Elementary Statistics Picturing the World, 7th Edition , Ron Larson , Betsy Farber Solution Manual by Ace Test Banks

Elementary Statistics Picturing the World, 7th Edition By Ron Larson , Betsy Farber

Email: Richard@qwconsultancy.com

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 ninth to twelfth graders in the United States. The sample consists of the responses of the 1501 ninth to twelfth graders in the survey. The sample data set consists of 1215 ninth to twelfth graders who said leaders today are more concerned with their own agenda than with achieving the overall goals of the organization they serve and 286 ninth to twelfth graders 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 43% is based on a subset of the population. 3a. The population consists of the responses of all U.S. adults, and the sample consists of the responses of the 1000 U.S. adults in the study. b. The part of this study that represents the descriptive branch of statistics involves the statement “three out of four adults will consult with their physician or pharmacist and only 8% visit a medicationspecific website [when they have a question about their medication].” c. A possible inference drawn from the study is that most adults consult with their physician or pharmacist when they have a question about their medication.

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 measure that describes 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. 1 Copyright © 2019 Pearson Education Ltd.

CHAPTER 1 │ INTRODUCTION TO STATISTICS

10. False. A sample statistic can differ from sample to sample. 11. Sample, because the collection of 95 shopkeepers is a subset within the population of the commercial complex’s 550 shopkeepers. 12. Population, because it is a collection of the energy collected from all the solar panels on a photo voltaic power plant. 13. Population, because it is a collection of the height of each athlete participating in the Summer Olympics. 14. Sample, because the collection of every sixth person is a subset within the population of the departmental store’s shoppers. 15. Sample, because the collection of the 10 patients is a subset of the population of 50 patients at the clinic. 16. Sample, because the collection of 25 households is a subset within the population of the neighborhood’s 75 households. 17. Population, because it is a collection of all the gamers’ scores in the tournament. 18. Population, because it is a collection of the ages of all presidents at the time of their election. 19. Sample, because the collection of top 10 taxpayers is a subset within the population of the country’s total tax payers. 20. Sample, because the collection of the 20 air contamination levels is a subset of the population. 21. Population: Parties of registered voters Sample: Parties of registered voters who respond to a survey 22. Population: Student donations at a food drive Sample: Student donations of canned goods 23. Population: Ages of adults in the United States who own automobiles Sample: Ages of adults in the United States who own Honda automobiles 24. Population: Incomes of home owners in Massachusetts Sample: Incomes of home owners in Massachusetts with mortgages 25. Population: Collections of the responses of all U.S. adults Sample: Collection of the responses of the 1020 U.S. adults surveyed Sample data set: 42% of adults who said they trust their political leaders and 58% who said they did not 26. Population: Collection of fetal tobacco exposure of all infants Sample: Collection of the fetal tobacco exposure of 203 infants Sample data set: Infants with fetal tobacco exposure and their focused attention levels 27. Population: Collection of the influenza immunization status of all adults in the United States Sample: Collection of the influenza immunization status of the 3301 U.S. adults surveyed Sample data set: 39% of U.S. adults who received an influenza vaccine and 61% who did not

CHAPTER 1 │ INTRODUCTION TO STATISTICS

28. Population: Collection of the responses of employees working in a foreign country Sample: Collection of the responses of the 1500 employees who have settled in foreign countries Sample data set: 62% of respondents said that they settle in the country where they work and 38% who said they did not 29. Population: Collection of the average hourly billing rates of all U.S. law firms Sample: Collection of the average hourly billing rates for partners of the 159 U.S. law firms surveyed Sample data set: The average hourly billing rate for partners of 159 U.S. law firms is $604. 30. Population: Collection of the responses of children Sample: Collection of the responses of the 328 children living in a locality Sample data set: 86% of respondents who said that they planned to visit their grandparents in vacations and 14% who said they did not 31. Population: Collection of blood donations collected globally Sample: Collection of the 112.5 million blood donations collected globally Sample data set: 50% of the donors who belong to high-income countries and 50% who do not 32. Population: Collection of the responses of laptop users Sample: Collection of the responses of the 1468 laptop users Sample data set: 81% of respondents who said that they preferred the use of mouse over touchpad and 19% who said they did not. 33. Population: Collection of the 1000 mutual funds listed on a recognized stock exchange Sample: Collection of the 134 mutual funds of the 1000 mutual funds listed on a recognized stock exchange Sample data set: Best mutual funds out of the 134 mutual funds listed on a recognized stock exchange 34. Population: Collection of parents of 13- to 17-year-olds Sample: Collection of responses of 1060 parents of 13- to 17-year-olds surveyed Sample data set: 636 parents who said they check their teen’s social media profile and 424 parents who did not 35. Population Parameter. Forty out of 500 total students is a numerical description of the students who received a Grade C. 36. Sample statistic. The value 56.3% is a numerical description of a sample of college board members 37. Sample statistic. The value two million is a numerical description of a sample of civilian casualties during World War II. 38. Population parameter. The value 62% is a numerical description of the total number of governors. 39. Population Parameter. The entire population of employees working in the organization has been reviewed. 40. Sample statistic. The value 16% is a numerical description of a sample of teachers. 41. Sample statistic. The value 80% is a numerical description of a sample of U.S. adults. 42. Population parameter. The score 20.6 is a numerical description of the ACT scores for all graduates.

CHAPTER 1 │ INTRODUCTION TO STATISTICS

43. The statement “50% are collected from high-income countries” is an example of descriptive statistics. Using inferential statistics, you may conclude that an association exists between income and the number of blood donations in a country. 44. The statement “81% preferred use of mouse over touchpad” is an example of descriptive statistics. Using inferential statistics, you may conclude that most laptop users prefer use of mouse over touchpad. 45. Answers will vary. 46. Answers will vary. 47. The inference may incorrectly imply that exercise increases a person’s cognitive ability. The study shows a slower decline in cognitive ability, not an increase. 48. The inference may incorrectly imply that obesity trends will continue in future years. Even though the obesity rates have been increasing, that does not mean the rates will continue to increase for eternity. 49. (a) The sample is the results on the standardized test by the participants in the study. (b) The population is the collection of all the results of the standardized test. (c) The statement “the closer that participants were to an optimal sleep duration target, the better they performed on a standardized test” is an example of descriptive statistics. (d) Individuals who obtain optimal sleep will be more likely to perform better on a standardized test then they would without optimal 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 │ INTRODUCTION TO STATISTICS

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 breeds of horses are attributes. 8. Qualitative, because ASCII codes are labels. 9. Quantitative, because blood pressure levels are numerical measurements. 10. Quantitative, because train speeds are numerical measurements. 11. Qualitative, because colors are attributes. 12. Quantitative, because width is a numerical measurement. 13. Quantitative, because weight is a numerical measurement. 14. Qualitative, because marital status is an attribute. 15. Ordinal. Data can be arranged in order, but the differences between data entries are not meaningful. 16. Ordinal. Data can be arranged in order, but differences between data entries are not meaningful. 17. Nominal. No mathematical computations can be made, and data are categorized using numbers. 18. Interval. Data can be ordered and meaningful differences can be calculated, but it does not make sense to say that one time is a multiple of another. 19. Ordinal. Data can be arranged in order, but the differences between data entries are not meaningful. 20. Ratio. A ratio of two data values can be formed, so one data value can be expressed as a multiple of another. 21. Horizontal: Nominal; Vertical: Ratio 22. Horizontal: Ordinal; Vertical: Ratio

CHAPTER 1 │ INTRODUCTION TO STATISTICS

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

(b) Ratio

(d) Interval

26. (a) Interval

(b) Ratio

(d) Ordinal

27. Quantitative. Ratio. A ratio of two data entries can be formed, so one data entry can be expressed as a multiple of another. 28. Quantitative. Ratio. A ratio of two data entries can be formed, so one data entry can be expressed as a multiple of another. 29. Qualitative. Ordinal. Data can be arranged in order, but the differences between data entries are not meaningful. 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 │ INTRODUCTION TO STATISTICS

(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. Observational study. The study does not apply a treatment to the adults. 12. Experiment. The study applies a treatment (intensive program to lower systolic blood pressure) to the subjects. 13. Experiment. The study applies a treatment (different photographs) to the subjects. 14. Observational study. The study does not apply a treatment to the motorists. 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.

CHAPTER 1 │ INTRODUCTION TO STATISTICS

17. Answers will vary. 18. Answers will vary. 19. (a) The experimental units are the 500 females ages 25 to 45 years old who suffer from migraine headaches. The treatment is the new drug used to treat migraine headaches. (b) A problem with the design is that the sample is not representative of the entire population because only females ages 25 to 45 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 drug or the placebo. 20. (a) The experimental units are the 31 patients with type 2 diabetes. The treatment is the dietary supplement designed to control metabolism in patients with type 2 diabetes. (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 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: Dennis, Alice, Arthur, Kate, Harry, Bertha, Bill, Mercer and Edgar. Control group: Lewis, Raj, William, Edwin, Zoya, Lara, Jennifer, Ahmed and Ronald. 22. Answers will vary. Sample answer: Using a random number table: Treatment group: 1, 3, 4, 6, 9, 11, 15, 16, 17, 18, 20, 23, 26, 27, 29, 31, 32, 33, 37, 39, 42, 45, 46, 49, 50, 52, 55, 56, 58, 59, 61, 62, 65, 69, 70, 73, 74, 78, 79, 80. Control group: 2, 5, 7, 8, 10, 12, 13, 14, 19, 21, 22, 24, 25, 28, 30, 34, 35, 36, 38, 40, 41, 43, 44, 47, 48, 51, 53, 54, 57, 60, 63, 64, 66, 67, 68, 71, 72, 75, 76, 77. 23. Cluster sampling is used because the constituency is divided into areas, and 12 areas are then entirely selected. A possible source of bias is that problems of the residents of one area might be different from that of the other area. 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. 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.

CHAPTER 1 │ INTRODUCTION TO STATISTICS

26. Stratified sampling is used because the organization is divided into departments and a sample executive is taken from each department. 27. Simple random sampling is used because each house number has an equal chance of being selected, and all samples of 1638 house numbers have an equal chance of being selected. The sample is unbiased. 28. Systematic sampling is used because every sixth customer entering the parlor is sampled. It is possible for bias to enter into the sample, if, for some reason, there is a regular pattern to customers entering the shop. 29. Sampling, because the population of mobile phone purchasers is too large for their most popular model of mobile phone to be easily recorded. Random sampling would be advised because it would be easy to select mobile phone purchasers randomly and then record their most popular model of mobile phone. 30. Census, because it is relatively easy to obtain the heights of the 264 students. 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 survey question is unbiased because it does not suggest that one should drink a lot of water to stay hydrated. 33. The question is biased because it already suggests that listening to music while studying increases the chances of retention. The question could be rewritten as “Does listening to music while studying have an effect on retention?” 34. The survey question is unbiased. 35. The households sampled represent various locations, ethnic groups, and income brackets. Each of these variables is considered a stratum. Stratified sampling ensures that each segment of the population is represented. 36. 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. 37. Answers will vary. 38. Answers will vary. 39. 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.

CHAPTER 1 │ INTRODUCTION TO STATISTICS

CHAPTER 1 REVIEW EXERCISE SOLUTIONS 1. Population: Collection of the responses of all U.S. adults Sample: Collection of the responses of the 4787 U.S. adults who were sampled Sample data set: 15% of adults who use ride-hailing applications and 85% who do not 2. Population: Collection of the opinions on the current educational policy of all professors in the Pennsylvania state. Sample: Collection of the opinions on educational policy of the 42 professors in the Pennsylvania state that were sampled. 3. Population: Collection of the responses of all U.S. adults Sample: Collection of the responses of the 2223 U.S. adults who were sampled Sample data set: 62% of adults who would encourage a child to pursue a career as a video game developer or designer and 38% who would not 4. Population: Collection of the responses of all U.S. children and adults ages 16 years and older Sample: Collection of the responses of the 1601 U.S. children and adults ages 16 and older who were sampled Sample data set: 48% of children and adults who have visited a public library or a bookmobile over a recent span of 12 months and 52% who did not 5. Population parameter. The value $22.7 million is a numerical description of the total infrastructurestrengthening investments. 6. Sample statistic. The value 29% is a numerical description of a sample of U.S. voters. 7. Parameter. The 12 students minoring in math is a numerical description of all physics majors at a university. 8. Sample statistic. The value 30% is a numerical description of a sample of U.S. workers. 9. The statement “62% would encourage a child to pursue a career as a video game developer or designer” is an example of descriptive statistics. An inference drawn from the sample is that a majority of people encourage children to pursue a career as a video game developer or designer. 10. The statement “48% have visited a public library or a bookmobile over a recent span of 12 months” is an example of descriptive statistics. An inference drawn from the sample is that about half of U.S. children and adults ages 16 years and older have visited a public library or a bookmobile over a recent span of 12 months. 11. Quantitative, because ages are numerical measurements. 12. Quantitative, because IQ levels are numerical measurements. 13. Quantitative, because revenues are numerical measures. 14. Qualitative, because genders are attributes.

CHAPTER 1 │ INTRODUCTION TO STATISTICS

15. Interval. The data can be ordered and meaningful differences can be calculated, but it does not make sense to say that 84 degrees is 1.05 times as hot as 80 degrees. 16. Ordinal. The data are qualitative and could be arranged in order of income level. 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 $53.2 million is 1.12 times as much as $47.5 million. 19. Experiment. The study applies a treatment (drug to treat hypertension in patients with obstructive sleep apnea) 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 professor sampled a convenient group of his students. The study may be biased toward the opinions of the professor’s students. 25. Cluster sampling is used because each district is considered a cluster and every pregnant woman in a selected district is surveyed. A potential source of bias is that the selected districts may not be representative of the entire area. 26. Systematic sampling is used because every tenth house is surveyed. A potential source of bias is that the locality the researcher is using may be posh. 27. Stratified sampling is used because the population is divided by religious groups and then 50 voters are randomly selected from each religious group. 28. Convenience sampling is used because of the convenience of surveying students in just one school. A potential source of bias is that the school is located in the downtown area where a lot of junk food may be available. 29. Answers will vary. Sample answer: Using the random number table in Appendix B, starting with the left-most number in row 7: 681/088/926/694/730/957/617/502/348/464/655/449/658/318/ The random numbers are 88, 502, 348, 464, 449, 318.

CHAPTER 1 │ INTRODUCTION TO STATISTICS

CHAPTER 1 QUIZ SOLUTIONS 1. Population: Collection of the school performance of all Korean adolescents Sample: Collection of the school performance of the 359,264 Korean adolescents in the study 2. (a) Sample statistic. The value 52% 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 25% 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 (multivitamin) 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 │ INTRODUCTION TO STATISTICS

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 72% is a numerical description of a sample of U.S. adults ages 18 years and older. (b) Population parameter. The average evidence based reading and writing score of 543 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 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 19,183 U.S. physicians who were sampled. (b) Both. Location, employment status, benefits received, and speciality are qualitative because they are attributes. Income and time spent seeing patients per week are quantitative because they are numerical measurements. (c) Nominal: location, employment status, benefits received, specialty Ratio: 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.

CHAPTER

Descriptive Statistics

2.1 FREQUENCY DISTRIBUTIONS AND THEIR GRAPHS 2.1 TRY IT YOURSELF SOLUTIONS 1. The number of classes is 6.

Range 55 -14 = = 6.83  7 Number of classes 6 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 = 14, Max = 55, Class width =

Lower limit 14 21 28 35 42 49

Upper limit 20 27 34 41 48 55

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 14-20 21-27 28-34 35-41 42-48 49-55

Frequency, f 8 15 14 7 4 3

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 │ DESCRIPTIVE STATISTICS

Class

Midpoint

14–20

21–27

28–34

35–41

42–48

49–55

14 + 20 = 17 2 21 + 27 = 24 2 28 + 34 = 31 2 35 + 41 = 38 2 42 + 48 = 45 2 49 + 55 = 52 2

∑f = 51

Relative frequency 8 » 0.1569 51 15 » 0.2941 51 14 » 0.2745 51 7 » 0.1373 51 4 » 0.0784 51 3 » 0.0588 51 f å =1 n

Cumulative frequency 8 8 + 15 = 23 23 + 14 = 37 37 + 7 = 44 44 + 4 = 48 48 + 3 = 51

Sample answer: The most common range of points scored by winning teams is 21 to 27. About 14% of the winning teams scored more than 41 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 14–20 21–27 28–34 35–41 42–48 49–55

Class Boundaries 13.5–20.5 20.5–27.5 27.5–34.5 34.5–41.5 41.5–48.5 48.5–55.5

Frequency, f 8 15 14 7 4 3

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 21 to 27. About 14% of the winning teams scored more than 41 points.

CHAPTER 2 │ 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 24 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 20.5 and 27.5. 7.

CHAPTER 2 │ 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. An ogive is a graph that displays cumulative frequencies. 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 135 -17 = = 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 247 - 54 = = 19.3  20 Number of classes 10 Lower class limits: 54, 74, 94, 114, 134, 154, 174, 194, 214, 234 Upper class limits: 73, 93, 113, 133, 153, 173, 193, 213, 233, 253

14. Class width =

CHAPTER 2 │ DESCRIPTIVE STATISTICS

15. (a) Class width = 11 - 0 = 1 1 (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) (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

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 │ 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 Midpoint f 188 5 372 16 264 27 205 38 83 49 76 60 32 71 å f = 1220

Relative frequency 0.15 0.30 0.22 0.17 0.07 0.06 0.03 æfö å çççè n ø÷÷÷ » 1

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 æfö å f = 365 å çççè n ø÷÷÷ » 1

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 │ 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 │ 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 that are 70 inches tall from the number in the sample: 77 – 47 = 30 29. Class width =

Range 30 - 8 = = 4.4  5 Number of classes 5

Class

Midpoint

8–12 13–17 18–22 23–27 28–32

Frequency, f 5 8 8 2 1 ∑f = 24

10 15 20 25 30

Relative frequency 0.22 0.33 0.33 0.08 0.04 æfö å çççè n ø÷÷÷ » 1

Cumulative frequency 5 13 21 23 24

Classes with greatest frequency: 13 – 17, 18 – 22 Class with least frequency: 28 – 32 Range 541 - 213 = » 54.7  55 Number of classes 6 Class Frequency, Midpoint Relative Cumulative f frequency frequency 213–267 4 240 0.133 4 268–322 6 295 0.200 10 323–377 2 350 0.067 12 378–432 10 405 0.333 22 433–487 1 460 0.033 23 488–542 7 515 0.234 30 ∑f = 30 æfö å çççè n ø÷÷÷ = 1 Class with greatest frequency: 378 – 432 Class with least frequency: 433 – 487

30. Class width =

CHAPTER 2 │ DESCRIPTIVE STATISTICS

Range 2804 - 985 = » 303.2  304 Number of classes 6 Class Frequency, Mid-point Relative f frequency 985–1288 3 1136.5 0.1429 1289–1592 6 1440.5 0.2857 1593–1896 2 1744.5 0.0952 1897–2200 4 2048.5 0.1905 2201–2504 2 2352.5 0.0952 2505–2808 4 2656.5 0.1905 ∑f = 21 æfö å çççè n ø÷÷÷ = 1

31. Class width =

Cumulative frequency 3 9 11 15 17 21

Sample answer: The graph shows that the production is evenly distributed between 1745 units to 2657 units. 32. Class width = Class 38–50 51–63 64–76 77–89 90–102

Range 99 - 38 = = 12.2  13 Number of classes 5

Frequency, f 8 4 3 4 5

å f = 24

Midpoint 44 57 70 83 96

Relative frequency 0.3333 0.1667 0.1250 0.1667 0.2083 æfö å çççè n ø÷÷÷ = 1

Cumulative frequency 8 12 15 19 24

CHAPTER 2 │ DESCRIPTIVE STATISTICS

33. Class width =

Range 23 - 2 = = 2.625  3 Number of classes 8

Class

Frequency, f

Midpoint

2–4 5–7 8–10 11–13 14–16 17–19 20–22 23–25

9 10 4 1 3 2 0 1

3 6 9 12 15 18 21 24

å f = 30

Relative frequency 0.31 0.33 0.13 0.03 0.10 0.07 0.00 0.03 æfö å çççè n ø÷÷÷ = 1

Cumulative frequency 9 19 23 24 27 29 29 30

Sample answer: The graph shows that the response time is evenly distributed between 9 days to 18 days or 2 to 7 days. 34. Class width = Class 100–107 108–115 116–123 124–131 132–139 140–147 148–155 156–163

Range 162 - 100 = = 7.75  8 Number of classes 8

Frequency, f 2 1 2 6 3 3 1 3

å f = 21

Midpoint 103.5 111.5 119.5 127.5 135.5 143.5 151.5 159.5

Relative frequency 0.10 0.05 0.10 0.28 0.14 0.14 0.05 0.14 æ f ö÷ å çççè n ø÷÷ = 1

Cumulative frequency 2 3 5 11 14 17 18 21

CHAPTER 2 │ DESCRIPTIVE STATISTICS

Sample answer: The graph shows that the most frequent bowling speeds were from 124 to 131 kmph. 35. Class width = Class 42 – 46 47 – 51 52 – 56 57 – 61 62 – 66 67 – 71

Range 70 - 42 = » 4.7  5 Number of classes 6

Frequency, f 4 11 14 9 4 3

Midpoint 44 49 54 59 64 69

å f = 45

Relative frequency 0.0889 0.2444 0.3111 0.2000 0.0889 0.0667 f å n =1

Cumulative frequency 4 15 29 38 42 45

Sample answer: The graph shows that the number of U.S. presidents who were 52 or older at inauguration was twice as many as those who were 51 and younger. 36. Class width = Class 1–11 12–22 23–33 34–44 45–55

Range 55 - 0 = = 10.8  11 Number of classes 5

Frequency, f 25 16 5 7 3

å f = 56

Midpoint 6 27 28 39 48

Relative frequency 0.4643 0.2679 0.0892 0.1250 0.0536 æfö å çççè n ÷÷÷ø » 1

Cumulative frequency 26 41 46 53 56

CHAPTER 2 │ DESCRIPTIVE STATISTICS

Sample answer: The graph shows that most of the monarchs had regnal years of less than 11 years. 37. Class width = Class 1-2 3-4 5-6 7-8 9-10

Range 10 - 1 = = 1.8  2 Number of classes 5

Frequency, f 7 6 14 4 5

å f = 36

Midpoint 1.5 3.5 5.5 7.5 9.5

Relative frequency 0.19 0.17 0.39 0.11 0.14 æfö å çççè n ÷÷÷ø » 1

Class with greatest relative frequency: 5 – 6 Class with least relative frequency: 7 – 8

Cumulative frequency 7 13 27 31 36

CHAPTER 2 │ DESCRIPTIVE STATISTICS

38. Class width = Class 9–11 12–14 15–17 18–20 21–23

Range 23 - 9 = = 2.8  3 Number of classes 5

Frequency, f 3 10 8 6 1

Midpoint 10 13 16 19 27

å f = 28

Relative frequency 0.11 0.36 0.28 0.21 0.04 æ f ö÷ å çççè n ÷ø÷ » 1

Cumulative frequency 3 13 21 27 28

Class with greatest relative frequency: 12 – 14 Class with least relative frequency: 21 – 23 39. Class width = Class 401–420 421–440 441–460 461–480 481–500

Range 499 - 402 = = 19.4  19 Number of classes 5

Frequency, f 10 5 4 5 4

å f = 28

Midpoint 410.5 430.5 450.5 470.5 490.5

Relative Cumulative frequency frequency 0.36 10 0.18 15 0.14 19 0.18 24 0.14 28 æfö å çççè n ÷÷÷ø » 1

CHAPTER 2 │ DESCRIPTIVE STATISTICS

Class with greatest relative frequency: 401 – 420 Classes with least relative frequency: 441 – 460 and 481 – 500 40. Class width = Class 95–112 113–130 131–148 149–166 167–184

Range 184 - 95 = = 17.8  18 Number of classes 5

Frequency, f 5 11 6 4 2

å f = 28

Midpoint 103.5 121.5 139.5 157.5 175.5

Relative frequency 0.18 0.39 0.22 0.14 0.07 æfö å çççè n ÷÷÷ø » 1

Class with greatest relative frequency: 113 − 130 Class with least relative frequency: 167 − 184

Cumulative frequency 5 16 22 26 28

CHAPTER 2 │ DESCRIPTIVE STATISTICS

41. Class width = Class 49–54 55–60 61–66 67–72 73–78 79–84

Range 83 - 49 = » 5.6  6 Number of classes 6

Frequency, f 4 12 8 6 3 2

Midpoint

å f = 35

51.5 57.5 63.5 69.5 75.5 81.5

Relative frequency 0.11 0.34 0.23 0.17 0.09 0.06 æfö å çççè n ÷÷÷ø » 1

Cumulative frequency 4 16 24 30 33 35

Location of the greatest increase in frequency: 55 – 60 42. Class width =

Range 16900 - 7800 = = 1820  1900 Number of classes 5

Class 7800–9699 9700–11599 11600–13499 13500–15399 15400–17299

Frequency, f 6 15 1 4 2

å f = 28

Midpoint 8749.5 10649.5 12549.5 14449.5 16349.5

Relative frequency 0.2143 0.5357 0.0357 0.1429 0.0714 æfö å çççè n ÷÷÷ø » 1

Cumulative frequency 6 21 22 26 28

CHAPTER 2 │ DESCRIPTIVE STATISTICS

Sample answer: The graph shows that majority of persons have daily calorie intake from 7800 kJ to 11599 kJ. 43. (a) Class width = Class 2–6 7–11 12–16 17–21 22–26 27–31

Range 30 - 2 = » 4.67  5 Number of classes 6

Frequency, f 5 10 6 2 0 1

Midpoint

Relative frequency 0.2083 0.4167 0.2500 0.0833 0.0000 0.0417 æfö å ççèç N ÷ø÷÷ » 1

4 9 14 19 24 29

å f = 24 (b)

(c)

(d)

(e)

Cumulative frequency 5 15 21 23 23 24

CHAPTER 2 │ DESCRIPTIVE STATISTICS

44. (a) Class width = Class 2–3 4–5 6–7 8–9 10–11 12–13

Range 55 - 1 = = 10.8  11 Number of classes 5

Frequency, f 7 8 2 2 3 1 å f = 23

Midpoint 2.5 4.5 6.5 8.5 10.5 12.5

Relative frequency 0.30 0.35 0.09 0.09 0.13 0.04 æfö å çç ÷÷÷ = 1 çè n ø

(b)

(c)

(d)

(e)

Cumulative frequency 7 15 17 19 22 23

CHAPTER 2 │ 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

å f = 30

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

(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

å 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 å n =1

CHAPTER 2 │ 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 14 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

4 6 6 6 7 Key: 1 | 4 = 14 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 2 2 3 4 4 4 4 5 5 5 7 8 8 9 2 3 6 8 9 2 5

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

CHAPTER 2 │ 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

4 Key: 1 | 4 = 14 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 2 2 3 4 4 4 4 5 5 5 7 8 8 9 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 455 1051 330 104

å f = 1940

Relative Frequency 0.235 0.542 0.170 0.054 f å n »1

Angle 85 195 61 19

å = 360

From 1990 to 2014, as percentages of the total degrees conferred, associate’s degrees increased by 2.9%, bachelor’s degrees decreased by 5.1%, master’s degrees increased by 2.8%, and doctoral degrees decreased by 0.7%. 5.

Cause Auto dealers (used cars) Insurance companies Mortgage brokers Collection agencies Travel agencies and bureaus

Frequency, f 16,281 8384 3634 19,277 6985

CHAPTER 2 │ DESCRIPTIVE STATISTICS

Collection agencies 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 represent the years and let the vertical axis represent the number of burglaries (in millions).

Sample answer: The number of burglaries remained about the same until 2012 and then decreased through 2015.

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.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

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. 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 Pinterest has the least. Tumblr and Instagram have about the same number of users. 14. Sample answer: The year 2010 had the most motor vehicle thefts and 2013 had the least. Motor vehicle thefts decreased the most between 2011 and 2012. 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: Food is the most costly aspect of pet care and live animal purchases is the least. The amounts spent on veterinarian care and supplies/OTC medicine are about the same. 17. Humidity (in percentages) 18 19 20 21 22

6 2 1 0 5

4 5 3 6

6 8 4 8

6 8 5

Key: 9 8 6

18|6 = 18.6

Sample answer: Most of the days had a humidity level of 19.2% to 21.8%. 18. Hours studied by students 1 2 3 4

1 0 0 2

5 1 2

6 3 5

6 3

Key: 7 4

8 4

8 5

1|1 = 11 8 5

9 6

CHAPTER 2 │ DESCRIPTIVE STATISTICS

19. Runs scored 7 8 9

0 1 0

Key:

0 2 0

1 5 1

3 6 1

3 7 4

7|0 = 70 5 8 8

Sample answer: Most runs scored by the batsman were in the 70s. 20. Drunk driving cases (in numbers) 1 2 3 4

0 0 0 0

1 2 0 1

3 3 1 2

5 5 2

6 5 2

7 6 4

8 7 6

Key:

1|0 = 10

9 8 6

Sample answer: Most drunk driving cases registered were from 20 to 28. 21. Incomes of Highest Paid tech CEOs 1 2 3 4

3 8 0 8 2 1

3 9 0 8 3 1

4 9 0

4 9 2

Key: 1|3 = 13

Sample answer: Most of the highest paid tech-CEOs have incomes that range from $13 million to $22 million. 22. x =

å x = 3134 = 314.1

10 228 296 298 310 361 368 410 298 + 310 median = = 304 2 mode = none The mode cannot be found because no data entry is repeated. n 159 225

486

23.

Sample answer: Blood glucose level tends to be between 94 to 101 milligrams per deciliter. 24.

Sample answer: The weight of a polar bear tends to be from 400 kg to 440 kg.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

25.

Balance Owed $1 to $10,000 $10,001 to $25,000 $25,001 to $50,000 $50,001 +

f 16.7 12.4 8.3 6.7 å f = 44.1

Relative Frequency 0.379 0.281 0.188 0.152 f å n =1

Angle 136 101 68 55 å = 360

Sample answer: The majority of student loan borrowers owe $25,000 or less. 26.

Category United States Italy Ethiopia South Africa Tanzania Kenya Mexico Morocco Great Britain Brazil New Zealand Eritrea

Frequency, f 15 4 2 2 1 12 4 1 1 2 1 1

å f = 46

Relative Frequency 0.326 0.087 0.043 0.043 0.022 0.261 0.087 0.022 0.022 0.043 0.022 0.022 f å n »1

Angle 117 31 15 15 8 94 31 8 8 15 8 8

å » 360

Sample answer: Most of the New York City Marathon winners are from the United States and Kenya.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

27.

Sample answer: Brazil won the FIFA World cup the most number of times out of the five countries, and Uruguay and Argentina won the least number of times. 28. Type of Vehicle Small sedan Medium sedan SUV Minivan

Cost $6579 $8604 $10,255 $9262

Sample answer: It costs the least to own and operate a small sedan. 29. Let the horizontal axis represent hours and let the vertical axis represent the hourly fees (in dollars).

Sample answer: It appears that there is a slight positive relation between fees and hours of coaching.

CHAPTER 2 │ 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 represent the years and let the vertical axis represent the number of degrees (in thousands).

Sample answer: The number of bachelor’s degrees in engineering conferred in the U.S. has increased from 2008 to 2015. 32. Let the horizontal axis represent the years and let the vertical axis represent the percent.

Sample answer: The percentage of the Egypt’s gross domestic product that comes from the travel and tourism sector has decreased from 2007 to 2016. 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 Copyright © 2019 Pearson Education Ltd.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

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. 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 3rd quarter are disproportionately larger than the other quarters. (b)

CHAPTER 2 │ DESCRIPTIVE STATISTICS

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%. (b)

39. (a) The graph is misleading because the angle makes it appear as though the 3rd quarter had a larger percent of sales than the others, when the 1st and 3rd 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.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

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

(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)

CHAPTER 2 │ DESCRIPTIVE STATISTICS

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

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

å x = 35 + 33 + 16 + 23 + 16 + ... + 28 + 24 + 34 = 1541

å x 1541 = » 30.2 n 51 The mean points scored by the 51 winning teams is about 30.2. x=

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. 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, 32, 32, 33, 34, 34, 34, 34, 35, 35, 35, 37, 38, 38, 39, 42, 43, 46, 48, 49, 52, 55 Because there are 51 entries (an odd number), the median is the middle, or 26th entry. So, the median is 30 points. 3. Order the data from smallest to largest. 17, 20, 21, 21, 24, 24, 27, 28, 29, 31, 31, 32, 34, 34, 43, 48 Because there are an even number of entries, the median is the mean of the two middle entries. 28 + 29 57 median = = = 28.5 2 2 The median points scored by the winning teams in the Super Bowls for the National Football League’s 2001 through 2016 seasons is 28.5 points. 4. Look at the ordered data from Try It Yourself 1 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, 32, 32, 33, 34, 34, 34, 34, 35, 35, 35, 37, 38, 38, 39, 42, 43, 46, 48, 49, 52, 55 The mode is the data entry that occurs with the greatest frequency. The entry 27 occurs the most, so the mode is 27. 5. “some” occurs with the greatest frequency (578). The mode is “some”. 6.

å 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 │ DESCRIPTIVE STATISTICS

Final Grade C C D A B B

å ( x ⋅ w)

Points, x 2 2 1 4 3 3

Weight, w 3 4 1 3 2 3

å w = 16

x∙w 6 8 1 12 6 9

å ( x ⋅ w) = 42

42 » 2.6 åw 16 The new weighted mean is about 2.6.

Class 14-20 21-27 28-34 35-41 42-48 49-55

Midpoint, x 17 24 31 38 45 52

Frequency, f

x·f

8 15 14 7 4 3

136 360 434 266 180 156

å f = 51 = n

å ( x ⋅ f ) = 1532

å( x ⋅ f ) 1532 = » 30.0 n 51 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: 1, 2, 2, 2, 3 6. Sample answer: 2, 4, 5, 5, 6, 8 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. 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.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

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. å x 347 = » 24.8 n 14 21 21 23 23 24 24 24 25 26 26 26 27 28 29   

17. x =

median =

24 + 25 49 = = 24.5 2 2

mode = 24, 26 å x 3141 = » 314.1 n 10 159 225 228 296 298 310 361 368 410 486  

18. x =

median =

298 + 310 = 304 2

mode = none The mode cannot be found because no data entry is repeated. å x 788 = » 37.52 n 21 12 18 18 22 22 28 28 28 28 32 34 34  38 42 46 50 54 60 62 64 68

19. x =

median = 34 mode = 28 The mode does not represent the center of the data set because they are small values compared to the rest of the data. å x 401 = = 40.1 n 10 33 37 37 39 39 41 43 43 44 45 

20. x =

39 + 41 = 40 2 mode = 37, 39, 41 (each occurs two times) median =

CHAPTER 2 │ DESCRIPTIVE STATISTICS

å x 697 = » 49.8 n 14 45 47 48 48 48 49 50 51 51 51 51 51 52 55 

21. x =

50 + 51 = 50.5 2 mode = 51 (occurs 5 times) median =

å 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 =

å x 119 = » 7.4 n 16 1 2 2 3 3 5 6 66 8 10 10 10 1117 19

23. x =

6+6 =6 2 mode = 6, 10 (both occur 3 times) median =

å 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.

24. x =

å 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.

25. x =

26. x =

å x 235 = » 7.12 n 33

median median = 5 mode = 5 (occurs 9 times)

CHAPTER 2 │ DESCRIPTIVE STATISTICS

27. x is not possible (nominal data) median = not possible (nominal data) mode = “Search and buy online” 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 = “Yes, since 2014 or earlier” The mean and median cannot be found because the data are at the nominal level of measurement. å 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 

31. x =

median =

30 + 31 = 30.5 2

mode = 23, 34 (both occur 3 times each) å 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 2 times) The mode does not represent the center of the data set because it is the largest entry in the data set. å 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 3 times)

33. x =

CHAPTER 2 │ DESCRIPTIVE STATISTICS

å 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 5 times) 35. Cluster around 275 – 425 36. Cluster around 450 – 1050, gap between 1950 and 2850, outlier at 3000 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 Score, x Weight, w Assignment 75 0.10 Class Participation 60 0.25 Practical 90 0.25 Theory exam 85 0.40 å w =1

å( x ⋅ w) 79 = = 79 1 åw

42. Source Score, x Weight, w Assignment 75 0.10 Class Participation 60 0.25 Practical 90 0.25 Theory exam 85 0.40

å w =1

x= 43.

x∙w 7.5 15 22.5 34 å ( x ⋅ w) = 79

å( x ⋅ w) 79 = = 79 åw 1

Balance, x $523 $2415 $250

x∙w 7.5 15 22.5 34 å ( x ⋅ w) = 79

å ( x ⋅ w) åw

Days, w 24 2 4

å w = 30 =

å ( x ⋅ w) = 18,382

18,382 » $612.73 30

x∙w 12,552 4830 1000

CHAPTER 2 │ DESCRIPTIVE STATISTICS

44.

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

x= 45.

åw

Score, x 85 81 90

å ( x ⋅ w)

2268 = 84 27

åw

Grade A B B C D

å( x ⋅ w) åw

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

13,925.64 » $449.21 31

Source Engineering Business Math

x= 46.

å( x ⋅ w)

Days, w 12 6 7 2 4 å w = 31

Weight, w 9 13 5 w å = 27

Points, x

Credits, w

x∙w

4 3 3 2 1

4 3 3 3 2 w å = 15

16 9 9 6 2 x ⋅ å( w) = 42

42 = 2.8 15

47. Source Score, x Weight, w Assignment 75 0.10 Class Participation 60 0.25 Practical 100 0.25 Theory exam 85 0.40 å w =1

x∙w 765 1053 450 x ⋅ ( å w) = 2268

å( x ⋅ w) 81.5 = = 81.5 åw 1

x∙w 7.5 15 25 34 å ( x ⋅ w) = 81.5

CHAPTER 2 │ DESCRIPTIVE STATISTICS

48. Grade A A B C D

å ( x ⋅ w) åw

49. Class 20–24 25–29 30–34 35–39

51.

x∙w 16 12 9 6 2 å( x ⋅ w) = 45

45 =3 15 Midpoint, x 22 27 32 37

Frequency, f 15 8 4 3 n = 30

x∙f 330 216 128 111 å ( x ⋅ f ) = 785

Midpoint, x 22 27 32 37

Frequency, f 8 16 5 1 n = 30

x∙f 176 432 160 37 ( x ⋅ å f ) = 805

å( x ⋅ f ) 805 = » 26.83 kilometers per hour 30 n

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

Credits, w 4 3 3 3 2 å w = 15

å( x ⋅ f ) 785 = » 26.178 kilometers per hour n 30

50. Class 20–24 25–29 30–34 35–39

Points, x 4 4 3 2 1

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

Frequency, f 40 72 78 90 84 42 31 22 18 4 n = 481

å( x ⋅ f ) 17,804.5 = » 37.02 years old 481 n

x∙f 180.0 1044.0 1911.0 3105.0 3738.0 2289.0 1999.5 1639.0 1521.0 378.0 å ( x ⋅ f ) = 17,804.5

CHAPTER 2 │ DESCRIPTIVE STATISTICS

52.

Class 0-49 50-99 100-149 150-199 200-249 250-299 300-349 350-399 400-449

Midpoint, x 24.5 74.5 124.5 174.5 224.5 274.5 324.5 374.5 424.5

Frequency, f 41 9 6 2 1 2 0 1 2 n = 64

å(x ⋅ f ) n

53. Class width = Class 127 – 161 162 – 196 197 – 231 232 – 266 267 – 301

x∙f 1004.5 670.5 747 349 224.5 549 0 374.5 849 å( x ⋅ f ) = 4768

4768 = 74.5 years old 64

Range 297 - 127 = = 34  35 Number of classes 5

Frequency, f 7 6 3 3 1

Midpoint 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 │ DESCRIPTIVE STATISTICS

Positively skewed

55. Class width = Class 40–45 46–51 52–57 58–63 64–69

Range 70 - 40 = =6 Number of classes 5 Midpoint Frequency, f 42.5 7 48.5 10 54.5 4 60.5 4 66.5 5 å f = 30

Shape: Positively skewed 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 å f = 30

CHAPTER 2 │ DESCRIPTIVE STATISTICS

å x 160 = = 20 n 8 18.0 19.4 19.6 20.2 20.4 

57. (a) x =

median =

20.5

20.9

20.5

20.9

20.2 + 20.4 = 20.3 2

å x 159 = = 19.875 n 8 18.0 19.4 19.6 20.0 20.2 

(b) x =

median =

20.4

20.0 + 20.2 = 20.1 2

(c) The median was affected more. å x 841.7 = » 46.76 n 18 9.2 9.3 10.9 12.5 15 15.5 17.4 17.7 21.7 23.3 28 28.3 30.4 30.9 60.7 68.9 74.8 367.2  

58. (a) x =

median =

21.7 + 23.3 = 22.5 2

å x 474.5 = » 27.91 n 17 9.2, 9.3, 10.9, 12.5, 15, 15.5, 17.4, 17.7, 21.7, 23.3, 28, 28.3, 30.4, 30.9, 60.7, 68.9, 74.8 The median is the middle value, 21.7. The mean was affected more.

(b) x =

å x 849 = = 44.68 n 19 7.3, 9.2, 9.3, 10.9, 12.5, 15, 15.5, 17.4, 17.7, 21.7, 23.3, 28, 28.3, 30.4, 30.9, 60.7, 68.9, 74.8, 367.2 The median is the middle value, 21.7. 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 │ 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. å x 3222 = = 358 n 9 147 177 336 360 375 393 408 504 522 median = 375

62. (a) x =

å 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 å x 152 x= = = 30.4 n 5 28 28 30 32 34 median = 30 mode = 28 (occurs 2 times) Car B å x 151 x= = = 30.2 n 5 29 29 31 31 31 median = 31 mode = 31 (occurs 3 times) Car C å x 151 x= = = 30.2 n 5 28 29 30 32 32 median = 30 mode = 32 (occurs 2 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 │ DESCRIPTIVE STATISTICS

å 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 median 22 8 36 6 6 7 7 7 8 41 3 4 6 7 mean 51 1 1 3 61 2 3 4 72 2 4 6 85 90 (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 3 observations (11, 13, 22). Delete the highest 10%, largest 3 observations (76, 85, 90). Find the 10% trimmed mean using the remaining 24 observations. å 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 │ DESCRIPTIVE STATISTICS

2.4 MEASURES OF VARIATION 2.4 TRY IT YOURSELF SOLUTIONS 1. Min = 23, or $23,000 and Max = 58, or $58,000, Range = Max – Min = 58 – 23 = 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. m = 41.5 , or $41,500 Salary, x

x–μ

(x – μ)2

–18.5

342.25

–12.5

156.25

–9.5

90.25

–1.5

2.25

–0.5

0.25

–0.5

0.25

7.5

56.25

8.5

72.25

10.5

110.25

16.5

272.25

å x = 415

å( x - m) = 0 å ( x - m ) = 1102.5 2

å ( x - m)

1102.5 » 110.3 N 10 1102.5 s = s2 = = 10.5, or $10,500 10 The population variance is about 110.3 and the population standard deviation is 10.5, or $10,500. 2

s =

å x 316 = = 39.5 n 8

Time, x

x-x

( x - x)

43 57 18 45 47 33 49 24 å x = 316

3.5 17.5 -21.5 5.5 7.5 -6.5 9.5 -15.5 å( x - m) = 0

12.25 306.25 462.25 30.25 56.25 42.25 90.25 240.25

å ( x - m ) = 1240 2

CHAPTER 2 │ DESCRIPTIVE STATISTICS

SS x = å ( x - x ) = 1240 2

å ( x - x)

s2 =

n -1

1240 » 177.1 7

1240 » 13.3 7

s = s2 =

4. Enter the data in a computer or a calculator. x » 19.8, s » 7.8 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 å x = 100 m=

x–μ –3 –3 –3 –3 –3 3 3 3 3 3 å ( x - m) = 0

å ( x - m ) = 90 2

å x 100 = = 10 N 10 2

(x – μ)2 9 9 9 9 9 9 9 9 9 9

å ( x - m) N

90 = 9 =3 10

6. 67.1 – 64.2 = 2.9 = 1 standard deviation Because 67.1 is one standard deviation above the mean height, the percent of heights between 64.2 inches and 67.1 inches is 34.13%. Approximately 34.13% of women ages 20-29 are between 64.2 and 67.1 inches tall. 7. 39.3 – 2(23.5) = 7.7 Because –7.7 does not make sense for an age, use 0. 39.3 + 2(23.5) = 86.3 1 1 1 1 - 2 = 1 - 2 = 1 - = 0.75 k 4 ( 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 86.3 years old. Because 80 < 86.3, and age of 80 lies within two standard deviations of the mean. So, the age is not unusual.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

x 0 1 2 3 4 5 6

f 10 19 7 7 5 1 1

xf 0 19 14 21 20 5 6 å xf = 85

n = 50 x=

å xf 85 = = 1.7 n 50

x-x

( x - x)

( x - x) f

–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

28.90 9.31 0.63 11.83 26.45 10.89 18.49

å( x - x) f = 106.5 2

å ( x - x) f 2

n -1

Class 1-99 100-199 200-299 300-399 400-499 500+

x 49.5 149.5 249.5 349.5 449.5 650

106.5 » 1.5 49

f 380 230 210 50 60 70 n= 1000

å xf 195,535 = » 195.5 n 1000

xf 18,810 34,385 52,395 17,475 26,970 45,500

å xf = 195,535

CHAPTER 2 │ DESCRIPTIVE STATISTICS

x-x

( x - x)

( x - x) f

–146.0 –46.0 54.0 154.0 254.0 454.5

21,316 2116 2916 23,716 64,516 206,570.25

8,100,080 486,680 612,360 1,185,800 3,870,960 14,459,917.5

å ( x - x) f = 28,715,797.5 2

å ( x - x) f 2

n -1

28,715,797.5 » 169.5 999

10. Los Angeles: x » 36.88 , s » 17.39 Dallas: x » 19.8 , s » 7.8 s 17.4 Los Angeles: CV = ⋅ 100% = ⋅100% » 47.2% x 36.9 s 7.8 Dallas: CV = ⋅100% = ⋅100% » 39.4% x 19.8 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 - m ) is the difference between an entry x and the mean of the data μ. 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 s 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. Copyright © 2019 Pearson Education Ltd.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

8. You must know that the distribution is approximately symmetric and bell-shaped. 9. Range = Max – Min = 75 – 40 = 35; Approximately 35, or $35,000 10. Range = Max – Min = 98 – 74 = 24 11. (a) Range = Max – Min = 938.5 – 870.7 = 67.8

(b) Range = Max – Min = 938.5 – 807.7 = 130.8 12. Changing the minimum value of the data set greatly affects the range. 13. Range = Max – Min = 14 – 1 = 13 å x 105 m= = » 5.83 N 18 x

x -m

( x - m)

8 13 2 11 5 5 14 1 6 3 13 3 5 4 2 4 3 3 å x = 105

2.17 7.17 −3.83 5.17 −0.83 −0.83 8.17 −4.83 0.17 −2.83 7.17 −2.83 −0.83 −1.83 −3.83 −1.83 −2.83 −2.83 å ( x - m) » 0

4.71 51.41 14.67 26.73 0.69 0.69 66.75 23.33 0.03 8.01 51.41 8.01 0.69 3.35 14.67 3.35 8.01 8.01

s =

å ( x - m)

N å ( x - m) N

å ( x - m) = 294.52

294.52 » 16.36 18

294.52 = 4.04 18

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

CHAPTER 2 │ DESCRIPTIVE STATISTICS

x -m

( x - m)

1.4 2330 2700 7870 1500 970 900 1740 4500 .09 å x = 22,511.5

2249.75 78.85 448.85 5618.85 751.15 1281.15 1351.15 511.15 2248.85 2251.06 å( x - m) » 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

s2 =

å ( x - m)

å ( x - m ) = 51, 257,584 2

51, 257,584 » 5,125,758.4 10

å ( x - m)

5,125,758.4 » 2264.0 10

15. Range = Max – Min = 23 – 15 = 8 å x 360 x= = = 18 n 20 x

x- x

( x - x)

18 18 19 15 16 18 18 19 21 21 23 18 16 19 15 16 15 17 20 18

0 0 1 −3 −2 0 0 1 3 3 5 0 −2 1 −3 −2 −3 −1 2 0

0 0 1 9 4 0 0 1 9 9 25 0 4 1 9 4 9 1 4 0

2 å x = 360 å ( x - x ) = 0 å ( x - x ) = 90

CHAPTER 2 │ DESCRIPTIVE STATISTICS 2

s =

å (x - x ) n -1

90 » 4.7 20 - 1

90 » 2.2 19

å( x - x ) n -1

16. Range = Max – Min = 29 – 21 = 8 å x 525 x= = = 25 n 21 x

x- x

( x - x)

21 21 21 23 23 24 24 24 24 25 25 25 25 26 26 26 27 28 29 29 29

−4 −4 −4 −2 −2 −1 −1 −1 −1 0 0 0 0 1 1 1 2 3 4 4 4

16 16 16 4 4 1 1 1 1 0 0 0 0 1 1 1 4 9 16 16 16

å x = 525

å (x - x) » 0

å( x - x) = 124

å ( x - x)

s =

n -1

å ( x - x)

124 = 6.2 21 - 1

n -1

124 » 2.49 20

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 A; An offer of $36,000 is within two standard deviations from the mean of Company A’s starting salaries, which makes it likely. The same offer is three standard deviations of the mean of Company B’s starting salaries, which makes the offer unlikely. Copyright © 2019 Pearson Education Ltd.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

20. Company D; An offer of $85,000 is within two standard deviations from the mean of Company D’s starting salaries, which makes it likely. The same offer is four standard deviations of the mean of Company C’s starting salaries, which makes the offer unlikely. 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: 1, 3, 3, 4, 7, 8, 9, 12, 16 27. Sample answer: 12, 12, 12, 12, 12

CHAPTER 2 │ DESCRIPTIVE STATISTICS

28. Sample answer: 4, 4, 8, 12, 12 29. (63, 71)  (67 - 1(4 ), 67 + 1(4 ))  ( x - s , x + s )

68% of the vehicles have speeds between 63 and 71 mph. 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 3 standard deviations from the mean. 34. $52 and $98 are unusual; $98 is very unusual because it is more than 3 standard deviations from the mean. 35. ( x - 3 s , x + 3 s )  (0, 6) are 3 standard deviations from the mean. 1 1 1 1 - 2 = 1 - 2 = 1 - = 0.89  At least 89% of the persons per house lie between 0 and 6. (3) 9 k If n = 60, at least (0.89)(60) ≈ 53 houses have between 0 and 6 persons. 36. ( x - 2 s , x + 2 s )  (16.18, 186.94) are 2 standard deviations from the mean. 1 1 1 1 - 2 = 1 - 2 = 1 - = 0.75  At least 75% of the eruption times lie between 16.18 and 186.94 (2) 4 k minutes. If n = 100, at least (0.75)(100) = 75 eruptions will lie between 16.18 and 186.94 minutes. 37. ( x - 2 s , x + 2 s )  (117, 133) are 2 standard deviations from the mean. 1 1 1 1 - 2 = 1 - 2 = 1 - = 0.75  At least 75% of the student heights are from 117 cm to 133 cm. (2) 4 k 1 1 1 = 1 - 2 = 1 - = 0.75 2 (2) 4 k At least 75% of the runs lie within 2 standard deviations of the mean.

38. 1 -

( x - 2 s , x + 2 s )  (- 2.86, 10.58)  (0, 10)

At least 75% of the runs per game scored by the Chicago Cubs during the 2016 World Series are from 0 to 10 (note that -2.86 and 10.58 do not make sense in the context of the data).

CHAPTER 2 │ DESCRIPTIVE STATISTICS

39.

xx

 x  x

 x  x f

0 1 2 3 4 5 6 7 8 9

10 12 9 2 2 3 1 4 2 5

0 12 18 6 8 15 6 28 16 45

−3.08 −2.08 −1.08 −0.08 0.92 1.92 2.92 3.92 4.92 5.92

9.4864 4.3264 1.1664 0.0064 0.8464 3.6864 8.5264 15.3664 24.2064 35.0464

94.864 51.9168 10.4976 0.0128 1.6928 11.0592 8.5264 61.4656 48.4128 175.232

n  50

 xf  154

40.

  x  x  f  463.38 2

å xf 154 = » 3.08 n 50 2

å(x - x ) f n -1

463.68 » 3.076 49

xx

 x  x

0 1

26 24

0 24  xf  24

–0.48 0.52

0.2304 0.2704

n  50

41.

 x  x f

5.9904 6.4896

 ( x  x ) f  12.48 2

å xf 24 = = 0.48 50 n 2

å(x - x ) f n -1

12.48 » 0.50 49

Class

xx

15,000 – 17,499 17,500 – 19,999 20,000 – 22,499 22,500 – 24,999 25,000 or more

16,249.5 18,749.5 21,249.5 23,749.5 26,249.5

9 10 16 11 6 n  52

146,245.5 187,495 339,992 261,244.5 157,497

4759.62 2259.62 240.38 2740.38 5240.38

å xf 1,092, 474 = » $21,009.12 n 52

 xf  1, 092, 474

 x  x 2

 x  x 2 f

22,653,982.54 203,885,842.9 5,105,882.54 51,058,825.4 57,782.54 924,520.64 7,509,682.54 82,606,507.94 27,461,582.54 164,769,495.20  ( x  x ) 2 f  503, 245,192.1

CHAPTER 2 │ DESCRIPTIVE STATISTICS

å ( x - x) f 2

s= 42.

n -1

Midpoint, x 2 7 12 17 22 27 32

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

503, 245,192.1 » $3141.27 51

f 5 12 24 17 16 11 5 n = 90

xf 10 84 288 289 352 297 160 å xf = 1480

å 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

å ( x - x) f = 5522.2240 2

å ( x - x) f 2

43.

n -1

5522.2240 » 7.9 89

xx

 x  x 2

 x  x 2 f

1 2 3 4

2 18 24 16 n  60

2 36 72 64

1.9 0.9 0.1 1.1

3.61 0.81 0.01 1.21

7.22 14.58 0.24 19.36

 xf  174

å xf 174 = » 2.9 n 60 å ( x - x) f 2

n -1

41.4 » 0.8 59

 ( x  x ) 2 f  41.4

CHAPTER 2 │ DESCRIPTIVE STATISTICS

Midpoint, x 70.5 92.5 114.5 136.5 158.5

44.

f 1 12 25 10 2

xf 70.5 1110.0 2862.5 1365.0 317.0

n = 50

å xf = 5725

å xf 5725 = = 114.5 n 50

x-x

( x - x)

( x - x) f

–44 –22 0 22 44

1936 484 0 484 1936

1936 5808 0 4840 3872

å ( x - x) f = 16, 456 2

å ( x - x) f 2

n -1

45. Denver: x =

16, 456 » 18.33 49

å x 552.3 = » 46.0 n 12

å ( x - x)

s =

n -1

å ( x - x)

220.89 » 20.08 11

220.89 » 4.48 11 s 4.48 CV = ⋅100% = ⋅100% » 9.7% x 46.0 å x 634.5 = » 52.9 Los Angeles: x = n 12 s=

n -1

å ( x - x)

s =

n -1

å ( x - x)

239.97 » 21.82 11

239.97 » 4.67 11 s 4.67 CV = ⋅100% = ⋅100% » 8.8% x 52.9 Salaries for entry level architects are more variable in Denver than in Los Angeles. s=

n -1

CHAPTER 2 │ DESCRIPTIVE STATISTICS

å x 52.7 = » 5.86 n 9 x -m ( x - m)2

12.2 7.7 7.2 6.8 5.3 4 4 2.8 2.7

6.34 1.84 1.34 0.94 −0.56 −1.86 −1.86 −3.06 −3.16

46. Africa: x =

s2 =

40.2 3.39 1.8 0.88 0.31 3.46 3.46 9.36 9.99 å ( x - x )2 = 72.85

å ( x - x )2 72.85 = » 9.11 n -1 8

å( x - x ) 2 72.85 = » 3.02 n -1 8 s 3.02 CV = = ⋅ 100% = 51.54% x 5.86 å x 206.4 Asia: x = = » 22.93 n 9 x -m ( x - m)2 x 33.3 10.37 107.54 24.8 1.87 3.5 24.2 1.27 1.61 22.7 −0.23 0.05 21.1 −1.83 3.35 21 −1.93 3.72 20.2 −2.73 7.45 20 −2.93 8.58 19.1 −3.83 14.67 s=

å ( x - x ) = 150.47 2

s2 =

å ( x - x ) 2 150.47 = » 18.81 n -1 8

å( x - x ) 2 150.47 = » 4.34 n -1 8 s 4.34 CV = = ⋅100% = 18.93% x 22.93 Wealth of billionaires is more variable for Africa than for Asia. s=

CHAPTER 2 │ DESCRIPTIVE STATISTICS

47. Ages: m =

å x 491 = » 22.32 N 22 2

s =

å ( x - m)

194.77 » 8.85 22

å ( x - x)

194.77 » 2.98 N 22 s 2.98 ⋅100% » 13.3% CV = ⋅100% = m 22.32 å x 1546 Heights: m = = » 70.27 N 22 2 å ( x - m) 134.36 s2 = = » 6.11 N 22 s=

å ( x - x)

134.36 » 2.47 N 22 s 2.47 ⋅100% » 3.5% CV = ⋅100% = m 70.27 Ages are more variable than heights for all members of the 2016 Women’s U.S. Olympic swimming team. s=

48. Ages: m =

å x 263 = = 26.3 N 10 2

s =

å ( x - m)

å ( x - x)

112.1 » 11.21 10

112.1 » 3.35 N 10 s 3.35 ⋅100% » 12.7% CV = ⋅100% = m 26.3 å x 853 Weights: m = = = 85.3 N 10 2 å ( x - m) 5850.1 2 s = = » 585.01 N 10 s=

å ( x - x)

5850.1 » 24.19 N 10 s 24.19 ⋅100% » 28.4% CV = ⋅100% = m 85.3 Weight classes are more variable than ages for all members of the 2016 Men’s U.S. Olympic wrestling team. s=

CHAPTER 2 │ DESCRIPTIVE STATISTICS

49. Male: x =

x 70 72 75 69 64 75 60 71 73 72

å x 701 = = 70.1 n 10 ( x - x )2 x-x −0.1 0.01 1.9 3.61 4.9 24.01 −1.1 1.21 −6.1 37.21 4.9 24.01 −10.1 102.01 0.9 0.81 2.9 8.41 1.9 3.61

å ( x - x ) = 204.9 2

s2 =

å ( x - x )2 204.9 = » 22.7667 n -1 9

å( x - x ) 2 204.9 = » 4.7714 n -1 9 s 4.7714 CV = ⋅100% = ⋅100% »~ 6.81% x 70.1 s=

å x 648 = = 64.8 n 10 ( x - x )2 x-x

65 66 68 61 64 69 65 63 62 65

0.2 1.2 3.2 −3.8 −0.8 4.2 0.2 −1.8 −2.8 0.2

Female: x =

0.04 1.44 10.24 14.44 0.64 17.64 0.04 3.24 7.84 0.04

å ( x - x ) = 55.6 2

s2 =

å ( x - x ) 2 55.6 = » 6.1778 n -1 9

å( x - x ) 2 55.6 = » 2.4855 n -1 9

CHAPTER 2 │ DESCRIPTIVE STATISTICS

CV =

s 2.4855 ⋅100% = ⋅100% » 3.84% x 64.8

Weight averages are more variable for males than for females. 50. Male: x =

x 167 165 160 181 190 175 178 164 168 155

å x 1703 = = 170.3 n 10 ( x - x )2 x-x −3.3 10.89 −5.3 28.09 −10.3 106.09 10.7 114.49 19.7 388.09 4.7 22.09 7.7 59.29 −6.3 39.69 −2.3 5.29 −15.3 234.09

å ( x - x ) = 1008.1 2

s2 =

å ( x - x ) 2 1008.1 = » 112.0111 n -1 9

å( x - x ) 2 1008.1 = » 10.5835 n -1 9 s 10.5835 CV = ⋅ 100% = ⋅ 100% » 6.21% x 170.3 å x 1518 Female: x = = = 151.8 n 10 ( x - x )2 x-x x 145 −6.8 46.24 170 18.2 331.24 138 −13.8 190.44 140 −11.8 139.24 151 −0.8 0.64 171 19.2 368.64 142 −9.8 96.04 151 −0.8 0.64 153 1.2 1.44 157 5.2 27.04 s=

å ( x - x ) = 1201.6 2

CHAPTER 2 │ DESCRIPTIVE STATISTICS

s2 =

å ( x - x ) 2 1201.6 = » 133.5111 n -1 9

å( x - x ) 2 1201.6 = » 11.5547 n -1 9 s 11.5547 CV = ⋅ 100% = ⋅100% » 7.61% x 151.8 Height averages are slightly more variable for females than for males. s=

51. (a) Answers will vary.

(b) x 18 18 19 15 16 18 18 19 21 21 23 18 16 19 15 16 15 17 20 18

å x = 360

x2 324 324 361 225 256 324 324 361 441 441 529 324 256 361 225 256 225 289 400 324 2 x å = 6570 2

å x2 s=

(å x)

n -1

(360)2 20 = 90 » 2.2 20 -1 19

(6570) =

CHAPTER 2 │ DESCRIPTIVE STATISTICS

52. (a)

x 18 18 19 15 16 18 18 19 21 21 23 18 16 19 15 16 15 17 20 18

x-x 0 0 1 −3 −2 0 0 1 3 3 5 0 −2 1 −3 −2 −3 −1 2 0

x-x

å x = 360

å (x - x ) = 0

å x - x = 32

å x-x

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

32 = 1.6 n 20 The mean absolute deviation is less than the standard deviation calculated in Exercise 15. MAD =

CHAPTER 2 │ DESCRIPTIVE STATISTICS

(b) x

x-x

21 21 21 23 23 24 24 24 24 25 25 25 25 26 26 26 27 28 29 29 29

−4 −4 −4 −2 −2 −1 −1 −1 −1

å x = 525

å (x - x ) = 0

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

å x- x

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

å x - x = 40

40 = 1.90 n 21 The mean absolute deviation is less than standard deviation calculated in Exercise 16. MAD =

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

CHAPTER 2 │ DESCRIPTIVE STATISTICS

(d) Adding a constant k to, or subtracting it from, each entry makes the new sample mean x + k , or x - k , with the sample standard deviation being unaffected.

55. (a) P =

(b) P =

(d) P = (e) P =

3( x - median ) s 3( x - median ) s 3( x - median ) s 3( x - median ) s 3( x - median ) s

= =

3(17 -19) 2.3 3(32 - 25) 5.1

» -2.61; The data are skewed left. » 4.12; The data are skewed right.

3(9.2 - 9.2) 1.8 3(42 - 40) 6.0

= 0; The data are symmetric.

= 1; The data are skewed right.

3(155 -175) 20.0

= -3; The data are skewed left.

1 1 1 1 = 0.99  1- 0.99 = 2  k 2 = k = = 10 2 0.01 0.01 k k At least 99% of the data in any data set lie within 10 standard deviations of the mean.

56. 1-

2.5 MEASURES OF POSITION 2.5 TRY IT YOURSELF SOLUTIONS 1. Order the data from least to greatest. The median (or Q 2 ) is 30. 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 Q 2 and the third quartile is the median of the data entries to the right of Q 2 . Q 1 = 2 3, Q 2 = 3 0, Q 3 = 3 5

About one-quarter of the winning scores were 23 points or less, about one-half were 30 points or less, and about three-quarters were 35 points or less. 2. Enter data Q1  23.5, Q2  30, Q3  41 About one-quarter of these universities charge tuition of $23,500 or less; about one-half charge $30,000 or less; and about three-quarters charge $41,000 or less. 3.

Q1 = 23, Q 3 = 35 IQ R = Q 3 - Q1 = 3 5 - 2 3 = 1 2 Q1 - 1.5(IQ R ) = 23 - 1.5(12) = 5 ; Q 3 + 1.5(IQ R ) = 35 + 1.5(12) = 53 The score 55 is greater than Q 3 + 1.5(IQ R ) . So, 55 is an outlier.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

4. Min = 14, Q1 = 23, Q 2 = 30, Q 3 = 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 10th percentile is 19.5. About 10% of the winning scores were 19.5 or less. 6 17,18,19,20,20,23,24,26,29,29,29,30,30,34,35,36,38,39,39,43,44,44,44,45,45 7 data entries are less than 26 number of data entries less than 26 7 Percentile of 26 = ⋅100 = ⋅100 = 28th percentile total number of data entries 25 The tuition cost of $26,000 is greater than 28% of the other tuition costs. 7.

m = 70, s = 8

x - m 60 - 70 = = -1.25 s 8 x - m 71 - 70 x = 71: z = = = 0.125 s 8 x - m 92 - 70 x = 92 : z = = = 2.75 s 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 = 60 : z =

5 feet = 5(12) = 60 inches x - m 60 - 69.9 x - m 60 - 64.3 Man: z = = = -3.3 ; Woman: z = = » -1.7 s s 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 scored higher than 89% of the students who took the Fundamentals of Engineering exam. 4. The student has a higher IQ score than 91% of the students in the same age group.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

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 a z-score of zero when the x-value equals the mean. 11. (a) 35 39 40 41 42 43 45 46 46 47 48 48 48 49 65   

(b) IQR = Q3 - Q1 = 48 - 41 = 7

(c) Q1 – 1.5(IQR) = 41 – 1.5(7) = 30.5, Q3 + 1.5(IQR) = 48 + 1.5(7) = 58.5. The data entry 65 is an outlier. 12. (a) 30 31 32 34 35 35 37 37 38 39 40 41 41 42 44 46 48 49 49 49    Q1 = 35 Q2 = 39.5 Q3 = 45

(b) IQR = Q3 - Q1 = 45.5 - 35 = 10.5 (c) Q1 – 1.5(IQR) = 35 – 1.5(10.5) = 19.25, Q3 + 1.5(IQR) = 45.5 + 1.5(10.5) = 61.25 There are no outliers. 13. Min = 0, Q1 = 2, Q2 = 5, Q3 = 8, Max = 10 14. Min = 500, Q1 = 580, Q2 = 605, Q3 = 630, Max = 720

upper half lower half       15. (a) 45 48 51 52 52 55 56 59 63 64 64 65 9 75 79    Q1 Q2 Q3 Min = 45, Q1 = 52, Q2 = 49, Q3 = 65, Max = 79

CHAPTER 2 │ DESCRIPTIVE STATISTICS

(b)

16. (a)

upper half lower half     220 228 228 230 232 238 240 242 250 252 260 262    Q1 Q2 Q3 Min = 220, Q1 = 229, Q2 = 239, Q3 = 251, Max = 262

(b)

17. (a)

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

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, Q 1 = 3 , Q 2 = 5 , Q 3 = 8 , Max = 9

Q3 = 8

CHAPTER 2 │ DESCRIPTIVE STATISTICS

(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, Q 2 = 3, Q 3 = 5.75, Max = 9

24. Min = 2, Q1 = 4, Q 2 = 6, Q 3 = 9, Max = 15

25. Min = 131, Q1 = 153.5, Q 2 = 171.5, Q 3 = 195.25, Max = 238

CHAPTER 2 │ DESCRIPTIVE STATISTICS

26. Min = 1.02, Q1 = 12.09, Q 2 = 16.14, Q 3 = 23.81, Max = 63.22

27. (a) 2 hours

(b) About 50%

28. (a) $16.14 thousand

(b) About 75%

(d) About 25%

29. About 158; About 70% of quantitative reasoning scores on the Graduate Record Examination

are less than 158. 30. About 150; About 40% of quantitative reasoning scores on the Graduate Record Examination

are less than 150. 31. About 8th percentile; About 8% of quantitative reasoning scores on the Graduate Record

Examination are less than 140. 32. About 97th percentile; About 97% of quantitative reasoning scores on the Graduate Record

Examination are less than 170. 33. Percentile of 47 =

number of data entries less than 47 12 ⋅100 = ⋅100 = 40 th percentile total number of data entries 30

34. Percentile of 57 =

number of data entries less than 57 24 ⋅100 = ⋅100 = 80 th percentile total number of data entries 30

35. 75th percentile = Q3 = 56; Ages below 56 are 28, 35, 38, 40, 41, 41, 42, 43, 43, 43, 44, 45, 47, 47, 48, 50, 50, 50, 53, 54, 54.

