Triola Elem ISM 12e TOC - Palm Beach State College | … does appear to be a potential to create a bias. 6. There does not appear to be a potential to create a bias. 7. There does

CONTENTS

Chapter 1 ....................................................................................................................... 1 Chapter 2 ....................................................................................................................... 9 Chapter 3 ..................................................................................................................... 29 Chapter 4 ..................................................................................................................... 45 Chapter 5 ..................................................................................................................... 59 Chapter 6 ..................................................................................................................... 73 Chapter 7 ................................................................................................................... 101 Chapter 8 ................................................................................................................... 117 Chapter 9 ................................................................................................................... 139 Chapter 10 ................................................................................................................. 159 Chapter 11 ................................................................................................................. 199 Chapter 12 ................................................................................................................. 211 Chapter 13 ................................................................................................................. 219 Chapter 14 ................................................................................................................. 235

Chapter 1: Introduction to Statistics 1

Copyright © 2014 Pearson Education, Inc.

Chapter 1: Introduction to Statistics Section 1-2

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

2. If the source of the data can benefit from the results of the study, it is possible that an element of bias is introduced so that the results are favorable to the source.

3. A voluntary response sample is a sample in which the subjects themselves decide whether to be included in the study. A voluntary response sample is generally not suitable for a statistical study because the sample may have a bias resulting from participation by those with a special interest in the topic being studied.

4. Even if we conduct a study and find that there is a correlation, or association, between two variables, we cannot conclude that one of the variables is the cause of the other.

5. There does appear to be a potential to create a bias.

6. There does not appear to be a potential to create a bias.

7. There does not appear to be a potential to create a bias.

8. There does appear a potential to create a bias.

9. The sample is a voluntary response sample and is therefore flawed.

10. The sample is a voluntary response sample and is therefore flawed.

11. The sampling method appears to be sound.

12. The sampling method appears to be sound.

13. Because there is a 30% chance of getting such results with a diet that has no effect, it does not appear to have statistical significance, but the average loss of 45 pounds does appear to have practical significance.

14. Because there is only a 1% chance of getting the results by chance, the method appears to have a statistical significance. The result of 540 boys in 1000 births is above the approximately 50% rate expected by chance, but it does not appear to be high enough to have practical significance. Not many couples would bother with a procedure that raises the likelihood of a boy from 50% to 54%.

15. Because there is a 23% chance of getting such results with a program that has no effect, the program does not appear to have statistical significance. Because the success rate of 23% is not much better than the 20% rate that is typically expected with random guessing, the program does not appear to have practical significance.

16. Because there is a 25% chance of getting such results with a program that has no effect, the program does not appear to have statistical significance. Because the average increase is only 3 IQ point, the program does not appear to have practical significance.

17. The male and female pulse rates in the same column are not matched in any meaningful way. It does not make sense to use the difference between any of the pulse rates that are in the same column.

18. Yes, the source of the data is likely to be unbiased.

19. The data can be used to address the issue of whether males and females have pulse rates with the same average (mean) value.

20. The results do not prove that the populations of males and females have the same average (mean) pulse rate. The results are based on a particular sample of five males and five females, and analyzing other samples might lead to a different conclusion. Better results would be obtained with larger samples.

2 Chapter 1: Introduction to Statistics


21. Yes, each IQ score is matched with the brain volume in the same column, because they are measurements obtained from the same person. It does not make sense to use the difference between each IQ score and the brain volume in the same column, because IQ scores and brain volumes use different units of measurement. For example, it would make no sense to find the difference between an IQ score of 87 and a brain volume of 1035 cm3.

22. The issue that can be addressed is whether there is a correlation, or association, between IQ score and brain volume.

23. Given that the researchers do not appear to benefit from the results, they are professionals at prestigious institutions, and funding is from a U.S. government agency, the source of the data appears to be unbiased.

24. No. Correlation does not imply causation, so a statistical correlation between IQ score and brain volume should not be used to conclude that larger brain volumes cause higher IQ scores.

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

26. The sample is a voluntary response sample, so there is a good chance that the results are not valid.

27. The correlation, or association, between two variables does not mean that one of the variables is the cause of the other. Correlation does not imply causation.

