,892.21 and $75,623.67. 34. Yes, $93,867 falls between $88,151.00 and $102,735.00.

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CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

35. Use a t-distribution because σ unknown and n ≥ 30 . s 6.12 x ± tc = 27.7 ± 2.009 ≈ 27.7 ± 1.74 ≈ (26.0, 29.4) n 50 With 95% confidence, you can say that the population mean BMI is between 26.0 and 29.4. 36. Use a t-distribution because σ is unknown and the interest rates are normally distributed. s 0.49 x ± tc = 3.95 ± 2.110 ≈ 3.95 ± 0.24 = (3.71, 4.19) n 18 With 95% confidence, you can say that the population mean interest rate is between 3.71% and 4.19%. 37. Use the standard normal distribution because σ is known and the weights are known to be normally distributed. σ 2.49 x ± zc = 13.68 ± 1.96 ≈ 13.68 ± 1.543 ≈ (12.14, 15.22) 10 n With 95% confidence, you can say that the population mean weight of two-year-old males is between 12.14 pounds and 15.22 pounds. 38. Neither distribution can be used because n < 30 and the data used are not normally distributed. 39. Neither distribution can be used because n < 30 and the mileages are not normally distributed. 40. Use the standard normal distribution because σ is known and the yards per carry are normally distributed. σ 1.76 x ± zc = 4.392 ± 1.96 ≈ 4.392 ± 0.957 ≈ (3.44, 5.35) 13 n With 95% confidence, you can say that the population mean yards per carry is between 3.44 and 5.35. 41. n = 25, x = 56.0, s = 0.25 x − μ 56.0 − 55.5 t= = = 10 s 0.25 n 25 −t0.99 = −2.797, t0.99 = 2.797 No; They are not making good tennis balls because for this sample the t-value is t = 10 , which is not between −t0.99 = −2.797 and t0.99 = 2.797 . 42. n = 16, x = 1015, s = 25 x − μ 1015 − 1000 t= = = 2.4 s 25 n 16 −t0.99 = −2.947, t0.99 = 2.947 Yes; They are making good light bulbs because for this sample the t-value is t = 2.4 , which is between −t0.99 = −2.947 and t0.99 = 2.947 .

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

6.3 CONFIDENCE INTERVALS FOR POPULATION PROPORTIONS 6.3 TRY IT YOURSELF SOLUTIONS 1.

x = 147, n = 147 + 420 + 378 + 105 = 1050 x 147 pˆ = = = 0.14 = 14% n 1050

pˆ = 0.14, qˆ = 1 − 0.14 = 0.86 npˆ = (1050)(0.14) = 147 > 5 nqˆ = (1050)(0.86) = 903 > 5 Distribution of p̂ is approximately normal. zc = 1.645

ˆˆ pq (0.14) (0.86) = 1.645 ≈ 0.018 n 1050 pˆ ± E ≈ 0.14 ± 0.018 = (0.122, 0.158)

E = zc

n = 800, pˆ = 0.52 qˆ = 1 − pˆ = 1 − 0.52 = 0.48 npˆ = 800 ⋅ 0.52 = 416 > 5 nqˆ = 800 ⋅ 0.48 = 384 > 5 Distribution of p̂ is approximately normal.

ˆˆ pq (0.52) (0.48) = 2.575 ≈ 0.045 n 800 pˆ ± E ≈ 0.52 ± 0.045 = (0.475, 0.565)

zc = 2.575, E = zc

4. (1) pˆ = 0.5, qˆ = 0.5, zc = 1.645, E = 0.02 2

§z · § 1.645 · ˆ ˆ ¨ c ¸ = (0.5)(0.5) ¨ n = pq ¸ ≈ 1691.266 → 1692 © 0.02 ¹ ©E¹ At least 1692 people should be included in the sample.

(2) pˆ = 0.622, qˆ = 0.378, zc = 1.645, E = 0.02 2

§z · § 1.645 · ˆ ˆ ¨ c ¸ = (0.622)(0.378) ¨ n = pq ¸ ≈ 1590.574 → 1591 © 0.02 ¹ ©E¹ At least 1591 people should be included in the sample.

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530

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

6.3 EXERCISE SOLUTIONS 1. True 2. False. To find the minimum sample size required for a confidence interval, solve for n in the formula for margin of error and round n up to the next whole number. 3.

pˆ =

x 62 = ≈ 0.060, qˆ = 1 − pˆ ≈ 0.940 n 1040

pˆ =

x 200 = ≈ 0.570, qˆ = 1 − pˆ ≈ 0.430 n 351

pˆ =

x 218 = ≈ 0.330, qˆ = 1 − pˆ ≈ 0.670 n 661

pˆ =

x 1025 = ≈ 0.240, qˆ = 1 − pˆ ≈ 0.760 n 4272

( 0.905, 0.933) → pˆ =

0.905 + 0.933 = 0.919 E = 0.933 − 0.919 = 0.014 2

( 0.245, 0.475 ) → pˆ =

0.245 + 0.475 = 0.360 E = 0.475 − 0.360 = 0.115 2

( 0.512, 0.596 ) → pˆ =

0.512 + 0.596 = 0.554 E = 0.596 − 0.554 = 0.042 2

10. ( 0.087, 0.263) → pˆ =

0.087 + 0.263 = 0.175 E = 0.263 − 0.175 = 0.088 2

x 1325 = ≈ 0.740, qˆ = 1 − pˆ ≈ 0.260 n 1790 ˆˆ pq (0.740)(0.260) 90% CI: pˆ ± zc ≈ 0.740 ± 1.645 ≈ 0.740 ± 0.017 = (0.723, 0.757) n 1790 ˆˆ (0.740)(0.260) pq 95% CI: pˆ ± zc ≈ 0.740 ± 1.96 ≈ 0.740 ± 0.020 = (0.720, 0.760) 1790 n With 90% confidence, you can say that the population proportion of the U.S. adults who say they have made a New Year’s resolution is between 72.3% and 75.7%. With 95% confidence, you can say that it is between 72.0% and 76.0%. The 95% confidence interval is slightly wider.

11. pˆ =

x 816 = ≈ 0.456, qˆ = 1 − pˆ ≈ 0.544 n 1790 ˆˆ pq (0.456)(0.544) 90% CI: pˆ ± zc ≈ 0.456 ± 1.645 ≈ 0.456 ± 0.019 = (0.437, 0.475) n 1790

12. pˆ =

ˆˆ pq (0.456)(0.544) ≈ 0.456 ± 1.96 ≈ 0.456 ± 0.023 = (0.433, 0.479) n 1790 With 90% confidence, you can say that the population proportion of the U.S. adults who say they made a New Year’s resolution related to their health is between 43.7% and 47.5%. With 95% confidence, you can say that it is between 43.3% and 47.9%. The 95% confidence interval is slightly wider.

95% CI: pˆ ± zc

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

x 364 = ≈ 0.360, qˆ = 1 − pˆ ≈ 0.640 n 1010 ˆˆ pq (0.360)(0.640) pˆ ± zc ≈ 0.360 ± 2.575 ≈ 0.360 ± 0.039 = (0.321, 0.399) n 1010 With 99% confidence, we can say that the population proportion of the U.S. adults who say they worry a great deal about the possibility of a future terrorist attack is between 32.1% and 39.9%.

13. pˆ =

x 556 = ≈ 0.550, qˆ = 1 − pˆ ≈ 0.450 n 1010 ˆˆ pq (0.550)(0.450) pˆ ± zc ≈ 0.550 ± 2.575 ≈ 0.550 ± 0.040 = (0.510, 0.590) n 1010 With 99% confidence, we can say that the population proportion of the U.S. adults who say they worry a great deal about hunger and homelessness is between 51.0% and 59.0%.

14. pˆ =

15. pˆ =

x 860 = ≈ 0.056, qˆ = 1 − pˆ ≈ 0.944 n 15,349

pˆ ± zc

ˆˆ (0.056)(0.944) pq ≈ 0.056 ± 1.96 ≈ 0.056 ± 0.004 = (0.052, 0.060) n 15,349

x 378 = ≈ 0.546, qˆ = 1 − pˆ ≈ 0.454 n 692 ˆˆ pq (0.546)(0.454) pˆ ± zc ≈ 0.546 ± 1.645 ≈ 0.546 ± 0.031 = (0.515, 0.577) n 692

16. pˆ =

531

532

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS 2

(c) Having an estimate of the population proportion reduces the minimum sample size needed. 21. 90% CI = (0.723, 0.757) 95% CI (0.720, 0.760) No; 0.725 falls within both confidence intervals. 22. 90% CI = (0.437, 0.475) 95% CI = (0.433, 0.479) Yes; 0.48 falls outside the confidence intervals. 23. Yes; The minimum sample size needed is 399 adults. 24. Yes; The minimum sample size needed is 3720 adults. 25. France: pˆ = 0.32, qˆ = 0.68, n = 1051 pˆ ± zc

ˆˆ pq (0.32)(0.68) = 0.32 ± 2.575 ≈ 0.32 ± 0.037 = (0.283, 0.357) n 1051

Germany: pˆ = 0.25, qˆ = 0.75, n = 1042 ˆˆ (0.25)(0.75) pq = 0.25 ± 2.575 ≈ 0.25 ± 0.035 = (0.215, 0.285) 1042 n United Kingdom: pˆ = 0.31, qˆ = 0.69, n = 1003 pˆ ± zc

ˆˆ (0.31)(0.69) pq = 0.31 ± 2.575 ≈ 0.31 ± 0.038 = (0.272, 0.348) 1003 n United States: pˆ = 0.35, qˆ = 0.65, n = 1000 pˆ ± zc

pˆ ± zc

ˆˆ pq (0.35)(0.65) = 0.35 ± 2.575 ≈ 0.35 ± 0.039 = (0.311, 0.389) n 1000

26. Germany and the United States. Their confidence intervals do not overlap.

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

27. Cyberterrorism: pˆ = 0.82, qˆ = 1 − pˆ = 0.18 ˆˆ (0.82)(0.18) pq = 0.82 ± 1.96 ≈ 0.82 ± 0.024 = (0.796, 0.844) 1021 n Development of nuclear weapons by Iran: pˆ = 0.75, qˆ = 1 − pˆ = 0.25 pˆ ± zc

ˆˆ pq (0.75)(0.25) = 0.75 ± 1.96 ≈ 0.75 ± 0.027 = (0.723, 0.777) n 1021 The spread of infectious diseases throughout the world: pˆ = 0.72, qˆ = 1 − pˆ = 0.28 pˆ ± zc

ˆˆ (0.72)(0.28) pq = 0.72 ± 1.96 ≈ 0.72 ± 0.028 = (0.692, 0.748) 1021 n Global warming or climate change: pˆ = 0.58, qˆ = 1 − pˆ = 0.42 pˆ ± zc

ˆˆ pq (0.58)(0.42) = 0.58 ± 1.96 ≈ 0.58 ± 0.030 = (0.550, 0.610) n 1021 The military power of Russia: pˆ = 0.44, qˆ = 1 − pˆ = 0.56 pˆ ± zc

ˆˆ pq (0.44)(0.56) = 0.44 ± 1.96 ≈ 0.44 ± 0.030 = (0.410, 0.470) n 1021 The conflict between the Israelis and Palestinians: pˆ = 0.32, qˆ = 1 − pˆ = 0.68 pˆ ± zc

pˆ ± zc

ˆˆ pq (0.32)(0.68) = 0.32 ± 1.96 ≈ 0.32 ± 0.029 = (0.291, 0.349) n 1021

28. “Development of nuclear weapons by Iran” and “the spread of infectious diseases throughout the world.” Their confidence intervals overlap. 29. 31% ± 2.9% → (28.1%, 33.9%) → (0.281, 0.339)

ˆˆ 1502 pq n → zc = E = 0.029 ≈→ zc = 2.43 ˆˆ (0.31)(0.69) n pq P(−2.43 < z < 2.43) = 0.9925 − 0.0075 = 0.985 (0.281, 0.339) is approximately a 98.5% CI. E = zc

30. 65% ± 7.9% → (57.1%, 72.9%) → (0.571, 0.729)

ˆˆ 220 pq n → zc = E = 0.079 ≈→ zc = 2.46 ˆ ˆ (0.65)(0.35) n pq P(−2.46 < z < 2.46) = 0.9931 − 0.0069 = 0.9862 (0.571, 0.729) is approximately a 98.6% CI. E = zc

533

534

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

31. 71% ± 3% → (68%, 74%) → (0.68, 0.74)

ˆˆ 1000 pq n → zc = E = 0.03 ≈→ zc = 2.09 ˆ ˆ (0.71)(0.29) n pq P(−2.09 < z < 2.09) = 0.9817 − 0.0183 = 0.9634 (0.68, 0.74) is approximately a 96.3% CI. E = zc

32. 73% ± 4.8% → (68.2%, 77.8%) → (0.682, 0.778)

ˆˆ 880 pq n → zc = E = 0.048 ≈→ zc = 3.21 ˆˆ (0.73)(0.27) n pq P(−3.21 < z < 3.21) = 0.9993 − 0.0007 = 0.9986 (0.682, 0.778) is approximately a 99.9% CI. E = zc

33. 57% ± 3.6% → (53.4%, 60.6%) → (0.534, 0.606)

ˆˆ 2094 pq n → zc = E = 0.036 ≈→ zc = 3.33 ˆˆ (0.57)(0.43) n pq P(−3.33 < z < 3.33) = 0.9996 − 0.0004 = 0.9992 (0.534, 0.606) is approximately a 99.9% CI. E = zc

34. 68% ± 3% → (65%, 71%) → (0.65, 0.71)

ˆˆ 1052 pq n → zc = E = 0.03 ≈→ zc = 2.09 ˆˆ (0.68)(0.32) n pq P(−2.09 < z < 2.09) = 0.9817 − 0.0183 = 0.9634 (0.65, 0.71) is approximately a 96.3% CI. 42% ± 3% → (39%, 45%) → (0.39, 0.45) E = zc

ˆˆ 1052 pq n → zc = E = 0.03 ≈→ zc = 1.97 ˆˆ (0.42)(0.58) n pq P(−1.97 < z < 1.97) = 0.9756 − 0.0244 = 0.9512 (0.39, 0.45) is approximately a 95.1% CI. E = zc

35. If npˆ < 5 or nqˆ < 5, the sampling distribution of p̂ may not be normally distributed, so zα 2 cannot

be used to calculate the confidence interval.