36. 25th percentile = Q1 = 43; Ages above 43 are 44, 45, 47, 47, 48, 50, 50, 50, 53, 54, 54, 56, 56, 56, 57, 57, 61, 61, 65, 66.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

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. 42. A  z =-1.54 B  z = 0.77 C  z = 1.54 None of the z-scores are unusual. 43. Christopher Froome: x = 31  z =

x - m 31 - 27.9 = » 0.94 s 3.3

Not unusual; The z-score is 0.94, so the age of 31 is about 0.94 standard deviation above the mean. 44. Jan Ullrich: x = 24  z =

x - m 24 - 27.9 = » -1.18 s 3.3

Not unusual; The z-score is 1.18, so the age of 24 is about 1.18 standard deviations below the mean. 45. Antonin Magne: x = 27  z =

x - m 27 - 27.9 = » -0.27 s 3.3

Not unusual; The z-score is 0.27, so the age of 27 is about 0.27 standard deviation below the mean. 46. Firmin Lambot: x = 36  z =

x - m 36 - 27.9 = » 2.45 s 3.3

Unusual; The z-score is 2.45, so the age of 36 is about 2.45 standard deviations above the mean.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

47. Henri Cornet: x = 20  z =

x - m 20 - 27.9 = » -2.39 s 3.3

Unusual; The z-score is 2.39, so the age of 20 is about 2.39 standard deviations below the mean. 48. Philippe Thys: x = 28  z =

x - m 28 - 27.9 = » 0.03 s 3.3

Not unusual; The z-score is 0.03, so the age of 28 is about 0.03 standard deviation above the mean. x - m 1250 -1000 = = 2.5 s 100 x - m 1175 - 1000 x = 1175  z = = = 1.75 s 100 x - m 950 - 1000 x = 950  z = = = -0.5 s 100 The lady bug with a life span of 1,250 days has an unusually long life span.

49. (a) x = 1250  z =

x - m 1150 - 1000 = = 1.5  about 93rd percentile s 100 x - m 910 - 1000 x = 910  z = = = -0.9  about 18th percentile s 100 x - m 845 - 1000 x = 845  z = = = -1.55  about 6th percentile s 100

(b) x = 1150  z =

x - m 13,500 -15,000 = » -1.2 s 1,250 x - m 17,000 -15,000 x = 17,000  z = = = 1.6 s 1, 250 x - m 18,500 -15,000 x = 18,500  z = = = 2.8 s 1,250 The bearing with a life span of 18,500 cycles has an unusually long life span.

50. (a) x = 13,500  z =

x - m 12,500 -15,000 = = -2  about 2nd percentile s 1,250 x - m 14,750 -15,000 x = 14,750  z = = = -0.2  about 42nd percentile s 1, 250 x - m 19,000 -15,000 x = 19,000  z = = = 3.2  about 99.9th percentile s 1, 250

(b) x = 12,500  z =

x - m 53 - 43.7 = » 1.07 s 8.7 x - m 46 - 50.4 Jack Nicholson: x = 46  z = = » -0.32 s 13.8

51. Robert Duvall: x = 53  z =

CHAPTER 2 │ DESCRIPTIVE STATISTICS

The age of Robert Duvall was about 1 standard deviation above the mean age of Best Actor winners, and the age of Jack Nicholson was less than 1 standard deviation below the mean age of Best Supporting Actor winners. Neither actor’s age is unusual. x - m 37 - 43.7 = » -0.77 s 8.7 x - m 67 - 50.4 Morgan Freeman: x = 67  z = = » 1.20 s 13.8 The age of Jamie Foxx was less than 1standard deviation below the mean age of Best Actor winners, and the age of Morgan Freeman was between 1and 2 standard deviations above the mean age of Best Supporting Actor winners. Neither actor’s age is unusual.

52. Jamie Foxx: x = 37  z =

x - m 62 - 43.7 = » 2.10 s 8.7 x - m 56 - 50.4 Gig Young: x = 56  z = = » 0.41 s 13.8 The age of John Wayne was more than 2 standard deviations above the mean age of Best Actor winners, which is unusual. The age of Gig Young was less than 1 standard deviation above the mean age of Best Supporting Actor winners, which is not unusual.

53. John Wayne: x = 62  z =

x - m 76 - 43.7 = » 3.71 s 8.7 x - m 77 - 50.4 John Gielgud: x = 77  z = = » 1.93 s 13.8 The age of Henry Fonda was more than 3 standard deviations above the mean age of Best Actor winners, which is very unusual. The age of John Gielgud was less than 2 standard deviations above the mean age of Best Supporting Actor winners, which is not unusual.

54. Henry Fonda: x = 76  z =

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.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

(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)

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)

CHAPTER 2 │ DESCRIPTIVE STATISTICS

61. (a) 1, 23, 29, 35, 37, 46, 46, 47, 49, 52, 53, 59, 83 Q 1 = 3 2 , Q 2 = 46 , Q 3 = 52.5 IQ R = Q 3 - 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 , Q 2 = 49 , Q 3 = 5 5 IQ R = Q 3 - 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 REVIEW EXERCISE SOLUTIONS

Class width =

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

Max - Min 55 - 26 = = 5.8  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 å =1 n

Cumulative frequency 5 9 15 18 20

CHAPTER 2 │ DESCRIPTIVE STATISTICS

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

å f = 25

Class width =

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

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

å f = 25

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

CHAPTER 2 │ DESCRIPTIVE STATISTICS

Class width =

Class

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

Max - Min 166 - 79 = = 14.5  15 Number of classes 6

Mid point 86 101 116 131 146 161

Frequency,

9 12 5 4 2 1

å f = 33

CHAPTER 2 │ DESCRIPTIVE STATISTICS

7. Because the data entries go from a low of 22 to a high of 65, 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 2 3 8 8 Key: 2 | 2 = 22 3 2 3 6 8 8 9 9 4 1 1 3 6 9 5 0 3 4 6 7 6 3 5 5 Sample answer: Most U.S. cities have a pollution index from 32 to 57. 8.

Sample answer: Most U.S. cities have a pollution index from 32 to 57. 9.

Location

Sleeping Leisure and Sports Working Educational Activities Other

Frequency

8.8 4.0 2.3 3.5 5.4

å f = 24

Relative frequency 0.3667 0.1667 0.0958 0.1458 0.2250 f å n =1

Degrees 132 60 34 52 81

Sample answer: Full-time university and college students spend the least amount of time working. 10.

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

CHAPTER 2 │ DESCRIPTIVE STATISTICS

11.

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

Sample answer: The real unemployment rate varied by a couple of percentage points from 2005 to 2008, then increased dramatically from 2008 to 2010, and then decreased from 2011 to 2016.

å x 295 = = 29.5 n 10 21.0 24.0 26.0 28.0 29.5 29.5 31.0 33.0 35.5 37.5

13. x =

median = 29.5 Mode = 29.5 (occurs 2 times) 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. 15.

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

å ( x ⋅ w) åw

Score, x 78 72 86 91 87 80

Weight, w 0.15 0.15 0.15 0.15 0.15 0.25

å w =1

82.1 = 82.1 1

x∙w 11.7 10.8 12.9 13.65 13.05 20

å ( x ⋅ w) = 82.1

16.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

Source Test 1 Test 2 Test 3 Test 4

x= 17.

å ( x ⋅ w) åw

Score, x 96 85 91 86

Weight, w 0.2 0.2 0.2 0.4

å w =1

n = 20

å( x ⋅ f ) n

x 0 1 2 3 4 5 6

18.

f 13 9 19 8 5 2 4

å( x ⋅ f ) n

x∙f 142.5 138 243 139.5 105

å ( x ⋅ f ) = 768

768 » 38.4 20

n = 60

å ( 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 19.2 17 18.2 34.4

x∙f 0 9 38 24 20 10 24

å ( x ⋅ f ) = 125

125 » 2.1 60

19. Skewed right 20. Skewed right 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.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

25. Range = Max – Min = 12 – 4 = 8 å x 113 = » 8.1 m= 14 N x-m x 7 –1.1 5 –3.1 12 3.9 12 3.9 6 –2.1 9 0.9 11 2.9 4 –4.1 7 –1.1 6 –2.1 8 –0.1 7 –1.1 10 1.9 9 0.9

( x - m) 2 1.21 9.61 15.21 15.21 4.41 0.81 8.41 16.81 1.21 4.41 0.01 1.21 3.61 0.81

å x = 113 å ( x - m ) » 0 å ( x - m ) = 82.94 2

s2 =

å( x - m)2 82.94 = » 5.92 N 14

å( x - m ) 2 82.94 = » 2.43 N 14

26. Range = Max – Min = 83 – 56 = 27 å x 554 m= = » 69.25 8 N

x-m

61 80 68 83 78 66 62 56

8.25 10.75 1.25 13.75 8.75 3.25 7.25 13.25

å x = 554 å ( x - m ) » 0 2

s =

å ( x - m)

689.5 » 86.19 8

å ( x - m) N

689.5 » 9.28 8

( x - m)

68.063 115.563 1.563 189.063 76.563 10.563 52.563 175.563

å ( x - m ) = 689.5 2

CHAPTER 2 │ DESCRIPTIVE STATISTICS

27. Range = Max – Min = $1847 – $585 = $1262 å x 16,595 x= = » $1,106.27 n 16

x 1514 1473 1847 1746 1545 994 883 705 612 1204 612 585 936 1122 1514 816

x- x

407.73 366.73 740.73 639.73 438.73 –112.27 –223.27 –401.27 –494.27 97.73 –494.27 –521.27 –170.27 15.73 407.73 –290.27

( x - x )2 166243.7529 134490.8929 548680.9329 409254.4729 192484.0129 12604.5529 49849.4929 161017.6129 244302.8329 9551.1529 244302.8329 271722.4129 28991.8729 247.4329 166243.7529 84256.6729

å x = 16,594 å ( x - x ) » 0 å ( x - x ) = 2,558,000.934 2

s2 =

å ( x - x )2 2,558, 000.934 = » 182, 714.35 n -1 14

å( x - x ) 2 2,558,000.934 = » $427.45 n -1 14

28. Range = Max – Min = 58,298 – 48,250 = $10,048 å x 3,12,496 x= = » $39,062 8 n

x 37,224 40,964 43,724 36,188 38,882 38,157 39,914 37,443

x- x

–1,838 1,902 4,662 –2,874 –180 –905 852 –1,619

( x - x )2 33,78,244 3617604 21734244 8259876 32400 819025 725904 2621161

å x = 3,12, 496 å ( x - x ) » 0 å ( x - x ) = 41,188, 458 2

s2 =

å ( x - x )2 41,188, 458 = » 5,884, 065.43 n -1 7

å( x - x ) 2 41,188,458 = » $2,425.71 n -1 7

CHAPTER 2 │ DESCRIPTIVE STATISTICS

29. 95% of the distribution lies within 2 standard deviations of the mean. x - 2 s = 18 - (2)(2.5) = 13 x + 2 s = 18 + (2)(2.5) = 23

95% of the distribution lies between $13 and $23. 30. (12.00, 21.00 )  (16.50 - 3(1.50 ),16 .50 + 3(1.50 ))  ( x - 3 s , x + 3 s ) 99.7% of the wages lie between $12.00 and $21.00 per day. 31. ( x - 2 s , x + 2 s )  (58, 92) are 2 standard deviations from the mean. 1 1 1 - 2 = 1 - 2 = 0.75 (2) k At least (20)(0.75) = 15 students scored between 58 and 92. 32. ( x - 2 s , x + 2 s )  (2.3, 17.5 ) are 2 standard deviations from the mean. 1-

1 1 1 = 1 - 2 = 1 - = 0.75 2 k 4 ( 2)

At least (135 )(0.75 ) » 101 shuttle flights lasted between 2.3 days and 17.5 days. 33. x =

å xf 99 = » 2.5 n 40

x- x

( x - x)

( x - x) f

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

6.25 18.00 3.25 2.50 11.25 18.75

n = 40

å xf = 99

å ( x - x) f 2

n -1

60 » 1.2 39

å( x - x) f = 60 2

CHAPTER 2 │ DESCRIPTIVE STATISTICS

34. x =

å xf 61 = » 2.4 n 25

x- x

( x - x)

( x - x) f

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

23.04 9.80 0.32 3.24 2.56 20.28 12.96

n = 25

å xf = 61

å ( x - x) f

å( x - x) f = 72.2 2

n -1

35. Company A: x = x

2.3 2.9 3.9 3.1 2.2 1.5 2.2 2.7 1.8 å x = 22.6

72.2 » 1.7 24

å x 22.6 = » 2.511 n 9

x-x –0.211 0.389 1.389 0.589 –0.311 –1.011 –0.311 0.189 –0.711 å(x - x ) = 0

å ( x - x ) = 4.2287

å( x - x ) 2 4.2287 = » 0.7271 n -1 8 s 0.7271 CV = = ⋅100% » 28.96% x 2.511 å x 25.7 = » 2.856 Company B: x = n 9 s=

( x - x )2 0.0445 0.1513 1.9293 0.3469 0.0967 1.0221 0.0967 0.0357 0.5055 2

CHAPTER 2 │ DESCRIPTIVE STATISTICS

x-x –0.056 –0.356 0.044 0.744 0.044 0.444 0.444 –0.056 –1.256 å(x - x ) = 0

x 2.8 2.5 2.9 3.6 2.9 3.3 3.3 2.8 1.6

å x = 25.7

( x - x )2 0.003136 0.126736 0.001936 0.553536 0.001936 0.197136 0.197136 0.003136 1.577536

å ( x - x ) = 2.6622 2

å( x - x ) 2 2.6622 = » 0.5769 n -1 8 s 0.5769 CV = = ⋅100% » 20.2% x 2.856 Dividends are more variable for Company A than Company B. s=

36. Heights: m =

å x 516 = = 64.5 inches 8 N

x 62 58 60 64 70 62 72 68

x-m

å x = 516

å (x - m) = 0

–2.5 –6.5 –4.5 –0.5 5.5 –2.5 7.5 3.5

( x - m)2 6.25 42.25 20.25 0.25 30.25 6.25 56.25 12.25 å ( x - m ) 2 = 174

å ( x - m) 2 174 = » 4.664 inches N 8 s 4.664 CV = = ⋅100% » 7.2% m 64.15 s=

Weights: m =

å x 876 = » 109.5 lbs 8 N

CHAPTER 2 │ DESCRIPTIVE STATISTICS

x 92 80 82 106 136 96 146 138

x-m

å x = 876

å (x - m) = 0

–17.5 –29.5 –27.5 –3.5 26.5 –13.5 36.5 28.5

( x - m)2 306.25 870.25 756.25 12.25 702.25 182.25 1332.25 812.25 ( x å - m ) 2 = 4974

å( x - m ) 2 4974 = » 24.935 lbs N 8 s 24.935 CV = = ⋅100% » 22.8% m 109.5 Weights are more variable than heights for all students of the secondary school. s=

37. 16, 16, 22, 22, 22, 22, 25, 25, 30, 30, 34, 34, 35, 35, 35, 41, 46, 50, 52, 56, 58, 107, 112, 119, 124, 136 Min = 16, Q1 = 25 , Q 2 = 35 , Q 3 = 5 6 , Max = 136 38. IQ R = Q 3 - Q1 = 56 - 25 = 31 miles per gallon 39.

40. Count the number of entries that are 56 or below: 20 vehicles 41. 21.0 24.0 26.0 28.0 29.5 29.5 31.0 33.0 35.5 37.5

median = 29.5 IQ R = Q 3 - Q1 = 33 - 26 = 7 inches

42. 145, 156, 167, 172, 173, 184, 185, 190, 190, 192, 195, 197, 205, 208, 212, 227, 228, 240 Min = 145, Q1 = 173 , Q 2 = 191 , Q 3 = 208 , Max = 240

43. The 15th percentile means that 15% had an income of $32 or less. So, 85% earned more than $32.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

44. If there are 115 stations with a larger daily audience, then this station has the 721 – 115 = 606th largest 606 ⋅100 = 84th percentile audience. The percentile of 606 = 721

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 =

11,300 -11,830 = -0.22 2370 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.

48. z =

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

Max - Min 157 -101 = = 11.2  12 Number of classes 5

Class

Midpoint

101-112 113-124 125-136 137-148 149-160

106.5 118.5 130.5 142.5 154.5

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

Frequency, f

3 11 8 3 2

å f = 27 (b)

Relative frequency 0.11 0.41 0.30 0.11 0.07 f å n =1

Cumulative frequency 3 14 22 25 27

CHAPTER 2 │ DESCRIPTIVE STATISTICS

(c)

(d) Skewed right (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 , Q 2 = 124 , Q 3 = 132 , Max = 157

CHAPTER 2 │ DESCRIPTIVE STATISTICS

å 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

å ( x - x) f = 4362.72 2

å xf = 3403.5

n = 27

å ( x - x) f

3. (a)

n -1

4362.72 » 13.0 26

Category Metals Metalloids Halogens Noble gases Rare earth elements Other nonmetals

Frequency 57 7 5 6 30 7 n = 112

(b)

Relative frequency 0.5089 0.0625 0.0446 0.0536 0.2679 0.0625 f å n =1

Degrees 183 23 16 19 96 23

100

CHAPTER 2 │ DESCRIPTIVE STATISTICS

å 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 =

(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

å ( x - x) = 706,815 2

å ( x - x)

s =

n -1

å ( x - x)

706,815 » 47,120.9 15

n -1

706,815 » 217.1 15

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

x - 2 s = 180,000 - 2 ⋅ 15,000 = $150,000 x + 2 s = 180,000 + 2 ⋅ 15,000 = $210,000 95% of the new home prices fall between $150,000 and $210,000.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

101

x - x 225,000 -180,000 = = 3.0 s 15,000 Unusual; The z-score is 3, so a new home price of $225,000 is about 3 standard deviations above the mean. x - x 80,000 -180,000 = » -6.67 (b) x = 80, 000 : z = s 15,000 Unusual; The z-score is 6.67, so a new home price of $80,000 is about 6.67 standard deviations below the mean.

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

x - x 200,000 -180,000 = » 1.33 s 15,000 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 s 15,000 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. 59 68 68 68 68 69 69 71 73 74 75 78 78 79 81 84 84 86 86 86 87 87 89 89 91 93 94 95 95 103 Q1 = 71

Q 2 = 8 2 .5

Q3 = 89

Min = 59, Q1 = 71 , Q 2 = 8 2 .5 , Q 3 = 8 9 , Max = 103

CUMULATIVE REVIEW FOR CHAPTERS 1 AND 2 1. Systematic sampling is used because every fortieth 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.

102

CHAPTER 2 │ DESCRIPTIVE STATISTICS

4. $68,232 is a parameter. The median salary is based on all marketing account executives. 5. 88% is a statistic. The percent is based on a subset of the population. 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 1500 s x - x 85,600 - 86,500 x = $85,600 : z = = = -0.6 s 1500 x - x 82,750 - 86,500 x = $82,750 : z = = = -2.5 1500 s The salaries of $93,500 and $82,750 are unusual.

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

7. Population: Collection of opinions of all college and university admissions directors and enrollment officers Sample: Collection of opinions of the 339 college and university admission directors and enrollment officers surveyed 8. Population: Reasons for pain reliever use of all Americans ages 12 or older Sample: Reasons for pain reliever use of the 67,901 Americans ages 12 or older 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.

CHAPTER 2 │ DESCRIPTIVE STATISTICS

13. (a)

103

Q1 Q2 0 0 0 0 0 0 0 1 1 1 2 2 2 2 2 3 3 3 3 4 4 6 6 7 9 11 11 12 15 16 16 23 23 27 31 31 32 32 40 44 45 46 47 48 50 55 67 87 90 99 Q3 Min = 0, Q1 = 2 , Q2 = 10 , Q3 = 32 , Max = 99

(b) The distribution of the number of tornadoes is skewed right. 14.

Source Test 1 Test 2 Test 3 Test 4 Test 5

å ( x ⋅ w) åw

15. (a) x =

Score, x 85 92 84 89 91

Weight, w 0.15 0.15 0.15 0.15 0.40

x∙w 12.75 13.80 12.60 13.35 36.40

å w =1

å ( 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

å( x - x) = 18.7289 2

104

CHAPTER 2 │ DESCRIPTIVE STATISTICS

å ( x - x)

s2 =

n -1

å ( x - x)

18.7289 » 2.34 8

n -1

18.7289 » 1.53 8

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. 17. Class width = Class limits 0-8 9-17 18-26 27-35 36-44 45-53 54-62 63-71

Max - Min 64 - 0 = =89 Number of classes 8

Midpoint

4 13 22 31 40 49 58 67

Class boundaries –0.5-8.5 8.5-17.5 17.5-26.5 26.5-35.5 35.5-44.5 44.5-53.5 53.5-62.5 62.5-71.5

Frequency

20 7 6 1 2 1 2 1

å f = 40

Relative frequency 0.500 0.175 0.150 0.025 0.050 0.025 0.050 0.025 f å n =1

18. The distribution is skewed right. 19.

Class with greatest frequency: 0 – 8 Classes with least frequency: 27 – 35, 45 – 53, and 63 – 71

Cumulative frequency 20 27 33 34 36 37 39 40

CHAPTER 2 │ DESCRIPTIVE STATISTICS

CHAPTER 2 TEST SOLUTIONS 1. (a) x =

å x 964 = » 80.3 12 n

63 67 68 72 73 81 85 87 87 88 94 99 

median = 83 mode = 87 (occurs twice) The median best represents the center of the data. (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

å ( x - x) » 0

å ( x - x) = 1458.68

å ( x - x)

s =

n -1

å ( x - x)

1458.68 » 132.6 11

1458.68 » 11.5 n -1 11 s 11.5 ⋅100% » 14.3% (c) CV = ⋅100% = 80.3 x s=

(d) Points scored

Key: 6 3 = 63

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

105

106

CHAPTER 2 │ DESCRIPTIVE STATISTICS

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

å f = 20 (b)

(c)

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

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

Cumulative frequency 2 6 11 16 19 20

CHAPTER 2 │ DESCRIPTIVE STATISTICS

å 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

å xf = 2254

n = 20

å ( x - x) f

å ( x - x) f » 41,847.2 2

n -1

41,847.2 » 46.9 19

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 5.

f 6 26 42 48

Certification Diamond Multi-Platinum Platinum Gold

å f = 122 (a)

(b)

Relative Frequency 0.049 0.213 0.344 0.393 f å n »1

Angle 18 77 124 141

å = 360

107

108

CHAPTER 2 │ DESCRIPTIVE STATISTICS

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  42 46 51 55  56 58 59 60  62 65 68

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

(b) About 75% of the professors are over the age of 40. 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.

110

CHAPTER 3 │ 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

(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 │ PROBABILITY

111

6. The event is “read only digital books.” The frequency is 91. Total Frequency = 1490 91 P(read only digital books) = » 0.061 1490 7. Frequency = 254 Total of the Frequencies = 975 254 P(age 36 to 49) = » 0.261 975 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 156 = 0.16 975 156 819 P(age is not 18 to 22) = 1 = = 0.84 975 975 819 21 or 0.84 = 975 25

9. P(age 18 to 22) =

10. There are 5 outcomes in the event: {T1, T2, T3, T4, T5}. 5 P(tail and less than 6) = » 0.313 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.

112

CHAPTER 3 │ PROBABILITY

(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. 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 tenth 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 19 23 19 4 = - = 23 23 23 23

18. P ( E ') = 1 - P ( E ) = 1 - 0.55 = 0.45 19. P ( E ') = 1 - P ( E ) = 1 - 0.03 = 0.97 2 7 2 5 20. P ( E ') = 1 - P ( E ) = 1 - = - = 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 3 4 3 1 23. P ( E ) = 1 - P ( E ') = 1 - = - = 4 4 4 4

14. c

15. a

16. e

CHAPTER 3 │ PROBABILITY

24. P ( E ) = 1 - P ( E ') = 1 -

113

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 27. {A♥, K♥, Q♥, J♥, 10♥, 9♥, 8♥, 7♥, 6♥, 5♥, 4♥, 3♥, 2♥, A♦, K♦, Q♦, J♦, 10♦, 9♦, 8♦, 7♦, 6♦, 5♦, 4♦, 3♦, 2♦, A♠, K♠, Q♠, J♠, 10♠, 9♠, 8♠, 7♠, 6♠, 5♠, 4♠, 3♠, 2♠, A♣, K♣, Q♣, J♣, 10♣, 9♣, 8♣, 7♣, 6♣, 5♣, 4♣, 3♣, 2♣}; 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

114

CHAPTER 3 │ PROBABILITY

33. 1 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. 36. 1 outcome; simple event because it is an event that consists of a single outcome. 37. (8)(11)(7) = 616 38. (4)(3)(4)(6)(3) = 864 39. (8)(10)(10)(10)(3) = 24000 40. (2)(2)(2)(2)(2)(2)(2)(2)(2)(2) = 1024 41. P ( A) =

1 » 0.083 12

42. P ( B ) =

1 » 0.083 12

43. P (C ) =

2 1 = » 0.17 12 6

44. P ( D ) =

1 » 0.08 12

45. P ( E ) =

3 1 = » 0.25 12 4

46. P ( F ) =

3 1 = » 0.25 12 4

47. P (no car ) =

960 » 0.24 4000

48. P (one car ) =

1800 » 0.45 4000

49. P (18 to 29) =

48.9 » 0.216 226.9

50. P (30 to 44) =

53.9 » 0.238 226.9

CHAPTER 3 │ PROBABILITY

51. P (45 to 64) =

115

78.1 » 0.344 226.9

52. P (65 years old and older ) =

46.0 » 0.203 226.9

53. Classical probability because each outcome in the sample space is equally likely to occur. 54. Subjective probability because it is most likely based on an educated guess. 55. Empirical probability because survey results were used to calculate the frequency of a person watching a movie every week. 56. Empirical probability because survey results were used to calculate the frequency of an adult favoring the ban. 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 -

123 » 1 - 0.158 » 0.842 777

60. P ( B ) = 1 - P ( B ') = 1 -

137 » 1 - 0.176 » 0.824 777

61. P (C ) = 1 - P (C ') = 1 -

173 » 1 - 0.223 » 0.777 777

62. P ( D ) = 1 - P ( D ') = 1 -

122 » 1 - 0.157 » 0.843 777

63-66. 1 R B

2 G Y

63. P ( A) =

R B

3 G Y

1 » 0.042 ; unusual 24

3 = 0.125 ; not unusual 24 5 65. P (C ) = » 0.208 ; not unusual 24

64. P ( B ) =

R B

4 G Y

R B

5 G Y

R B

6 G Y

R B

G Y

116

CHAPTER 3 │ PROBABILITY

66. P ( D) =

67. (a)

1 » 0.042 ; unusual 24

1 1 999 = 0.001 (b) 1 = = 0.999 1000 1000 1000

68. Number of possible codes: 26 ⋅ 26 = 676

(a)

1 » 0.0015 676

(b) 1 -

1 675 = » 0.9985 676 676

69. There are 8 outcomes in the sample space: {SSS, SSR, SRS, SRR, RSS, RSR, RRS, RRR} 1 P (SSS) = = 0.125 8 70. There are 8 outcomes in the sample space: {SSS, SSR, SRS, SRR, RSS, RSR, RRS, RRR} 1 P ( RRR ) = = 0.125 8 71. There are 8 outcomes in the sample space: {SSS, SSR, SRS, SRR, RSS, RSR, RRS, RRR} 3 P (SSR, SRS, RSS) = = 0.375 8 72. There are 8 outcomes in the sample space: {SSS, SSR, SRS, SRR, RSS, RSR, RRS, RRR} 7 P (SSR, SRS, RSS, RRS, RSR, SRR, RRR ) = = 0.875 8 1 OR: 1 - P (SSS) = 1 - = 0.875 8 73. P (voted Republican) =

74. P (did not vote) =

75. ( P (doctorate) =

1.8 = 0.450 1.8 + 2.2

6.1 » 0.404 6.1 + 9.0

3 » 0.033 91

76. P (associate's degree) =

23 » 0.253 91

CHAPTER 3 │ PROBABILITY

77. P (master's degree) =

117

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 Exercise 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) =

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

1 = 0.25 4

8 = 0.5 16

83. Total workers (in thousands) = 120, 221 + 2422 + 15,338 + 10,852 = 148,833 120, 221 P (services industry ) = » 0.808 148,833 84. P(not in services industry) = 1 - P(services industry) » 1 - 0.808 = 0.192 85. Total workers (in thousands) = 120, 221 + 2422 + 15,338 + 10,852 = 148,833 15,338 P (manufacturing industry ) = » 0.103 148,833 86. Total workers (in thousands) = 120, 221 + 2422 + 15,338 + 10,852 = 148,833 P (not in agriculture, forestry, fishing, and hunting industry) = 1 - P (agriculture, forestry, fishing, and hunting industry ) = 1 -

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

2422 » 0.984 148,833

40 » 0.313 128

(b) P (between 20 and 30 inclusive) = (c) P (more than 63) =

10 = 0.078 1 28

4 » 0.031 ; This event is unusual because its probability is less than or 128

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

118

CHAPTER 3 │ PROBABILITY

(c) P ($30 or more) = 0.5 ; median = 30 89. The probability of randomly choosing a cricket player who did not play for his school team. 90. The probability of randomly choosing a smoker whose mother did not smoke 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)

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

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

3.2 CONDITIONAL PROBABILITY AND THE MULTIPLICATION RULE

CHAPTER 3 │ PROBABILITY

119

3.2 TRY IT YOURSELF SOLUTIONS 1.

P (female│not offended by something on social media ) =

2. (1) Dependent

549 = 0.488 1125

(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} æ 13 ö æ12 ö P ( A and B ) = P ( A) ⋅ P ( B A) = çç ÷÷÷ ⋅ çç ÷÷÷ » 0.059 çè 52 ø çè 51ø

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.

120

CHAPTER 3 │ 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} 191,310 172, 489 » 0.526 » 0.159 (a) P ( M | B ) = (b) P ( B | F ) = 363,799 1,082, 265 8.

A = {employee is executive}; B = {employee goes on leisure trip} 150 125 (a) P ( B | A¢) = (b) P ( A | B ¢) = = 0.462 » 0.455 325 275

9. Dependent. The outcome of the first card drawn affects 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. Independent. The outcome of returning a library book before its due date does not affect the outcome of issuing a new book. 14. Independent. The outcome of the first ball drawn does not affect the outcome of the second ball drawn. 15. Events: having more cone cells, perceiving color differently; Independent. Having more cone cells does not lead to perceiving colors differently. 16. Events: certain components in coffee, stomach ulcers; Independent. Coffee only irritates existing stomach ulcers. 17. Events: eating chocolates, improvement in cardiovascular health; Dependent. Eating chocolates improves cardiovascular health. 18. Events: high engagement with mobile technology for escapism, depression, and anxiety; Dependent. High engagement with mobile technology for escapism may cause depression and anxiety in collegeage students. 19. Let A = {Card is a king} and B = {Card is a queen}.

CHAPTER 3 │ PROBABILITY

P ( A) =

121

4 1 4 1 4 and P ( B | A) = . Thus, P ( A and B ) = ⋅ » 0.006 = 52 13 51 13 51

20. Let A = {coin is head} and B = {number on die is less than 3}. 1 2 1 1 1 P( A) = and P ( B ) = = . Thus, P ( A and B ) = ⋅ » 0.167 2 6 3 2 3 21. Let A = {woman carries mutation of BRCA1 gene} and B = {woman develops breast cancer}. æ 1 ö÷ æ 3 ö÷ 1 6 3 and P ( B | A) = = . Thus, P ( A and B ) = P ( A) ⋅ P ( B | A) = çç P ( A) = ÷ ⋅ çç ÷ = 0.0 01 èç 600 ÷ø èç 5 ø÷ 600 10 5 22. Let A = {adult drives a Ford} and B = {adult drives a pickup truck}. æ 3 ö æ 2ö 3 2 and P ( B | A) = . Thus, P ( A and B ) = P ( A) ⋅ P ( B | A) = çç ÷÷÷ ⋅ çç ÷÷÷ » 0.067 P ( A) = èç10 ø èç 9 ø 10 9 23. Let A = {first adult thinks that most Hollywood celebrities are good role models} and B = {second adult thinks that most Hollywood celebrities are good role models}. 200 199 and P ( B | A) = P ( A) = 1000 999 æ 200 ö÷ æ 199 ÷ö (a) P ( A and B ) = P ( A) ⋅ P ( B | A) = çç ⋅ç » 0.040 çè1000 ÷÷ø ççè 999 ÷÷ø æ 800 ö÷ æ 799 ÷ö (b) P ( A ' and B ') = P ( A ') ⋅ P ( B ' | A ') = çç ⋅ç » 0.640 çè1000 ÷÷ø ççè 999 ÷÷ø

(c) P(at least one of the two adults thinks that most Hollywood celebrities are good role models) = 1 - P ( A ' and B ') » 1 - 0.640 » 0.360 24. Let A = {first adult said they know a murder victim}, B = {second adult said they know a murder victim}, C = {third adult said they know a murder victim }, and D = {fourth adult said they know a murder victim} 300 299 298 297 , P ( B | A) = , P (C | A and B ) = , and P ( D | A and B and C ) = P ( A) = 1000 999 998 997

(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 ) æ 300 ö÷ æ 299 ÷ö æ 298 ÷ö æ 297 ö÷ = çç ⋅ç ⋅ç ⋅ç » 0.008 çè1000 ÷÷ø ççè 999 ÷÷ø ççè 998 ÷÷ø ççè 997 ø÷÷

(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 ') æ 700 ö÷ æ 699 ö÷ æ 698 ö÷ æ 697 ö÷ = çç ÷ ⋅ çç ÷ ⋅ çç ÷ ⋅ çç ÷ » 0.239 èç1000 ÷ø èç 999 ø÷ èç 998 ø÷ èç 997 ø÷

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CHAPTER 3 │ PROBABILITY

25. Let A = {first adult said John Kennedy was the best president} and B = {second adult said John Kennedy was the best president}. 217 216 and P ( B | A) = . P ( A) = 1446 1445 æ 217 ö÷ æ 216 ö÷ (a) P ( A and B ) = P ( A) ⋅ P ( B | A) = çç ⋅ç » 0.022 çè1446 ÷÷ø ççè1445 ÷÷ø æ1229 ö÷ æ1228 ö÷ (b) P ( A ' and B ') = P ( A ') ⋅ P ( B ' | A ') = çç ⋅ç » 0.722 çè1446 ÷÷ø ççè1445 ÷÷ø

(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 Richard Nixon worst president}, B = {second adult said Richard Nixon worst president}, and C = {third adult said Richard Nixon worst president}. 188 187 186 , P ( B | A) = , and P (C | A and B ) = P ( A) = 1446 1445 1444 æ 188 ö÷ æ 187 ÷ö æ 186 ÷ö (a) P ( A and B and C ) = P ( A) ⋅ P ( B | A) ⋅ P (C | A and B ) = çç ⋅ç ⋅ç » 0.002 çè1446 ÷÷ø ççè1445 ÷÷ø ççè1444 ÷÷ø æ1258 ö÷ æ1257 ö÷ æ1256 ö÷ (b) P ( A ' and B 'and C ') = P ( A ') ⋅ P ( B ' | A ') ⋅ P (C ' | A ' and B ') = çç ⋅ç ⋅ç » 0.658 çè1446 ø÷÷ èçç1445 ø÷÷ èçç1444 ÷÷ø

(d) The event in part (a) is unusual because its probability is less than or equal to 0.05. 27. (a) P (all six have O+) = (0.47) ⋅ (0.47) ⋅ (0.47) ⋅ (0.47) ⋅ (0.47) ⋅ (0.47) » 0.011

(b) P (none have O+) = (0.53) ⋅ (0.53) ⋅ (0.53) ⋅ (0.53) ⋅ (0.53) ⋅ (0.53) » 0.022 (c) P (at least one has O+) = 1 - P (none have O+) » 1 - 0.022 = 0.978 (d) The events in parts (a) and (b) are unusual because their probabilities are less than or equal to 0.05.

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123

28. (a) P (all four have AB-) = (0.01) ⋅ (0.01) ⋅ (0.01) ⋅ (0.01) » 0.00000001

(b) P (none have AB-) = (0.99) ⋅ (0.99) ⋅ (0.99) ⋅ (0.99) » 0.961 (c) P (at least one has AB-) = 1 - P (none have AB-) » 1- 0.961 = 0.039 (d) The events in parts (a) and (c) are unusual because their probabilities are less than or equal to 0.05. 29. Let A = {IVF procedure resulted in pregnancy} and B = {IVF pregnancy resulted in multiple birth}. P ( A) = 0.016 and P ( B | A) = 0.411

(a) P ( A and B ) = P ( A) ⋅ P ( B | A) = (0.016)⋅ (0.411) » 0.007 (b) P ( B ' | A) = 1 - P ( B | A) = 1 - 0.411 = 0.589 (c) Yes, this is usual because the probability is less than or equal to 0.05. 30. Let A = {purchased a lottery ticket} and B = {annual income of less than $36,000}. P ( A) = 0.49 and P ( B | A) = 0.275

(a) P ( A and B ) = P ( A) ⋅ P ( B | A) = (0.49)⋅ (0.275) » 0.135 (b) P( B ' | A) = 1 - P ( B | A) = 1 - 0.275 = 0.725 (c) No, this is not unusual because the probability is not less than or equal to 0.05. 31. Let A = {use digital content} and B = {use as part of curriculum}. 4 P ( A) = 0.80 and P ( B | A) = = 0.4 10 P ( A and B ) = P ( A) ⋅ P ( B | A) = (0.8)⋅ (0.4) = 0.32 32. Let A = {gets a job} and B = {does not resign in an year}. P ( A) = 0.75 and P ( B | A) = 0.60

P ( A and B ) = P ( A) ⋅ P ( B A) = (0.75) ⋅ (0.6) = 0.45 33. P( A B) =

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

æ 2 ÷ö æ 1 ÷ö 2 çç ÷ ⋅ çç ÷ çè 3 ÷ø çè 5 ÷ø 4 = = 15 = » 0.444 æ 2 ö÷ æ 1 ÷ö æ 1 ÷ö æ 1 ÷ö 3 9 çç ÷ ⋅ çç ÷ + çç ÷ ⋅ çç ÷ èç 3 ÷ø çè 5 ÷ø çè 3 ÷ø çè 2 ÷ø 10

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CHAPTER 3 │ PROBABILITY

34. P( A B) =

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

æ 3ö÷ æ 2 ÷ö 1 çç ÷ ⋅ çç ÷ èç 8 ø÷ èç 3 ÷ø 2 = = 4 = = 0.4 æ 3ö÷ æ 2 ö÷ æ 5 ö÷ æ 3 ö÷ 5 5 çç ÷⋅ çç ÷ + çç ÷ ⋅ çç ÷ çè 8 ÷ø çè 3 ø÷ çè 8 ø÷ çè 5 ø÷ 8 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 │ 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 P ( A or B ) = P ( A) + P ( B ) =

1 1 + » 0.667 6 2

(2) Not mutually exclusive Let A = {face card} and B = {heart}. 12 13 3 , P ( B ) = , and P ( A and B ) = P ( A) = 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 and P ( B ) = P ( A) = 36 36 3 5 P ( A or B ) = P ( A) + P ( B ) = + » 0.222 36 36 4. (1) Let A = {type B} and B = {type AB}. Copyright © 2019 Pearson Education Ltd.

125

126

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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 and P ( B ) = P ( A) = 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( B) = , and P ( A and B ) = P ( A) = 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( B) = , and P ( A and B ) = P ( A) = 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}. 34 15 P ( A or B ) = P ( A) + P ( B ) = + » 0.194 253 253 P(not a linebacker or quarterback) = 1 – P(A or B) » 1- 0.194 = 0.806

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 6-sided die. Let A = {head} and B = {6 on die}. P ( B A) = = P ( B ) implies A and B 6 1 are independent events. However, P(A and B) = implies A and B are not mutually exclusive. 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) .

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127

6. True 7. Not mutually exclusive. A presidential candidate can lose the popular vote and win the election. 8. Mutually exclusive. A movie cannot have two ratings. 9. Mutually Exclusive. A student cannot study more than 5 hours but less than 2 hours daily. 10. Mutually exclusive. A student cannot have a birthday in both months. 11. Not Mutually Exclusive. A badminton player can be female and 25 years old. 12. Not mutually exclusive. A member of Congress can be a male senator. 13. Let A = {male} and B = {can play an instrument }. 35 20 5 , P( B) = , and P ( B | A) = P ( A) = 60 60 35 35 5 1 Thus, P ( A and B ) = P ( A) ⋅ P ( B | A) = ⋅ = 60 35 12 35 20 1 50 P ( A or B ) = P ( A) + P ( B ) - P ( A and B ) = + - = = 0.833 60 60 12 60 14. Let A = {college professor} and B = {male}. 3120 3595 1505 , P( B) = , and P ( B | A) = P ( 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 = {does not have poor battery life} and B = {does not have corrupt OS}. P ( A) = 0.97 , P ( B ) = 0.95 , and P ( A and B ) = 0.876 . P ( A or B ) = P ( A) + P ( B ) - P ( A and B ) = 0.97 + 0.95 - 0.935 = 0.985 16. Let A = {flash problem} and B = {focus malfunction}. P ( A) = 0.08 , P ( B ) = 0.12 , and P ( A and B ) = 0.009 . P ( A or B) = P ( A) + P ( B ) - P ( A and B) = 0.08 + 0.12 - 0.009 = 0.191 17. (a) P (6 or greater than 4) = P (6) + P (greater than 4) – P (6 and greater than 4) =

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

1 3 1 + - » 0.5 6 6 6

1 2 1 + - » 0.33 6 6 6

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(c) P (less than 5 or odd ) = P (less than 5) + P (odd ) – P (less than 5 and odd ) = 18. (a) P ( black or king) = P (black ) + P (king) – P ( black and king) =

4 3 2 + - » 0.833 6 6 6

26 4 2 + - » 0.538 52 52 52

(b) P (diamond or face card ) = P (diamond ) + P (face card ) – P (diamond and face card ) =

13 12 3 + - » 0.423 52 52 52

13 13 + - 0 » 0.50 52 52

19. (a) P (under 5) = 0.06

(b) P (45 +) = P (45 - 64) + P (65 - 74) + P (75 +) = 0.236 + 0.107 + 0.083 = 0.426 (c) P (not 65 +) = 1 - P (65 +) = 1- P(65 - 74) - P(75+) = 1 - 0.107 - 0.083 = 0.81 (d) P (between 20 and 34) = P (20 - 24) + P (25 - 34) = 0.064 + 0.137 = 0.201 20. (a) P ( never used ) = 0.603

(b) P (used ) = 1 - P (never used ) = 1 - 0.603 = 0.397 (c) P (used between 1 and 29) = P (1 to 9) + P (10 to 29) = 0.12 + 0.04 = 0.16 (d) P (used on at least 1 of last 30 days) = P (1 to 9) + P (10 to 29) + P (all 30) = 0.12 + 0.04 + 0.027 = 0.187 21. (a) P (not A or B) = 1 – P ( A or B) = 1 -

276 978 = » 0.780 1254 1254

(b) P (better than D) = P (A or B) + P (C) = (c) P (D or F) = P (D ) + P (F) =

263 477 740 + = » 0.590 1254 1254 1254

(d) P (C or D ) = P (C) + P (D ) =

276 238 514 + = » 0.410 1254 1254 1254

238 263 501 + = » 0.400 1254 1254 1254

CHAPTER 3 │ PROBABILITY

22. (a) P (somewhat important ) =

(b) P (not at all important ) =

312 » 0.310 1007

141 » 0.140 1007

191 141 332 + = » 0.330 1007 1007 1007

(d) P (extremely important or very important ) = P (extremely important ) + P ( very important ) =

131 212 343 + = » 0.341 1007 1007 1007

23. Let A = {male}; B = {business major}

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

812,669 363,799 191,310 + » 0.520 1,894,934 1,894,934 1, 894, 934

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

1,082, 265 1,531,135 909,776 + » 0.899 1,894,934 1,894,934 1,894, 934

812,669 1,531,135 621,359 + » 0.909 1, 894,934 1,894,934 1, 894, 934

24. Let A = {opposes tax}; B = {supports tax}; C = {unsure about tax}; D = {female}

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

164 259 65 + » 0.708 506 506 506

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

301 247 128 + » 0.830 506 506 506

301 + 164 259 173 + 65 + » 0.960 506 5 06 506

25. A = {frequently}; B = {occasionally}; C = {not at all}; D = {male}

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

1472 428 221 + » 0.589 2850 2850 2850

(b) P ( D' or C ) = P ( D ') + P (C ) – P ( D ' andC ) =

1378 1536 741 + » 0.762 2850 2850 2850

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CHAPTER 3 │ PROBABILITY

428 886 + » 0.461 2850 2850

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

1378 886 + 1536 430 + 741 + » 0.922 2850 2850 2850

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

(a) P ( E ¢ or C ) = P ( E ¢) + P (C ) - P ( E ¢ and C ) = (b) P ( E or D ) = P ( E ) + P ( D ) - P ( E and D ) =

2499 818 552 + » 0.575 4807 4807 4807

2308 1706 684 + » 0.693 4807 4807 4807

2308 1903 + 1706 1262 684 + » 0.826 4807 4807 4807 4807

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

380 1903 + » 0.475 4807 4807

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.

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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. 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!

C3 =

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 P2 =

15! = 3003 5!(15 - 5)! 54! = 3,162,510 54 C5 = 5!(54 - 5)! C5 =

C5 3003 == » 0.0009 3,162, 510 54 C5

P (5 diamonds from deck with 2 jokers) = 15

C3 ⋅7 C0 =

C3 =

5! 7! ⋅ = 10 ⋅1 = 10 3!2! 0!7!

12! = 220 3!9!

P (all three are men ) =

10 » 0.045 220

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

5. True

P5 =

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

P3 =

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

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! C 70 (8 - 4)!4! = 4!4! = » 0.076 11. 8 4 = 12! 12! 924 12 C6 (12 - 6)!6! 6!6! 10! 10! C 120 (10 7)!7! = 3!7! = » 0.035 12. 10 7 = 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! P 13 13 1 (13 -1)! 12!

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133

7! 7! P 7 ⋅ 6⋅ 5 (7 3)! = 4! = » 0.018 14. 7 3 = 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. 19. 9! = 362880

20. 8! = 40,320

21. 6! = 720

22. 12! = 479,001,600

23. 20 P3 =

20! 20! = = 6,840 (20 - 3)! 17!

24. 16 P5 =

16! 16! = = 524,160 (16 - 5)! 11!

25. 24 P6 =

24! 24! = = 96,909,120 (24 - 6)! 18!

26. 38 P4 =

38! 38! = = 1,771,560 (38 - 4)! 34!

27.

15! = 360,360 5!7!3!

29. 3 S’s, 3 T’s, 1 A, 2 I’s, 1 C

28.

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

30.

8! = 56 5!3!

10! = 50, 400 3!3!1!2!1!

32. 36 C12 =

31. 20 C 4 = 4845

33. 50 C3 =

50! = 2,118,760 (50 - 5)!5!

35. 10 ⋅ 8 ⋅13 C2 = 10 ⋅ 8 ⋅

36. 8 C6 ⋅6 C3 ⋅7 C3 =

34. 52 C6 =

13! = 6240 (13 - 2)!2!

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

36! = 1, 251,677,700 (36 - 12)!12! 52! = 20,358,520 (52 - 6)!6!

134

CHAPTER 3 │ PROBABILITY

37. 6 C2 ⋅ 69 C3 =

6! 69! ⋅ = 785,910 2!4! 3!66!

38. 6 C3 ⋅18 C5 =

6! 18! ⋅ = 171,360 3!3! 13!5!

39.

1 » 0.022 ( 10 C2 )

40.

1 1 = = 0.05 P 20 5 2

41.

1 1 = » 0.005 220 12 C3

42.

1 1 = » 0.003 330 11 C 4

43. (a)

C3 455 = » 0.016 27,720 56 C3

(b)

C3 10,660 = » 0.385 27,720 56 C3

44. (a)

C4 15 = » 0.015 1001 14 C4

(b)

C4 70 = » 0.070 1001 14 C4

45. (3%)(1500) = (0.03)(1500) = 45 of the 1500 say most have food allergies. C2 990 = » 0.0009 C 1,12 4, 250 1500 2

P (2 say most ) = 45

46. (33%)(1500) = (0.33)(1500) = 495 of 1500 say none have food allergies. C3 » 0.036 1500 C3

P (3 say none) = 495

47. (21%)(1500) = (0.21)(1500) = 315 of the 1500 say some have food allergies. 1500 - 315 = 1185 do not say some have food allergies. C P (none of the 6 say some) = 1185 6 » 0.242 1500 C6 48. (43%)(1500) = (0.43)(1500) = 645 of 1500 say only a few have food allergies  1500 - 645 = 855 say none have only a few food allergies. C P (none of 4 say only a few ) = 855 4 » 0.105 1500 C4 49. 70 C 7 = 1,198, 774, 720 50. (a)

200

P (win) =

C 5 = 2,535, 650, 040

1 » 0.000000001 1,198,774,720

(b) 144 C5 = 481, 008,528

C5 » 0.190 200 C5

(d) Yes. The probability that no minorities are selected is very small.

CHAPTER 3 │ PROBABILITY

51.

( 24 C5 )( 30 C3 ) 54

52.

( 17 C6 )( 37 C2 ) 54

53.

( 13 C4 )( 41 C4 ) 54

54.

(42,504)(4060) » 0.166 1,040, 465,790

(12,376)(666) » 0.0079 1,040, 465,790

(715)(101,270) » 0.070 1,040, 465,790

( 17 C2 )( 13 C2 )( 24 C4 ) 54

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

55.

C3 ( 8 C2 )( 2 C1 ) 56 (28)(2) 112 + = + = » 0.933 120 120 120 10 C3 10 C3

56.

C4 ( 95 C3 )( 5 C1 ) 3,183,545 (138,415)(5) + = + » 0.988 3,921, 225 3,921,225 100 C4 100 C4

æ( C )( C )( C )( C )ö (1)(1)(1)(1) 57. 4 C2 ççç 2 2 2 2 2 0 2 0 ÷÷÷ = 6 ⋅ » 0.086 ÷ çè 70 ø 8 C4 58.

59.

4! çæ( 2 C2 )( 2 C1 )( 2 C1 )( 2 C0 )ö÷ (1)(2)(2)(1) ÷÷ = 12 ⋅ » 0.686 çç 2!1!1!çè 70 ø÷ 8 C4

( 13 C2 )( 13 C1 )( 13 C1 )( 13 C1 ) 52

(78)(13)(13)(13) » 0.066 2,598,960

æ( C )( C )ö æ (1)(48) ö÷ 60. ( 13 C1 )ççç 4 4 48 1 ÷÷÷ = 13 ⋅ çç ÷ » 0.0002 çè 2,598,960 ÷ø èç ø÷ 52 C5 æ( C )( C )ö æ (4)(6) ÷ö 61. ( 13 C2 )ççç 4 3 4 2 ÷÷÷ = 78çç ÷ » 0.001 çè 2,598,960 ÷ø ÷ø èç 52 C5 62.

( 13 C1 ) æç( 4 C3 )( 48 C1 )( 44 C1 )ö÷ 13 æç (4)(48)(44) ö÷ ÷ » 0.021 ç ÷÷ = ⋅ ç 2

ççè

÷ø

2 çè 2,598,960 ÷ø

135

136

CHAPTER 3 │ 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 4 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 4 or 5 {(1, 3), (1, 4), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)} There are 7 outcomes.

CHAPTER 3 │ PROBABILITY

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

Event: The family has two boys. {BGB, BBG, GBB} There are 3 outcomes. 5. (3)(5)(2) = 30

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

7. Empirical probability because an experiment was used to calculate the frequency of a component lasting for 3 hours. 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 observations obtained from 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 observations obtained from probability experiments. 13. P (business or psychology) = P ( business ) + P (psychology ) =

361 114 + » 0.258 1840 1840

14. P (not in health professions or social sciences/history) æ 181 178 ÷ö + » 0.805 = 1- [ P (health professions ) + P (social sciences/history )] = 1- çç çè1840 1840 ÷÷ø

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)

137

138

CHAPTER 3 │ PROBABILITY

17. P (first failed) =

13,194 » 0.555 23,775

18. P (passed | repeated) =

6454 » 0.379 17,035

19. Independent. The outcomes of the first three rolls do not affect the outcome of the fourth roll. 20. Dependent. The outcome of participating in a training-camp affects the outcome of successfully completing the marathon run. 21. Dependent. The outcome of regularly attending lectures affects the outcome of passing the course. 22. No. You do not know whether events A and B are independent or dependent. 23. P (black balls in both the draws) = P (black ball in first draw )⋅ P (black ball in second draw )

5 5 ⋅ = 0.04 25 25 The event is unusual because its probability is less than or equal to 0.05. =

12 18 ⋅ » 0.248 30 29 The event is not unusual because its probability is not less than or equal to 0.05.

24. P ( white and (red or black )) = P ( white) ·P (red or black | white) =

25. Mutually exclusive. A jelly bean cannot be both completely red and completely yellow. 26. Not mutually exclusive. A person who is a professional singer can also be a professional dancer. 27. Mutually exclusive. A person cannot simultaneously be of Indian and Chinese origin. 28. P (black or sedan ) = P (black ) + P (sedan ) – P (black and sedan ) =

1560 3120 1170 + – = 0.54 6500 6500 6500

29. P (golf or soccer ) = P (golf ) + P (soccer ) – P (golf and soccer ) = 0.40 + 0.60 – 0.30 = 0.7 30. P (English or Spanish ) = P (English ) + P (Spanish ) – P (English and Spanish )

= 0.60 + 0.30 – 0.15 = 0.75 31. P (7 -10 or black ) = P (7 - 10) + P (black ) – P (7 - 10 and black ) =

16 26 8 + - » 0.654 52 52 52

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

CHAPTER 3 │ PROBABILITY

33. P (even or greater than 6) = P (even ) + P (greater than 6) - P (even and greater than 6)

5 4 2 + - = 0.7 10 10 10

34. P (300 or more) = 1 - P (fewer than 300) = 1 - 0.302 = 0.698 35. P (Middle class or Upper-middle class) =

36. P (Working class or Lower class) =

1329 468 + » 0.584 3078 3078

929 254 + » 0.384 3078 3078

37. P (not Middle class) = 1 - P (Middle class) = 1 -

1329 » 0.568 3078

38. P(not Upper or Lower class) = 1 - [ P(Upper class) + P(Lower class)] æ 98 254 ö÷ = 1- çç + » 0.886 çè 3078 3078 ÷÷ø 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 10 (5 3)!3! = 2!3! = » 0.083 44. 5 3 = 10! 10! 120 10 C3 (10 - 3)!3! 7!3! 45. Order is important: 15 P3 = 46. 6! = 720

15! 15! = = 2730 (15 - 3)! 12!

139

140

CHAPTER 3 │ PROBABILITY

47. Order is not important: 17 C4 =

17! = 2380 (17 - 4)!4!

48. Order is not important: 12 C4 =

12! = 495 8!4!

( C1 )⋅ ( 4 C2 )⋅ ( 4 C3 )

49. P (1 ace, 2 kings and 3 queens) = 4

50.

4⋅ 6⋅ 4 » 0.000005 20,358,520

1 1 = » 0.000002 (26)(25)(10)(10)(8) 520,000 C3 138, 415 = » 0.856 161,700 100 C3

51. (a) P (no smartphones) = 95

C3 10 = » 0.000062 161, 700 100 C3

(b) P (all smartphones ) = 5

(c) P (at least one smartphone) = 1 – P ( no smartphones) » 1- 0.856 = 0.144 (d) P (at least one non-smartphone) = 1 – P (all smartphones) » 1 - 0.000062 » 0.999938 C3 1, 254,890 = » 0.955 1,313, 400 200 C3

52. (a) P (no defective laptops) = 197

C3 1 = » 0.0000007 1,313, 400 200 C3

(b) P (all defective laptops) = 3

( C1 )( 3 C2 )

200

(197)(3) » 0.0004 1,313, 400

(d) P (at least one defective laptop) = 1 - P (nodefective laptop) » 1- 0.955 » 0.045 C3 35 = » 0.292 120 10 C3

53. (a) P (3 men ) = 7

C3 1 = » 0.008 120 10 C3

(b) P (all women ) = 3

( C2 )( 3 C1 )

(21)(3) = 0.525 120

CHAPTER 3 │ PROBABILITY

( C1 )( 3 C2 )

(d) P (1 man and 2 women ) = 7

141

(7)(3) = 0.175 120

CHAPTER 3 QUIZ SOLUTIONS 1. (9)(10)(10)(10)(10)(5) = 450,000 2. (a) P ( bachelor’s degree) =

319.2 » 0.713 447.4

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

164.3 » 0.662 248.3

154.9 » 0.778 199.1

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

319.2 100.1 + » 0.937 447.4 447.4

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

12.1 » 0.049 248.3

(f) P (master’s or degree in natural sciences/mathematics) = P ( master’s) + P (degree in natural sciences/mathematics) - P ( master’s and degree in natural sciences/mathematics)

100.1 199.1 28.2 271 + = » 0.606 447.4 447.4 447.4 447.4

(g) P ( bachelor’s and degree in natural sciences/mathematics) =

154.9 » 0.346 447.4

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

164.3 » 0.515 319.2

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.

P4 =

30! = 657,720 (30 - 4)!

142

CHAPTER 3 │ PROBABILITY

6. (a)

247

C3 =

(b) 3 C3 =

247! = 2, 481,115 (247 - 3)!3!

3! =1 (3 - 3)!3!

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

C3 2, 481,115 = » 0.964 2,573,000 250 C3

7. (a)

247

(c)

250

(b)

C3 1 = » .0000004 2,573,000 250 C3 3

C3 -3 C3 2,572,999 = » 0.9999996 2,573,000 250 C3

CHAPTER 3 TEST SOLUTIONS

12! 220 (12 - 3)!3! 12 C3 = = » 0.005 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. 192.6 » 0.327 588.4 192.6 + 197.9 390.5 (b) P (sixth or seventh) = = » 0.664 588.4 588.4

4. (a) P (sixth grade) =

62.9 » 0.334 188.3

CHAPTER 3 │ PROBABILITY

(d) P (Ohio | seventh) =

134.3 » 0.679 197.9

(e) P(seventh or Minnesota) = P(seventh) + P(Minnesota) – P(seventh and Minnesota) 197.9 188.3 63.6 322.6 = + = » 0.548 588.4 588.4 588.4 588.4 (f) P (sixth and Ohio) =

130.8 » 0.222 588.4

5. None are unusual because none of the events have a probability of 0.05 or less. 6. The events are dependent because P (sixth grade) ¹ P (sixth | Minnesota) . 192.6 P (sixth grade) = » 0.327 588.4 61.8 P (sixth | Minnesota) = » 0.328 188.3 7. (a) 16 P3 =

(b)

16! = (16)(15)(14) = 3360 (16 - 3)!

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

143

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. 2.

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 │ DISCRETE PROBABILITY DISTRIBUTIONS

145

   xP ( x )  2.6 On average, a new employee makes 2.6 sales per day. 6. From 5,   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 2.6 1.6 0.6 0.4 1.4 2.4 3.4 4.4

(x  )2 6.76 2.56 0.36 0.16 1.96 5.76 11.56 19.36

P(x)(x  )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   )2  3.72

 2  3.7

   2  3.7  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

E ( x) 

P(x) 1 2000 1 2000 1 2000 1 2000 1 2000 1995 2000  P ( x)  1

 xP( x)  $3.08

xP(x) 1995 2000 995 2000 495 2000 245 2000 95 2000 9975  2000  xP( 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.

146

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

2. A discrete probability distribution lists each possible value a random variable can assume, together with its probability. Condition 1: 0  P ( x)  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. Continuous; The speed of the soccer hits is a random variable that must be measured. 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. Discrete; The number of times a book is issued from the library is a random variable that is countable. 16. Discrete; The number of tornadoes in the month of May in Oklahoma is a random variable that is countable. 17. Continuous; The weight of a student’s school bag is a random variable that must 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 │ DISCRETE PROBABILITY DISTRIBUTIONS

19. (a)

x 0 1 2 3

f 598 462 381 259 n  1700

P(x) 0.35 0.27 0.23 0.15  P ( x)  1

(b)

Skewed right 20. (a)

x 0 1 2 3 4 5

f 15 40 45 64 88 148 n  400

P(x) 0.0375 0.1000 0.1125 0.1600 0.2200 0.3700  P ( x)  1

(b)

Skewed left 21. (a) P (1 or 2)  0.27  0.23  0.50

(b) P (less than 2)  0.35  0.27  0.62

147

148

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

(c) P (1  x  3)  0.27  0.23  0.15  0.65 (d) P ( x  2)  0.23  0.15  0.38 22. (a) P (2 or 3)  0.1125  0.1600  0.2725

(b) P(more than 3)  0.22  0.37  0.59 (c) P(1  x  4)  0.1000  0.1125  0.1600  0.2200  0.5925 (d) P(2  x  5)  0.1125  0.1600  0.2200  0.3700  0.8625 (e) P (3 or less)  0.0375  0.1000  0.1125  0.1600  0.41 23. P(3)  0.15 . No, because the probability is greater than 0.05. 24. P (0)  0.0375 . Yes, because the probability is less than 0.05. 25.  P ( x )  1  P (2)  1  (0.09  0.15  0.26  0.17)  0.33 26.  P ( x )  1  P (4)  1  (0.09  0.11  0.16  0.25  0.08  0.02)  0.29 27. No.  P ( x )  0.40  0.21  0.11  0.15  0.18  1.05 28. Yes, because each P(x) is between 0 and 1, and  P ( x )  1 , the distribution is a probability

distribution. 29. (a)

P(x)

0 1 2 3 4 5

0.002 0.018 0.054 0.199 0.715 0.012  P ( x)  1

xP(x) 0 0.018 0.108 0.597 2.860 0.060  xP( x)  3.643

   xP ( x )  3.643  3.6

(x  ) −3.643 −2.643 −1.643 −0.643 0.351 1.351

(x  )2 13.271 6.985 2.699 0.413 0.127 1.841

(x  )2P(x) 0.027 0.126 0.146 0.082 0.091 0.022  ( x   )2 P ( x)  0.494

 2   ( x   ) 2 P ( x )  0.494  0.5

   2  0.494  0.7 (b) The mean is 3.6, so the average number of books per shelf is about 3 or 4 books. The standard deviation is 0.7, so most of the shelves differ from the mean by no more than 1 book.

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

30. (a)

x 4 5 7 8 10 11

P(x) 0.1 0.2 0.1 0.4 0.1 0.1

   xP ( x)  7.4

xP(x) 0.4 1.0 0.7 3.2 1.0 1.1  xP( x)  7.4

(x  ) −3.4 −2.4 −0.4 0.6 2.6 3.6

(x  )2 11.56 5.76 0.16 0.36 6.76 12.96

149

(x  )2P(x) 1.156 1.152 0.016 0.144 0.676 1.296  ( x   )2 P( x)  4.44

 2   ( x   ) 2 P ( x )  4.44  4.4

   2  4.44  2.1 (b) The mean is 7.4, so the average countries played per Women’s World cup was about 7. The standard deviation is 2.1, so most of the Women’s World Cups differ from the mean by no more than about 2 countries. 31. (a)

x 0 1 2 3 4 5

P(x) 0.441 0.285 0.163 0.076 0.019 0.017

   xP ( x)  1

xP(x) 0 0.285 0.326 0.228 0.076 0.085  xP( x)  1

(x  ) −1 0 1 2 3 4

(x  )2 1 0 1 4 9 16

(x  )2P(x) 0.44 0 0.163 0.304 0.171 0.272  ( x   )2 P( x)  1.35

 2   ( x   ) 2 P ( x )  1.35  1.4

   2  1.35  1.2 (b) The mean is 1, so the average batch of LEDs has about 0 or 1 defect. The standard deviation is 1.2, so most of the batches differ from the mean by no more than about 1.2 defects. 32. (a)

x 0 1 2 3 4 5 6 7

P(x) 0.008 0.099 0.185 0.161 0.289 0.121 0.081 0.056

xP(x) 0 0.099 0.370 0.483 1.156 0.605 0.486 0.392  xP( x)  3.591

   xP ( x )  3.591  3.6

(x  ) −3.591 −2.591 −1.591 −0.591 0.409 1.409 2.409 3.409

(x  )2 12.895 6.713 2.531 0.349 0.167 1.985 5.803 11.621

(x  )2P(x) 0.103 0.665 0.468 0.056 0.048 0.240 0.470 0.651  ( x   )2 P( x)  2.701

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CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

   2  2.701  1.6 (b) The mean is 3.6, so the average student has received about 4 Grade As. The standard deviation is 1.6, so most of the students differ from the mean by no more than about 2 Grade As. 33. (a)

x 1 2 3 4 5

P(x) 0.411 0.279 0.223 0.077 0.010

xP(x) 0.411 0.558 0.669 0.308 0.050  xP( x)  1.996

(x  ) 0.996 0.004 1.004 2.004 3.004

(x  )2 0.992 0.000 1.008 4.016 9.024

   xP ( x )  1.996  2.0

(x  )2P(x) 0.408 0.000 0.225 0.309 0.090  ( x   )2 P( x)  1.032

 2   ( x   ) 2 P ( x )  1.032  1.0

   2  1.032  1.0 (b) The mean is 2.0, so the average hurricane that hits the U.S. mainland is expected to be a category 2 hurricane. The standard deviation is 1.0, so most of the hurricanes differ from the mean by no more than 1 category level. 34. (a)

X 1 2 3 4 5

P(x) 0.019 0.086 0.212 0.238 0.445

xP(x) 0.019 0.172 0.636 0.952 2.225  xP ( x)  4.004

(x  ) 3.004 2.004 1.004 0.004 0.996

   xP ( x)  4.004  4.0

(x  )2 9.024 4.016 1.008 0.000 0.992

(x  )2P(x) 0.171 0.345 0.214 0.000 0.441  ( x   )2 P( x)  1.171

 2   ( x   ) 2 P ( x )  1.171  1.2

   2  1.171  1.1 (b) The mean is 4.0, so the average reviewer rating is 4. The standard deviation is 1.1, so most of the ratings differ from the mean by no more than about 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. 37. E ( x)   

 37 

 1 

 xP( x)  ($1)   38   ($35)   38   $0.05

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

38. E ( x )     xP ( x )

 1   1   18   3480   ($5450)     ($486)     ($90)      $10     $7.78  3500   3500   3500   3500  39.  y  a  b x  600  1.03(46,000)  $47,980 40.  y  b  x  (1.035) 18, 000, 000  $4391.13 41. x y  x   y  524  494  1018

x y  x   y  524  494  30 42.  x2 y   x2   y2  (126) 2  (116) 2  29,332   x  y   x2 y  171.3

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 uses the social media platform Instagram Failure: selecting an adult who does not use the social media platform Instagram n  5, p  0.28, q  0.72, x  0,1, 2,3, 4,5

P(0)  5 C0 (0.28)0 (0.72)5  0.193 P(1)  5 C1 (0.28)1 (0.72)4  0.376 P(2)  5 C2 (0.28)2 (0.72)3  0.293 P(3)  5 C3 (0.28)3 (0.72)2  0.114 P(4)  5 C4 (0.28)4 (0.72)1  0.022

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CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

P(5)  5 C5 (0.28)5 (0.72)0  0.002 x 0 1 2 3 4 5

P(x) 0.193 0.376 0.293 0.114 0.022 0.002  xP ( x)  1

n  150, p  0.52, x  65 P(65)  0.007 2

5. (1) P(2)  5 C2 (0.27) (0.73)  0.284 3

(2) P(2)  5 C2 (0.27) (0.73)  0.284 ; P(3)  5 C3 (0.27) (0.73)  0.105

P(4)  5 C4 (0.27)4 (0.73)1  0.019 ; P(5)  5 C5 (0.27)5 (0.73)0  0.001

P( x  2)  P(2)  P(3)  P(4)  P(5)  0.284  0.105  0.019  0.001  0.409 0

(3) P(0)  5 C0 (0.27) (0.73)  0.2073 ; P(1)  5 C1 (0.27) (0.73)  0.3834 ; P( x  2)  P(0)  P(1)  0.2073  0.3834  0.591 6.

n  6, p  0.05, x  2 From Table 2 in Appendix B, the probability is 0.031.