28. The correlation, or association, between two variables does not mean that one of the variables is the cause of the other. Correlation does not imply causation.

29. a. The number of people is (0.39)(1018) 397.02=

b. No. Because the result is a count of people among 1018 who were surveyed, the result must be a whole number.

c. The actual number is 397 people

d. The percentage is 255

0.25049 25.049%1018

= =

30. a. The number of women is (0.38)(427) 162.26=

b. No. Because the result is a count of women among 427 who were surveyed, the result must be a whole number.

b. The actual number is 162 women.


0.07026 7.026%427

= =

31. a. The number of adults is (0.14)(2302) 322.28=

b. No. Because the result is a count of adults among 2302 who were surveyed, the result must be a whole number.

c. The actual number is 322 adults.


0.01998 1.998%2302

= =

32. a. The number of adults is (0.76)(2513) 1909.88=

b. No. Because the result is a count of adults among 2513 who were surveyed, the result must be a whole number.

b. The actual number is 1910 adults.


0.13012 13.012%2513

= =

33. Because a reduction of 100% would eliminate all of the size, it is not possible to reduce the size by 100% or more.



34. If the Club eliminated all car thefts, it would reduce the odds of car theft by 100%, so the 400% figure is impossible.

35. If foreign investment fell by 100% it would be totally eliminated, so it is not possible for it to fall by more than 100%.

36. Because a reduction of 100% would eliminate all plague, it is not possible to reduce it by more than 100%.

37. Without our knowing anything about the number of ATVs in use, or the number of ATV drivers, or the amount of ATV usage, the number of 740 fatal accidents has no context. Some information should be given so that the reader can understand the rate of ATV fatalities.

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

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

Section 1-3

1. A parameter is a numerical measurement describing some characteristic of a population, whereas a statistic is a numerical measurement describing some characteristic of a sample.

2. Quantitative data consist of numbers representing counts or measurements, whereas categorical data can be separated into different categories that are distinguished by some characteristic that is not numerical.

3. Parts (a) and (c) describe discrete data.

4. The values of 1010 and 55% are both statistics because they are based on the sample. The population consists of all adults in the United States.

5. Statistic

6. Parameter

7. Parameter

8. Statistic

9. Parameter

10. Parameter

11. Statistic

12. Statistic

13. Continuous

14. Discrete

15. Discrete

16. Continuous

17. Discrete

18. Discrete

19. Continuous

20. Continuous

21. Nominal

22. Ratio

23. Interval

24. Ordinal

25. Ratio

26. Nominal

27. Ordinal

28. Interval

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

30. The flight numbers do not count or measure anything. They are at the nominal level of measurement, and it does not make sense to compute the average (mean) of them.

31. The numbers are used as substitutes for the categories of low, medium, and high, so the numbers are at the ordinal level of measurement. It does not make sense to compute the average (mean) of such numbers.

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



33. a. Continuous, because the number of possible values is infinite and not countable.

b. Discrete, because the number of possible values is finite.

c. Discrete, because the number of possible values is finite.

d. Discrete, because the number of possible values is infinite and countable.

34. Either ordinal or interval is a reasonable answer, but ordinal makes more sense because differences between values are not likely to be meaningful. For example, the difference between a food rated 1 and a food rated 2 is not necessarily the same as a difference between a food rated 9 and a food rated 10.

35. With no natural starting point, temperatures are at the interval level of measurement, so ratios such as “twice” are meaningless.

Section 1-4

1. No. Not every sample of the same size has the same chance of being selected. For example, the sample with the first two names has no chance of being selected. A simple random sample of (n) items is selected in such a way that every sample of same size has the same chance of being selected.

2. In an observational study, you would examine subjects who consume fruit and those who do not. In the observational study, you run a greater risk of having a lurking variable that affects weight. For example, people who consume more fruit might be more likely to maintain generally better eating habits, and they might be more likely to exercise, so their lower weights might be due to these better eating and exercise habits, and perhaps fruit consumption does not explain lower weights. An experiment would be better, because you can randomly assign subjects to the fruit treatment group and the group that does not get the fruit treatment, so lurking variables are less likely to affect the results.