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

535

36. Sample answer:

E = zα 2 E zα 2

ˆˆ pq n

Write original equation. Divide each side by zα 2 .

Square each side. Solve for n.

ˆˆ p̂ qˆ = 1 − pˆ pq p̂ 0.0 1.0 0.00 0.9 0.1 0.9 0.09 1.0 0.2 0.8 0.16 0.45 0.3 0.7 0.21 0.46 0.4 0.6 0.24 0.47 0.5 0.5 0.25 0.48 0.6 0.4 0.24 0.49 0.7 0.3 0.21 0.50 0.8 0.2 0.16 0.51 ˆp = 0.5 gives the maximum value of pq ˆ ˆ.

qˆ = 1 − pˆ 0.1 0.0 0.55 0.54 0.53 0.52 0.51 0.50 0.49

ˆˆ pq 0.09 0.00 0.2475 0.2484 0.2491 0.2496 0.2499 0.2500 0.2499

p̂ 0.52 0.53 0.54 0.55

qˆ = 1 − pˆ 0.48 0.47 0.46 0.45

ˆˆ pq 0.2496 0.2491 0.2484 0.2475

6.4 CONFIDENCE INTERVALS FOR VARIANCE AND STANDARD DEVIATION 6.4 TRY IT YOURSELF SOLUTIONS 1.

d.f. = n − 1 = 29 level of confidence = 0.90 Area to the right of χ R2 is 0.05. Area to the right of χ L2 is 0.95.

χ R2 = 42.557 , χ L2 = 17.708 For a chi-square distribution curve with 29 degrees of freedom, 90% of the area under the curve lies between 17.708 and 42.557. 2. 90% CI: χ R2 = 42.557 , χ L2 = 17.708

95% CI: χ R2 = 45.722 , χ L2 = 16.047 § (n − 1) s 2 (n − 1) s 2 · § 29 ⋅ (1.2) 2 29 ⋅ (1.2) 2 · , , 90% CI for σ 2 : ¨ ¸=¨ ¸ ≈ (0.98, 2.36) 2 17.708 ¹ χ L2 ¹ © 42.557 © χR Copyright © 2023 Pearson Education, Inc.

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CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

§ (n − 1) s 2 (n − 1) s 2 · § 29 ⋅ (1.2) 2 29 ⋅ (1.2) 2 · , , 95% CI for σ 2 : ¨ ¸=¨ ¸ ≈ (0.91, 2.60) 2 16.047 ¹ χ L2 ¹ © 45.722 © χR 90% CI for σ : ( 0.981, 2.358 ) ≈ ( 0.99, 1.54 )

95% CI for σ : ( 0.913, 2.602 ) ≈ ( 0.96, 1.61) With 90% confidence, you can say that the population variance is between 0.98 and 2.36 and that the population standard deviation is between 0.99 and 1.54. With 95% confidence, you can say that the population variance is between 0.91 and 2.60, and that the population standard deviation is between 0.96 and 1.61.

6.4 EXERCISE SOLUTIONS 1. Yes. 2. It approaches the shape of the normal curve. 3.

χ R2 = 14.067 , χ L2 = 2.167

χ R2 = 31.319 , χ L2 = 4.075

χ R2 = 32.852 , χ L2 = 8.907

χ R2 = 44.314 , χ L2 = 11.524

χ R2 = 52.336 , χ L2 = 13.121

χ R2 = 63.167 , χ L2 = 37.689

§ (n − 1) s 2 ( n − 1) s 2 · § 29 ⋅ (11.56) 29 ⋅ (11.56) · , , 9. (a) ¨ ¸=¨ ¸ ≈ (7.33, 20.89) 2 16.047 ¹ χ L2 ¹ © 45.722 © χR (b) ( 7.3321,

20.8911 ) ≈ (2.71, 4.57)

§ ( n − 1) s 2 ( n − 1) s 2 · § 6 ⋅ (0.64) 6 ⋅ (0.64) · , , 10. (a) ¨ ¸=¨ ¸ ≈ (0.21, 5.68) 2 0.676 ¹ χ L2 ¹ © 18.548 © χR (b) ( 0.2070,

5.6805 ) ≈ (0.46, 2.38)

§ (n − 1) s 2 (n − 1) s 2 · § 17 ⋅ (35) 2 17 ⋅ (35) 2 · , , 11. (a) ¨ ¸=¨ ¸ ≈ (755, 2401) 2 8.672 ¹ χ L2 ¹ © 27.587 © χR (b) ( 754.885,

2401.407 ) ≈ (27, 49)

§ (n − 1) s 2 (n − 1) s 2 · § 40 ⋅ (278.1) 2 40 ⋅ (278.1)2 · , , 12. (a) ¨ ¸=¨ ¸ ≈ (48,571.8, 139,577.0) 2 22.164 ¹ χ L2 ¹ © 63.691 © χR (b) ( 48,571.77 ,

139,576.99 ) ≈ (220.4, 373.6)

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

537

13. (a) s 2 ≈ 0.0756 § (n − 1) s 2 (n − 1) s 2 · § 17 ⋅ (0.0756) 17 ⋅ (0.0756) · , , ¨ ¸≈¨ ¸ ≈ (0.0426, 0.1699) 2 7.564 ¹ χ L2 ¹ © 30.191 © χR

(b) ( 0.04257 , 0.16991 ) ≈ (0.2063, 0.4122) With 95% confidence, you can say that the population variance is between 0.0426 and 0.1699, and the population standard deviation is between 0.2063 and 0.4122 inch. 14. (a) s 2 ≈ 18.116 § (n − 1) s 2 (n − 1) s 2 · § 14 ⋅ (18.116) 14 ⋅ (18.116) · , , ¨ ¸≈¨ ¸ ≈ (10.71, 38.60) 2 6.571 ¹ χ L2 ¹ © 23.685 © χR

(b) ( 10.71, 38.60 ) ≈ (3.27, 6.21) . With 90% confidence, you can say that the population variance is between 10.71 and 38.60, and the population standard deviation is between 3.27 and 6.21 minutes. 15. (a) s 2 = 104.811 § ( n − 1) s 2 ( n − 1) s 2 · § 20 ⋅ (104.811) 20 ⋅ (104.811) · , , ¨ ¸≈¨ ¸ ≈ (52.41, 281.98) 2 7.434 χ L2 ¹ © 39.997 ¹ © χR

(b) ( 52.41, 281.98 ) ≈ (7.24, 16.79) With 99% confidence, you can say that the population variance is between 52.41 and 281.98, and the population standard deviation is between 7.24 and 16.79 thousand dollars. 16. (a) s 2 ≈ 253.913 § (n − 1) s 2 (n − 1) s 2 · § 23 ⋅ (253.913) 23 ⋅ (253.913) · , , ¨ ¸≈¨ ¸ ≈ (153.4, 499.6) 2 11.689 χ L2 ¹ © 38.076 ¹ © χR

(b) ( 153.4, 499.6 ) ≈ (12.4, 22.4) With 95% confidence, you can say that the population variance is between 153.4 and 499.6, and the population standard deviation is between 12.4 and 22.4. § (n − 1) s 2 (n − 1) s 2 · § 13 ⋅ (3.54) 2 13 ⋅ (3.54) 2 · , , 17. (a) ¨ ¸≈¨ ¸ ≈ (5.46, 45.70) 2 3.565 ¹ χ L2 ¹ © 29.819 © χR (b) ( 5.46, 45.70 ) ≈ (2.34, 6.76) With 99% confidence, you can say that the population variance is between 5.46 and 45.70, and the population standard deviation is between 2.34 and 6.76 days. § (n − 1) s 2 (n − 1) s 2 · § 14 ⋅ (26.1) 2 14 ⋅ (26.1) 2 · , , 18. (a) ¨ ¸=¨ ¸ ≈ (452.76, 1224.25) 2 7.790 ¹ χ L2 ¹ © 21.064 © χR

538

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

(b) ( 452.76, 1224.25 ) ≈ (21.28, 34.99) With 80% confidence, you can say that the population variance is between 452.76 and 1224.25, and the population standard deviation is between 21.28 and 34.99 service aces. § ( n − 1) s 2 ( n − 1) s 2 · § 18 ⋅ (15) 2 18 ⋅ (15) 2 · , , 19. (a) ¨ ¸=¨ ¸ ≈ (128, 492) 2 8.231 ¹ χ L2 ¹ © 31.526 © χR (b) ( 128.465, 492.042 ) ≈ (11, 22) With 95% confidence, you can say that the population variance is between 128 and 492, and the population standard deviation is between 11 and 22 grains per gallon. § ( n − 1) s 2 ( n − 1) s 2 · § 29(3600) 2 29(3600) 2 · , , 20. (a) ¨ ¸=¨ ¸ ≈ (8,831, 450, 21,224,305) 2 17.708 ¹ χ L2 ¹ © 42.557 © χR (b) ( 8,831, 450, 21, 224,305 ) ≈ (2972, 4607) With 90% confidence, you can say that the population variance is between 8,831,450 and 21,224,305, and the population standard deviation is between $2972 and $4607. § (n − 1) s 2 (n − 1) s 2 · § 17 ⋅ (0.25) 2 17 ⋅ (0.25) 2 · , , 21. (a) ¨ ¸=¨ ¸ ≈ (0.04, 0.11) 2 10.085 ¹ χ L2 ¹ © 24.769 © χR (b) ( 0.0429, 0.1054 ) ≈ (0.21, 0.32) With 80% confidence, you can say that the population variance is between 0.04 and 0.11, and the population standard deviation is between 0.21 and 0.32 day. § (n − 1) s 2 (n − 1) s 2 · § 60 ⋅ (6.46) 2 60 ⋅ (6.46) 2 · , , 22. (a) ¨ ¸=¨ ¸ ≈ (28.33, 66.80) 2 37.485 ¹ χ L2 ¹ © 88.379 © χR (b) ( 28.33,

66.80 ) ≈ (5.32, 8.17)

With 98% confidence, you can say that the population variance is between 28.33 and 66.80, and the population standard deviation is between 5.32 and 8.17 inches. § ( n − 1) s 2 (n − 1) s 2 · § 27 ⋅ (56.1) 2 27 ⋅ (56.1) 2 · , , 23. (a) ¨ ¸=¨ ¸ ≈ (1809.40, 6597.92) 2 12.879 ¹ χ L2 ¹ © 46.963 © χR (b) ( 1809.40, 6597.92 ) ≈ (42.54, 81.23) With 98% confidence, you can say that the population variance is between 1809.40 and 6597.92, and the population standard deviation is between 42.54 and 81.23 seconds. § (n − 1) s 2 (n − 1) s 2 · § 19 ⋅ (0.91) 2 19 ⋅ (0.91) 2 · , , 24. (a) ¨ ¸=¨ ¸ ≈ (0.52, 1.56) 2 10.117 ¹ χ L2 ¹ © 30.144 © χR

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

539

(b) ( 0.52, 1.56 ) ≈ (0.72, 1.25) With 90% confidence, you can say that the population variance is between 0.52 and 1.56, and the population standard deviation is between 0.72 and 1.25 seconds. 25. 95% CI for σ : (0.2063, 0.4122) Yes, because all of the values in the confidence interval are less than 0.5. 26. 90% CI for σ : (3.27, 6.21) Yes, because all the values in the confidence interval are less than 10. 27. 80% CI for σ : (0.21, 0.32) No, because 0.25 is contained in the confidence interval. 28. 98% CI for σ : (42.54, 81.23) No, because 1 minute is contained in the confidence interval. 29. Answers will vary. Sample answer: Unlike a confidence interval for a population mean or proportion, a confidence interval for a population variance does not have a margin of error. The left and right endpoints must be calculated separately.

CHAPTER 6 REVIEW EXERCISE SOLUTIONS 1. (a) x ≈ 103.5

(b) E = zc

σ n

= 1.645

45 40

≈ 11.7

2. (a) x ≈ 9.5

(b) E = zc

3. (a) x ± zc

σ n

= 1.96

8 30

≈ 2.9

= 103.5 ± 1.645

≈ 103.5 ± 11.7 = (91.8, 115.2) 40 n With 90% confidence, you can say that the population mean waking time is between 91.8 and 115.2 minutes past 5:00 A.M.

(b) x ± 0.10 x = 103.5 ± 0.10(103.5) = 103.5 ± 10.35 = (93.15, 113.85) Yes; If the population mean is within 10% of the sample mean, then it falls inside the confidence interval. 4. (a) x ± zc

= 9.5 ± 1.96

≈ 9.5 ± 2.9 = (6.6, 12.4) 30 n With 95% confidence, you can say that the population mean driving distance is between 6.6 and 12.4 miles.

540

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

(b) x ± 0.10 x = 9.5 ± 0.125(9.5) = 9.5 ± 1.19 = (8.31, 10.69) No; 12.5 is outside the confidence interval. 5.

( 20.75, 24.10 ) → x =

20.75 + 24.10 = 22.425 → E = 24.10 − 22.425 = 1.675 2

( 7.428, 7.562 ) → x =

7.428 + 7.562 = 7.495 → E = 7.562 − 7.495 = 0.067 2

tc = 1.383

13. (a) E = tc

(b) x ± tc

14. (a) E = tc

(b) x ± tc

15. (a) E = tc

(b) x ± tc

16. (a) E = tc

(b) x ± tc

17. x ± tc

10. tc = 2.069

11. tc = 2.624

12. tc = 2.756

s 25.6 = 1.753 ≈ 11.2 16 n s ≈ 72.1 ± 11.2 = (60.9, 83.3) n s 1.1 = 2.064 ≈ 0.5 25 n s ≈ 3.5 ± 0.5 = (3.0, 4.0) n

s ≈ 6.8 ± 0.7 = (6.1, 7.5) n

s ≈ 25.2 ± 10.6 = (14.6, 35.8) n = 165 ± 1.690

≈ 165 ± 18.87 ≈ (146, 184) 36 n With 90% confidence, you can say that the population mean height is between 146 and 184 feet.