P(0)  4 C0 (0.28)0 (0.72)4  0.269 P(1)  4 C1 (0.28)1 (0.72)3  0.418 P(2)  4 C2 (0.28)2 (0.72)2  0.244 P(3)  4 C3 (0.28)3 (0.72)1  0.063 P(4)  4 C4 (0.28)4 (0.72)0  0.006 x 0 1 2 3 4

P(x) 0.269 0.418 0.244 0.063 0.006

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

153

8. n = 31, p = 0.44, q = 0.56   np  (31)(0.44)  13.6

 2  npq  (31)(0.44)(0.56)  7.6  

npq 

(31)(0.44)(0.56)  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.

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. Unusual values have probabilities that are less than 0.05. (4) x  0,5 (5) x  4,5 (3) x  0,1 10. Unusual values have probabilities that are less than 0.05. (7) x  0 (8) x  0,1,2,8 (6) x  0,1,2,3,4,11,12

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CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

11.   np  (50)(0.4)  20

12.   np  (84)(0.65)  54.6

 2  npq  (50)(0.4)(0.6)  12

 2  npq  (84)(0.65)(0.35)  19.1

  npq  (50)(0.4)(0.6)  3.5

  npq  (84)(0.65)(0.35)  4.4

13.   np  (124)(0.26)  32.2

14.   np  (316)(0.82)  259.1

  npq  (124)(0.26)(0.74)  23.9

 2  npq  (316)(0.82)(0.18)  46.6

  npq  (124)(0.26)(0.74)  4.9

  npq  (316)(0.82)(0.18)  6.8

15. It is a binomial experiment. Success: frequent gamer who plays video games on smartphone n  10 , p  0.36 , q  0.64 , x  0,1,2,3,4,5,6,7,8,9,10 16. It is a binomial experiment. 17. It is a binomial experiment. Success: card drawn is a diamond n = 4, p = 0.25, q = 0.75, x = 0, 1, 2, 3, 4 18. It is a binomial experiment. Success: Woman age 18-33 who is a mother n  8 , p  0.42 , q  0.58 , x  0,1,2,3,4,5,6,7,8 19. n  7 , p  0.52 (a) P(4) = 0.283

(b) P(x ≥ 5) = P(5) + P(6) + P(7) ≈ 0.184 + 0.066 + 0.010 ≈ 0.260 (c) P(x < 4) = P(0) + P(1) + P(2) + P(3) ≈ 0.006 + 0.045 + 0.145 + 0.261 ≈ 0.457 20. n  12 , p  0.33 (a) P(6) = 0.108

(b) P ( x  6)  P ( 7 )  P (8)  P (9)  P (10)  P (11)  P (12)  0.046  0.014  0.003  0.000  0.000  0.000  0.063

(c) P ( x  6)  P ( 0)  P (1)  P (2)  P (3)  P (4)  P (5)  P (6)  0.000  0.000  0.000  0.003  0.014  0.046  0.108  0.171

21. n  10 , p  0.56 (a) P (4)  0.150

(b) P( x  5)  P(5)  P(6)  P(7)  P(8)  P(9)  P (10)  0.2289  0.2427  0.1765  0.0843  0.0238  0.0030  0.759

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

(c) P ( x  7)  1  P ( x  7)  1  [ P (7)  P(8)  P (9)  P(10)]  1  (0.1765  0.0843  0.0238  0.0030)  0.712 22. n  12 , p  0.11 (a) P (5)  0.006

(b) P( x  2)  1  P( x  2)  1  [ P(0)  P (1)]  1  (0.2470  0.3663)  0.387 (c) P( x  3)  P(0)  P(1)  P(2)  0.2470  0.3663  0.2490  0.862

23. n  11 , p  0.40 (a) P (5)  0.221

(b) P( x  5)  P(6)  P (7)  P(8)  P(9)  P(10)  P(11)  0.1471  0.0701  0.0234  0.0052  0.0007  0.0000  0.247

(b) P ( x  1)  1  P (0)  P (1)  1  0.028  0.113  0.859 (c) P ( x  1)  P (0)  P (1)  0.028  0.113  0.141 25. n  14 , p  0.04 (a) P (2)  0.089

(b) P( x  2)  1  P( x  2)  1  [ P(0)  P(1)  P(2)]  1  (0.5647  0.3294  0.0892)  0.017 (c) P(2  x  5)  P(2)  P(3)  P(4)  P(5)  0.0892  0.0149  0.0017  0.0001  0.106 26. n  15 , p  0.44 (a) P (7)  0.199

(b) P ( x  7)  P (8)  P (9)    P (15)  0.1561  0.0954  0.0450  0.0161  0.0042  0.0007  0.0001  0.0000  0.318

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CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

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)

Approximately symmetric (c) The values 0, 6, and 7 are unusual because their probabilities are less than 0.05. 28. (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)

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

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. 30. (a)

P(x)

0 1 2 3 4 5 6 7 8

0.00117 0.01239 0.05751 0.15246 0.25262 0.26790 0.17756 0.06725 0.01114

(b)

Approximately symmetric (c) The values 0, 1, and 8 are unusual because their probabilities are less than 0.05.

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CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

31.   np  (7)(0.71)  5.0

 2  npq  (7)(0.71)(0.29)  1.4   npq  (7)(0.71)(0.29)  1.2 On average, 5 out of every 7 U.S. adults think that political correctness is a problem in America today. The standard deviation is 1.2, so most samples of 7 U.S. adults would differ from the mean by at most 1.2 U.S. adults. 32.   np  (4)(0.50)  2

 2  npq  (4)(0.50)(0.50)  1  

npq 

(4)(0.50)(0.50)  1

On average, 2 out of every 4 adults are offended by how men portray women in rap and hip-hop music. The standard deviation is 1, so most samples of 4 adults would differ from the mean by at most 1 adult. 33.   np  (8)(0.79)  6.3

 2  npq  (8)(0.79)(0.21)  1.3  

npq 

(8)(0.79)(0.21)  1.2

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. 34.   np  (5)(0.36)  1.8

 2  npq  (5)(0.36)(0.64)  1.2  

npq 

(5)(0.36)(0.64)  1.1

On average, 1.8 out of every 5 likely U.S. voters believe that the federal government should get more involved in fighting local crime. The standard deviation is 1.1, so more samples of 5 likely U.S. voters would differ from the mean by at most 1.1 likely U.S. voters. 35.   np  (6)(0.32)  1.9

 2  npq  (6)(0.32)(0.68)  1.3  

npq 

(6)(0.32)(0.68)  1.1

On average, 1.9 out of every 6 U.S. employees who are late for work blame oversleeping. The standard deviation is 1.1, so most samples of 6 U.S. employees who are late would differ from the mean by at most 1.1 U.S. employees. 36.   np  (5)(0.10)  0.5

 2  npq  (5)(0.10)(0.90)  0.5  

npq 

(5)(0.10)(0.90)  0.7

On average, 0.5 out of every 5 college graduates think that Judge Judy serves on the Supreme Court. The standard deviation is 0.7, so most samples of 5 college graduates would differ from the mean by at most 0.7 graduate. 37. P  5, 2, 2,1 

10!  9   3   3   1           0.033 5!2!2!1!  16   16   16   16 

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

38. P  5, 2, 2,1 

10!  5   4   1   6           0.002 5!2!2!1!  16   16   16   16 

39. (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)     ...   0.107 10, 000 9999 9998 9991

(b) n  10 , p  0.20 P (0)  0.107 (c) The results are the same.

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

p  0.14, q  0.86, x  6

P(6)  (0.14)(0.86)61  0.066 2.

30 (2.71828)3 31 (2.71828)3 P(1)   0.050  0.149 0! 1! 32 (2.71828)3 33 (2.71828)3 P(2)  P(3)   0.224  0.224 2! 3! 34 (2.71828)3 P(4)   0.168 4! P(0)  P(1)  P(2)  P(3)  P (4)  0.050  0.149  0.224  0.224  0.168  0.815 P ( x  4)  1  P ( x  4)  1  0.815  0.185

P(0) 

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.



4.3 EXERCISE SOLUTIONS

P(3)  (0.65)(0.35)31  0.080

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CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

P(1)  (0.45)(0.55)11  0.45

P(5)  (0.09)(0.91)51  0.062

P(8)  (0.28)(0.72)81  0.028

P (4) 

(5) 4 (2.71828) 5  0.175 4!

P (3) 

(6)3 (2.71828) 6  0.089 3!

P(2) 

(1.5)2 (2.71828)1.5  0.251 2!

P(5) 

(9.8)5 (2.71828) 9.8  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  0.26 41 (a) P(4)  (0.26)(0.74)  0.105 st nd rd (b) P(clearing in 1 , 2 ,or 3 attempt)  P(1)  P(2)  P(3)

 (0.26)(0.74)11  (0.26)(0.74)21  (0.26)(0.74)31  0.595 st nd rd (c) P(not clearing in first three attempts)  1  P(clearing in 1 , 2 ,or3 attempt)  1  0.595  0.405

12. p 

3  0.03 100

31 (a) P(8)  (0.03)(0.97)  0.024 . This event is unusual because its probability is less than 0.05.

(b) P (defective on 1st, 2nd, or 3rd part sold )  P(1)  P(2)  P (3)

 (0.03)(0.97)11  (0.03)(0.97)21  (0.03)(0.97)31  0.03  0.029  0.028  0.087 This event is unusual because its probability is less than 0.05.

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

 1  P(1)  P(2)  P(3)  P(4)  P(5)  P(6)  P(7)  P(8)  1  (0.03)(0.97)11  (0.03)(0.97)21  (0.03)(0.97)31  (0.03)(0.97)41  (0.03)(0.97)51  (0.03)(0.97)61  (0.03)(0.97)71  (0.03)(0.97)81  1  0.030  0.029  0.028  0.027  0.027  0.026  0.025  0.024  0.784 13.   2

(2)5 (2.71828) 2  0.036 5! This event is unusual because its probability is less than 0.05.

(a) P (5) 

(b) P( x  5)  1   P(0)  P(1)  P(2)  P(3)  P(4) 

 1   0.135  0.271  0.271  0.180  0.090  0.053

This event is unusual because its probability is less than 0.05. (c) P( x  5)  1  P( x  5)  1   P(0)  P(1)  P(2)  P(3)  P(4)  P(5) 

 1   0.135  0.271  0.271  0.180  0.090  0.036  0.017 This event is unusual because its probability is less than 0.05 14.   6

(a) P(4) 

(6) 4 (2.71828) 6  0.134 4!

(b) P ( x  4)  P (0)  P(1)  P (2)  P(3)  P(4)  0.002  0.015  0.045  0.089  0.134  0.285 (c) P ( x  4)  1  P ( x  4)  1  0.285  0.715 15. p  0.641 21 (a) P(2)  (0.641)(0.359)  0.230 . st

(b) P(completes 1 or 2

pass)  P(1)  P(2)  (0.641)(0.359)11  (0.641)(0.359)21  0.641  0.230  0.871

st st nd (c) P(does not complete his 1 two passes)  1  P(completes 1 or 2 pass)  1  0.871  0.129

16. p  0.67 11 (a) P(1)  (0.67)(0.33)  0.67

(b) P(2)  (0.67)(0.33)

21

 0.221

161

162

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

1  0.0025 400

81 (a) P(8)  (0.0025)(0.9975)  0.0025 This event is unusual because its probability is less than 0.05.

(b) P (1st , 2nd , or 3rd item is faded )  P (1)  P(2)  P(3)

 (0.0025)(0.9975)11  (0.0025)(0.9975)21  (0.0025)(0.9975)31  0.0025  0.0025  0.0025  0.0075 This event is unusual because its probability is less than 0.05. (c) P( x  8)  1  P(1)  P(2)  P(3)  P(4)  P(5)  P(6)  P(7)  P(8)  (0.0025)(0.9975)11  (0.0025)(0.9975) 21  (0.0025)(0.9975)31  (0.0025)(0.9975) 4 1  1    (0.0025)(0.9975)51  (0.0025)(0.9975)6 1  (0.0025)(0.9975)7 1  (0.0025)(0.9975)81     1  0.0025  0.0025  0.0025  0.0025  0.0025  0.0025  0.0025  0.0025  1  0.02  0.98 18. p 

1  0.20 5

51 (a) P(5)  (0.20)(0.80)  0.082

(b) P( toy with 1st , 2nd , 3rd , or 4th purchase)  P(1)  P (2)  P(3)  P(4)

 (0.20)(0.80)11  (0.20)(0.80)21  (0.20)(0.80)31  (0.20)(0.80)41  0.200  0.160  0.128  0.102  0.59 (c) P( x  5)  1  P(1)  P(2)  P(3)  P(4)  P(5)



 1  (0.20)(0.80)11  (0.20)(0.80) 2 1  (0.20)(0.80)31  (0.20)(0.80) 4 1  (0.20)(0.80)51  1   0.200  0.160  0.128  0.102  0.082   1  0.672  0.328

19.   3.24

(a) P(2) 

(3.24)1 (2.71828) 3.24  0.206 2!

(b) P ( x  2)  P (0)  P (1)  P (2)  0.039  0.127  0.206  0.372 (c) P ( x  2)  1  P ( x  2)  1  0.372  0.628



CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

20.   16

(a) P(12) 

(16)12 (2.71828) 16  0.066 12!

(b) P( x  8)  1  P ( x  7)  1   P(0)  P(1)  P(2)  P (3)  P(4)  P(5)  P(6)  P (7) 

 1   0.0000  0.0000  0.0000  0.0001  0.0003  0.0010  0.0026  0.0060   1  0.010  0.990

(c) P( x  10)  P(0)  P (1)  P (2)  P(3)  P(4)  P (5)  P (6)  P(7)  P (8)  P (9)  P (10)  0.0000  0.0000  0.0000  0.0001  0.0003  0.0010  0.0026  0.0060

0.0120  0.0213  0.0341  0.077 21. n  6 , p  0.52 6! (a) P (5)  (0.52)5 (0.48)1  0.109 5!1!

(b) P ( x  4)  1  P ( x  4)  1   P (4)  P (5)  P (6)   1   0.253  0.109  0.020   1  0.382  0.618

(c) P( x  4)  P(4)  P(5)  P(6)  0.253  0.109  0.020  0.382 22. n  7 , p  0.63 7! (a) P (2)  (0.63) 2 (0.37)5  0.058 5!2!

(b) P( x  3)  1  P( x  3)  1   P(0)  P(1)  P(2)  P(3) 

 1   0.0009  0.0113  0.0578  0.1640  1  0.234  0.766 (c) P (1  x  4)  P(1)  P (2)  P (3)  P(4)  0.0113  0.0578  0.1640  0.2793  0.512 23. n  5 , p  0.68 5! (a) P (3)  (0.68)3 (0.32) 2  0.322 2!3!

(b) P( x  4)  P( x  3)  P(0)  P (1)  P(2)  P (3)  0.0034  0.0357  0.1515  0.3220  0.513 (c) P( x  3)  P(3)  P(4)  P(5)  0.322  0.342  0.145  0.809 24. n  6 , p  0.68 6! (a) P (4)  (0.68) 4 (0.32) 2  0.328 2!4!

(b) P( x  2)  1  P( x  2)  1   P(0)  P(1)  P(2)  1   0.0011  0.0137  0.0727  0.913

163

164

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

(a) P(17) 

(14)17 (2.71828)14  0.071 17!

(b) P ( x  17)  P (0)  P (1)    P (17)  0.827 (c) P ( x  17)  1  P ( x  17)  1  0.827  0.173 26.   8

(a) P(8) 

(8)8 (2.71828)8  0.140 8!

(b) P ( x  3)  P (0)  P (1)  P (2)  P (3)  0.000  0.003  0.011  0.029  0.042 This is an unusual event because the probability is less than 0.05. (c) P( x  8)  1  P( x  8)  1   P(0)  P(1)    P(8) 

 1   0.000  0.003  0.011  0.029  0.057  0.092  0.122  0.140  0.140  0.407 27. (a) n  6000 , p  P (4) 

1  0.0004 2500

6000! (0.0004) 4 (0.9996)5996  0.12542 5996!4!

6000  2.4 cars with defects per 6000. 2500 (2.4)4 (2.71828)2.4 P(4)   0.12541 4! The results are approximately the same.

(b)  

28. (a) P(0) 

(b) P(1) 

 2 C0 13 C3   (1)(286)  0.629 455

15 C3

 2 C1  13 C2  15 C3

 2 C2  13 C1  15 C3



(2)(78)  0.343 455



(1)(13)  0.029 455

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

29. p 

165

1  0.001 1000

1 1   1000 p 0.001 q 0.999 2  2   999,000 p (0.001)2

(a)  

   2  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. 30. p  0.005 1 1  200 (a)    p 0.005 q 0.995 2  2   39,800 p (0.005)2

   2  199.5 (b) 200 records, because it is the mean. 31.   4.1 2 (a)     4.1

   2  2.0 The standard deviation is 2.0 strokes, so most of Steven’s scores per hole differ from the mean by no more than 2 strokes. (b) For 18 holes, Steven’s average would be (18)(4.1)  73.8 strokes. So,   73.8 . P ( x  72)  1  P ( x  72)  1  0.447  0.553 32.   2.8 2 (a)     2.8

   2  1.7 The standard deviation is 1.7 bankruptcies, so most of the bankruptcies filed per hour differ from the mean by no more than 1.7 bankruptcies. (b) P ( x  5)  P (0)  P (1)  P(2)  P (3)  P (4)  P (5)  0.935

166

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

CHAPTER 4 REVIEW EXERCISE SOLUTIONS 1. Discrete. 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

(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 )  1.

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

7. (a)

x 0 1 2 3 4 5 6

P(x) 0.150 0.165 0.265 0.195 0.110 0.090 0.025

xP(x) 0 0.165 0.530 0.585 0.440 0.450 0.150  xP( x)  2.32

(x  ) –2.32 –1.32 –0.32 0.68 1.68 2.68 3.68

(x  )2 5.382 1.742 0.102 0.462 2.822 7.182 13.542

167

(x  )2P(x) 0.807 0.287 0.027 0.090 0.310 0.646 0.339  ( x   )2 P( x)  2.506

   xP ( x )  2.32  2.3  2   ( x   ) 2 P ( x)  2.506  2.5 ;

   2  2.506  1.6

(b) The mean is 2.3, so the average number of accidents on a weekday is about 2. The standard deviation is 1.6, so most of the accidents differ from the mean by no more than about 1 accident. 8. (a)

x 5 10 20 25 40

P(x) 0.250 0.325 0.200 0.100 0.125

xP(x) 1.25 3.25 4.00 2.50 5.00  xP( x)  16

(x  ) –11 –6 4 9 24

(x  )2 121 36 16 81 576

(x  )2P(x) 30.25 11.70 3.20 8.10 72.00  ( x   )2 P ( x)  125.25

   xP ( x )  16

 2   ( x   ) 2 P ( x )  125.25  125.3

   2  125.25  11.2 (b) The mean is 16, so the average weight of a sold packet is 16 lbs. The standard deviation is 11.3, so most of the packets that are sold differ from the mean by no more than about 5 lbs. 9. $1 10.

x 190 90 40 10

P(x) 0.0025 0.0100 0.0400 0.9475

xP(x) 0.475 0.900 1.600 9.475  xP( x)  6.500

   xP ( x )  $6.50 11. It is a binomial experiment. Success: a green candy is selected n  12 , p  0.16 , q  0.84 , x  0,1,2,3,4,5,6,7,8,9,10,11,12

168

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

12. Not a binomial experiment because the experiment is not repeated for a fixed number of trials. 13. n  8 , p  0.53 (a) P (3)  0.191

(b) P( x  3)  1  P  x  3  1   P(0)  P(1)  P(2) 

 1   0.0024  0.0215  0.0848  1  0.109  0.891 (c) P( x  3)  1  P  x  3  1   P(0)  P(1)  P(2)  P(3) 

 1   0.0024  0.0215  0.0848  0.1912  1  0.300  0.700 14. n  12 , p  0.39 (a) P (2)  0.072

(b) P( x  2)  1  P  x  2  1   P(0)  P(1) 

 1   0.0027  0.0204  1  0.023  0.977 (c) P( x  2)  1  P  x  2  1   P(0)  P(1)  P(2) 

 1   0.0027  0.0204  0.0716  1  0.095  0.905 15. n  9 , p  0.88 (a) P (6)  0.067

(b) P ( x  6)  P (6)  P (7)  P (8)  P (9)  0.0674  0.2119  0.3884  0.3165  0.984 (c) P( x  6)  P(7)  P(8)  P(9)  0.2119  0.3884  0.3165  0.917 16. n  5 , p  0.62 (a) P (2)  0.211

(b) P( x  2)  1  P  x  2  1   P(0)  P(1) 

 1   0.008  0.065  1  0.073  0.927 (c) P( x  2)  1  P  x  2  1   P(0)  P(1)  P(2) 

 1   0.0079  0.0646  0.2109  1  0.283  0.717

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

17. (a)

P(x)

0 1 2 3 4 5

0.0008 0.0126 0.0798 0.2529 0.4003 0.2536

(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

0.000003 0.000131 0.002409 0.023552 0.129534 0.379967 0.464404

(b)

Skewed left (c) The values 0, 1, 2, and 3 are unusual because their probabilities are less than 0.05. 19. n  8 , p  0.13   np  (8)(0.13)  1.0

169

170

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

  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. 20. n  5 , p  0.56   np  (5)(0.56)  2.8

 2  npq  (5)(0.56)(0.44)  1.2   npq  (5)(0.56)(0.44)  1.1 On average, 2.8 out of every 5 college student-athletes receive athletics scholarships. The standard deviation is 1.1, so most samples of 5 college student-athletes would differ from the mean by at most 1.1 college student-athletes. 21. p  0.82 1 (a) P(2)  (0.82)(0.18)  0.148 th

(b) P(4 or 5 )  P(4)  P(5)  (0.82)(0.18)  (0.82)(0.18)  0.006 This event is unusual because its probability is less than 0.05. (c) P(not one of the 2nd through 7th )  P(1st or 8th or 9th or 10th )  0.820  0.000  0.000  0.000  0.820 22.   0.27

(a) P(0) 

(0.27)0 (2.71828)0.27  0.763 0!

(b) P( x  2)  P(0)  P(1)  P(2)  0.763  0.206  0.028  0.997 (c) P  x  1  1  P( x  1)  1   P(0)  P(1)   1   0.763  0.206  0.031 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  0.011 This event would be unusual because its probability is less than 0.05. 24. p 

27  0.474 57

0 (a) P(1)  (0.474)(0.526)  0.474 1

(b) P(2)  (0.474)(0.526)  0.249

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

171

0 1 2 (c) P( x  3)  P(1)  P(2)  P(3)  (0.474)(0.526)  (0.474)(0.526)  (0.474)(0.526)  0.854

(d) 1  P ( x  3)  1  0.854  0.146 25.   6.1

(a) P(3) 

(6.1)3 (2.71828)6.1  0.085 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   0.002  0.014  0.042  0.085  0.129  0.158  0.160  0.410 (c) P( x  5)  P(0)  P(1)  P(2)  P (3)  P (4)  P(5)  0.002  0.014  0.042  0.085  0.129  0.158  0.430 26. n  10 , p  0.82 (a) P (8)  0.298

(b) P ( x  6)  P (7)  P (8)  P (9)  P (10)  0.1745  0.2980  0.3017  0.1374  0.912 (c) P ( x  6)  1  P ( x  6)  1  0.912  0.088

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 number of die rolls required for an individual to roll a five 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

P(x) 0.238 0.405 0.209 0.090 0.040 0.019  P ( x)  1

172

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

(b)

Skewed right (c)

xP(x) 0.000 0.405 0.418 0.271 0.158 0.095  xP( x)  1.345

(x  ) 1.345 0.345 0.655 1.655 2.655 3.655

(x  )2 1.809 0.119 0.429 2.739 7.049 13.359

   xP ( x )  1.345  1.3

(x  )2P(x) 0.430 0.048 0.090 0.247 0.279 0.252  ( x   )2 P( x)  1.346

 2   ( x   ) 2 P ( x )  1.346  1.3

   2  1.346  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  0.36

(a) P (3)  0.269 (b) P( x  4)  P(0)  P(1)  P(2)  P(3)  P(4)  0.0180  0.0912  0.2052  0.2693  0.2272  0.811 (c) P ( x  5)  P (6)  P (7)  P (8)  P (9)  0.0479  0.0116  0.0016  0.0001  0.061 4.

n  6 , p  0.86

(a)

P(x)

0 1 2 3 4 5 6

0.000008 0.000278 0.004262 0.034907 0.160820 0.395159 0.404567

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

173

(b)

Skewed left (c)   np  (6)(0.86)  5.2

 2  npq  (6)(0.86)(0.14)  0.7   npq  (6)(0.86)(0.14)  0.8 On average, 5.2 out of every 6 patients have a successful surgery. The standard deviation is 0.8, so most samples of 6 surgeries would differ from the mean by at most 0.8 surgery. 5.

 5 (5)5 (2.71828)5  0.175 5!

(a) P(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  0.56 3 (a) P(4)  (0.56)(0.44)  0.048 nd

(b) P(2

or 3rd )  P(2)  P(3)  (0.56)(0.44)1  (0.56)(0.44)2  0.355





 1  (0.56)(0.44)0  (0.56)(0.44)1  (0.56)(0.44)2  0.085 7. Event (a) is unusual because its probability is less than 0.05.

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CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

CHAPTER 4 TEST SOLUTIONS

p

1  0.0084 119

(a) P (25)  (0.0084)(0.9916) 24  0.007 This event is unusual because its probability is less than 0.05. (b) P (1st or 2 nd )  P (1)  P (2)  (0.0084)(0.9916) 0  (0.0084)(0.9916)1  0.017 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   0.0084  0.0083  0.0083  0.0082  0.0081  1  0.0413  0.959

n  6 , p  0.60 (a) P (2)  0.138 (b) P ( x  3)  1  P ( x  3)  1  ( P (0)  P (1)  P(2))  1  (0.004  0.037  0.138)  0.821 (c) P ( x  4)  P (0)  P (1)  P (2)  P (3)  0.0041  0.0369  0.1382  0.2765  0.456

4 (a) P (6) 

(4) 6 (2.71828) 4  0.104 6!

(b) P ( x  8)  1  P ( x  8)  1   P (0)  P (1)  P (2)  P (3)  P (4)  P (5)  P (6)  P (7)  P (8)   1   0.018  0.073  0.147  0.195  0.195  0.156  0.104  0.060  0.030 

 1  0.979  0.021

(b) No,  P ( x )  1 . 5. (a)

x 17 18 19 20 21 22

f 2 13 4 3 2 1 n  25

P(x) 0.08 0.52 0.16 0.12 0.08 0.04  P ( x)  1

CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

175

(b)

Skewed right (c)

xP(x) 1.36 9.36 3.04 2.40 1.68 0.88  xP( x)  18.72

(x  ) 1.72 0.72 0.28 1.28 2.28 3.28

(x  )2 2.9584 0.5184 0.0784 1.6384 5.1984 10.7584

(x  )2P(x) 0.2367 0.2696 0.0125 0.1966 0.4159 0.4303  ( x   )2 P( x)  1.5616

   xP ( x )  18.72  18.7

 2   ( x   ) 2 P ( x )  1.5616  1.6

   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  0.77 (a) x P(x) 0 0.0006 1 0.0108 2 0.0721 3 0.2415 4 0.4043 5 0.2707

 P ( x)  1

(b)

Skewed left

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CHAPTER 4 │ DISCRETE PROBABILITY DISTRIBUTIONS

 2  npq  (5)(0.77)(0.23)  0.9   npq  (5)(0.77)(0.23)  0.9 On average, 3.9 out of every 5 U.S. college students pay their bills on time. The standard deviation is 0.9, so most samples of 5 U.S. college students would differ from the mean by at most 0.9 U.S. college student.

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: m = 300 Inflection points are approximately: 263 and 337 Because the inflection points are one standard deviation from the mean, s = 37 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. Answers will vary. 2. Neither. In a normal distribution, the mean and median are equal. 3. 1 4. Points at which the curve changes from curving upward to curving downward; m -s and m + s 177 Copyright © 2019 Pearson Education Ltd.

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CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

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.

m = 0 , s =1

8. Transform each data value x into a z-score by subtracting the mean from x and dividing the result by x-m the standard deviation. In symbols, z = . s 9. “The” standard normal distribution is used to describe one specific normal distribution ( m = 0 , s =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. m » 18.5 , s » 2 13. No, the graph crosses the x-axis. 14. No, the graph is not symmetric. 15. Yes, the graph fulfills the properties of the normal distribution. m » 11.5 , s » 1.5 16. No, the graph is skewed to the right. 17. (Area left of z = 1.3) = 0.9032 18. (Area left of z = -1.5) = 0.0668 19. (Area right of z = 2) = 1 - (Area leftt of z = 2) = 1 - 0.9772 = 0.0228 20. (Area right of z = -2.3) = 1 - (Area leftt of z = -2.3) = 1 - 0.0107 = 0.9893 21. (Area left of z = 1.2) - (Area left of z = -0.7) = 0.8849 - 0.2420 = 0.6429 22. (Area left of z = 0) - (Area left of z = -2.25) = 0.5000 - 0.0122 = 0.4878 23. 0.6026 Copyright © 2019 Pearson Education Ltd.

24. 0.0021

25. 0.0311

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

26. 0.8820

27. 1 – 0.1611 = 0.8389

28. 1 – 0.9960 = 0.0040

29. 1 – 0.1698 = 0.8302

30. 1 – 0.8800 = 0.1200

31. 0.9979 - 0.5000 = 0.4979

32. 0.5000 – 0.0066 = 0.4934

33. 0.9878 – 0.0122 = 0.9756

34. 0.9750 – 0.0250 = 0.9500

35. 0.4404 + 0.4404 = 0.8808

36. 0.0052 + 0.1056 = 0.1108

179

37. (a)

It is reasonable to assume that the life spans are not normally distributed because the histogram is asymmetric towards the right. (b) x = 69,836; s = 16, 405 (c) The sample mean of 69,836 runs is less than the claimed mean, so, on average, the flash drives in the sample lasted for a shorter time. The sample standard deviation of 16,405 is greater than the claimed standard deviation, so the flash drives in the sample had greater variation in running cycles than the manufacturer’s claim. 38. (a)

It is reasonable to assume that the monthly cocoa consumptions are not normally distributed because the histogram is slightly skewed towards the left.

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(b) x = 19.5; s = 4.8 (c) The mean of your sample is 1.5 grams less than that of the previous study, so the average cocoa consumption from the sample is less than in the previous study. The standard deviation is about 0.3 grams less than that of the previous study, so cocoa consumptions are slightly less spread out than in the previous study. x - m 66.50 - 66.25 = » 0.1 s 2.50 x - m 69.75 - 66.25 x = 69.75  z = = » 1.4 s 2.50 x - m 72.50 - 66.25 x = 72.50  z = = » 2.5 s 2.50 x - m 60.75 - 66.25 x = 60.75  z = = » -2.2 s 2.50

39. (a) x = 66.50  z =

(b) x = 72.5 is unusual because its corresponding z-score (2.5) lies more than 2 standard deviations from the mean. x = 60.75 is unusual because its corresponding z-score (−2.2) lies more than 2 standard deviations from the mean. x - m 67.50 - 63.75 = » 2.14 s 1.75 x - m 66.50 - 63.75 x = 66.50  z = = » 1.57 s 1.75 x - m 61.25 - 63.75 x = 61.25  z = = » -1.43 s 1.75 x - m 63.75 - 63.75 x = 63.75  z = = =0 s 1.75

40. (a) x = 67.50  z =

(b) x = 67.50 is unusual because its corresponding z-score (2.14) lies more than 2 standard deviations from the mean. 41. 0.9750

42. 0.2660

43. 1- 0.0168 = 0.9832

44. 1- 0.8997 = 0.1003

45. 0.8413 - 0.1587 = 0.6826

46. 0.9535 - 0.5000 = 0.4535

47. P(z < 2.32) = 0.9898

48. P(z < –0.26) = 0.3974

49. Using technology: P(z > 3.285) = 1 – P(z < 3.285) = 1 – 0.9995 = 0.0005 50. P(z > –2.55) = 1 – P(z < –2.55) = 1 – 0.0054 = 0.9946 51. P(–1.35 < z < 0) = 0.5 – 0.0885 = 0.4115 52. Using technology: P(0 < z < 2.315) = 0.9897 – 0.5 = 0.4897 53. P(–2.96 < z < 2.96) = 0.9985 – 0.0015 = 0.9970

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

181

54. P(–0.18 < z < 0.18) = 0.5714 – 0.4286 = 0.1428 55. P(z < –3.1 or z > 3.1) = 2(0.0007) = 0.0014 56. P(z < –2.46 or z > 2.46) = 2(0.0069) = 0.0139 57.

The normal distribution curve is centered at its mean (60) and has 2 points of inflection (48 and 72) representing m  s . 58.

The normal distribution curve is centered at its mean (450) and has 2 points of inflection (400 and 500) representing m  s . æ 1 ö÷ b - a 59. The area under the curve is (b - a ) çç = =1. çè b - a ÷ø÷ b - a (Because a < b, you do not have to worry about division by 0.) æ 1 ö÷ 60. (a) P (2 < x < 8) = (8 - 2) çç = 0.25 çè 25 - 1÷÷ø æ 1 ö÷ (b) P (4 < x < 12) = (12 - 4) çç = 0.333 çè 25 - 1÷ø÷ æ 1 ö÷ (c) P (5 < x < 17) = (17 - 5) çç = 0.5 çè 25 - 1ø÷÷ æ 1 ö÷ (d) P (8 < x < 14) = (14 - 8) çç = 0.25 çè 25 - 1÷ø÷

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CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

5.2 NORMAL DISTRIBUTIONS: FINDING PROBABILITIES 5.2 TRY IT YOURSELF SOLUTIONS 1.

x - m 70 - 67 = » 0.857 s 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. z=

x - m 31 - 43 = = -1 s 12 x - m 58 - 43 z= = = 1.25 s 12 P (31 < x < 58) = P (-1 < z < 1.25) = P ( z < 1.25) - P ( z < -1) z=

= 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 31 and 58 minutes. 3.

P (100 < x <150) = P(0.12 < z < 2.12) = 0.9830 - 0.5478 = 0.4352 The probability that a randomly selected person’s triglyceride level is between 100 and 150 is 0.4352.

5.2 EXERCISE SOLUTIONS 1.

P ( x <180) = P ( z < 0.3) = 0.6179

P ( x <160) = P ( z <-0.7) = 0.2420

P ( x > 185) = P ( z > 0.55) = 1- 0.7088 = 0.2912

P ( x > 170) = P ( z >-0.2) = 1- 0.4207 = 0.5793

P (170 < x <195) = P (-0.2 < z <1.05) = 0.8531- 0.4207 = 0.4324

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

P (155 < x < 172) = P (-0.95 < z <-0.1) = 0.4602 - 0.1711 = 0.2891

183

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CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

7. (a) Using technology: P ( x < 4) » P ( z < -0.6364) = 0.2623 (b) Using technology: P (4 < x < 6) = P (-0.6364 < z < 0.2727) = 0.6075 - 0.2623 = 0.3452 (c) Using technology: P ( x > 8) = P ( z > 1.1818) = 1 - P ( z < 1.1818) = 1 - 0.8814 = 0.1186 None of these events are unusual because their probabilities are greater than 0.05. 8. (a) Using technology: P ( x < 6.5) = P ( z <-1.28) = 0.1003 (b) Using technology: P (7 < x < 7.5) = P (-0.28 < z < 0.72) = 0.7642 - 0.3897 = 0.3745 (c) Using technology: P ( x > 8) = P ( z > 1.72) = 1- P ( z < 1.72) = 1- 0.9573 = 0.0427 Yes, the event in part (c) is unusual because its probability is less than 0.05. 9. (a) Using technology: P ( x < 50) = P ( z <-0.71) = 0.2389 (b) Using technology: P (55 < x < 60) = P (-0.28 < z < 0.16) = 0.5636 - 0.3897 = 0.1739 (c) Using technology: P ( x > 70) = P ( z > 1.02) = 1- P ( z < 1.02) = 1- 0.8461 = 0.1539 No, none of these events are unusual because their probabilities are greater than 0.05. 10. (a) Using technology: P( x < 120) = P( z < -1.6333) = 0.0512 (b) Using technology: P (122 < x < 128) = P (-0.9667 < z < 1.0333) = 0.84927 - 0.16685 = 0.6824 (c) Using technology: P ( x > 130) = P ( z > 1.7) = 1 - P ( z < 1.7) = 1 - 0.9554 = 0.0446 The event in part (c) is unusual because its probability is less than 0.05. 11. (a) Using technology: P ( x < 200) = P ( z <-2.0) = 0.0228 (b) Using technology: P (300 < x < 600) = P (-1.3333 < z < 0.6667) = 0.74754 - 0.09124 = 0.6563 (c) Using technology: P ( x > 800) = P ( z > 2.0) = 1- P ( z < 2.0) = 1- 0.9772 = 0.0228 12. (a) Using technology: P ( x < 7) = P ( z <-1.5) = 0.0668 (b) Using technology: P (10 < x < 15) = P (0 < z <-2.5) = 0.9938 - 0.5 = 0.4938 (c) Using technology: P ( x > 16) = P ( z > 3) = 1- 0.9987 = 0.0013 Copyright © 2019 Pearson Education Ltd.

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

185

13. Using technology: P (750 < x < 1000) = P (-1.7254 < z < -0.4301) = 0.3336 - 0.0422 = 0.2914 14. Using technology: P (28 < x < 33) = P (1.2857 < z < 2.1786) = 0.9853 - 0.9007 = 0.0846 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 <1200) = P ( z < 0.6062) = 0.7278  72.78% (b) Using technology: P ( x > 1300) = P ( z > 1.1244) = 1- P ( z < 1.1244) = 1- 0.8696 = 0.1304

(500)(0.1304) = 65.2  65 scores 18. (a) Using technology: P ( x < 16) = P ( z <-0.8571) = 0.1957  19.57% (b) Using technology: P ( x > 23) = P ( z > 0.3929) = 1- P ( z < 0.3929) = 1- 0.6528 = 0.3472

(1000)(0.3472) = 347.2  347 scores 19. (a) Using technology: P ( x < 280) = P ( z <1.3000) = 0.9032  90.32% (b) Using technology: (250 < x < 275) = P (-1.7000 < z < 0.8000) = 0.7881- 0.0446 = 0.7435  74.35% P (c) Using technology: P ( x > 290) = P ( z > 2.3) = 1- P ( z < 2.3) = 1- 0.9893 = 0.0107

(500)(0.0107) = 5.35  5 mothers 20. (a) Using technology: P ( x < 5.75) = P ( z < 0.8750) = 0.8092  80.92% (b) Using technology: P (4.75 < x < 5.25) = P (-1.6250 < z <-0.3750) = 0.3538 - 0.0521 = 0.3017  30.17% (c) Using technology: P ( x > 5.00) = 1- P ( x < 5.00) = 1- P ( z <-1.0000) = 1- 0.1587 = 0.8413

(500)(0.8413) = 420.65  420 men 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.

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24. In control, because none of the three warning signals detected a change.

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

187

5.3 NORMAL DISTRIBUTIONS: FINDING VALUES 5.3 TRY IT YOURSELF SOLUTIONS 1. (1)

(2)

z = -1.77 2. (1)

(2)

use area = 0.1003

z = -1.28

z =1.96 (3)

use area = 0.2005

z = -0.84

use area = 0.9901

z = 2.33

m = 52, s = 15

(1) z = -2.33  x = m + zs = 52 + (-2.33)(15) = 17.05 (2) z = 3  x = m + zs = 52 + (3)(15) = 97 (3) z = 0.58  x = m + zs = 52 + (0.58)(15) = 60.70 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.

z =-2.33 x = m + zs = 129 + (-2.33)(5.18) » 116.93 So, the longest braking distance one of these cars could have and still be in the bottom 1% is about 117 feet.

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CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

z = -1.28 x = m + zs = 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. z = 0.98

z = 0.63

z = –1.53

4. z = –0.41

z = –1.905

z = 0.878

7. z = 1.205

z = –2.155

z = –0.842

10. z = –0.126

11. z = 1.645

12. z = 0.524

13. z = –1.405

14. z = –0.10

15. z = 0.954

16. z = 0.126

17. z = -0.38

18. z = 0.25

19. z =1.99

20. z = -0.58

21. z = 1.96

22. z = 1.645 23. z = 1.011

24. z = −1.221

25. z = −0.16

26. z = −0.4817

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

27. z = 1.656

28. z = 1.49

29. z = ±1.036

30. z = ±0.1891

31. (a) 90th percentile  Area = 0.90  z » 1.2816 (using technology) x = m + zs = 7.14 + (1.2816)(0.5) » $ 7.78 million (b) 32nd percentile  Area = 0.32  z »-0.4677 (using technology) x = m + zs = 7.14 + (-0.4677)(0.5) » $ 6.91 million (c) 75th percentile  Area = 0.75  z »-0.6745 (using technology) x = m + zs = 7.14 + (0.6745)(0.5) » $ 7.48 million 32. (a) 88th percentile  Area = 0.88  z »1.17499 (using technology) x = m + zs = 5.4 + (1.17499)(2.2) » 7.98 (b) 61st percentile  Area = 0.61  z » 0.2793 (using technology) x = m + zs = 5.4 + (0.2793)(2.2) » 6.01 (c) 1st quartile  Area = 0.25  z »-0.6745 (using technology) x = m + zs = 5.4 + (-0.6745)(2.2) » 3.92 33. (a) 5th percentile  Area = 0.05  z »-1.64485 (using technology) x = m + zs = 2277 + (-1.64485)(584.2) » 1316.08 kilowatt-hours (b) 17th percentile  Area = 0.17  z »-0.954165 (using technology) x = m + zs = 2277 + (-0.954165)(584.2) » 1719.58 kilowatt-hours (c) 3rd quartile  Area = 0.75   z » 0.6745 (using technology) x = m + zs = 2277 + (0.6745)(584.2) » 2671.04 kilowatt-hours

189

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CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

34. (a) 80th percentile  Area = 0.80  z » 0.8416 (using technology) x = m + zs = 1.64 + (0.8416)(2.84) » 4.03 mega gallons (b) 29th percentile  Area = 0.29  z »-0.5534 (using technology) x = m + zs = 1.64 + (-0.5534)(2.84) » 0.07 mega gallon (c) 3rd quartile  Area = 0.75   z » 0.6745 (using technology) x = m + zs = 1.64 + (0.6745)(2.84) » 3.56 mega gallons 35. (a) Top 5%  Area = 0.95  z » 1.6449 (using technology) x = m + zs = 3.36 + (1.6449)(0.18) » 3.66 (b) Middle 50%  Area = 0.25 to 0.75  z » 0.6745 (using technology) x = m + zs = 3.36 + (0.6745)(0.18) » 3.24 to 3.48 36. (a) Bottom 10%  Area = 0.10  z »-1.2816 (using technology) x = m + zs = 3.5 + (-1.2816)(0.87) » 2.39 (b) Middle 80%  Area = 0.10 to 0.90  z » 1.2816 (using technology) x = m + zs = 3.5 + (1.2816)(0.87) » 2.39 to 4.62 37. (a) Upper 15%  Area = 0.85  z » 1.0364 (using technology) x = m + zs = 13.5 + (1.0364)(0.5) » 14.02 grams per deciliter (b) Bottom 25%  Area = 0.25  z »-0.6745 (using technology) x = m + zs = 13.5 + (-0.6745)(0.5) » 13.16 grams per deciliter 38. (a) Upper 20%  Area = 0.80  z » 0.8416 (using technology) x = m + zs = 80,000 + (0.8416)(1,500) » 81,262 kilometers (b) Bottom 10%  Area = 0.10  z »-1.2816 (using technology) x = m + zs = 80,000 + (-1.2816)(1,500) » 78,078 kilometers 39. Upper 2%  Area = 0.98  z » 2.0537 (using technology) x = m + zs = 500 + (2.0537)(25) » 551.34 grams 40. Bottom 15%  Area = 0.15  z »-1.0364 (using technology) x = m + zs = 8.5 + (-1.0364)(0.75) » 7.72 years Thermostats that do not wear out by 7.72 years will be replaced free of charge. 41. Top 1%  Area = 0.99  z » 2.3264 (using technology)

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x = m + zs  8 = m + (2.3264)(0.03)  m » 7.93 ounces 42. (a) A: Top 10%  Area = 0.90  z » 1.2816 (using technology) x = m + zs = 72 + (1.2816)(9) » 83.53 (b) B: Top 30%  Area = 0.70  z » 0.5244 (using technology) x = m + zs = 72 + (0.5244)(9) » 76.72 (c) C: Top 70%  Area = 0.30  z »-0.5244 (using technology) x = m + zs = 72 + (-0.5244)(9) » 67.28 (d) D: Top 90%  Area = 0.10  z »-1.2816 (using technology) x = m + zs = 72 + (-1.2816)(9) » 60.47

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

x 1 1.67 2.33 3 3.67 4.33 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

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 

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

191

192

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

m x = 3, s x2 » 0.889, s x » 0.943

mx = m = 3, 8 s2 = 3 » 0.889, n 3 8 s 3 » 0.943 sx = = n 3 s x2 =

mx = m = 6.8, s x =

1.4

» 0.18

With a smaller sample size, the mean stays the same but the standard deviation increases. s

mx = m = 3.5, s x =

mx = m = 20.7, s x =

x = 19.4 : z =

x = 22.5: z =

s n

0.2

= 0.05

6.5 100

= 0.65

x - m 19.4 - 20.7 1.3 = == -2 s 6.5 0.65 n 100

x - m 22.5 - 20.7 1.8 = =» 2.77 s 6.5 0.65 n 100

P(19.4 < x < 22.5) = P(-2 < z < 2.77) » 0.9972 - 0.0228 = 0.9744 Of the samples of 100 drivers ages 16 to 19, about 97.44% will drive a mean distance each day that is between 19.4 and 22.5 miles.

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

mx = m = 235,500, s x =

x = 225,000 : z =

s n

50,000 12

193

» 14, 433.76

x - m 225,000 - 235,500 -10,500 = » » -0.73 s 50,000 14, 433.76 n 12

P( x > 225,000) = P( z >-0.73) = 1- P( z <-0.73) = 1- 0.2327 = 0.7673 About 77% of samples of 12 single-family houses will have a mean sales price greater than $225,000. x - m 200 - 190 6. x = 200 : z = = » 0.21 s 48 x - m 200 -190 10 x = 200 : z = = » » 0.66 s 48 15.18 n 10 P ( z < 0.21) = 0.5832 P ( z < 0.66) = 0.7454 There is about a 58% chance that an LCD computer monitor will cost less than $200. There is about a 75% chance that the mean of a sample of 10 LCD computer monitors is less than $200.

5.4 EXERCISE SOLUTIONS 1.

mx = m = 225 sx =

2. mx = m = 99

40 s = » 4.619 75 n

sx =

mx = m = 1022 sx =

12 s = = 0.8 225 n

4. mx = m = 4848

144 s = » 7.589 360 n

sx =

24 s = » 0.693 1200 n

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 ³ 30 or the population is normal. 8. True 9. (c) Because mx = 16.5, s x =

s n

11.9 100

= 1.19, and the graph approximates a normal curve.

194

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

10. (b) Because mx = 5.8, s x =

s n

2.3 100

» 0.23, and the graph approximates a normal curve.

11. (a) The mean and the standard deviation of the population (64, 48, 19, 79, 56) are: m = 53.2, s » 19.9

(b)

Sample

Mean

Sample

Mean

Sample

Mean

19, 19

48, 79

63.5

64, 64

19, 48

33.5

56, 19

37.5

64, 79

71.5

19, 56

37.5

56, 48

79, 19

19, 64

41.5

56, 56

79, 48

63.5

19, 79

56, 64

79, 56

67.5

48, 19

33.5

56, 79

67.5

79, 64

71.5

48, 48

64, 19

41.5

79, 79

48, 56

64, 48

48, 64

64, 56

19.9

» 14.1 2 The means are equal, but the standard deviation of the sampling distribution is smaller.

12. (a) The mean and the standard deviation of the population (1.000, 1.004, 1.001, 1.003) are: m = 1.002, s » 0.0016

(b)

Sample

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

Mean

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

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

0.0016

» 0.0011 2 The means are equal, but the standard deviation of the sampling distribution is smaller.

13. (a) The mean and the standard deviation of the population (350, 399, 418) are: m = 389, s » 28.65 (b) Sample Mean Sample Mean Sample

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 (c) mx = 389, s x »

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

28.65

Mean

389 395.33 389 405.33 411.67 395.33 411.67 418

» 16.54 3 The means are equal, but the standard deviation of the sampling distribution is smaller.

14. (a) The mean and the standard deviation of the population (70, 85, 81, 67) are: m = 75.75, s » 7.46

(b)

Sample

Mean

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

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

195

196

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

70, 67, 81 70, 67, 85 70, 70, 67 70, 70, 70

72.67 74 69 70

Sample

Mean

Sample

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

7.46

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

» 4.31 3 The means are equal, but the standard deviation of the sampling distribution is smaller.

0.8 x - m 37.2 - 38 = » » -2.58 s 2.17 0.31 n 49 P ( x < 37.2) = P ( z < -2.58) = 0.0049 The probability is unusual because it is less than 0.05.

15. z =

0.8 x - m 37.2 - 38 = » » -5.53 s 2.17 0.1447 n 225 P ( x > 37.2) = P ( z > -5.53) = 1 - P ( z < -5.53) = 1 - 0 = 1 The probability is not unusual because it is greater than 0.05.

16. z =

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

2 x - m 132 -130 = » » 0.96 s 16.1 2.079 n 60 P ( x > 132) = P ( z > 0.96) = 1 - P ( z < 0.96) = 1 - 0.8315 = 0.1685 (using technology) The probability is not unusual because it is greater than 0.05.

17. z =

2 x - m 102 - 100 = = » 2.22 s 4.5 0.9 n 25 P ( x < 100 or x > 102) = P ( z < 0 or z > 2.22) » 0.5 + 0.0131 = 0.5131 (using technology) The probability is not unusual because it is greater than 0.05.

18. z =

19. m x = 495 sx =

120 s = » 26.83 20 n

21. mx = 23

sx =

1.3 s = = 0.26 25 n

20. mx = 493

sx =

s n

114 36

= 19

22. mx = 87

sx =

14.5 s = » 2.65 30 n

197

198

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

23. mx = 1.64

sx =

24. mx = 196

2.89 s = » 0.83 12 n

sx =

25. mx = 132,000

sx =

s n

18,000 35

» 3042.56

26. mx = 111,000

sx =

13,000 s = » 1876.39 48 n

27. n = 15 : mx = 495, s x =

n = 10 : mx = 495, s x =

s n s

= =

120 15 120 10

» 30.98 » 37.95

70 s = » 18.07 15 n

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

199

As the sample size decreases, the standard deviation of the sample mean increases, while the mean of the sample means remains constant. 28. n = 18 : mx = 493, s x =

n = 12 : mx = 493, s x =

s n s n

= =

114 18 114 12

» 26.87 » 32.91

As the sample size decreases, the standard deviation of the sample mean increases, while the mean of the sample means remains constant. x - m 200 - 456 = » -1.1919 s 1215 n 32 x - m 500 - 456 z= = » 0.2049 s 1215 n 32 P (200 < x < 500) = P (-1.1919 < z < 0.2049) » 0.5812 - 0.1167 = 0.4645 (using technology) About 46% of samples of 32 years will have a mean gain between 200 and 500.

29. z =

x - m 9.1 - 10.72 = » -0.536901 s 18.6 n 38 x - m 10.3 - 10.72 z= = » -0.139196 s 18.6 n 38 P (9.1 < x < 10.3) = P (-0.536901 < z < -0.139196) = 0.44465 - 0.29567 » 0.1490 (using technology) About 15% of samples of 38 years will have a mean return between 9.1% and 10.3%.

30. z =

x - m 2.6 - 2.25 = » 1.4746 s 1.30 n 30 P ( x > 2.6) = P ( z > 1.4746) » 1 - 0.9298 = 0.0702 (using technology) About 7% of samples of 30 Chinese cities will have a mean childhood asthma rate greater than 2.6%.

31. z =

200

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x - m 28 - 25.5 = » 0.04197 s 395.1 n 44 P ( x > 28) = P ( z > 0.04197) = 1 - P ( z < 0.04197) » 1 - 0.5167 » 0.4833 (using technology) About 48% of samples of 44 countries will have a mean per capita carbon dioxide emission greater than 28 metric tons.

32. z =

x - m 3.2 - 2.25 = » 0.7308 s 1.30 P ( x < 3.2) = P ( z < 0.7308) = 0.7675 (using technology) x - m 3.2 - 2.25 z= = » 2.3109 1.30 s n 10 P ( x < 3.2) = P ( z < 2.3109) » 0.9896 (using technology) It is more likely to select a sample of 10 cities with a mean childhood asthma prevalence less than 3.2% because the sample of 10 has a higher probability.

33. z =

x - m 30 - 25.5 = » 0.01139 s 395.1 P ( x < 30) = P( z < 0.01139) = 0.5045 (using technology) x - m 30 - 25.5 z= = » 0.04411 s 395.1 n 15 P ( x < 30) = P ( z < 0.04411) » 0.5176 (using technology) It is more likely to select a sample of 15 countries with mean carbon dioxide emissions less than 30 metric tons because the sample of 15 has a higher probability.

34. z =

0.01 x - m 4.99 - 5 = »» -4.35 s 0.02 0.002 n 75 P ( x < 4.99) = P ( z < -4.35) » 0.0000 (using technology) No, it is not likely that we would have randomly sampled 75 cans with a mean less than or equal to 5 kilograms because it is more than 4 standard deviations from the mean of the sample means. So, the machine needs to be reset.

35. z =

0.5 x - m 125.5 - 125 = »» 0.5 s 4 1 n 16 P ( x < 125.5) = P ( z < 0.5) = 1 - P ( z < 0.5) = 1 - 0.6915 = 0.3085 (using technology) Yes, it is very likely that we would have randomly sampled 16 packs with a mean greater than or equal to 125.5 grams because it is less than 1 standard deviation from the mean of the sample means.

36. z =

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

37. (a) m = 3; s = 0.25 x - m 3.2 - 3 = = 0.80 z= s 0.25 P( x > 3.2) = P( z > 0.8) = 1- P( z < 0.8) » 1- 0.7881 = 0.2119 (using technology)

x - m 3.2 - 3 0.2 = » » 6.25 0.35 s 0.032 60 n P( x > 3.2) = P( z > 6.25) = 1 - P( z < 6.25) » 1-1 = 0 (using technology)

(b) z =

38. (a) m = 1.25; s = 0.1 x - m 1.3 -1.25 0.05 = = » 0.5 z= s 0.1 0.1 P( x > 1.3) = P( z > 0.5) = 1- P( z < 0.5) » 1- 0.6915 = 0.3085 (using technology)

x - m 1.3 -1.25 0.05 = » »5 s 0.1 0.01 n 100 P( x > 1.3) = P( z > 5) = 1- P( z < 5) = 1-1 = 0 (using technology)

(b) z =

39. Use the finite correction factor because n = 55 > 0.05(1000) = 50. x -m 50 - 47.12 = » 0.4552 z= s N -n 48.24 1000 - 55 1000 - 1 n N -1 55 P ( x < 50) » P ( z < 0.4552) » 0.6755 (using technology) 40. Use the finite correction factor since n = 30 > 0.05(100) = 5. x -m 95 -101.56 z= = » -1.0009 s N - n 42.69 100 - 30 n N -1 30 100 -1 x -m 110 -101.56 z= = » 1.2878 s N - n 42.69 100 - 30 n N -1 30 100 -1 P(9.5 < x < 110) » P (-1.0009 < z < 1.2878) » 0.9011 - 0.1584 » 0.7427 (using technology)

201

202

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

pq (0.63)(0.37) = » 0.0471 n 105 pˆ - p -0.08 0.55 - 0.63 z= = = = -1.6979 pq 0.63(0.37) 0.0471 n 105 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. m = p = 0.63; s =

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. m = p = 0.74; s =

5.5 NORMAL APPROXIMATIONS TO BINOMIAL DISTRIBUTIONS 5.5 TRY IT YOURSELF SOLUTIONS 1.

n = 100, p = 0.29, q = 0.71 np = 100(0.29) = 29, nq = 100(0.71) = 71 Because np ³ 5 and nq ³ 5 , the normal distribution can be used. m = np = 100(0.29) = 29

s = npq = 100(0.29)(0.71) » 4.54 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.

m = np = 100(0.29) = 29 s = npq = 100(0.29)(0.71) » 4.54 P ( x > 30.5)

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x - m 30.5 - 29 » » 0.330 s 4.54 P ( x > 30.5) = P ( z > 0.330) = 1 - P ( z < 0.330) = 1 - 0.6293 = 0.3707 z=

P( x < 80.5)

x-m 80.5 - 94 » » -1.91 s 200(0.47)(0.53) P ( x < 80.5) = P ( z < -1.91) = 0.0281 z=

n = 75, p = 0.32 np = 75(0.32) = 24 ³ 5 and nq = 75(0.68) = 51 ³ 5 The normal distribution can be used. m = np = 75(0.32) = 24

s = npq = 75(0.32)(0.68) » 4.04

P (14.5 < x < 15.5) x - m 14.5 - 24 z= = » -2.35 s 4.04 x - m 15.5 - 24 z= = » -2.10 s 4.04 P( z < -2.35) = 0.0094 P( z < -2.10) = 0.0179 P(-2.35 < z <-2.10) = 0.0179 - 0.0094 = 0.0085

5.5 EXERCISE SOLUTIONS 1.

np = (24)(0.85) = 20.4 ³ 5

2. np = (15)(0.70) = 10.5 ³ 5

nq = (24)(0.15) = 3.6 < 5

nq = (15)(0.30) = 4.5 < 5

Cannot use normal distribution.

np = (18)(0.90) = 16.2 ³ 5

4. np = (20)(0.65) = 13 ³ 5

nq = (18)(0.10) = 1.8 < 5

nq = (20)(0.35) = 7 ³ 5

Cannot use normal distribution.

Can use normal distribution.

5. a

6. d

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)

203

204

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

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 £ x £ 7) = P( x = 5) + P( x = 6) + P( x = 7) 5

=16 C5 (0.4) (0.6) +16 C6 (0.4) (0.6) +16 C7 (0.4) (0.6) » 0.549 Normal: m = np = 16(0.4) = 6.4, s = npq = 16(0.4)(0.6) » 1.9596 x - m 4.5 - 6.4 z= » » -0.9696 s 1.9596 x - m 7.5 - 6.4 z= » » 0.5613 s 1.9596 P (5 £ x £ 7) » P (4.5 £ x £ 7.5) = P (-0.9696 £ z £ 0.5613) » 0.71272 - 0.16613 = 0.5466 (using technology) The results are about the same. 16. Binomial: P(3 £ x £ 5) = P( x = 3) + P( x = 4) + P( x = 5) 3

=12 C3 (0.5) (0.5) +12 C4 (0.5) (0.5) +12 C5 (0.5) (0.5) » 0.3679 Normal: m = np = 12(0.5) = 6, s = npq = 12(0.5)(0.5) » 1.7321 x - m 2.5 - 6 z= » » -2.0207 s 1.7321 x - m 5.5 - 6 z= » » -0.2887 s 1.7321 P (3 £ x £ 5) » P (2.5 £ x £ 5.5) = P (-2.0207 £ z £ -0.2887) » 0.38641 - 0.02166 » 0.3648 (using technology) The results are about the same. 17. n = 30, p = 0.31, q = 0.69 np = 30(0.31) = 9.3 ³ 5, nq = 30(0.69) = 20.7 ³ 5 Can use normal distribution. m = np = (30)(0.31) = 9.3 s = npq = (30) (0.31) (0.69) » 2.533

18. n = 10, p = 0.65, q = 0.35

np = 10(0.65) = 6.5 ³ 5, nq = 10(0.35) = 3.5 < 5 Cannot use normal distribution because nq < 5.

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

19. n = 200, p = 0.38 np = 200(0.38) = 76 ³ 5, nq = 200(0.62) = 124 ³ 5

m = np = 200(0.38) = 76, s = npq = 200(0.38)(0.62) » 6.8644 Can use normal distribution. x - m 79.5 - 76 » » 0.5099 s 6.864 x - m 80.5 - 76 z= » » 0.6556 s 6.864 P ( x = 80) = P (79.5 < x < 80.5) » P (0.5099 < z < 0.6556)

(a) z =

» 0.7440 - 0.6949 » 0.0491 (using technology)

(b) P ( x ³ 80) = P ( x > 79.5) = P( z > 0.5099) = 1 - P ( z < 0.5099) » 1- 0.6949 = 0.3051 (using technology)

(d) Double charging exactly 80 cases is an unusual event as the probability is less than 0.05. 20. n = 150, p = 0.73 np = 150(0.73) = 109.5 ³ 5, nq = 150(0.27) = 40.5 ³ 5

205

206

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

(a) P ( x ³ 100) = P ( x > 99.5) = P ( z > -1.8391) = 1 - P ( z < -1.8391) » 1 - 0.0330 = 0.9670 ( using technology)

(b) P ( x < 100) = P ( x < 99.5) = P ( z <-1.8391) » 0.0330 (using technology)

(c) P ( x ³120) = P ( x > 119.5) = P ( z > 1.8391) = 1- P ( z <1.8391) » 1- 0.9670 = 0.0330 (using technology) 21. n = 150, p = 0.28 np = 150(0.28) = 42 ³ 5, nq = 150(0.72) = 108 ³ 5

m = np = 150(0.28) = 42, s = npq = 150(0.28)(0.72) » 5.499 Can use normal distribution. x - m 40.5 - 42 » » -0.2728 s 5.499 P ( x £ 40) » P ( x < 40.5) = P ( z < -0.2728) » 0.3925 (using technology)

(a) z =

x - m 50.5 - 42 » » 1.5457 s 5.499 P ( x > 50) » P ( x > 50.5) » P ( z > 1.5457) = 1 - P ( z < 1.5457) (using technology) » 1 - 0.9389 » 0.0611

(b) z =

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x - m 19.5 - 42 » » -4.0917 s 5.499 x - m 30.5 - 42 z= » » -2.0913 s 5.499 P (20 £ x £ 30) » P (19.5 < x < 30.5) » P (-4.0917 < z < -2.0913) (using technology) » 0.01825 - 0.00002 » 0.0182

The event in part (c) is unusual because its probability is less than 0.05. 22. n = 500, p = 0.39, q = 0.61 np = 500(0.39) = 195 ³ 5, nq = 500(0.61) = 305 ³ 5

m = np = 500(0.39) = 195, s = npq = 500(0.39)(0.61) » 10.9064 Can use normal distribution. x - m 174.5 - 195 » » -1.8796 s 10.9064 x - m 175.5 - 195 z= » » -1.7879 s 10.9064 P ( x = 175) » P (174.5 < x < 175.5) » P (-1.8796 < z < -1.7879) (using technology) » 0.0369 - 0.0301 » 0.0068

(a) z =

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CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x - m 225.5 - 195 » » 2.7965 s 10.9064 P ( x £ 225) » P ( x < 225.5) » P ( z < 2.7965) » 0.9974 (using technology)

(b) z =

x - m 200.5 - 195 » » 0.5043 s 10.9064 P ( x £ 200) » P ( x < 200.5) » P ( z < 0.5043) » 0.6930 (using technology)

The event in part (a) is unusual because its probability is less than 0.05. 23. n = 14, p = 0.33

np = 14(0.33) = 4.62 < 5, nq = 14(0.67) = 9.38 ³ 5 Cannot use normal distribution because np < 5. 8

(a) P( x = 8) =14 C8 (0.33) (0.67) = 0.0382 (b) P ( x £ 4) = P ( x = 0) + P ( x = 1) + P ( x = 2) + P ( x = 3) + P ( x = 4) 0

=14 C0 (0.33) (0.67) +14 C1 (0.33) (0.67) +14 C2 (0.33) (0.67) 3

+14 C3 (0.33) (0.67) +14 C4 (0.33) (0.67)

» 0.00367 + 0.02533 + 0.08109 + 0.15976 + 0.21639 » 0.4862 (c) P ( x < 6) = P ( x = 0) + P ( x = 1) + P ( x = 2) + P ( x = 3) + P ( x = 4) + P ( x = 5) 0

=14 C0 (0.33) (0.67) +14 C1 (0.33) (0.67) +14 C2 (0.33) (0.67) 3

+14 C3 (0.33) (0.67) +14 C4 (0.33) (0.67) +14 C5 (0.33) (0.67)

» 0.00367 + 0.02533 + 0.08109 + 0.15976 + 0.21639 + 0.21316 » 0.6994 The event is part (a) is unusual because its probability is less than 0.05.

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

24. n = 25, p = 0.38 np = 25(0.38) = 9.5 ³ 5, nq = 25(0.62) = 15.5 ³ 5

m = np = 25(0.38) = 9.5, s = npq = 25(0.38)(0.62) » 2.4269 Can use normal distribution. . x - m 9.5 - 9.5 (a) z = » =0 s 2.4269 P ( x ³ 10) » P ( x > 9.5) = P ( z < 0) = 0.5000 (using technology)

x - m 12.5 - 9.5 » = 1.2361 s 2.4269 P( x ³ 13) » P ( x > 12.5) » P ( z > 1.2361) = 1 - P ( z < 1.2361) (using technology) » 1 - 0.8918 » 0.1082

(b) z =

x - m 5.5 - 9.5 » = -1.6482 s 2.4269 x - m 8.5 - 9.5 z= » = -0.4120 s 2.4269 P (6 £ x £ 8) » P (5.5 < x < 8.5) » P (-1.6482 < z < -0.4120) (using technology) » 0.3402 - 0.0497 » 0.2905

No unusual events because all of the probabilities are greater than 0.05.

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210

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25. n = 250, p = 0.51, q = 0.49 np = 250(0.51) = 127.5 ³ 5, nq = 250(0.49) = 122.5 ³ 5

m = np = 250(0.51) = 127.5, s = npq = 250(0.51)(0.49) » 7.9041 Can use normal distribution. x - m 125.5 -127.5 » » -0.2530 s 7.9041 P ( x £ 125) » P ( x < 125.5) » P ( z < -0.2530) » 0.4001 (using technology)

(a) z =

x - m 134.5 - 127.5 » » 0.88562 s 7.9041 P ( x ³ 135) » P ( x > 134.5) » P ( z > 0.88562) » 1 - P ( z < 0.88562) » 1 - 0.8121 » 0.1879

(b) z =

x - m 99.5 - 127.5 » » -3.5425 s 7.9041 x - m 125.5 -127.5 z= » » -0.2530 s 7.9041 P (100 < x < 125) » P (99.5 < x < 125.5) » P (-3.5425 < z < -0.2530) » 0.4001 - 0.0002 » 0.3999 (using technology)

No unusual events because all of the probabilities are greater than 0.05.

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

26. n = 1500, p = 0.14, q = 0.86 np = 1500(0.14) = 210 ³ 5, nq = 1500(0.86) = 1290 ³ 5

m = np = 1500(0.14) = 210, s = npq = 1500(0.14)(0.86) » 13.4387 Can use normal distribution. x - m 174.5 - 210 » » -2.6416 s 13.4387 x - m 175.5 - 210 z= » » -2.5672 s 13.4387 P ( x = 175) » P (174.5 < x < 175.5) » P (-2.6416 < z < -2.5672) (using technology) » 0.0051 - 0.0041 » 0.0010

(a) z =

x - m 225.5 - 210 » » 1.1534 s 13.4387 P ( x £ 225) » P ( x < 225.5) » P( z < 1.1534) » 0.8756 (using technology)

(b) z =

x - m 200.5 - 210 » » -0.7069 s 13.4387 P( x £ 200) » P ( x < 200.5) » P ( z < -0.7069) » 0.2398 (using technology)

The event in part (a) is unusual because its probability is less than 0.05.