3. The population consists of the adult friends on the list. The simple random sample is selected from the population of adult friends on the list , so the results are not likely to be representative of the much larger general population of adults in the United States.

4. Because there is nothing about left-handedness or right-handedness that would affect being in the author’s classes, the results are likely to be typical of the population. The results are likely to be good, but convenience samples in general are not likely to be so good.

5. Because the subjects are subjected to anger and confrontation, they are given a form or treatment, so this is an experiment, not an observational study.

6. Because the subjects were given a treatment consisting of Lipitor, this is an experiment.

7. This is an observational study because the therapists were not given any treatment. Their responses were observed.

8. This is an observational study because the survey subjects were not given any treatment. Their responses were observed.

9. Cluster

10. Convenience

11. Random

12. Systematic

13. Convenience

14. Random

15. Systematic

16. Cluster

17. Random

18. Cluster

19. Convenience

20. Systematic

21. The sample is not a simple random sample. Because every 1000th pill is selected, some samples have no chance of being selected. For example, a sample consisting of two consecutive pills has no chance of being selected, and this violates the requirement of a simple random sample.

22. The sample is not a simple random sample. Not every sample of 1500 adults has the same chance of being selected. For example, a sample of 1500 women has no chance of being selected.

23. The sample is a simple random sample. Every sample of size 500 has the same chance of being selected.



24. The sample is a simple random sample. Every sample of the same size has the same chance of being selected.

25. The sample is not a simple random sample. Not every sample has the same chance of being selected. For example, a sample that includes people who do not appear to be approachable has no chance of being selected.

26. The sample is not a simple random sample. Not all samples of the same size have the same chance of being selected. For example, a sample would not be selected which included people who do not appear to be approachable.

27. Prospective study

28. Retrospective study

29. Cross-sectional study

30. Prospective study

31. Matched pairs design

32. Randomized block design

33. Completely randomized design

34. Matched pairs design

35. Blinding is a method whereby a subject (or a person who evaluates results) in an experiment does not know whether the subject is treated with the DNA vaccine or the adenoviral vector vaccine. It is important to use blinding so that results are not somehow distorted by knowledge of the particular treatment used.

36. Prospective: The experiment was begun and results were followed forward in time. Randomized: Subjects were assigned to the different groups through the process of random selection, and whereby they had the same chance of belonging to each group. Double-blind: The subjects did not know which of the three groups they were in, and the people who evaluated results did not know either. Placebo-controlled: There was a group of subjects who were given a placebo, by comparing the placebo group to the two treatment groups, the effect of the treatments might be better understood.

Chapter Quick Quiz

1. No. The numbers do not measure or count anything.

2. Nominal

3. Continuous

4. Quantitative data

5. Ratio

6. False

7. No

8. Statistic

9. Observational study

10 False

Review Exercises

1. a. Discrete

b. Ratio

c. Stratified

d. Cluster

e. The mailed responses would be a voluntary response sample, so those with strong opinions are more likely to respond. It is very possible that the results do not reflect the true opinions of the population of all costumers.

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

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



4. a. It uses a voluntary response sample, and those with special interests are more likely to respond, so it is very possible that the sample is not representative of the population.

b. Because the statement refers to 72% of all Americans, it is a parameter (but it is probably based on a 72% rate from the sample, and the sample percentage is a statistic).

c. Observational study.

5. a. If they have no fat at all, they have 100% less than any other amount with fat, so the 125% figure cannot be correct.

b. The exact number is (0.58)(1182) 685.56= . The actual number is 686.

c. 331

0.28003 28.003%1182

= =

6. The Gallop poll used randomly selected respondents, but the AOL poll used a voluntary response sample. Respondents in the AOL poll are more likely to participate if they have strong feelings about the candidates, and this group is not necessarily representative of the population. The results from the Gallop poll were more likely to reflect the true opinions of American voters.

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

8. a. Random

b. Stratified

c. Nominal

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

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

9. a. Systematic

b. Random

c. Cluster

d. Stratified

e. Convenience

f. No, although this is a subjective judgment.