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

541

18. Yes; 160 feet falls within the confidence interval, (146, 184). 19. (a) pˆ =

x 383 = ≈ 0.420, qˆ = 1 − pˆ ≈ 0.580 n 912

(b) 90% CI: pˆ ± zc 95% CI: pˆ ± zc

ˆˆ pq 0.420 ⋅ 0.580 ≈ 0.420 ± 1.645 ≈ 0.420 ± 0.027 = (0.393, 0.447) 912 n ˆˆ pq 0.420 ⋅ 0.580 ≈ 0.420 ± 1.96 ≈ 0.420 ± 0.032 = (0.388, 0.452) 912 n

x 5362 = ≈ 0.390, qˆ = 1 − pˆ ≈ 0.610 n 13,749

(b) 90% CI: pˆ ± zc

ˆˆ 0.390 ⋅ 0.610 pq ≈ 0.390 ± 1.645 ≈ 0.390 ± 0.007 = (0.383, 0.397) n 13,749

95% CI: pˆ ± zc

ˆˆ 0.390 ⋅ 0.610 pq ≈ 0.390 ± 1.96 ≈ 0.390 ± 0.008 = (0.382, 0.398) n 13,749

x 23,991 = ≈ 0.325, qˆ = 1 − pˆ ≈ 0.675 n 73,901

(b) 90% CI: pˆ ± zc

ˆˆ pq 0.325 ⋅ 0.675 ≈ 0.325 ± 1.645 ≈ 0.325 ± 0.003 = (0.322, 0.328) n 73,901

95% CI: pˆ ± zc

ˆˆ 0.325 ⋅ 0.675 pq ≈ 0.325 ± 1.96 ≈ 0.325 ± 0.003 = (0.322, 0.328) 73,901 n

x 1334 = ≈ 0.600, qˆ = 1 − pˆ ≈ 0.400 n 2223

542

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

(b) 90% CI: pˆ ± zc 95% CI: pˆ ± zc

ˆˆ pq 0.600 ⋅ 0.400 ≈ 0.600 ± 1.645 ≈ 0.600 ± 0.017 = (0.583, 0.617) 2223 n ˆˆ pq 0.600 ⋅ 0.400 ≈ 0.600 ± 1.96 ≈ 0.600 ± 0.020 = (0.580, 0.620) 2223 n

§z · § 1.96 · ˆ ˆ ¨ c ¸ = 0.50 ⋅ 0.50 ¨ 25. (a) n = pq ¸ ≈ 384.16 → 385 adults © 0.05 ¹ ©E¹ §z · § 1.96 · ˆ ˆ ¨ c ¸ = 0.32 ⋅ 0.68 ¨ (b) n = pq ¸ ≈ 334.4 → 335 adults © 0.05 ¹ ©E¹ (c) Having an estimate of the population proportion reduces the minimum sample size needed. 26. Yes, because the minimum sample size needed is 335 adults. 27. χ R2 = 23.337 , χ L2 = 4.404 28. χ R2 = 42.980 , χ L2 = 10.856 29. χ R2 = 24.996 , χ L2 = 7.261 30. χ R2 = 23.589 , χ L2 = 1.735 31. s 2 ≈ 25.833

§ (n − 1) s 2 ( n − 1) s 2 · § 12 ⋅ (25.833) 12 ⋅ (25.833) · , , (a) 95% CI for σ 2 : ¨ ¸≈¨ ¸ ≈ (13.28, 70.39) 2 4.404 ¹ χ L2 ¹ © 23.337 © χR (b) 95% CI for σ :

( 13.28, 70.39 ) ≈ (3.64, 8.39)

With 95% confidence, we can say that the population variance is between 13.28 and 70.39, and the population standard deviation is between 3.64 and 8.39.

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

543

32. s 2 ≈ 0.011

§ (n − 1) s 2 ( n − 1) s 2 · § (30)(0.011) (30)(0.011) · , , (a) 98% CI for σ 2 : ¨ ¸≈¨ ¸ ≈ (0.006, 0.022) 2 14.954 ¹ χ L2 ¹ © 50.892 © χR § (30)(0.011) (30)(0.011) · (b) 98% CI for σ : ¨ , ¸ ≈ (0.081, 0.149) ¨ 50.892 14.954 ¸¹ © With 98% confidence, you can say that the population variance is between 0.006 and 0.022, and the population standard deviation is between 0.081 and 0.149.

CHAPTER 6 QUIZ SOLUTIONS

1. (a) x =

¦ x = 49.17 ≈ 2.46 n

(b) E = zc (c) x ± zc

= 1.96

0.068 20

≈ 0.030

≈ 2.46 ± 0.030 = (2.43, 2.49) n With 95% confidence, you can say that the population mean winning time is between 2.43 and 2.49 hours.

(d) No; It falls outside the confidence interval. 2

3. (a) x = 6.61, s ≈ 3.376 s 3.376 ≈ 6.61 ± 1.833 ≈ 6.61 ± 1.957 ≈ (4.65, 8.57) n 10 With 90% confidence you can say that the population mean amount of time is between 4.65 and 8.57 minutes.

(b) x ± tc

≈ 6.61 ± 1.645

3.5

≈ 6.61 ± 1.82 = (4.79, 8.43) n 10 With 90% confidence you can say that the population mean amount of time is between 4.79 and 8.43 minutes. This confidence interval is narrower than the one found in part (b).

544

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

s 36,729 = 133,326 ± 2.201 ≈ 133,326 ± 23,336 = (109,990, 156,662) n 12 With 95% confidence you can say that the population mean annual earnings is between $109,990 and $156,662. x ± tc

5. Yes. $131,935 falls within the confidence interval, (109,990, 156,662). 6. (a) pˆ =

x 838 = ≈ 0.830 n 1010

ˆˆ 0.830 ⋅ 0.170 pq ≈ 0.830 ± 1.645 ≈ 0.830 ± 0.019 = (0.811, 0.849) 1010 n With 90% confidence, you can say that the population proportion of the U.S. adults who say that the energy situation in the United States is very or fairly serious is between 81.1% and 84.9%.

(b) pˆ ± zc

§z · § 2.575 · ˆ ˆ ¨ c ¸ ≈ 0.830 ⋅ 0.170 ¨ (d) n = pq ¸ ≈ 584.74 → 585 adults © 0.04 ¹ ©E¹ § (n − 1) s 2 (n − 1) s 2 · § 9 ⋅ (3.38) 2 9 ⋅ (3.38) 2 · , , 7. (a) ¨ ¸≈¨ ¸ ≈ (5.41, 38.08) 2 2.700 ¹ χ L2 ¹ © 19.023 © χR (b)

( 5.4050, 38.0813 ) ≈ (2.32, 6.17) With 95% confidence you can say that the population standard deviation is between 2.32 and 6.17 minutes.

CHAPTER 6 TEST SOLUTIONS

1. (a) pˆ =

x 1740 = ≈ 0.830 n 2096

ˆˆ 0.830 ⋅ 0.170 pq ≈ 0.830 ± 1.96 ≈ 0.830 ± 0.016 = (0.814, 0.846) 2096 n With 95% confidence, you can say that the population proportion of the U.S. adults who think football teams of all levels should require players who suffer a head injury to take a set amount of time off from playing to recover is between 81.4% and 84.6%.

(b) pˆ ± zc

CHAPTER 6 ALTERNATE Ň CONFIDENCE INTERVALS

545

2. (a) x = 214.1, s ≈ 51.1 s 51.07 ≈ 214.1 ± 2.262 ≈ 214.1 ± 36.53 ≈ (177.6, 250.6) n 10 With 95% confidence you can say that the population mean weight is between 177.6 and 250.6 pounds.

(b) x ± tc

§ (n − 1) s 2 (n − 1) s 2 · § (9)(51.1) 2 (9)(51.1)2 · ¸ ¨ ¸ ≈ (31.6, 116.4) , , ≈ (c) 99% CI for σ : ¨ 2 2 ¨ ¸ ¨ 23.589 1.735 ¸ χ χ R L © ¹ © ¹ With 99% confidence, you can say that the population standard deviation is between 31.6 and 116.4 pounds. 3. (a) x = 605

(b) x ± zc

= 605 ± 1.645

108

≈ 605 ± 51.29 = (553.71, 656.29) n 12 With 90% confidence, you can say that the population mean test score is between 553.71 and 656.29.

§ z σ · § 1.96 ⋅ 108 · (d) n = ¨ c ¸ = ¨ ¸ ≈ 448.08 → 449 students ¹ © E ¹ © 10 4. (a) Use the t-distribution because σ is unknown and n ≥ 30 . s 7.5 x ± tc = 20 ± 2.023 ≈ 20 ± 2.4 = (17.6, 22.4) n 40 With 95% confidence, you can say that the population mean waiting time is between 17.6 and 22.4 minutes.

(b) Use the standard normal distribution because σ is known and the weights are normally distributed. σ 0.05 x ± zc = 11.89 ± 1.645 ≈ 11.89 ± 0.021 ≈ (11.87, 11.91) n 15 With 90% confidence, you can say that the population mean weight is between 11.87 and 11.91 ounces.

Alternative Presentation of the Standard Normal Distribution TRY IT YOURSELF SOLUTIONS 1. (1) 0.4857 (2) z o2.17 2.

The area corresponding to z 2.13 is 0.4834 Area = 0.5 + 0.4834 = 0.9834 3.

The area corresponding to z 2.16 is 0.4846 Area = 0.50+00.4846 = 0.9846 4.

The area corresponding to z 1.35 is 0.0885 The area corresponding to z 2.165 is 0.0152 Area = 0.0885 − 0.0152 = 0.0733

APPENDIX

Normal Probability Plots

TRY IT YOURSELF SOLUTIONS 1.

The points show a curve, so you can conclude that the sample data come from a population that is not normally distributed. APPENDIX C EXERCISE SOLUTIONS 1. The observed values are usually plotted along the horizontal axis. The expected z-scores are plotted along the vertical axis. 2. If the plotted points in a normal probability plot are approximately linear, then you can conclude that the data come from a normal distribution. If the plotted points are not approximately linear or follow some type of pattern that is not linear, then you can conclude that the data come from a distribution that is not normal. Multiple outliers or clusters of points indicate a distribution that is not normal. 3. Because the points appear to follow a nonlinear pattern, you can conclude that the data do not come from a population that has a normal distribution. 4. Because the points are approximately linear, you can conclude that the data come from a population that has a normal distribution. 5.

Because the points are approximately linear, you can conclude that the data come from a population that has a normal distribution. 6.

548

ACTIVITIES Ň

ACTIVITY 1.3 1. Sample answer: 1, 17, 17, 17, 14, 10, 15, 6 This list is a random sample taken with replacement, because 17 appears three times. When you uncheck “Allow repeats,” the samples will not have duplicate numbers. This will give you samples that are taken without replacement. 2. The minimum is 1, the maximum is 731, and the number of samples is 8. Sample answer: 565, 718, 305, 75, 364, 221, 230, 231 The random number generator is easier to use than the random number table.

ACTIVITIES Ň 549

ACTIVITY 2.3 1. When the mean and median are very close (or the same), the distribution of the plotted data is symmetric. When the mean and median are not close, the distribution of the plotted points is skewed, either negatively or positively. 2.

Sample answer: 2, 4, 4, 5, 5, 5, 5, 6, 6, 8. The mean and median are the same value, or 5. Changing a data set by adding a data point will change the mean, but it may or may not change the median. This is true of all data sets.

ACTIVITY 2.4 1. When a point with a value of 15 is added, both the mean and standard deviation increase. When a point with a value of 20 is added, the mean and standard deviation both increase by even more. 2. To get the largest standard deviation, plot four of the points at the lower limit and four of the points at the upper limit. To get the smallest standard deviation, plot all of the points at the same number. That way, each x x is 0, so the standard deviation will be 0.

550

ACTIVITIES Ň

ACTIVITY 3.1 1. Answers will vary. 2. Answers will vary. P market goes up on day 36 0.5

ACTIVITY 3.3 1. Answers will vary. 2. The theoretical probability is 0.5, so the green line should be placed there.

ACTIVITIES Ň 551

ACTIVITY 4.2 1. Answers will vary. Sample answer: {7, 8, 7, 7, 7, 7, 8, 8, 9, 10} (a) P x 5

0 10

(b) P( x p 8)

5 10

(b) P( x p 5)

1 10

2. Answers will vary. Sample answer: {3, 1, 2, 2, 1, 2, 2, 2, 5, 3} (a) P x 4

0 10

9 10

3. Answers will vary. Sample answer: P x 5 x 0.103 When using N 100 , the estimated probability of exactly five is closer to 0.103.

552

ACTIVITIES Ň

ACTIVITY 5.4 1. Answers will vary. The mean of the sampling distribution of a uniform, bell-shaped, and skewed distribution will be approximately 25. 2. Answers will vary. The estimated standard deviation of the sampling distribution will be approximately

σ 50

ACTIVITIES Ň 553

ACTIVITY 6.2 1. Answers will vary. Approximately 95% of the z and t confidence intervals will contain the mean of 25. 2. Answers will vary. Approximately 95% of the z and t confidence intervals will contain the mean of 7.26. Because σ is not known, we should use a t-CI.

ACTIVITY 6.3 1. Answers will vary. Approximately 95% and 99% of the CI’s will contain 0.6. 2. Answers will vary. Approximately 95% and 99% of the CI’s will contain 0.6.

554

ACTIVITIES Ň

ACTIVITY 7.3 1. Answers will vary. At B 0.05 , approximately 50 of the 1000 null hypotheses will be rejected. At B 0.01 , approximately 10 of the 1000 null hypotheses will be rejected. 2. Answers will vary. If the null hypothesis is rejected at the 0.01 level, the P-value will be smaller than 0.01. So, it would also be rejected at the 0.05 level. Suppose a null hypothesis is rejected at the 0.05 level. It will not necessarily be rejected at the 0.01 level because it is possible that 0.01 < P-value < 0.05. 3.

H 0 : N p 27

Answers will vary. The proportion of rejected null hypotheses will be larger than 0.05 and 0.01 because the true mean is 25 (i.e. N 27 ).

ACTIVITY 7.4 1. Answers will vary. Approximately 50 null hypotheses will be rejected at the 0.05 level while approximately 10 null hypotheses will be rejected at the 0.01 level. 2.

H 0 : p p 0.4 H a : p 0.4

Answers will vary. The proportion of null hypotheses rejected will be less than 0.05 and 0.01 because the true p = 0.6.

ACTIVITIES Ň 555

ACTIVITY 9.1 1. Answers will vary. 2. Create a data set that is linear with a positive slope. 3. Create a data set that is nonlinear. 4.

r ≈ −0.9 has a negative slope while using r ≈ 0.9 will have a positive slope.

ACTIVITY 9.2 1. Answers will vary. 2. Answers will vary. The regression line is influenced by the new point. 3. Answers will vary. The regression line is not as influenced by the new point due to more points being used to calculate the regression line. 4. Answers will vary. As the sample size increases, the slope changes less.