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212

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27. (a) n = 700, p = 0.033

np = 700(0.033) = 23.1 ³ 5, nq = 700(0.967) = 676.9 ³ 5 Can use normal distribution. m = np = 700(0.033) = 23.1, s = npq = 700(0.033)(0.967) » 4.7263 P ( x ³ 30) » P( x > 29.5) » P ( z > 1.35413) = 1 - P ( z < 1.35413) (using technology) » 1 - 0.9122 » 0.0878 (b) n = 750, p = 0.033 np = 750(0.033) = 24.75 ³ 5, nq = 750(0.967) = 725.25 ³ 5 Can use normal distribution. m = np = 750(0.033) = 24.75, s = npq = 750(0.033)(0.967) » 4.8922 P ( x ³ 30) » P( x > 29.5) » P ( z > 0.97094) = 1 - P ( z < 0.97094) (using technology) » 1 - 0.8342 » 0.1658 (c) n = 1000, p = 0.033 np = 1000(0.033) = 33 ³ 5, nq = 1000(0.967) = 967 ³ 5 Can use normal distribution. m = np = 1000(0.033) = 33, s = npq = 1000(0.033)(0.967) » 5.64898 P ( x ³ 30) » P( x > 29.5) » P ( z > -0.61958) = 1 - P ( z < -0.61958) (using technology) » 1 - 0.2678 » 0.7322 28. (a) n = 1000, p = 0.08

np = 1000(0.08) = 80 ³ 5, nq = 1000(0.92) = 920 ³ 5 Can use normal distribution. m = np = 1000(0.08) = 80, s = npq = 1000(0.08)(0.92) » 8.5790 P ( x > 100) » P ( x > 100.5) » P ( z > 2.3895) = 1 - P ( z < 2.3895) (using technology) » 1 - 0.9916 » 0.0084 (b) n = 1250, p = 0.08 np = 1250(0.08) = 100 ³ 5, nq = 1250(0.92) = 1150 ³ 5 Can use normal distribution. m = np = 1250(0.08) = 100, s = npq = 1250(0.08)(0.92) » 9.59166 P ( x > 100) » P ( x > 100.5) » P ( z > 0.0521) = 1 - P ( z < 0.0521) (using technology) » 1 - 0.5208 » 0.4792 (c) n = 1500, p = 0.08 np = 1500(0.08) = 120 ³ 5, nq = 1500(0.92) = 1380 ³ 5 Copyright © 2019 Pearson Education Ltd.

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

213

Can use normal distribution. m = np = 1500(0.08) = 120, s = npq = 1500(0.08)(0.92) » 10.5071 P ( x > 100) » P ( x > 100.5) » P ( z > -1.8559) = 1 - P ( z < -1.8559) (using technology). » 1 - 0.0317 » 0.9683 29. n = 250, p = 0.70, m = np = 250(0.70) = 175, s = npq = 250(0.70)(0.30) » 7.2457 60% say no  n = 250(0.60) = 150 say no while 100 say yes. x - m 100.5 - 175 z= » » -10.28 s 7.2457 P (less than or equal to 100 say yes) = P ( x £ 100) » P ( x < 100.5) » P ( z < -10.28) » 0 It is highly unlikely that 60% responded no. Answers will vary. 30. n = 200, p = 0.11, m = np = 200(0.11) = 22, s = npq = 200(0.11)(0.89) » 4.4249 9% of 200 = 18 people x - m 18.5 - 22 z= » » -0.7910 s 4.4249 P ( x £ 18) » P ( x < 18.5) » P ( z < -0.791) = 0.2145 It is probable that 18 of the 200 people responded that they participate in hiking. Answers will vary. 31. n = 100, p = 0.75, m = np = 100(0.75) = 75, s = npq = 100(0.75)(0.25) » 4.3301 x - m 69.5 - 75 z= » » -1.2702 s 4.3301 P (reject claim) = P ( x < 70) » P ( x < 69.5) » P ( z < -1.2702) » 0.1020 32. n = 100, p = 0.65, m = np = 100(0.65) = 65, s = npq = 100(0.65)(0.35) » 4.7697

x - m 69.5 - 65 » » 0.9435 s 4.7697 P(accept claim) = P( x ³ 70) » P( x > 69.5) z=

» P( z > 0.9435) = 1- P( z < 0.9435) = 1- 0.8273 = 0.1727

CHAPTER 5 REVIEW EXERCISE SOLUTIONS 1.

m = 15, s = 3

2. m = 56, s = 5

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. 0.6368

8. 0.0256

214

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

9. –0.46

10. 0.05208 (using technology)

11. 1 – 0.9956 = 0.0044

12. 1 – 0.5438 = 0.4562

13. 0.4750

14. 0.85083 - 0.06057 » 0.7903 (using technology)

15. 0.3899 16. 0.99632 - 0.00368 » 0.9926 (using technology) 17. 0.1052 18. 0.73891 + (1 - 0.99968) » 0.7392 (using technology) x - m 17 - 21.3 19. x = 17  z = = » -0.66 s 6.5 x - m 29 - 21.3 x = 29  z = = » 1.18 s 6.5 x - m 8 - 21.3 x=8 z = = » -2.05 s 6.5 x - m 23 - 21.3 x = 23  z = = » 0.26 s 6.5 20. x = 8 is unusual because its corresponding z-score (-2.05) lies more than 2 standard deviations from the mean. 21. P ( z < 1.36) = 0.9131 22. P ( z > -0.74) = 1 - P(z < -0.74) = 1 - .2296 = 0.7704 23. 0.8285 24. P (0.42 < z < 3.15) = 0.99918 - 0.66276 » 0.3364 (using technology) 25. 0.1336 26. P ( z < 0 or z > 1.68) = 0.5 + (1 - .9535) = 0.5465 x - m 82 - 74 = =1 s 8 P ( x < 84) = P ( z < 1) = 0.8944

27. z =

x - m 68.4 - 74 = = -0.7 s 8 P ( x > 68.4) = P ( z > -0.7) = 1 - P ( z < -0.7) = 1 - 0.242 = 0.758

28. z =

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

215

x - m 70 - 74 = = -0.5 s 8 P ( x > 70) = P ( z > -0.5) = 1 - P ( z < -0.5) = 0.6915

29. z =

x - m 57 - 74 = = -2.125 s 8 P ( x < 57) = P ( z < -2.125) = 0.0168 (using technology)

30. z =

x - m 60 - 74 = = -1.75 s 8 x - m 70 - 74 z= = = -0.5 s 8 P (60 < x < 70) = P (-1.75 < z < -0.5) » 0.30854 - 0.04001 » 0.2685 (using technology)

31. z =

x - m 72 - 74 = = -0.25 s 8 x - m 82 - 74 z= = = 1.00 s 8 P (72 < x < 82) = P (-0.25 < z < 1.00) » 0.84135 - 0.40129 » 0.4401 (using technology)

32. z =

x - m 12.3 - 14.7 = » -0.2087 s 11.5 P ( x < 12.3) » P ( z < -0.2087) » 0.4173 (using technology)

33. (a) z =

x - m 15.4 -14.7 = » 0.0609 s 11.5 x - m 19.6 -14.7 z= = » 0.4261 s 11.5 P (15.4 < x < 19.6) » P (0.0609 < z < 0.4261) » 0.6650 - 0.5243 » 0.1407 (using technology)

(b) z =

x - m 17.7 -14.7 = » 0.2609 s 11.5 P ( x > 17.7) » P ( z > 0.2609) = 1 - P ( z < 0.2609) » 1 - 0.6029 = 0.3971 (using technology)

x - m 5.1 - 9.1 = = -1.1429 s 3.5 x - m 15.7 - 9.1 z= = = 1.8857 s 3.5 P (5.1 < x < 15.7) = P (-1.1429 < z < 1.8857) » 0.9703 - 0.1265 » 0.8438 (using technology)

34. (a) z =

(b) z =

x - m 10.5 - 9.1 = = 0.4 s 3.5

216

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x - m 12.3 - 9.1 = » 0.9143 s 3.5 P (10.5 < x < 12.3) = P(0.4 < z < 0.9143) » 0.8197 - 0.6554 » 0.1643 (using technology) z=

x - m 11.1 - 9.1 = » 0.5714 s 3.5 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 =-0.16

38. z = -1.28

39. z = 2.457 (using technology)

40. z = -2.05

41. z = 0.44

42. z = -0.10

43. 0.81 44. 94% between  97% below, z = 1.88 45. x = m + zs = 132 + (-2.75)(4.53) » 119.54 feet 46. x = m + zs = 132 + (1.6)(4.53) » 139.25 feet 47. 90th percentile  Area = 0.90  z = 1.2816 x = m + zs = 132 + (1.2816)(4.53) » 137.81 feet 48. 1st quartile  Area = 0.25  z » -0.6745 (using technology) x = m + zs = 132 + (-0.6745) (4.53) » 128.94 feet 49. Top 15%  Area = 0.85  z = 1.036 (using technology) x = m + zs = 132 + (1.036)(4.53) » 136.70 feet 50. Bottom 20%  Area = 0.20  z = -0.8416 x = m + zs = 132 + (-0.8416)(4.53) = 128.19 feet 51. (a) m = 1.5, s » 1.118

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

(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

1.118

» 0.791 2 The means are equal, but the standard deviation of the sampling distribution is smaller.

52. (a) m = 140, s » 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

50.990 3

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

» 29.439

The means are equal, but the standard deviation of the sampling distribution is smaller. 53. mx = 471.5, s x =

s n

187.9

» 31.761

217

218

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

54. mx = 155.69, s x =

s n

5.05 40

» 0.798

x - m 12.3 -14.7 = » -0.29514 11.5 s n 2 P ( x < 12.3) » P ( z < -0.29514) » 0.3839 (using technology)

55. (a) z =

x - m 15.4 -14.7 = » 0.0861 11.5 s n 2 x - m 19.6 -14.7 z= = » 0.6026 11.5 s n 2 P (15.4 < x < 19.6) » P(0.0861 < z < 0.6026) » 0.7266 - 0.5343 » 0.1923 (using technology)

(b) z =

x - m 17.7 -14.7 = » 0.3689 11.5 s n 2 P ( x > 17.7) = P ( z > 0.3689) = 1 - P ( z < 0.3689) » 1 - 0.6439 » 0.3561 (using technology)

The probabilities in parts (a) and (c) are smaller, and the probability in part (b) is larger. This is to be expected because the standard error of the sample mean is smaller. x - m 5.1 - 9.1 = » -2.7994 3.5 s n 6 x - m 15.7 - 9.1 z= = » 4.6190 3.5 s n 6 P (5.1 < x < 15.7) » P (-2.7994 < z < 4.6190) » 0.999999 - 0.002560 » 0.9974 (using technology)

56. (a) z =

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x - m 10.5 - 9.1 = » 0.9798 3.5 s n 6 x - m 12.3 - 9.1 z= = » 2.2395 3.5 s n 6 P (10.5 < x < 12.3) » P (0.9798 < z < 2.2395) » 0.9874 - 0.8364 = 0.1510 (using technology)

(b) z =

x - m 11.1 - 9.1 = » 1.3997 3.5 s n 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.

x - m 21.6 - 20.8 = » 0.8571 5.6 s n 36 P ( x < 21.6) = P( z < 0.8571) = 0.8043

57. (a) z =

x - m 19.8 - 20.8 = » -1.0714 5.6 s n 36 P ( x > 19.8) = P ( z > -1.0714) » 1 - 0.1420 = 0.8580

(b) z =

x - m 20.5 - 20.8 = » -0.3214 5.6 s n 36 x - m 21.5 - 20.8 z= = = 0.75 5.6 s n 36 P (20.5 < x < 21.5) = P(-0.3214 < z < 0.75) » 0.7734 - 0.3740 » 0.3994 (using technology)

x - m 503 - 500 = » 1.601 10.6 s n 32 P ( x < 503) » P ( z < 1.601) = 0.9453 (using technology)

58. (a) z =

x - m 502 - 500 = » 1.0673 10.6 s n 32 P ( x > 502) » P ( z > 1.0673) » 1- 0.8571 = 0.1429 (using technology)

(b) z =

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220

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x - m 498 - 500 = » -1.0673 10.6 s n 32 x - m 501 - 500 z= = » 0.5337 10.6 s n 32 P (498 < x < 501) » P (-1.0673 < z < 0.5337) » 0.7032 - 0.1429 » 0.5603 (using technology)

x - m 60,000 - 61,000 = » -0.6098 11,000 s n 45 P ( x < 60,000) » P ( z < -0.6098) » 0.2710 (using technology)

59. (a) z =

x - m 63,000 - 61,000 = » 1.2197 11,000 s n 45 P ( x > 63,000) » P ( z > 1.2197) » 1 - 0.8887 » 0.1113 (using technology)

(b) z =

x - m 71,500 - 72,000 = » -0.35355 10,000 s n 50 P ( x < 71,500) » P ( z < -0.35355) » 0.3618 (using technology)

60. (a) z =

x - m 74,500 - 72,000 = » 1.7678 10,000 s n 50 P ( x > 74,500) » P ( z > 1.7678) » 1 - 0.9615 = 0.0385 (using technology)

(b) z =

61. n = 20, p = 0.75, q = 0.25 np = 20(0.75) = 15 ³ 5 , nq = 20(0.25) = 5 ³ 5 Can use the normal distribution.

m = np = 20(0.75) = 15, s = npq = 20(0.75)(0.25) » 1.936 62. n = 45, p = 0.11, q = 0.89 np = 45(0.11) = 4.95 < 5 , nq = 45(0.89) = 40.05 ³ 5 Cannot use the normal distribution because np < 5 63. The probability of getting at least 28 successes; P ( x > 27.5) 64. The probability of getting more than 18 successes; P ( x > 18) » P ( x > 18.5) 65. The probability of getting exactly 30 successes; P (29.5 < x < 30.5)

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

66. The probability of getting at most 32 successes; P ( x £ 32) » P ( x < 32.5) 67. The probability of getting less than 50 successes; P ( x < 49.5) 68. The probability of getting between 54 and 64 successes; P (54 < x < 64) » P (54.5 < x < 63.5) 69. n = 70, p = 0.32, q = 0.68 np = 70(0.32) = 22.4 ³ 5 , nq = 70(0.68) = 47.6 ³ 5 Can use the normal distribution.

m = np = 70(0.32) = 22.4 s = npq = 70(0.32)(0.68) » 3.9028 x - m 15.5 - 22.4 » » -1.7680 s 3.9028 P ( x £ 15) » P ( x < 15.5) = P ( z < -1.7680) » 0.0385 (using technology)

(a) z =

(b) z =

x - m 24.5 - 22.4 » » 0.5381 s 3.9028

x - m 25.5 - 22.4 » » 0.7943 s 3.9028 P ( x = 25) » P (24.5 < x < 25.5) » P (0.5381 < z < 0.7943) » 0.7865 - 0.7047 » 0.0818 (using technology) z=

x - m 30.5 - 22.4 » » 2.0754 s 3.9028 P ( x > 30) » P ( x > 30.5) » P ( z > 2.0754) » 1 - 0.9810 » 0.0190 (using technology)

The events in parts (a) and (c) are unusual because their probabilities are less than 0.05.

221

222

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

70. n = 20, p = 0.65, q = 0.35 np = 20(0.65) = 13 ³ 5 , nq = 20(0.35) = 7 ³ 5 Can use normal distribution.

m = np = 20(0.65) = 13, s = npq = 20(0.65)(0.35) » 2.1331 x - m 14.5 - 13 » » 0.7032 s 2.1331 x - m 15.5 - 13 z= » » 1.1720 s 2.1331 P ( x = 15) » P (14.5 < x < 15.5) = P (0.7032 < z < 1.1720) » 0.8794 - 0.7590 » 0.1204 (using technology)

(a) z =

x - m 9.5 -13 » » -1.6408 s 2.1331 P ( x < 10) » P ( x < 9.5) » P ( z < -1.6408) » 0.0504 (using technology)

(b) z =

x - m 20.5 - 13 » » 3.5160 s 2.1331 x - m 34.5 - 13 z= » » 10.0792 s 2.1331 P (20 < x < 35) » P (20.5 < x < 34.5) » P (3.5160 < z < 10.0792) » 1 - 0.999781 » 0.000219 (using technology)

The event in part (c) is unusual because its probability is less than 0.05.

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

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) x - m 5.97 - 9.2 = » -1.9938 s 1.62 P ( x < 5.97) » P ( z < -1.9938) » 0.0231 (using technology)

2. (a) z =

x - m 40.5 - 87 = » -2.4474 s 19 P ( x > 40.5) » P( z > -2.4474) » 1 - 0.0072 » 0.9928 (using technology)

(b) z =

x - m 5.36 - 5.5 = = -1.75 s 0.08 x - m 5.64 - 5.5 z= = = 1.75 s 0.08 P (5.36 < x < 5.64) = P (-1.75 < z < 1.75) » 0.95994 - 0.04006 » 0.9199 (using technology)

x - m 19.6 -18.5 = » 0.2588 s 4.25 x - m 26.1 - 18.5 z= = » 1.7882 s 4.25 P (19.6 < x < 26.1) » P (0.2588 < z < 1.7882) » 0.96313 - 0.60211 » 0.3610 (using technology)

(d) z =

x - m 125 - 100 = » 1.6667 s 15 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=

x - m 95 - 100 = » -0.3333 s 15 x - m 105 - 100 z= = » 0.3333 s 15 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. z=

223

224

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x - m 112 - 100 = = 0.80 s 15 P ( x > 112) = P ( z > 0.80) = 1 - .7881 = 0.2119  21.19% (using technology) z=

x - m 90 - 100 = » -0.6667 s 15 P ( x < 90) = P ( z < -0.6667) = 0.2525 (using technology) (2000)(0.2525) = 505 people z=

7. Top 5%  z » 1.6449 m + zs = 100 + (1.6449)(15) » 124.7  125 8. Bottom 10%  z »-1.2816 m + zs = 100 + (-1.2816)(15) » 80.8  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 - m 105 -100 5 = » » 2.5820 15 s 1.9365 n 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. z=

x - m 105 - 100 = » 0.3333 s 15 P ( x > 105) » P ( z > 0.3333) » 1 - 0.6306 » 0.3694 (using technology) x - m 105 -100 5 z= = » » 1.2910 15 s 3.873 n 15 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.

10. z =

11. n = 250, p = 0.16, q = 0.84 np = 250(0.16) = 40 ³ 5 , nq = 250(0.84) = 210 ³ 5 Can use normal distribution. m = np = 250(0.16) = 40, s = npq = 250(0.16)(0.84) » 5.797 x - m 40.5 - 40 » » 0.08625 s 5.797 P ( x £ 40) » P ( x < 40.5) » P ( z < 0.08625) » 0.5344 (using technology)

12. (a) z =

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

x - m 44.5 - 40 » » 0.7763 s 5.797 P ( x < 45) » P( x < 44.5) » P ( z < 0.7763) » 0.7812 (using technology)

(b) z =

x - m 47.5 - 40 » » 1.2938 s 5.797 x - m 48.5 - 40 z= » » 1.4663 s 5.797 P ( x = 48) » P (47.5 < x < 48.5) » P (1.2938 < z < 1.4663) » 0.9287 - 0.9021 » 0.0266 (using technology) The event in part (c) is unusual because its probability is less than 0.05.

CUMULATIVE REVIEW, CHAPTERS 3-5 1. (a) np = 30(0.61) = 18.3 ³ 5, nq = 30(0.39) = 11.7 ³ 5 Can use normal distribution.

(b) m = np = 30(0.61) = 18.3, s = npq = 30(0.61)(0.39) » 2.6715 x - m 14.5 - 18.3 » » -1.4224 s 2.6715 P ( x £ 14) » P ( x < 14.5) » P ( z < -1.4224) » 0.0775 (using technology) z=

x - m 13.5 - 18.3 » » -1.7967 s 2.6715 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.

x 1 2 3 4 5 6 7

P(x) 0.281 0.340 0.154 0.129 0.060 0.022 0.013

xP(x) 0.281 0.680 0.462 0.516 0.300 0.132 0.091

å xP ( x ) = 2.462

(a) m = å xP ( x ) » 2.46 (b) s 2 = å ( x - m) P( x) » 1.95 2

x -m –1.462 -0.462 0.538 1.538 2.538 3.538 4.538

( x -m)

( x -m) P(x)

2.1374 0.2134 0.2894 2.3654 6.4414 12.5174 20.5934

0.6006 0.0726 0.0446 0.3051 0.3865 0.2754 0.2677

å ( x - m) P( x) » 1.9525 2

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CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

(c) s = s 2 » 1.40 (d) E ( x ) = å xP ( x ) » 2.46 The size of a household on average is about 2.5 persons. The standard deviation is 1.4, so most households differ from the mean by no more than about 1 person. 3.

x 0 1 2 3 4 5 6

P(x) 0.113 0.188 0.188 0.288 0.150 0.038 0.038

xP(x) 0.000 0.188 0.376 0.864 0.600 0.190 0.228

å xP ( x ) = 2.446

x -m -2.446 -1.446 -0.446 0.554 1.554 2.554 3.554

( x -m)

( x -m) P(x)

5.9829 2.0909 0.1989 0.3069 2.4149 6.5229 12.6309

0.6761 0.3931 0.0374 0.0884 0.3622 0.2479 0.4800

å ( x - m) P( x) » 2.285 2

(a) m = å xP ( x ) » 2.45 (b) s 2 = å( x - m) P( x) » 2.29 2

(c) s = s 2 » 1.51 (d) E ( x ) = m » 2.45 The number of fouls per game for Garrett Temple is about 2.45 fouls. The standard deviation is 1.5, so most of Temple’s games differ from the mean by no more than about 1 or 2 fouls. 4. (a) P ( x < 4) = 0.113 + 0.188 + 0.188 + 0.288 = 0.777

(b) P ( x ³ 3) = 0.288 + 0.150 + 0.038 + 0.038 = 0.514 (c) P (2 £ x £ 4) = 0.188 + 0.288 + 0.150 = 0.626 5. (a) (16)(15)(14)(13) = 43,680

(b)

(7)(6)(5)(4) 840 = » 0.0192 (16)(15)(14)(13) 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

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

12. n = 25, p = 0.28

(a) P ( x = 3) = 0.0367 (using technology) (b) P (8 £ x £ 11) = 0.15345 + 0.11272 + 0.07014 + 0.03719 » 0.3735 (using technology) (c) P ( x < 2) = 0.00027 + 0.00264 » 0.0029 (using technology) (d) The events in parts (a) and (c) are unusual because their probabilities are less than 0.05. 13. p =

1 = 0.005 200 4

(a) P( x = 5) = (0.005)(0.995) » 0.0049 0

(b) P( x £ 3) » (0.005)(0.995) + (0.005)(0.995) + (0.005)(0.995) » 0.005 + 0.004975 + 0.004950 » 0.0149

1,127, 205 » 0.886 1, 272,750

(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, 272,750 + 464,535 -145,545 » 0.413 3,849,535

15. (a) mx = 70

sx =

s n

1.2 40

» 0.1897

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228

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

æ ö÷ çç ÷ çç 69 - 70 ÷÷÷ » P( z <-3.2275) » 0.0006 (using technology) (b) P( x £ 69) = P ç z £ çç 1.2 ÷÷÷ ÷ çç è 15 ø÷

æ è

16. (a) P( x < 36) = P ççç z <

36 - 44 ö÷ ÷ = P( z <-1.6) = 0.0548 (using technology) 5 ÷ø

æ 42 - 44 60 - 44 ö÷ <z< (b) P(42 < x < 60) = P ççç ÷ è 5 5 ÷ø = P (-0.4 < z < 3.2) » 0.99931 - 0.34458 » 0.6547 (using technology) (c) Top 5%  z = 1.6449 x = m + zs = 44 + (1.6449)(5) » 52.2 months 17. (a) 12 C4 = 495

(b)

18. (a)

(1)(1)(1)(1) = 0.0020 12 C4

x 0 1 2 3 4 5 6 7 8 9 10

P(x) 0.000006 0.0001 0.0014 0.0090 0.0368 0.1029 0.2001 0.2668 0.2335 0.1211 0.0282

(b)

Skewed left (c) The values 0, 1, 2, 3, 4, and 10 are unusual because their probabilities are less than 0.05.

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

CHAPTER 5 TEST SOLUTIONS

1. (a)  x    83,  x 

 n



11.9 50

 1.683

x - m 85 - 83 = » 1.1884 11.9 s n 50 P ( x > 85) » P ( z > 1.1884) » 1 - 0.8827 » 0.1173 (using technology)

(b) z =

x - m 80 - 83 = » -1.7826 11.9 s n 50 x - m 82 - 83 z= = » -0.5942 11.9 s n 50 P (80 < x < 82) » P (-1.7826 < z < -0.5942) » 0.27619 - 0.03733 » 0.2389 (using technology)

x - m 20 - 18 » » 0.2632 s 7.6 P ( x > 20) » P ( z > 0.2632) » 1 - 0.6038 » 0.3962 (using technology)

2. (a) z =

x - m 0 - 18 » » -2.3684 s 7.6 x - m 5 - 18 z= » » -1.7105 s 7.6 P (0 < x < 5) » P (-2.3684 < z < -1.7105) » 0.04358 - 0.00893 » 0.0347 (using technology)

(b) z =

x - m 9 - 18 » » -1.1842 s 7.6 x - m 27 - 18 z= » » 1.1842 s 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)

88.3%  z » 1.1901 (using technology) x = m + zs = 18 + (1.1901)(7.6) » 27.045

64.8% to right  35.2% to left  z » -0.3799 (using technology) x = m + zs = 18 + (-0.3799)(7.6) » 15.113

229

230

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

n = 18, p = 0.69, q = 0.31 np = 18(0.69) = 12.42 ³ 5 , and nq = 18(0.31) = 5.58 ³ 5 Can use the normal distribution.

m = np = 18(0.69) = 12.42, s = npq = 18(0.69)(0.31) » 1.9622 x - m 9.5 - 12.42 » » -1.4881 s 1.9622 x - m 10.5 - 12.42 z= » » -0.9785 s 1.9622 P ( x = 10) » P (9.5 < x < 10.5) » P (-1.4881 < z < -0.9785) » 0.16391 - 0.06836 » 0.0956 (using technology)

(a) z =

x - m 6.5 - 12.42 » » -3.0170 s 1.9622 P ( x < 7) » P ( x < 6.5) » P ( z < -3.0170) » 0.0013 (using technology)

(b) z =

x - m 14.5 - 12.42 » » 1.0600 s 1.9622 P ( x ³ 15) » P ( x > 14.5) » P( z > 1.0600) » 1 - 0.8554 » 0.1446 (using technology)

The event in part (b) is unusual because its probability is less than 0.05.

CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS

n = 30, p = 0.86, q = 0.14 np = 30(0.86) = 25.8 ³ 5 , and nq = 30(0.14) = 4.2 < 5 Cannot use the normal distribution because nq < 5 . 25

(a) P ( x = 25) =30 C25 (0.86) (0.14) = 0.1766 (b) P ( x > 25) = P ( x ³ 26) 26

=30 C26 (0.86) (0.14) +30 C27 (0.86) (0.14) +30 C28 (0.86) (0.14) 29

+30 C29 (0.86) (0.14) +30 C30 (0.86) (0.14)

= 0.20859 + 0.18983 + 0.12494 + 0.05293 + 0.01084 » 0.5871

(c) P ( x < 25) = 1 - P ( x ³ 25) 25 5 26 4 27 3 æ ö÷ çç 30 C25 (0.86) (0.14) +30 C26 (0.86) (0.14) +30 C27 (0.86) (0.14) ÷÷ = 1-ç 28 2 29 1 30 0÷ çç +30 C28 (0.86) (0.14) +30 C29 (0.86) (0.14) +30 C30 (0.86) (0.14) ÷÷ø è

= 1 - (0.17658 + 0.20859 + 0.18983 + 0.12494 + 0.05293 + 0.01084) = 1- 0.76371 » 0.2363 No unusual events because all of the probabilities are greater than 0.05. 7.

x - m 45,000 - 44,000 » » 0.4082 s 2450 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=

x - m 40,000 - 44,000 » » -1.6327 s 2450 x - m 42,000 - 44,000 z= » » -0.8163 s 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 z=

Middle 60%  20% to left and 20% to right  z »  - 0.8416 (using technology) x = m + zs = 44,000 + (-0.8416)(2450) » 41,938.08 x = m + zs = 44,000 + (0.8416)(2450) » 46,061.92 hours 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.

231

CHAPTER

Confidence Intervals

6.1 CONFIDENCE INTERVALS FOR THE MEAN (LARGE SAMPLES) 6.1 TRY IT YOURSELF SOLUTIONS

x

 x  626  20.9

n 30 Another point estimate for the population mean is about 20.9. zc  1.96, n  30,   2.3



2.3  0.8 n 30 You are 95% confident that the margin of error for the population mean is about 0.8 hour. E  zc

 1.96

x  20.9, E  0.8

x  E  20.9  0.8  20.1 x  E  20.9  0.8  21.7

(20.1, 21.7) This confidence interval is wider than the one found in Example 3. 4. 75% CI: (20.632, 21.468)  (20.6, 21.5) 85% CI: (20.526, 21.574)  (20.5, 21.6) 90% CI: (20.452, 21.648)  (20.5, 21.7) As the confidence level increases, so does the width of the interval. n  30, x  22.9,   1.5, z c  1.645  1.5 E  zc  1.645  0.5 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. z c  1.96, E  0.75,   2.3

 z    1.96  2.3  n c     36.13  37  E   0.75  Because of the larger margin of error, the sample size needed is smaller.

CHAPTER 6 │ CONFIDENCE INTERVALS

231

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    3.8  4.27  0.47

11. x    26.43  24.67  1.76 13. E  zc

15. E  zc

 n

7. 1.15

8. 2.17

10. x    9.5  8.76  0.74 12. x    46.56  48.12  1.56

 1.96

5.2  1.861 30

14. E  zc

 1.28

1.3  0.192 75

16. E  zc

 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  zc

22. x  zc

23. x  zc

 n

 12.3  1.645

1.5  12.3  0.349  (12.0, 12.6) 50

 31.39  1.96

0.8  31.3  0.173  (31.22, 31.56) 82

 10.5  2.575

2.14  10.5  0.821  (9.7, 11.3) 45

232

CHAPTER 6 │ CONFIDENCE INTERVALS

24. x  zc

 n

 20.6  1.28

25. (12.0, 14.8)  x 

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  20.6  0.602  (20.0, 21.2) 100

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  z c  1.645 2

 z    (1.645)(6.8)  n c     125.13  126 1   E   30. c  0.95  zc  1.96 2

 z    (1.96)(2.5)  n c     24.01  25 1   E   31. c  0.80  z c  1.28 2

 z    (1.28)(4.1)  n c     6.89  7 2   E   32. c  0.98  zc  2.33 2

 z    (2.33)(10.1)  n c     138.45  139 2   E   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



2.79  59.23  0.66  (58.57, 59.89) 48  2.79 95% CI: x  zc  59.23  1.96  59.23  0.79  (58.44, 60.02) n 48 With 90% confidence, you can say that the population mean price is between $58.57 and $59.89. With 95% confidence, you can say that the population mean price is between $58.44 and $60.02. The 95% CI is wider.

35. 90% CI: x  zc

 59.23  1.645

CHAPTER 6 │ CONFIDENCE INTERVALS



233

1.91  5.62  0.524  (5, 6) n 36  1.91 95% CI: x  zc  5.62  1.96  5.62  0.624  (5, 6) n 36 With 90% confidence and with 95% confidence, you can say that the population mean concentration is between EGP 5 and 6. When rounded to the nearest whole number, both confidence intervals have the same width.

36. 90% CI: x  zc



 5.62  1.645

15.77  10.22  3.243  (6.98, 13.46) n 64  15.77 95% CI: x  zc  10.22  1.96  10.22  3.863  (6.36, 14.08) n 64 With 90% confidence, you can say that the population mean rainfall is between 6.98 mm and 13.46 mm. With 95% confidence, you can say that the population mean rainfall is between 6.36 mm and 14.08 mm. The 95% CI is wider.

37. 90% CI: x  zc



 10.22  1.645

20469  43202  6873  (36329, 50075) n 24  20469 95% CI: x  zc  43202  1.96  43202  8189  (35013, 51391) n 24 With 90% confidence, you can say that the population mean number of accidents is between 36329 and 50075. With 95% confidence, you can say that the population mean number of accidents is between 35013 and 51391. The 95% CI is wider.

38. 90% CI: x  zc

 43202  1.645

39. Yes; The margin of error is small (E = 0.67). 40. No; The margin of error is large (E = 0.62). 41. Yes; The right endpoint of the 95% CI is 14.08. 42. No; The left endpoint of both confidence intervals is more. Therefore, it is not possible for the population mean to be 35,000 or less. 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.

234

CHAPTER 6 │ CONFIDENCE INTERVALS

45. x 

 x  5180  172.7 30



70.7  172.7  21.234  (151.5, 193.9) n 30  70.7 99% CI: x  zc  172.7  2.575  172.7  33.238  (139.5, 205.9) n 30 With 90% confidence, you can say that the population mean quantity is between 151.5 and 193.9 milligrams. With 99% confidence, you can say that the population mean quantity is between 139.5 and 205.9 milligrams. The 99% CI is wider.

90% CI: x  zc

46. x 

 172.7  1.645

 x  795.71  22.10

36  7.61 90% CI: x  zc  22.10  1.645  22.10  2.09  (20.01, 24.19) n 36  7.61 99% CI: x  zc  22.10  2.575  22.10  3.27  (18.83, 25.37) 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. n

 z    1.96  4.8  47. n   c      88.510  89 1   E   2

 z    2.575  1.4  48. n   c      3.249  4 2   E   2

 z    1.96  3.10  49. (a) n   c      65.6  66 servings  E   0.75 



3.10  29  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  zc

 29  1.96

 z    1.645  1.6  50. (a) n   c      3.1  4 students   E   1.5 (b) 90% CI: x  zc

 n

 20  1.645

1.6  20  1.316  (18.684, 21.316) 4

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235

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. 2

 z    1.645  0.75  51. (a) n   c      6.1  7 cans 0.5   E  



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.

(b) 90% CI: x  zc

 127.75  1.645

 z    1.96  1  52. (a) n   c      61.5  62 bottles  E   0.25 



1  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.

(b) 95% CI: x  zc

 64  1.96

 z    2.575  0.5  53. (a) n   c      73.7  74 balls  E   0.15 



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.

(b) 99% CI: x  zc

 27.5  2.575

 z    2.575  0.10  54. (a) n   c      26.5  27 balls 0.05   E  



0.10  8.3  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  zc

 8.3  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.

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57. (a)

N n 1000  500   0.707 N 1 1000  1

(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 400  50   0.937 N 1 400  1

(g)

(b)

N n 1000  100   0.949 N 1 1000  1

N n 700  50 N n 1200  50   0.964   0.979 (h) 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  zc

(b) 90% CI: x  zc

(d) 80% CI: x  zc

 n

N n 4.9  8.6  2.575 N 1 25

200  25  8.6  2.366  (6.2, 11.0) 200  1

N n 2.8  10.9  1.645 N 1 50

500  50  10.9  0.619  (10.3, 11.5) 500  1

N n 0.5  40.3  1.96 N 1 68

300  68  40.3  0.105  (40.2, 40.4) 300  1

N n 9.8  56.7  1.28 N 1 36

400  36  56.7  1.997  (54.7, 58.7) 400  1

59. Sample answer: z E c Write original equation. n E n  zc

Multiply each side by n .

z n= c E z n c   E 

Divide each side by E. 2

Square each side.

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237

6.2 CONFIDENCE INTERVALS FOR THE MEAN (SMALL SAMPLES) 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.

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4. Is  known? No Is n  30? No Is the population normally distributed? Yes Use the t-distribution because  is not known and the population is normally distributed.

6.2 EXERCISE SOLUTIONS 1.

tc  1.833

tc  2.947

E  tc

s 5  2.131  2.664 n 16

E  tc

s 3  4.032  4.938 n 6

E  tc

s 2.4  1.691  0.686 n 35

E  tc

s 4.7  2.896  4.537 n 9

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

s 0.34  4.3  2.650  4.3  0.241  (4.1, 4.5) n 14

12. x  tc

s 4.6  24.7  2.678  24.7  1.742  (23.0, 26.4) n 50

t c  2.201

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. (64.6, 83.6)  x 

64.6  83.6  74.1  E  83.6  74.1  9.5 2

16. (16.2, 29.8)  x 

16.2  29.8  23  E  29.8  23  6.8 2

s 7.2  2.365  6.0 n 8 x  E  35.5  6.0  (29.5, 41.5)

17. E  tc

tc  2.426

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239

With 95% confidence, you can say that the population mean commute time to work is between 29.5 and 41.5 minutes. 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

s 184.00  2.365  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.

19. E  tc

s 33.61  2.201  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.

20. E  tc



9.3  6.4 n 8 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 n 5 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

26. (a) x  2.51

(b) s  0.87

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CHAPTER 6 │ CONFIDENCE INTERVALS

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 1.755  12.19  2.898  12.19  1.20  (10.99, 13.39) n 18

29. No, 1020 does not fall between 1034.3 and 1335.7. 30. No, 7.8 does not fall between 10.99 and 13.39. 31. (a) x  278, 430

(b) s  56, 769 56,769 s x  tc  278, 430  2.453  278, 430  24,616.93  ( 253,813.07, 303, 046.93) 32 n 32. (a) x  4,173

(b) s  800 (c) x  tc

800 s  4,173  2.426  4,173  306.87  (3,866.13, 4, 479.87 ) 40 n

33. No, 305,046 Yen does not fall between 253,813.07 Yen and 303,046.93 Yen. 34. No, 3,750 Yen does not fall between 3,866.13 Yen and 4,479.87 Yen. 35. Use a t-distribution because  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 known and the GDP growth rates are normally distributed. s 1.306 x  tc  0.403  2.110  0.403  0.65  ( 0.247, 1.053) n 18 With 95% confidence, you can say that the GDP mean growth rate is between −0.247 and 1.053.

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241

37. Neither distribution can be used because n < 30 and the mileages are not normally distributed. 38. Use the standard normal distribution because σ is known and the deliveries per fifty are normally distributed. s 6.37 x  tc  27  1.96  27  3.95  (23.05, 30.95) n 10 With 95% confidence, you can say that the population mean deliveries per fifty is between 16.82 deliveries and 24.72 deliveries. 39. Yes; Half the sample mean is 0.202, which falls within the confidence interval. 40. Yes; For the population mean to be within 10% of the sample mean, its value would be in the interval (23.05, 30.95). This interval falls within the confidence interval. 41. n  25, x  56.0, s  0.25 x   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   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 .

6.3 CONFIDENCE INTERVALS FOR POPULATION PROPORTIONS 6.3 TRY IT YOURSELF SOLUTIONS 1.

x  717, n  717  1338  1769  956  4780 x 717 pˆ    0.15  15% n 4780

pˆ  0.15, qˆ  1  0.15  0.85 npˆ  (4780)(0.15)  717  5 nqˆ  (4790)(0.85)  4063  5

Distribution of p̂ is approximately normal.

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CHAPTER 6 │ CONFIDENCE INTERVALS

zc  1.645

E  zc

ˆˆ pq 0.15  0.85  1.645  0.01 n 4780

pˆ  E  0.15  0.01  (0.14, 0.16)

n  800, pˆ  0.50 qˆ  1  pˆ  1  0.50  0.50 npˆ  800  0.50  400  5 nqˆ  800  0.50  400  5

Distribution of p̂ is approximately normal.

zc  2.575, E  zc

ˆˆ pq 0.50(0.50)  2.575  0.046 n 800

pˆ  E  0.50  0.046  (0.454, 0.546)

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  E At least 1692 people should be included in the sample. (2) pˆ  0.063, qˆ  0.937, z c  1.645, E  0.02 2

z   1.645  ˆ ˆ  c   (0.063)(0.937)  n  pq   399.3  400  0.02  E At least 400 people should be included in the sample.

6.3 EXERCISE SOLUTIONS 1. False. To estimate the value of p, the population proportion of successes, use the point estimate x pˆ  . n 2. True 3.

pˆ 

x 62   0.060, qˆ  1  pˆ  0.940 n 1040

pˆ 

x 478   0.460, qˆ  1  pˆ  0.540 n 1040

pˆ 

x 1310   0.650, qˆ  1  pˆ  0.350 n 2016

pˆ 

x 665   0.330, qˆ  1  pˆ  0.670 n 2016

 0.905, 0.933  pˆ 

0.905  0.933  0.919  E  0.933  0.919  0.014 2

CHAPTER 6 │ CONFIDENCE INTERVALS

 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

243

x 1322   0.590, qˆ  1  pˆ  0.410 n 2241 ˆˆ pq (0.590)(0.410)  0.590  1.645  0.590  0.017  (0.573, 0.607) 90% CI: pˆ  zc n 2241 ˆˆ pq (0.590)(0.410)  0.590  1.96  0.590  0.020  (0.570, 0.610) 95% CI: pˆ  zc n 2241 With 90% confidence, you can say that the population proportion of U.S. adults who say they have made a New Year’s resolution is between 57.3% and 60.7%. With 95% confidence, you can say it is between 57.0% and 61.0%. The 95% confidence interval is slightly wider.

11. pˆ 

x 650   0.290, qˆ  1  pˆ  0.710 n 2241 ˆˆ pq (0.290)(0.710)  0.290  1.645  0.290  0.016  (0.274, 0.306) 90% CI: pˆ  zc n 2241

12. pˆ 

ˆˆ pq (0.290)(0.710)  0.290  1.96  0.290  0.019  (0.271, 0.309) n 2241 With 90% confidence, you can say that the population proportion of U.S. adults who say they made a New Year’s resolution to eat healthier is between 27.4% and 30.6%. With 95% confidence, you can say it is between 27.1% and 30.9%. The 95% confidence interval is slightly wider. 95% CI: pˆ  zc

x 700   0.7, qˆ  1  pˆ  0.3 n 1000 ˆˆ pq (0.7)(0.3) pˆ  zc  0.7  2.575  0.7  0.037  (0.663, 0.737) n 1000 With 99% confidence, you can say that the population proportion of U.S. adults who say they think police officers should be required to wear body cameras while on duty is between 66.3% and 73.7%.

13. pˆ 

x 226   0.377, qˆ  1  pˆ  0.623 n 600 ˆˆ pq (0.377)(0.623) pˆ  zc  0.377  2.575  0.377  0.051  (0.326, 0.428) n 600 With 99% confidence, you can say that the population proportion of United Kingdom teachers who say they would wear a body camera in school is between 32.6% and 42.8%.

14. pˆ 

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CHAPTER 6 │ CONFIDENCE INTERVALS

15. pˆ 

x 49,311   0.0303, qˆ  1  pˆ  0.9697 n 1,626,773

pˆ  zc

ˆˆ pq (0.0303)(0.9697)  0.0303  1.96  0.0303  0.0003  (0.030, 0.031) n 1,626,773

x 490   0.49, qˆ  1  pˆ  0.51 n 1000 ˆˆ pq (0.49)(0.51) pˆ  zc  0.49  1.645  0.49  0.026  (0.464, 0.516) n 1000

16. pˆ 

z   1.96  ˆ ˆ  c   0.5  0.5  17. (a) n  pq   600.25  601 adults  0.04  E 2

z   1.96  ˆ ˆ  c   0.25  0.75  (b) n  pq   450.2  451 adults  0.04  E (c) Having an estimate of the population proportion reduces the minimum sample size needed. 2

z   2.575  ˆ ˆ  c   0.5  0.5  18. (a) n  pq   4144.14  4145 adults  0.02  E 2

z   2.575  ˆ ˆ  c   0.75  0.25  (b) n  pq   3108.1  3109 adults  0.02  E (c) Having an estimate of the population proportion reduces the minimum sample size needed. 2

z   1.645  ˆ ˆ  c   0.5  0.5  19. (a) n  pq   751.67  752 adults  0.03  E 2

z   1.645  ˆ ˆ  c   0.11 0.89  (b) n  pq   294.4  295 adults  0.03  E (c) Having an estimate of the population proportion reduces the minimum sample size needed. 2

z   1.96  ˆ ˆ  c   0.5  0.5  20. (a) n  pq   384.16  385 adults  0.05  E 2

z   1.96  ˆ ˆ  c   0.31 0.69  (b) n  pq   328.7  329 adults  0.05  E (c) Having an estimate of the population proportion reduces the minimum sample size needed. Copyright © 2019 Pearson Education Ltd.

CHAPTER 6 │ CONFIDENCE INTERVALS

245

21. 90% CI  (0.573, 0.607) 95% CI  (0.570, 0.610) Yes; 0.59 falls within both confidence intervals. 22. pˆ  0.01 pˆ  0.377  0.00377  (0.373, 0.381) Yes; For the population proportion to be within 1% of the point estimate, its value would be in the interval (0.373, 0.381). This interval falls within the confidence interval. 23. No; The minimum sample size needed is 451 adults. 24. Yes; The minimum sample size needed is 329 adults. 25. United States: pˆ  0.32, qˆ  0.68, n  1003

pˆ  zc

ˆˆ pq (0.32)(0.68)  0.32  2.575  0.32  0.038  (0.282, 0.358) n 1003

Canada: pˆ  0.21, qˆ  0.79, n  1020

pˆ  zc

ˆˆ pq (0.21)(0.79)  0.21  2.575  0.21  0.033  (0.177, 0.243) n 1020

France: pˆ  0.25, qˆ  0.75, n  999

pˆ  zc

ˆˆ pq (0.25)(0.75)  0.25  2.575  0.25  0.035  (0.215, 0.285) n 999

Japan: pˆ  0.50, qˆ  0.50, n  1000

pˆ  zc

ˆˆ pq (0.50)(0.50)  0.50  2.575  0.50  0.041  (0.459, 0.541) n 1000

Australia: pˆ  0.13, qˆ  0.87, n  1000

pˆ  zc

ˆˆ pq (0.13)(0.87)  0.13  2.575  0.13  0.027  (0.103, 0.157) n 1000

26. Yes; It is possible that the population proportion for the United States is the same as the population proportion for France, and/or the population proportion for France is the same as the population proportion for Canada, because the confidence intervals overlap. 27. (a) Expect to stay at first employer for 3 or more years: pˆ  0.69, qˆ  1  pˆ  0.31

ˆˆ pq (0.69)(0.31)  0.69  1.96  0.69  0.020  (0.670, 0.710) n 2000 Completed an apprenticeship or internship: pˆ  zc

pˆ  0.68, qˆ  1  pˆ  0.32

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CHAPTER 6 │ CONFIDENCE INTERVALS

ˆˆ pq (0.68)(0.32)  0.68  1.96  0.68  0.020  (0.660, 0.700) n 2000 Employed in field of study: pˆ  zc

pˆ  0.65, qˆ  1  pˆ  0.35

ˆˆ pq (0.65)(0.35)  0.65  1.96  0.65  0.021  (0.629, 0.671) n 2000 Feel underemployed: pˆ  zc

pˆ  0.51, qˆ  1  pˆ  0.49

ˆˆ pq (0.51)(0.49)  0.51  1.96  0.51  0.022  (0.488, 0.532) n 2000 Prefer to work for a large company: pˆ  zc

pˆ  0.14, qˆ  1  pˆ  0.86

pˆ  zc

ˆˆ pq (0.14)(0.86)  0.14  1.96  0.14  0.015  (0.125, 0.155) n 2000

(b) Expect to stay at first employer for 3 or more years: pˆ  0.69, qˆ  1  pˆ  0.31

ˆˆ pq (0.69)(0.31)  0.69  2.575  0.69  0.027  (0.663, 0.717) n 2000 Completed an apprenticeship or internship: pˆ  zc

pˆ  0.68, qˆ  1  pˆ  0.32

ˆˆ pq (0.68)(0.32)  0.68  2.575  0.68  0.027  (0.653, 0.707) n 2000 Employed in field of study: pˆ  zc

pˆ  0.65, qˆ  1  pˆ  0.35

ˆˆ pq (0.65)(0.35)  0.65  2.575  0.65  0.027  (0.623, 0.677) n 2000 Feel underemployed: pˆ  zc

pˆ  0.51, qˆ  1  pˆ  0.49

ˆˆ pq (0.51)(0.49)  0.51  2.575  0.51  0.029  (0.481, 0.539) n 2000 Prefer to work for a large company: pˆ  zc

pˆ  0.14, qˆ  1  pˆ  0.86

pˆ  zc

ˆˆ pq (0.14)(0.86)  0.14  2.575  0.14  0.020  (0.120, 0.160) n 2000

28. Yes; The confidence intervals in parts (a) and (b) for “expect to stay at first employer for 3 or more years,” “completed an apprenticeship or internship,” and “employed in field of study” overlap. 29. 70%  3.4%  (66.6%, 73.4%)  (0.666, 0.734) E  zc

ˆˆ 1003 pq n  zc  E  0.034  zc  2.35 ˆˆ (0.70)(0.30) n pq

CHAPTER 6 │ CONFIDENCE INTERVALS

P(2.35  z  2.35)  0.9906  0.0094  0.9812 (0.666, 0.734) is approximately a 98.1% CI. 30. 79%  2.9%  (76.1%, 81.9%)  (0.761, 0.819) E  zc

ˆˆ 1503 pq n  zc  E  0.029  zc  2.76 ˆˆ (0.79)(0.21) n pq

P (2.76  z  2.76)  0.9971  0.0029  0.9942 (0.761, 0.819) is approximately a 99.4% CI. 31. 71%  3%  (68%, 74%)  (0.68, 0.74) E  zc

ˆˆ pq  zc  E n

1000 n  0.03  zc  2.09 ˆˆ (0.71)(0.29) pq

P(2.09  z  2.09)  0.9817  0.0183  0.9634 (0.68, 0.74) is approximately a 96.3% CI. 32. 37%  4%  (33%, 41%)  (0.33, 0.41) E  zc

ˆˆ 1035 pq n  zc  E  0.04  zc  2.67 ˆˆ (0.37)(0.63) n pq

P(2.67  z  2.67)  0.9962  0.0038  0.9924 (0.33, 0.41) is approximately a 99.2% CI. 33. 47%  2%  (45%, 49%)  (0.45, 0.49) E  zc

ˆˆ 3539 pq n  zc  E  0.02  zc  2.38 ˆ ˆ (0.47)(0.53) n pq

P (2.38  z  2.38)  0.9913  0.0087  0.9826 (0.45, 0.49) is approximately a 98.3% CI. 53%  2%  (51%, 55%)  (0.51, 0.55) E  zc

ˆˆ 3539 pq n  zc  E  0.02  zc  2.38 ˆ ˆ (0.53)(0.47) n pq

P (2.38  z  2.38)  0.9913  0.0087  0.9826 (0.51, 0.55) is approximately a 98.3% CI. 34. 68%  3%  (65%, 71%)  (0.65, 0.71) E  zc

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 pq  zc  E n

1052 n  0.03  zc  1.97 ˆˆ (0.42)(0.58) pq

P (1.97  z  1.97)  0.9756  0.0244  0.9512

247

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CHAPTER 6 │ CONFIDENCE INTERVALS

(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

Write original equation.

ˆˆ E n  zc pq n

ˆˆ pq

zc E

z  ˆ ˆ c  n  pq E

37.

Multiply each side by n . Divide each side by E. 2

Square each side.

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 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 R is 0.05. 2

Area to the right of L is 0.95. 2

R2  42.557 , 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:  R  42.557 , L  17.708 2

CHAPTER 6 │ CONFIDENCE INTERVALS

249

95% CI:  R  45.722 , L  16.047 2

2 (n  1)s 2   29  (1.2)2 29  (1.2)2  2  (n  1) s , , 90% CI for  :     (0.98, 2.36) 2 17.708   L2   42.557  R 2 (n  1)s 2   29  (1.2)2 29  (1.2)2  2  (n  1) s , , 95% CI for  :     (0.91, 2.60) 2 16.047   L2   45.722  R

90% CI for  :  0.981,

2.358    0.99, 1.54

95% CI for  :  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.

R2  14.067 , L2  2.167

R2  31.319 , L2  4.075

R2  32.852 , L2  8.907

R2  44.314 , L2  11.524

R2  52.336 , L2  13.121

R2  63.167 , L2  37.689

 (n  1)s 2 (n  1)s 2   29  (11.56) 29  (11.56)   (7.33, 20.89) , , 9. (a)   2 16.047   L2   45.722  R (b)  7.3321,

20.8911   (2.71, 4.57)

 (n  1)s 2 (n  1)s 2   6  (0.64) 6  (0.64)   (0.21, 5.68) , , 10. (a)   2 0.676   L2   18.548  R (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   L2   27.587  R (b)  754.885,

2401.407   (27, 49)

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CHAPTER 6 │ CONFIDENCE INTERVALS

 (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   L2   63.691  R (b)

 48,571.77, 139,576.99   (220.4, 373.6)

13. (a) s 2  0.262  (n  1)s 2 (n  1)s 2   17  (0.262) 17  (0.262)   (0.1615, 0.5136) , ,   2 8.672   L2   27.587  R

(b)  0.16145, 0.51361   (0.4018, 0.7167) With 90% confidence, you can say that the population variance is between 0.1615 and 0.5136, and the population standard deviation is between 0.4018 and 0.7167 centimeters. 14. (a) s 2  0.266  (n  1)s 2 (n  1)s 2   14  (0.266) 14  (0.266)   (0.1426, 0.6616) , ,   2 5.629   L2   26.119  R

(b)  0.142578, 0.661574   (0.3776, 0.8134) With 95% confidence, you can say that the population variance is between 0.1426 and 0.6616, and the population standard deviation is between 0.3776 and 0.8134 thousand IUs. 15. (a) s 2  362.98  (n  1)s 2 (n  1)s 2   20  (362.98) 20  (362.98)   (181.50, 976.54) , ,   2 7.434   L2   39.997  R

(b)  181.50, 976.54   (13.47, 31.25) With 99% confidence, you can say that the population variance is between 181.50 and 976.54, and the population standard deviation is between 13.47 and 31.25 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   L2   38.076  R

(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   L2   29.819  R (b)  5.46,

45.70   (2.34, 6.76)

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251

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   10  (9.35)2 10  (9.35)2  , , 18. (a)     (54.68, 179.70) 2 2 15.987 4.865    R L    (b)  54.68, 179.70   (7.39, 13.41) With 80% confidence, you can say that the population variance is between 54.68 and 179.70, and the population standard deviation is between 7.39 and 13.41 touchdowns.

 (n  1) s 2 (n  1) s 2   18  (15)2 18  (15)2  , , 19. (a)     (128, 492) 2 8.231   L2   31.526  R (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   L2   42.557  R (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  (0.40)2 17  (0.40)2  , , 21. (a)     (0.0986, 0.3137) 2 8.672   L2   27.587  R (b)  0.098597, 0.313653   (0.314, 0.560) With 90% confidence, you can say that the population variance is between 0.0986 and 0.3137, and the population standard deviation is between 0.314 and 0.560 hours.

 (n  1)s 2 (n  1)s 2   63  (16.8)2 63  (16.8)2    22. (a)     (193.3, 446.1) 2 2 92.010 39.855    R L    (b)  193.3,

446.1   (13.9, 21.1)

With 98% confidence, you can say that the population variance is between 193.3 and 446.1, and the population standard deviation is between 13.9 °F and 21.1 °F.

 (n  1)s 2 (n  1)s 2   25  (4.9)2 25  (4.9)2    23. (a)     (12.8, 57.1) 2 10.520   L2   46.928  R

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CHAPTER 6 │ CONFIDENCE INTERVALS

(b)  12.79, 57.06   (3.6, 7.6) With 99% confidence, you can say that the population variance is between 12.8 and 57.1, and the population standard deviation is between 3.6 and 7.6 minutes.

 (n  1)s 2 (n  1)s 2   22  (1200)2 19  (1200)2    24. (a)     (1028137,2256250) 2 14.014   L2   30.813  R (b)  1028137.474 , 2256249.555   (1014,1502) With 80% confidence, you can say that the population variance is between 1,028,137 and 2,256,250, and the population standard deviation is between $1014 and $1502. 25. 90% CI for σ: (0.4018, 0.7167) No, because all of the values in the confidence interval are greater than 0.35. 26. 95% CI for σ: (0.3776, 0.8134) No, because 0.50 is contained in the class interval. 27. 90% CI for σ: (0.314, 0.560) No, because 0.50 is contained in the class interval. 28. 99% CI for σ: (3.6, 7.6) Yes, because all of the values in the confidence interval are less than 7.9. 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  119.05

(b) E  zc

 n

 1.645

25 40

 6.5

2. (a) x  23.6

(b) E  zc

x  zc



 n

 1.96

10 30

 119.05  1.645

 3.6 25

 119.05  6.5  (112.6, 125.6) n 40 With 90% confidence, you can say that the systolic blood pressure mean is between 112.6 and 125.6 mmHg.

CHAPTER 6 │ CONFIDENCE INTERVALS

x  zc



 23.6  1.96

 23.6  3.6  (20, 27.2) n 30 With 95% confidence, you can say that the population mean age of participants is between 20 and 27.2 years.

 30.25, 42.50  x 

30.25  42.50  36.375  E  42.50  36.375  6.125 2

 7.428, 7.562  x 

7.428  7.562  7.495  E  7.562  7.495  0.067 2

 z    1.96  45  n c     77.79  78 people  E   10 

 z    2.575 10  n c     73.7  74 people 3   E  

tc  1.796

13. E  tc

s n

14. (a) E  tc

(b) x  tc

15. E  tc

10. tc  2.069  1.704 s n s n

27.4 36

 2.064

11. tc  1.341

12. tc  2.756

 5.96 1.1 25

 0.5

 3.5  0.5  (3.0, 4.0)

 3.6   1.726    1.39 n  20 

16. (a) E  tc

(b) x  tc

17. x  tc

 16.5  s  2.861   10.6 n  20  s n

 25.2  10.6  (14.6, 35.8)

 165  1.690

 165  18.87  (146, 184) n 36 With 90% confidence, you can say that the population mean height is between 146 and 184 feet.

18. Yes; 160 feet falls within the confidence interval, (146, 184).

253

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CHAPTER 6 │ CONFIDENCE INTERVALS

19. (a) pˆ 

x 745   0.720, qˆ  1  pˆ  0.280 n 1035

(b) 90% CI: pˆ  zc 95% CI: pˆ  zc

ˆˆ pq 0.720  0.280  0.720  1.645  0.720  0.023  (0.697, 0.743) n 1035 ˆˆ pq 0.720  0.280  0.720  1.96  0.720  0.027  (0.693, 0.747) n 1035

(c) With 90% confidence, you can say that the population proportion of U.S. adults who say they want the U.S. to play a leading or major role in global affairs is between 69.7% and 74.3%. With 95% confidence, you can say it is between 69.3% and 74.7%. The 95% confidence interval is slightly wider. 20. (a) pˆ 

x 451   0.450, qˆ  1  pˆ  0.550 n 1003

(b) 90% CI: pˆ  zc 95% CI: pˆ  zc

ˆˆ pq 0.450  0.550  0.450  1.645  0.450  0.026  (0.424, 0.476) n 1003 ˆˆ pq 0.450  0.550  0.450  1.96  0.450  0.031  (0.419, 0.481) n 1003

(c) With 90% confidence, you can say that the population proportion of U.S. adults who believe that for a person to be considered truly American, it is very important that he or she share American customs and traditions is between 42.4% and 47.6%. With 95% confidence, you can say it is between 41.9% and 48.1%. The 95% confidence interval is slightly wider.

x 1167   0.530, qˆ  1  pˆ  0.470 n 2202 ˆˆ pq 0.530  0.470  0.530  1.645  0.530  0.017  (0.513, 0.547) (b) 90% CI: pˆ  zc n 2202 ˆˆ pq 0.530  0.470  0.530  1.96  0.530  0.021  (0.509, 0.551) 95% CI: pˆ  zc n 2202

21. (a) pˆ 

(c) With 90% confidence, you can say that the population proportion of U.S. adults who think antibiotics are effective against viral infections is between 51.3% and 54.7%. With 95% confidence, you can say it is between 50.9% and 55.1%. The 95% confidence interval is slightly wider. 22. (a) pˆ 

x 1334   0.600, qˆ  1  pˆ  0.400 n 2223

(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) n 2223 ˆˆ pq 0.600  0.400  0.600  1.96  0.600  0.020  (0.580, 0.620) n 2223

CHAPTER 6 │ CONFIDENCE INTERVALS

255

(c) With 90% confidence, you can say that the population proportion of 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. No; it falls outside both confidence intervals. 24. pˆ  0.01 pˆ  0.600  0.01(0.600)  0.600  0.006  (0.594, 0.606) Yes; If the population proportion is within 1% of the point estimate, then it falls inside both confidence intervals. 2

z   1.96  ˆ ˆ  c   0.50  0.50  25. (a) n  pq   384.16  385 adults  0.05  E z   1.96  ˆ ˆ  c   0.32  0.68  (b) n  pq   334.4  335 adults  0.05  E (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.  R2  23.685,  L2  6.571 28.  R2  42.980,  L2  10.856 29.  R2  32.852,  L2  8.907 30.  R2  23.589,  L2  1.735 31. s 2  359.936 2 (n  1)s 2   12  (359.936) 12  (359.936)  2  (n  1) s , , (a) 95% CI for  :     (185.1, 980.8) 2 4.404  L2   23.337   R

(b) 95% CI for  :

 185.1, 980.8   (13.6, 31.3)

With 95% confidence we can say that the population variance is between 185.1 and 980.8, and the population standard deviation is between 13.6 and 31.3 knots. 32. s 2  8.2249 2 (n  1)s 2   (32)(8.2249) (32)(8.2249)  2  (n  1) s  (4.92, 16.09) , , (a) 98% CI for  :   2 16.3622   L2   53.4858  R

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CHAPTER 6 │ CONFIDENCE INTERVALS

(b) 98% CI for  :

 4.92, 16.09   (2.22, 4.01)

With 98% confidence, you can say that the population variance is between 4.92 and 16.09, and the population standard deviation is between 2.22 and 4.01 seconds.

CHAPTER 6 QUIZ SOLUTIONS

1. (a) x 

 x  77.94  2.598 30



(b) E  zc (c) x  zc

 1.96

0.344 30

 0.123



 2.598  0.123  (2.475, 2.721) n With 95% confidence, you can say that the population mean winning time is between 2.475 and 2.721 hours.

(d) No; It falls outside the confidence interval. 2

 z    2.575  0.325  n c     41.4  42 champions 0.13   E  

3. (a) x  6.61, s  3.376

(b) x  tc

3.376

 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.  3.5 (c) x  zc  6.61  1.645  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).

x  tc

 6.61  1.833

 133,326  2.201

36,729

 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.

5. Yes. $131,935 falls within the confidence interval, (109,990, 156,662). 6. (a) pˆ 

x 753   0.740 n 1018

CHAPTER 6 │ CONFIDENCE INTERVALS

257

ˆˆ pq 0.740  0.260  0.740  1.645  0.740  0.023  (0.717, 0.763) n 1018 With 90% confidence you can say that the population proportion of U.S. adults who say that the energy situation in the United States is very or fairly serious is between 71.7% and 76.3%.

(b) pˆ  zc

z   2.575  ˆ ˆ  c   0.740  0.260  (d) n  pq   797.3  798 adults  0.04  E  (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   L2   19.023  R (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 pq 0.830  0.170  0.830  1.96  0.830  0.016  (0.814, 0.846) n 2096 With 95% confidence you can say that the population proportion of 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

z   2.575  ˆ ˆ  c   0.83  0.17  (d) n  pq   1039.5  1040 adults  0.03  E 2. (a) x  214.1, s  51.1

(b) x  tc

 214.1  2.262

51.07

 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.

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CHAPTER 6 │ CONFIDENCE INTERVALS

 (n  1) s 2 (n  1) s 2   (9)(51.1)2 (9)(51.1)2      (31.6, 116.4)  , , 2 2    23.589 1.735    R L     With 99% confidence, you can say that the population standard deviation is between 31.6 and 116.4 pounds.

3. (a) x  579.2

(b) x  zc



 579.2  1.645

111

 579.2  52.71  (631.91, 526.49) n 12 With 90% confidence, you can say that the population mean test score is between 526.49 and 631.91.

 z    1.96 111  (d) n   c      473.3  473 students  E   10  4. (a) Use the t-distribution because  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  is known and the weights are normally distributed.  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.

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; m ¹ 74 . H 0 : m = 74; H a : m ¹ 74 (claim) (2) The variance is less than or equal to 2.7; s 2 £ 2.7 .

s2 > 2.7 H0 : s 2 £ 2.7 (claim); Ha : s2 > 2.7 (3) The proportion is more than 24%; p > 0.24 . H 0 : p £ 0.24; H a : p > 0.24 (claim) 2.

H 0 : 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 : m = 74; H a : m ¹ 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  2 £ 2.7; H a : 2 > 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%. H a : The proportion of homeowners who feel their house is too small for their family is greater than 24%. H 0 : p £ 0.24; H a : p > 0.24 259 Copyright © 2019 Pearson Education Ltd.

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CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

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 : m ³ 650; H a : m < 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 : m = 98.6 (claim); H a : m ¹ 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. 10. False. If you want to support a claim, write it as your alternative hypothesis.

CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

11. H 0 : m £ 645 (claim); H a : m > 645

12. H 0 : m ³ 128 ; H a : m < 128 (claim)

13. H 0 : s = 5 ; H a : s ¹ 5 (claim)

14. H0 : s ³1.2 (claim); Ha : s <1.2

15. H 0 : p ³ 0.45 ; H a : p < 0.45 (claim)

16. H 0 : p = 0.21 (claim); H a : p ¹ 0.21

17. c; H 0 : m £ 3

18. d; H a : m ³ 3

19. b; H 0 : m = 3

20. a; H a : m £ 2

21. Right-tailed

22. Left tailed

25. s £ 225 H 0 : s £ 225 (claim); H 0 : s > 225

27. m > 4 H 0 : s £ 4 ; H a : s > 4 (claim)

29. p = 0.73 H 0 : p = 0.73 (claim); H a : p ¹ 0.73

23. Two-tailed

261

24. Two-tailed

26. m ³ 300 H 0 : m ³ 300 (claim); H a : m < 300 28. s < 2 H 0 : s ³ 2 ; H 0 : s < 2 (claim)

30. p = 0.52 H 0 : p = 0.52 (claim); H a : p ¹ 0.52

31. A type I error will occur if the actual proportion of new customers who return to place their next order is at least 0.75, but you reject H 0 : p ³ 0.75 . A type II error will occur if the actual proportion of new customers who return to place their next order is less than 0.75, but you fail to reject H 0 : p ³ 0.75 . 32. A type I error will occur if the actual mean flow rate of the hose is 40 liters per minute, but you reject H0: μ = 40. A type II error will occur if the actual mean flow rate of the hose is not 40 liters per minute, but you fail to reject H0: μ = 40. 33. A type I error will occur if the actual standard deviation of the length of time to bowl an over is less than or equal to 4 minutes, but you reject H 0 : s £ 4 . A type II error will occur if the actual standard deviation of the length of time to bowl an over is greater than 4 minutes, but you fail to reject H 0 : s £ 4 . 34. A type I error will occur if the actual proportion of Egyptian adults who own a surround sound system is 0.40, but you reject H 0 : p = 0.40 . A type II error will occur if the actual proportion of Egyptian adults who own a surround sound system is not 0.40, but you fail to reject H 0 : p = 0.40 .

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CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

35. A type I error will occur if the actual proportion of encrypted data remains safe is at least 0.98, but you reject H 0 : p ³ 0.98 . A Type II error will occur if the actual proportion of encrypted data remains safe is at less than 0.98, but you fail to reject H 0 : p ³ 0.98 . 36. A type I error will occur if the actual mean cost of repairing a laptop is greater than or equal to $125, but you reject H0: μ ≥ 125. A type II error will occur if the actual mean cost of repairing a laptop is less than $125, but you fail to reject H0: μ ≥ 125. 37. H0: The mean number of glasses that break per production cycle is less than or equal to 3 glasses. Ha: The mean number of glasses that break per production cycle is greater than 3 glasses. H 0 : m £ 3, H a : m > 3

Right-tailed because the alternative hypothesis contains >.

38. H0: The proportion of business who have a fire insurance policy is greater than or equal to 65%. Ha: The proportion of business who have a fire insurance policy is less than 65%. H 0 : p ³ 0.65, H a : p < 0.65

Left-tailed because the alternative hypothesis contains <.

CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

263

39. H0: The percentage of IT students getting employment is 95%. Ha: The percentage of IT students getting employment is not 95%. H 0 : p = 0.95; H a : p ¹ 0.95

Two-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 standard deviation of the jockey clearing the obstacles is greater than or equal to 2 obstacles. Ha: The standard deviation of the jockey clearing the obstacles is less than 2 obstacles. H 0 : s ³ 2; H a :s < 2

Left-tailed because the alternative hypothesis contains <.

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CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

42. H0: The mean increase in sales in response to free samples distributed is equal to 5,000. Ha: The mean increase in sales in response to free samples distributed is less than 5,000. H 0 : m ³ 5, 000; H a :m < 5, 000

Left-tailed because the alternative hypothesis contains <.