10. a. ( )0.52 1500 780= adults

b. 345

0.23 23%1500

= =

c. Men: 727

0.485 48.5%1500

= = ;

Women: 773

0.515 51.5%1500

= =

Cumulative Review Exercises

1. The mean is 11. Because the flight numbers are not measures or counts of anything, the result does not have meaning.

2. The mean is 101, and it is reasonably close to the population mean of 100.

3. ( )247 176

11.836

−= is an unusually high value.

4. (175 172)

0.4629

20

−=

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

5. ( )2

2

1.96 0.251067

0.03

×=

6. ( )288 88.57

0.003788.57

−=



7. ( ) ( ) ( )( )

( )

2 2 296 100 106 100 98 100

28.03 1

− + − + −=

−

8. ( ) ( ) ( )( )

( )

2 2 296 100 106 100 98 100

28 5.33 1

− + − + −= =

−

9. 140.6 0.00078364164=

10. 128 68719476736=

11. 147 678223072849=

12. 100.3 0.0000059049=

Chapter 2: Summarizing and Graphing Data 9


Chapter 2: Summarizing and Graphing Data Section 2-2

1. No. For each class, the frequency tells us how many values fall within the given range of values, but there is no way to determine the exact IQ scores represented in the class.

2. If percentages are used, the sum should be 100%. If proportions are used, the sum should be 1.

3. No. The sum of the percentages is 199% not 100%, so each respondent could answer “yes” to more than one category. The table does not show the distribution of a data set among all of several different categories. Instead, it shows responses to five separate questions.

4. The gap in the frequencies suggests that the table includes heights of two different populations: students and faculty/staff.

5. Class width: 10.

Class midpoints: 24.5, 34.5, 44.5, 54.5, 64.5, 74.5, 84.5.

Class boundaries: 19.5, 29.5, 39.5, 49.5, 59.5, 69.5, 79.5, 89.5.

6. Class width: 10.

Class midpoints: 24.5, 34.5, 44.5, 54.5, 64.5, 74.5.

Class boundaries: 19.5, 29.5, 39.5, 49.5, 59.5, 69.5, 79.5.

7. Class width: 10.

Class midpoints: 54.5, 64.5, 74.5, 84.5, 94.5, 104.5, 114.5, 124.5.

Class boundaries: 49.5, 59.5, 69.5, 79.5, 89.5, 99.5, 109.5, 119.5, 129.5.

8. Class width: 5.

Class midpoints: 2, 7, 12, 17, 22, 27, 32, 37.

Class boundaries: –0.5, 4.5, 9.5, 14.5, 19.5, 24.5, 29.5, 34.5, 39.5.

9. Class width: 2.

Class midpoints: 3.95, 5.95, 7.95, 9.95, 11.95.

Class boundaries: 2.95, 4.95, 6.95, 8.95, 10.95, 12.95.

10. Class width: 2.

Class midpoints: 3.95, 5.95, 7.95, 9.95, 11.95.

Class boundaries: 2.95, 4.95, 6.95, 8.95, 10.95, 12.95, 14.95.

11. No. The frequencies do not satisfy the requirement of being roughly symmetric about the maximum frequency of 34.

12. Yes. The frequencies start low, increase to the maximum frequency of 43, and then decrease. Also, the frequencies are approximately symmetric about the maximum frequency of 43.

13. 18, 7, 4

14. 12, 12, 6, 2

10 Chapter 2: Summarizing and Graphing Data


15. On average, the actresses appear to be younger than the actors.

Age When Oscar Was Won Relative Frequency

(Actresses)

Relative Frequency

(Actors)

20 – 29 32.9% 1.2%

30 – 39 41.5% 31.7%

40 – 49 15.9% 42.7%

50 – 59 2.4% 15.9%

60 – 69 4.9% 7.3%

70 – 79 1.2% 1.2%

80 – 89 1.2% 0.0%

16. The differences are not substantial. Based on the given data, males and females appear to have about the same distribution of white blood cell counts.