556

CHAPTER 6 ALTERNATE Ň ACTIVITIES

ACTIVITY 6.3 1. Answers will vary. Approximately 95% and 99% of the CI’s will contain 0.6. 2. Answers will vary. Approximately 95% and 99% of the CI’s will contain 0.6.

Ň CASE STUDY

557

CHAPTER 1 CASE STUDY: REPUTATIONS OF COMPANIES IN THE U.S. 1.

34, 026 x 0.000132 º 0.0132% 257, 900, 000

2. Company Name, Reputation, Industry 3. Company Name; The data can be ordered from best to worst or from worst to best. 4. Year Company Formed; ascending order 5. Number of employees 6. A decision of whether to update a company’s business operations or not can be made based on the Harris Poll.

558

CASE STUDY Ň

CHAPTER 2 CASE STUDY: Business Size 1. California, because it has the most total number of businesses. 2. California: Midpoint, x 2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500 N

xf 38,660.0 48,503.0 82,505.0 181,228.5 164,794.0 279,025.5 170,772.0 120,669.5 127,500 xf 1, 213, 657.5

xf 1, 213, 657.5 x 32.066 x 32.1 N 37, 849

Illinois: Midpoint, x 2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500 N

f 15,464 6929 5690 5253 2212 1599 456 161 85 N 37,849

f 4513 2239 2000 2105 1074 839 257 92 35 N 13,154

xf 572, 697 x 43.538 x 43.5 N 13,154

xf 11,282.5 15,673.0 29,000.0 72,622.5 80,013.0 146,405.5 96,246.5 68,954.0 52,500.0 xf 572, 697

Ň CASE STUDY

Indiana: Midpoint, x 2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500 N

f 4007 2062 2043 1952 1039 858 282 105 52 N 12, 400

xf 10,017.5 14,434.0 29,623.5 67,344.0 77,405.5 149,721.0 105,609.0 78,697.5 78,000.0 xf 610,852.0

xf 610,852.0 x 49.262 x 49.3 N 12, 400

New York: Midpoint, x 2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500 N

xf 5532.5 9016.0 18,357.0 47,265.0 59,376.5 122,324.5 91,378.0 89,940.0 69,000.0 xf 512,189.5

xf 512,189.5 x 63.666 x 63.7 N 8045

Michigan: Midpoint, x 2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500 N

f 2213 1288 1266 1370 797 701 244 120 46 N 8045

f 6809 2726 2245 2030 838 565 188 60 27 N 15, 488

xf 455, 591.5 x 29.416 x 29.4 N 15, 488

xf 17,022.5 19,082.0 32,552.5 70,035.0 62,431.0 98,592.5 70,406.0 44,970.0 40,500.0 xf 455, 591.5

559

560

CASE STUDY Ň

Ohio: Midpoint, x 2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500 N

f 4323 2374 2213 2269 1054 879 274 82 34 N 13, 502

xf 10,807.5 16,618.0 32,088.5 78,280.5 78,523.0 153,385.5 102,613.0 61,459.0 51,000.0 xf 584, 775.0

xf 584, 775.0 x 43.310 x 43.3 N 13, 502

Texas: Midpoint, x 2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500 N

xf 10,197.5 16,870.0 33,103.5 81,765.0 94,317.0 172,580.5 122,087.0 94,437.0 79,500.0 xf 704,857.5

xf 704,857.5 x 50.702 x 50.7 N 13, 902

Pennsylvania: Midpoint, x 2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500 N

f 4079 2410 2283 2370 1266 989 326 126 53 N 13, 902

f 7171 3472 2990 3157 1376 1079 312 140 67 N 19, 764

xf 807, 574.5 x 40.861 x 40.9 N 19, 764

xf 17,927.5 24,304.0 43,355.0 108,916.5 102,512.0 188,285.5 116,844.0 104,930.0 100,500.0 xf 807, 574.5

Ň CASE STUDY

Wisconsin: Midpoint, x 2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500 N

f 2581 1371 1311 1520 900 757 241 102 34 N 8817

xf 6452.5 9597.0 19,009.5 52,440.0 67,050.0 132,096.5 90,254.5 76,449.0 51,000.0 xf 504, 349.0

xf 504,349.0 x 57.202 x 57.2 N 8817

3. Indiana had the greatest number of employees per manufacturing business (63.7). 4. California: N x 32.066 Midpoint, x

x N

2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500

í29.566 í25.066 í17.566 2.434 42.434 142.434 342.434 717.434 1467.934

874.148 628.304 308.564 5.924 1800.644 20,287.444 117,261.044 514,711.544 2,154,830.228

x N

x N N

375, 581, 002.500 x 99.6 37, 849

13,517,824.67 4,353,518.416 1,755,729.16 31,118.772 3,983,024.528 32,439,622.960 53,471,036.060 82,868,558.580 183,160,569.400

x N T

f 375, 581, 002.500

561

562

CASE STUDY Ň

Illinois: N x 43.538 Midpoint, x

x N

2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500

−41.038 −36.538 −29.038 −9.038 30.962 130.962 330.962 705.962 1456.462

1684.117 1335.025 843.205 81.685 958.645 17,151.045 109,535.845 498,382.345 2,121,281.557

x N

7,600,420.021 2,989,120.975 1,686,410.000 171,946.925 1,029,584.730 14,389,726.760 28,150,712.170 45,851,175.740 74,244,854.500

x N x N

f 176,113, 951.800

176,113, 951.800 x 115.7 13,154

Indiana: N x 63.666 Midpoint, x

x N

2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500

í61.166 −56.666 −49.166 −29.166

3741.280 3211.036 2417.300 850.656 117.376 12,284.176 96,617.776 470,368.276 2,063,055.36

10.834 110.834 310.834 685.834 1436.334

x N

x N N

200, 265, 200.600 x 157.8 8, 045

8,279,452.640 4,135,814.368 3,060,301.800 1,165,398.720 93,548.672 8,611,207.376 23,574,737.340 56,444,193.120 94,900,546.560

x N T

f 200, 265, 200.600

Ň CASE STUDY

Michigan: N x 49.262 Midpoint, x

x N

2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500

−46.762 −42.262 −34.762 −14.762

2186.685 1786.077 1208.397 217.917 636.957 15,684.557 105,779.757 490,333.257 2,104,640.745

25.238 125.238 325.238 700.238 1450.738

x N

8,762,046.795 3,682,890.774 2,468,755.071 425,373.984 661,798.323 13,457,349.910 29,829,891.470 51,484,991.990 109,441,318.700

x N x N

f 220, 214, 417.000

220, 214, 417.000 x 133.3 12, 400

New York: N x 29.416 Midpoint, x

x N

2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500

−26.916 −22.416 −14.916

724.471 502.477 222.487 25.847 2032.567 21,049.367 119,082.967 518,520.967 2,162,617.301

5.084 45.084 145.084 345.084 720.084 1470.584

x N

132, 340, 334.500 x 92.4 15, 488

4,932,923.039 1,369,752.302 499,483.315 52,469.410 1,703,291.146 11,892,892.360 22,387,597.800 31,111,258.020 58,390,667.130

x N x N

f 132, 340, 334.500

563

564

CASE STUDY Ň

Ohio: N x 50.702 Midpoint, x

x N

2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500

−48.202 −43.702 −36.202 −16.202 23.798

2323.433 1909.865 1310.585 262.505 566.345 15,325.945 104,845.145 488,318.645 2,100,464.693

123.798 323.798 698.798 1449.298

x N

9,477,283.207 4,602,774.650 2,992,065.555 622,136.850 716,992.770 15,157,359.610 34,179,517.270 61,528,149.270 111,324,628.700

x N x N

f 240, 600, 907.900

240, 600, 907.900 x 131.6 13, 902

Pennsylvania: N x 43.310 Midpoint, x

x N

2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500

−40.810 −36.310 −28.810 −8.810

1665.456 1318.416 830.016 77.616 972.816 17,210.816 109,686.816 498,704.316 2,121,945.756

31.190 131.190 331.190 706.190 1456.690

x N

171, 590, 374.500 x 112.7 13, 502

7,199,766.288 3,129,919.584 1,836,825.408 176,110.704 1,025,348.064 15,128,307.260 30,054,187.580 40,893,753.910 72,146,155.700

x N x N

f 171, 590, 374.500

Ň CASE STUDY

565

Texas: N x 40.861 Midpoint, x

x N

2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500

−38.361 −33.861 −26.361 í6.361

1471.566 1146.567 694.902 40.462 1131.582 17,859.382 111,314.982 502,169.232 2,129,086.621

33.639 133.639 333.639 708.639 1459.139

x N

10,552,599.790 3,980,880.624 2,077,756.980 127,738.534 1,557,056.832 19,270,273.180 34,730,274.380 70,303,692.480 142,648,803.600

x N x N

f 285, 249, 076.400

285, 249, 076.400 x 120.1 19, 764

Wisconsin: N x 57.202 Midpoint, x

x N

2.5 7.0 14.5 34.5 74.5 174.5 374.5 749.5 1500

−54.702 −50.202 −42.702 −22.702

2992.309 2520.241 1823.461 515.381 299.221 13,758.821 100,678.021 479,276.521 2,081,666.069

17.298 117.298 317.298 692.298 1442.798

x N

7,723,149.529 3,455,250.411 2,390,557.371 783,379.120 269,298.900 10,415,427.500 24,263,403.060 48,886,205.140 70,776,646.350

x N x N

f 168, 963, 317.400

168, 963, 317.400 x 138.4 8,817

5. Indiana has the greatest standard deviation (157.8). 6. The number of manufacturing businesses decreases as the number of employees at the manufacturing businesses increases.

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CASE STUDY Ň

CHAPTER 3 CASE STUDY: UNITED STATES CONGRESS 1.

P female representative P female senator

118 x 0.273 432

24 0.24 100

2. The probabilities are close, but there is a slightly higher probability of selecting a female from the House of Representatives than from the Senate.

314 x 0.727 432 211 x 0.488 (b) P Republican representative 432 182 x 0.863 (c) P male Republican representative 211 89 x 0.206 (d) P female and Democrat 432

3. (a) P male representative

4. Dependent. The outcome of selecting a Democrat affects the outcome of selecting a female. 118 89 P female representative x 0.273 and P female | Democrat x 0.403 432 221 5. (a) P male senator

76 0.76 100 48 0.52 100 P Republican P female and Republican

(b) P senator is not a Democrat 1 P senator is a Democrat 1 (c) P female or Republican P female

24 100

(d) P male or Democrat P male

76 100

50 8 0.66 100 100 P Democrat P male and Democrat

48 32 0.92 100 100

6. Mutually exclusive. In the 117th Congress, a senator cannot be both female and Independent because there are no female Independent senators. 7. U.S. Congress

Gender

Male Female Total

Republican 224 37 261

Political Party Democrat Independent 164 2 105 0 269 2

Total 390 142 532

Ň CASE STUDY

2 x 0.004 532 37 x 0.070 (b) P(female and Republican) 532 (c) P male or Democrat P male P Democrat P male and Democrat

8. (a) P Independent

390 532

269 164 x 0.930 532 532

567

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CASE STUDY Ň

CHAPTER 4 CASE STUDY: DISTRIBUTION OF NUMBER OF HITS IN BASEBALL GAMES 1. (a) n = 3 x 0 1 2 3

P(x) 7/13 § 0.538 4/13 § 0.308 2/13 § 0.154 0/13 = 0.000

(b) n = 4 x 0 1 2 3 4

P(x) 5/29 § 0.172 17/29 § 0.586 2/29 § 0.069 3/29 § 0.103 2/29 § 0.069

P(x) 0/15 = 0.000 4/15 § 0.267 9/15 § 0.600 2/15 § 0.133 0/15 = 0.000 0/15 = 0.000

2. (a) n = 3, p = 0.317 P(x) x 0 0.319 1 0.444 2 0.206 3 0.032

(b) n = 4, p = 0.317 P(x) x 0 0.218 1 0.404 2 0.281 3 0.087 4 0.010

Ň CASE STUDY

569

(c) n = 5, p = 0.317 P(x) x 0 0.149 1 0.345 2 0.320 3 0.149 4 0.034 5 0.003 3. Answers will vary. The distributions in Exercises 1 and 2 are fairly similar. The distributions for part b are the most similar, because the number of games is the largest (29 versus 13 and 15). The binomial distribution should be a fairly good model for determining the number of hits in a baseball game. However, there are some problems with using a binomial distribution. For a binomial, the trials have to be independent. If a batter is facing the same pitcher in each at bat, the trials may not be independent. There are two possible outcomes for each at bat—either a hit or not. The probability of success has to be the same for each trial. This may not be true. A batter may hit better off a righthanded pitcher than off a left-handed pitcher. If there are a very large number of at bats, then the binomial distribution should be a fairly good model for determining the number of hits in a baseball game. 4. (a) Mookie Betts P(x) x 0 3/11 § 0.273 1 5/11 § 0.455 2 3/11 § 0.273 3 0/11 = 0.000 Answers will vary. Sample answer: Betts had the higher batting average.

(b) No, Answers will vary.

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CASE STUDY Ň

CHAPTER 5 CASE STUDY: BIRTH WEIGHTS IN AMERICA 1. (a) 39 to 40 weeks ( μ = 7.59 )

(b) 32 to 33 weeks ( μ = 4.62 ) (c) 34 to 36 weeks ( μ = 5.97 ) 2. (a) P ( x < 5.5) ≈ P ( z < 1.9901) ≈ 0.9767 → 97.67% (using technology)

(b) P ( x < 5.5) ≈ P ( z < 0.9263) ≈ 0.8229 → 82.29% (using technology) (c) P ( x < 5.5) ≈ P ( z < −2.4881) ≈ 0.0064 → 0.64% (using technology) (d) P ( x < 5.5) ≈ P ( z < −2.5446) ≈ 0.0055 → 0.55% (using technology) 3. (a) Top 10% 90th percentile z ≈ 1.2816 (using technology) x = μ + zσ = 1.90 + (1.2816)(0.72) ≈ 2.82 Birth weights must be at least 2.82 pounds to be in the top 10%.