43. Null hypothesis (a) There is enough evidence to reject the claim that the standard deviation of the life of the air conditioners is at most 4.6 years.

(b) There is not enough evidence to reject the claim that the standard deviation of the life of the air conditioners is at most 4.6 years. 44. Alternative hypothesis (a) There is enough evidence to support the report’s claim that more than 40% of households in a New York county struggle to afford basic necessities.

(b) There is not enough evidence to support the report’s claim that more than 40% of households in a New York county struggle to afford basic necessities. 45. Alternative hypothesis (a) There is enough evidence to support the scientist’s claim that the mean number of migratory birds is less than 3,800.

(b) There is not enough evidence to support the scientist’s claim that the mean number of migratory birds is less than 3,800. 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 models is 5.6 kilometers per liter.

(b) There is not enough evidence to reject the automotive manufacturer’s claim that the standard deviation for the gas mileage of its models is 5.6 kilometers per liter. 47. Null hypothesis (a) There is enough evidence to reject the report’s claim that at least 65% of individuals convicted of terrorism or terrorism-related offenses in the United States are foreign born.

(b) There is not enough evidence to reject the report’s claim that at least 65% of individuals convicted of terrorism or terrorism-related offenses in the United States are foreign born.

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265

48. Alternative hypothesis (a) There is enough evidence to reject the NGO’s claim that none of the children in a slum area get balanced diet.

(b) There is not enough evidence to reject the NGO’s claim that the children in a slum area get balanced diet. 49. H 0 : m ³ 75; H a : m < 75

50. H 0 : m = 7; H a : m ¹ 7

51. (a) H0: μ ≥ 46; Ha: μ < 46

52. (a) H 0 : m £ 39; H a : m > 39

(b) H0: μ ≤ 46; Ha: μ > 46

(b) H 0 : m ³ 39; H a : m < 39

53. If you decrease a, you are decreasing the probability that you reject H 0 . Therefore, you are increasing the probability of failing to reject H 0 . This could increase b, the probability of failing to reject H 0 when H 0 is false. 54. If a = 0, the null hypothesis cannot be rejected and the hypothesis test is useless. 55. Yes; If the P-value is less than a = 0.05, it is also less than a = 0 .1 0 . 56. Not necessarily; A P-value less than a = 0 .1 0 may or may not also be less than a = 0 .0 5 . 57. (a) Fail to reject H 0 because the CI includes values greater than 70.

(b) Reject H 0 because the CI is located below 70. (c) Fail to reject H 0 because the CI includes values greater than 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 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 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 (LARGE SAMPLES) 7.2 TRY IT YOURSELF SOLUTIONS 1. (1) Fail to reject H 0 because 0 .0 7 4 5 > 0 .0 5 . (2) Reject H 0 because 0 .0 7 4 5 < 0 .1 0 . 2.

P = 0 .0 4 3 6

Reject H 0 because P = 0 .0 4 3 6 < 0 .0 5 . 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 .1 0 1 0 > 0 .1 0 . 4. The claim is “the mean speed is greater than 35 miles per hour.” H 0 : m £ 35; H a : m > 35 (claim) a = 0 .0 5

x - m 36 - 35 = = 2.5 s 4 n 100 P-value = {Area right of z = 2 .5 0 } = 0 .0 0 6 2 Reject H 0 because P-value = 0 .0 0 6 2 < 0 .0 5 . 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. z=

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267

5. The claim is “the mean time number of workdays missed due to illness or injury in the past 12 months is 3.5 days.” H 0 : m = 3.5 (claim); H a : m ¹ 3.5 a = 0 .0 1

x - m 4 - 3.5 = » 1.67 s 1.5 n 25 P-value = 2{Area to the right of z = 1 .6 7 } = 2(0.0475) = 0.0950 z=

Fail to reject H 0 because P-value = 0 .0 9 5 0 > 0 .0 1 . 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 3.5 days. 6.

P = 0 .0 4 4 0 > 0 .0 1 = a .

Fail to reject H 0 . 7.

Area = 0.1003

0.0401 and 0.9599

z 0 = - 1.28

- z 0 = - 1.75 and z0 = 1.75 Rejection region: z < - 1.75, z > 1.75

Rejection region: z < - 1 .2 8

9. The claim is “the mean work day of the company’s mechanical engineers is less than 8.5 hours.” H 0 : m ³ 8.5; H a : m < 8.5 (claim) a = 0 .0 1

z0 = - 2.33; Rejection region: z < - 2 .3 3

x - m 8.2 - 8.5 = = -3.00 s 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. a = 0 .0 1 - z 0 = - 2.575, z 0 = 2.575; Rejection regions: z < - 2.575, z > 2.575

Because z » - 1 .9 8 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 (age 2 and under) by married-couple families in the United States is $14,050.

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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 a. 2. No; Both involve comparing the test statistic’s probability with the level of significance. The P-value method converts the standardized test statistic to a probability (P-value) and compares this with the level of significance, whereas the critical value method converts the level of significance to a z-score and compares this with the standardized test statistic. 3. (a) Fail to reject H0 because P = 0 .0 4 6 1 > 0 .0 1 = a .

(b) Reject H0 because P = 0 .0 4 6 1 < 0 .0 5 = a . (c) Reject H0 because P = 0 .0 4 6 1 < 0 .1 0 = a . 4. (a) Fail to reject H0 because P = 0 .0 6 9 1 > 0 .0 1 = a .

(b) Fail to reject H0 because P = 0 .0 6 9 1 > 0 .0 5 = a . (c) Reject H0 because P = 0 .0 6 9 1 < 0 .1 0 = a . 5. (a) Fail to reject H0 because P = 0 .1 2 7 1 > 0 .0 1 = a .

(b) Fail to reject H0 because P = 0 .1 2 7 1 > 0 .0 5 = a . (c) Fail to reject H0 because P = 0 .1 2 7 1 > 0 .1 0 = a . 6. (a) Fail to reject H0 because P = 0 .0 1 0 7 > 0 .0 1 = a .

(b) Reject H0 because P = 0 .0 1 0 7 < 0 .0 5 = a . (c) Reject H0 because P = 0 .0 1 0 7 < 0 .1 0 = a . 7. (a) Fail to reject H0 because P = 0 .0 8 3 8 > 0 .0 1 = a .

(b) Fail to reject H0 because P = 0 .0 8 3 8 > 0 .0 5 = a . (c) Reject H0 because P = 0 .0 8 3 8 < 0 .1 0 = a . 8. (a) Reject H0 because P = 0 .0 0 6 2 < 0 .0 1 = a .

(b) Reject H0 because P = 0 .0 0 6 2 < 0 .0 5 = a . (c) Reject H0 because P = 0 .0 0 6 2 < 0 .1 0 = a .

CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

P = 0 .0 9 3 4 ; Reject H 0 because P = 0 .0 9 3 4 < 0 .1 0 .

10.

P = 0 .0 6 0 6 ; Fail to reject H 0 because P = 0 .0 6 0 6 > 0 .0 5 .

11.

P = 0 .0 0 6 9 ; Reject H 0 because P = 0 .0 0 6 9 < 0 .0 1 .

12.

P = 0.1093 ; Fail to reject H 0 because P = 0.1093 > 0.10 . 13.

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P = 2(Area) = 2(0.0465) = 0.0930 ; Fail to reject H 0 because P = 0.0930 > 0.05 . 14.

P = 2(Area) = 2(0.0256) = 0.0512 ; Reject H 0 because P = 0.0512 < 0.08 15. (a) P = 0.0089 (b) P = 0.3050 The larger P-value corresponds to the larger area.

(b) P = 0.0688 16. (a) P = 0.2802 The larger P-value corresponds to the larger area. 17. Fail to reject H 0 , ( P = 0.0628 > 0.05). 18. Reject H 0 , ( P = 0.0065 < 0.01). 19. Critical value: z 0 = - 1.88 ; Rejection region: z <-1.88

20. Critical value: z 0 = - 1.34 ; Rejection region: z <-1.34

21. Critical value: z 0 = 1.645 ; Rejection region: z > 1.645

22. Critical value: z0 = 1.41 ; Rejection region: z > 1.41

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23. Critical values: - z0 = - 2.33, z0 = 2.33 ; Rejection regions: z < -2.33, z > 2.33

24. Critical values: -z0 = 1.555, z0 = 1.555 Rejection regions: z < 1.555, z > 1.555

25. (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 . 26. (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 . 27. H 0 : m = 40 (claim); H a : m ¹ 40 a = 0.05  z 0 =  1.96

x - m 39.2 - 40 = » -2.030 s 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. z=

28. H 0 : m ³ 1475 (claim); H a : m < 1475 a = 0.07  z 0 = - 1.48

x - m 1468 -1475 = » -1.231 s 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. z=

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a = 0.03  z 0 =  2.17

x - m 5771- 5880 = » -2.16 s 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=

30. H 0 : m £ 22,500 (claim); H a : m > 22,500 a = 0.01  z 0 = 2.33

x - m 23,500 - 22,500 = » 5.590 s 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. z=

31. (a) The claim is “the mean total score for the school’s applicants is more than 499.” H 0 : m £ 499; H a : m > 499 (claim)

x - m 502 - 499 = » 2.83 s 10.6 n 100 Area = 0.9977

(b) z =

(c) P-value = {Area to right of z = 2.83 } = 0.0023 (d) Because P = 0.0023 < 0.01 = a , reject H 0 . (e) There is 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 499. 32. (a) The claim is “the average coverage area is at least 60 square meters.” H 0 : m ³ 60 (claim); H a : m < 60

x - m 58 - 60 = » -2.68 s 3.5 n 22 Area = 0.0037

(b) z =

(c) P-value = {Area to left of z = -2.68 } = 0.0037 (d) Because P = 0.0037 < 0.10 = a , reject H 0 . (e) There is enough evidence at the 10% level of significance to reject the claim that the average coverage area is at least 60 square meters.

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33. (a) The “the mean winning times for Boston Marathon women’s open division champions is at least 2.68 hours.” H 0 : m ³ 2.68 (claim); H a : m < 2.68

x - m 2.60 - 2.68 = » -1.37 s 0.32 n 30 Area = 0.0853

(b) z =

(c) P-value = {Area to the left of z = -1.37 } = 0.0853 (d) Because P = 0.0853 > 0.05 = a , fail to reject H 0 . (e) There is not 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 2.68 hours. 34. (a) The claim is “the mean acceleration time from 0 to 60 miles per hour for a sedan is 6.3 seconds.” H 0 : m = 6.3 (claim); H a : m ¹ 6.3

x - m 7.2 - 6.3 = » 2.07 s 2.5 n 33 Area to right of 2.07 = 0.0192

(b) z =

(c) P-value = 2{Area to right of z = 2.07 } = 2(0.0192) = 0.0384 (d) Because P = 0.0384 < 0.05 = a , reject H 0 . (e) There is enough evidence at the 5% level of significance to reject the consumer group’s claim that the mean acceleration time from 0 to 60 miles per hour for a sedan is 6.3 seconds. 35. (a) The claim is “the mean height of top-rated roller coasters is 160 feet.” H 0 : m = 160 (claim); H a : m ¹ 160

(b) x » 164.61 x - m 164.61 -160 z= = » 0.3863 » 0.39 s 71.6 n 36 Area to the right of 0.3863= 0.3496 (c) P-value = 2{Area to right of z = 0.3863 } = 2(0.3496) = 0.6992 (d) Because P = 0.6992 > 0.05 = a , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to reject the claim that the mean height of top-rated roller coasters is 160 feet.

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36. (a) The claim is “the mean annual salary for intermediate level architects in Wichita, Kansas, is more than the national mean, $52,000.” H 0 : m £ 52, 000; H a : m > 52, 000 (claim)

(b) x » 50,065.6 x - m 50,065.6 - 52,000 = » -1.11 z= s 8000 n 21 Area to the left of -1.11= 0.1335 (c) P-value = {Area to right of z = -1.11 } » 1- 0.1335 = 0.8665 (d) Because P = 0.8665 > 0.09 = a , fail to reject H 0 . (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 architects in Wichita, Kansas, is more than the national mean, $52,000. 37. (a) The claim is “the mean caffeine content per 12-ounce bottle of a population of caffeinated soft drinks is 37.7 milligrams.” H 0 : m = 37.7 (claim); H a : m ¹ 37.7

(b) - z 0 = - 2.575, z 0 = 2.575; Rejection regions: z <-2.575, z > 2.575 (c) z =

x - m 36.4 - 37.7 = » -0.72 s 10.8 n 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. 38. (a) The claim is “the mean high school graduation rate per state in the United States is 80%.” H 0 : m = 80 (claim); H a : m ¹ 80

(b) - z 0 = - 1.96, z 0 = 1.96; Rejection regions: z <-1.96, z >1.96 (c) z =

x - m 82 - 80 = » 2.15 s 5.1 n 30

(d) Because z > 1.96 , reject H 0 .

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(e) There is enough evidence at the 5% level of significance to reject the education researcher’s claim that the mean high school graduation rate per state in the United States is 80%. 39. (a) The claim is “the mean sugar content in each of the cookies produced by a manufacturer is no more than 18%.” H 0 : m £ 18 (claim); H a : m > 18

(b) z0 = 2.05; Rejection region: z > 2.05 (c) 1.87 (d) Fail to reject H 0 . (e) There is enough evidence at the 2% level of significance to support the claim that the mean sugar content in each of their cookies is no more than 18%. 40. (a) The claim is “the mean life of a certain type of LED is at least 25,000 hours.” H 0 : m ³ 25, 000 (claim); H a : m < 25, 000

(b) z0 = - 1.645; Rejection region: z <-1.645 (c) z =

x - m 24800 - 25000 = = -2.81 s 500 n 49

(d) Because z >-2.81 , fail to reject H 0 . (e) (e) There is enough evidence at the 5% level of significance to reject the claim that the mean life of a LED is at least 25,000 hours. 41. (a) The claim is “the mean life of a fluorescent lamp is at least 10,000 hours.” H 0 : m ³ 10, 000 (claim); H a : m < 10, 000

(b) z0 = - 1.23; Rejection region: z <-1.23 (c) x » 9580.9 x - m 9580.9 -10,000 z= » » -1.28 s 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 fluorescent lamps is at least 10,000 hours.

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42. (a) The claim is “the mean carbon dioxide emissions per country in a recent year are greater than 150 megatons.” H 0 : m £ 150; H a : m > 150 (claim)

(b) z 0 = 1.555; Rejection region: z > 1.555 (c) x » 77.33 x - m 77.33 -150 » » -0.577 z= s 816 n 42 (d) Because z < 1.555 , fail to reject H 0 . (e) There is not enough evidence at the 6% level of significance to support the scientist’s claim that the mean carbon dioxide emissions per country in a recent year are greater than 150 megatons. 43. Outside; When the standardized test statistic is inside the rejection region, P < a . 44. To the right; When P < a , the standardized test statistic is inside the rejection region.

7.3 HYPOTHESIS TESTING FOR THE MEAN (SMALL SAMPLES) 7.3 TRY IT YOURSELF SOLUTIONS 1.

d.f = n -1 = 14 -1 = 13 and a = 0.01 , One tail (left-tailed) t0 = - 2.650

2. d.f = n-1 = 9 -1 = 8 and a = 0.10 , One tail (right-tailed) t 0 = 1.397

3. d.f = n -1 = 16 -1 = 15 and a = 0.05 , Two tail -t0 = - 2.131, t 0 = 2.131

4. The claim is “the mean age of a used car sold in the last 12 months is less than 4.1 years.” H 0 : m ³ 4.1; H a : m < 4.1 (claim)

a = 0.10 and d.f. = n -1 = 25 -1 = 24 t0 = - 1.318; Rejection region: t <-1.318 x - m 3.7 - 4.1 t= = » -1.54 s 1.3 n 25 Because t <-1.318 , reject H 0 . There is enough evidence at the 10% level of significance to support the claim that the mean age of a used car sold in the last 12 months is less than 4.1 years. Copyright © 2019 Pearson Education Ltd.

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5. The claim is “the mean conductivity of the river is 1890 milligrams per liter.” H 0 : m = 1890 (claim); H a : m ¹ 1890

a = 0.01 and d.f. = n -1 = 38 -t 0 = - 2.712, t 0 = 2.712; Rejection regions: t <-2.712, t > 2.712 x - m 2350 -1890 t= = » 3.192 s 900 n 39 Because t > 2.712 , reject H 0 . 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 : m £ 18 minutes (claim); H a : m > 18 minutes

P - value = 0.9997 P - value = 0.9997 > 0.05 = a 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. Identify the level of significance a 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. If the hypothesis test is: (1) left-tailed, use the “One Tail, a ” column with a negative sign.

(2) right-tailed, use the “One Tail, a ” column with a positive sign. (3) two-tailed, use the “Two Tail, a ,” 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

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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) Fail to reject H 0 because t >-2.086 . (d) Reject H 0 because t <-2.086 . 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 : m = 15 (claim); H a : m ¹ 15 a = 0.01 and d.f. = n -1 = 35 t 0 =  2.724

x - m 13.9 -15 = » -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. t=

14. H 0 : m £ 25; H a : m > 25 (claim) a = 0.05 and d.f. = n -1 = 16 t 0 = 1.746

x - m 26.2 - 25 = » 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. t=

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15. H 0 : m ³ 8000 (claim); H a : m < 8000 a = 0.01 and d.f. = n -1 = 24 t0 = - 2.492

x - m 7700 - 8000 = » -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. t=

16. H 0 : m £ 1600 (claim); H a : m > 1600 a = 0.02 and d.f. = n -1 = 45 t 0 = 2.115

x - m 1550 -1600 = » -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. t=

17. H 0 : m ³ 4915; H a : m < 4915 (claim) a = 0.02 and d.f. = n -1 = 50 t0 = - 2.109

x - m 5017 - 4915 = » 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. t=

18. H 0 : m = 52, 200; H a : m ¹ 52, 200 (claim) a = 0.05 and d.f. = n -1 = 33 t 0 =  2.035

x - m 53,220 - 52,200 = » 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. t=

19. (a) The claim is “the mean rental of a warehouse (with basic amenities) is $2500.” H 0 : m = 2500 (claim); H a : m ¹ 2500

(b) -t0 = - 2.715, t0 = 2.715; Rejection regions: t <-2.715, t > 2.715 (c) t =

x - m 2850 - 2500 = » 3.082 s 700 n 38

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(d) Because t > 2.715 , reject H 0 . (e) There is enough evidence at the 1% level of significance to reject the claim that the mean rental of a warehouse (with basic amenities) is $2,500. 20. (a) The claim is “the wait time for customers to connect with customer care executives is at most 2.5 minutes.” H 0 : m £ 2.5 (claim); H a : m > 2.5

(b) t0 = 1.69; Rejection regions: t >1.69 (c) t =

x - m 2.75 - 2.5 = » 1.183 s 1.25 n 35

(d) Because t <1.69 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to support the claim that the mean wait time for customers to connect with customer care executives was at most 2.5 minutes. 21. (a) The claim is “the mean credit card debt by state is greater than $5500 per person.” H 0 : m £ 5500 ; H a : m > 5500 (claim)

(b) t0 = 1.699; Rejection region: t >1.699 (c) t =

x - m 5594 - 5500 = » 0.86 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 credit reporting agency’s claim that the mean credit card debt by state is greater than $5500 per person. 22. (a) The claim is “the mean number of runs of flash drives is at least 25000 hours.” H 0 : m ³ 25000 (claim); H a : m < 25000

(b) t0 = - 2.624; Rejection regions: t <-2.624 (c) t =

x - m 24500 - 25000 = » -2.582 s 750 n 15

(d) Because t >-2.602 , fail to reject H 0 . (e) There is enough evidence at the 1% level of significance to accept the claim that the mean number of runs of their flash drives is at least 25,000 runs. Copyright © 2019 Pearson Education Ltd.

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23. (a) The claim is “the mean amount of carbon monoxide in the air in U.S. cities is less than 2.34 parts per million.” H 0 : m ³ 2.34; H a : m < 2.34 (claim)

(b) t0 = - 1.295; Rejection region: t <-1.295 (c) t =

x - m 2.37 - 2.34 = » 0.11 s 2.11 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 U.S. cities is less than 2.34 parts per million. 24. (a) The claim is “the mean amount of lead in the air in U.S. cities is less than 0.036 microgram per cubic meter.” H 0 : m ³ 0.036; H a : m < 0.036 (claim)

(b) t0 = - 2.396; Rejection region: t <-2.396 (c) t =

x - m 0.039 - 0.036 = » 0.33 s 0.069 n 56

(d) Because t >-2.396 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to support the claim that the mean amount of lead in the air in U.S. cities is less than 0.036 microgram per cubic meter. 25. (a) The claim is “the mean annual salary for senior-level product engineers is $98,000.” H 0 : m = 98, 000 (claim); H a : m ¹ 98, 000

(b) -t0 = - 2.131, t0 = 2.131; Rejection region: t <-2.131, t > 2.131

s »12,671.6 (c) x » 92,068.9, x - m 92,068.9 - 98,000 t= » » -1.87 s 12,671.6 n

(d) Because -2.131 < t < 2.131 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to reject the employment information service’s claim that the mean annual salary for senior-level product engineers is $98,000.

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26. (a) The claim is “the mean annual salary for home care physical therapists is more than $80,000.” H 0 : m £ 80, 000; H a : m > 80, 000 (claim)

(b) t0 = 1.363; Rejection region: t > 1.363

s » 9370.1 (c) x » 81, 666.9, x - m 81,666.9 - 80,000 t= » » 0.62 s 9370.1 n 12 (d) Because t <1.363 , fail to reject H 0 . (e) There is not enough evidence at the 10% level of significance to support the employment information service’s claim that the mean annual salary for home care physical therapists is more than $80,000. 27. (a) The claim is “the mean minimum time it takes for a sedan to travel a quarter mile is greater than 14.7 seconds.” H 0 : m £ 14.7; H a : m > 14.7 (claim)

x - m 15.4 -14.7 = » 1.56 s 2.10 n 22 P-value = {Area to right of t = 1.56 } = 0.0664

(b) t =

(c) Because P < 0.10 = a , reject H 0 . (d) There is 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 14.7 seconds. 28. (a) The claim is “the mean dive duration of a North Atlantic right whale is 11.5 minutes.” H 0 : m = 11.5 (claim); H a : m ¹ 11.5

x - m 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 = a , 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 deliveries per delivery person is more than 28 packets per day.” H 0 : m ³ 28 (claim); H a : m < 28

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283

s » 3.632 (b) x » 27.389 x - m 27.389 - 28 t= = » -0.714 s 3.632 n 18 P-value = {Area to the left of t = -0.714 } = 0.2425 (c) Because P > 0.05 = a , fail to reject H 0 . (d) There is enough evidence at the 5% level of significance to accept the claim that the mean number of deliveries per delivery person is more than 28 packets per day. 30. (a) The claim is “the mean number of hours spent by delivery persons per day is 6.0.” H 0 : m = 6.0 (claim); H a : m ¹ 6.0

s » 0.501 (b) x = 5.775 x - m 5.775 - 6.0 t= » » -1.270 s 0.501 n 8 P-value = 2{Area to the left of t = -1.270 } = 2(0.12235) = 0.2447 (c) Because P > 0.01 = a , fail to reject H 0 . (d) There is not enough evidence at the 1% level of significance to reject the claim that the mean number of hours spent by delivery persons per day is 6.0. 31. Because s is unknown, n < 30, the sample is random, and the gas mileage is normally distributed, use the t-distribution. H 0 : m ³ 23 (claim); H a : m < 23

x - m 22 - 23 » » -0.559 s 4 n 5 P-value = {Area left of t = t = -0.559 } = 0.3030 Because P > 0.05 = a , fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the claim that the mean gas mileage for the luxury sedan is at least 23 miles per gallon. t=

32. Because s is known, the sample is random, and n ³ 30, use the standard normal distribution. H 0 : m £ 9000; H a : m > 9000 (claim)

x - m 9231 - 9000 » » 0.532 s 2380 n 30 P-value = {Area right of z = 0.532 } = 0.2974 Because P > 0.01 = a , 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 $9000. z=

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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, 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).

7.4 HYPOTHESIS TESTING FOR PROPORTIONS 7.4 TRY IT YOURSELF SOLUTIONS 1.

np = (150)(0.90) =135 > 5, nq = (150)(0.10) =15 > 5 The claim is “more than 90% of U.S. adults have access to a smartphone.” H 0 : p £ 0.90; H a : p > 0.90 (claim)

a = 0.01 z0 = 2.33; Rejection region: z > 2.33 p - p 0.87 - 0.90 z= = » -1.22 pq (0.90)(0.10) n 150

Fail to reject H 0 . There is not enough evidence at the 1% level of significance to support the claim that more than 90% of U.S. adults have access to a smartphone. 2.

np = (1768)(0.67) »1185 > 5, nq = (1768)(0.33) » 583 > 5 The claim is “67% of U.S. adults believe that doctors prescribing antibiotics for viral infections for which antibiotics are not effective is a significant cause of drug-resistant superbugs.” H 0 : p = 0.67 (claim); H a : p ¹ 0.67

a = 0.10 - z 0 = - 1.645, z 0 = 1.645; Rejection region: z <-1.645, z >1.645 æ1150 ÷ö ç p - p ççè1768 ÷÷ø - 0.67 = » -1.75 z= pq (0.67) (0.33) n 1768

Reject H 0 . There is enough evidence at the 10% level of significance to reject the claim that 67% of U.S. adults believe that doctors prescribing antibiotics for viral infections for which antibiotics are not effective is a significant cause of drug-resistant superbugs.

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285

7.4 EXERCISE SOLUTIONS 1. If np ³ 5 and nq ³ 5, the normal distribution can be used. 2. Verify that np ³ 5 and nq ³ 5. State H 0 and H a . Specify the level of significance a . 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. 3.

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 ³ 0.48 (claim); H a : p < 0.48 z 0 = - 1.405; Rejection region: z <-1.405 p - 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 ¹ 0.15 (claim) - z 0 = - 1.96, z 0 = 1.96; Rejection region: z <-1.96, z >1.96 p - p 0.12 - 0.15 z= = » -1.88 pq (0.15)(0.85) n 500 Fail to reject H 0 . There is not enough evidence at the 5% level of significance to support the claim.

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) z 0 = 1.75; Rejection region: z > 1.75 p - 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.

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7. (a) The claim is “less than 80% of U.S. adults think that healthy children should be required to be vaccinated.” H 0 : p ³ 0.80; H a : p < 0.80 (claim)

(b) z 0 = - 1.645; Rejection region: z <-1.645 (c) z =

p - p

0.82 - 0.80

» 0.707 pq (0.80)(0.20) n 200 (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 medical researcher’s claim that less than 80% of U.S. adults think that healthy children should be required to be vaccinated. 8. (a) The claim is “at least 27% of U.S. adults think that the IRS will audit their taxes.” H 0 : p ³ 0.27 (claim); H a : p < 0.27

(b) z0 = - 2.33; Rejection region: z <-2.33 (c) z =

p - p pq n

0.23 - 0.27

(0.27 )(0.73)

» -2.85

1000

(d) Because z <-2.33 , reject H 0 . (e) There is enough evidence at the 1% level of significance to reject the research center’s claim that at least 27% of U.S. adults think that the IRS will audit their taxes. 9. (a) The claim is “at most 3% of working college students are employed as teachers or teaching assistants.” H 0 : p £ 0.03 (claim); H a : p > 0.03

(b) z 0 = 2.33; Rejection region: z > 2.33 (c) z =

p - p pq n

0.04 - 0.03

(0.03)(0.97)

» 0.83

200

(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 education researcher’s claim that at most 3% of working college students are employed as teachers or teaching assistants.

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287

10. (a) The claim is “57% of college students work year-round.” H 0 : p = 0.57 (claim); H a : p ¹ 0.57

(b) - z 0 = - 1.645, z 0 = 1.645; Rejection region: z <-1.645, z >1.645 æ 171 ö÷ ÷ - 0.57 ççèç p - p 300 ÷ø (c) z = = =0 pq (0.57 )(0.43) n 300 (d) Because -1.645 < z <1.645 , fail to reject H 0 .

(e) There is not enough evidence at the 10% level of significance to reject the education researcher’s claim that 57% of college students work year-round. 11. (a) The claim is “85% of Americans think they are unlikely to contract the Zika virus.” H 0 : p = 0.85 (claim); H a : p ¹ 0.85

(b) - z 0 = - 1.96, z 0 = 1.96; Rejection region: z <-1.96, z >1.96 æ 225 ÷ö ÷ - 0.85 ççèç p - p 250 ÷ø (c) z = = » 2.21 pq (0.85)(0.15) n 250

(d) Because z > 1.96 , reject H 0 . (e) There is enough evidence at the 5% level of significance to reject the medical researcher’s claim that 85% of Americans think they are unlikely to contract the Zika virus. 12. (a) The claim is “more than 29% of U.S. employees have changed jobs in the past three years.” H 0 : p £ 0.29; H a : p > 0.29 (claim)

(b) z 0 = 1.28; Rejection region: z > 1.28 æ 63 ö÷ çç - 0.29 p - p çè180 ÷ø÷ (c) z = = » 1.77 pq (0.29)(0.71) n 180

(d) Because z > 1.28 , reject H 0 . (e) There is enough evidence at the 10% level of significance to support the research center’s claim that more than 29% of U.S. employees have changed jobs in the past three years.

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CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

13. (a) The claim is “27% of U.S. adults would travel into space on a commercial flight if they could afford it.” H 0 : p = 0.27 (claim); H a : p ¹ 0.27

(b) z =

p - p

0.30 - 0.27

» 2.14 pq (0.27)(0.73) n 1000 P-value = 2{Area to right of z = 2.14 } = 2(0.0162) » 0.03

(c) Because P < 0.05 = a , 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 18% of U.S. adults’ online food purchases are for snacks.” H 0 : p £ 0.18 (claim); H a : p > 0.18

(b) z =

p - p

0.20 - 0.18

» 2.33 pq (0.18)(0.82) n 1995 P-value = Area to right of z = 2.33 » 0.01

(c) Because P < 0.10 = a , reject H 0 . (d) There is enough evidence at the 10% level of significance to reject the research center’s claim that at most 18% of U.S. adults’ online food purchases are for snacks. 15. (a) The claim is “less than 67% of U.S. households own a pet.” H 0 : p ³ 0.67; H a : p < 0.67 (claim)

æ 390 ö÷ çç - 0.67 p - p çè 600 ÷ø÷ (b) z = = » -1.04 pq (0.67 )(0.33) n 600 P-value = Area to left of z = -1.04 » 0.15

(c) Because P > 0.10 = a , fail to reject H 0 . (d) There is not enough evidence at the 10% level of significance to support the humane society’s claim that less than 67% of U.S. households own a pet.

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289

16. (a) The claim is “5% of U.S. households have taken in a stray dog.” H 0 : p = 0.05 (claim); H a : p ¹ 0.05

æ 12 ö÷ çç - 0.05 p - p çè 200 ÷÷ø (b) z = = » 0.65 pq (0.05)(0.95) n 200 P-value = 2{Area to right of z = 0.65 } = 2(0.2578) » 0.52

(c) Because P > 0.05 = a , fail to reject H 0 . (d) There is not enough evidence at the 5% level of significance to reject humane society’s claim that 5% of U.S. households have taken in a stray dog. 17. H 0 : p ³ 0.63 (claim); H a : p < 0.63 z0 = - 1.645 : Rejection region: z <-1.645 p - p 0.59 - 0.63 z= = » -0.83 pq (0.63)(0.37) 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 63% 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 “less than 80% of U.S. adults think that healthy children should be required to be vaccinated.” H 0 : p ³ 0.80; H a : p < 0.80 (claim)

(b) z 0 = - 1.645; Rejection region: z <-1.645 (c) z =

164 - 200(0.80) x - np = » 0.707 200(0.80)(0.20) npq

(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 medical researcher’s claim that less than 80% of U.S. adults think that healthy children should be required to be vaccinated. The results are the same.

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CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

20. Answers will vary. Sample answer: p - p z= Write the original equation pq n æ x ö÷ çç ÷ - p çè n ø÷ x = Substitute for pˆ n pq n x - np n = Multiply by n pq n n x - np = Use the fact that n = n 2 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, a = 0.01 c02 = 33.409

d.f. = 29, a = 0.05 c02 = 17.708

d.f. = 50, a = 0.01 cR2 = 79.490 cL2 = 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 : s 2 £ 0.40 (claim); H a : s 2 > 0.40 a = 0.01 and d.f. = n -1 = 30 c02 = 50.892; Rejection region: c2 > 50.892 c2 =

(n - 1) s 2 s

(30)(0.75) 0.40

= 56.250

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291

Because c > 50.892 , reject H 0 . There is enough evidence at the 1% level of significance to reject the 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 : s ³ 3.7; H a : s < 3.7 (claim)

a = 0.05 and d.f. = n -1 = 8 c02 = 2.733; Rejection region: c2 < 2.733 2

c =

(n -1) s 2 s2

(8)(3.0) = » 5.259 2 (3.7)

Because c > 2.733 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to support the 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.”

H0 : s 2 = 25.5 (claim); H a : s 2 ¹ 25.5 a = 0.10 and d.f. = n -1 = 12 cL2 = 5.226 and cR2 = 21.026; Rejection region: c2 > 21.026, c2 < 5.226 c2 =

(n - 1) s 2

(12)(10.8)

» 5.082 25.5 2 Because c < 5.226 , reject H 0 . There is enough evidence at the 10% level of significance to reject the claim that the variance of the weight losses of users is 25.5. s

7.5 EXERCISE SOLUTIONS 1. Specify the level of significance a . Determine the degrees of freedom. Determine the critical values 2 using the c -distribution . For a right-tailed test, use the value that corresponds to d.f. and a . For a left-tailed test, use the value that corresponds to d.f. and 1 - a . For a two-tailed test, use the value 1 1 that corresponds to d.f. and a , and d.f. and 1- a . 2 2 2

2. No; In a c -distribution , all c -values are greater than or equal to 0 because anything squared is greater than or equal to 0. 3. 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. 4. 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

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CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

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. 2

5. Critical value: c0 = 38.885 ; Rejection region: c > 38.885 6. Critical value: c0 = 14.684 ; Rejection region: c >14.684 7. Critical value: c0 = 0.872 ; Rejection region: c < 0.872 8. Critical value: c0 = 13.091 ; Rejection region: c <13.091 2

9. Critical values: cL = 60.391, cR = 101.879 ; Rejection regions: c < 60.391, c >101.879 10. Critical values: cL = 35.534, cR = 91.952 ; Rejection regions: c < 35.534, c > 91.952 2

11. Critical value: c0 = 49.588 ; Rejection region: c > 49.588 2

12. Critical values: cL = 16.791, cR = 46.979 ; Rejection regions: c <16.791, c > 46.979 2

13. (a) Fail to reject H 0 because c < 6.251 . 2

(b) Fail to reject H 0 because c < 6.251 . 2

(d) Reject H 0 because c > 6.251 . 2

14. (a) Fail to reject H 0 because 8.547 < c < 22.307 . 2

(b) Reject H 0 because c > 22.307 . 2

(d) Fail to reject H 0 because 8.547 < c < 22.307 . 2

15. H0 : s = 0.52 (claim); H a : s ¹ 0.52

cL2 = 7.564, cR2 = 30.191; Rejection regions: c2 < 7.564, c2 > 30.191 c2 =

(n - 1) s 2 s

(17 )(0.508) » 16.608 (0.52) 2

Because 7.564 < c < 30.191 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the claim.

CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 2

293

16. H0 : s ³ 8.5 (claim); H a : s < 8.5

c02 = 12.338; Rejection region: c2 <12.338 c2 =

(n - 1) s 2

(22)(7.45)

» 19.28 8.5 2 Because c >12.338 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to reject the claim. s

17. H0 : s £17.6 (claim); H a : s >17.6

c02 = 63.691; Rejection region: c2 > 63.691 c2 =

(n - 1) s 2

(40)(28.33)

» 64.39 17.6 2 Because c > 63.691 , reject H 0 . There is enough evidence at the 1% level of significance to reject the claim. s

18. H 0 : s 2 £ 19; H a : s 2 > 19 (claim)

c02 = 23.542; Rejection region: c2 > 23.542 c2 =

(n -1) s 2 s

(16)(28) » 23.58 (19)

Because c > 23.542 , reject H 0 . There is enough evidence at the 10% level of significance to support the claim. 2

19. H 0 : s = 32.8; H a : s ¹ 32.8 (claim)

cL2 = 77.929, cR2 = 124.342; Rejection regions: c2 < 77.929, c2 >124.342 2

c =

(n -1) s 2 s

(100)(40.9) » 124.7 (32.8)

Because c >124.342 , Reject H 0 . There is enough evidence at the 10% level of significance to support the claim. 2

20. H0 : s = 63 (claim); Ha : s ¹ 63

cL2 = 12.461, cR2 = 50.993; Rejection regions: c2 <12.461, c2 > 50.993 c2 =

(n -1) s 2 s

(28)(58) » 25.78 (63) 2

Because 12.461 < c < 50.993 , Fail to reject H 0 . There is not enough evidence at the 1% level of significance to reject the claim. 21. H 0 : s ³ 40; H a : s < 40 (claim)

c02 = 3.053; Rejection region: c2 < 3.053 2

c =

(n -1) s 2 s2

(11)(40.8) = » 11.444 2 (40)

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CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE 2

Because c > 3.053 , fail to reject H 0 . There is not enough evidence at the 1% level of significance to support the claim. 22. H 0 : s = 24.9 (claim); H a : s ¹ 24.9

cL2 = 34.764, cR2 = 67.505; Rejection regions: c2 < 34.764, c2 > 67.505 c2 =

(n -1) s 2 s2

(50)(29.1) » 68.29 2 (24.9)

Because c > 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 thickness in a certain helmet model is 7.5.”

H0 : s 2 = 7.5 (claim); H a : s 2 ¹ 7.5 2

(b) cL = 6.571, cR = 23.685; Rejection regions: c < 6.571, c > 23.685 (c) c 2 =

(n - 1) s 2 s

(14)(2.7 ) 7.5

= 5.04

(d) Because 6.571> c , reject H 0 . (e) There is enough evidence at the 10% level of significance to reject the claim that the variance of the thickness in a certain helmet model is 7.5. 24. (a) The claim is “the variance of the gas mileages in a model of hybrid vehicle is 0.16.”

H0 : s 2 = 0.16 (claim); H a : s 2 ¹ 0.16 2

(b) cL =16.047, cR = 45.722; Rejection regions: c <16.047, c > 45.722 (c) c 2 =

(n - 1) s 2 s

(29)(0.26) 0.16

= 47.125

(d) Because c > 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 35 points.” H 0 : s ³ 35; H a : s < 35 (claim) 2

(b) c0 = 18.114; Rejection region: c <18.114

CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

(n -1) s 2 s2

295

(27)(34) = » 25.48 2 (35)

(d) Because c >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 35 points. 26. (a) The claim is “the standard deviation for grade 12 students on a vocabulary assessment test is greater than 45 points.” H 0 : s £ 45; H a : s > 45 (claim) 2

(b) c0 = 42.98; Rejection region: c > 42.98 (c) c2 =

(n -1) s 2 s2

(24)(46) » 25.08 2 (45)

(d) Because c < 42.98 , 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 vocabulary assessment test is greater than 45 points. 27. (a) The claim is “the standard deviation of waiting times experienced by patients is no more than 0.5 minute.” H 0 : s £ 0.5 (claim); H a : s > 0.5 2

(b) c0 = 33.196; Rejection region: c > 33.196 (c) c2 =

(n -1) s 2 s2

(24)(0.7) » 47.04 2 (0.5)

(d) Because c > 33.196 , reject H 0 . (e) There is enough evidence at the 10% level of significance to reject the 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 : s ³ 68 (claim); H a : s < 68 2

(b) c02 = 6.408; Rejection regions: c < 6.408

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CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

(n -1) s 2

(17)(40) = » 5.88 2 (68)

s2 2

(d) Because c < 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 is different from $10,300.” H 0 : s = 10,300; H a : s ¹ 10,300 (claim) 2

(b) cL = 5.629, cR = 26.119 Rejection regions: c < 5.629, c > 26.119 (c) s » 14,530.3 2

c =

(n -1) s 2 s2

(14)(14,530.3) = » 27.86 2 (10,300)

(d) Because c > 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 $10,300. 30. (a) The claim is “the standard deviation of the annual salaries of nursing supervisors is $16,500.” H 0 : s = 16,500 (claim); H a : s ¹ 16,500 2

(b) cL = 4.575, cR = 19.675 ; Rejection regions: c < 4.575, c >19.675 (c) s » 14,516.6 2

c =

(n -1) s 2 s2

(11)(14,516.6) = » 8.51 2 (16,500) 2

(d) Because 4.575 <c <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 $16,500. 2

31. c = 25.48 , d.f. = n -1 = 27 2

P-value = {Area left of c = 25.48 } = 0.4524 Fail to reject H 0 because P-value = 0.4524 > 0.10 .

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297

32. c = 25.08 , d.f. = n -1 = 24 2

P-value = {Area right of c = 25.08 } = 0.4014 Fail to reject H 0 because P-value = 0.4014 > 0.01 . 2

33. c = 67.5 2

P-value = {Area right of c = 67.5} = 0.0001 Reject H 0 because P-value = 0.0001 < 0.10 . 2

34. c = 5.88 , d.f. = n -1 = 17 2

P-value = {Area left of c = 5.88 } = 0.0060 Reject H 0 because P-value = 0.0060 < 0.01 .

CHAPTER 7 REVIEW EXERCISE SOLUTIONS 1.

H 0 : m £ 100 (claim); H a : m > 100

2. H 0 : m = 48; H a : m ¹ 48 (claim)

H 0 : p ³ 0.205; H a : p < 0.205 (claim)

4. H 0 : m = 16 (claim); H a : m ¹ 16

H 0 : s £ 2.5; H a : s > 2.5 (claim)

6. H 0 : p ³ 0.4 (claim); H a : p < 0.4

7. (a) H0: p = 0.65 (claim); Ha: p  0.65

(b) A type I error will occur when the actual proportion of U.S. adults who have volunteered their time or donated money to help clean up the environment is 65%, but you reject H0: p = 0.65. A type II error will occur when the actual proportion is not 65%, but you fail to reject H0: p = 0.65. (c) Two-tailed because the alternative hypothesis contains . (d) There is enough evidence to reject the polling organization’s claim that the proportion of U.S. adults who have volunteered their time or donated money to help clean up the environment is 65%. (e) There is not enough evidence to reject the polling organization’s claim that the proportion of U.S. adults who have volunteered their time or donated money to help clean up the environment is 65%. 8. (a) H 0 : m ³ 10, 000 (claim); H a : m < 10, 000

(b) A type I error will occur if the actual mileage on an average is at least 10,000 miles, but you reject H 0 : m ³ 10, 000 . A type II error will occur if the actual mileage on an average is less than 10,000 miles, but you fail to reject H 0 : m ³ 10, 000 .

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(c) Left-tailed because the alternative hypothesis contains <. (d) There is enough evidence to reject the car rental company’s claim that its cars are driven for at least 10,000 miles, on an average. (e) There is not enough evidence to reject the car rental company’s claim that its cars are driven for at least 10,000 miles, on an average. 9. (a) H0: σ ≤ 9.5 (claim); Ha: σ > 9.5

(b) A type I error will occur when the actual standard deviation of the fuel economies is no more than 9.5 miles per gallon, but you reject H0: ≤ 9.5. A type II error will occur when the actual standard deviation of the fuel economies is more than 9.5 miles per gallon, but you fail to reject H0: σ ≤ 9.5. (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 fuel economies of its top-rated vehicles for a recent year is no more than 9.5 miles per gallon. (e) There is not enough evidence to reject the nonprofit consumer organization’s claim that the standard deviation of the fuel economies of its top-rated vehicles for a recent year is no more than 9.5 miles per gallon. 10. (a) H 0 : m ³ 25; H a : m < 25 (claim)

(b) A type I error will occur if the actual mean number of milligrams of sodium in one serving is at least 25, but you reject H 0 : m ³ 25 . A type II error will occur if the actual mean number of milligrams of sodium in one serving is less than 25, but you fail to reject H 0 : m ³ 25 . (c) Left-tailed because the alternative hypothesis contains <. (d) There is enough evidence to support the company’s claim that the mean number of milligrams of sodium in one serving is less than 25. (e) There is not enough evidence to support the company’s claim that the mean number of milligrams of sodium in one serving is less than 25. 11. 0.1190; Fail to reject H 0 . 12. P-value = 2{Area to right of z = 2.57 } = 2(0.0051) = 0.0102 Reject H 0 .

CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

13. Critical value: z 0 » - 2.05 ; Rejection region: z <-2.05

14. Critical values: - z0 = - 2.81, z 0 = 2.81 ; Rejection regions: z < -2.81, z > 2.81

15. Critical value: z0 = 1.96 ; Rejection region: z > 1.96

16. Critical values: - z 0 = - 2.17, z 0 = 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 : m = 3725 (claim); H a : m ¹ 3725 - z 0 = - 1.645, z 0 = 1.645; Rejection regions: z < -1.645, z > 1.645 x - m 3748 - 3725 z= = » 1.041 s 121 n 30 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. 22. H 0 : m ³ 8.25; H a : m < 8.25 (claim) z 0 = - 2.33; Rejection region: z <-2.33 x - m 8.246 - 8.25 z= = » -1.488 s 0.017 n 40

299

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CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

Because z <-2.33 , fail to reject H 0 . There is not enough evidence at the 1% level of significance to support the claim. 23. H 0 : m = 21.25; H a : m ¹ 21.25 (claim) - z0 = - 2.17, z 0 = 2.17; Rejection regions: z <-2.17, z > 2.17 x - m 19.7 - 21.25 z= = » -2.76 s 4.35

60 n Because z <-2.17 , fail to reject H 0 . There is enough evidence at the 3% level of significance to support the claim. 24. H 0 : m = 930 (claim); H a : m ¹ 930 - z 0 = - 1.645, z 0 = 1.645; Rejection regions: z <-1.645, z >1.645 x - m 937 - 930 = » 1.278 z= s 30 n 30 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 annual production of cotton is 3.5 million bales per country.” H 0 : m = 3.5 (claim); H a : m ¹ 3.5

(b) z =

x - m 2.1 - 3.5 = » -2.06 s 4.5 n 44

(c) P-value = 2{Area to left of z = -2.06 } = 2(0.0197) = 0.0394 (d) Because P-value < 0.05 = a , reject H0. (e) There is enough evidence at the 5% level of significance to reject the researcher’s claim that the mean annual production of cotton is 3.5 million bales per country. 26. (a) The claim is “the mean annual consumption of cotton is greater than 1.1 million bales per country.” H 0 : m £ 1.1; H a : m > 1.1 (claim)

(b) z =

x - m 1.0 -1.1 = » -0.19 s 4.3 n 67

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301

(e) There is not enough evidence at the 1% level of significance to support the researcher’s claim that the mean annual consumption of cotton is greater than 1.1 million bales per country. 27. (a) The claim is “the mean amount of sulfur dioxide in the air in U.S. cities is 1.15 parts per billion.” H 0 : m = 1.15 (claim); H a : m ¹ 1.15

(b) - z 0 = - 2.575, z 0 = 2.575; Rejection regions: z <-2.575, z > 2.575 (c) z =

x - m 0.93 -1.15 = » -0.97 s 2.62 134 n

(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 environmental researcher’s claim that the mean amount of sulfur dioxide in the air in U.S. cities is 1.15 parts per billion. 28. (a) The claim is “the mean price of a round-trip flight from New York City to Los Angeles is less than $507.” H 0 : m ³ 507; H a : m < 507 (claim)

(b) - z 0 = - 1.645; Rejection region: z <-1.645 (c) z =

x - m 502 - 507 = » -0.33 s 111 n 55

(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 travel analyst’s claim that the mean price of a round-trip flight from New York City to Los Angeles is less than $507. 29. Critical values: -t0 = - 2.11, t0 = 2.11 ; Rejection region: t <-2.11, t > 2.11 30. Critical value: t0 = - 2.024 ; Rejection region: t <-2.024 31. Critical value: t0 = 1.796 ; Rejection region: t >1.796 32. Critical value: -t0 = - 2.861, t0 = 2.861; ; Rejection region: -t < -2.861, t > 2.861 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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CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

35. H 0 : m £ 12, 700; H a : m > 12, 700 (claim) t0 = 2.845; Rejection region: t > 2.845 x - m 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 : m ³ 0 (claim); H a : m < 0 t0 = - 1.310; Rejection region: t <-1.310 x - m -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 : m £ 51 (claim); H a : m > 51 t0 = 2.426; Rejection region: t > 2.426 x - m 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 : m ³ 850; H a : m < 850 (claim) t0 = - 2.160; Rejection region: t <-2.160 x - m 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 : m = 195 (claim); H a : m ¹ 195 -t 0 = - 1.660; t 0 = 1.660 Rejection regions: t < - 1.660; t > 1.660 x - m 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. 40. H 0 : m = 3,330, 000; H a : m ¹ 3,330, 000 (claim) -t 0 = - 2.032; t 0 = 2.032 Rejection regions: t < -2.032; t > 2.032

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303

x - m 3, 293,995 - 3,330,000 = » -16.64 s 12,801 n 35 Because t <-2.032 , reject H 0 . There is enough evidence at the 5% level of significance to support the claim. t=

41. (a) The claim is “the mean monthly cost of joining a health club is $25.” H 0 : m = 25 (claim); H a : m ¹ 25

(b) -t0 = - 1.740, t0 = 1.740; Rejection regions: t <-1.740, t >1.740 (c) t =

x - m 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 : m £ 14 (claim); H a : m > 14

(b) t0 = 2.040; Rejection region: t > 2.040 (c) t =

x - m 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 : m £ 145; H a : m > 145 (claim)

s » 43.93 (b) x » 155.39 x - m 155.39 -145 t= = » 1.419 s 43.93 n 36 P-value = {Area to right of t = 1.419 } » 0.0824 (c) Because P-value < 0.10 = a , reject H 0 .

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CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

(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 494.” H 0 : m = 494 (claim); H a : m ¹ 494

s » 53.295 (b) x » 468.030 x - m 468.030 - 494 t= = » -2.799 s 53.295 n 33 P-value = 2{Area to left of t = -2.799 } » 2(0.0043) = 0.0086 (c) Because P-value < 0.05 = a , reject H 0 . (d) There is enough evidence at the 5% level of significance to reject the education researcher’s claim that the overall average score of 15-year-old students on an international mathematics literacy test is 494. 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 ¹ 0.15 - z 0 = - 1.96, z 0 = 1.96; Rejection regions: z <-1.96, z >1.96 p - 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 ¹ 0.65 - z 0 = - 2.17, z 0 = 2.17; Rejection regions: z <-2.17, z > 2.17 p - 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. 47. np = (68)(0.70) = 47.6 > 5 nq = (68)(0.30) = 20.4 > 5  can use normal distribution H 0 : p ³ 0.70; H a : p < 0.70 (claim) z0 = - 2.33; Rejection region: z <-2.33

CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

p - p

305

0.50 - 0.70

» -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 40% of U.S. adults say they are less likely to travel to Europe in the next six months for fear of terrorist attacks.” H 0 : p £ 0.40; H a : p > 0.40 (claim)

(b) z 0 = 2.33; Rejection region: z > 2.33 (c) z =

p - p pq n

0.42 - 0.40

(0.40)(0.60)

» 1.29

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 polling agency’s claim that over 40% of U.S. adults say they are less likely to travel to Europe in the next six months for fear of terrorist attacks. 50. (a) The claim is “6% of U.S. employees say it is likely they will be laid off in the next year.” H 0 : p = 0.06 (claim); H a : p ¹ 0.06

(b) - z 0 = - 1.96, z 0 = 1.96; Rejection regions: z <-1.96, z >1.96

x 44 » 0.08044 (c) p = = n 547 p - p 0.08044 - 0.06 z= = » 2.013 pq (0.06)(0.94) n 547 (d) Because z > 1.96 , reject H 0 . (e) There is enough evidence at the 5% level of significance to reject the labor researcher’s claim that 6% of U.S. employees say it is likely they will be laid off in the next year. 2

51. Critical value: c0 = 30.144 ; Rejection region: c > 30.144 2

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CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

53. Critical values: cL = 26.509, cR = 55.758 ; Rejection regions: c < 26.509, c > 55.758 2

54. Critical value: c0 =1.145 ; Rejection region: c <1.145 2

55. H0 : s £ 2; H a : s > 2 (claim)

c02 = 24.769; Rejection region: c2 > 24.769 c2 =

(n -1) s 2 s

(17)(2.95) = 25.075 ( 2)

Because c > 24.769 , reject H 0 . There is enough evidence at the 10% level of significance to support the claim. 2

56. H0 : s £ 60 (claim); Ha : s > 60

c02 = 26.119; Rejection region: c2 > 26.119 c2 =

(n -1) s 2 s

(14)(72.7) » 16.963 (60)

Because c < 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 : s = 1.25 (claim); H a : s ¹ 1.25

cL2 = 0.831, cR2 = 12.833; Rejection regions: c2 < 0.831, c2 >12.833 2

c =

(n -1) s 2 s2

(5)(1.03) = » 3.395 2 (1.25) 2

Because 0.831 < c <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 : s = 0.035; H a : s ¹ 0.035 (claim)

cL2 = 4.601, cR2 = 32.801; Rejection regions: c2 < 4.601, c2 > 32.801 2

c =

(n -1) s 2 s2

(15)(0.026) = » 8.278 2 (0.035) 2

Because 4.601 < c < 32.801, fail to reject H 0 . There is not enough evidence at the 1% level of significance to support the claim. 59. (a) The claim is “the variance of the bolt widths is at most 0.01.”

H0 : s 2 £ 0.01 (claim); H a : s 2 > 0.01 2

(b) c0 = 49.645; Rejection region: c > 49.645 (c) c 2 =

(n -1) s 2 s

(27)(0.064) = 172.8 (0.01)

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307

(d) Because c > 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.” H 0 : s = 3 (claim); H a : s ¹ 3 2

(b) cL = 11.160, cR = 48.290; Rejection regions: c <11.160, c > 48.290 2

(n -1) s 2 s2

(26)(3.9) = = 43.94 2 (3) 2

(d) Because 11.160 < c < 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. 2

61. c0 = 46.963; Rejection region: c > 46.963 2

From Exercise 59, c =172.8 . 2

You can reject H 0 at the 1% level of significance because c = 172.8 > 46.963 . 2

62. cL = 13.844, cR = 41.923; Rejection regions: c <13.844, c > 41.923 2

From Exercise 60, c = 43.94 . 2

You can reject H 0 at the 5% level of significance because c = 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.” H 0 : m ³ 7.25 (claim); H a : m < 7.25

(b) Left-tailed because the alternative hypothesis contains <; z-test because s is known and the population is normally distributed. (c) Sample answer: z0 = - 2.33; Rejection region: z <-2.33 x - m 7.15 - 7.25 z= = » -1.283 s 0.27 n 12 (d) Because z >-2.33 , fail to reject H 0 .

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CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

(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, Washington, is more than $36.” H0: µ ≤ 36 Ha: µ > 36 (claim)

(b) Right-tailed because the alternative hypothesis contains >; z-test because σ is known and n ≥ 30. (c) Sample answer: z 0 = 1.28; Rejection region: z > 1.28 x - m 42 - 36 = » 1.997 z= s 19 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, Washington, is more than $36. 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 $47,254.” H0: µ = 47,254 (claim); Ha: µ  47,254

(b) Two-tailed because the alternative hypothesis contains ; t-test because σ 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 - m 50,781 - 47, 254 = » 2.58 s 5290 n 15 (d) Because t > 2.145 , reject H 0 . t=

(e) There is not enough evidence at the 5% level of significance to support the government agency’s claim that the mean amount of earnings for full-time workers ages 18 to 24 with a bachelor’s degree is a recent year is $47,254. 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 σ is unknown and n ≥ 30.

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309

(c) Sample answer: t0 = - 2.426; Rejection region: t <-2.426 x - m 9.233 -10.5 t= = » -3.09 s 2.593 n 40 (d) Because t <-2.426 , 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 18% of the vehicles a nonprofit consumer organization rated in a recent year have an overall score of 78 or more.” H0: p ≥ 0.18; Ha: p < 0.18 (claim)

(b) Left-tailed because the alternative hypothesis contains <; z-test because np ≥ 5 and nq ≥ 5. (c) Sample answer: z 0 = - 1.645; Rejection region: z <-1.645 p - p 0.20 - 0.18 z= = » 0.49 pq (0.18)(0.82) n 90 (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 nonprofit consumer organization’s claim that less than 18% of the vehicles a nonprofit consumer organization rated in a recent year have an overall score of 78 or more. 6. (a) The claim is “the standard deviation of vehicle rating scores is 11.90.” H0: σ = 11.90 (claim).18; Ha: σ  11.90.

(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. 2

c =

(n -1) s 2 s2

(89)(11.96) = » 89.90 2 (11.90) 2

(d) Because 68.249 < c <112.022 , fail to reject H 0 . (e) There is not enough evidence at the 10% level of significance to reject the nonprofit consumer organization’s claim that the standard deviation of vehicle rating scores is 11.90.

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CHAPTER 7 │ HYPOTHESIS TESTING WITH ONE SAMPLE

CHAPTER 7 TEST SOLUTIONS 1. (a) The claim is “more than 30% of adults have purchased a meal kit in a recent year.” H0: p  0.30; Ha: p  0.30 (claim)

(b) Right-tailed because the alternative hypothesis contains ; z-test because np  5 and nq  5. (c) Sample answer: z 0 = 1.28; Rejection region: z > 1.28 p - p 0.25 - 0.30 z= = » -0.65 pq (0.30)(0.70) n 36 (d) Because z < 1.28 , fail to reject H 0 . (e) There is not enough evidence at the 10% level of significance to support the retail grocery chain’s claim that more than 30% 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: μ  134 (claim); Ha: μ  134

(b) Two-tailed because the alternative hypothesis contains ; z-test because s is known and n  30. (c) Sample answer: - z 0 = - 1.645, z 0 = 1.645; Rejection regions: z < -1.645, z > 1.645 x - m 143 -134 z= = » 1.82 s 30 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: μ  100 (claim); Ha: μ  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 - m 110 -100 t= = » 3.02 s 19 n 33 (d) Because t > 2.449 , reject H 0 .

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311

(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: z 0 = 1.645; Rejection region: z > 1.645 p - 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.” H 0 : s = 1.11 (claim); H a : s ¹ 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. 2

c =

(n -1) s 2 s2

(25)(1.19) = » 28.733 2 (1.11) 2

(d) Because 13.120 < c < 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 price of the vehicles the nonprofit consumer organization rated in a recent year is at least $41,000.” H0:   41,000 (claim); Ha:   41,000

(b) Left-tailed because the alternative hypothesis contains ; t-test because s is unknown and n  30. (c) Sample answer: t0 = - 2.352; Rejection region: t <-2.352 x - m 40,600 - 41,000 t= = » -0.28 s 17,300 n 150 (d) Because t >-2.352 , fail to reject H 0 . Copyright © 2019 Pearson Education Ltd.

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(e) There is not enough evidence at the 1% level of significance to reject the nonprofit consumer organization’s claim that the mean price of the vehicles the organization rated in a recent year is at least $41,000. 7. (a) The claim is “the mean age of the residents of a small town is more than 38 years.” H0:   38 (claim); Ha:   41,000

(b) Right-tailed because the alternative hypothesis contains ; z-test because  is known and n  30. (c) Sample answer: z 0 = 1.28; Rejection region: z > 1.28 x - m 42.0667 - 38 = » 2.47 z= s 9 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.

CHAPTER

Hypothesis Testing with Two Samples

8.1 TESTING THE DIFFERENCE BETWEEN MEANS (LARGE INDEPENDENT SAMPLES) 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.” H 0 : 1  2 ; H a : 1  2 (claim)  = 0.10  z0  1.645, z0  1.645; Rejection regions: z  1.645, z  1.645 ( x  x )  ( 1  2 ) (60,680  59, 430)  (0) z 1 2   1.499  12  22 (6200) 2 (5575) 2   100 100 n1 n2

Since 1.645  z  1.645 , fail to reject H 0 . There is not enough evidence at the 10% level of significance to support the claim that there is a difference in the mean annual wages for forensic science technicians working for local and state governments. 3.

z

( x1  x2 )  ( 1  2 )



2 1





2 2



(310  306)  (0) (25) 2 (20) 2  15 20

 0.509

P-value = {area right of z  0.509 }  0.305 Because P-value  0.05   , fail to reject H 0 . There is not enough evidence at the 5% level of significance to support the travel agency’s claim that the average daily cost of meals and lodging for vacationing in Alaska is greater than the average daily cost for vacationing in Colorado.

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

8.1 EXERCISE SOLUTIONS 1. Two samples are dependent if 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 if the sample selected from one population is not related to the sample 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. H 0 : 1  2 (claim); H a : 1  2 Rejection region: z  2.575, z  2.575 ( x  x )  ( 1   2 ) (16  14)  (0)   2.89 z 1 2 2 2 1  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.

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

315

12. H 0 : 1  2 ; H a : 1  2 (claim) Rejection region: z  1.28 ( x  x )  ( 1  2 ) (500  495)  (0)   1.15 z 1 2  12  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. H 0 : 1  2 ; H a : 1  2 (claim) Rejection region: z  1.645 ( x  x )  ( 1  2 ) (2435  2432)  (0)   0.18 z 1 2  12  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. H 0 : 1  2 (claim); H a : 1  2 Rejection region: z  1.88 ( x  x )  ( 1  2 ) (5004  4895)  (0)   5.29 z 1 2  12  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 makes of automobiles.” H 0 : 1  2 ; H a : 1  2 (claim)

(b)  z0  1.645, z0  1.645; Rejection regions: z  1.645, z  1.645 (c) z 

( x1  x2 )  ( 1   2 )



2 1





2 2



(137  132)  (0) (5.5) 2 (6.7) 2  23 24

 2.80

(d) Because z  1.645 , reject H0 . (e) There is enough evidence at the 10% level of significance to support the safety engineer’s claim that the mean braking distances are different for the two makes of automobiles. 16. (a) The claim is “the mean customer rating of online retailers is greater than the mean customer rating of walk-in retailers.” H 0 : 1  2 ; H a : 1  2 (claim)

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

( x1  x2 )  ( 1   2 )

 12 n1



 22



(90  88)  (0) (3.4) 2 (3.5) 2  30 31

 2.26

(d) Because z  2.33 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to support the researcher’s claim that the mean customer rating of online retailers is greater than the mean customer rating of walk-in retailers. 17. (a) The claim is “the rainfall in region A is less than the rainfall in Region B.” H 0 : 1  2 ; H a : 1  2 (claim)

(b) z0  2.33; Rejection region: z  2.33 (c) z 

( x1  x2 )  ( 1  2 )



2 1





2 2



(700  725)  (0) (60) 2 (66) 2  60 60

 2.171

(d) Because, z  2.33 fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to conclude that the rainfall in Region A is more than the rainfall in Region B. 18. (a) The claim is “the mean running costs for Model A and Model B are equal.” H 0 : 1  2 (claim); H a : 1  2

(b)  z0  1.96, z0  1.96; Rejection regions: z  1.96, z  1.96 (c) z 

( x1  x2 )  ( 1   2 )

 12 n1



 22 n2



(5  6.5)  (0) (1.50) 2 (2.50) 2  24 26

 2.59

(d) Because, z  1.96 , reject H0 . (e) There is enough evidence at the 5% level of significance to reject the claim that the mean running costs for Model A and Model B are equal. 19. (a) The claim is “ACT mathematics and science scores are equal.” H 0 : 1  2 (claim); H a : 1  2

(b)  z0  2.575, z0  2.575; Rejection regions: z  2.575, z  2.575

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

( x1  x2 )  ( 1  2 )

12 n1



 22



(20.6  20.8)  (0) (5.4) 2 (5.6) 2  60 75

317

 0.21

(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 1 represents English and 2 represents reading. H 0 : 1  2 ; H a : 1  2 (claim)

(b) z0  1.28; Rejection region: z  1.28 (c) z 

( x1  x2 )  ( 1  2 )

 12 n1



 22



(20.1  21.3)  (0) (6.8) 2 (6.5) 2  120 150

 1.47

(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. The claim is “the mean home sales price in Casper, Wyoming, is the same as in Cheyenne, Wyoming.”

(a) H 0 : 1  2 (claim); H a : 1  2 (b)  z0  2.575, z0  2.575; Rejection regions: z  2.575, z  2.575 (c) z 

( x1  x2 )  ( 1  2 )



2 1





2 2



(294, 220  287,984)  (0) (135,387) 2 (151,996) 2  25 25

 0.15

(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 real estate agency’s claim that the mean home sales price in Casper, Wyoming, is the same as in Cheyenne, Wyoming. 22. The claim is “the mean home sales price in Casper, Wyoming, is the same as in Cheyenne, Wyoming.”

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

(b)  z0  2.575, z0  2.575; Rejection regions: z  2.575, z  2.575 (c) z 

( x1  x2 )  ( 1  2 )



2 1





2 2



(231,581  315,706)  (0) (135,387) 2 (151,996) 2  50 50

 2.92

(d) Because z  2.575 , reject H0 . (e) There is enough evidence at the 1% level of significance to reject the real estate agency’s 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 is Seattle, Washington, was greater than in Birmingham, Alabama.” H 0 : 1  2 ; H a : 1  2 (claim)

(b) z0  1.645; Rejection region: z  1.645 (c) x1  0.157, n1  30 x2  0.081, n2  30 ( x  x )  ( 1  2 ) (0.157  0.081)  (0)   1.02 z 1 2  12  22 (0.24)2 (0.33) 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 climatologist’s claim that the precipitation is Seattle, Washington, was greater than in Birmingham, Alabama. 24. (a) The claim is “the temperature in Seattle, Washington, was lower than in Birmingham, Alabama.” H 0 : 1  2 ; H a : 1  2 (claim)

(b) z0  2.33; Rejection region: z  2.33 (c) x1  61.87, n1  30 x2  75.83, n2  30 ( x  x )  ( 1  2 ) (61.87  75.83)  (0)   3.72 z 1 2  12  22 (13.6) 2 (15.4) 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. Copyright © 2019 Pearson Education Ltd.

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

319

25. They are equivalent through algebraic manipulation of the equation. 1  2  1  2  0 26. They are equivalent through algebraic manipulation of the inequality. 1  2  1  2  0 27. H 0 : 1  2  2000; H a : 1  2  2000 (claim) z0  1.645; Rejection region: z  1.645 ( x  x )  ( 1  2 ) (64, 270  62,610)  (2000)   0.14 z 1 2  12  22 (10,850) 2 (10,970) 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 Raleigh, North Carolina, and Wichita, Kansas, is more than $2000. 28. H 0 : 1  2  10,000 (claim); H a : 1  2  10,000  z0  2.575, z0  2.575; Rejection regions: z  2.575, z  2.575 ( x  x )  ( 1  2 ) (50, 410  54,640)  (10,000)   8.18 z 1 2  12  22 (6520) 2 (7130)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 Los Angeles, California, is equal to $10,000. 29.

( x1  x2 )  zc (64, 270  62, 610)  1.96

12 n1



 22 n2

 1  2  ( x1  x2 )  zc

12 n1



 22 n2

(10,850) 2 (10,970) 2 (10,850) 2 (10,970) 2   1  2  (64, 270  62, 610)  1.96  42 38 42 38

1660  1.96 5,969, 782.456  1  2  1660  1.96 5,969, 782.456 $3129  1  2  $6449

30.

( x1  x2 )  zc (50, 410  54, 640)  2.575

12 n1



 22 n2

 1  2  ( x1  x2 )  zc

12 n1



 22 n2

(6520) 2 (7130) 2 (6520) 2 (7130) 2   1  2  (50, 410  54, 640)  2.575  32 30 32 30

4230  2.575 3, 023, 013.333  1  2  4230  2.575 3, 023, 013.333 $8707  1  2  $247

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

8.2 TESTING THE DIFFERENCE BETWEEN MEANS (SMALL INDEPENDENT SAMPLES) 8.2 TRY IT YOURSELF SOLUTIONS 1. The claim is “there is a difference in the mean annual earnings based on level of education.” H 0 : 1  2 ; H a : 1  2 (claim)

  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 ( x  x )  ( 1  2 ) (36,875  44,900)  (0)   3.33 t 1 2 s12 s22 (5475) 2 (8580) 2   25 16 n1 n2

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 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.” H 0 : 1  2 ; H a : 1  2 (claim)

  0.10; d.f.  n1  n2  2  34  38  2  70 t0  1.294; Rejection regions: t  1.294 ( x1  x2 )  ( 1  2 ) (0.52  0.54)  (0)  t 2 2 (n1  1) s1  (n2  1) s2 1 1 (34  1)(0.08) 2  (38  1)(0.07) 2  34  38  2 n1  n2  2 n1 n2

1 1  34 38

 1.13

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.