White Blood Cell Counts Relative Frequency

(Males)

Relative Frequency

(Females)

3.0 – 4.9 20.0% 15.0%

5.0 – 6.9 37.5% 40.0%

7.0 – 8.9 27.5% 22.5%

9.0 – 10.9 12.5% 17.5%

11.0 – 12.9 2.5% 0.0%

13.0 – 14.9 0.0% 5.0%

17. The cumulative frequency table is

Age (years) of Best Actress When Oscar Was Won Cumulative Frequency

Less than 30 27

Less than 40 61

Less than 50 74

Less than 60 76

Less than 70 80

Less than 80 81

Less than 90 82

18. The cumulative frequency table is

Age (years) of Best Actor When Oscar Was Won Cumulative Frequency

Less than 30 1

Less than 40 27

Less than 50 62

Less than 60 75

Less than 70 81

Less than 80 82



19. Because there are disproportionately more 0s and 5s, it appears that the heights were reported instead of measured. Consequently, it is likely that the results are not very accurate.

x Frequency

0 9

1 2

2 1

3 3

4 1

5 15

6 2

7 0

8 3

9 1

20. Because there are disproportionately more 0s and 5s, it appears that the heights were reported instead of measured. Consequently, it is likely that the results are not very accurate.

x Frequency

0 26

1 1

2 1

3 2

4 2

5 12

6 1

7 0

8 4

9 1

21. Yes, the distribution appears to be a normal distribution.

Pulse Rate (Male) Frequency

40 – 49 1

50 – 59 7

60 – 69 17

70 – 79 9

80 – 89 5

90 – 99 1



22. Yes. The pulse rates of males appear to be generally lower than the pulse rates of females.

Pulse Rate (Females) Frequency

50 – 59 1

60 – 69 8

70 – 79 18

80 – 89 5

90 – 99 6

100 – 109 2

23. No, the distribution does not appear to be a normal distribution.

Magnitude Frequency

0.00 – 0.49 5

0.50 – 0.99 15

1.00 – 1.49 19

1.50 – 1.99 7

2.00 – 2.49 2

2.50 – 2.99 2

24. No, the distribution does not appear to be a normal distribution.

Depth (km) Frequency

1.00 – 4.99 7

5.00 – 8.99 21

9.00 – 12.99 4

13.00 – 16.99 12

17.00 – 20.99 6

25. Yes, the distribution appears to be roughly a normal distribution.

Red Blood Cell Count Frequency

4.00 – 4.39 2

4.40 – 4.79 7

4.80 – 5.19 15

5.20 – 5.59 13

5.60 – 5.99 3

26. Yes, the distribution appears to be roughly a normal distribution.

Red Blood Cell Count Frequency

3.60 – 3.99 2

4.00 – 4.39 13

4.40 – 4.79 15

4.80 – 5.19 7

5.20 – 5.59 2

5.60 – 5.99 1



27. Yes. Among the 48 flights, 36 arrived on time or early, and 45 of the flights arrived no more than 30 minutes late.

Arrival Delay (min) Frequency

(–60) – (–31) 11

(–30) – (–1) 25

0 – 29 9

30 – 59 1

60 – 89 0

90 – 119 2

28. No. The times vary from a low of 12 minutes to a high of 49 minutes. It appears that many flights taxi out quickly, but many other flights require much longer times, so it would be difficult to predict the taxi-out time with reasonable accuracy.

Taxi-Out Time (min) Frequency

10 – 14 10

15 – 19 20

20 – 24 9

25 – 29 1

30 – 34 2

35 – 39 2

40 – 44 2

45 – 49 2

29.

Category Relative Frequency

Male Survivors 16.2%

Males Who Died 62.8%

Female Survivors 15.5%

Females Who Died 5.5%

30.

Cause Relative Frequency

Bad Track 46%

Faulty Equipment 18%

Human Error 24%

Other 12%

31. Pilot error is the most serious threat to aviation safety. Better training and stricter pilot requirements can improve aviation safety.