(b) Top 10% 90th percentile z ≈ 1.2816 (using technology) x = μ + zσ = 5.97 + (1.2816)(1.02) ≈ 7.28 Birth weights must be at least 7.28 pounds to be in the top 10%. (c) Top 10% 90th percentile z ≈ 1.2816 (using technology) x = μ + zσ = 7.92 + (1.2816)(0.87) ≈ 9.03 Birth weights must be at least 9.03 pounds to be in the top 10%. (d) Top 10% 90th percentile z ≈ 1.2816 (using technology) x = μ + zσ = 8.07 + (1.2816)(1.01) ≈ 9.36 Birth weights must be at least 9.36 pounds to be in the top 10%. 4. (a) P (6 < x < 9) ≈ P (5.6944 < z < 9.8611) = 0 (using technology)

(b) P (6 < x < 9) ≈ P (1.4526 < z < 4.6105) ≈ 1 − 0.9268 = 0.0732 (using technology) (c) P (6 < x < 9) ≈ P (−1.0978 < z < 2.1630) ≈ 0.9847 − 0.1361 = 0.8486 (using technology) (d) P (6 < x < 9) ≈ P (−2.2069 < z < 1.2414) ≈ 0.8928 − 0.0137 = 0.8791 (using technology) 5. (a) P( x < 3.25) = P( z < 1.875) ≈ 0.9696 (using technology)

(b) P ( x < 3.25) ≈ P ( z < −0.2376) ≈ 0..4061 (using technology)

Ň CASE STUDY

(c) P ( x < 3.25) ≈ P ( z < −1.4421) ≈ 0.0746 (using technology) (d) P ( x < 3.25) ≈ P( z < −5.1667) ≈ 0.0000001 (using technology)

571

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CASE STUDY Ň

CHAPTER 6 CASE STUDY: MARATHON TRAINING 1. (a) x = 167.9 (b) x = 189.0 2. (a) s ≈ 8.36 (b) s ≈ 7.90 s

3. (a) x ± tc

≈ 167.9 ± 2.045

8.36 30

≈ 167.9 ± 3.12 ≈ (164.8, 171.0)

s 7.90 ≈ 189.0 ± 2.045 ≈ 189.0 ± 2.95 ≈ (186.1, 191.9) n 30

(b) x ± tc

4. (a) With 95% confidence, you can say the mean male training time is between 164.8 and 171.0 minutes. (b) With 95% confidence, you can say the mean female training time is between 186.1 and 191.9 minutes. 5.

x = 178.45

σ ≈ 13.35 x ± tc

s n

≈ 178.45 ± 2.000

13.35 60

≈ 178.45 ± 3.45 ≈ (175.0, 181.9)

With 95% confidence, you can say the mean training time for all runners is between 175.0 and 181.9 minutes. The confidence interval is in between the confidence intervals of the male and female runners. 2

§ z σ · § (2.575)(8.9) · 6. (a) n = ¨ c ¸ = ¨ ¸ ≈ 131.3 → 132 runners 2 ¹ © E ¹ © § z σ · § (2.575)(8.4) · (b) n = ¨ c ¸ = ¨ ¸ ≈ 117.0 → 117 runners 2 ¹ © E ¹ ©

Ň CASE STUDY

573

CHAPTER 7 CASE STUDY: HUMAN BODY TEMPERATURE: WHAT’S NORMAL? 1. (a) (see part d) (b) z0 = ±1.96 (see part d) (c) Rejection regions: z < −1.96 and z > 1.96 (see part d) (d) x ≈ 98.25, s ≈ 0.73 x − μ 98.25 − 98.6 −0.35 = = ≈ −5.467 z= 0.73 s 0.0640 n 130

(e) Reject H 0 . (f) There is sufficient evidence at the 5% level of significance to reject the claim that the mean body temperature of adult humans is 98.6° F. 2. No, α = 0.01 → z0 = ±2.575 → Reject H 0 3.

H 0 : μm = 98.6 and H a : μm ≠ 98.6 α = 0.01 → z0 = ±2.575 x ≈ 98.105, s ≈ 0.699

x − μ 98.105 − 98.6 −0.495 = = ≈ −5.709 s 0.699 0.0867 n 65 Reject H 0 . There is sufficient evidence to reject the claim. z=

H 0 : μw = 98.6 and H a : μw ≠ 98.6 α = 0.01 → z0 = ±2.575 s ≈ 0.743 x ≈ 98.394, x − μ 98.394 − 98.6 −0.206 z= = = ≈ −2.234 s 0.743 0.0922 n 65 Fail to reject H 0 . There is insufficient evidence to reject the claim.

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CASE STUDY Ň

x ≈ 98.25, s ≈ 0.73 s 0.73 x ± zc = 98.25 ± 2.575 ≈ 98.25 ± 0.165 ≈ (98.085, 98.415) n 130

6. Answers will vary. Sample answer: Wunderlich may have sampled more women than men, thus causing an overestimated body temperature.

Ň CASE STUDY

575

CHAPTER 8 CASE STUDY: HOW PROTEIN AFFECTS WEIGHT GAIN IN OVEREATERS 1. The claim is “the weight gains of the low protein group and the normal protein group are different.” H 0 : μ1 = μ 2 ; H a : μ1 ≠ μ 2 (claim) d.f. = n1 + n2 − 2 = 26 −t0 = −2.056, t0 = 2.056; Rejection regions: t < −2.056, t > 2.056 ( x1 − x2 ) − ( μ1 − μ2 ) (6.8 − 13.5) − (0) t= ≈ ≈ −7.98 2 2 ( n1 − 1) s1 + (n2 − 1) s2 1 1 (12 − 1)(1.7) 2 + (16 − 1)(2.5) 2 1 1 + + 12 + 16 − 2 12 16 n1 + n2 − 2 n1 n2

Because t < −2.056 , reject H 0 . There is enough evidence at the 5% level of significance to support the claim that the weight gains of the low protein group and the normal protein group are different. 2. The claim is “the weight gains of the low protein group and the high protein group are different.” H 0 : μ1 = μ3 ; H a : μ1 ≠ μ3 (claim) d.f. = n1 + n3 − 2 = 25 −t0 = −2.060, t0 = 2.060; Rejection regions: t < −2.060, t > 2.060 ( x1 − x3 ) − ( μ1 − μ3 ) (6.8 − 14.2) − (0) t= ≈ ≈ −9.88 (n1 − 1) s12 + (n3 − 1) s32 1 1 (12 − 1)(1.7) 2 + (15 − 1)(2.1) 2 1 1 + + 12 + 15 − 2 12 15 n1 + n3 − 2 n1 n3

Because t < −2.060 , reject H 0 . There is enough evidence at the 5% level of significance to support the claim that the weight gains of the low protein group and the high protein group are different. 3. The claim is “the weight gains of the normal protein group and the high protein group are different.” H 0 : μ 2 = μ3 ; H a : μ 2 ≠ μ3 (claim) d.f. = n2 + n3 − 2 = 29 −t0 = −2.045, t0 = 2.045; Rejection regions: t < −2.045, t > 2.045 ( x2 − x3 ) − ( μ2 − μ3 ) (13.5 − 14.2) − (0) t= ≈ ≈ −0.84 (n2 − 1) s22 + (n3 − 1) s32 1 (16 − 1)(2.5) 2 + (15 − 1)(2.1) 2 1 1 1 + + 16 + 15 − 2 16 15 n2 + n3 − 2 n2 n3

Because −2.045 < t < 2.045 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to support the claim that the weight gains of the normal protein group and the high protein group are different. 4. The experiment results support that there is a difference in weight gains for the low and normal protein groups and for the low and high protein groups, but the experiment results do not support that there is a difference in weight gains for the normal and high protein groups.

576

CASE STUDY Ň

5. σˆ =

(n1 − 1) s12 + (n2 − 1) s22 (16 − 1)(2.5) 2 + (15 − 1)(2.1) 2 = ≈ 2.316 n1 + n2 − 2 16 + 15 − 2

( x1 − x2 ) ± tcσˆ

1 1 1 1 + → (13.5 − 14.2) ± 2.045 ⋅ 2.316 + 16 15 n1 n2

→ −0.7 ± 1.702 → −2.4 < μ1 − μ 2 < 1.0

Ň CASE STUDY

CHAPTER 9 CASE STUDY: CORRELATION OF BODY MEASUREMENTS 1. Answers will vary. 2. a b c d e f g h i j k l

r 0.698 0.746 0.351 0.953 0.205 0.580 0.844 0.798 0.116 0.954 0.710 0.710

3. d, g, and j have strong correlations (r > 0.8). (d) y = 1.129 x − 8.817 (g) y = 1.717 x + 23.144 (j) y = 1.279 x + 10.451 4. (a) y = 0.086 (180 ) + 21.923 = 37.403 (b) y = 1.026 (100 ) − 15.323 = 87.277 5. (weight, chest) (weight, hip) (neck, wrist) (chest, hip) (chest, thigh) (chest, knee) (chest, ankle) (abdom, thigh) (abdom, knee) (hip, thigh) (hip, ankle) (thigh, knee) (thigh, ankle) (ankle, knee) (forearm, wrist)

r = 0.888 r = 0.930 r = 0.849 r = 0.953 r = 0.936 r = 0.855 r = 0.908 r = 0.863 r = 0.903 r = 0.894 r = 0.908 r = 0.954 r = 0.874 r = 0.857 r = 0.854

577

578

CASE STUDY Ň

CHAPTER 10 CASE STUDY: LIVING ON MARS 1.

Expected Frequencies: Response Yes, I would No, I would not Don’t know Total

18–34 60.191 105.566 28.243 194

35–54 42.506 74.549 19.945 137

Age 55+ 27.303 47.885 12.811 88

Total 130 228 61 419

The age group 18–34 exceeded the expected number of “Yes, I would” responses; the age groups 35– 54 and 55+ did not. 2. The age groups 35–54 and 55+ exceeded the expected number of “No, I would not” responses; the age group 18–34 did not. 3.

H 0 : Response is independent of age. H a : Response is dependent on age.

α = 0.01, d.f = (r − 1)((c − 1) = 4 χ 02 = 13.277 O

OíE

(O − E ) 2

87 33 10 80 78 70 27 26 8

60.191 42.506 27.303 105.566 74.549 47.885 28.243 19.945 12.811

26.809 í9.506 í17.303 í25.566 3.451 22.115 í1.243 6.055 í4.811

718.722481 90.364036 299.393809 653.620356 11.909401 489.073225 1.545049 36.663025 23.145721

(O − E ) 2 E 11.9406968 2.125912483 10.96560118 6.191580206 0.15975526593 10.21349535 0.0547055554 1.838206317 1.806706814 45.30

χ 2 ≈ 45.30 Reject H 0 . There is enough evidence at the 1% level of significance to support the claim that response and age are dependent.

H 0 : The distribution of responses by 18-to-34-year-olds is 25% “Yes I would,” 62% “No I would not,” and 13% “Don’t know.” H a : The distribution of responses of 18-to-34-year-olds differs from the claimed or expected distribution. α = 0.05, d.f = k − 1 = 2 χ 02 = 5.991

Ň CASE STUDY

Response

Distribution

Yes, I would No, I would not Don’t know Total

25% 62% 13%

O 87 80 27

OíE

(O − E ) 2

48.5 120.28 25.22

38.5 í40.28 1.78

1482.25 1622.4784 3.1684

579

(O − E ) 2 E 30.56185567 13.489178583 0.12563045 44.18

χ 2 ≈ 44.18 Reject H 0 . There is enough evidence at the 5% level of significance to conclude that the distribution of responses of 18-to-34-year-olds differs from the expected distribution.

H 0 : The distribution of responses by 35-to-54-year-olds is 25% “Yes I would,” 62% “No I would not,” and 13% “Don’t know.” H a : The distribution of responses of 35-to-54-year-olds differs from the claimed or expected distribution.

α = 0.05, d.f. = k − 1 = 2 χ 02 = 5.991 Response

OíE

(O − E ) 2

25%

34.25

í1.25

1.5625

62%

84.94

í6.94

48.1636

13%

17.81

8.19

67.0761

Distribution

Yes, I would No, I would not Don’t know Total

(O − E ) 2 E 0.045620 438 0.567030 8453 3.766204 38 4.38

χ 2 ≈ 4.38 Fail to reject H 0 . There is not enough evidence at the 5% level of significance to conclude that the distribution of responses of 35-to-54-year-olds differs from the expected distribution.

6. Answers will vary. Sample answer: Other variables that are important to consider when studying the distribution of U.S. adults’ feelings about living on Mars in a colony are gender, degree, and region.

580

CASE STUDY Ň

CHAPTER 11 CASE STUDY: COLLEGE RANKS 1.

The populations of students for the Midwest and West have similar ranges. The interquartile range (middle 50% of the data) for the Northeast, Midwest, and West are similar. The median populations for the Northeast, Midwest, and West are below 10,000. The median population for the South is far above the medians of the Northeast, Midwest, and West. 2. H0: median ≤ 7000 (claim) Ha: median > 7000 The critical value is 1 The test statistic, x = 4 Fail to reject H0 There is not enough evidence at the 5% level of significance to reject the claim that the median student population at a college in the Northeast is less than or equal to 7000. 3. H0: median ≤ 8000 Ha: median > 8000 (claim) The critical value is 1 The test statistic, x = 4 Fail to reject H0 There is not enough evidence at the 5% level of significance to support the claim that the median student population at a college in the Midwest is greater than 8000. 4. H0: median = 10,000 (claim) Ha: median ≠ 10,000 The critical value is 1 The test statistic, x = 3 Copyright © 2023 Pearson Education, Inc.