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

3. (a) d.f.  n1  n2  2  23 t0  1.714, t0  1.714

(b) d.f.  min{n1  1, n2  1}  10 t0  1.812, t0  1.812 5. (a) d.f.  n1  n2  2  16 t0  1.746

(b) d.f.  min{n1  1, n2  1}  6 t0  1.943 7. (a) d.f.  n1  n2  2  19 t0  1.729

(b) d.f.  min{n1  1, n2  1}  7 t0  1.895 9.

321

4. (a) d.f.  n1  n2  2  25 t0  2.485

(b) d.f.  min{n1  1, n2  1}  11 t0  2.718 6. (a) d.f.  n1  n2  2  39 t0  2.708, t0  2.708

(b) d.f.  min{n1  1, n2  1}  18 t0  2.878, t0  2.878 8. (a) d.f.  n1  n2  2  60 t0  1.296

(b) d.f.  min{n1  1, n2  1}  29 t0  1.311

H 0 : 1  2 (claim); H a : 1  2 d.f.  n1  n2  2  27 Rejection regions: t  2.771, t  2.771 ( x1  x2 )  ( 1  2 ) (33.7  35.5)  (0)   1.70 t 2 2 (n1  1) s1  ( n2  1) s2 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. H 0 : 1  2 ; H a : 1  2 (claim) d.f.  n1  n2  2  18 Rejection region: t  1.330 ( x1  x2 )  ( 1  2 )  t (n1  1) s12  (n2  1) s22 1 1  n1  n2  2 n1 n2

(0.345  0.515)  (0) (11  1)(0.305) 2  (9  1)(0.215) 2 1 1  11  9  2 11 9

 1.41

Because t  1.330 , reject H0 . There is enough evidence at the 10% level of significance to support the claim. 11. H 0 : 1  2 (claim); H a : 1  2 d.f.  min{n1  1, n2  1}  9 Rejection region: t  1.833

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

t

( x1  x2 )  ( 1  2 ) s12 s22  n1 n2



(2410  2305)  (0) (175) 2 (52) 2  13 10

 2.05

Because t  1.833 , reject H0 . There is enough evidence at the 5% level of significance to reject the claim. 12. H 0 : 1  2 ; H a : 1  2 (claim) d.f.  min{n1  1, n2  1}  31 Rejection region: t  2.453 ( x  x )  ( 1  2 ) (52  50)  (0)   2.30 t 1 2 2 2 s1 s2 (4.8) 2 (1.2) 2   32 40 n1 n2

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.” H 0 : 1  2 (claim); H a : 1  2

(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 )  ( 1  2 ) (n1  1) s12  ( n2  1) s22 n1  n2  2

1 1  n1 n2



(263  183)  (0) (16  1)(30) 2  (18  1)(27) 2 1 1  16  18  2 16 18

 8.19

(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 family at movies is greater than the mean amount spent by a family at amusement parks.” H 0 : 1  2 ; H a : 1  2 (claim)

(b) d.f.  n1  n2  2  60 t0  2.390; Rejection regions: t  2.390 (c) t 

( x1  x2 )  ( 1   2 ) ( n1  1) s12  ( n2  1) s22 n1  n2  2

1 1  n1 n2

(d) Because t  2.390 , fail to reject H0 .



(86  82)  (0) (26  1)(9) 2  (36  1)(7) 2 26  36  2

1 1  26 36

 1.97

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323

(e) There is not enough evidence at the 1% level of significance to support the claim that the mean amount spent by a family at movies is greater than the mean amount spent by a family at amusement parks. 15. (a) The claim is “the units produced at Site A contain more defects than the unit produced at Site B.” H 0 : 1  2 ; H a : 1  2 (claim)

(b) t0  1.684; Rejection regions: t  1.684 (c) 1.86 (d) Reject H0 . (e) There is enough evidence to support the claim at the 5% level of significance that the units produced at Site A have more defects than units produced at Site B. 16. (a) The claim is “the mean fork length of yellowfin tuna is different in two zones in the eastern tropical Pacific Ocean.” H 0 : 1  2 ; H a : 1  2 (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 )  ( 1   2 ) (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 Cuyahoga County, Ohio, than it is in Wayne County, Michigan.” H 0 : 1  2 ; H a : 1  2 (claim)

(b) d.f.  min{n1  1, n2  1}  14 t0  1.761; Rejection region: t  1.761 ( x  x )  (0) (45,600  41,500)  (0)   5.65 (c) t  1 2 (2800) 2 (1310) 2 s12 s22   19 15 n1 n2 (d) Because t  1.761 , reject H0 .

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

(e) There is enough evidence at the 5% level of significance to support the demographics researcher’s claim that the mean household income in a recent year is greater in Cuyahoga County, Ohio, than it is in Wayne County, Michigan. 18. (a) The claim is “the mean household income in a recent year is the same in Ada County, Idaho, and Cameron Parish, Louisiana.” H 0 : 1  2 (claim); H a : 1  2

(b) d.f.  min{n1  1, n2  1}  17 t0  1.740; t0  1.740 Rejection regions: t  1.740, t  1.740 (c) t 

( x1  x2 )  (0) s12 n1



s22 n2



(58,300  56,600)  (0) (9000) 2 (15,600) 2  18 20

 0.42

(d) Because 1.740  t  1.740 , fail to reject H0 . (e) There is not enough evidence at the 10% level of significance to reject the demographics researcher’s claim that the mean household income in a recent year is the same in Ada County, Idaho, and Cameron Parish, Louisiana. 19. (a) The claim is “an experimental method makes a difference in the tensile strength of steel bars.” H 0 : 1  2 ; H a : 1  2 (claim)

(b) d.f.  n1  n2  2  22 t0  2.819, t0  2.819; Rejection regions: t  2.819, t  2.819 (c) x1  368.3, s1  22.301, n1  10 x2  388.214, s2  14.797, n2  14 ( x1  x2 )  ( 1  2 )  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. 20. (a) The claim is “the experimental method produces steel with greater mean tensile strength.” H 0 : 1  2 ; H a : 1  2 (claim)

(b) d.f.  min{n1  1, n2  1}  13 t0  1.350; Rejection region: t  1.350

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325

(c) x1  403.167, s1  11.137, n1  18 x2  384, s2  17.698, n2  14 ( x  x )  ( 1  2 ) (403.167  384)  (0)   3.543 t 1 2 2 2 s1 s2 (11.137) 2 (17.698) 2   18 14 n1 n2 (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” H 0 : 1  2 ; H a : 1  2 (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 )  ( 1  2 )  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.” H 0 : 1  2 ; H a : 1  2 (claim)

(b) d.f.  n1  n2  2  39 t0  2.426; Rejection region: t  2.426 (c) x1  79.091, s1  6.900, n1  22 x2  83, s2  7.645, n2  19 ( x1  x2 )  ( 1   2 )  t (n1  1) s12  (n2  1) s22 1 1  n1  n2  2 n1 n2 (d) Because t  2.426 , fail to reject H0 . Copyright © 2019 Pearson Education Ltd.

(79.091  83)  (0) (22  1)(6.900) 2  (19  1)(7.645)2 22  19  2

1 1  22 19

 1.72

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(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 )  tc

s12 s22 (0.17) 2 (0.04) 2   (0.73  0.75)  1.363  20 12 n1 n2

 0.02  0.054  0.074  1   2  0.034  0.07  1   2  0.03 24. ( x1  x2 )  tc

s12 s22 (6) 2 (12) 2   (267  244)  2.132  9 5 n1 n2

 23  12.21  10.78  1  2  35.21  10.8  1   2  35.2 25. ˆ 

( n1  1) s12  ( n2  1) s22  n1  n2  2

( x1  x2 )  tcˆ

(20  1)(39.7) 2  (17  1)(42.4) 2  40.956 20  17  2

1 1 1 1   (28  26)  1.69  40.956  20 17 n1 n2

 2  22.833  1  2  2  22.833  20.8  1  2  24.8 26. ˆ 

( n1  1) s12  ( n2  1) s22  n1  n2  2

( x1  x2 )  tcˆ

(20  1)(13.7) 2  (12  1)(14.4) 2  13.961 20  12  2

1 1 1 1   (41  41)  2.042  13.961  20 12 n1 n2

 0  10.4  10.4  1  2  10.4

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 “athletes can decrease their times in the 40-yard dash.” H 0 : d  0; H a : d  0 (claim)   0.05 , d.f.  n  1  11 t0  1.796; Rejection region: t  1.796 Before 4.85 4.90 5.08 4.72 4.62 4.54 5.25 5.18 4.81 4.57 4.63 4.77

d

After 4.78 4.90 5.05 4.65 4.64 4.50 5.24 5.27 4.75 4.43 4.61 4.82

 d  0.047 2

 d 

t

d2 0.0049 0.0000 0.0009 0.0049 0.0004 0.0016 0.0001 0.0081 0.0036 0.0196 0.0004 0.0025

 d  0.28  0.0233 2

sd 

d 0.07 0.00 0.03 0.07 0.02 0.04 0.01 0.09 0.06 0.14 0.02 0.05  d  0.28

 

2  d      0.282  n  0.047     12  0.0607 n 1 11

d  d 0.0233  0   1.330 0.0607 sd 12 n

Because t  1.796 , fail to reject H0 . There is not enough evidence at the 5% level of significance to support the claim that athletes can decrease their times in the 40-yard dash using new strength shoes.

327

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2. The claim is “the drug changes the body’s temperature.” H 0 : d  0; H a : d  0 (claim)   0.05 , d.f.  n  1  6 t0  2.447, t0  2.447; Rejection regions: t  2.447, t  2.447 Before 101.8 98.5 98.1 99.4 98.9 100.2 97.9

d

t

d 2.6 0.1 0.1 0.4 0.3 0.5 0.1  d  3.9

d2 6.76 0.01 0.01 0.16 0.09 0.25 0.01

 d  7.29 2

 d  3.9  0.5771 7

 sd 

After 99.2 98.4 98.2 99 98.6 99.7 97.8

 

2  d    2 d  3.9    n  7.29     7  0.9235 6 n 1 2

d  d 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.

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329

H 0 : d  0; H a : d  0 (claim)

  0.05 and d.f.  n  1  13 t0  1.771 ; Rejection region: t  1.771 t

d  d 1.5  0   1.754 3.2 sd 14 n

Because t  1.771 , fail to reject H0 . There is not enough evidence at the 5% level of significance to support the claim. 4.

H 0 : d  0 (claim); H a : d  0

  0.01 and d.f.  n  1  7 t0  3.499, t0  3.499; Rejection regions: t  3.499, t  3.499 t

d  d 3.2  0   1.071 8.45 sd 8 n

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.

H 0 : d  0 (claim); H a : d  0

  0.10 and d.f.  n  1  15 t0  1.341; Rejection region: t  1.341 t

d  d 6.5  0   2.725 9.54 sd 16 n

Because t  1.341 , reject H0 . There is enough evidence at the 10% level of significance to reject the claim. 6.

H 0 : d  0; H a : d  0 (claim)

  0.05 and d.f.  n  1  27 t0  1.703; Rejection region: t  1.703 t

d  d 0.55  0   2.940 0.99 sd 28 n

Because t  1.703 , reject H0 . There is enough evidence at the 5% level of significance to support the claim. 7.

H 0 : d  0 (claim); H a : d  0

  0.01 and d.f.  n  1  14

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t0  2.624; Rejection region: t  2.624 t

d  d 2.3  0   7.423 1.2 sd 15 n

Because t  2.624 , reject H0 . There is enough evidence at the 1% level of significance to reject the claim. 8.

H 0 : d  0; H a : d  0 (claim)

  0.10 and d.f.  n  1  19 t0  1.729, t0  1.729; Rejection regions: t  1.729, t  1.729 t

d  d 1  0   1.626 2.75 sd 20 n

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 “seven of the stocks that make up the Dow Jones Industrial Average lost value from one hour to the next on one business day." H 0 : d  0; H a : d  0 (claim)

(b) t0  3.143; Rejection region: t  3.143 (c) d  0.087 and sd  0.405 (d) t 

d  d 0.0871  0   0.569 0.4053 sd 7 n

(e) Because t  3.143 , fail to reject H0 . (f) There is not enough evidence at the 1% level of significance to support the stock market analyst’s claim that seven of the stocks that make up the Dow Jones Industrial Average lost value from one hour to the next on one business day. 10. (a) The claim is “an SAT preparation course improves the test scores of students.” H 0 : d  0; H a : d  0 (claim)

(b) t0  2.821; Rejection region: t  2.821 (c) d  60 and sd  27.889

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

(d) t 

331

d  d 60  0   6.803 27.889 sd 10 n

(e) Because t  2.821 , reject H0 . (f) There is enough evidence at the 1% level of significance to support the claim that the SAT preparation course improves the test scores of students. 11. (a) The claim is “caffeine ingestion improves repeated freestyle sprints in trained male swimmers.” H 0 : d  0; H a : d  0 (claim)

(b) t0  3.365; Rejection region: t  3.365 (c) d  0.533 and sd  0.350 (d) t 

d  d 0.533  0   3.730 0.350 sd 6 n

(e) Because t  3.365 , reject H0 . (f) There is enough evidence at the 1% level of significance to support the researcher’s claim that caffeine ingestion improves repeated freestyle sprints in trained male swimmers. 12. (a) The claim is “a baseball clinic will help players raise their batting average.” H 0 : d  0; H a : d  0 (claim)

(b) t0  1.771; Rejection region: t  1.771 (c) d  0.002 and sd  0.015 (d) t 

d  d 0.002  0   0.499 0.015 sd 14 n

(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. 13. (a) The claim is “soft tissue therapy helps to reduce the numbers of days per week patients suffer from headaches.” H 0 : d  0; H a : d  0 (claim)

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

(b) t0  2.567; Rejection region; t  2.567 (c) d  1.5 and sd  1.249 (d) t 

d  d 1.5  0   5.095 1.249 sd 18 n

(e) Because t  2.567 , reject H0 . (f) There is enough evidence at the 1% level of significance to support the claim that soft tissue therapy helps to reduce the numbers of days per week patients suffer from headaches. 14. (a) The claim is “the use of a specific type of therapeutic tape reduces pain in patients with chronic tennis elbow.” H 0 : d  0; H a : d  0 (claim)

(b) t0  1.761; Rejection region: t  1.761 (c) d  2.133 and sd  1.885 (d) t 

d  d 2.133  0   4.383 1.885 sd 15 n

(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 “student housing rates have increased from one academic year to the next.” H 0 : d  0; H a : d  0 (claim)

(b) t0  1.796; Rejection region: t  1.796 (c) d  254.5 and sd  291.767 (d) t 

d  d 254.5  0   3.022 291.767 sd 12 n

(e) Because t  1.796 , reject H0 .

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333

(f) There is enough evidence at the 5% level of significance to support the college administrator’s claim that student housing rates have increased from one academic year to the next. 16. (a) The claim is “stipends for PhD students have increased from one academic year to the next.” H 0 : d  0; H a : d  0 (claim)

(b) t0  1.943; Rejection region: t  1.943 (c) d  664.4 and sd  2472.373 (d) t 

d  d 664.4  0   0.711 2472.373 sd 7 n

(e) Because t  1.943 , fail to reject H0 . (f) There is not enough evidence at the 5% level of significance to support the education researcher’s claim that stipends for PhD students have increased from one academic year to the next. 17. (a) The claim is “the product ratings have changed from last year to this year.” H 0 : d  0; H a : d  0 (claim)

(b) t0  2.365, t0  2.365; Rejection regions: t  2.365, t  2.365 (c) d  1 and sd  1.309 (d) t 

d  d 1  0   2.161 1.309 sd 8 n

(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 “the pass completion percentages have changed.” H 0 : d  0; H a : d  0 (claim)

(b) t0  1.833, t0  1.833; Rejection regions: t  1.833, t  1.833 (c) d  1.7 and sd  4.020

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

(d) t 

x  d 1.7  0   1.337 4.020 sd 10 n

(e) Because 1.833  t  1.833 , fail to reject H0 . (f) There is not enough evidence at the 10% level of significance to support the claim that the pass completion percentages have changed. 19. (a) The claim is “the cookies manufactured by a manufacturer has more fibre content than his competitors’ cookies.” H 0 : d  0; H a : d  0 (claim)

(b) t0  3.143; Rejection regions: t  3.143 (c) d  0.43 and sd  3.69 (d) t 

d  d 0.43  0   0.308 3.69 sd 7 n

(e) Because t  3.143 , fail to reject H0 . (f) There is not enough evidence at the 1% level of significance to support the claim that the cookies manufactured by a manufacturer has more fibre content than his competitors’ cookies. 20. (a) The claim is “horses’ times have changed.” H 0 : d  0; H a : d  0 (claim)

(b) t0  2.365; Rejection regions: t  2.365, t  2.365 (c) d  0.49 and sd  10.85 (d) t 

x  d 0.49  0   0.128 10.85 sd 8 n

(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 horses’ timings have changed. 21. Yes; P  0.0058  0.05, so you reject H0 .

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335

22. Yes; P  0.2139  0.10, so you fail to reject H0 . 23. d  1.525 and sd  0.542 s s d  t /2 d  d  d  t /2 d n n

 0.542   0.542  1.525  1.753    d  1.525  1.753    16   16  1.525  0.238  d  1.525  0.238 1.76  d  1.29 24. d  0.436 and sd  0.677 s s d  t /2 d  d  d  t /2 d n n

 0.677   0.677  0.436  2.160    d  0.436  2.160    14   14  0.436  0.391  d  0.436  0.391 0.83  d  0.05

8.4 TESTING THE DIFFERENCE BETWEEN PROPORTIONS 8.4 TRY IT YOURSELF SOLUTIONS 1.

p

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 40- to 49-year olds who are yoga users and the proportion of 40- to 49-year-olds who are non-yoga users.” H 0 : p1  p2 ; H a : p1  p2 (claim)   0.05  z0  1.96, z0  1.96; Rejection regions: z  1.96, z  1.96 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.230  0.210)  (0) z 1   1.91 1  1 1  1 0.211(0.789)   pq      1593 29,948   n1 n2 

*********

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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 40- to 49-year olds who are yoga users and the proportion of 40- to 49-year-olds who are non-yoga users. 2.

p

x1  x2 239  5990   0.1975, q  1  p  0.8025 n1  n2 1593  29948

n1 p  1593(0.1975)  314.6  5, n1q  1593(0.8025)  1278.4  5 n2 p  29,948(0.1975)  5914.7  5, n2 q  29,948(0.8025)  24,033.3  5 The claim is “the proportion of yoga users with incomes of $20,000 to $34,499 is less than the proportion of non-yoga users with incomes of $20,000 to $34,499.” H 0 : p1  p2 ; H a : p1  p2 (claim)   0.05 z0  1.645; Rejection region: z  1.645 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.15  0.20)  (0) z 1   4.88 1  1 1  1 0.1975  0.8025   pq     n n  1593 29,948  2   1

Because z  1.645 , reject H0 . There is enough evidence at the 5% level of significance to support the claim that the proportion of yoga users with incomes of $20,000 to $34,499 is less than the proportion of non-yoga users with incomes of $20,000 to $34,499.

8.4 EXERCISE SOLUTIONS 1. (1) The samples are randomly selected.

(2) The samples are independent. (3) n1 p  5, n1q  5, n2 p  5, n2 q  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.

p

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

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

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. H 0 : p1  p2 ; H a : p1  p2 (claim)  z0  2.575, z0  2.575; Rejection regions: z  2.575, z  2.575 z

( pˆ1  pˆ 2 )  ( p1  p2 ) 1 1 pq     n1 n2 



(0.5  0.6)  (0) 1   1 0.5462  0.4538     70 60 

 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.

p

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 Because all conditions are met above, the normal sampling distribution can be used.

H 0 : p1  p2 ; H a : p1  p2 (claim) z0  1.645; Rejection region: z  1.645 z

( pˆ1  pˆ 2 )  ( p1  p2 ) 1 1 pq     n1 n2 



(0.600  0.800)  (0) 1   1 0.6744  0.3256     785 465 

 7.293

Because z  1.645 , reject H0 . There is enough evidence at the 5% level of significance to support the claim. 5.

p

x1  x2 42  76   0.337, q  1  p  0.663 n1  n2 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 2 q  200(0.663)  132.58  5 Because all conditions are met above, the normal sampling distribution can be used.

H 0 : p1  p2 (claim); H a : p1  p2  z0  1.645, z0  1.645; Rejection regions: z  1.645, z  1.645 z

( pˆ1  pˆ 2 )  ( p1  p2 ) 1 1 pq     n1 n2 



(0.28  0.38)  (0) 1   1 (0.337)(0.663)     150 200 

Because z  1.645 , reject H0 .

 1.96

337

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There is enough evidence at the 10% level of significance to reject the claim. 6.

p

x1  x2 64   0.2, q  1  p  0.8 n1  n2 20  30

n1 p  20(0.2)  4  5, n1q  20(0.8)  16  5 . n2 p  30(0.2)  6  5, and n 2 q  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.” H 0 : p1  p2 ; H a : p1  p2 (claim)

(b)  z0  2.575, z0  2.575; Rejection regions: z  2.575, z  2.575 (c) p 

z

x1  x2 327  148   0.649, q  1  p  0.351 n1  n2 488  244

( pˆ1  pˆ 2 )  ( p1  p2 ) 1 1 pq     n1 n2 



(0.670  0.607)  (0) 1   1  (0.649)(0.341)    488 244 

 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.” H 0 : p1  p2 ; H a : p1  p2 (claim)

(b) z0  1.28; Rejection region: z  1.28 (c) p 

z

x1  x2 293  268   0.738 q  1  p  0.262 n1  n2 384  376

( pˆ1  pˆ 2 )  ( p1  p2 ) 1 1 pq     n1 n2 



(0.763  0.713)  (0) 1   1 (0.738)(0.262)     384 376 

(d) Because z  1.28 , reject H0 .

 1.6

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339

(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 females ages 20 to 24 and males ages 20 to 24.” H 0 : p1  p2 ; H a : p1  p2 (claim)

(b)  z0  2.575, z0  2.575; Rejection regions: z  2.575, z  2.575 x1  x2 1127  1464   0.691, q  1  p  0.309 n1  n2 1750  2000 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.644  0.732)  (0) z 1   5.82 1  1 1  1 (0.691)(0.309)   pq      1750 2000   n1 n2 

(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 females ages 20 to 24 and males ages 20 to 24. 10. (a) The claim is “the proportion of males ages 20 to 24 who were neither in school nor working is less than the proportion of females ages 20 to 24 who were neither in school nor working.” H 0 : p1  p2 ; H a : p1  p2 (claim)

(b) z0  1.645; Rejection region: z  1.645 x1  x2 79  89   0.168, q  1  p  0.832 n1  n2 500  500 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.158  0.178)  (0)   0.85 z 1 1  1 1  1  (0.168)(0.832)  pq    500 500   n n 2   1

(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 proportion of males ages 20 to 24 who were neither in school nor working is less 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.” H 0 : p1  p2 ; H a : p1  p2 (claim)

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x1  x2 934  909   0.9215, q  1  p  0.0785 n1  n2 1000  1000 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.934  0.909)  (0) z 1   2.08 1  1 1  1 (0.9215)(0.0785)   pq      1000 1000   n1 n2 

(d) Because z  1.645 , reject H0 . (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.” H 0 : p1  p2 ; H a : p1  p2 (claim)

(b) z0  1.28; Rejection region: z  1.28 x1  x2 855  909   0.882, q  1  p  0.118 n1  n2 1000  1000 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.855  0.909)  (0)   3.74 z 1 1  1 1   1  (0.882)(0.118)  pq      1000 1000   n1 n2 

(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 newlywed Asians who have a spouse of a different race or ethnicity is the same as the proportion of newlywed Hispanics who have a spouse of a different race or ethnicity.” H 0 : p1  p2 (claim); H a : p1  p2

 z0  1.96, z0  1.96; Rejection regions: z  1.96, z  1.96 x  x2 290  270 p 1   0.28, q  1  p  0.72 n1  n2 1000  1000 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.29  0.27)  (0) z 1   0.996 1  1 1  1 (0.28)(0.72)   pq      1000 1000   n1 n2  Because 1.96  z  1.96 , fail to reject H0 . No, there is not enough evidence at the 5% level of significance to reject the claim that the proportion of newlywed Asians who have a spouse of a different race or ethnicity is the same as the proportion of newlywed Hispanics who have a spouse of a different race or ethnicity.

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341

14. The claim is “the proportion of newlywed blacks who have a spouse of a different race or ethnicity is less than the proportion of newlywed Asians who have a spouse of a different race or ethnicity.” H 0 : p1  p2 ; H a : p1  p2 (claim)

z0  2.33; Rejection region: z  2.33 x  x2 180  290 p 1   0.235, q  1  p  0.765 n1  n2 1000  1000 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.18  0.29)  (0)   5.80 z 1 1  1 1   1  (0.235)(0.765)  pq      1000 1000   n1 n2  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 newlywed blacks who have a spouse of a different race or ethnicity is less than the proportion of newlywed Asians who have a spouse of a different race or ethnicity. 15. The claim is “the proportion of newlywed Asians who have a spouse of a different race or ethnicity is greater than the proportion of newlywed whites who have a spouse of a different race or ethnicity.” H 0 : p1  p2 ; H a : p1  p2 (claim)

z0  2.33; Rejection region: z  2.33 x  x2 290  110 p 1   0.2, q  1  p  0.8 n1  n2 1000  1000 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.29  0.11)  (0)   10.1 z 1 1  1 1  1  (0.2)(0.8)  pq      1000 1000   n1 n2  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 newlywed Asians who have a spouse of a different race or ethnicity is greater than the proportion of newlywed whites who have a spouse of a different race or ethnicity. 16. The claim is “the proportion of newlywed Hispanics who have a spouse of a different race or ethnicity is different from the proportion of newlywed blacks who have a spouse of a different race or ethnicity.” H 0 : p1  p2 ; H a : p1  p2 (claim)

 z0  1.96, z0  1.96; Rejection regions: z  1.96, z  1.96 x  x2 270  180 p 1   0.225, q  1  p  0.775 n1  n2 1000  1000 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.27  0.18)  (0) z 1   4.82 1  1 1  1 (0.225)(0.775)   pq      1000 1000   n1 n2  Because z  1.96 , reject H0 .

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Yes, there is enough evidence at the 5% level of significance to support the claim that the proportion of newlywed Hispanics who have a spouse of a different race or ethnicity is different from the proportion of newlywed blacks who have a spouse of a different race or ethnicity. 17. The claim is “the proportion of newly wed whites who have a spouse of a different race or ethnicity is less than the proportion of newlywed blacks who have a spouse of a different race or ethnicity.” H 0 : p1  p2 ; H a : p1  p2 (claim)

z0  2.33; Rejection region: z  2.33 x  x2 110  180 p 1   0.145, q  1  p  0.855 n1  n2 1000  1000 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.11  0.18)  (0) z 1   4.45 1  1 1   1 (0.145)(0.855)   pq      1000 1000   n1 n2  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 newlywed whites who have a spouse of a different race or ethnicity is less than the proportion of newlywed blacks who have a spouse of a different race or ethnicity. 18. The claim is “the proportion of newlywed Hispanics who have a spouse of a different race or ethnicity is greater than the proportion of newlywed whites who have a spouse of a different race or ethnicity.” H 0 : p1  p2 ; H a : p1  p2 (claim)

z0  1.645; Rejection region: z  1.645 x  x2 270  110 p 1   0.19, q  1  p  0.81 n1  n2 1000  1000 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.27  0.11)  (0) z 1   9.12 1  1 1  1 (0.19)(0.81)   pq    1000 1000   n n 2   1 Because z  1.645 , reject H0 . Yes, there is enough evidence at the 5% level of significance to support the claim that the proportion of newlywed Hispanics who have a spouse of a different race or ethnicity is greater than the proportion of newlywed whites who have a spouse of a different race or ethnicity. 19. The claim is “the proportion of men who work 40 hours per week is the same as the proportion of men who work more than 40 hours per week.” H 0 : p1  p2 (claim); H a : p1  p2

 z0  2.576; z0  2.576 Rejection regions: z  2.576; z  2.576 x  x2 141  141 p 1   0.47, q  1  p  0.53 n1  n2 300  300 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.47  0.47)  (0) z 1  0 1  1 1  1 (0.47)(0.53)   pq    300 300   n n 1 2  

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

343

Because 2.576  z  2.576 , fail to reject H0 . No, there is not enough evidence at the 1% level of significance to reject claim that the proportion of men who work 40 hours per week is the same as the proportion of men who work more than 40 hours per week. 20. The claim is “the proportion of women who work 40 hours per week is greater than the proportion of women who work more than 40 hours per week.” H 0 : p1  p2 ; H a : p1  p2 (claim)

z0  1.645; Rejection region: z  1.645 x  x2 140  75 p 1   0.43, q  1  p  0.57 n1  n2 250  250 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.56  0.30)  (0)   5.87 z 1 1  1 1  1  (0.43)(0.57)  pq      250 250   n1 n2  Because z  1.645 , reject H0 . Yes, there is enough evidence at the 5% level of significance to support the claim that the proportion of women who work 40 hours per week is greater than the proportion of women who work more than 40 hours per week. 21. The claim is “the proportion of the U.S. workforce that works 40 hours per week is greater for women than for men.” H 0 : p1  p2 ; H a : p1  p2 (claim)

z0  1.645; Rejection region: z  1.645 x  x2 140  141 p 1   0.511, q  1  p  0.489 n1  n2 250  300 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.56  0.47)  (0) z 1   2.10 1  1 1  1 (0.511)(0.489)   pq    250 300   n n 2   1 Because z  1.645 , reject H0 . Yes, there is enough evidence at the 5% level of significance to support the claim that the proportion of the U.S. workforce that works 40 hours per week is greater for women than for men. 22. The claim is “the proportion of the U.S. workforce that works more than 40 hours per week is less for women than for men.” H 0 : p1  p2 ; H a : p1  p2 (claim)

z0  1.28; Rejection region: z  1.28 x  x2 140  141 p 1   0.511, q  1  p  0.489 n1  n2 250  300 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.30  0.47)  (0) z 1   3.97 1  1 1  1 (0.511)(0.489)   pq      250 300   n1 n2  Because z  1.28 , reject H0 .

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

Yes, there is enough evidence at the 10% level of significance to support the claim that the proportion of the U.S. workforce that works more than 40 hours per week is less for women than for men. ( pˆ1  pˆ 2 )  zc

23.

pˆ1qˆ1 pˆ 2 qˆ2   p1  p2  ( pˆ1  pˆ 2 )  zc n1 n2

pˆ1qˆ1 pˆ 2 qˆ2  n1 n2

(0.07)(0.93) (0.09)(0.91) (0.07)(0.93) (0.09)(0.91)   p1  p2  (0.07  0.09)  1.96  10, 000 8000 10, 000 8000

(0.07  0.09)  1.96

0.02  0.008  p1  p2  0.02  0.008 0.028  p1  p2  0.012 ( pˆ1  pˆ 2 )  zc

24.

pˆ1qˆ1 pˆ 2qˆ2   p1  p2  ( pˆ1  pˆ 2 )  zc n1 n2

pˆ1qˆ1 pˆ 2qˆ2  n1 n2

(0.07)(0.93) (0.03)(0.97) (0.07)(0.93) (0.03)(0.97)   p1  p2  (0.07  0.03)  1.645  10,000 8000 10,000 8000

(0.07  0.03)  1.645

0.04  0.005  p1  p2  0.04  0.005 0.035  p1  p2  0.045

( pˆ1  pˆ 2 )  zc

25.

pˆ1qˆ1 pˆ 2 qˆ2   p1  p2  ( pˆ1  pˆ 2 )  zc n1 n2

pˆ1qˆ1 pˆ 2 qˆ2  n1 n2

(0.69)(0.31) (0.65)(0.35) (0.69)(0.31) (0.65)(0.35)   p1  p2  (0.69  0.65)  1.96  2000 2000 2000 2000 0.04  0.029  p1  p2  0.04  0.029

(0.69  0.65)  1.96

0.011  p1  p2  0.069

Answers will vary. ( pˆ1  pˆ 2 )  zc

26. (0.69  0.65)  2.575

pˆ1qˆ1 pˆ 2 qˆ2   p1  p2  ( pˆ1  pˆ 2 )  zc n1 n2

pˆ1qˆ1 pˆ 2 qˆ2  n1 n2

(0.69)(0.31) (0.65)(0.35) (0.69)(0.31) (0.65)(0.35)   p1  p2  (0.69  0.65)  2.575  2000 2000 2000 2000 0.04  0.038  p1  p2  0.04  0.038 0.002  p1  p2  0.078

Answers will vary.

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

345

CHAPTER 8 REVIEW EXERCISE SOLUTIONS 1. Dependent because the same adults were sampled. 2. Independent because different species were sampled. 3. Independent because different vehicles were sampled. 4. Dependent because the same cars were sampled. 5.

H 0 : 1  2 (claim); H a : 1  2 z0  1.645; Rejection region: z  1.645 z

( x1  x2 )  ( 1  2 )

 12 n1



 22



(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. 6.

H 0 : 1  2 (claim); H a : 1  2  z0  2.575, z0  2.575; Rejection regions: z  2.575, z  2.575 z

( x1  x2 )  ( 1  2 )

 12 n1



 22



(5595  5575)  (0) (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.

H 0 : 1  2 ; H a : 1  2 (claim) z0  1.28; Rejection regions: z  1.28 z

( x1  x2 )  ( 1  2 )

 12 n1



 22



(0.28  0.33)  (0) (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.

H 0 : 1  2 ; H a : 1  2 (claim)  z0  1.96, z0  1.96; Rejection regions: z  1.96, z  1.96

346

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

z

( x1  x2 )  ( 1  2 )

 12 n1



 22



(87  85)  (0)

(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.” H 0 : 1  2 ; H a : 1  2 (claim)

(b) z0  1.645; Rejection region: z  1.645 (c) z 

( x1  x2 )  ( 1  2 )



2 1





2 2



(670  690)  (0) (20) 2 (30) 2  22 28

 2.82

(d) Because z  1.645 , reject H0 . (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 salary of entry-level paralegals in Peoria, Illinois, and Gary, Indiana, is the same.” H 0 : 1  2 (claim); H a : 1  2

(b)  z0  1.645, z0  1.645; Rejection regions: z  1.645, z  1.645 (c) z 

( x1  x2 )  ( 1  2 )



2 1





2 2



(50, 410  47,350)  (0)

(9320) 2 (9330) 2  40 35

 1.42

(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 career counselor’s claim that the mean annual salary of entry-level paralegals in Peoria, Illinois, and Gary, Indiana, is the same. 11. H 0 : 1  2 (claim); H a : 1  2

d.f.  n1 + n2  2  31 t0  2.040, t0  2.040; Rejection regions: t  2.040, t  2.040

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

t

( x1  x2 )  ( 1   2 ) ( n1  1) s12  ( n2  1) s22 n1  n2  2

1 1  n1 n2



(228  207)  (0) (20  1)(27) 2  (13  1)(25) 2 20  13  2

1 1  20 13

347

 2.25

Because t  2.040 , reject H0 . There is enough evidence at the 5% level of significance to reject the claim. 12. H 0 : 1  2 ; H a : 1  2 (claim)

d.f.  min{n1  1, n2  1}  5 t0  1.476; Rejection region: t  1.476 t

( x1  x2 )  ( 1  2 ) s12 n1



s22

(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. 13. H 0 : 1  2 (claim); H a : 1  2

d.f.  min{n1  1, n2  1}  39 t0  1.304; Rejection region: t  1.304 t

( x1  x2 )  ( 1  2 ) s12 s22  n1 n2



(664.5  665.5)  (0) (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. H 0 : 1  2 (claim); H a : 1  2

d.f.  n1  n2  2  33 t0  2.445; Rejection region: t  2.445 t

( x1  x2 )  ( 1  2 ) (n1  1) s12  (n2  1) s22 n1  n2  2

1 1  n1 n2



(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.

348

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

15. H 0 : 1  2 ; H a : 1  2 (claim)

d.f.  n1  n2  2  10 t0  3.169, t0  3.169; Rejection regions: t  3.169, t  3.169 t

( x1  x2 )  ( 1  2 ) ( n1  1) s12  (n2  1) s22

1 1   n1 n2

n1  n2  2



(61  55)  (0) (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. H 0 : 1  2 ; H a : 1  2 (claim)

d.f.  min{n1  1, n2  1}  5 t0  1.476; Rejection region: t  1.476 t

( x1  x2 )  ( 1  2 ) s12 s22  n1 n2



(520  500)  (0) (25) 2 (55) 2  7 6

 0.821

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: 1 represents the new curriculum and 2 represents the old curriculum. H 0 : 1  2 ; H a : 1  2 (claim)

(b) d.f.  n1  n2  2  70 t0  1.667; Rejection region: t  1.667 (c) x1  62.806, s1  22.845, n1  36 x2  48.694, s2  28.592, n2  36 ( x1  x2 )  ( 1  2 ) t  (n1  1) s12  (n2  1) s22 1 1  n1  n2  2 n1 n2

(62.806  48.694)  (0) (36  1)(22.845) 2  (36  1)(28.592) 2 36  36  2

1 1  36 36

 2.31

(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.

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

349

18. (a) The claim is “there is no difference between the mean household incomes of two neighborhoods.” H 0 : 1  2 (claim); H a : 1  2

(b) d.f.  n1  n2  2  20 t0  2.845, t0  2.845; Rejection regions: t  2.845, t  2.845 (c) t 

( x1  x2 )  ( 1  2 ) (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 two neighborhoods. 19. H 0 : d  0 (claim); H a : d  0   0.01 and d.f.  n  1  15 t0  2.947, t0  2.947; Rejection regions: t  2.947, t  2.947

d  d 8.5  0   3.178 10.7 sd 16 n Because t  2.947 , reject H0 . There is enough evidence at the 1% level of significance to reject the claim. t

20. H 0 : d  0; H a : d  0 (claim)   0.10 and d.f.  n  1  24 t0  1.318 ; Rejection region: t  1.318

d  d 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. H 0 : d  0 (claim); H a : d  0   0.10 and d.f.  n  1  33 t0  1.308 ; Rejection region: t  1.308

d  d 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

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

22. H 0 : d  0; H a : d  0 (claim)   0.05 and d.f.  n  1  36 t0  2.028, t0  2.028; Rejection regions: t  2.028, t  2.028

d  d 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 passing yards for college football quarterbacks change from their junior to their senior years.” H 0 : d  0; H a : d  0 (claim)

(b) t0  2.262, t0  2.262; ; Rejection regions: t  2.262, t  2.262 (c) d  12.3 and sd  553.0877 (d) t 

d  d 12.3  0   0.070 553.0877 sd 10 n

(e) Because 2.262  t  2.262 , fail to reject H0 . (f) There is not enough evidence at the 5% level of significance to support the sports statistician’s claim the numbers of passing yards for college football quarterbacks change from their junior to their senior years. 24. (a) The claim is “the weight loss supplement will help users lose weight after two weeks.” H 0 : d  0; H a : d  0 (claim)

(b) t0  1.397 ; Rejection region: t  1.397 (c) d  1.444 and sd  2.744 (d) t 

d  d 1.444  0   1.579 2.744 sd 9 n

(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.

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

25. p 

351

x1  x2 425  410   0.522, q  1  p  0.478 n1  n2 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 2 q  760(0.478)  363.28  5 . Can use normal sampling distribution.

H 0 : p1  p2 (claim); H a : p1  p2  z0  1.96, z0  1.96; Rejection regions: z  1.96, z  1.96 z

( pˆ1  pˆ 2 )  ( p1  p2 ) 1 1 pq     n1 n2 



(0.506  0.539)  (0) 1   1 0.522(0.478)     840 760 

 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 36  46   0.273, q  1  p  0.727 n1  n2 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. H 0 : p1  p2 (claim) : H a : p1  p2 z0  2.33; Rejection region: z  2.33 z

( pˆ1  pˆ 2 )  ( p1  p2 ) 1 1 pq     n1 n2 



(0.36  0.23)  (0) 1   1 0.273(0.727)     100 200 

 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 261  207   0.450, q  1  p  0.550 n1  n2 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 2 q  483(0.550)  265.65  5 . Can use normal sampling distribution. H 0 : p1  p2 ; H a : p1  p2 (claim) z0  1.28; Rejection region: z  1.28

352

CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

z

( pˆ1  pˆ 2 )  ( p1  p2 ) 1 1 pq     n1 n2 



(0.469  0.429)  (0) 1   1  0.450(0.550)    556 483 

 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 86  107   0.092, q  1  p  0.908 n1  n2 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. H 0 : p1  p2 ; H a : p1  p2 (claim) z0  1.645; Rejection region: z  1.645 z

( pˆ1  pˆ 2 )  ( p1  p2 ) 1 1 pq     n1 n2 



(0.096  0.089)  (0) 1   1 0.092(0.908)     900 1200 

 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.” H 0 : p1  p2 (claim); H a : p1  p2

(b)  z0  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 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.2794  0.0294)  (0) z 1   4.03 1  1 1  1 0.1544(0.8456)    pq     68 68  n n 1 2  

(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.

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353

30. (a) The claim is “the proportion of motorcyclists who use helmets that are compliant with federal safety regulations increased from the first year to the second year.” H 0 : p1  p2 ; H a : p1  p2 (claim)

(b) z0  2.33 Rejection region: z  2.33 (c) p 

z

x1  x2 607  653   0.63, q  1  p  0.37 n1  n2 1000  1000

( pˆ1  pˆ 2 )  ( p1  p2 ) 1 1 pq     n1 n2 



(0.607  0.653)  (0) 1   1 0.63(0.37)    1000 1000  

 2.13

(d) Because z  2.33 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to support the traffic safety research team’s claim that the proportion of motorcyclists who use helmets that are compliant with federal safety regulations increased from the first year to the second year.

CHAPTER 8 QUIZ SOLUTIONS 1. (a) The claim is “the mean score on the reading assessment test for male high school students

is greater than the mean score for female high school students.” H 0 : 1  2 ; H a : 1  2 (claim) (b) Right-tailed because H a contains >; z-test because 1 and  2 are known, the samples are random samples, the samples are independent, and n1  30 and n2  30 . (c) z0  1.645; Rejection region: z  1.645 (d) z 

( x1  x2 )  ( 1  2 )



2 1





2 2



(279  278)  (0) (41) 2 (39) 2  49 50

 0.12

(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 greater than for the female high school students. 2. (a) The claim is “the mean scores on a music assessment test for eighth grade boys and girls are equal.” H 0 : 1  2 (claim); H a : 1  2

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(b) Two-tailed because H a contains ; t-test because 1 and  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  1.706, t0  1.706; Rejection regions: t  1.706, t  1.706 (d) t 

( x1  x2 )  ( 1  2 ) ( n1  1) s12  (n2  1) s22 n1  n2  2

1 1  n1 n2



(142  156)  (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 teacher’s claim that the mean scores on the music assessment test are the same for eighth grade boys and girls. 3. (a) The claim is “the seminar helps adults increase their credit scores.” H 0 : d  0; H a : d  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  d 51.167  0   5.073 34.938 sd 12 n

(e) Because t  2.718 , reject H0 . (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 U.S. adults who approve of the job the Supreme Court is doing is less than it was 3 years prior.” H 0 : p1  p2 ; H a : p1  p2 (claim)

(b) Left-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

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355

x1  x2 459  694   0.4557, q  1  p  0.5443 n1  n2 1020  1510 ( pˆ  pˆ 2 )  ( p1  p2 ) (0.4500  0.4596)  (0)   0.48 z 1 1  1 1  1  0.4557(0.5443)  pq    1020 1510   n n 2   1

(d) p 

(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 proportion of U.S. adults who approve of the job the Supreme Court is doing is less than it was 3 years prior.

CUMULATIVE REVIEW, CHAPTERS 68 1. (a) pˆ  0.80, qˆ  0.20 pˆ  zc

ˆˆ (0.80)(0.20) pq  0.80  1.96  0.80  0.014  (0.786, 0.814) 3015 n

(b) H 0 : p  0.75; H a : p  0.75 (claim) z0  1.645; Rejection region: z  1.645 0.80  0.75 pˆ  p z   6.34 pq (0.75)(0.25) n 3015 Because z  1.645 , reject H0 . There is enough evidence at the 5% level of significance to support the researcher’s claim that more than 75% of U.S. adults say their household contains a desktop or a laptop computer. 2.

H 0 : d  0; H a : d  0 (claim)   0.10 d.f.  n  1  7 t0  1.415; Rejection region: t  1.415 d  1.575 and sd  0.803 d  d 1.575  0   5.548 0.803 sd 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. 3.

x  zc

 n

 26.97  1.96

3.4  26.97  1.03  (25.94, 28.00); z-distribution 42

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

x  tc

s 1.63  3.46  2.131  3.46  0.87  (2.59, 4.33); t-distribution n 16

x  tc

s 2.64  12.1  2.787  12.1  1.4  (10.7, 13.5); t-distribution n 26

x  zc

 n

 8.21  1.645

H 0 :   33 H a :   33 (claim)

H 0 : p  0.19 (claim) H a : p  0.19

0.62  8.21  0.36  (7.85, 8.57); z-distribution 8

H 0 :   0.63 (claim) H a :   0.63

10. H 0 :   2.28 H a :   2.28 (claim)

11. H 0 : 1  2 ; H a : 1  2 (claim) z0  1.28; Rejection region: z  1.28 ( x  x )  0 (3086  2263)  (0)   8.464 z 1 2 (563) 2 (624) 2  12  22   85 68 n1 n2

Because z  1.28 , reject H0 . There is enough evidence at the 10% level of significance to support the 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.  ( 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   L2   46.928  R (b)

 511.95, 2283.74   (22.63, 47.79)

 02  44.314; Rejection region:  2  44.314

2 

 n  1 s 2 2



(25)(31)2  26.7 (30)2

Because  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.

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357

13. H 0 : 1  2 ; H a : 1  2 (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)

t

(n1  1) s12  (n2  1) s22 n1  n2  2

1 1  n1 n2



1189  1376 (26  1)(218) 2  (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  53,518.03 , s  9429.71 s 9429.71 x  tc  53,518.03  2.045 n 30  (49,997, 57,039) (b) H 0 :   53,000 (claim); H a :   40,000

t0  2.045, t0  2.045; Rejection regions: t  2.045, t  2.045 t

x   53,518.03  53,000   0.30 9429.71 s 30 n

Because 2.045  t  2.045 , fail to reject H0 . There is not enough evidence at the 5% level of significance to reject the researcher’s claim that the mean annual earnings for locksmiths is $53,000. 15. H 0 : p 1  p2 ; H a : p1  p2 (claim) p

x1  x2 6 1   0.065, q  1  p  0.935 n1  n2 52  56

z0  1.28; Rejection regions: z  1.28; z0 

( pˆ1  pˆ 2 )  0 1 1 pq     n1 n2 



(0.1154  0.0179) 1   1 (0.065)(0.935)    52 56  

 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  z

 n

 42  1.96

1.6  42  0.496  (41.5, 42.5) 40

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(b) H 0 :   45 (claim); H a :   45

z0  1.645; Rejection region: z  1.645; z

x   42  45   11.86 1.6 s 40 n

Because z  1.645; , reject H0 . There is enough evidence at the 5% level of significance to reject the 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.” H 0 : p1  p2 (claim); H a : p1  p2

(b) Two-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, z0  1.645; Rejection regions: z  1.645, z  1.645 (d) p 

z

x1  x2 350  360   0.0418, q  1  p  0.9582 n1  n2 5000  12,000

( pˆ1  pˆ 2 )  ( p1  p2 ) 1 1 pq     n1 n2 



(0.07  0.03)  (0) 1   1 0.0418(0.9582)     5000 12,000 

 11.9

(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.” H 0 : 1  2 ; H a : 1  2 (claim)

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359

(b) Right-tailed because H a contains >; z-test because 1 and  2 are known, the samples are random samples, the samples are independent, and n1  30 and n2  30 . (c) z0  1.645; Rejection region: z  1.645 (d) z 

( x1  x2 )  ( 1  2 )



2 1





2 2



(356,889  189,389)  (0) (537, 407) 2 (113,555) 2  64 36

 2.40

(e) Because z  1.645 , reject H0 . (f) There is enough evidence at the 5% level of significance to support the real estate agent’s 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.” H 0 : d  0; H a : d  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  d 1.233  0   4.33 1.207 sd 18 n

(e) Because t  1.740 , reject H0 . (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 in Polk County, Iowa, than it is in Woodward County, Oklahoma.” H 0 : 1  2 ; H a : 1  2 (claim)

(b) Two-tailed because H a contains ; t-test because 1 and  2 are unknown, the samples are random samples, the samples are independent, and the populations are normally distributed. (c) d.f.  min{n1  1, n2  1}  12 t0  3.055, t0  3.055; Rejection regions: t  3.055, t  3.055

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CHAPTER 8 │ HYPOTHESIS TESTING WITH TWO SAMPLES

(d) t 

( x1  x2 )  ( 1  2 ) s12 s22  n1 n2



(61,300  59,800)  (0) (1770) 2 (8350) 2  13 15

 0.68

(e) Because 3.055  t  3.055 , fail to reject H0 . (f) There is not enough evidence at the 1% level of significance to support the demographics researcher’s claim that the mean household income in a recent year is different in Polk County, Iowa, than it is in Woodward County, Oklahoma.

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 attendance per home game tends to increase.

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CHAPTER 9 │ CORRELATION AND REGRESSION

n=7

x 1 10 5 15 3 24 30 å x = 88

r= = =

y 12.5 8.7 14.6 5.2 9.9 3.1 2.7 å y = 56.7

x2 1 100 25 225 9 576 900 2 å x =1836

xy 12.5 87.0 73.0 78.0 29.7 74.4 81.0 å xy = 435.6

y2 156.25 75.69 213.16 27.04 98.01 9.61 7.29 2 å y = 587.05

n å xy - (å x)(å y ) 2

n å x 2 - (å x)

n å y 2 - (å y)

(7)(435.6) - (88)(56.7) 2

(7)(1836) - (88) (7)(587.05) - (56.7) -1940.4 » -0.908 5108 894.46

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. 0.775; Because r is close to 1, this suggest a strong positive linear correlation. As the team salaries increase, the average attendance per home game tends to increase. 6.

n = 7 , a = 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 attendances per home game 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 .

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363

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 r 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 :   0 (no significant linear correlation) H a :   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. Strong negative linear correlation 10. No linear correlation 11. No linear correlation 12. Perfect positive linear correlation 13. Explanatory variable: Amount of water consumed Response variable: Weight loss 14. Explanatory variable: Hours of safety classes Response variable: Number of driving accidents 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 │ CORRELATION AND REGRESSION

20. Answers will vary. Sample answer: People who use one substance are likely to use another substance, since they aren’t as concerned about the impact of a substance on their health. 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  x  41

r r 

y 3 220 540 1,100 2,100 2,600 730 2,200 260 1,200 2,500  y  13, 453

xy 3 440 1,620 4,400 10,500 15,600 2,190 11,000 520 4,800 15,000  xy  66,073

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

 y  25,850,509

 x 181

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, 230 310 103,372,390

 0.979

(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.

CHAPTER 9 │ CORRELATION AND REGRESSION

24. (a)

(b)

96.9 73.1 83.1 96.5 64.1 118.8 71.3 122.7

y 107.8 97.2 95.1 112 78 112 95 118

xy 10,445.82 7,105.32 7,902.81 10,808.00 4,999.80 13,305.60 6,773.50 14,478.60

x2 9,389.61 5,343.61 6,905.61 9,312.25 4,108.81 14,113.44 5,083.69 15,055.29

y2 11,620.84 9,447.84 9,044.01 12,544.00 6,084.00 12,544.00 9,025.00 13,924.00

 x  726.5

 y  815.1

 xy  75,819.45

 x  69,312.31

 y  84,233.69

r r 

n  xy    x   y  n  x2    x 

n  y2    y 

8(74,699.45)  (716.5)(815.1) 8(67, 482.31)  (716.5) 2 8(84, 233.69)  (815.1) 2 13576.45 162.75 97.37

 0.857

(c) Strong positive linear correlation. As the weight increases, the waist tends to increase. (d) There is enough evidence at the 1% level of significance to conclude that there is a significant linear correlation between the weights of males and the circumferences of their waists.

365

366

CHAPTER 9 │ CORRELATION AND REGRESSION

25. (a)

(b)

x 190 185 155 180 175 170 150 160 160 180 190 210 å x =2105 r= = =

y 60 57 54 60 56 64 52 51 49 57 59 64 å y =683

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 2 å x =372,575

xy 11,400 10,545 8370 10,800 9800 10,880 7800 8160 7840 10,260 11,210 13,440 å xy =120,505

y2 3600 3249 2916 3600 3136 4096 2704 2601 2401 3249 3481 4096 2 å y =39,129

n å xy - (å x)(å y ) 2

n å x 2 - (å x)

n å y 2 - (å y )

12 (120,505) - (2105)(683) 2

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.

CHAPTER 9 │ CORRELATION AND REGRESSION

367

26. (a)

(b)

x 175 180 155 210 150 190 185 160 190 180 160 170 å x =2105

r= = =

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 å y =21.41

xy 315 318.6 317.75 298.2 306 305.9 314.5 305.6 304 293.4 316.8 323 å 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 å x2 =372,575

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 å y =38.6309

n å xy - (å x)(å y ) 2

n å x 2 - (å x)

n å y 2 - (å y )

12(3718.75) - (2105)(21.41) 2

12(372,575) - (2105) -443.05 39,875 5.1827

12(38.6309) - (21.41)

» -0.975

(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.

368

CHAPTER 9 │ CORRELATION AND REGRESSION

27. (a)

(b)

x 1.22 4 3.53 8.21 1.74 3.14  x  21.84

r r 

y 0.9 0.31 2.1 1 0.55 1.48  y  6.34

xy 1.098 1.24 7.413 8.21 0.957 4.6472  xy  23.5652

x2 1.4884 16 12.4609 67.4041 3.0276 9.8596  x 2 110.2406

y2 0.81 0.0961 4.41 1 0.3025 2.1904  y 2  8.809

n  xy    x   y  n  x2    x 

n  y2    y 

6(23.5652)  (21.84)(6.34) 6(110.2406)  (21.84) 2 6(8.809)  (6.34)2 2.9256  0.061 184.458 12.6584

(c) No linear correlation; The earnings per share for the companies do not appear to be related to their dividends per share. (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)

CHAPTER 9 │ CORRELATION AND REGRESSION

(b)

x 0 5 10 15 20 25 30 35 40 45 50 å x =275

r= = =

y 1116.3 1096.9 1077.3 1057.2 1036.8 1015.8 994.5 969.0 967.7 967.7 967.7 å 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 å xy =272,636

x2 0 25 100 225 400 625 900 1225 1600 2025 2500 å x2 =9625

369

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 å y =11,571,687.43

n å xy - (å x )(å y ) 2

n å x 2 - (å x)

n å y 2 - (å y )

11(272,636) - (275)(11, 266.9) 2

11(9625) - (275)

11(11,571,687.43) - (11, 266.9)

-99, 401.5 30, 250 345,526.12

» -0.972

(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 correlation coefficient gets stronger, going from r ≈ 0.979 to r ≈ 0.985. 30. The correlation coefficient gets weaker, going from r ≈ 0.857 to r ≈ 0.800. 31. The correlation coefficient gets weaker, going from r ≈ 0.756 to r ≈ 0.666. 32. The correlation coefficient gets stronger, going from r ≈ –0.975 to r ≈ –0.963. 33. r » 0.623 n = 8 and a = 0.01

critical value = 0.834 r » 0.623 < 0.834  The correlation is not significant. OR

370

CHAPTER 9 │ CORRELATION AND REGRESSION

H 0 : r = 0; H a : r ¹ 0 a = 0.01 d.f. = n - 2 = 6 -t0 = -3.707, t0 = 3.707; Rejection regions: t < -3.707 or t > 3.707.

r 2

1- r n-2

0.623 2

1 - (0.623) 8- 2

» 1.951

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. 34. r » 0.955 n = 8 and a = 0.05

critical value = 0.707 r = 0.955 > 0.707  The correlation is significant. OR H 0 : r = 0; H a : r ¹ 0 a = 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 2 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.

1- r 2 n-2

35. r » 0.756 n = 12 and a = 0.05 critical value = 0.576 r =| 0.756 |> 0.576  The correlation is significant.

OR H 0 : r = 0; H a : r ¹ 0 a = 0.05 d.f. = n - 2 = 10 -t0 = -2.228, t0 = 2.228; Rejection regions: t < -2.228 or t > 2.228.

r 2

1- r n-2

0.756 2

1 - (0.756) 12 - 2

» 3.652

CHAPTER 9 │ CORRELATION AND REGRESSION

371

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. 36. r » - 0.975 n = 12 and a = 0.01

critical value = 0.708 r =| -0.975 |> 0.708  The correlation is significant. OR H 0 : r = 0; H a : r ¹ 0 a = 0.01 d.f. = n - 2 = 10 -t0 = -3.169, t0 = 3.169; Rejection regions: t < -3.169 or t > 3.169. t=

-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.

37.

1- r n-2

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 å =21.41 r=

y 175 180 155 210 150 190 185 160 190 180 160 170 y å =2105

xy 315 318.6 317.75 298.2 306 305.9 314.5 305.6 304 293.4 316.8 323 xy å =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 å =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 å 2 =372,575

n å xy - (å x )(å y ) 2

n å x 2 - (å x )

n å y 2 - (å y )

12 (3718.75) - (21.41)(2105) 2

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. =

372

CHAPTER 9 │ CORRELATION AND REGRESSION

9.2 LINEAR REGRESSION 9.2 TRY IT YOURSELF SOLUTIONS 1.

n = 7, å x = 88, å y = 56.7, å xy = 435.6, å x 2 = 1836

(7)(435.6) -(88)(56.7) -1940.4 = »-0.379875 2 5108 n å x - ( å x) (7)(1836) -(88) (88) åy å x (56.7) b = y - mx = -m » - (-0.379875) » 12.8756 m=

n å xy -(å x)(å y) 2

y = -0.380 x +12.876 2.

yˆ  108.022 x  16,586.282

3. (1) y = 12.481(2) + 33.683

(2)

y = 58.645 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

9. e

10. c

11. f

CHAPTER 9 │ CORRELATION AND REGRESSION

373

14. c 15. d 16. a 17.

x 84 96 48 90 79 81 58 64 29

y 13 41 8 38 5 25 1 37 2

xy 1092 3936 384 3420 395 2025 58 2368 58

x2 7056 9216 2304 8100 6241 6561 3364 4096 841

 x  629

 y 170

 xy 13,736

 x 2  47,779

n å xy - (å x)(å y ) 2

n å x - ( å x) 2

9(13,736) - (629)(170) 16,694 = » 0.485714 9(47,779) - (629)2 34,370

æ170 ÷ö æ 629 ÷ö b = y - m x » çç - (0.485714) çç » - 15.0571 çè 9 ÷÷ø çè 9 ÷÷ø

yˆ  0.486 x  15.057

(a) yˆ  0.486(75)  15.057  21 medals (b) yˆ  0.486(80)  15.057  24 medals (c) yˆ  0.486(85)  15.057  26 medals (d) It is not meaningful to predict the value of y for x = 120 because x = 120 is outside the range of the original data.

374

CHAPTER 9 │ CORRELATION AND REGRESSION

x 36 71 100 40 65 90 85

y 1782.5 380 450 525 750 900 1150

xy 64170 26980 45000 21000 48750 81000 97750

x2 1296 5041 10000 1600 4225 8100 7225

 x  487

 y  5,937.5

 xy  384,650

2  x  37, 487

18.

n å xy - (å x)(å y ) 2

n å x - ( å x) 2

(7)(384650) - (487)(5937.5) -199013 = » -7.88481 (7)(37487) - (487)2 25240

æ 5937.5 ö÷ æ 48 7 ö÷ b = y - m x » çç - (- 7.88481) çç » 1 396.771 ÷ ÷ çè 7 ÷÷ø çè 7 ø

yˆ  1396.771  7.885x

(a) y  1396.771  7.885(45)  EGP 1041.9 (b) y  1396.771  7.885(55)  EGP 963.1 (c)

y  1396.771  7.885(95)  EGP 647.7

(d) It is not meaningful to predict the value of y for x = 125 because x = 125 is outside the

range of the original data.

CHAPTER 9 │ CORRELATION AND REGRESSION

19.

x 0 1 1 2 4 6 7 7 8 å x = 36

y 45 55 60 64 68 69 85 94 89 å y = 629

n å xy -(å x)(å y) 2

xy 0 55 60 128 272 414 595 658 712 å xy = 2,894

375

x2 0 1 1 4 16 36 49 49 64 2 å x = 220

(9)(2894) - (36)(629) 3402 = » 4.973684 (9)(220) - (36)2 684

n å x - ( å x) æ 629 ö÷ æ 36 ö - ( 4.973684) çç ÷÷÷ » 49.9942 b = y - mx » çç çè 9 ÷÷ø çè 9 ø y = 4.974 x + 49.994 2

(a) y = 4.974(3.5) + 49.994 » 67 (b) y = 4.974(5) + 49.994 » 75 (c) y = 4.974(7.5) + 49.994 » 87 (d) It is not meaningful to predict the value of y for x = 14 because x = 14 is outside the range of the original data.

376

20.

CHAPTER 9 │ CORRELATION AND REGRESSION

x 85 86 80 78 77 54 62 41 55 43  x  661

y 30 26 23 22 23 18 17 12 12 12  y 195

xy 2,550 2,236 1,840 1,716 1,771 972 1,054 492 660 516  xy 13,807

x2 7,225 7,396 6,400 6,084 5,929 2,916 3,844 1,681 3,025 1,849 2  x  46,349

n å xy -(å x)(å y ) 2

n å x 2 - ( å x) 10(13,807)  (661)(195) 9,175 m   0.345 ; 10(46,349)  (661) 2 26,569

b = y - mx =

b

åy åx -m n n

195  9,175   661      3.326 10  26,569   10 

y  0.345x  3.326

(a) y  0.345(50)  3.326  14 wins (b) y  0.345(70)  3.326  21 wins (c) y  0.345(75)  3.326  23 wins (d) It is not meaningful to predict the value of y from x  95 because x  95 is outside the range of the observed data.

CHAPTER 9 │ CORRELATION AND REGRESSION

21.

x 60 75 62 68 84 97 66 65 86 78 93 75 88 å x =997

m= =

y 403 363 381 367 341 317 401 384 342 377 329 377 349 å 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 å xy =359,204

377

x2 3600 5625 3844 4624 7056 9409 4356 4225 7396 6084 8649 5625 7744 2 å x =78, 237

x å xy - (å x)(å y ) 2

n å x 2 - (å x)

(13)(359, 204) - (997)(4731) 2 (13)(78, 237) - (997)

-47,155 » -2.0438 23,072 æ 4731÷ö æ 997 ö÷ - (-2.04382)çç b = y - mx » çç ÷ ÷ » 520.6683 ÷ èç 13 ø èç 13 ÷ø 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

378

CHAPTER 9 │ CORRELATION AND REGRESSION

22.

x 137 168 152 145 159 159 124 137 155 148 147 146 å x =1777

y 106 130 116 106 125 119 103 104 120 110 107 109 å y =1355

x å xy -(å x)(å y) 2

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 å xy =201,703

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 2 å x =264,683

(12)(201,703) -(1777)(1355) 12,601 = » 0.682352 2 18,467 (12)(264,683) -(1777)

n å x 2 - ( å x) æ1355 ö÷ æ1777 ö÷ - (0.682352)çç b = y - mx » çç ÷ ÷ » 11.87166 ÷ èç 12 ø èç 12 ø÷ 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

CHAPTER 9 │ CORRELATION AND REGRESSION

23.

x 180 220 230 90 160 190 150 110 110 160 140 150

y 510 740 740 280 530 580 490 480 330 640 480 460

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

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  1,890

 y  6, 260

 xy  1, 046, 200

 x  317,900

379

n å xy -(å x)(å y ) 2

n å x 2 - ( å x) 12(1,046, 200)  (1,890)(6, 260) 723,000 m   2.979 ; 12(317,900)  (1,890) 2 242,700

b = y - mx =

b

åy åx -m n n

6,260  723,000   1,890     52.476  12  242,700   12 

y  2.979 x  52.476

(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. (d) y  2.979(210)  52.476  678 milligrams

380

CHAPTER 9 │ CORRELATION AND REGRESSION

x 1,800 8,300 45,000 7,300 52,000 10,000 20,200 13,700 5,200 24,600 19,900 4,000 18,800

y 925 1,271 4,429 1,006 9,455 1,811 4,519 1,452 1,601 4,466 2,184 1,309 3,034

xy 1,665,000 10,549,300 199,305,000 7,343,800 491,660,000 18,110,000 91,283,800 19,892,400 8,325,200 109,863,600 43,461,600 5,236,000 57,039,200

x2 3,240,000 68,890,000 2,025,000,000 53,290,000 2,704,000,000 100,000,000 408,040,000 187,690,000 27,040,000 605,160,000 396,010,000 16,000,000 353,440,000

 x  230,800

 y  37, 462

 xy  1, 063, 734,900

 x  6,947,800,000

24.

m

n å xy -(å x)(å y ) 2

n å x 2 - ( å x)

13(1,063,734,900)  (230,800)(37,462) 5,182,324,100   0.140 ; 37,052,760,000 13(6,947,800,000)  (230,800)2

b = y - mx =

b

åy åx -m n n

37,462  5,182,324,100   230,800     398.579  13  37,052,760,000   13 

y  0.140 x  398.579

(a) y  0.140(32,500)  398.579  4949 or $4,949,000,000 (b) y  0.140(6000)  398.579  1239 or $1,239,000,000 (c) It is not meaningful to predict the value of y for x  1350 because x  1350 is outside the

range of the observed data. (d) y  0.140(47,000)  398.579  6978.579 or $6,979,000,000

CHAPTER 9 │ CORRELATION AND REGRESSION

25.

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 å =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 å y =993

x å xy -(å x)(å y) 2

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 å =10,427

381

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 å 2 =1552.25

(14)(10,427) -(146.5)(993) 503.5 = » 1.870009 2 269.25 (14)(1552.25) -(146.5)

n å x - ( å x) æ 993.0 ö÷ æ146.5 ö÷ - 1.870)çç b = y - mx » çç ÷ » 51.3603 çè 14 ø÷÷ ( èç 14 ø÷ y = 1.870 x + 51.360 2

(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

382

CHAPTER 9 │ CORRELATION AND REGRESSION

x 0.1 0.2 0.4 0.7 0.6 0.9 0.1 0.2 0.4 0.9

y 14.5 14.3 14.1 13.9 13.9 13.7 14.3 14.2 14 13.8

xy 1.45 2.86 5.64 9.73 8.34 12.33 1.43 2.84 5.6 12.42

x2 0.01 0.04 0.16 0.49 0.36 0.81 0.01 0.04 0.16 0.81

 x  4.5

 y  140.7

 xy  62.64

 x  2.89

26.

n å xy -(å x)(å y ) 2

n å x 2 - ( å x) 10(62.64)  (4.5)(140.7) 6.75 m   0.780 ; 10(2.89)  (4.5) 2 8.65

åy åx -m n n 140.7  6.75  4.5  b     14.421 10  8.65  10 

b = y - mx =

y  0.780 x  14.421

(a) y  0.780(0.3)  14.421  14.2 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  14.0 hours (d) y  0.780(0.8)  14.421  13.8 hours 27. Strong positive linear correlation; As the years of experience of the registered nurses increase, their salaries tend to increase.

CHAPTER 9 │ CORRELATION AND REGRESSION

x 0.5 2 4 5 7 9 10 12.5 13 16 18 20 22 25

y 45.2 49.9 54.7 59.3 61.4 62.9 66 67.1 65.3 68.4 70.6 69.5 73.9 71.6

xy 22.6 99.8 218.8 296.5 429.8 566.1 660 838.75 848.9 1,094.4 1,270.8 1,390 1,625.8 1,790

x2 0.25 4 16 25 49 81 100 156.25 169 256 324 400 484 625

 x  164

 y  885.8

 xy  11,152.25

 x  2,689.5

28.