Cause Relative Frequency

Pilot Error 50.5%

Other Human Error 6.1%

Weather 12.1%

Mechanical 22.2%

Sabotage 9.1%



32. The digit 0 appears to have occurred with a higher frequency than expected, but in general the differences are not very substantial, so the selection process appears to be functioning correctly. The digits are qualitative data because they do not represent measures or counts of anything. The digits could be replaced by the first 10 letters of the alphabet, and the lottery would be essentially the same.

Digit Relative Frequency

0 16.7%

1 8.3%

2 10.0%

3 10.0%

4 6.7%

5 9.2%

6 7.5%

7 8.3%

8 7.5%

9 15.8%

33. An outlier can dramatically affect the frequency table.

Weight (lb) With Outlier Without Outlier

200 – 219 6 6

229 – 239 5 5

240 – 259 12 12

260 – 279 36 36

280 – 299 87 87

300 – 319 28 28

320 – 339 0

340 – 359 0

360 – 379 0

380 – 399 0

400 – 419 0

420 – 439 0

440 – 459 0

460 – 479 0

480 – 499 0

500 – 519 1

34.

Number of Data Values Ideal Number of Classes

16 – 22 5

23 – 45 6

46 – 90 7

91 – 181 8

182 – 362 9

363 – 724 10

725 – 1448 11

1449 – 2896 12



Section 2-3

1. It is easier to see the distribution of the data by examining the graph of the histogram than by the numbers in the frequency distribution.

2. Not necessarily. Because those with special interests are more likely to respond, and the voluntary response sample is likely to consist of a group having characteristics that are fundamentally different than those of the population.

3. With a data set that is so small, the true nature of the distribution cannot be seen with a histogram. The data set has an outlier of 1 minute. That duration time corresponds to the last flight, which ended in an explosion that killed seven crew members.

4. When referring to a normal distribution, the term normal has a meaning that is different from its meaning in ordinary language. A normal distribution is characterized by a histogram that is approximately bell-shaped. Determination of whether a histogram is approximately bell-shaped does require subjective judgment.

5. Identifying the exact value is not easy, but answers not too far from 200 are good answers.

6. Class width of 2 inches. Approximate lower limit of first class of 43 inches. Approximate upper limit of first class of 45 inches.

7. The tallest person is about 108 inches, or about 9 feet tall. That tallest height is depicted in the bar that is farthest to the right in the histogram. That height is an outlier because it is very far from all of the other heights. The height of 9 feet must be an error, because the height of the tallest human ever recorded was 8 feet 11 inches.

8. The first group appears to be adults. Knowing that the people entered a museum on a Friday morning, we can reasonably assume that there were many school children on a field trip and that they were accompanied by a smaller group of teachers and adult chaperones and other adults visiting the museum by themselves.

9. The digits 0 and 5 seem to occur much more than the other digits, so it appears that the heights were reported and not actually measured. This suggests that the results might not be very accurate.

10. The digits 0 and 5 seem to occur much more often than the other digits, so it appears that the heights were reported and not measured. This suggests that the results might not be very accurate.

11. The histogram does appear to depict a normal distribution. The frequencies increase to a maximum and then tend to decrease, and the histogram is symmetric with the left half being roughly a mirror image of the right half.



11. (continued)

12. The histogram appears to roughly approximate a normal distribution. The frequencies generally increase to a maximum and then tend to decrease, and the histogram is symmetric with the left half being roughly a mirror image of the right half.

13. The histogram appears to roughly approximate a normal distribution. The frequencies increase to a maximum and then tend to decrease, and the histogram is symmetric with the left half being roughly a mirror image of the right half.

14. No, the histogram does not appear to approximate a normal distribution. The frequencies do not increase to a maximum and then decrease, and the histogram is not symmetric with the left half being a mirror image of the right half.



14. (continued)





17. The two leftmost bars depict flights that arrived early, and the other bars to the right depict flights that arrived late.

18. Yes, the entire distribution would be more concentrated with less spread.

19. The ages of actresses are lower than those of actors.

20. a. 107 inches to 109 inches; 8 feet 11 inches to 9 feet 1 inch.

b. The heights of the bars represent numbers of people, not heights. Because there are many more people between 43 inches tall and 55 inches tall, they have the tallest bars in the histogram, but they have the lowest actual heights. They have the tallest bars because there are more of them.