Ň CASE STUDY

581

Fail to reject H0 There is not enough evidence at the 5% level of significance to reject the claim that the median student population at a college in the South is 10,000. 5. H0: median = 8000 Ha: median ≠ 8000 (claim) The critical value is 1 The test statistic, x = 5 Fail to reject H0 There is not enough evidence at the 5% level of significance to support the claim that the median student population at a college in the West is different from 8000. 6. H0: There is no difference between student populations for colleges in the Midwest and West. (claim) Ha: There is a difference between student populations for colleges in the Midwest and West. Use Wilcoxon rank sum test. The critical values are ±2.575 . Ordered data Sample Rank 895 W 1 1109 W 2 1445 MW 3 1475 MW 4 1717 W 5 2098 MW 6 2214 W 7 3004 MW 8 3570 MW 9 3906 MW 10 5319 W 11 8847 W 12 14,297 MW 13 17,019 W 14 21,003 W 15 21,946 MW 16 22,517 W 17 31,240 MW 18 40,445 W 19 44,551 MW 20 R = sum ranks of Midwest = 107 (or R = 103) n (n + n + 1) 10(10 + 10 + 1) μR = 1 1 2 = = 105 2 2 n n (n + n + 1) (10)(10)(10 + 10 + 1) = ≈ 13.229 σR = 1 2 1 2 12 12

582

CASE STUDY Ň

R − μR

107 − 105 ≈ 0.151 (or z ≈ −0.151) 13.229 σR Fail to reject H0 There is not enough evidence at the 1% level of significance to reject the claim that there is no difference between student populations for colleges in the Midwest and West. z=

≈

7. H0: There is no difference between student populations for colleges in the Northeast and colleges in the South. Ha: There is a difference between student populations for colleges in the Northeast and colleges in the South. (claim) Use Wilcoxon rank sum test. The critical values are ±2.575 . Ordered data Sample Rank 1835 NE 1 2134 NE 2 2264 S 3 2519 NE 4 2631 NE 5 2692 S 6 3697 NE 7 4524 S 8 6608 NE 9 14,171 NE 10 14,318 S 11 17,811 S 12 19,593 S 13 22,207 NE 14 24,027 NE 15 28,079 S 16 33,720 NE 17 36,383 S 18 46,148 S 19 69,525 S 20 R = sum ranks of Northeast = 84 (or R = 126) n (n + n + 1) 10(10 + 10 + 1) μR = 1 1 2 = = 105 2 2 n n (n + n + 1) (10)(10)(10 + 10 + 1) = ≈ 13.229 σR = 1 2 1 2 12 12 R − μ R 84 − 105 z= ≈ ≈ −1.587 (or z ≈ 1.587) 13.229 σR Fail to reject H0 There is not enough evidence at the 1% level of significance to support the claim that there is a difference between student populations for colleges in the Northeast and colleges in the South. Copyright © 2023 Pearson Education, Inc.

CHAPTER 6 ALTERNATE Ň CASE STUDY

583

CHAPTER 6 ALTERNATE CASE STUDY: MARATHON TRAINING 1. (a) x = 167.9 (b) x = 189.0 2. (a) s ≈ 8.36 (b) s ≈ 7.90 3. (a) x ± tc

(b) x ± tc

s n s n

≈ 167.9 ± 2.045

≈ 189.0 ± 2.045

8.36 30 7.90 30

≈ 167.9 ± 3.12 ≈ (164.8, 171.0)

≈ 189.0 ± 2.95 ≈ (186.1, 191.9)

4. (a) With 95% confidence, you can say the mean male training time is between 164.8 and 171.0 minutes.

(b) With 95% confidence, you can say the mean female training time is between 186.1 and 191.9 minutes. 5.

x = 178.45 σ ≈ 13.35 s 13.35 x ± tc ≈ 178.45 ± 2.000 ≈ 178.45 ± 3.45 ≈ (175.0, 181.9) n 60

§ z σ · § (2.575)(8.9) · 6. (a) n = ¨ c ¸ = ¨ ¸ ≈ 131.3 → 132 runners 2 ¹ © E ¹ © § z σ · § (2.575)(8.4) · (b) n = ¨ c ¸ = ¨ ¸ ≈ 117.0 → 117 runners 2 ¹ © E ¹ ©

584

USES AND ABUSES Ň

CHAPTER 1 USES AND ABUSES 1. Answers will vary. 2. Answers will vary.

Ň USES AND ABUSES

CHAPTER 2 USES AND ABUSES 1. Answers will vary. 2. No, it is not ethical because it misleads the consumer to believe that drinking red wine is more effective at preventing heart disease than it may actually be.

585

586

USES AND ABUSES Ň

CHAPTER 3 USES AND ABUSES 1. P(winning Tuesday and Wednesday) = P(winning Tuesday) · P(winning Wednesday) 1 ¬ 1 ¬ ¸ 0.000001 1000 ® 1000 ® 2. P(winning Wednesday given won Tuesday) = P(winning Wednesday) 1 0.001 1000 3. P(winning Wednesday given didn’t win Tuesday) = P(winning Wednesday) 1 0.001 1000

Ň USES AND ABUSES

587

CHAPTER 4 USES AND ABUSES 1. 11 incidents; The probability of 11 incidents is about 0.119, whereas there is about a 0.093 chance of at least 16 incidents. 2. 10 to 12 incidents; The probability of 10 to 12 incidents is about 0.348, whereas there is about a 0.341 chance of fewer than 10 incidents. 3. Yes. The probability of 17 incidents is about 0.024, which is less than 0.05.

588

USES AND ABUSES Ň

CHAPTER 5 USES AND ABUSES 1.

μ = 100, σ = 15 (a) z =

x−μ

112 − 100 ≈ 1.39 15 3

x = 112 is not an unusual sample mean because it is less than 2 standard deviations from the population mean.

(b) z =

x−μ

105 − 100 ≈ 2.89 15 75

x = 105 is an unusual sample mean because it is more than 2 standard deviations from the population mean.

2. The ages of students at a high school may not be normally distributed. 3. Answers will vary.

Ň USES AND ABUSES

CHAPTER 6 USES AND ABUSES 1. Answers will vary 2. Answers will vary. 3. (a) No. Trump: 0.3% + 4.0% = 4.3% , which is more than Clinton’s 3.4% (b) Yes. Trump: 0.7% + 4.0% = 4.7% , which is less than Clinton’s 6.5%

589

590

USES AND ABUSES Ň

CHAPTER 7 USES AND ABUSES 1.

H 0 : p 0.50

Answers will vary. We cannot prove that p 0.50 . We can only show that there is not enough evidence in our sample to reject H 0 : p 0.50 . 2. Answers will vary. If we gather enough evidence to reject the null hypothesis when it is really true, then a Type I error would occur. 3. Answers will vary. If we fail to gather enough evidence to reject the null hypothesis when it is really false, then a Type II error would occur. Pearson correlation coefficient

Ň USES AND ABUSES

591

CHAPTER 8 USES AND ABUSES 1. Age and health. (Answers will vary.) 2. Blind: The patients do not know which group (medicine or placebo) they belong to. Double Blind: Both the researcher and patient do not know which group (medicine or placebo) that the patient belongs to.

592

USES AND ABUSES Ň

CHAPTER 9 USES AND ABUSES 1. Answers will vary. 2. Answers will vary. Sample answer: One example would be temperature and output of sulfuric acid. When sulfuric acid is manufactured, the amount of acid that is produced depends on the temperature at which the industrial process is run. As the temperature increases, so does the output of acid, up to a point. Once the temperature passes that point, the output begins to decrease.

Ň USES AND ABUSES

593

CHAPTER 10 USES AND ABUSES 1. Answers will vary. Sample answer: An example would be a biologist who is analyzing effectiveness of three types of pesticides. The biologist sprays one pesticide on five different acres during week one. At the end of week one, he calculates the mean number of insects in the acres. He sprays the second pesticide on the same five acres during week two. At the end of week two, he calculates the mean number of insects in the acres. He follows a similar procedure with the third pesticide during week three. He wants to determine whether there is a difference in the mean number of insects per acre. Since the same five acres are treated each time, it is unclear as to the effectiveness of each of the pesticides. 2. Answers will vary. Sample answer: An example would be another biologist who is analyzing the effectiveness of three types of pesticides. He has fifteen acres to test. He randomly assigns each pesticide to five acres and treats all of the acres for three weeks. At the end of three weeks, the biologist calculates the mean number of insects for each of the pesticides. He wants to determine whether there is a difference in the mean number of insects per acres. Rejecting the null hypothesis would mean that at least one of the means differs from the others. In this example, rejecting the null hypothesis means that there is at least one pesticide whose mean number of insects per acre differs from the other pesticides.

594

USES AND ABUSES Ň

CHAPTER 11 USES AND ABUSES 1. Answers will vary. Sample answer: H 0 : median ≥ 10 H a : median < 10 Using the sign test with n = 8 and α = 0.05 , the critical value is 0. Thus, every item in the sample would have to be less than 10 in order to reject the H 0 . (In other words, 100% if the sample would need to be less than 10.) However, using n = 20 and α = 0.05, the critical value is 5. Not only 5 items (or 75%) of the sample would need to be less than 10 in order to reject H 0 . 2. Nonparametric test

Parametric test

(a) Sign test

z-test or t-test

(b) Paired sample sign test

t-test

paired t-test

(d) Wilcoxon rank sum test

two-sample t-test or z-test

(e) Kruskal-Wallis test

one-way ANOVA

(f) Spearman rank correlation coefficient

Pearson correlation coefficient

CHAPTER 6 ALTERNATE Ň USES AND ABUSES

CHAPTER 6 ALTERNATE USES AND ABUSES 1. Answers will vary 2. Answers will vary. 3. (a) No. Trump: 0.3% + 4.0% = 4.3% , which is more than Clinton’s 3.4% (b) Yes. Trump: 0.7% + 4.0% = 4.7% , which is less than Clinton’s 6.5%

595

596

REAL STATISTICS–REAL DECISIONS Ň

CHAPTER 1 REAL STATISTICS—REAL DECISIONS 1. (a)–(b) Answers will vary. (c) Sample answer: Use surveys. (d) Sample answer: You may take too large a percentage of your sample from a subgroup of the population that is relatively small. 2. (a) Sample answer: Qualitative, because questions will ask for demographics and the sample questions have nonnumerical categories. (b) Sample answer: Nominal and ordinal, because the results can be put in categories and the categories can be ranked. (c) Sample (d) Statistics 3. (a) Sample answer: Sample includes only members of the population with access to the Internet. (b) Answers will vary.

Ň REAL STATISTICS–REAL DECISIONS 597

CHAPTER 2 REAL STATISTICS–REAL DECISIONS 1. (a) Find the average cost of renting an apartment for each area and do a comparison. (b) The mean would best represent the data sets for the four areas of the city. (c) Area A: x $1131.58 Area B: x $998.33 Area C: x $991.58 Area D: x $1064.17 2. (a) Construct a Pareto chart, because the data are quantitative and a Pareto chart positions data in order of decreasing height, with the tallest bar positioned at the left. (b)

(c) Yes. From the Pareto chart you can see that Area A has the highest average cost of monthly rent, followed by Area D, Area B, and Area C. 3. Sample answer (a) You could use the range and sample standard deviation for each area. (b) Area A: s x $138.45 Range = $467 Area B: s x $163.11 Range = $474 Area C: s x $164.51 Range = $518 Area D: s x $156.26 Range = $560 (c) No. Area A has the lowest range and standard deviation, so the rents in Areas B–D are more spread out. There could be one or two inexpensive rents that lower the means for these areas. It is possible that the population means of Areas B–D are close to the populations mean of Area A. 4. (a) Answers will vary. (b) Location, weather, population

598

REAL STATISTICS–REAL DECISIONS Ň

CHAPTER 3 REAL STATISTICS—REAL DECISIONS 1. (a) Sample answer: Investigate the number of possible passwords when different sets of characters, such as lowercase and capital letters, numbers, and special characters, are used. (b) You could use the definition of theoretical probability, the Fundamental Counting Principal, and the Multiplication Rule. 2. (a) Sample answer: Allow lowercase letters, uppercase letters, and numerical digits. (b) Sample answer: Because there are 26 lowercase letters, 26 uppercase letters, and 10 numerical digits, there are 26 26 10 62 choices for each digit. So, there are 6 2 8 eight-digit 1 passwords and the probability of guessing a password correctly on one try is 8 , which is less 62 1 than 8 . 60 3. (a) Without the requirement, the number of possible PINs is 10 5 100, 000 . With the requirement, the possible number of PINs is (10)(9)(8)(7)(6) 30, 240 . (b) Sample answer: No, although the requirement would likely discourage customers from choosing predictable PINs, the numbers of possible PINs would significantly decrease, and the most popular PIN, 12345, would still be allowed.

Ň REAL STATISTICS–REAL DECISIONS 599

CHAPTER 4 REAL STATISTICS—REAL DECISIONS 1. (a) Answers will vary. For instance, calculate the probability of obtaining zero live births out of six randomly selected ART cycles. (b) Binomial. The distribution is discrete because the number of live births is countable. 2.

n = 6, p = 0.52 x P(x) 0 0.01223 1 0.07950 2 0.21531 3 0.31100 4 0.25269 5 0.10950 6 0.01977 Sample answer: Because P ( 0 ) ≈ 0.012 , this event is unusual but not impossible.

3. (a) Not suspicious, because the probability is greater than 0.05. (b) Suspicious, because the probability is less than 0.05.

600

REAL STATISTICS–REAL DECISIONS Ň

CHAPTER 5 REAL STATISTICS–REAL DECISIONS x−μ

9.65 − 9.96 = −6.2 σ 0.05 x − μ 9.95 − 9.96 = = −0.2 z= σ 0.05 P (9.65 < x < 9.95) = P( −6.2 < z < −0.2) ≈ 0.42074 − 0.0000 ≈ 0.4207 (using technology)

1. (a) z =

(b) n = 15, p = 0.4207, q = 0.5793 np = 15(0.4207) = 6.3105 ≥ 5, nq = 15(0.5793) = 8.6895 ≥ 5 Can use normal distribution μ = np = 15(0.4207) = 6.3105, σ = npq = 15(0.4207)(0.5793) ≈ 1.9120

x−μ

0.5 − 6.3105 = −3.0390 σ 1.9120 P( x ≥ 1) ≈ P( x > 0.5) = 1 − P( x < 0.5) ≈ 1 − P( z < −3.0390) ≈ 1 − 0.0012 ≈ 0.9988 (using technology) z=

x −μ

9.65 − 9.96 ≈ −13.8636 σ 0.05 n 5 x − μ 9.95 − 9.96 = ≈ −0.4472 z= 0.05 σ n 5 P(9.65 < x < 9.65) = P(−13.8636 < z < −0.4472) ≈ 0.3274 − 0.0000 ≈ 0.3274 (using technology)

2. (a) z =

(b) n = 3, p = 0.3274, q = 0.6726

P( x ≥ 1) = 1 − P( x = 0) = 1 −3 C0 (0.3274)0 (0.6726)3 = 1 − 0.3043 = 0.6957 (c) The mean is more sensitive. 3. Answers will vary.

Ň REAL STATISTICS–REAL DECISIONS 601

CHAPTER 6 REAL STATISTICS–REAL DECISIONS 1. (a) Yes, there has been a change in the mean concentration level because the confidence interval for Year 1 does not overlap the confidence interval for Year 2. (b) No, there has not been a change in the mean concentration level because the confidence interval for Year 2 overlaps the confidence interval for Year 3. (c) Yes, there has been a change in the mean concentration level because the confidence interval for Year 1 does not overlap the confidence interval for Year 3. 2. Due to the fact the CIs from Years 2 and 3 do not overlap the CI from Year 1, it is very likely that concentrations of cyanide in the drinking water have increased over the three-year period. 3. The width of the confidence interval for Year 2 may have been caused by greater variation in the levels of cyanide than in other years, which may be the result of outliers. 4. The water department can increase the sample size to decrease the size of the confidence intervals. 5. Answers will vary. (a) Sample answer: The sampling distribution of the sample means was used because the “mean concentration” was used. The sample mean is the most unbiased point estimate of the population mean. (b) Sample answer: No, because typically σ is unknown. They could have used the sample standard deviation.