383

n å xy -(å x)(å y ) 2

n å x 2 - ( å x) 14(11,152.25)  (164)(885.8) 10,860.3 m   1.010 14(2,689.5)  (164) 2 10,757

b = y - mx =

b

åy åx -m n n

885.8  10,860.3   164     51.445  14  10,757   14 

y  1.010 x  51.445

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. Since the calculated t-value of approximately 8.12 is greater than the critical t-value for a two-tailed test at 0.01 level of significance and 12 degrees of freedom, namely 0.661, the null hypothesis of no significant.

384

CHAPTER 9 │ CORRELATION AND REGRESSION

31. (a)

(b)

y = -4.297 x + 94.200

y =-0.141x +14.763

(b)

y = 0.453x - 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 │ CORRELATION AND REGRESSION

34. (a) m » 1.711

385

(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 = 1.711 x + 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.

386

CHAPTER 9 │ 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 y log y 1 165 2.218 2 280 2.447 3 468 2.670 4 780 2.892 5 1310 3.117 6 1920 3.283 7 4900 3.690 log y = 0.233 x + 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 │ CORRELATION AND REGRESSION

42.

x 1 2 3 4 5 6 7 8

y 695 410 256 110 80 75 68 74

log x 0 0.301 0.477 0.602 0.699 0.778 0.845 0.903

387

log y 2.842 2.613 2.408 2.041 1.903 1.875 1.833 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.

388

CHAPTER 9 │ CORRELATION AND REGRESSION

9.3 MEASURES OF REGRESSION AND PREDICTION INTERVALS 9.3 TRY IT YOURSELF SOLUTIONS

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. r » 0.979

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 å = 231.947818

n=8

se =

(

å yi -  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, t c = 2.306, se » 141.935

( y  199.535(4)  56.036  854.176 ) 2 n ( x - x) 1 E = tc se 1 + + n n å x 2 - ( å x )2

(

E  (2.306)(141.935) 1 

)

1 10(4  2.29) 2   376.624 10 10(65.49)  (22.9) 2

y  E  854.176  376.624  477.552 y  E  854.176  376.624  1230.8 You can be 95% confident that when the gross domestic product is $4 trillion, the carbon dioxide emissions will be between 477.552 and 1230.8 million metric tons.

CHAPTER 9 │ CORRELATION AND REGRESSION

389

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 å ( yi - y ) . 2

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 å 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 å y - y .

(

4. The coefficient of determination r 2 =

)

(

å y i - y

) is the ratio of the explained variation to the total 2

å ( yi - y )

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. 2

r 2 = (-0.328) » 0.108; About 10.8% of the variation is explained. About 89.2% of the variation is unexplained. 2 9. r 2 = (-0.957) » 0.916; About 91.6% of the variation is explained. About 8.4% of the variation is unexplained. 8.

10. r 2 = (0.881) » 0.776; About 77.6% of the variation is explained. About 22.4% of the variation is unexplained.

( 11. (a) r = 2

å y i - y

) » 0.798 2

å ( yi - y )

About 79.8% of the variation in proceeds can be explained by the relationship between the number of offerings and proceeds, and about 20.2% of the variation is unexplained.

390

CHAPTER 9 │ CORRELATION AND REGRESSION

(b) se =

(

å yi -  yi

) = 648,721,935 » 8054.328 2

n-2 10 The standard error of estimate of the proceeds for a specific number of offerings is about $8,054,328,000.

12. (a)

( r2 =

å yi - y

) » .947

2 å ( yi - y )

About 94.7% of the variation in the median annual earnings of female workers can be explained by the relationship between the median annual earnings of male and female workers, and about 5.3% of the variation is unexplained

(b) se =

(

yi å yi - 

) = 15,607,927.36 » 1396.779

8 n-2 The standard error of estimate of the median annual earnings of female workers for a specific median annual earnings of a male worker is about $1396.78.

13. (a)

( r2 =

å yi - y

) » .730

2 å ( yi - y )

About 73% of the variation in points earned can be explained by the relationship between the number of goals allowed and points earned, and about 27% of the variation is unexplained.

(b) se =

(

yi å yi - 

) = 1068.938017 » 9.438

12 n-2 The standard error of estimate of the points earned for a specific number of goals allowed is about 9.438.

( 14. (a) r = 2

å y i - y

) » 0.670 2

å ( yi - y )

About 67.0% of the variation in the trunk diameters can be explained by the relationship between tree heights and the trunk diameters, and about 33.0% of the variation is unexplained.

(b) se =

(

å yi -  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 │ CORRELATION AND REGRESSION

15. (a)

( r2 =

å yi - y

391

) » 0.651

2 å ( yi - y )

About 65.1% of the variation in mean annual wages can be explained by the relationship between percentages of employment in STEM occupations and mean annual wages, and about 34.9% of the variation is unexplained.

(b) se 



yi  yi  

  927.7893389  8.141 .

14 n2 The standard error of estimate of the mean annual wages for a specific percentage of employment in STEM occupations is about $8141.

16. (a) r 2 =

(

å y i - y

) » 0.995 2

å ( yi - y )

About 99.5% of the variation in the turnout for democratic elections can be explained by the relationship between the voting age population and turnout for democratic elections, and about 0.5% of the variation is unexplained.

(b) se 



  0.0896  0.1131 . 2

 yi   yi

7 n2 The standard error of estimate of the turnout in democratic elections for a specific voting age population is about 113,100 people.

17. (a) r =

(

å y i - y

) » 0.642 2

å ( yi - y )

About 64.2% of the variation in the quantity of wheat exported can be explained by the relationship between the quantity of wheat produced and quantity of wheat exported, and about 35.8% of the variation is unexplained.

(b) se 



 yi   yi

  13672345  1653.623 . 2

n2 5 The standard error of estimate of the quantity of wheat exported for a specific quantity of wheat produced is about 1,653,623,000 kilograms per year.

( ) » 0.916 18. (a) r = å ( y - y) 2

å y i - y

About 91.6% of the variation in asses in federal pension plans can be explained by the relationship between IRA assets and federal DB plan assets, and 8.4% of the variation is unexplained.

392

CHAPTER 9 │ CORRELATION AND REGRESSION

(b) se 



  32,679.04329  63.913 . 2

 yi   yi

8 n2 The standard error of estimate of assets in federal DB plans for a specific total of IRA assets is about $63,913,000,000.

( ) » 0.816 19. (a) r = å ( y - y) 2

å y i - y

About 81.6% of the variation in the new-vehicle sales of General Motors can be explained by the relationship between the new-vehicle sales of Ford and General Motors, and about 18.4% of the variation is unexplained.

(b) se 



  1,079,571.213  346.341 . 2

 yi   yi

n2 9 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 346,341 new vehicles.

( ) » 0.894 20. (a) r = å ( y - y) 2

å y i - y

About 89.4% of the variation in the new-vehicle sales of Honda can be explained by the relationship between the new-vehicle sales of Toyota and Honda, and 10.6% of the variation is unexplained.

(b) se 



 yi   yi

  27,834.71877  55.612 . 2

9 n2 The standard error of estimate of the new-vehicle sales of Honda for a specific amount of newvehicle sales of Toyota is about 55,612 new vehicles.

21. n = 12, d.f. = 10, tc = 2.228, se » 8054.328 y = 104.965x +14,093.666 = 104.965(450) +14,093.666 = 61,327.916

n ( x - x) 1 E = tc se 1 + + n n (å x 2 ) - (å x)2 2

» (2.228)(8054.328) 1 +

12(450 - 2138 / 12) 1 + 12 12 (613,126) - (2138)2

» 21, 244.664

y  E  61,327.916  21,244.664  (40,083.251, 82,572.581) You can be 95% confident that the proceeds will be between $40,083,251,000 and $82,572,581,000 when the number of initial offerings is 450.

CHAPTER 9 │ CORRELATION AND REGRESSION

393

22. n = 10, d.f. = 8, tc = 2.306, se » 1396.779

y  1.005(45,637)  10,770.313  35,094.872 ). 2 n ( x - x) 1 E = tc se 1 + + n n å x 2 - ( å x )2

(

)

E  (2.306)(1396.779) 1 



10(45, 637  48, 406.3) 2

10 10(23, 710, 959, 939)  (484, 063) 2

 3420.093067 )

y  E  35, 094.872  3420.093067  31,674.779 ; y  E  35, 094.872  3420.093067  38,514.965 . 31,674.779  y  38,514.965 You can be 95% confident that the median annual earnings of female workers will be between $31,674.779 and $38 514.965 when the median annual earnings of a male worker is $45,637. 23. n = 14, d.f. = 12, tc = 1.782, se » 9.438

y  0.573(250)  220.087  76.837 2 n ( x - x) 1 E = tc se 1 + + n n å x 2 - ( å x )2

(

E  (1.782)(9.438) 1 

)

1 14



14(250  228.6428571) 2  17.826 14(740, 657)  (3, 201) 2

y  E  76.837  17.826  59.011 ; y  E  76.837  17.826  94.663 . 59.011  y  94.663 You can be 90% confident that the total points earned will be between 59.011 and 94.663 when the number of goals allowed by the team is 250. 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

n ( x - x) 1 E = tc se 1 + + n n (å x 2 ) - (å x)2 2

8(80 - 615 / 8) 1 » (1.943)(1.780) 1 + + 8 8(47447) - (615)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.

394

CHAPTER 9 │ CORRELATION AND REGRESSION

25. n = 16, d.f. = 14, tc = 2.977, se » 8.141 y  1.153(13)  46.374  61.363 n ( x - x) 1 E = tc se 1 + + n n (å x 2 ) - (å x )2 2

1 16(13  9.38125)2   25.1 16 16(2,712.05)  (150.1)2 y  E  61.363  25.1  36.263 ; y  E  61.363  24.974  86.462

E  (2.977)(8.141) 1 

36.263  y  86.462 You can be 99% confident that the mean annual wage is between $36,263 and $86,462 when the percentage of employment in STEM occupations is 13% in the industry. 26. n = 9, d.f. = 7, tc = 3.499, se » 0.1131 y  0.8194 x  1.842  0.8194 15  1.842  14.133 . n ( x - x) 1 1 9(15 -137.2 / 9) 2 » + + » 0.418 (3.499)(0.1131) 1 E = tc se 1 + + 9 9(2,116.04) - (137.2) 2 n n (å x 2 ) - (å x )2 2

y  E  14.133  0.418  (13.715,14.551) . You can be 99% confident that the voter turnout in federal election will be in between 13.715 million and 14.551 million when the voting age population is 15 million. 27. n = 7, d.f. = 5, tc = 1.476, se » 1653.623 y  0.249 x  19023  0.249  99000  19023  5628 .

n ( x - x) 1 E = tc se 1 + + n n (å x 2 ) - (å x )2 2

1 7(99000 - 566971 / 7) 2 » (1.476)(1653.623) 1 + + » 3418.581 7 7(46318135431) - (566971) 2 y  E  5628  3418.581   2209.419,9046.581 . You can be 80% confident that the quantity of wheat exported will be in between 2209.419 million kilograms and 9046.581 million kilograms when the quantity of wheat produced is 99000 million kilograms per year.

28. n = 10, d.f. = 8, tc = 1.86, se » 63.913 y  0.140(6200)  453.959  1321.959

CHAPTER 9 │ CORRELATION AND REGRESSION

395

n ( x - x) 1 E = tc se 1 + + n n (å x 2 ) - (å x )2 2

1 10(6200  5817.4)2   125.137 10 10(356,554,650)  (58,174)2 y  E  1321.959  125.137  1196.822 ; y  E  1321.959  125.137  1447.096 .

E  (1.860)(63.913) 1 

1196.822  y  1447.096 You can be 90% confident that the total assets in federal defined benefit plans is between $1,196,822,000,000 and $1,447,096,000,000 when the total assets in IRAs is $6200 billion. 29. n = 11, d.f. = 9, tc = 2.262, se » 346.341 y  1.624(2628)  747.304  3520.568 . n ( x - x) 1 E = tc se 1 + + n n (å x 2 ) - (å x )2 2

1 11(2628  2334.454545)2   835.856 . 11 11(61,763,389)  (25,679)2 y  E  3520.568  835.856  2684.712 ; y  E  3520.568  835.856  4356.424 . E  (2.262)(346.341) 1 

2684.712  y  4356.424 You can be 95% confident that the new-vehicle sales for General Motors is between 2,684,712 and 4,356,424 when the number of new vehicles sold by Ford is 2628 thousand. 30. n = 11, d.f. = 9, tc = 3.250, se » 55.612 y  0.460(2359)  410.839  1495.979 .

1 11(2359  2183)2  191.17 . E  (3.250)(55.612) 1   11 11(53,533,697)  (24,013)2 y  E  1495.979  191.17  1304.809 ; y  E  1495.979  191.17  1687.149 . 1304.809  y  1687.149 You can be 99% confident that new-vehicle sales for Honda is between 1, 304,809 and 1, 687,149 when the number of new vehicles sold by Toyota is 2359 thousand. 31.

396

CHAPTER 9 │ CORRELATION AND REGRESSION

32. y  1.45 x  5.21

33. r 2 » 0.987; About 98.7% of the variation in the median ages of light trucks can be explained by the relationship between the median ages of cars and light trucks, and about 1.3% of the variation is unexplained.

34. se 



 yi   yi

  0.0355  0.077 . The standard error of estimate of the median age of light 2

6 n2 trucks for a specific median age of cars is about 0.077 year.

35. critical value =  3.707 m » -0.205

se » 0.554 2

(å x) -0.205 (79.5) m å x2 » » -2.578 838.55 se n 0.554 8 Because - 3.707 < t < 3.707 , fail to reject H 0 : 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. t=

36. critical value =  2.306 m » 1.328

se » 5.086 2

(å x) 1.328 (374) m å x2 » » 6.771 t= 14,660 se n 5.086 10 Because t > 2.306 , reject H 0 : 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 (2.29)2 2.306 (141.935)     231.8716672 . 37. E  tc se   (22.9)2 n  x 2    x  2 / n  10 65.49    10 b  E  56.036  231.8716672  287.908 ; b  E  56.036  231.8716672  175.836 . tc se (2.306)(141.935) E   90.60667419 2 (22.9) 2  x  2 65.49  x  10 n

CHAPTER 9 │ CORRELATION AND REGRESSION

m  E  199.535  90.60667419  290.142 ; m  E  199.535  90.60667419  108.928 . Confidence intervals for B is 175.836  B  287.908 ; Confidence intervals for M is 108.928  M  290.142 2

1 x 1 (2.29)2 3.355 (141.935)     337.350149 . 38. E  tc se   (22.9)2 n  x 2    x  2 / n  10 65.49    10 b  E  56.036  337.350149  393.386 ; b  E  56.036  337.350149  281.314 . tc se (3.355)(141.935) E   131.8236739 2 2 (22.9)  x 65.49   x2  10 n m  E  199.535  131.8236739  331.359 ; m  E  199.535  131.8236739  67.711 . Confidence intervals for B is 281.314  B  393.386 ; Confidence intervals for M is 67.711  M  331.359

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

y = 24,791 + 4.508 x - 4.723x 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

397

398

CHAPTER 9 │ CORRELATION AND REGRESSION

(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 2.

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

y = -52.2 + 0.3x + 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

y = -4016 +11.5x + 7.55x +12.5x 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 (d) y = -4016 +11.5(231) + 7.55(135) +12.5(86) = 734.75 kilograms

5. (a) y  17,899  606.58 x1  52.9 x2

(b) se = 564.314; ; The standard error of estimate of the predicted price given specific age and mileage of pre-owned Honda Civic Sedans is about $564.31.

CHAPTER 9 │ CORRELATION AND REGRESSION

399

(c) r 2 = 0.966; ; The multiple regression model explains about 96.6% of the variation. 6. (a) y  29.69  0.229 x1  0.004 x2

(b) se = 1.847; The standard error of estimate of the predicted shareholder’s equity given specific net sales and total assets for Wal-Mart is about $1.847 billion. (c) r 2 = 0.843; The multiple regression model explains about 84.3% of the variation. 7.

n = 9 k = 2, r 2 = 0.966  (1  0.966)(9  1)  r 2 adj  1     0.955 9  2 1   About 95.5% of the variation in y can be explained by the relationship between variables; r 2 adj  r 2 .

n = 6 k = 2, r 2 = 0.843  (1  0.843)(6  1)  r 2 adj  1     0.738 6  2 1   About 73.8% 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)

(b)

x 610 545 567 552 432 486 403

y 4,428 4,240 4,090 3,877 3,554 3,401 2,710

xy 2,701,080 2,310,800 2,319,030 2,140,104 1,535,328 1,652,886 1,092,130

x2 372,100 297,025 321,489 304,704 186,624 236,196 162,409

 x  3595

 y  26,300

 xy 13,751,358

 x 1,880,547

y2 19,607,184 17,977,600 16,728,100 15,031,129 12,630,916 11,566,801 7,344,100 2

 y 100,885,830

400

CHAPTER 9 │ CORRELATION AND REGRESSION

r

n  xy    x   y  n  x2    x 

n  y2    y 



7(13,751,358)  (3595)(26,300) 7(1,880,547)  (3595)2 7(100,885,830)  (26,300)2



1,711,006  0.917 239,804 14,510,810

(c) Strong positive linear correlation; As the number of pass attempts increase, the number of passing yards tends to increase 2. (a)

(b)

x 78.8 72 74.1 67.8 47.6 63.3 68.2 67.7

r  

y 5.9 3.4 8.7 9.3 4.3 3.6 10.1 5.5

xy 464.92 244.8 644.67 630.54 204.68 227.88 688.82 372.35

 x  539.5  y  50.8  xy  3,478.66 n  xy    x   y  n  x2    x 

n  y2    y 

x2 6,209.44 5,184 5,490.81 4,596.84 2,265.76 4,006.89 4,651.24 4,583.29 2

 x  36,988.27

y2 34.81 11.56 75.69 86.49 18.49 12.96 102.01 30.25 2

 y  372.26

8(3478.66)  (539.5)(50.8) 8(36988.27)  (539.5)2 8(372.26)  (50.8) 2 422.68  0.305 4845.91 397.44

(c) Weak positive linear correlation; The number of wildland acres burned does not appear to be related to the number of wildland fires.

CHAPTER 9 │ CORRELATION AND REGRESSION

3. (a)

(b)

x 138 140 96 83 101 135 85 77 88 å x =943 r= = =

y 991 856 879 865 808 791 799 794 894 å y =7677

x2 19,044 1960 9216 6889 10,201 18,225 7225 5929 7744 2 å x =104,073

xy 136,758 119,840 84,384 71,795 81,608 106,785 67,915 61,138 78,672 å xy =808,895

y2 982081 732736 772641 748225 652864 625681 638401 630436 799236 2 å y =6,582,301

n å xy - (å x )(å y ) 2

n å x 2 - (å x )

n å y 2 - (å y )

9 (808,865) - (943)(7677) 2

9 (104,073) - (943)

9 (6,582,301) - (7677)

40,644 » 0.338 47, 408 304,380

401

402

CHAPTER 9 │ CORRELATION AND REGRESSION

(b)

x 2.1 5.0 6.3 6.5 7.7 8.7 11.6 å x =47.9

r= = =

y 0.59 1.51 1.55 1.70 2.18 2.10 2.73 å y =12.36

xy 1.239 7.55 9.765 11.05 16.786 18.27 31.668 å xy =96.328

x2 4.41 25.00 39.69 42.25 59.29 75.69 134.56 å x2 =380.89

y2 0.3481 2.2801 2.4025 2.8900 4.7524 4.4100 7.4529 å y 2 =24.536

n å xy - (å x)(å y ) 2

n å x 2 - (å x)

n å y 2 - (å y )

7 (96.328) - (47.9)(12.36) 2

7 (380.89) - (47.9)

7 (24.536) - (12.36)

82.252 » 0.979 371.82 18.9824

H 0 : r = 0; H a : r ¹ 0 a = 0.05, d.f. = n - 2 = 5 -t0 = -2.571, t0 = 2.571; Rejection regions: t < -2.571 or t > 2.571 t=

1- r n-2

0.917

» 5.14 2 1 - (0.917) 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. 2

H 0 : r = 0; H a : r ¹ 0 a = 0.05, d.f. = n - 2 = 6 -t0 = -2.447, t0 = 2.447; Rejection regions: t < - 2.447 or t > 2.447 t=

r 2

1- r n-2

0.305 2

1 - (0.305) 8- 2

» 0.784

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.

CHAPTER 9 │ CORRELATION AND REGRESSION

403

H 0 : r = 0; H a : r ¹ 0

a = 0.01, d.f. = n - 2 = 7 -t 0 = - 3.499, t0 = 3.499; Rejection regions: t < -3.499 or t > 3.499

0.338

» 0.950 2 1- r 1 - (0.338) n-2 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 and brain size. 2

H 0 : r = 0; H a : r ¹ 0

a = 0.01, d.f. = n - 2 = 5 -t0 = - 4.032, t 0 = 4.032; Rejection regions: t < - 4.032 or t > 4.032

0.979

» 10.738 2 1- r 1 - (0.979) n-2 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. 2

y  0.106 x  781.327

(a) It is not meaningful to predict the value of y for x  9080 because x  9080 is outside the range of the observed data. (b) y  0.106(9230)  781.327  197.053 billions of pounds (c) y  0.106(9250)  781.327  199.173 billions of pounds (d) y  0.106(9300)  781.327  204.473 billions of pounds

404

CHAPTER 9 │ CORRELATION AND REGRESSION

10. y  0.866 x  0.059

(a) y  0.866(2.85)  0.059  2.4091 hours (b) y  0.866(2.97)  0.059  2.51302 hours (c) y  0.866(3.04)  0.059  2.57364 hours (d) It is not meaningful to predict the value of y when x = 3.13 because x = 3.13 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 │ CORRELATION AND REGRESSION

405

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. 2

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.

 17. (a) r  2

 yi  y



 yi  y

  0.690 . 2



About 69.0% of the variation in top speed for hybrid and electric cars can be explained by the relationship between their fuel efficiencies and top speeds, and about 31.0% of the variation is unexplained.

406

CHAPTER 9 │ CORRELATION AND REGRESSION



 yi   yi

(b) se 

  239.665646  5.851 . 2

7 n2 The standard error of estimate of the top speed for hybrid and electric cars for a specific fuel efficiency is about 5.851 miles per hour.

 18. (a) r  2

 yi  y



 yi  y

  0.565 2



About 56.5% of the variation in the price of gas grills can be explained by the relationship between the cooking areas and their prices, and about 43.5% of the variation is unexplained.

(b) se 



 yi   yi

  2,077,616.611  360.35 . 2

16 n2 The standard error of estimate of the price of a gas grill for a specific cooking area is about $360.35.

19. y  0.106(9275)  781.327  201.823 n ( x - x) 1 1 8(9275 - 9235.5) 2 = + + (1.943)(4.014) 1 E = tc se 1 + + 8 8(682,386,108) - (73,884) 2 n n (å x 2 ) - (å x )2 2

» 8.459

y + E = 201.823 + 8.459 = 210.282 ; y - E = 201.823 - 8.459 = 193.364 You can be 90% confident that the amount of milk produced will be between 193.364 billion pounds and 210.282 billion pounds when the average number of cows is 9275 thousand. 20. y  0.866(3.08)  0.059  2.60828 .





n xx 1 1 10(3.08  2.984)2 E  tc se 1       0.080 (1.860)(0.039) 1 n n   x 2     x 2 10 10(89.1164)  (29.84)2

y  E  2.608  0.080  2.528 ; y  E  2.608  0.080  2.688 . 2.528  y  2.688 You can be 90% confident that the average time women spend per day watching television will be between 2.528 and 2.688 hours when the average time men spend per day watching television is 3.08 hours. 21. y =-0.086(45) +10.450 = 6.580 2 n ( x - x) 7 (45 - 337 / 7) 1 1 E = tc se 1 + + » + + 2.571 0.622 1 ( )( ) n n (å x 2 ) - (å x )2 7 7 (18,563) - (337)2 2

» 1.714

CHAPTER 9 │ CORRELATION AND REGRESSION

407

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 E = tc se 1 + + » (2.571)(1.476) 1 + + n n (å x 2 ) - (å x )2 7 7 (369,382) - (1516)2 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  0.465(90)  139.433  97.583

1 9(90  103)2 E  (3.499)(5.851) 1    22.234 9 9(97,955)  (927)2 y  E  97.583  22.234  75.349 ; y  E  97.583  22.234  119.817 . 75.349  y  119.817 You can be 99% confident that the top speed of a hybrid or electric car will be between 75.349 and 119.817 miles per hour when the combined city and highway fuel efficiency is 90 miles per gallon equivalent. 24. y  2.335(900)  853.278  1248.222 n ( x - x) 1 1 18(900 - 624.1667) 2 E = tc se 1 + + » + + (2.921)(360.348) 1 n n (å x 2 ) - (å x )2 18 18(7,508,057) - (11, 235) 2 2

» 1157.40

y  E 1248.222 1157.40 (90.822,2405.622) You can be 99% confident that the price of a gas grill will be between $90.822 and $2405.622 when the cooking area is 900 square inches. 25. (a) y = 3.6738 + 1.2874 x1 - 7.531x2

(b) se » 0.710; The standard error of estimate of the carbon monoxide content given specific tar and nicotine contents is about 0.710 milligram. (c) r 2 » 0.943; The multiple regression model explains about 94.3% of the variation in y.

408

CHAPTER 9 │ CORRELATION AND REGRESSION

26. (a) y = 4021 + 2.264 x1 -1.9933x2

(b) se » 1205.22; The standard error of estimate of the annual yield of spinach given specific acres planted and acres harvested is about 1205.22 pounds. (c) r 2 » 0.705; The multiple regression model explains about 70.5% of the variation in y. 27. (a) 21.705 miles per gallon

(b) 25.21 miles per gallon

(d) 25.86 miles per gallon

28. y  3.67  1.29 x1  7.53x2 : (a) 11.3 milligrams; y  3.67  1.29(10)  7.53(0.7)  11.3

(b) 14.7 milligrams; y  3.67  1.29(15)  7.53(1.1)  14.7 (c) 12.9 milligrams; y  3.67  1.29(13)  7.53(1.0)  12.9 (d) 9.3 milligrams; y  3.67  1.29(9)  7.53(0.8)  9.3

CHAPTER 9 QUIZ SOLUTIONS 1.

The data appear to have a positive linear correlation. As x increases, y tends to increase. 2.

r » 0.992; ; 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.

H 0 : r = 0; H a : r ¹ 0 a = 0.05, d.f. = n - 2 = 9 -t0 = -2.262, t0 = 2.262; Rejection regions: t < - 2.262 or t > 2.262 t=

1- r n-2

0.992

» 23.57 2 1 - (0.992) 11 - 2 Because t > 2.262 , reject H 0 . 2

CHAPTER 9 │ CORRELATION AND REGRESSION

409

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. 4.

x 51.2 52.5 54.4 55.2 56 56.8 57.8 58.3 59.3 60.4 61.4 å x = 623.3

y 48.7 50 52.2 53.2 54.3 55.3 56.1 56.3 56.8 57.7 59

xy 2493.44 2625 2839.68 2936.64 3040.8 3141.04 3242.58 3282.29 3368.24 3485.08 3622.6

x2 2621.44 2756.25 2959.36 3047.04 3136 3226.24 3340.84 3398.89 3516.49 3648.16 3769.96

å y = 599.6

å xy = 34077.39

å x 2 = 35420.67

n å xy -(å x)(å y) 2

n å x 2 - ( å x)

b = y - mx =

11(34,077.39) - (623.3)(599.6) » 0.997 11(35,420.67) - (623.3)2

æ å x ÷ö 599.6 æç1,120.61ö÷æ 623.3 ÷ö åy - m çç » -ç ÷çç ÷ » -1.960 çè n ÷÷ø n 11 èç1,124.48 ÷øçè 11 ÷ø

y  0.997 x  1.960

y  0.997(52.5)  1.960  50.3825 $ 50,382.5

r 2 » (0.9921) » 0.984 About 98.4% 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 1.6% of the variation is unexplained.

se » 0.422;

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 $422.

410

CHAPTER 9 │ CORRELATION AND REGRESSION

y  0.997(52.5)  1.960  50.3825 . se 

1.61271403  0.423 . 9

E  (2.262)(0.423) 1 

1 11(52.5  56.66363636)2   1.074 11 11(35,420.67)  (623.3)2

y  E  50.3825  1.074  49.31 ; y  E  50.3825  1.074  51.46 49.31  y  51.46 You can be 95% confident that the average annual salary of elementary school teachers is between $49,310 and $51,460 when the average annual salary of secondary school teachers is $52,500. 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 2.

CHAPTER 9 │ CORRELATION AND REGRESSION

411

r » 0.949; ; strong positive linear correlation; As the average annual salaries of librarians increase, the average annual salaries of library science teachers tend to increase.

H 0 : r = 0; H a : r ¹ 0 a = 0.05, d.f. = n - 2 =10 -t0 = -3.169, t0 = 3.169; Rejection regions: t < - 3.169 or t > 3.169 r 0.949 t= » » 9.52 2 2 1- r 1 - (0.949) n-2 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. x 49.1 50.9 52.9 54.7 55.7 56.4 57 57.2 57.6 58.1 58.9 59.9

y 56.6 57.6 59.7 61.6 64.3 67 70 70.8 73.3 72.4 73 72.3

xy 2779.06 2931.84 3158.13 3369.52 3581.51 3778.8 3990 4049.76 4222.08 4206.44 4299.7 4330.77

x2 2410.81 2590.81 2798.41 2992.09 3102.49 3180.96 3249 3271.84 3317.76 3375.61 3469.21 3588.01

 x  668.4

 y  798.6

 xy  44697.61

 x  37347

m

n  xy    x   y  12(44,697.61)  (668.4)(798.6)   1.841 ; 2 12(37,347)  (668.4)2 n  x2    x 

b  y  mx 

y   x  798.6  2,587.08   668.4   m     35.980  n 12  n   1, 405.44   12 

y  1.841x  35.98

412

CHAPTER 9 │ CORRELATION AND REGRESSION

y  1.841(56)  35.98  67.116 $67,116.

r 2 » (0.9497 ) » 0.902 About 90.2% 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 9.8% of the variation is unexplained.

se » 2.082; The standard error of estimate of the average annual salary of library science teachers for a specific average annual salary of librarians is about $2082.

9. y  1.841(56)  35.98  67.116 1 12(67.116  55.7)2   6.896 . 12 12(37,347)  (668.4)2 y  E  67.116  6.896  60.22 ; y  E  67.116  6.896  74.012 .

E  (3.169)(2.082) 1 

60.22  y  74.012 ; You can be 99% confident that the average annual salary of library science teachers is between $60,220 and $74,012 when the average annual salary of librarians is $56,000.

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.

  0.05 d.f.  n  1  7 2  0  14.067 ; Rejection region:  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 1.800 413

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CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

 2  6.694

Because  2  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.   0.05 d.f.  n  1  5

 02  11.071 ; Rejection region:  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

 2  12.933

(O - E )2 E 2.133

0.300 2.133 4.033 4.033

0.300 12.933

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415

Because  2  11.071 , reject H 0 . There is enough evidence at the 5% level of significance to dispute 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  (125)(0.4)  50

Ei  npi  (800 )(0.7)  560

Ei  npi  (350 )( 0.35)  122.5

Ei  npi  ( 610 )( 0.89)  542.9

7. (a) Claimed Distribution: Age

217 1824 2539 4049 50+

Distribution 23% 20% 22% 9% 26%

H 0 : The distribution of the ages of moviegoers is 23% ages 217, 20% ages 1824, 22% ages

2539, 9% ages 4049, and 26% ages 50+ (claim) H a : The distribution of the ages differs from the claimed or expected distribution.

(b)  02  7.779 , Rejection region:  2  7.779

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(c)

Age

Distribution

Observed

Expected

217 1824 2539 4049 50+

23% 20% 22% 9% 26%

240 209 203 106 242

230 200 220 90 260

(O - E ) 2 E 0.4348 0.405 1.3136 2.8444 1.2462 6.244

 2  6.244

(d) Because  2  7.779 , 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)  02  9.488 , Rejection region:  2  9.488 (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

 2  7.5882

(d) Because  2  9.488 , 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 © 2019 Pearson Education Ltd.

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417

9. (a) Claimed distribution: Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday

Distribution 7% 4% 6% 13% 10% 36% 24%

H 0 : The distribution of the days people order food for delivery is 7% Sunday, 4% Monday, 6%

Tuesday, 13% Wednesday, 10% Thursday, 36% Friday, and 24% Saturday. H a : The distribution of days differs from the claimed or expected distribution.

(b)  02  16.812 , Rejection region:  2  16.812 (c)

Distribution

Observed

Expected

7% 4% 6% 13% 10% 36% 24%

43 16 25 49 46 168 153

35 20 30 65 50 180 120

Sunday Monday Tuesday Wednesday Thursday Friday Saturday

(O - E )2 E 1.8286 0.8000 0.8333 3.9385 0.3200 0.8000 9.0750 17.5954

 2  17.5954

(d) Reject H 0 . (e) There is enough evidence at the 1% level of significance to conclude that the distribution of delivery days has changed. 10. (a) Claimed Distribution: Response All purchases with cash Most purchases with cash Half of purchases with cash Some purchases with cash No purchases with cash

Distribution 19% 17% 20% 33% 11%

H 0 : The distribution of people who use cash to make their purchases is 19% all purchases, 17%

most purchases, 20% half of purchases, 33% some purchases, and 11% no purchases.

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CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

H a : The distribution of people who use cash to make their purchases differs from the expected

distribution. (claim) (b)  02  13.277 , Rejection region:  2  13.277 (c)

Response

Distribution

Observed

Expected

All purchases with cash Most purchases with cash Half of purchases with cash Some purchases with cash No purchases with cash

19% 17% 20% 33% 11%

60 84 132 252 72

114 102 120 198 66

(O - E ) 2 E 25.5789 3.1765 1.2 14.7273 0.5455 45.2282

 2  45.2282

(d) Because  2  13.277 , reject H 0 . (e) There is enough evidence at the 1% level of significance to conclude that the distribution of people who use cash to make purchases 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) H a : The distribution of homicides by county is not uniform.

(b)  02  30.578 , Rejection region:  2  30.578

CHAPTER 10 │ 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%

116 55 57 62 101 58 30 65 90 89 45 51 62 39 37 43

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

419

(O - E ) 2 E

45.796 0.9 0.484 0.004 23.716 0.324 16.9 0.1 12.1 11.236 4.9 2.116 0.004 8.836 10.404 6.084 143.904

 2  143.904

(d) Because  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: Year 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014

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 violence crimes in England and Wales by years is

uniform. (claim) H  : The distribution of violence by years is not uniform.

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(b)  02  19.675 , Rejection region:  2  19.675 (c) O

O–E

(O - E ) 2

2307 2213 2010 1984 2103 1815 1774 1687 1896 1744 1666 1327

1877 1877 1877 1877 1877 1877 1877 1877 1877 1877 1877 1877

430 336 133 107 226 -62 -103 -190 19 -133 -211 -550

184900 112896 17689 11449 51076 3844 10609 36100 361 17689 44521 302500

(O - E ) 2 E

98.50826 60.14704 9.424081 6.099627 27.21151 2.047949 5.652104 19.23282 0.192328 9.424081 23.71923 161.1614 422.8205

 2  422.8205

(d) Because  2  19.675 , reject H 0 . (e) There is enough evidence at the 5% level of significance to reject the claim that the distribution of number of violence crimes in England and Wales by year is uniform. 13. (a) Claimed distribution: Month Strongly agree Somewhat agree Neither agree nor disagree Somewhat disagree Strongly disagree

Distribution 55% 30% 5% 6% 4%

H 0 : The distribution of the opinions of U.S. parents on whether a college education is worth the

expense is 55% strongly agree, 30% somewhat agree, 5% neither agree nor disagree, 6% somewhat disagree, 4% strongly disagree. H a : The distribution of opinions differs from the expected distribution. (claim) (b)  02  9.488 , Rejection region:  2  9.488

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(c)

Month Strongly agree Somewhat agree Neither agree nor disagree Somewhat disagree Strongly disagree

Distribution

Observed

Expected

55% 30% 5% 6% 4%

86 62 34 14 4

110 60 10 12 8

421

(O - E )2 E 5.2364 0.0667 57.6000 0.3333 2.0000 65.2364

 2  65.236

(d) Because  2  9.488 , reject H 0 . (e) There is enough evidence at the 5% level of significance to conclude that the distribution of opinions of U.S. parents on whether a college education is worth the expense differs from the claimed or expected distribution. 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) The distribution of how much married U.S. female adults trust their spouses to manage their Ha : finances differs from the distribution of how much married U.S. male adults trust their spouses to manage their finances. (b)  02  6.251 , Rejection region:  2  6.251 (c) Response Completely trust Trust with certain aspects Do not trust Not sure

 2  29.057 (d) Because  2  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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(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: Types of Tourist Students Patients Tourists

Distribution 33.3% 33.3% 33.3%

H 0 : The distribution of prospective tourists by the purpose of their visit is uniform. H  : The distribution of prospective tourists by the purpose of their visit is not uniform. (claim)

(b)  02  9.210 , Rejection region:  2  9.210 (c)

Types of Visitors

Distribution

Observed

Expected

33.3% 33.3% 33.3%

518 497 985

666.66 666.66 666.66

Students Patients Tourists

(O - E )2 E 33.15 43.18 152.01 228.34

 2  228.34 (d) Because  2  9.210 , reject H 0 . (e) There is enough evidence at the 1% level of significance to conclude that the distribution of prospective tourists by the purpose of their visit 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.

(b)  02  10.645 , Rejection region:  2  10.645

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(c)

Day

Distribution

Observed

Expected

Sunday Monday Tuesday Wednesday Thursday Friday Saturday

14.29% 14.29% 14.29% 14.29% 14.29% 14.29% 14.29%

68 108 115 113 111 108 77

100 100 100 100 100 100 100

423

(O - E ) 2 E

10.24 0.64 2.25 1.69 1.21 0.64 5.29 21.96

 2  21.96 (d) Because  2  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:   69.435;   8.337 Lower Boundary 49.5 58.5 67.5 76.5 85.5

Upper Boundary 58.5 67.5 76.5 85.5 94.5

Lower z-score 2.39 1.31 0.23 0.85 1.93

Upper z-score 1.31 0.23 0.85 1.93 3.01

Area 0.0867 0.3139 0.3933 0.1709 0.0255

(O  E )2 E

Class Boundaries

Distribution

Frequency

Expected

49.558.5 58.567.5 67.576.5 76.585.5 85.594.5

8.67% 31.39% 39.33% 17.09% 2.55%

19 61 82 34 4 200

17 63 79 34 5

H 0 : Test scores have a normal distribution. (claim) H a : Test scores do not have a normal distribution.

(b)  02  13.277 ; Rejection region .  2  13.277 (c)  2  0.613 (d) Because  2  13.277 , fail to reject H 0 .

0.2353 0.0635 0.1139 0 0.2000 0.6127

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(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:   74.775;   9.822 Lower Upper Lower Boundary Boundary z-score 50.5 60.5 2.47 60.5 70.5 1.45 70.5 80.5 0.44 80.5 90.5 0.58 90.5 100.5 1.60 Class Boundaries

Distribution

Frequency

Upper z-score 1.45 0.44 0.58 1.60 2.62

Area 0.0668 0.2564 0.3891 0.2262 0.0504

Expected

(O  E )2 E

6.68% 25.64% 38.91% 22.62% 5.04%

28 106 151 97 18 400 H 0 : Test scores have a normal distribution. (claim) H a : Test scores do not have a normal distribution. 50.560.5 60.570.5 70.580.5 80.590.5 90.5100.5

27 103 156 90 20

0.0370 0.0874 0.1603 0.5444 0.2000 1.0291

(b)  02  9.488 ; Rejection region .  2  9.488 (c)  2  1.029 (d) Because  2  9.488 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to reject the claim that the test scores are normally distributed.

10.2 INDEPENDENCE 10.2 TRY IT YOURSELF SOLUTIONS 1. Business Leisure Total

Hotel 36 38 74

Leg Room 108 54 162

Rental Size 14 14 28

Other 22 14 36

n  300

Business Leisure

Hotel 44.4 29.6

Leg Room 97.2 64.8

Rental Size 16.8 11.2

Other 21.6 14.4

Total 180 120 300

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425

2. The claim is that “the travel concerns depend on the purpose of travel.” H 0 : Travel concern is independent of travel purpose. H a : Travel concern is dependent on travel purpose. (claim) (r  1)(c  1)  3   0.01

 02  11.345; Rejection region:  2  11.345 O

OE

O - E 

36 108 14 22 38 54 14 14

44.4 97.2 16.8 21.6 29.6 64.8 11.2 14.4

8.4 10.8 2.8 0.4 8.4 10.8 2.8 0.4

70.56 116.64 7.84 0.16 70.56 116.64 7.84 0.16

(O - E )2 E 1.5892 1.2000 0.4667 0.0074 2.3838 1.8000 0.7000 0.0111 8.1582

 2  8.158

Because  2  11.345 , fail to reject H 0 . There is not enough evidence at the 1% level of significance for the consultant to conclude that travel concern is dependent on travel purpose. 3.

H 0 : Whether or not a tax cut would influence an adult to purchase a hybrid vehicle is

independent of age. H a : Whether or not a tax cut would influence an adult to purchase a hybrid vehicle is dependent on age. (claim) Enter the data.  02  9.210; Rejection region:  2  9.210

 2  15.306 Because  2  9.210 , reject H 0 . There is enough evidence at the 1% level of significance to conclude that whether or not a tax cut would influence an adult to purchase a hybrid vehicle is dependent on age.

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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 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. 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 Injury No injury

Athlete has Stretched Not stretched 25 33 220 156 245 189

(b) Result Injury No injury

Athlete has Stretched 32.74 212.26

Not stretched 25.26 163.74

Drug 42 212 254

Treatment Placebo 33 239 272

8. (a) Result Nausea No nausea (b) Result Nausea No Nausea

Total 58 376 434

Drug 36.22 217.78

Treatment Placebo 38.78 233.22

Total 75 451 526

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

9. (a) Students High School Middle School

Marking System 225 150 375

Preference Grading System No Preference 95 35 75 10 170 45

(b) Students High school Middle School 10. (a) Size of restaurant Seats 100 or fewer Seats over 100 (b) Size of restaurant Seats 100 or fewer Seats over 100

Marking System 225.64 149.36

Preference Grading System 102.29 67.71

No Preference 27.08 17.92

Rating Excellent 210 215 425

Fair 156 134 290

Poor 174 196 370

Total 540 545 1085

Total 170 160 330

Rating Excellent 211.52 213.48

11. (a) Gender Male Female

Fair 144.33 145.67

Poor 184.15 185.85

Science Fiction 65 55 120

Crime 55 60 115

Type of book Romance 20 30 50

Mythology 30 15 45

Science Fiction 61.82 58.18

Type of book Crime Romance 59.24 25.76 55.76 24.24

Mythology 23.18 21.82

(b) Gender Male Female

Total 355 235 590

12. (a)

Age Type of movie rented Comedy Action Drama Total

1824 38 15 12 65

2534 30 17 11 58

3544 24 16 19 59

4564 10 9 25 44

65 and older 8 5 13 26

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

Total 110 62 80 252

427

(b)

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CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

13. (a) The claim is “an athlete’s allergy is independent of whether or not the athlete has stretched.” H 0 : An athlete’s injury result is independent on 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

 02  3.841 ; Rejection region:  2  3.841 (c)

OE

(O - E ) 2

25 33 220 156

32.74 25.26 212.26 163.74

−7.74 7.74 7.74 −7.74

59.9076 59.9076 59.9076 59.9076

(O - E )2 E 1.8298 2.3716 0.2822 0.3659 4.8495

 2  4.8495 (d) Because  2  3.841 , reject H 0 . (e) There is enough evidence at the 5% 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 dependent on 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

 02  6.635 ; Rejection region:  2  6.635 (c)

OE

(O - E ) 2

42 33 212 239

36.22 38.78 217.78 233.22

5.78 −5.78 −5.78 5.78

33.4084 33.4084 33.4084 33.4084

(O - E )2 E 0.9224 0.8615 0.1534 0.1432 2.0805

 2  2.0805 (d) Because  2  6.635 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to conclude that nausea result is dependent on treatment.

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429

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

 02  2.706 ; Rejection region:  2  2.706 (c)

O−E

(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

(O  E ) 2 E 1.9029 1.9029 4.3361 4.3361 12.478

 2  12.478 (d) Because  2  2.706 , reject H 0 . (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

02  6.635 ; Rejection region:  2  6.635 (c) O

O−E

(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

 2  8.691 (d) Because   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. 2

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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

02  5.991 ; Rejection region:  2  5.991 (c) O

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

(O  E )2 E 0 0.0003 0.0003 0 0.0006 0.0006 0.0018

 2  0.002 2 (d) Because   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

02  9.210 ; Rejection region:  2  9.210 (c) O

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

 2  0.297 2 (d) Because   9.210 , fail to reject H 0 .

(O - E )2 E 0.0851 0.0002 0.0966 0.0540 0.0001 0.0613 0.2973

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431

(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

02  9.210 ; Rejection region:  2  9.210 (c) O

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

 2  7.326 2 (d) Because   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

02  7.779 ; Rejection region:  2  7.779

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CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

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

 2  5.757 2 (d) Because   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 borrowing money for college is related to race.” H 0 : A family borrowing money for college is independent of race. H a : A family borrowing money for college is dependent on race. (claim) (b) d.f.  (r  1)(c  1)  2

 02  9.210 ; Rejection region:  2  9.210 (c)

OE

(O - E ) 2

49 64 85 123 85 180

42.23 70.77 77.73 130.27 99.04 165.96

6.77 –6.77 7.27 –7.27 –14.04 14.04

45.8329 45.8329 52.8529 52.8529 197.1216 197.1216

(O - E )2 E 1.0853 0.6476 0.6800 0.4057 1.9903 1.1878 5.9967

 2  5.9967 (d) Because  2  9.210 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to conclude that a family borrowing money for college is dependent on race. 22. (a) The claim is “who borrows money for college in a family is related to the income of the family..” H 0 : Who borrows money for college in a family is independent of the family’s income. H a : Who borrows money for college in a family is dependent on the family’s income. (claim)

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433

(b) d.f.  (r  1)(c  1)  6

 02  16.812 ; Rejection region:  2  16.812 (c)

OE

(O - E ) 2

149 34 10 311 181 68 58 421 69 40 14 238

126.24 44.93 25.94 306.89 182.34 64.89 37.47 443.29 90.42 32.18 18.58 219.82

22.76 –10.93 –15.94 4.11 –1.34 3.11 20.53 –22.29 –21.42 7.82 –4.58 18.18

518.0176 119.4649 254.0836 16.8921 1.7956 9.6721 421.4809 496.8441 458.8164 61.1524 20.9764 330.5124

(O - E )2 E 4.1034 2.6589 9.7951 0.0550 0.0098 0.1491 11.2485 1.1208 5.0743 1.9003 1.1290 1.5036 38.7478

 2  38.7478 (d) Because  2  16.812 , reject H 0 . (e) There is enough evidence at the 1% level of significance to conclude that who borrows money for college in a family is dependent on the family’s income. 23. (a) The claim is “the type of crash is depends on the type of vehicle.” H 0 : Type of crash is independent of the type of vehicle. H a : Type of crash is dependent on the type of vehicle. (claim) (b) d.f.  (r  1)(c  1)  2

 02  5.991 ; Rejection region:  2  5.991 (c)

OE

(O - E ) 2

1059 507 491 1476 354 383

1221.19 414.77 421.03 1313.81 446.23 452.97

–162.19 92.23 69.97 162.19 –92.23 –69.97

26,305.5961 8,506.3729 4,895.8009 26,305.5961 8,506.3729 4,895.8009

 2  103.5713 (d) Because  2  5.991 , reject H 0 .

(O - E )2 E 21.5410 20.5087 11.6282 20.0224 19.0628 10.8082 103.5713

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(e) There is enough evidence at the 5% level of significance to conclude that the type of crash is dependent on the type of vehicle. 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

 02  11.071 ; Rejection region:  2  11.071 (c)

OE

(O - E ) 2

31 147 95 67 57 42 9 36 25 17 15 9

31.93 146.07 95.78 67.05 57.47 40.71 8.07 36.93 24.22 16.95 14.53 10.29

–0.93 0.93 –0.78 –.0.05 –0.47 1.29 0.93 –0.93 0.78 0.05 0.47 –1.29

0.8649 0.8649 0.6084 0.0025 0.2209 1.6641 0.8649 0.8649 0.6084 0.0025 0.2209 1.6641

(O - E )2 E 0.0271 0.0060 0.0064 0.0000 0.0038 0.0409 0.1072 0.0234 0.0251 0.0001 0.0152 0.1617 0.4169

 2  0.4169 (d) Because  2  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 “preference for method of assessment is dependent upon the school level.” H 0 : Preferred method of assessment is independent of the school level. H a : Preferred method of assessment is dependent on the school level. (claim) (b) d.f.  (r  1)(c  1)  2

 02  9.210 ; Rejection region:  2  9.210

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(c)

OE

(O - E ) 2

225 95 35 150 75 10

225.64 102.29 27.08 149.36 67.71 17.92

−0.64 −7.29 7.92 0.64 7.29 −7.92

0.4096 53.1441 62.7264 0.4096 53.1441 62.7264

435

(O - E )2 E 0.001815 0.519543 2.316337 0.002742 0.784878 3.500357 7.125673

 2  7.125673 (d) Because  2  9.210 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to conclude that preferred method of assessment is dependent on the school level. 26. (a) The claim is “restaurant rating is dependent on 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

02  5.991 ; Rejection region:  2  5.991 (c)

OE

(O - E ) 2

210 156 174 215 134 196

211.52 144.33 184.15 213.48 145.67 185.85

−1.52 11.67 −10.15 1.52 −11.67 10.15

2.3104 136.1889 103.0225 2.3104 136.1889 103.0225

(O - E )2 E 0.0109 0.9436 0.5594 0.0108 0.9349 0.5543 3.0139

 2  3.0139 (d) Because  2  5.991 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to conclude that restaurant rating is dependent on size of the restaurant. 27. (a) The claim is “type of book is independent of the gender of reader.” H 0 : Type of book is dependent on gender of reader. H a : Type of book is independent of gender of reader. (claim) (b) d.f.  (r  1)(c  1)  3

 02  7.815 ; Rejection region:  2  7.815

436

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(c)

OE

(O  E )2

65 55 20 10 25 45 85 95

33.75 37.5 39.375 39.375 56.25 62.5 65.625 65.625

31.25 17.5 −19.375 −29.375 −31.25 −17.5 19.375 29.375

976.5625 306.25 375.3906 862.8906 976.5625 306.25 375.3906 862.8906

(O  E ) 2 E 28.935 8.167 9.534 21.915 17.361 4.9 5.72 13.149 109.681

 2  109.681 (d) Because  2  7.815 , fail to reject H 0 . (e) There is enough evidence at the 5% level of significance to conclude that type of book is dependent on gender of the reader. 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

 02  13.362 ; Rejection region:  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

 2  29.0471 (d) Because  2  13.362 , reject H 0 .

(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

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437

(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

 02  14.067 ; Rejection region:  2  14.067 O 96 98 72 80 74 44 25 12 39 33 25 29 26 21 16 10

OE

(O  E )2

96.6214 93.7586 69.4243 78.0129 71.5714 46.5214 29.3443 15.7457 38.3786 37.2414 27.5757 30.9871 28.4286 18.4786 11.6557 6.2543

–0.6214 4.2414 2.5757 1.9871 2.4286 2.5214 –4.3443 –3.7457 0.6214 –4.2414 –2.5757 –1.9871 –2.4286 2.5214 4.3443 3.7457

0.3861 17.9895 6.6342 3.9486 5.8981 6.3575 18.8729 14.0303 0.3861 17.9895 6.6342 3.9486 5.898 6.3575 18.8729 14.0303

(O  E ) 2 E 0.0040 0.1919 0.0956 0.0506 0.0824 0.1367 0.6432 0.8911 0.0101 0.4831 0.2406 0.1274 0.2075 0.3440 1.6192 2.2433 7.3705

 2  7.3705 Because  2  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

02  2.706 ; Rejection region:  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

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CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

 2  5.106 Because   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. 2

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. Educational Attainment

33.

Status Employed

Not a High school Graduate 0.046

High School Graduate 0.156

Some College, No Degree 0.101

Associate’s, Bachelor’s, or Advanced Degree 0.312

Unemployed

0.004

0.010

0.005

0.009

Not in the labor force

0.059

0.123

0.061

0.114

34. Several of the expected frequencies are less than 5. 35. (a) 0.9% (b) 6.1% 36. (a) 15.6% (b) 35.7% (c) 10.9% Educational Attainment

37.

Status Employed

Not a High School Graduate 0.076

High School Graduate 0.253

Some College, No Degree 0.164

Associate’s, Bachelor’s, or Advanced Degree 0.507

Unemployed

0.150

0.350

0.183

0.317

Not in the labor force

0.164

0.344

0.172

0.320

38. 50.7% 39. 17.2%

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

40.

439

Educational Attainment

Status Employed

Not a High School Graduate 0.426

High School Graduate 0.540

Some College, No Degree 0.603

Associate’s, Bachelor’s, or Advanced Degree 0.717

Unemployed

0.038

0.034

0.031

0.020

Not in the labor force

0.536

0.426

0.367

0.263

41. 26.3% 42. 3.8%

10.3 COMPARING TWO VARIANCES 10.3 TRY IT YOURSELF SOLUTIONS 1.   0.05

F0  2.45

2.   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.” H 0 :  12   22 ; H a :  12   22 (claim)   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 s2 56 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 :  1   2 (claim); H a :  1   2   .01 d.f.N  n1  1  15 d.f.D  n2  1  21 F0  3.43; Rejection region: F  3.43

440

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

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.

10.3 EXERCISE SOLUTIONS 1. Specify the level of significance  . 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 . (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 n2 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

F0  2.06

12. F0  2.42

13. H 0 :  12   22 ; H a :  12   22 (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.

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441

14. H 0 :  12   22 (claim); H a :  12   22 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. 15. H 0 :  12   22 (claim); H a :  12   22 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. 16. H 0 :  12   22 ; H a :  12   22 (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. 17. H 0 :  12   22 (claim); H a :  12   22 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. 18. H 0 :  12   22 ; H a :  12   22 (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.

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19. Population 1: City B; Population 2: City A (a) The claim is “the variance of drunk driving accidents of City A is less than the variance of the drunk driving accidents of City B.” H 0 :  12   22 ; H a :  12   22 (claim) (b) d.f.N  20 , d.f.D  15 F0  3.53; Rejection region: F  3.53

s12 3.6  1.895 (c) F  2  s2 1.9 (d) Because F  3.53 , fail to reject H 0 . (e) There is not enough evidence at the 1% level of significance to support City A’s claim that the variance of drunk driving accidents is less than the variance of drunk driving accidents of City B. 20. Population 1: Competitor; Population 2: Auto Manufacturer (a) The claim is “the variance of the life of the fuel consumption for the company’s hybrid vehicles is less than that of the competitor’s hybrid vehicles.” H 0 :  12   22 ; H a :  12   22 (claim) (b) d.f.N  20 , d.f.D  18 F0  1.84 Rejection region: F  1.84 (c) F 

s12 s22



0.34  1.62 0.21

(d) Because F  1.84 , fail to reject H 0 . (e) There is not enough evidence at the 10% level of significance to conclude that the variance of the fuel consumption for the company’s hybrid vehicles is less than that of the competitor’s hybrid vehicles. 21. Population 1: Age group 35-49; Population 2: Age group 18-34 (a) The claim is “the variances of the waiting times differ between the two age groups.” H 0 :  12   22 ; H a :  12   22 (claim) (b) d.f.N  5 , d.f.D  8 F0  4.82 Rejection region: F  4.82

s2 s2

458.7  2.58 177.7

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

443

(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 variance of the waiting times differ between the two age groups. 22. Population 1: Golfer 1; Population 2: Golfer 2 (a) The claim is “the variances of the driving distances differ between the two golfers.” H 0 :  12   22 ; H a :  12   22 (claim) (b) d.f.N  9 , d.f.D  9 F0  3.18; Rejection region: F  3.18 (c) s12  169.7333, s22  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. 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.” H 0 :  12   22 (claim); H a :  12   22 (b) d.f.N  11 , d.f.D  13 F0  2.635; Rejection region: F  2.635

s 2 (36.8)2  1.282 (c) F  12  s2 (32.5)2 (d) Because F  2.635 , 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.” H 0 :  12   22 (claim); H a :  12   22 (b) d.f.N  9 , d.f.D  12

444

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

F0  5.20; Rejection region: F  5.20

s 2 (30.9) 2  1.29 (c) F  12  s2 (27.2) 2 (d) Because F  5.20 , 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.” H 0 :  12   22 ; H a :  12   22 (claim) (b) d.f.N  40 , d.f.D  60 F0  1.59 Rejection region: F  1.59

s12 (37,100)2 F    1.31 (c) s22 (32,400)2 (d) Because F  1.59 , fail to reject H 0 . (e) There is not 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: Louisiana; Population 2: Florida (a) The claim is “the standard deviation of the annual salaries for public relations managers is less in Florida than in Louisiana.” H 0 :  12   22 ; H a :  12   22 (claim) (b) d.f.N  23 , d.f.D  27 F0  1.94 Rejection region: F  1.94 s 2 (42, 000) 2 (c) F  12   1.34 s2 (36, 300) 2

(d) Because F  1.94 , fail to reject H 0 . (e) There is not 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 Florida than in Louisiana.

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

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  2.20

Critical value is

1 1   0.45 . F 2.20

29.

2 s12 1  12 s12 1 12  10.89  1  1  10.89          3.422 (3.02) 0.340  9.61  3.33    2 s22 FR  22 s22 FL  22     2  9.61 

30.

2 s12 1  12 s12 1 12  5.29  1   1  5.29          4.425 (3.02) 0.440      2 s22 FR  22 s22 FL  22  3.61  3.33   2  3.61 

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 : 1   2   3   4 H a : At least one mean is different from the others. (claim)   0.05 d.f.N  3 ; d.f.D  14 F0  3.34 ; Rejection region: F  3.34 Variation Between Within

Sum of Squares 549.8 608.0

Degrees of Freedom 3 14

F  4.22

Because F  3.34 , reject H 0 .

Mean Squares 183.3 43.4

F 4.22

445

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CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

There is enough evidence at the 5% level of significance 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 : 1   2   3   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

Degrees of Freedom 3 30

Mean Squares 0.195 0.145

F 1.34

F  1.34  P -value  0.280 0.280  0.05 Because P -value  0.05 , fail to reject H 0 . There is not enough evidence to conclude that there is a

difference in the means of the GPAs.

10.4 EXERCISE SOLUTIONS 1.

H 0 : 1   2     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.

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

447

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 “the mean costs per ounce are different.” H 0 : 1   2   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  12 F0  3.89 ; Rejection region: F  3.89 (c) Variation Between Within

Sum of Squares 0.07406 0.09267

Degrees of Freedom 2 12

Mean Squares 0.037032 0.007723

F 4.80

F  4.80

(d) Because F  3.89 , reject H 0 . (e) There is enough evidence at the 5% level of significance to conclude that at least one mean cost per ounce is different from the others. 6. (a) The claim is that “at least one mean battery price is different from the others.” H 0 : 1   2   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  12 F0  3.89 ; Rejection region: F  3.89 (c) Variation Between Within

Sum of Squares 13083.3333 23460

Degrees of Freedom 2 12

Mean Squares 6541.6667 1955

F 3.35

F  3.35

(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 battery price 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 : 1   2   3

448

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

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 : 1   2   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 4878.734 1918.441

Degrees of Freedom 2 27

Mean Squares 2439.367 71.0534

F 34.33

F  34.33

(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 : 1   2   3   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  48 F0  2.84 ; Rejection region: F  2.84 (c) Variation Between Within

Sum of Squares 10.07692 262.1538

F  0.62

Degrees of Freedom 3 48

Mean Squares 3.358974 5.461538

F 0.62

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

449

(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 : 1   2   3   4  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 2211.5615 1182.9

Degrees of Freedom 4 21

Mean Squares 552.8904 56.3286

F 9.82

F  9.82

(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 : 1   2   3   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  3 0 F0  2.28 ; Rejection region: F  2.28 (c) Variation Between Within

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 : 1   2   3   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  29

450

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

F0  4.54 ; Rejection region: F  4.54

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 : 1   2   3   4   5   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 335,665,495 975,795,651

Degrees of Freedom 5 30

Mean Squares 67,133,099 32,526,522

F 2.06

F  2.06

(d) Because F  2.53 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to conclude that the mean salary is different in at least one of the areas. 14. (a) The claim is that “the mean sale price is different in at least one of the cities.” H 0 : 1   2   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 8 F0  2.50 ; Rejection region: F  2.50 (c) Variation Between Within

Sum of Squares 3218.0979 48337.7698

Degrees of Freedom 2 28

F  0.93

(d) Because F  2.50 , fail to reject H 0 .

Mean Squares 1609.049 1726.3489

F 0.93

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

451

(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. Source Type of vehicle

d.f. 2

SS 84.083

Gender Interaction Error Total

1 2 18 23

0.375 12.250 22.250 118.958

MS 42.04 2 0.375 6.125 1.236

F 34.01

P 0.000

0.30 4.96

0.589 0.019

There is an interaction effect between type of vehicle and gender on the mean number of vehicles sold. Also, the type of vehicle has an effect on the mean number of vehicles sold. 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.

452

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

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 disk drive at the 10% level of significance 19. Pop 1 Pop2 Pop3

Mean 0.455 0.606 0.460

Size 6 5 4

SSW  0.09267

 (n  1)  N  k  12 i

F0  3.89  CVScheffe  3.89(3  1)  7.78

( x1  x2 ) 2  8.05  Significant difference 1 1  SSW    (ni  1)  n1 n2 



( x1  x3 ) 2  0.01  No difference 1 1 SSW    ( ni  1)  n1 n3 



CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

( x2  x3 ) 2  6.13  No difference 1 SSW 1    ( ni  1)  n2 n3 

 20.

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 1 1     ( ni  1)  n1 n2 

SSW



( x1  x3 ) 2 1 1    ( ni  1)  n1 n3 

SSW



( x2  x3 ) 2



1 1    ( ni  1)  n2 n3 

SSW

21. Pop 1 Pop2 Pop3

 4.95  No difference

 11.62  No difference

 31.73  Significant difference

Mean 77.86 55.82 47.67

Size 10 10 10

SSW  1918.441

 (n  1)  N  k  27 i

F0  5.49  CVScheffe  5.49(3  1)  10.98

( x1  x2 ) 2



1 1     ( ni  1)  n1 n2 

SSW

( x1  x3 ) 2



1 1    ( ni  1)  n1 n3 

SSW

 34.18  Significant difference

 64.14  Significant difference

( x2  x3 ) 2  4.67  No difference 1 1 SSW    ( ni  1)  n2 n3 



453

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CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

22. Pop 1 Pop2 Pop3 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 1 1     ( ni  1)  n1 n2 

SSW



( x1  x3 ) 2



1 1    ( ni  1)  n1 n3 

SSW

( x1  x4 ) 2



1 1     ( ni  1)  n1 n4 

SSW

( x2  x3 ) 2



1 1    ( ni  1)  n2 n3 

SSW

( x2  x4 ) 2



1 1     ( ni  1)  n2 n4 

SSW

( x3  x4 ) 2



1 1    ( ni  1)  n3 n4 

SSW

 0.11  No difference

 4.23  No difference

 0.55  No difference

 3.46  No difference

 1.37  No difference

 10.17  Significant difference

455

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

CHAPTER 10 REVIEW EXERCISE SOLUTIONS 1. (a)

Response Less than 9 10-12 13-16 17-20 21-24 25 or more

Distribution 4% 24% 26% 22% 6% 18%

H 0 : The distribution of the lengths of office visits is 4% less than 9 minutes, 24% 10-12

minutes, 26% 13-16 minutes, 22% 17-20 minutes, 6% 21-24 minutes, and 18% 25 or more minutes. H a : The distribution of the lengths differ from the claimed or expected distribution. (claim) (b)  02  15.086 ; Rejection region:  2  15.086 (c)

Response

Distribution

Observed

Expected

Less than 9 10-12 13-16 17-20 21-24 25 or more

4% 24% 26% 22% 6% 18%

20 80 113 91 40 56

16 96 104 88 24 72

(O - E ) 2 E 1 2.667 0.779 0.102 10.667 3.556 18.771

 2  18.771 (d) Because  2  15.086 , reject H 0 . (e) There is not enough evidence at the 1% level of significance to conclude that the distribution of the lengths differs from the expected distribution. 2. (a)

Response Less than $10 $10 to $20 More than $21 Don’t give one/other

Distribution 29% 16% 9% 46%

H 0 : The distribution of the allowance amounts is 29% less than $10, 16% $10 to $20,

more than $21, and 46% don’t give one/other. H a : The distribution of amounts differ from the claimed or expected distribution. (claim) (b)  02  6.251 ; Rejection region:  2  6.251

456

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(c)

Response

Distribution

Observed

Expected

Less than $10 $10 to $20 More than $21 Don’t give one/other

29% 16% 9% 46%

353 167 94 489

319.87 176.48 99.27 507.38

(O - E )2 E 3.431 0.509 0.280 0.666 4.886

 2  4.886 (d) Because  2  6.251 , fail to reject H 0 . (e) There is not enough evidence at the 10% level of significance to conclude that there has been a change in the claimed or expected distribution. 3. (a)

Response Distribution Short-game shots 65% Approach and swing 22% Driver shots 9% Putting 4% H 0 : The distribution of responses from golf students about what they need the most help with is 22% approach and swing, 9% driver shots, 4% putting and 65% short-game shots. (claim) H a : The distribution of responses differs from the claimed or expected distribution .

(b)  02  7.815 ; Rejection region:  2  7.815 (c)

Response Short-game shots Approach and swing Driver shots Putting

Distribution

Observed

Expected

65% 22% 9% 4%

276 99 42 18

282.75 95.70 39.15 17.40

(O - E )2 E 0.161 0.114 0.207 0.021 0.503

 2  0.503 (d) Because  2  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 golf students’ responses is the same as the claimed or expected distribution.

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

4. (a)

Response 15,000-17,499 17,500-19,999 20,000-22,499 22,500-24,499 25,000+

457

Distribution 20% 20% 20% 20% 20%

H 0 : The distribution of charges for tuition, fees, room, and board at 4-year degree granting

postsecondary instiutions is uniform. (claim) H a : The distribution of charges for tuition, fees, room, and board not uniform.