602

REAL STATISTICS–REAL DECISIONS Ň

CHAPTER 7 REAL STATISTICS–REAL DECISIONS 1. Answers will vary. 2. (a) The sample is not random. Sample answer: Randomly select students from the student directory. (b) The sample is random. (c) The sample is random. 3. The alternative hypothesis should contain the claim because you cannot use a hypothesis test to support your claim if the null hypothesis contains the claim. 4.

H 0 : N p 25, 000; H a : N 25, 000 (claim) B 0.05 and d.f. n 1 39 t 0 1.685 ; Rejection region: t 1.685 x N 23, 475 25,000 t x 1.21 s 8000

n 40 Because t 1.685 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to support the claim.

H 0 : p p 0.80 (claim); H a : p 0.80

lp x 54 0.9 n 60 z0 1.645; Rejection region: z 1.645 lp p 0.9 0.8 z 1.94 pq 0.8 0.2 n 60 Because z 1.645 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the claim. 6. Answers will vary.

Ň REAL STATISTICS–REAL DECISIONS 603

CHAPTER 8 REAL STATISTICS–REAL DECISIONS 1. (a) Answers will vary. Sample answer: Divide the records into groups according to the inpatients’ ages, and then randomly select records from each group. (b) Answers will vary. Sample answers: Divide the records into groups according to geographic regions, and then randomly select several groups and examine all the records in the selected groups. (c) Answer will vary. Sample answer: Assign a different number to each record, randomly choose a starting number, and then select every 50th record. (d) Answers will vary. Sample answer: Assign a different number to each record and then use a table of random numbers to generate a sample of numbers. 2. (a) Answers will vary. (b) Answers will vary. 3. Use a t-test. The samples are independent because both samples are randomly selected. Yes, you need to know if the population distributions are normal so you can use a t-test. Yes, you need to know if the population variances are equal or not. 4. The claim is “there is a difference in the mean length of hospital stays for inpatients.” H 0 : μ1 = μ 2 ; H a : μ1 ≠ μ 2 (claim) d.f. = n1 + n2 − 2 = 56 −t0 = −2.003, t0 = 2.003; Rejection regions: t < −2.003, t > 2.003 ( x1 − x2 ) − ( μ1 − μ2 ) (6 − 4.97) − (0) t= ≈ ≈ 2.300 (n1 − 1) s12 + ( n2 − 1) s22 1 1 (28 − 1)(1.63) 2 + (30 − 1)(1.77) 2 1 1 + + 28 + 30 − 2 28 30 n1 + n2 − 2 n1 n2

Because t > 2.003 , reject H 0 . There is enough evidence at the 5% level of significance to support the claim that there is a difference in the mean length of hospital stays for inpatients. This decision does support the claim.

604

REAL STATISTICS–REAL DECISIONS Ň

CHAPTER 9 REAL STATISTICS–REAL DECISIONS 1. (a)

It appears there is a positive linear correlation. As the annual average of the daily maximum sulfur dioxide concentration increases, the annual average of the daily maximum nitrogen dioxide concentration tends to increase. (b) r ≈ 0.980; There is a strong positive linear correlation. (c) Critical value = 0.482 r ≈ 0.980 > 0.482 There is enough evidence at the 5% level of significance to conclude that there is a significant linear correlation between annual averages of the daily maximum concentrations of sulfur dioxide and nitrogen dioxide. (d) y = 0.242 x + 39.403

The regression line appears to fit the data. (e) Yes; Since the regression line fits the data and the data show a very strong correlation ( r ≈ 0.980 ), the regression equation may be used to make predictions about the y values, for the given x values. (f) r 2 ≈ 0.961 . About 96.1% of the variation in the average nitrogen dioxide concentration can be explained by the variation in sulfur dioxide concentration and about 3.9% of the variation is unexplained. se ≈ 1.106 ; The standard error of estimate of the annual averages of the daily maximum concentration of nitrogen dioxide for a specific annual average of the daily maximum sulfur dioxide concentration is about 1.106 parts per billion. Copyright © 2023 Pearson Education, Inc.

Ň REAL STATISTICS–REAL DECISIONS 605

43.742 < y < 48.616 You can be 95% confident that the annual average of the daily maximum nitrogen dioxide concentration will be between 43.742 and 48.616 parts per billion when the annual average of the daily maximum sulfur dioxide concentration is 28 parts per billion.

606

REAL STATISTICS–REAL DECISIONS Ň

CHAPTER 10 REAL STATISTICS–REAL DECISIONS 1.

H 0 : The distribution of ages of fraud victims is 0.71% ages under 18, 13.39% ages 18-25, 15.54% ages 26-35, 16.38% ages 36-45, 14.00% ages 46-55, 15.94% ages 56-65, and 24.05% ages 65+. (claim) H a : The distribution of the ages differs from the claimed or expected distribution.

χ 02 = 16.812 , Rejection region: χ 2 > 16.812 Ages

Distribution

Observed

Expected

Under 18 18-25 26-35 36-45 46-55 56-65 65+

0.71% 13.39% 15.54% 16.38% 14.00% 15.94% 24.05%

8 148 166 185 131 153 209

7.1 133.9 155.4 163.8 140 159.4 240.5

(O − E )2 E 0.114084507 1.48476475 0.723037323 2.743833944 0.5785714286 0.2569636136 4.125779626 10.03

χ 2 ≈ 10.03 Because χ 2 < 16.812 , fail to reject H 0 . There is not enough evidence at the 1% level of significance to conclude that the distribution of responses differs from the expected distribution.

2. (a) Expected frequencies: Age Type of Fraud Sweepstakes Credit cards Total

Under 20 15 15 30

20-29 120 120 240

30-39 165 165 330

40-49 185 185 370

50-59 135 135 270

60-69 115 115 230

(b) H 0 : Age and type of fraud are independent. H a : Age and type of fraud are dependent. (claim)

d.f. = (r − 1)(c − 1) = 7 ; χ 02 = 18475 ; Rejection region: χ 2 > 18.475

70-79 155 155 310

80+ 110 110 220

Total 1000 1000 2000

Ň REAL STATISTICS–REAL DECISIONS 607

OíE

(O − E )2

10 60 70 130 90 160 280 200 20 180 260 240 180 70 30 20

15 120 165 185 135 115 155 110 15 120 165 185 135 115 155 110

5 60 95 55 45 −45 −125 −90 −5 −60 −95 −55 −45 45 125 90

25 3600 9025 3025 2025 2025 15,625 8100 25 3600 9025 3025 2025 2025 15,625 8100

(O − E )2 E 1.6667 30.0000 54.6970 16.3514 15.0000 17.6087 100.8065 73.6364 1.6667 30.0000 54.6970 16.3514 15.0000 17.6087 100.8065 73.6364 619.533

χ 2 ≈ 619.533 Because χ 2 > 18.475 , reject H 0 . There is enough evidence at the 1% level of significance to conclude that the ages of the victims are related to the type of fraud.

608

REAL STATISTICS–REAL DECISIONS Ň

CHAPTER 11 REAL STATISTICS–REAL DECISIONS 1. (a) Answers will vary. (b) Answers will vary. (c) Answers will vary. 2. (a) Answers will vary. For example, ask “Are the data a random sample?” and “Are the data normally distributed?” (b) Sign test; You need to use the nonparametric test because nothing is known about the shape of the population. (c) H0: median ≥ 4.1 Ha: median < 4.1 (claim) (d) α = 0.05 n = 20 The critical value is 5. x=7 Because x > 5 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to support the claim that the median tenure for employees from the representative’s district is less than 4.1 years. 3. (a) Wilcoxon rank sum test; You need to use the nonparametric test because nothing is known about the shape of the population. (b) H 0 : There is no difference between the median tenures for male employees and female employees. H a : There is a difference between the median tenures for male employees and female employees.(claim) (c)

Ordered data 0.9 1.3 1.8 2.2 2.3 3 3.3 3.6 3.8 3.9 4 4 4.3

Sample Rank M 1 F 2 F 3 F 4 M 5 F 6 F 7 M 8 M 9 M 10 F 11.5 F 11.5 M 13.5

Ordered data 4.3 4.4 4.4 4.7 4.7 4.9 4.9 5.1 5.1 5.1 5.2 5.4 6.5

Sample Rank F 13.5 M 15.5 F 15.5 M 17.5 M 17.5 M 19.5 F 19.5 M 22 F 22 F 22 F 24 F 25 M 26

Ň REAL STATISTICS–REAL DECISIONS 609

R = sum ranks of male workers = 164.5 n (n + n + 1) 12(12 + 14 + 1) μR = 1 1 2 = = 162 2 2 n n (n + n + 1) (12)(14)(12 + 14 + 1) = ≈ 19.442 σR = 1 2 1 2 12 12 R − μ R 164.5 − 162 z= = ≈ 0.129 19.442 σR Using α = 0.05 , z0 = ±1.96. Because z < 1.96 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to support the claim that there is a difference between the median tenures for male employees and female employees.

610

CHAPTER 6 ALTERNATE Ň REAL STATISTICS–REAL DECISIONS

CHAPTER 6 ALTERNATE REAL STATISTICS–REAL DECISIONS 1. (a) Yes, there has been a change in the mean concentration level because the confidence interval for Year 1 does not overlap the confidence interval for Year 2. (b) No, there has not been a change in the mean concentration level because the confidence interval for Year 2 overlaps the confidence interval for Year 3. (c) Yes, there has been a change in the mean concentration level because the confidence interval for Year 1 does not overlap the confidence interval for Year 3. 2. Due to the fact the CIs from Years 2 and 3 do not overlap the CI from Year 1, it is very likely that concentrations of cyanide in the drinking water have increased over the three-year period. 3. The width of the confidence interval for Year 2 may have been caused by greater variation in the levels of cyanide than in other years, which may be the result of outliers. 4. The water department can increase the sample size to decrease the size of the confidence intervals. 5. Answers will vary. (a) Sample answer: The sampling distribution of the sample means was used because the “mean concentration” was used. The sample mean is the most unbiased point estimate of the population mean. (b) Sample answer: No, because typically σ is unknown. They could have used the sample standard deviation.

Ň TECHNOLOGY

611

CHAPTER 1 TECHNOLOGY: USING TECHNOLOGY IN STATISTICS 1. Use technology or a table to generate a list of 10 random numbers from 1 to 86. Answers will vary. 2. Use technology or a table to generate a list of 25 random numbers from 1 to 300. Answers will vary. 3. Answers will vary. 4.

2 " 38 41 Answers will vary. 0 1

820 20 . The average of the whole numbers from 0 to 40 is 20. 41

5. Answers will vary. 6. No, you would expect to get 10 of each number. An inference that you might draw from the results is that the die is not a fair die. 7. Answers will vary. 8. No, you would expect 50 heads and 50 tails. An inference that we might draw from the results is that the coin is not fair. 9. The analyst could survey all the registered voters from 10 counties by assigning each county a number from 1 to 47 and using a random number generator to find 10 numbers that correspond to certain counties.

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TECHNOLOGY Ň

CHAPTER 2 TECHNOLOGY: PARKING TICKETS 1.

x x $52.76

s x $37.35

3. Min = 30, Q1 30 , Q2 40 , Q3 50 , Max = 250 4.

Class

Midpoint

30–44 45–59 60–74 75–89 90–104 105–119 120–134 135–149 150–164 165–179 180–194 195–209 210–224 225–239 240–254

37 52 67 82 97 112 127 142 157 172 187 202 217 232 247

Class boundaries 29.5–44.5 44.5–59.5 59.5–74.5 74.5–89.5 89.5–104.5 104.5–119.5 119.5–134.5 134.5–149.5 149.5–164.5 164.5–179.5 179.5–194.5 194.5–209.5 209.5–224.5 224.5–239.5 239.5–254.5

Frequency, f 64 20 1 0 14 0 0 0 5 0 0 0 0 0 1

f 105 5.

Relative frequency 0.610 0.190 0.010 0.000 0.133 0.000 0.000 0.000 0.048 0.000 0.000 0.000 0.000 0.000 0.010 f n x1

Cumulative frequency 64 84 85 85 99 99 99 99 104 104 104 104 104 104 105

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613

x s 52.76 37.35 15.41 ; x s 52.76 37.35 90.11 94.3% of the entries were within one standard deviation of the mean. x 2s 52.76 2(37.35) 21.94 ; x 2 s 52.76 2(37.35) 127.46 94.3% of the entries were within two standard deviation of the mean. x 3s 52.76 3(37.35) 59.29 ; x 3s 52.76 3(37.35) 164.81 99.1% of the entries were within three standard deviations of the mean. 7. No. The percentage within one standard deviation of the mean is much greater than the percentage for the Empirical Rule of 68%. The percentage within two standard deviations of the mean is smaller than the percentage for the Empirical Rule of 95%. The percentage within three standard deviations of the mean is smaller than the percentage for the Empirical Rule of 99.7%.

8. Yes. At least

3 0.75 75% lie within two standard deviations of the mean and at least 4

8 x 0.889 x 89% lie within three standard deviations of the mean. 9 9.

Midpoint, x 37 52 67 82 97 112 127 142 157 172 187 202 217 232 247

Frequency, f 64 20 1 0 14 0 0 0 5 0 0 0 0 0 1 n 105

xf 5865 x $55.86 105 n

xf 2368 1040 67 0 1358 0 0 0 785 0 0 0 0 0 247 xf 5865

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TECHNOLOGY Ň 2

x x

−18.86 −3.86 11.14 26.14 41.14 56.14 71.14 86.14 101.14 116.14 131.14 146.14 161.14 176.14 191.14

355.6996 14.8996 124.0996 683.2996 1692.4996 3151.6996 5060.8996 7420.0996 10,229.2996 13,488.4996 17,197.6996 21,356.8996 25,966.0996 31,025.2996 36,534.4996

x x

22,764.7744 297.9920 124.0996 0.0 23,694.9944 0.0 0.0 0.0 51,146.4980 0.0 0.0 0.0 0.0 0.0 36,534.4996

x x x x

f 134, 562.8580

134, 562.8580 x $35.97 n 1 104 No, the formulas for grouped data do not give results that are as accurate as the individual entry formulas. s

10. Answers will vary.

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615

CHAPTER 3 TECHNOLOGY: SIMULATION: COMPOSING MOZART VARIATIONS WITH DICE 11 13 phrases were written.

(11)7 ¸ 2 ¸ (11)7 ¸ 2 1.518999334333q1015

3. (a)

1 x 0.091 11

(b) Results will vary.

3 x 0.273 11 14 3¬ P option 6, 7, or 8 for all 14 bars x 0.0000000126 11®

4. (a) P option 6, 7, or 8 for first bar

(b) Results will vary.

1 36 5 P(5) 36 3 P(9) 36

5. (a) P (1)

2 36 6 P(6) 36 2 P (10) 36 P(2)

3 36 5 P(7) 36 1 P (11) 36

P(3)

(b) Results will vary.

6 5 4 15 x 0.417 36 36 36 36 14 15 ¬ P option 6, 7, or 8 for all 14 bars x 0.00000475 36 ®

6. (a) P option 6, 7, or 8 for first bar

(b) Results will vary.

4 36 4 P(8) 36 P(4)

616 TECHNOLOGY Ň

CHAPTER 4 TECHNOLOGY: USING POISSON DISTRIBUTIONS AS QUEUING MODELS 1.

μ =4 P(x) 0.0183 0.0733 0.1465 0.1954 0.1954 0.1563 0.1042 0.0595 0.0298 0.0132 0.0053

x 0 1 2 3 4 5 6 7 8 9 10

x 11 12 13 14 15 16 17 18 19 20

P(x) 0.0019 0.0006 0.0002 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

The results look very similar to the graph. 2. (a) (See part b) 1, 2, 4, 7 (b)

Minute 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Customers Arriving at Checkout 3 3 3 3 5 5 6 7 3 6 3 5 6 3 4 6 2 2 4 1

3. Answers will vary. 4. Answers will vary.

Total Customers at Checkout 3 3 3 3 5 6 8 11 10 12 11 12 14 13 13 15 13 11 11 8

Customers Serviced 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4

Customers Remaining 0 0 0 0 1 2 4 7 6 8 7 8 10 9 9 11 9 7 7 4

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P (10) =

617

510 (2.71828) −5 ≈ 0.018 10!

6. (a) It makes no difference if the customers were arriving during the 1st minute or the 3rd minute. The mean number of arrivals will still be 4 customers per minute. P (3 ≤ x ≤ 5) = P (3) + P (4) + P (5) ≈ 0.1954 + 0.1954 + 0.1563 ≈ 0.5471

(b) P ( x > 4) = 1 − P ( x ≤ 4) = 1 − ( P(0) + P (1) + P (2) + P(3) + P(4) )

≈ 1 − ( 0.0183 + 0.0733 + 0.1465 + 0.1954 + 0.1954 ) ≈ 1 − 0.6289 ≈ 0.3711

(c) P ( x > 4 during each of the first four minutes ) ≈ (0.3711) 4 ≈ 0.019. 7. (a) Using the table in Problem 1: P ( no customers are waiting in line after one minute ) ≈ 0.0183

(b) P ( one customer is waiting in line after one minute ) ≈ 0.0733 § one customer is waiting in line after one minute · (c) P ¨ ¸ ≈ (0.0733)(0.0183) ≈ 0.0013 and no customers are waiting in line after the second mi n ut e © ¹

(d) P ( no customers are waiting in line after two minutes ) ≈ (0.0183) 2 ≈ 0.0003

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TECHNOLOGY Ň

CHAPTER 5 TECHNOLOGY: AGE DISTRIBUTION IN CALIFORNIA 1.

μ ≈ 38.09

x of the 36 sample means ≈ 38.03 This mean is close to the mean age in California found in Exercise 1. This agrees with the Central Limit Theorem.

3. No, the distribution of ages appears to be positively skewed. 4.

The distribution is approximately bell-shaped and symmetrical. This agrees with the Central Limit Theorem. 5. σ ≈ 22.807 6. σ of the 36 sample means § 2.9 The standard deviation of the set of 36 sample means is less than that of the standard deviation of the ages of people in California. σ 22.807 = ≈ 3.8 n 36 This agrees with the Central Limit Theorem.

Ň TECHNOLOGY

CHAPTER 6 TECHNOLOGY: UNITED STATES FOREIGN POLICY POLLS 1. (a) pˆ = 0.63; qˆ = 0.37; z = 1.96; n = 1021; (0.600, 0.659) (b) pˆ = 0.58; qˆ = 0.42; z = 1.96; n = 1021; (0.550, 0.610) (c) pˆ = 0.93; qˆ = 0.07; z = 1.96; n = 1021; (0.915, 0.946) (d) pˆ = 0.37; qˆ = 0.63; z = 1.96; n = 1021; (0.341, 0.400) (e) pˆ = 0.17; qˆ = 0.83; z = 1.96; n = 1021; (0.147, 0.193) 2

3. (a) Answers will vary.

(b) Answers will vary. 4. Answers will vary.

619

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TECHNOLOGY Ň

CHAPTER 7 TECHNOLOGY: THE CASE OF THE VANISHING WOMEN 1. Reject H 0 with a P-value less than 0.01. The level of significance is greater than 0.000. 2. Type I error 3. Sampling process was non-random. 4. (a) H 0 : p 0.291 (claim); H a : p v 0.291 (b) Test and CI for One Proportion Test of p = 0.291 vs p not = 0.291 Sample 1

X 9

N 100

Sample p 0.090000

99% CI (0.016284, 0.163716)

Z-Value -4.43

P-Value 0.000

Using the normal approximation.

(c) Reject H 0 . (d) It was highly unlikely that random selection produced a sample of size 100 that contained only 9 women.

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621

CHAPTER 8 TECHNOLOGY: TAILS OVER HEADS 1. Test and CI for One Proportion Test of p = 0.5 vs p not = 0.5 Sample 1

X 5772

N 11902

Sample p 0.484961

99% CI (0.473161, 0.496760)

Z-Value -3.28

P-Value 0.001

Using the normal approximation.

Because p-value < α = 0.01 , reject H 0 . There is enough evidence at the 1% level of significance to reject the hypothesis that the probability that a “found coin” will be lying heads up is 0.5. 2. (Answers will vary). Sample answer: Yes, obtaining 5772 heads is a very uncommon occurrence if the probability of lying head up is 0.5 (see sampling distribution). The coins might not be fair. 3. 4.

1 = 0.002 = 0.2% 500 d.f. = min{n1 − 1, n2 − 1} = 4725 ; −t0 = −1.96; t0 = 1.96 Rejection regions: t < −1.96, t > 1.96 ( x − x ) − (0) (1984.8 − 1983.4) − (0) t= 1 2 = ≈ 8.802 s12 s22 (8.6) 2 (8.4) 2 + + 7133 4726 n1 n2

Because t > 1.96 , reject H 0 . There is enough evidence at the 5% level of significance to support the claim that there is a difference in the mean mint dates of coins minted in Philadelphia and Denver. 5.

d.f. = min{n1 − 1, n2 − 1} = 4725 ; −t0 = −1.96; t0 = 1.96 Rejection regions: t < −1.96, t > 1.96 ( x − x ) − (0) (0.034 − 0.033) − (0) = ≈ 1.010 t= 1 2 s12 s22 (0.054) 2 (0.052) 2 + + 7133 4726 n1 n2

Because −1.96 < t < 1.96 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to support the claim that there is a difference in the mean values of coins minted in Philadelphia and Denver.

622 TECHNOLOGY Ň

CHAPTER 9 TECHNOLOGY: NUTRIENTS IN BREAKFAST CEREALS 1. (a)

(b)

(c)

(d)

(e)

(f)

2. (calories, sugar), (calories, carbohydrates), (sugar, carbohydrates) 3. (calories, sugar): (calories, fat): (calories, carbohydrates): (sugar, fat): (sugar, carbohydrates): (fat, carbohydrates): Largest r:

r ≈ 0.766 r ≈ 0.415 r ≈ 0.913 r ≈ 0.461 r ≈ 0.793 r ≈ 0.230

(calories, carbohydrates)

4. (a) y = −12.836 + 0.177 x (b) y = 1.072 + 0.213 x 5. (a) y = −12.836 + 0.177 (120 ) = 8.404 grams (b) y = 1.072 + 0.213 (120 ) = 26.632 grams

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6. (a) C = 12.929 − 0.268 S + 7.567 F + 3.886 R (b) C = 23.279 + 0.496 S + 3.518 R 7.

C = 12.929 − 0.268 ( 7 ) + 7.567 ( 0.5 ) + 3.886 ( 31) ≈ 135.303 calories

623

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TECHNOLOGY Ň

CHAPTER 10 TECHNOLOGY: TEACHER SALARIES 1. Yes, because the salaries are from different districts, it is safe to assume that the samples are independent. 2. Using a technology tool, it appears that all three samples were taken from approximately normal populations. 3.

H 0 : σ 12 = σ 22 , H a : σ 12 ≠ σ 22 ( F0 = 3.28) (Under 500, 1000–2999) → F ≈ 1.33 → Fail to reject H 0 . (Under 500, At least 12,000) → F ≈ 1.14 → Fail to reject H 0 . (1000–2999, At least 12,000) → F ≈ 1.51 → Fail to reject H 0 .

4. The three conditions for a one-way ANOVA test are satisfied. H 0 : μ1 = μ 2 = μ3 (claim) H a : At least one mean is different from the others. Variation

Between Within

Sum of squares 325099608 609025769

Degrees of Freedom 2 36

Mean squares 162549804 16917382.5

9.61

0.000453

F ≈ 9.61 Reject H 0 . There is enough evidence at the 5% level of significance to reject the claim that the mean salaries are the same.

5. The samples are independent. Using a technology tool, it appears that two of the samples were taken from approximately normal populations. H 0 : σ12 = σ 22 , H a : σ12 ≠ σ 22 ( F0 = 3.28) (IFT-AFT, None/Other) → F ≈ 1.00 → Fail to reject H 0 .

The three conditions for a one-way ANOVA test are not satisfied.

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CHAPTER 11 TECHNOLOGY: U.S. INCOME AND ECONOMIC RESEARCH 1.

The median annual incomes do appear to differ among regions. 2.

H 0 : median ≤ 30,000 H a : median > 30,000 (claim) The critical value is 2. x=3 Because x > 2 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to support the claim that the median annual income in the Midwest is greater than $30,000.

H 0 : There is no difference in incomes in the Northeast and South. H a : There is a difference in incomes in the Northeast and South. (claim) The critical values are z0 = ±1.96.

R = 144 n (n + n + 1) 12(12 + 12 + 1) μR = 1 1 2 = = 150 2 2 n n ( n + n + 1) (12)(12)(12 + 12 + 1) σR = 1 2 1 2 = ≈ 17.3 12 12 R − μ R 144 − 150 z= ≈ ≈ −0.347 17.3 σR Because −1.96 < z < 1.96 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to support the claim that the median annual incomes in the Northeast and South are different.

625

626 TECHNOLOGY Ň

H 0 : There is no difference in the incomes for all four regions. (claim) H a : There is a difference in the incomes for all four regions.

χ 02 = 7.815 H ≈ 0.73 Because H < 7.815 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the claim that the distributions of annual incomes for all four regions are the same.

H 0 : There is no difference in the incomes for all four regions. (claim) H a : There is a difference in the incomes for all four regions. Analysis of Variance

Sum of Variation squares Between 653,838,161 Within 81,964,435,085

Degrees of freedom 3 44

Mean squares 217,946,054 1,862,828,070

F 0.12

P 0.950

F ≈ 0.12 → P − value = 0.950 Because P − value > 0.05 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the claim that the average annual incomes for all four regions are the same. The Kruskal-Wallis test and the one-way ANOVA test give the same result. With both, the decision is to fail to reject the null hypothesis.

6. (Box-and-whisker plot)

The median family incomes appear to be higher in the Northeast and lower in the Midwest. (Wilcoxon rank sum test) H 0 : There is no difference in incomes in the Northeast and South. H a : There is a difference in incomes in the Northeast and South. (claim) The critical values are z0 = ±1.96 .

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627

R = 263 n (n + n + 1) 15(15 + 15 + 1) μR = 1 1 2 = = 232.5 2 2 n n (n + n + 1) (15)(15)(15 + 15 + 1) σR = 1 2 1 2 = ≈ 24.1 12 12 R − μ R 263 − 232.5 z= ≈ ≈ 1.266 24.1 σR

Because −1.96 < z < 1.96. , fail to reject H 0 . There is not enough evidence at the 5% level of significance to support the claim that the median annual incomes of families in the Northeast and South are different. (Kruskal–Wallis test) H 0 : There is no difference in incomes of families in all four regions. (claim) H a : There is a difference in incomes of families in all four regions.

χ 02 = 7.815 H = 2.63 Fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the claim that the distributions of annual incomes of families for all four regions are the same.

(ANOVA test) H 0 : There is no difference in the incomes of families for all four regions. (claim) H a : There is a difference in the incomes of families for all four regions. Analysis of Variance Sum of Variation squares Between 5,758,228,180 Within 0.000000000017319256

Degrees of freedom 3 56

Mean squares 1,919,409,393 3,092,708,386

F 0.62

P 0.605

F ≈ 0.62 → P − value = 0.605 P − value >0.05 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the claim that the average annual incomes of families for all four regions are the same. The Kruskal-Wallis test and the one-way ANOVA test give the same result. With both, the decision is to fail to reject the null hypothesis.

628

CHAPTER 6 ALTERNATE Ň TECHNOLOGY

CHAPTER 6 ALTERNATE TECHNOLOGY: UNITED STATES FOREIGN POLICY POLLS 1. (a) pˆ = 0.63; qˆ = 0.37; z = 1.96; n = 1021; (0.600, 0.659) (b) pˆ = 0.58; qˆ = 0.42; z = 1.96; n = 1021; (0.550, 0.610) (c) pˆ = 0.93; qˆ = 0.07; z = 1.96; n = 1021; (0.915, 0.946) (d) pˆ = 0.37; qˆ = 0.63; z = 1.96; n = 1021; (0.341, 0.400) (e) pˆ = 0.17; qˆ = 0.83; z = 1.96; n = 1021; (0.147, 0.193) 2

3. (a) Answers will vary.

(b) Answers will vary. 4. Answers will vary.

Instructor Manual for Elementary Statistics: Picturing the World, 8th Edition by Larson & Farber – F