(b)  02  9.488 , Rejection region:  2  9.488 (c)

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

 2  78.037 (d) Because  2  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:

Gender Male Female Total

Less than 3 years 95.4 222.6 318

Years of full-time teaching experience 20 years or more 39 years 1020 years 349.2 383.4 222 814.8 894.6 518 1164 1278 740

Total 1050 2450 3500

(b) The claim is “gender is related to the years of full-time teaching experience.” H 0 : Years of full-time teaching experience is independent of gender. H a : Years of full-time teaching experience is dependent on gender. (claim) (c) d.f.  (r  1)(c  1)  3 ;  02  11.345 ; Rejection region:  2  11.345

458

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(d)

OE

(O - E ) 2

102 339 402 207 216 825 876 533

95.4 349.2 383.4 222.0 222.6 814.8 894.6 518.0

6.6 –10.2 18.6 –15 –6.6 10.2 –18.6 15

43.56 104.04 345.96 225 43.56 104.04 345.96 225

(O - E )2 E 0.4566 0.2979 0.9023 1.0135 0.1957 0.1277 0.3867 0.4344 3.8148

 2  3.815 (e) Because  2  11.345 , fail to reject H 0 . (f) There is not enough evidence at the 1% level of significance to conclude that the years of fulltime teaching experience is dependent on gender. 6. (a) Expected frequencies: Car Gender Male Female Total

94.25 100.75 195

Type of Vehicle Owned Truck SUV

Van

81.2 86.8 168

5.8 6.2 12

50.75 54.25 105

(b) The claim is “type of vehicle owned is related to gender.” H 0 : The type of vehicle owned is independent of gender. H a : The type of vehicle owned is dependent on gender. (claim) (c) d.f.  ( r  1)(c  1)  3 ;  02  11.345 ; Rejection region:  2  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

–9.25 13.8 –6.75 2.2 9.25 –13.8 6.75 –2.2

85.5625 190.44 45.5625 4.84 85.5625 190.44 45.5625 4.84

 2  9.649

(O - E )2 E 0.9078 2.3453 0.8978 0.8345 0.8493 2.1940 0.8399 0.7806 9.6492

Total 232 248 480

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

459

(e) Because  2  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 gender. 7. (a) Expected frequencies: Mammals Status Endangered Threatened Total

136.8 37.2 174

Vertebrate Group Birds Reptiles Amphibians

Fish

Total

121 33 154

65.2 17.8 83

393 107 500

46.4 12.6 59

23.6 6.4 30

(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. (c) d.f.  ( r  1)(c  1)  4 ;  02  9.488 ; Rejection region:  2  9.488 (d)

OE

(O - E ) 2

151 137 37 18 50 23 17 22 12 33

136.8 121.0 46.4 23.6 65.2 37.2 33.0 12.6 6.4 17.8

14.2 16 –9.4 –5.6 –15.2 –14.2 –16 9.4 5.6 15.2

201.64 256 88.36 31.36 231.04 201.64 256 88.36 31.36 231.04

(O - E )2 E 1.4740 2.1157 1.9043 1.3288 3.5436 5.4204 7.7576 7.0127 4.9 12.9798 48.4369

 2  48.437 (e) Because  2  9.488 , reject H 0 . (f) There is enough evidence at the 5% level of significance to reject the claim that a species status is independent of vertebrate group. 8. (a) Expected frequencies: Gender

12 A.M.-5:59 A.M.

Male Female Total

552.6 318.4 871

6 A.M.11:59 A.M. 602.1 346.9 949

Time of Day 12 P.M.5:59 P.M.

6 P.M.-11:59 P.M.

Total

918.1 528.9 1447

928.2 534.8 1463

3001 1729 4730

460

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(b) The claim is “the time of day of a collision is related to gender.” H 0 : The time of day of a collision is independent of gender. H a : The time of day of a collision is dependent on gender. (claim) (c) d.f.  ( r  1)(c  1)  3 ;  02  6.251 ; Rejection region:  2  6.251 (d)

OE

(O - E ) 2

611 595 884 911 260 354 563 552

552.6 602.1 918.1 928.2 318.4 346.9 528.9 534.8

58.4 –7.1 –34.1 –17.2 –58.4 7.1 34.1 17.2

3410.56 50.41 1162.81 295.84 3410.56 50.41 1162.81 295.84

(O - E )2 E 6.1718 0.0837 1.2665 0.3187 10.7116 0.1453 2.1985 0.5532 21.4493

 2  21.449 (e) Because  2  6.251 , reject H 0 . (f) There is enough evidence at the 10% level of significance to conclude that the time of day of a collision is dependent on gender. 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 San Francisco, CA and Sacramento, CA are the same.” H 0 :  12   22 (claim); H a :  12   22 (b) d.f.N  35 ; d.f.D  30 F0  2.575; Rejection region: F  2.575

s 2 (75)2  2.905 (c) F  12  s2 (44)2 (d) Because F  2.575 , reject H 0 . (e) There is enough evidence at the 1% level of significance to reject the claim that the standard deviations of hotel room rates for San Francisco, CA and Sacramento, CA are the same.

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

461

18. (a) The claim is “the variation in wheat production is greater in Garfield County than in Kay County.” H 0 :  12   22; H a :  12   22 (claim) (b) d.f.N  20 ; d.f.D  15 ; F0  1.92; Rejection region: F  1.92

s 2 (0.76) 2  1.72 (c) F  12  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 critical reading scores for females is different than the variance of SAT critical reading scores for males” H 0 :  12   22 ; H a :  12   22 (claim) (b) d.f.N  11 ; d.f.D  11 F0  5.32 Rejection region: F  5.32 s 2 20790.15 (c) F  12   1.37 15169.7 s2

(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 females is different than the test score variance for males. 20. (a) The claim is “the new mold produces inserts that are less variable in diameter than the inserts the current mold produces.” H 0 :  12   22; H a :  12   22 (claim) (b) d.f.N  11 ; d.f.D  11 F0  2.82; Rejection region: F  2.82

s 2 0.00146  2.91 (c) F  12  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.

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CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

21. (a) The claim is “the mean amount spent on energy in one year is different in at least one of the regions.” H 0 : 1   2   3   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  28 F0  2.29 ; Rejection region: F  2.29 (c) Variation Between Within

Sum of Squares 2,207,334.3 3,329,531.8

Degrees of Freedom 3 28

Mean Squares 735,778.08 118,911.85

F 6.19

F  6.19

(d) Because F  2.29 , reject H 0 . (e) There is enough evidence at the 10% level of significance to conclude that at least one mean amount spent on energy is different from the others. 22. (a) The claim is “at least one of the mean incomes is different from the others.” H 0 : 1   2   3   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  20 F0  3.10 ; Rejection region: F  3.10 (c) Variation Between Within

Sum of Squares 554081012.1345 1995742029.8238

Degrees of Freedom 3 20

Mean Squares 184693670.7115 99787101.4912

F 1.85

F  1.85

(d) Because F  3.10 , 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 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

4.7% none-8th grade, 6.9% 9th-11th grade, 29.5% high school graduates, 16.6% some college, no degree, 9.8% associate’s degree, 20.5% bachelor’s degree, 8.7% master’s degree, and 3.3% 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)

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

463

(b)  02  14.067 ; Rejection region:  2  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

4.7% 6.9% 29.5% 16.6% 9.8% 20.5% 8.7% 3.3%

10 23 80 56 34 75 31 11

15.04 22.08 94.4 53.12 31.36 65.6 27.84 10.56

320

(O - E )2 E 1.6889 0.0383 2.1966 0.1561 0.2222 1.3470 0.3587 0.0183

6.0261

 2  6.026 (d) Because  2  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. 1. (a) H 0 : Age and educational attainment are independent. H a : Age and educational attainment are dependent. (claim) (b) d.f.  ( r  1)(c  1)  7 ;  02  18.475 ; Rejection region:  2  18.475 (c)

O 10 23 80 56 34 75 31 11 23 26 136 77 41 78 41 18

OE

13.89 20.63158 90.95 56 31.58 64.42 30.32 12.21 19.11 28.37 125.05 77 43.42 88.58 41.68 16.79

–3.89 2.36842 –10.95 0 2.42 10.58 0.68 –1.21 3.89 –2.37 10.95 0 –2.42 –10.58 –0.68 1.21

 2  8.183

(O - E ) 2 15.1321 5.609413 119.9025 0 5.8564 111.9364 0.4624 1.4641 15.1321 5.6169 119.9025 0 5.8564 111.9364 0.4624 1.4641

(O - E )2 E 1.0894 .271885 1.3183 0 0.1854 1.7376 0.0153 0.1200 0.7918 0.1980 0.9588 0 0.1349 1.2637 0.0111 0.0872 8.183

464

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(d) Because  2  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. 3. Population 1: Ithaca  s12  72.59 Population 2: Little Rock  s22  52.48 (a) H 0 :  12   22; H a :  12   22 (claim) (b) d.f.N  12 , d.f.D  14 ; F0  4.43 ; Rejection region: F  4.43

s12 72.59  1.38 (c) F  2  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. 4. (a) H 0 : 1   2   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 699.2111 2264.2159

Degrees of Freedom 2 40

Mean Squares 349.6055 56.6054

F 6.18

F  6.18

(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.

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

CUMULATIVE REVIEW, CHAPTERS 9-10 1. (a) r  0.827 ; 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  1.216 x  1.088

(d) 10.95 seconds 2.

H 0 : 1   2   3   4 (claim) H a : At least one mean is different from the others d.f.N  3 , d.f.D  28 F0  2.29 ; Rejection region: F  2.29

465

466

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

Variation Between Within

Sum of Squares 1,658,955.625 2,331,130.250

Degrees of Freedom 3 28

Mean Squares 552,985.2083 83,254.65179

F 6.642

F  6.64

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  16, 212  0.227(110,000)  0.221(100,000)  12, 442

(b)

y  16, 212  0.227(125,000)  0.212(115,000)  12, 217

H0 : 12   22 (claim); Ha : 12   22 d.f.N  15 , d.f.D  14 F0  2.46; Rejection region: F  2.46

s 2 34.62 F  12   1.086 s2 33.22 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 Colorado and Utah. 5.

H 0 : The distribution of credit card debts for college students are distributed as 32.8% $0, 42.2% $1-$500,

11.4% $501-$1000, 5.1% $1001-$2000, 5.3% $2001-$4000, and 3.2% More than $4000. (claim) H a : The distribution of credit card debts for college students differs from the claimed distribution.

02  11.071 ; Rejection region:  2  11.071 Distribution

Observed

Expected

.328 .422 .114 .051 .053 .032

290 397 97 54 40 22

295.2 379.8 102.6 45.9 47.7 28.8

$0 $1-$500 $501-$1000 $1001-$2000 $2001-$4000 More than $4000

(O  E )2 E 0.0916 0.7789 0.3057 1.4294 1.2430 1.6056 5.4541

 2  5.45 Because   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. 2

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

467

H 0 : Rating and gender are independent. H a : Rating and gender are dependent. (claim) 2 2 d.f.  (r  1)(c  1)  3 ; 0  7.815 ; Rejection region:   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

2 -4.5 -0.5 3 -2 4.5 0.5 -3

4.00 20.25 0.25 9.00 4.00 20.25 0.25 9.00

(O - E )2 E 0.0404 0.5400 0.0098 1.1250 0.0404 0.5400 0.0098 1.1250 3.4304

 2  3.43 Because   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 gender. 2

( 7. (a) r =

å y i - y

) » 0.751 2

å ( yi - y )

About 75.1% of the variation in height can be explained by the relationship between metacarpal bone length and height, and about 24.9% of the variation is unexplained.

(b) se =

(

å yi -  yi

) = 149.4 » 3.87 2

10 n-2 The standard error of estimate of the height for a specific metacarpal bone length is about 3.876 centimeters.

n ( x - x) 1 E = tc se 1 + + n n (å x 2 ) - (å x)2 2

» (2.228)(3.87) 1 +

12(50 - 44.5833) 1 + 12 12(24007) - (535)2

» 9.7 y  E  179.73  9.7  (170.03, 189.43) You can be 95% confident that the height will be between 170.03 centimeters and 189.43 centimeters when the metacarpal bone length is 50 centimeters.

468

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

CHAPTER 10 TEST SOLUTIONS 1. (a) H 0 :  12   22 (claim); H a :  12   22 (b) d.f.N  11 , d.f.D  10 ; F0  3.66 ; Rejection region: F  3.66 s 2 12.11 (c) F  12   1.03 s2 11.73

(d) Because F  3.66 , 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. (a) H 0 :  12   22 ; H a :  12   22 (claim) (b) d.f.N  9 , d.f.D  11 ; F0  4.63 ; Rejection region: F  4.63 (c) F 

s12 s22



13.16  1.09 12.11

(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 : 1   2   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

Sum of Squares 322.7713 368.9232

Degrees of Freedom 2 30

Mean Squares 161.3856 12.2974

F 13.12

F  13.12

(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.

CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

469

4. (a) H 0 : The distribution of ages of workers in Oklahoma is 6.1% ages 16-19, 11.8% ages 20-24, 41.8% ages 25-44, 21.2% ages 45-54, 10.1% ages 55-59, and 9.0% ages 60+ H a : The distribution of the ages of workers in Oklahoma differs from the distribution of ages of workers in Maine. (claim) (b)  02  9.236 , Rejection region:  2  9.236 (c)

Ages

Distribution

Observed

Expected

16-19 20-24 25-44 45-54 55-59 60+

6.1% 11.8% 41.8% 21.2% 10.1% 9.0%

11 27 96 37 14 15

12.2 23.6 83.6 42.4 20.2 18

(O - E ) 2 E 0.1180 0.4898 1.8392 0.6877 1.9030 0.5 5.538

 2  5.538 (d) Because  2  9.236 , fail to reject H 0 . (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 6.1% ages 16-19, 11.8% ages 2024, 41.8% ages 25-44, 21.2% ages 45-54, 10.1% ages 55-59, and 9.0% ages 60+. (claim) H a : The distribution of the ages of workers in Massachusetts differs from the distribution of ages of workers in Maine. (b)  02  15.086 , Rejection region:  2  15.086 (c)

Ages

Distribution

Observed

Expected

16-19 20-24 25-44 45-54 55-59 60+

6.1% 11.8% 41.8% 21.2% 10.1% 9.0%

15 22 86 42 17 18

12.2 23.6 83.6 42.4 20.2 18

 2  1.33 (d) Because  2  15.086 , fail to reject H 0 .

(O - E ) 2 E 0.6426 0.1085 0.0689 0.0038 0.5069 0 1.3307

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CHAPTER 10 │ CHI-SQUARE TESTS AND THE F-DISTRIBUTION

(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. (b) d.f.  ( r  1)(c  1)  5 ;  02  11.071 ; Rejection region:  2  11.071 (c)

OE

(O - E ) 2

11 27 96 37 14 15 15 22 86 42 17 18

13 24.5 91 39.5 15.5 16.5 13 24.5 91 39.5 15.5 16.5

–2 2.5 5 –2.5 –1.5 –1.5 2 –2.5 –5 2.5 1.5 1.5

4 6.25 25 6.25 2.25 2.25 4 6.25 25 6.25 2.25 2.25

(O - E )2 E 0.3077 0.2551 0.2747 0.1582 0.1452 0.1364 0.3077 0.2551 0.2747 0.1582 0.1452 0.1364 2.5546

 2  2.555 (d) Because  2  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

11.1 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 greater than 120.” H 0 : median 120; H a : median > 120 (claim)   0.025 n  24 The critical value is 6. x6 Because x  6 , reject H0 . There is enough evidence at the 2.5% level of significance to support the agency’s claim that the median number of days a home is on the market in its city is greater than 120. 2. The claim is “the median age of museum workers in the United States is 46 years old.” H 0 : median  46 (claim)

H a : median  46   0.10 n  91 The critical value is –1.645. x  34 z

( x  0.5)  0.5n n 2



(34  0.5)  0.5(91) 91 2



11  2.306 4.77

Because z  1.645 , reject H0 . There is enough evidence at the 10% level of significance to reject the claim that the median age of museum workers in the United States is 46 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)

  0.05

n  11 The critical value is 2. x2 Because x  2 R, 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.

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CHAPTER 11 │ NONPARAMETRIC TESTS

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  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 $300.” H 0 : median  $300; H a : median > $300 (claim)

(b) The critical value is 1. (c) x  5 (d) Because x  1 , fail to reject H0 . (e) There is not enough evidence at the 1% level of significance to support the accountant’s claim that the median credit card balance of college students is more than $300. 8. (a) The claim is “the median daily high temperature for the month of July in Pittsburgh is 83 Fahrenheit.” H 0 : median = 83 (claim); H a : median  83

(b) The critical value is 1. (c) x  4 (d) Because x  1 , fail to reject H0 .

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(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 83 Fahrenheit. 9. (a) The claim is “the median sales price of new privately owned one-family homes sold in a recent month is $253,000 or less.” H 0 : median  $253, 000 (claim); H a : median  $253, 000

(b) The critical value is 1 (c) x  4 (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 $253,000 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  66

(b) The critical value is 2. (c) x  0 (d) Because x  2 , reject H 0 . (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 amount of credit card debt for families holding such debt is at least $2300.” H 0 : m edian  $2300 (claim); H a : median  $2300

(b) The critical value is z 0   2.05 (c) x  44 ( x  0.5)  0.5n (44  0.5)  0.5(104) 7.5 z    1.47 5.099 n 104 2 2 (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 institution’s claim that the median amount of credit card debt for families holding such debts is at least $2300. 12. (a) The claim is “the median amount of financial debt for families holding such debt is less than $60,000.” Copyright © 2019 Pearson Education Ltd.

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H 0 : median  $60, 000 ; H a : median  $60, 000 (claim)

(b) The critical value is z0  1.96 (c) x  24 ( x  0.5)  0.5( n) (24  0.5)  0.5(70) 10.5 z    2.51 4.183 n 70 2 2 (d) Because z   1.96 , reject H 0 . (e) There is enough evidence at the 2.5% level of significance to support the accountant’s claim that the median amount of financial debt for families holding such debts is less than $60,000. 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  30; H a : median > 30 (claim)

(b) The critical value is 4. (c) x  10 (d) Because x  4 , fail to reject H 0 . (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  32; H a : median < 32 (claim)

(b) The critical value is 5. (c) x  5 (d) Because x  5 , reject H 0 . (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  4

(b) The critical value is z0  1.96. (c) x  29

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z

( x  0.5)  0.5(n) n 2



(29  0.5)  0.5(82) 82 2



475

11.5  2.54 4.53

(d) Because z   1.96 , reject H 0 . (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  1000

(b) The critical value is 5 (c) x  7 (d) Because x  5 , fail to reject H 0 . (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  $41.93

(b) The critical value is –2.575 (c) x  18 3 ( x  0.5)  0.5(n) (18  0.5)  0.5(43) z    0.91 3.28 n 43 2 2 (d) Because z   2.575 , fail to reject H 0 . (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 $60.01.” H 0 : median  $60.01 (claim); H a : median < $60.01

(b) The critical value is 6 (c) x  5 (d) Because x  6 , reject H 0 . (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 $60.01. 19. (a) The claim is “the lower back pain intensity scores will decrease after acupuncture treatment.” Copyright © 2019 Pearson Education Ltd.

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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  0 (d) Because x  1 , reject H 0 . (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  4 (d) Because x  2 , fail to reject H 0 . (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  1 , reject H 0 . (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)

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(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 than their real age is equal to the proportion of adults who feel younger than their real age. (claim) H a : The proportion of adults who feel older than their real age is different from the proportion of adults who feel younger than their real age. The critical value is 3. x3

Because x  3 , reject H 0 . (b) There is enough evidence at the 5% level of significance to reject the claim that 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. 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 H 0 . (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 female workers is less than or equal to $765.” H 0 : median  $765 (claim); H a : median  $765

(b) The critical value is z0  2.33. (c) x  29 ( x  0.5)  0.5( n) (29  0.5)  0.5(47) 5 z    1.46 3.428 n 47 2 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 weekly earnings of female workers is less than or equal to $765. 26. (a) The claim is “the median weekly earnings of male workers is greater than $950.” H 0 : median  $950 ; H a : median  $950 (claim) Copyright © 2019 Pearson Education Ltd.

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(b) The critical value is z0  2.33. (c) x  45 ( x  0.5)  0.5( n) (45  0.5)  0.5(68) 10.5 z    2.55 4.123 n 68 2 2 (d) Because z  2.33 , reject H 0 . (e) There is enough evidence at the 1% level of significance to support the organization’s claim that the median weekly earnings of male workers is greater than $950. 27. (a) The claim is “the median age of brides at the time of their first marriage is less than or equal to 27 years old.” H 0 : median  27 (claim); H a : median > 27

(b) The critical value is z0  1.645. (c) x  35 ( x  0.5)  0.5(n) (35  0.5)  0.5(59) 5 z    1.30 3.841 n 59 2 2 (d) Because z  1.645 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to reject the counselor’s claim that the median age of brides 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 grooms at the time of their first marriage is greater than 28 years.” H 0 : median  28; H a : median > 28 (claim)

(b) The critical value is z0  1.645. (c) x  33 ( x  0.5)  0.5(n) (33  0.5)  0.5(56) 4.5 z    1.20 3.742 n 56 2 2 (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 counselor’s claim that the median age of grooms at the time of their first marriage is greater than 28 years.

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11.2 THE WILCOXON TESTS 11.2 TRY IT YOURSELF SOLUTIONS 1. The claim is “a spray-on water repellant is effective.” H 0 : The water repellent does not increase the water repelled. H a : The water repellent increases the water repelled. (claim)   0.01 n  11

The critical value is 7. No Repellent repellent applied Difference 8 15 7 7 12 5 7 11 4 4 6 2 6 6 0 10 8 2 9 8 1 5 6 1 9 12 3 11 8 3 8 14 6 4 8 4 Sum of negative ranks = 55.5 Sum of positive ranks = 10.5

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

ws  10.5 Because ws  7 , fail to reject H 0 .

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 paid by the companies.” H 0 : There is no difference in the claims paid by paid by the companies. H a : There is a difference in the claims paid by paid by the companies. (claim)   0.05 The critical values are z0  1.96. n1  12 and n2  12

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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

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) n (n  n  1) 12(12  12  1)   150 R  1 1 2 2 2

n1n2 (n1  n2  1) (12)(12)(12  12  1)   17.321 12 12 R  R 120.5 150 z   1.703 (or z  1.703) R 17.321

R 

Because  1.96  z  1.96 , 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 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.” H 0 : There is no reduction in diastolic blood pressure. (claim) H a : There is a reduction in diastolic blood pressure.

(b) Wilcoxon signed-rank test (c) The critical value is 10.

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(d)

Before treatment

After treatment

108 109 120 129 112 111 117 135 124 118 130 115 Sum of negative ranks = 17 Sum of positive ranks = 61

99 115 105 116 115 117 108 122 120 126 128 106

Difference

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

8 4.5 12 10.5 2 4.5 8 10.5 3 6 1 8

481

ws  17

(e) Because ws  10 , fail to reject H 0 . (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.” H 0 : There is no difference in salaries. (claim) H a : There is a difference in salaries.

(b) Wilcoxon rank sum test (c) The critical value is z0   1.645. (d)

Ordered data 52

Sample

Rank

Ordered data

Sample

Rank

16.5

7.5

16.5

7.5

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R = sum ranks of Wholesale Trade = 129 (or R  81)

n (n1  n2  1) 10(10  10  1)   105 2 2 n n (n  n  1) (10)(10)(10  10  1) R  1 2 1 2   13.229 12 12 R  R 129 105 z   1.81 R 13.229

R  1

(e) Because z  1.645 , reject H 0 . (f) There is 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.” H 0 : There is no difference in the earnings. H a : There is a difference in the earnings. (claim)

(b) Wilcoxon rank sum test (c) The critical values are z0  1.96. (d)

Ordered data 52 58 58 60 62 62 64 68 71 78 84 85 87 88 90 91 91 95 98 98 99

sample B B B B B B B B B B B A A A A A A A A A A

R = sum ranks of advanced degree = 165

Rank 1 2.5 2.5 4 5.5 5.5 7 8 9 10 11 12 13 14 15 16.5 16.5 18 19.5 19.5 21

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n (n1  n2  1) 10(10  11  1)   110 2 2 n n (n  n  1) (10)(11)(10  11  1) R  1 2 1 2   14.201 12 12 R  R 165 110 z   3.87 R 14.201

R  1

(e) Because z  1.96 , reject H 0 . (f) There is enough evidence at the 5% 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.” H 0 : The new drug does not affect the number of headache hours. H a : The new drug affects the number of headache hours. (claim)

(b) Wilcoxon signed-rank test (c) The critical value is 2. (d)

Headache hours Headache before hours after 0.8 1.6 2.4 1.3 2.8 1.6 2.6 1.4 2.7 1.5 0.9 1.6 1.2 1.7 Sum of negative ranks = 6 Sum of positive ranks = 22

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

3 4 6 6 6 2 1

ws  6

(e) Because ws  2 , fail to reject H 0 . (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 Wisconsin and Michigan.” H 0 : There is no difference in salaries. H a : There is a difference in salaries. (claim)

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(d)

Ordered data sample Rank 47 1 W 49 2 W 51 3 W 52 4 W 53 5.5 M 53 5.5 W 55 8 M 55 8 W 55 8 W 56 10 W 57 11 M 58 12 M 59 13.5 W 59 13.5 M 61 16.5 W 61 16.5 W 61 16.5 M 61 16.5 M 62 19 M 65 20 M 67 21.5 M 67 21.5 M 76 23 M R = sum ranks of Wisconsin = 88 n (n  n  1) 11(11  12  1) R  1 1 2   132 2 2

n1n2 (n1  n2  1) (11)(12)(11  12  1)   16.248 12 12 R  R 88 132 z   2.71 R 16.248

R 

(e) Because z   1.96 , 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 Wisconsin and Michigan. 8. (a) The claim is “the experimental medication affects an individual’s heart rate.” H 0 : The experimental medication does not affect an individual’s heart rate. H a : The experimental medication affects an individual’s heart rate. (claim)

(b) Wilcoxon signed-rank test

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Before After Difference 72 73 1 81 80 1 75 75 0 76 79 3 79 74 5 74 76 2 65 73 8 67 67 0 76 74 2 83 77 6 66 70 4 75 77 2 76 76 0 78 75 3 68 74 6 Sum of negative ranks = 46.5 Sum of positive ranks = 31.5

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

ws  31.5

(e) Because ws  14 , fail to reject H 0 . (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.” H 0 : The fuel additive does not improve gas mileage. H a : The fuel additive does improve gas mileage. (claim) The critical value is z0  1.28.

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Before After Difference 36.4 36.7 0.3 36.4 36.9 0.5 36.6 37.0 0.4 36.6 37.5 0.9 36.8 38.0 1.2 36.9 38.1 1.2 37.0 38.4 1.4 37.1 38.7 1.6 37.2 38.8 1.6 37.2 38.9 1.7 36.7 36.3 0.4 37.5 38.9 1.4 37.6 39.0 1.4 37.8 39.1 1.3 37.9 39.4 1.5 37.9 39.4 1.5 38.1 39.5 1.4 38.4 39.8 1.4 40.2 40.0 0.2 40.5 40.0 0.5 40.9 40.1 0.8 35.0 36.3 1.3 32.7 32.8 0.1 33.6 34.2 0.6 34.2 34.7 0.5 35.1 34.9 0.2 35.2 34.9 0.3 35.3 35.3 0 35.5 35.9 0.4 35.9 36.4 0.5 36.0 36.6 0.6 36.1 36.6 0.5 37.2 38.3 1.1 Sum of negative ranks = 484.5 Sum of positive ranks = 43.5

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

ws  43.5

z

n(n  1) 4  n(n  1)(2n  1) 24 ws 

32(32  1) 220.5 4   4.123 2860 32(32  1)  (2)32  1 24 43.5 

Note: n = 32 because one of the differences is zero and should be discarded. Because z   1.28 , reject H 0 . There is enough evidence at the 10% level of significance for the engineer to conclude that the gas mileage is improved.

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10. The claim is “a fuel additive improves gas mileage.” H 0 : The fuel additive does not improve gas mileage. H a : The fuel additive does improve gas mileage. (claim) The critical value is z0   1.645. Before After Difference 34.0 36.6 2.6 34.2 36.7 2.5 34.4 37.2 2.8 34.4 37.2 2.8 34.6 37.3 2.7 34.8 37.4 2.6 35.6 37.6 2 35.7 37.7 2 30.2 34.2 4 31.6 34.9 3.3 32.3 34.9 2.6 33.0 34.9 1.9 33.1 35.7 2.6 33.7 36.0 2.3 33.7 36.2 2.5 33.8 36.5 2.7 35.7 37.8 2.1 36.1 38.1 2 36.1 38.2 2.1 36.6 38.3 1.7 36.6 38.3 1.7 36.8 38.7 1.9 37.1 38.8 1.7 37.1 38.9 1.8 37.2 39.1 1.9 37.9 39.1 1.2 37.9 39.2 1.3 38.0 39.4 1.4 38.0 39.8 1.8 38.4 40.3 1.9 38.8 40.8 2 42.1 43.2 1.1 Sum of negative ranks = 528 Sum of positive ranks = 0 ws  0

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

487

488

CHAPTER 11 │ NONPARAMETRIC TESTS

n(n  1) 32(32  1) 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 H 0 . There is enough evidence at the 5% level of significance to support the engineer’s claim that the fuel additive improves gas mileage. ws 

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 of the three states is different from the others.” H 0 : The distribution of the salaries is the same in all three states. H a : The distribution of the salaries in at least one state is different from the others. (claim)   0.05 d.f.  k  1  2  02  5.991 ; Rejection region:  2  5.991

State CA TX TX FL CA FL TX TX TX FL TX TX FL FL CA CA FL TX FL CA CA FL

Rank 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

CHAPTER 11 │ NONPARAMETRIC TESTS

143.3 146.1 147.7 149.9 154 160.2

FL CA TX TX CA CA

489

23 24 25 26 27 28

R1  121 , R2  128 , R3  157

H 

 R12 R2 2 R3 2  12      3( N  1) N ( N  1)  n1 n2 n3   (121) 2 (128) 2 (157) 2  12      3(28  1) 28(28  1)  10 9 9 

 2.015 Because H  5.991 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to support the claim that the distribution for 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 annual premiums in at least one state is different from the others.” H 0 : The distribution of the annual premiums is the same in all three states. H a : The distribution of the annual premiums in at least one state is different from the others. (claim)

(b)  02  5.991 ; Rejection region:  2  5.991 (c)

Ordered data 755 766 838 845 950 1034 1035 1098

Sample VA VA VA VA VA MA VA CT

Rank 1 2 3 4 5 6 7 8

Ordered data 1257 1263 1302 1303 1320 1382 1387 1413

Sample MA CT MA CT CT MA MA CT

Rank 12 13 14 15 16 17 18 19

490

CHAPTER 11 │ NONPARAMETRIC TESTS

1132 1166 1179

VA MA CT

9 10 11

1538 1572

CT MA

20 21

R1  102, R2  98, R3  31

H 

 R12 R2 2 R32  12      3( N  1) N ( N  1)  n1 n2 n3   (102) 2 (98) 2 (31) 2  12      3(21  1) 21(21  1)  7 7 7 

 11.807 (d) Because H  5.991 , reject H 0 . (e) There is enough evidence at the 5% level of significance to conclude that the distribution of annual premiums in at least one state 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.” H 0 : The distribution of the hourly pay rates is the same in all three states. H a : The distribution of the hourly pay rates in at least one state is different from the others. (claim)

(b) 0  5.991 ; Rejection region:  2  5.991 2

(c)

Ordered data 25.42 26.13 26.4 27.34 27.68 27.77 27.84 28.43 28.83 28.92 29.28 29.37 29.89 30.47 31.02 31.27 32.24

Sample KY IN KY OH IN KY OH IN IN KY IN KY OH IN KY IN OH

Rank 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

CHAPTER 11 │ NONPARAMETRIC TESTS

32.42 33.64 33.91

KY OH OH

491

18 19 20

R1  65, R2  65, R3  80

H

 R12 R2 2 R32  12      3( N  1) N ( N  1)  n1 n2 n3 

H

 (65)2 (65)2 (80)2  12      3(20  1) 20(20  1)  7 7 6 

 1.966 (d) Because H  5.991 , fail to reject H 0 . (e) There is not enough evidence at the 5% 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 the annual salaries of private industry workers in at least one state is different from the others.” H 0 : The distribution of the annual salaries is the same in all four states. H a : The distribution of the annual salaries in at least one state is different from the others. (claim)

(b)  02  6.251 ; Rejection region:  2  6.251 (c)

Ordered data 32.9 34.8 34.8 35.4 35.4 36.6 38.1 38.3 39.9 39.9 40.3 40.8 41.6 41.7

Sample KY WV WV SC SC WV WV KY KY KY SC NC KY SC

R1  95, R2  136, R3  95, R4  80

Rank 1 2.5 2.5 4.5 4.5 6 7 8 9.5 9.5 11 12 13 14

Ordered data 41.9 43 44.9 45.1 45.9 47.2 48.5 48.8 48.8 49.1 50.3 50.5 59.6 62.1

Sample NC SC NC WV WV NC SC NC NC SC WV KY NC KY

Rank 15 16 17 18 19 20 21 22.5 22.5 24 25 26 27 28

492

CHAPTER 11 │ NONPARAMETRIC TESTS

H

 R12 R2 2 R32  12      3( N  1) N ( N  1)  n1 n2 n3 

H

 (95)2 (136)2 (95)2 (80)2  12       3(28  1) 28(28  1)  7 7 7 7 

 3.667 (d) Because H  6.251 , fail to reject H 0 . (e) There is not enough evidence at the 10% level of significance to conclude that the distribution of annual salaries in at least one state 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.” H 0 : The distribution of the amounts of caffeine is the same for all four beverages. H a : The distribution of the amounts of caffeine in at least one beverage is different from the others. (claim)

(b)  02  1 1 .3 4 5 ; Rejection region:  2  1 1 .3 4 5 (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

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

R1  105, R2  56, R3  104, R4  35

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

CHAPTER 11 │ NONPARAMETRIC TESTS

H 

493

 R12 R22 R32 R42  12       3( N  1) N ( N  1)  n1 n2 n3 n4   (105) 2 (56) 2 (104) 2 (35) 2  12       3(24  1) 24(24  1)  5 7 6 6 

 18.197 (d) Because H 11.345 , reject H 0 . (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. 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.” H 0 : The number of days spent in the hospital is the same for all four regions. H a : The number of days spent in the hospital is different in at least one region. (claim)

02 11.345 ; Rejection region:  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

Sample S W W NE NE NE MW W W W MW S NE NE S S MW NE

R1  220.5, R2  171.5, R3  159.5, R4  151.5

H 

12  R12 R22 R32 R42        3( N  1) N ( N  1)  n1 n2 n3 n4   (220.5)2 (171.5) 2 (159.5)2 (151.5)2  12       3(37  1) 37(37  1)  10 10 8 9 

 1.507

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

494

CHAPTER 11 │ NONPARAMETRIC TESTS

Because H 11.345 , fail to reject H 0 . 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)

Sum of Degrees of Mean Variation squares freedom squares F p Between 9.17 3 3.06 0.52 0.673 Within 194.72 33 5.90 Because   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 H 0 . 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.

(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.” H 0 : The energy consumed is the same in all four regions. H a : The energy consumed is different in at least one region. (claim)

02 11.345 ; Rejection region:  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

R1  251.5, R2  271.5, R3  116.5, R4  140.5

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

CHAPTER 11 │ NONPARAMETRIC TESTS

H 

495

12  R12 R22 R32 R42        3( N  1) N ( N  1)  n1 n2 n3 n4   (251.5)2 (271.5) 2 (116.5) 2 (140.5) 2  12       3(39  1) 39(39  1)  10 11 8 10 

 8.438 Because H 11.345 , fail to reject H 0 . 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

Sum of Degrees of Mean squares freedom squares F p Between 11,993 3 3998 3.06 0.041 Within 45,794 35 1308 Because   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 H 0 . 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.

(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 :  s  0 ; H a :  s  0 (claim) The critical value is 0.714. Oat

Rank

Wheat

Rank

1.84 1.97 2.03 2.25 2.35 2.31 2.40

1 2 3 4 6 5 7

3.67 3.49 3.68 3.88 3.91 4.02 4.15

2 1 3 4 5 6 7

1 1 0 0 1 1 0

1 1 0 0 1 1 0

d  4 2

rs  1 

 d  1  6(4)  0.929

n(n2  1) 7(72  1) Because rs  0.714 , reject H 0 . There is enough evidence at the 10% level of significance to conclude that there is a significant correlation between the oat and wheat prices. Copyright © 2019 Pearson Education Ltd.

496

CHAPTER 11 │ NONPARAMETRIC TESTS

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. 4. The value of rs represents the correlation coefficient of the sample data, whereas  s 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.” H 0 : :  s  0 ; H a :  s  0 (claim)

(b) The critical value is 0.738 (c)

Seed expenses 490 1530 490 266 741 380 879 360

Rank 4.5 8 4.5 1 6 3 7 2

Fertilizer/lime expense 480 2060 480 402 642 470 858 560

Rank 3.5 8 3.5 1 6 2 7 5

d 1 0 1 0 0 1 0 3

d2 1 0 1 0 0 1 0 9

 d  12 2

 d  1  6(12)  0.857 r  1 6

n(n2  1)

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 :  s  0 ; H a :  s  0 (claim)

CHAPTER 11 │ NONPARAMETRIC TESTS

(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

497

d2 0 1 4 16 1 2.25 25 2.25 0

d  51.5 2

rs  1 

d

n(n  1)

1

6(51.5)  0.571 9(92  1)

(d) Because rs  0.700 , fail to reject H 0 . (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 the barley and corn prices.” H 0 :  s  0 ; H a :  s  0 (claim)

(b) The critical value is 0.700 (c)

Barley 4.89 4.52 4.85 4.97 5.12 4.91 5.08 4.98 4.87

Rank 4 1 2 6 9 5 8 7 3

Corn 3.21 3.22 3.29 3.23 3.33 3.4 3.44 3.49 3.43

Rank 1 2 4 3 5 6 8 9 7

d 3 1 2 3 4 1 0 2 4

d2 9 1 4 9 16 1 0 4 16

 d  60 2

rs  1 

 d  1  6(60)  0.5 2

n(n 2  1)

9(92  1)

(d) Because rs  0.700 , fail to reject H 0 . (e) There is not enough evidence at the 5% level of significance to conclude that there is a significant correlation between the barley and corn prices. 8. (a) The claim is “there is a significant correlation between the overall score and the price.” Copyright © 2019 Pearson Education Ltd.

498

CHAPTER 11 │ NONPARAMETRIC TESTS

H 0 :  s  0 ; H a :  s  0 (claim)

(b) The critical value is 0.497 (c)

Score

Rank

Price

Rank

65 71 69 47 55 38 47

9.5 12 11 4.5 7 2 4.5

150 200 550 350 470 90 80

4 6 12 10 11 2 1

5.5 6 1 5.5 4 0 3.5

30.25 36 1 30.25 16 0 12.25

4.5

130

1.5

2.25

47 57 34 65

4.5 8 1 9.5

210 190 300 260

7 5 9 8

2.5 3 8 1.5

6.25 9 64 2.25

d  209.5 2

rs  1 

 d  1  6(209.5)  0.267 2

n(n2  1)

12(122  1)

(d) Because rs  0.497 , fail to reject H 0 . (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.” H 0 :  s  0 ; H a :  s  0 (claim) Science average 528 495 509 481 538 416 493

Rank 8 5 7 2 9 1 3.5

GNI 1529 2458 3437 1815 4549 1143 1192

Rank 4 6 7 5 8 2 3

d 4 1 0 3 1 1 0.5

d2 16 1 0 9 1 1 0.25

493

3.5

503

2.5

6.25

496

18496

3

d  43.5 2

 d  1  6(43.5)  0.638 r  1 6

n(n2  1)

9(92  1)

CHAPTER 11 │ NONPARAMETRIC TESTS

499

Because rs  0.600 , reject H 0 . There is enough evidence at the 10% level of significance to conclude that there is a significant correlation between science achievement scores and GNI. 10. The claim is “there is a significant correlation between mathematics achievement scores and GNI.” H 0 :  s  0 ; H a :  s  0 (claim) The critical value is 0.600 Math average Rank GNI Rank d d2 516 8 1529 4 4 16 1 493 5 2458 6 1 506 7 3437 7 0 0 1 490 4 1815 5 1 532 9 4549 8 1 1 1 408 1 1143 2 1 486 3 1192 3 0 0 494 6 503 1 5 25 49 470 2 18496 9 7

 d  94 2

 d  1  6(94)  0.217 r  1 2

n(n2  1) 9(92  1) Because rs  0.600 , fail to reject H 0 . There is not enough evidence at the 10% level of significance to conclude that there is a significant correlation between mathematics achievement scores and GNI. s

11. The claim is “there is a significant correlation between science and mathematics achievement scores.” H 0 :  s  0 ; H a :  s  0 (claim) The critical value is 0.600 Science average Rank Math average Rank d d2 528 8 516 8 0 0 495 5 493 5 0 0 509 7 506 7 0 0 4 481 2 490 4 2 538 9 532 9 0 0 416 1 408 1 0 0 493 3.5 486 3 0.5 0.25 6.25 493 3.5 494 6 2.5 496 6 470 2 4 16

d  26.5 2

rs  1 

 d  1  6(26.5)  0.779

n(n2  1) 9(92  1) Because rs  0.600 , reject H 0 . There is enough evidence at the 10% level of significance to conclude that there is a significant correlation between science and mathematics achievement scores.

500

CHAPTER 11 │ NONPARAMETRIC TESTS

12. Sample answer: Although there is a strong correlation between math and science scores, and science scores and GNI, there is not a significant correlation between math scores and GNI. 13. The claim is “there is a significant correlation between average hours worked and the number of onthe-job injuries.” H 0 :  s  0 ; H a :  s  0 (claim)

The critical values are  z

n 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

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



 1.645 33  1

  0.291.

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

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 2

rs  1 

d 2

n( n  1)

1

6(246)  0.959 33(332  1)

CHAPTER 11 │ NONPARAMETRIC TESTS

501

Because rs  0.291 , reject H 0 . 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 :  s  0 ; H a :  s  0 (claim)

The critical values are Hours worked 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

z n 1

Rank 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



 1.96 35  1

  0.336.

Injuries 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

Rank 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

D 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

D2 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 2

502

CHAPTER 11 │ NONPARAMETRIC TESTS

d

6(121.5)  0.983 n( n  1) 35(352  1) Because rs  0.336 , reject H 0 . 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  1 

1

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 genders is not random.” H 0 : The sequence of genders is random. H a : The sequence of genders is not random. (claim)

  0.05

M FFF MM FF M F MM FFF n 1 = number of F’s = 9 n 2 = number of M’s = 6 G = number of runs = 8 lower critical value = 4 upper critical value = 13 G=8 Because 4  G  13 , fail to reject H 0 . There is not enough evidence at the 5% level of significance to support the claim that the sequence of genders is not random. 3. The claim is “the sequence of weather conditions is not random.” H 0 : The sequence of weather conditions is random. H a : The sequence of weather conditions is not random. (claim)

  0.05

n 1 = number of N’s = 21 n 2 = number of S’s = 10

G = number of runs = 17 The critical values are z0  1.96.

G  G 

2n1n2 2(21)(10) 1  1  14.55 n1  n2 2110 2n1n2 (2n1n2  n1  n2 ) 2(21)(10)(2(21)(10)  21  10)   2.38 2 (n1  n2 ) (n1  n2  1) (21  10)2 (21  10  1)

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503

G  G 17 14.55   1.03 G 2.38 Because z 1.96 , fail to reject H 0 . z

There is not enough evidence at the 5%level of significance to support the claim that the sequence of weather conditions is not random.

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 n 1 and n 2 are less than or equal to 20, the test statistic is equal to the number of runs G.

When either n 1 or n 2 is greater than 20, the test statistic is equal to

G  G

G

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

n 1 = number of T’s = 6 n 2 = number of F’s = 6

n 1 = number of U’s = 8 n 2 = number of D’s = 6

n 1 = number of M’s = 10 n 2 = number of F’s = 10

10. n 1 = number of A’s = 13 n 2 = number of B’s = 9

11. n 1 = number of T’s = 6 n 1 = number of F’s = 6 too high: 11; too low: 3

12. n 1 = number of M’s = 9 n 2 = number of F’s = 3 too high: 8; too low: 2

13. n 1 = number of N’s = 11 n 1 = number of S’s = 7 too high: 14; too low: 5

14. n 1 = number of X’s = 7 n 2 = number of Y’s = 14 too high: 15; too low: 5

15. (a) The claim is “the tosses were not random.” H 0 : The coin tosses were random. H a : The coin tosses were not random. (claim)

(b) n 1 = number of H’s = 7 n 2 = number of T’s = 9 lower critical value = 4 upper critical value = 14

504

CHAPTER 11 │ NONPARAMETRIC TESTS

(c) G = 9 runs (d) Because 4  G  14 , fail to reject H 0 . (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.” H 0 : The sequence of majority parties is random. H a : The sequence of majority parties is not random. (claim)

(b) n 1 = number of R’s = 20 n 2 = number of D’s = 34 The critical values are z 0   1.96 . (c) G = 15 runs G 

2 n1 n 2 1 n1  n 2

G 

2(20)(34)  1  25.185 20  34

G  G  Z 

Z

2 n1n2  2 n1n2  n1  n2 

 n1  n2   n1  n2  1 2(20)(34)  2(20)(34)  20  34   3.39 2  20  34   20  34  1 2

G  G

G 15  25.185

3.39

 3.3

(d) Because z 1.96 , reject H 0 . (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.” H 0 : The sequence of leagues of winning teams is random. H a : The sequence of leagues of winning teams is not random. (claim)

(b) n 1 = number of N’s = 22 n 2 = number of A’s = 25 The critical values are z 0   1.96 (c) G = 31 runs

CHAPTER 11 │ NONPARAMETRIC TESTS

505

2n n 1 2 1 n n 1 2 2(22)(25) G   1  24.404 22  25 2 n1n2  2 n1n2  n1  n2  G  2  n1  n2   n1  n2  1

G 

G  Z 

2(22)(25)  2(22)(25)  22  25 

 22  25   22  25  1 2

 3.376

G  G

G 31  24.404 Z  1.95 3.376

(d) Because z 1.96 , fail to reject H 0 . (e) There is not 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.” H 0 : The sequence of digits was randomly generated. H a : The sequence of digits was not randomly generated. (claim)

(b) n 1 = number of O’s = 16 n 2 = number of E’s = 16 lower critical value = 11; upper critical value = 23 (c) G = 9 runs (d) Because G  9 , reject H 0 . (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.” H 0 : The microchips are random by gender. H a : The microchips are not random by gender. (claim)

(b) n 1 = number of M’s = 9 n 2 = number of F’s = 20 lower critical value = 8; upper critical value = 18 (c) G = 12 runs

506

CHAPTER 11 │ NONPARAMETRIC TESTS

(d) Because 8  G  18 , fail to reject H 0 . (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.” H 0 : The sequence of past winners is random. H a : The sequence of past winners is not random. (claim)

(b) n 1 = number of F’s = 42 n 2 = number of A’s = 16 The critical values are z 0   1.96 . (c) G = 32 runs

G  G 

z

2n1n2 2(42)(16) 1  1  24.17 n1  n2 42 16 2n1n2 (2n1n2  n1  n2 ) 2(42)(16)(2(42)(16)  42  16)   3.0 2 (n1  n2 ) (n1  n2  1) (42  16)2 (42  16  1)

G  G

G



32  24.17  2.61 3.00

(d) Because z 1.96 , reject H 0 . (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.” H 0 : Daily high temperatures occur randomly. H a : Daily high temperatures do not occur randomly. (claim) median = 87 n 1 = number of above = 15 n 2 = number of below = 13 lower critical value = 9 upper critical value = 21 G = 11 runs Because 9  G  21, fail to reject H 0 , 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.” H 0 : The sequence of exam scores is random. (claim) H a : The sequence of exam scores is not random. median = 87 n 1 = number above median = 14

CHAPTER 11 │ NONPARAMETRIC TESTS

507

n 2 = number below median = 15

lower critical value = 9 upper critical value = 22 G = 15 runs Because 9  G  22 , fail to reject H 0 , 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  650 (claim) Ha: median  650

(b) The critical value is 2. (c) x  7 (d) Because x  2 , fail to reject H 0 . (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.” H 0 : median  710 (claim); H a : median < 710

(b) The critical value is 2. (c) x5 (d) Because x  2 , fail to reject H 0 . (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.” H 0 : median = 2 (claim); H a : median  2

(b) The critical value is z0   1.645. (c) x 65

508

CHAPTER 11 │ NONPARAMETRIC TESTS

z

( x  0.5)  0.5( n ) n 2



(65  0.5)  0.5(174) 174 2



21.5  3.26 6.595

(d) Because z 0   1.645 , reject H 0 . (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.” H 0 : There is no reduction in diastolic blood pressure. (claim) H a : There is a reduction in diastolic blood pressure.

(b) The critical value is 1. (c) x  4 (d) Because x  1 , fail to reject H 0 . (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.” H 0 : There is no reduction in diastolic blood pressure. (claim) H a : There is a reduction in diastolic blood pressure.

(b) The critical value is 2. (c) x3 (d) Because x  2 , fail to reject H 0 . (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 salary of lawyers after graduation from law school is $118,160.” H0: median  $118,160 (claim) Ha: median  $118,160

(b) The critical value is z 0   1.96 (c) x 49 z

( x  0.5)  0.5 n n 2

CHAPTER 11 │ NONPARAMETRIC TESTS

z

(49  0.5)  0.5(125) 125



 13 5.59

509

  2.33

(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 median annual salary is $118,160. 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   2.575. (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

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

n ( n1  n2  1) 12(12  12  1)   150 2 2

R  1

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

510

CHAPTER 11 │ NONPARAMETRIC TESTS

n1 n 2 ( n1  n 2  1) (12)(12)(12  12  1)  17.32 12 12 R   R 173.5  150 z   1.357 R 17.32

R 

z = –1.357 or 1.357. (e) Because z  2.575 , fail to reject H 0 . (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.” H 0 : The new drug does not affect the number of headache hours experienced. H a : 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 After Difference 0.9 1.4 0.5 2.3 1.5 0.8 2.7 1.4 1.3 2.4 1.8 0.6 2.9 1.3 1.6 1.9 0.6 1.3 1.2 0.7 0.5 3.1 1.9 1.2 Sum of negative ranks = 1.5 Sum of positive ranks = 34.5

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

ws  1.5

(e) Because ws  4 , reject H 0 . (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.” H 0 : The distribution of the ages of doctorate recipients is the same in all three fields of study. H a : The distribution of the ages of doctorate recipients in at least one field of study is different from the others.(claim)

(b) 0  9.210 ; Rejection region:   9.210 . 2

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

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

511

Rank 20 20 20 20 20 20 25 25 25 28 28 28 31 31 31 33

R1  191.5, R2  126, R3  243.5

H 

 R12 R22 R32  12      3( N  1) N ( N  1)  n1 n2 n3   (191.5) 2 (126)2 (243.5)2  12      3(33  1) 33(33  1)  11 11 11 

 6.741

(d) Because H  9.210 , fail to reject H 0 . (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)  02  7.815 ; Rejection region:  2  7.815

512

(c)

CHAPTER 11 │ NONPARAMETRIC TESTS

Ordered data 62.7 63.7 64.3 64.4 64.8 64.9 65.3 65.5 65.5 65.6 65.6 65.7 65.8 65.9 66.1 66.4 66.4 66.9 67 67.1 67.3 67.4 67.5 67.6 67.6 68.1 68.1 68.2 68.4 68.5 68.6 68.9 69.4 69.5 69.9 70.2 70.5 71.1 71.7 72.6

Sample Mech Mech Chem Elec Mech Mech Mech Elec Mech Mech Mech Chem Comp Chem Elec Comp Comp Elec Comp Elec Elec Elec Chem Comp Elec Elec Mech Comp Chem Comp Mech Elec Chem Comp Chem Comp Chem Chem Chem Comp

Rank 1 2 3 4 5 6 7 8.5 8.5 10.5 10.5 12 13 14 15 16.5 16.5 18 19 20 21 22 23 24.5 24.5 26.5 26.5 28 29 30 31 32 33 34 35 36 37 38 39 40

CHAPTER 11 │ NONPARAMETRIC TESTS

513

R1  263, R2  257.5, R3  191.5, R4  108

H

12  R12 R2 2 R32 R42        3( N  1) N ( N  1)  n1 n2 n3 n4 

H

 (263)2 (257.5)2 (191.5)2 (108)2  12       3(40  1) 40(40  1)  10 10 10 10 

 11.496 (d) Because H  7.815 , reject H 0 . (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 overall score and the price.” H 0 :  s  0 ; H a :  s  0 (claim)

(b) The critical value is 0.829. (c)

Score 93 91 90 87 85 69

Rank 6 5 4 3 2 1

Price 500 300 500 150 250 130

Rank 5.5 4 5.5 2 3 1

d 0.5 1 1.5 1 1 0

d2 0.25 1 2.25 1 1 0

 d  5.5 2

rs = 1 -

6 (5.5) 6åd2 = 1= 0.8429 2 n (n -1) 6 (6 2 - 1)

(d) Because rs  0.829 , reject H 0 . (e) There is enough evidence at the 10% level of significance to conclude that there is a significant correlation between overall score and the price. 12. (a) The claim is “there is a correlation between the overall score and the price.” H 0 :  s  0 ; H a :  s  0 (claim)

(b) The critical value is 0.714.

514

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(c)

Score 46 73 64 56 94 86 50

Rank 1 5 4 3 7 6 2

Price 24 40 25 24 40 38 26

Rank 1.5 6.5 3 1.5 6.5 5 4

d 0.5 1.5 1 1.5 0.5 1 2

d2 0.25 2.25 1 2.25 0.25 1 4

 d  11 2

rs = 1 -

6 (11) 6å d2 = 1= 0.804 2 n (n - 1) 7 (7 2 - 1)

(d) Because rs  0.714 , reject H 0 . (e) There is enough evidence at the 10% level of significance to conclude that there is a significant correlation between the overall score and the price. 13. (a) The claim is “the stops were not random by gender.” H 0 : The traffic stops were random by gender. H a : The traffic stops were not random by gender. (claim)

(b) n 1 = number of F’s = 12 n 2 = 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 H 0 . (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.” H 0 : The departure status of buses is random. H a : The departure status of buses is not random. (claim)

(b) n 1 = number of T’s = 11 n 2 = number of L’s = 7 lower critical value = 5 upper critical value = 14 (c) G = 5 runs (d) Because G  5 , reject H 0 .

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515

(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 52.” H0: median  52 (claim) Ha: median  52

(b) Sign test (c) The critical values are z 0   1.96 (d) x=23

( x + 0.5) - 0.5n (23 + 0.5) - 0.5(70) n 2

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 52. 2. (a) The claim is “there is a difference in the hourly earnings of union and nonunion workers in state and local governments.” H 0 : There is no difference in the hourly earnings. H a : There is a difference in the hourly earnings. (claim)

(b) Wilcoxon rank sum test (c) The critical values are z 0   1.645 . (d)

Ordered data 20.45 21.20 21.40 22.05 22.25 22.50 23.10 24.75 26.15 26.75

Sample N N N N N N N N N U

Rank 1 2 3 4 5 6 7 8 9 10

Ordered data 26.95 27.35 27.60 27.85 28.15 29.05 29.75 32.30 32.88 35.52

Sample N U U U U U U U U U

Rank 11 12 13 14 15 16 17 18 19 20

516

CHAPTER 11 │ NONPARAMETRIC TESTS

n ( n1  n2  1) 10(10  10  1)   105 2 2

R  1

n1n2 (n1  n2 1) (10)(10)(10 10 1)  13.229 12 12 R  R 56 105   3.70 (or 3.70) z R 13.229

R 

(e) Because z  1.645 , reject H 0 . (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 and nonunion workers 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  2  1 1 .3 4 5 (d)

Ordered data

Sample

Rank

Ordered data

Sample

Rank

148.5 153.9

MW MW

1 2

242.7 243.4

NE NE

17 18

155.6

250.3

156.7 163.3 165.1 166.8 166.9 169.9 170.4 175.3 178.4 178.9 181.3 183.1 196.3

S MW MW S MW MW S S S MW S MW S

4 5 6 7 8 9 10 11 12 13 14 15 16

254.8 257.3 264.2 270.7 275 291.6 303.6 308 320.2 321.7 327.4 331.7 357.4

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

R1  164, R2  59, R3  77, R4  228

CHAPTER 11 │ NONPARAMETRIC TESTS

H 

517

 R12 R22 R32 R42  12       3( N  1) N ( N  1)  n1 n2 n3 n4   (164) 2 (59) 2 (77) 2 (228) 2  12       3(32  1) 32(32  1)  8 8 8 8 

 26.412

(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 :  s  0 ; H a :  s  0 (claim)

(b) Spearman rank correlation coefficient. (c) The critical value is 0.833. Emails Rank Emails sent received 30 8.5 32 30 8.5 36 25 4.5 21 26 6 22 24 3 20 18 1.5 20 18 1.5 22 25 4.5 23 28 7 23

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  20.5 2

rs  1 

d

n(n  1)

1

6(20.5)  0.829 9(92  1)

(e) Because rs  0.833 , fail to reject H 0 . (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.” H 0 : The days with rain are random. H a : The days with rain are not random. (claim)

(b) Runs test

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CHAPTER 11 │ NONPARAMETRIC TESTS

(c) n 1 = number of N’s = 15 n 2 = 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 H 0 . (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.” H 0 : The selection of members is random. H a : The selection of members is not random. (claim)

(b) Runs test (c) n 1 = number of R’s = 18 n 2 = 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 H 0 . (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  2  1 1 .3 4 5

CHAPTER 11 │ NONPARAMETRIC TESTS

(d)

Ordered data

Sample

Rank

Ordered data

Sample

Rank

46.4 49.3 50.5 51.3 51.9 51.9 54.1 54.1 55.2 56 57 57.4 58.5 58.6

S S S S MW S S S MW MW NE MW MW W

1 2 3 4 5.5 5.5 7.5 7.5 9 10 11 12 13 14

58.7 59.6 59.9 60.7 61.1 61.2 61.5 61.9 62.4 63.1 64 64.2 64.7 65.6

MW W NE W MW W NE W NE W W NE NE NE

15 16 17 18 19 20 21 22 23 24 25 26 27 28

519

R1  153, R2  83.5, R3  30.5, R4  139

 R2 R2 R2 R2  12  1  2  3  4   3( N  1) H N ( N  1)  n1 n2 n3 n4     (153) 2 (83.5) 2 (30.5) 2 (139) 2  12    3(28  1)     28(28  1)  7 7 7 7     19.895

(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 people with mutual funds is 51 years.” H 0 : median = 51 (claim); H a : median  51

(b) Sign test (c) The critical value is 3. (d) x = 4 (e) Because x > 4 , fail to reject H 0 . (f) There is not enough evidence at the 1% level of significance to reject the company’s claim that the median age of people with mutual funds is 51 years.

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CHAPTER 11 │ NONPARAMETRIC TESTS

4. (a) The claim is “there is a difference in the weekly earnings of workers who are union members and workers 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 z 0   1.96 (d)

Ordered data

Sample

Rank

649 N 1 687 N 2 730 N 3 747 N 4 783 N 5 788 M 6 790 N 7 850 N 8 890 M 9 895 N 10 896 M 11 906 N 12 919 M 13 951 M 14 954 N 15 980 M 16 1000 M 17 1026 M 18 1087 M 19 1090 M 20 1136 M 21 R = sum ranks of nonmember workers = 67 n ( n  n  1) 10(11  10  1)   110 R  1 1 2 2 2

R 

n1n2 (n1  n2 1) (10)(11)(10 11 1)   14.201 12 12 R  R 67 110 z   3.03 (or 3.03) R 14.201

(e) Because z 1.96 , reject H0.

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521

(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 workers who are union members and workers who are not union members. 5. (a) The claim is “there is a correlation between the overall score and the price.” H 0 :  s  0 ; H a :  s  0 (claim)

(b) Spearman rank correlation coefficient. (c) The critical value is 0.738. Score 90 85 81 78 72 68 64 61

Rank 8 7 6 5 4 3 2 1

Price 495 230 190 160 350 230 260 200

Rank 8 4.5 2 1 7 4.5 6 3

d 0 2.5 4 4 3 1.5 4 2

d2 0 6.25 16 16 9 2.25 16 4

d  69.5 2

d r 1 6

n(n  1)

1

6(69.5)  0.173 8(82  1)

(e) Because rs  0.738 , fail to reject H 0 . (f) There is not enough evidence at the 5% level of significance to conclude that there is a correlation between the overall score and the price.

Alternative Presentation of the Standard Normal Distribution TRY IT YOURSELF SOLUTIONS 1. (1) 0.4857 (2) z = 2.17 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.

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583

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.

é 0 + 1 + 2 +  + 38 + 39 + 40 ù 820 ê ú= = 20 . The average of the whole numbers from 0 to 40 is 20. 41 41 ëê ûú Answers will vary.

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.

584

TECHNOLOGY │

CHAPTER 2 TECHNOLOGY: PARKING TICKETS 1.

x » $44.19

s » $34.03

3. Min = 15, Q1 = 30 , Q2 = 30 , Q3 = 40 , Max = 250 4.

Class

Midpoint

15-29 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

22 37 52 67 82 97 112 127 142 157 172 187 202 217 232 247

Class boundaries 14.5-29.5 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 4 84 2 7 0 2 0 0 0 5 0 0 0 0 0 1

å f = 105

Relative frequency 0.038 0.800 0.019 0.067 0.000 0.019 0.000 0.000 0.000 0.048 0.000 0.000 0.000 0.000 0.000 0.010 f å n »1

The distribution does not appear to be bell-shaped. 6.

x - s = 44.19 - 34.03 = 10.16 ; x + s = 44.19 + 34.03 = 78.22 92.4% of the entries were within one standard deviation of the mean. x - 2 s = 44.19 - 2(34.03) = -23.87 ; x + 2 s = 44.19 + 2(34.03) = 112.25 94.3% of the entries were within two standard deviation of the mean. x - 3s = 44.19 - 3(34.03) = -57.90 ; x + 3s = 44.19 + 3(34.03) = 146.28 94.3% of the entries were within three standard deviations of the mean.

Cumulative frequency 4 88 90 97 97 99 99 99 99 104 104 104 104 104 104 105

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585

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 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 » 0.889 » 89% lie within three standard deviations of the mean. 9 9.

Midpoint, x 22 37 52 67 82 97 112 127 142 157 172 187 202 217 232 247

Frequency, f 4 84 2 7 0 2 0 0 0 5 0 0 0 0 0 1 n = 105

4995 å xf = » $47.57 105 n

xf 88 3108 104 469 0 194 0 0 0 785 0 0 0 0 0 247 å xf = 4995

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TECHNOLOGY │

x- x

( x - x)

( x - x) f

25.57 10.57 4.43 19.43 34.43 49.43 64.43 79.43 94.43 109.43 124.43 139.43 154.43 169.43 184.43 199.43

653.8249 111.7249 19.6249 377.5249 1185.4249 2443.3249 4151.2249 6309.1249 8917.0249 11,974.9249 15,482.8249 19,440.7249 23,848.6249 28,706.5249 34,014.4249 39,772.3249

2615.2996 9384.8916 39.2498 2642.6743 0.0 4886.6498 0.0 0.0 0.0 59,874.6245 0.0 0.0 0.0 0.0 0.0 39,772.3249

å( x - x) f = 119, 215.7145 2

å ( x - x) f 2

119, 215.7145 » $33.86 104 n -1 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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CHAPTER 3 TECHNOLOGY: SIMULATION: COMPOSING MOZART VARIATIONS WITH DICE 1.

2 + 11 = 13 phrases were written.

(11)7 ⋅ 2 ⋅ (11)7 ⋅ 2 = 1.518999334333´1015

3. (a)

1 » 0.091 11

(b) Results will vary.

3 » 0.273 11 14 æ3ö P (option 6, 7, or 8 for all 14 bars) = çç ÷÷÷ » 0.0000000126 çè11ø

4. (a) P (option 6, 7, or 8 for 1st 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 + + = » 0.417 36 36 36 36 14 æ 15 ö÷ ç P (option 6, 7, or 8 for all 14 bars) = ç ÷÷ » 0.00000475 çè 36 ø

6. (a) P (option 6, 7, or 8 for 1st bar ) =

(b) Results will vary.

4 36 4 P(8) = 36 P(4) =

588

TECHNOLOGY │

CHAPTER 4 TECHNOLOGY: USING POISSON DISTRIBUTIONS AS QUEUING MODELS 1.

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

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

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.

Customers Remaining 0 0 0 0 1 2 4 7 6 8 7 8 10 9 9 11 9 7 7 4

│ TECHNOLOGY

P (10) 

589

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 minute  (d) P  no customers are waiting in line after two minutes   (0.0183)2  0.0003

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CHAPTER 5 TECHNOLOGY: AGE DISTRIBUTION IN THE UNITED STATES 1.

m » 35.96

x of the 36 sample means » 36.209 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.

s » 22.238

s of the 36 sample means = 3.552 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. s 22.238 = » 3.706 n 36 This agrees with the Central Limit Theorem.

│ TECHNOLOGY

CHAPTER 6 TECHNOLOGY: UNITED STATES FOREIGN POLICY POLLS 1. (a) pˆ  0.51; qˆ  0.49; z  1.96; n  1527; (0.485, 0.535)

(b) pˆ  0.62; qˆ  0.38; z  1.96; n  1035; (0.590, 0.650) (c) pˆ  0.22; qˆ  0.78; z  1.96; n  1035; (0.195, 0.245) (d) pˆ  0.16; qˆ  0.84; z  1.96; n  485; (0.127, 0.193) (e) pˆ  0.19; qˆ  0.81; z  1.96; n  1035; (0.166, 0.214) 2

z   1.96  ˆ ˆ  c   (0.22)  (0.78)  n  pq   1649 adults  0.02  E

3. (a) Answers will vary.

(b) Answers will vary. 4. Answers will vary.

591

592

TECHNOLOGY │

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 ¹ 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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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    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)   8.802 t 1 2 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) t 1 2   1.010 2 2 s1 s2 (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.

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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):

r » 0.766 r » 0.415 r » 0.913 r » 0.461 r » 0.793 r » 0.230

Largest r: (calories, carbohydrates) 4. (a) y = -12.836 + 0.177 x

(b) y = 1.072 + 0.213x 5. (a) y = -12.836 + 0.177 (120) = 8.404 grams

(b) y = 1.072 + 0.213(120) = 26.632 grams

│ TECHNOLOGY

6. (a) C = 12.929 - 0.268S + 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

595

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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 : 12   22 , H a : 12   22 ( F0  3.28) (Under 500, 1000-2999)  F  1.418  Fail to reject H 0 . (Under 500, At least 12,000)  F  1.583  Fail to reject H 0 . (1000-2999, At least 12,000)  F  2.245  Fail to reject H 0 .

4. The three conditions for a one-way ANOVA test are satisfied. H 0 : 1   2  3 (claim) H a : At least one mean is different from the others. Variation

Between Within

Sum of squares 327271650 534950363

Degrees of Freedom 2 36

Mean squares 163635825 14859732

11.01

0.000186

F  11.01 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 all three samples were taken from approximately normal populations. H 0 : 12   22 , H a : 12   22 ( F0  3.28) (Northeast, Northwest)  F  1.264  Fail to reject H 0 . (Northeast, Southwest)  F  2.017  Fail to reject H 0 . (Northwest, Southwest)  F  1.595  Fail to reject H 0 . The three conditions for a one-way ANOVA test are satisfied. H 0 : 1   2  3 (claim) H a : At least one mean is different from the others. Variation

Between Within

Sum of squares 337764117 748358501

Degrees of Freedom 2 36

Mean squares 168882058 20787736

8.12

0.001

F  8.12 Reject H 0 . There is enough evidence at the 5% level of significance to reject the claim that the mean salaries are the same.

│ TECHNOLOGY

CHAPTER 11 TECHNOLOGY: U.S. INCOME AND ECONOMIC RESEARCH 1.

The median annual incomes do not appear to differ among regions. 2.

H 0 : median  30,000 H a : median > 30,000 (claim) The critical value is 2. x6. 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. (claim) H a : There is a difference in incomes in the Northeast and South. The critical values are z0  1.96. R  159

n (n1  n2  1) 12(12  12  1)   150 2 2 n n (n  n  1) (12)(12)(12  12  1) R  1 2 1 2   17.3 12 12 R   R 159  150 z   0.520 R 17.3 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 that the median annual incomes in the Northeast and South are the same.

R  1

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.

597

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TECHNOLOGY │

H  0.450 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 62,625,869.2 Within 2,472,051,265

Degrees of freedom 3 44

Mean squares 20,875,289.7 56,182,983.3

F 0.37

P 0.774

F  0.37  P  value  0.774 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.

6. (Box-and-whisker plot)

The median family incomes appear to be higher in the Northeast and lower in the South. (Wilcoxon rank sum test) H 0 : There is no difference in incomes in the Northeast and South. (claim) H a : There is a difference in incomes in the Northeast and South. The critical values are z0  1.96 . R  289 n (n  n  1) 15(15  15  1) R  1 1 2   232.5 2 2

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n1n2 ( n1  n2  1) (15)(15)(15  15  1)   24.1 12 12 R   R 289  232.5 z   2.344 R 24.1 Because z  1.96. , reject H 0 . There is enough evidence at the 5% level of significance to reject the claim that the median annual incomes of families in the Northeast and South are the same. (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.

R 

02  7.815 H  7.878 Reject H 0 . There is 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 2,454,721,399 Within 16,158,508,049

Degrees of freedom 3 56

Mean squares 818,240,466 288,544,787

F 2.83

P 0.046

F  2.83  P  value  0.046 P  value < 0.05 , reject H 0 . There is 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.