Section 2-4

1. In a Pareto chart, the bars are arranged in descending order according to frequencies. The Pareto chart helps us understand data by drawing attention to the more important categories, which have the highest frequencies.



2. A scatter plot is a plot of paired quantitative data, and each pair of data is plotted as a single point. The scatterplot requires paired quantitative data. The configuration of the plotted points can help us determine whether there is some relationship between two variables.

3. The data set is too small for a graph to reveal important characteristics of the data. With such a small data set, it would be better to simply list the data or place them in a table.

4. The sample is a voluntary response sample since the students report their scores to the website. Because the sample is a voluntary response sample , it is very possible that it is not representative of the population, even if the sample is very large. Any graph based on the voluntary response sample would have a high chance of showing characteristics that are not actual characteristics of the population.

5. Because the points are scattered throughout with no obvious pattern, there does not appear to be a correlation.

6. The configuration of the points does not support the hypothesis that people with larger brains have larger IQ scores.



7. Yes. There is a very distinct pattern showing that bears with larger chest sizes tend to weigh more.

8. Yes. There is a very distinct pattern showing that cans of Coke with larger volumes tend to weigh more. Another notable feature of the scatterplot is that there are five groups of points that are stacked above each other. This is due to the fact that the measured volumes were rounded to one decimal place, so the different volume amounts are often duplicated, with the result that points are stacked vertically.

9. The first amount is highest for the opening day, when many Harry Potter fans are most eager to see the movie; the third and fourth values are from the first Friday and the first Saturday, which are the popular weekend days when movie attendance tends to spike.



10. The numbers of home runs rose from 1990 to 2000, but after 2000 there was a very gradual decline.

11. Yes, because the configuration of the points is roughly a bell shape, the volumes appear to be from a normally distributed population. The volume of 11.8 oz. appears to be an outlier.

12. No, because the configuration of points is not at all a bell shape, the amounts do not appear to be from a normally distributed population.

13. No. The distribution is not dramatically far from being a normal distribution with a bell shape, so there is not strong evidence against a normal distribution.

4 | 5 5 | 3 3 3 5 5 7 9 6 | 1 1 1 6 7 7 | 1 1 1 1 5 5 6 8 8 | 4

14. There are no outliers. The distribution is not dramatically far from being a normally distribution with a bell shape, so there is not strong evidence against a normal distribution.

12 | 6 8 13 | 1 2 3 4 5 5 6 6 6 7 7 8 9 4 14 | 0 0 0 3 3 5



15.

16. To remain competitive in the world, the United States should require more weekly instruction time.

17.

18. Because there is not a single total number of hours of instruction time that is partitioned among the five countries, it does not make sense to use a pie chart for the given data.



19. The frequency polygon appears to roughly approximate a normal distribution. The frequencies increase to a maximum and then tend to decease, and the graph is symmetric with the left half being roughly a mirror image of the right half.

20. No, the frequency polygon does not appear to approximate a normal distribution. The frequencies do not increase to a maximum and then decrease, and the graph is not symmetric with the left half being a mirror image of the right half.

21. The vertical scale does not start at 0, so the difference is exaggerated. The graphs make it appear that Obama got about twice as many votes as McCain, but Obama actually got about 69 million votes compared to 60 million to McCain.

22. The fare doubled from $1 to $2, but when the $2 bill is shown with twice the width and twice the height of the $1 bill, the $2 bill has an area that is four times that of the $1 bill, so the illustration greatly exaggerates the increase in fare.

23. China’s oil consumption is 2.7 times (or roughly 3 times) that of the United States, but by using a larger barrel that is three times as wide and three times as tall (and also three times as deep) as the smaller barrel, the illustration has made it appear that the larger barrel has a volume that is 27 times that of the smaller barrel. The actual ratio of US consumption to China’s consumption is roughly 3 to 1, but the illustration makes it appear to be 27 to 1.

24. The actual braking distances are 133 ft., 136 ft., and 143 ft., so the differences are relatively small, but the illustration has a scale that begins at 130 ft., so the differences are grossly exaggerated.



25. The ages of actresses are lower than those of actors.

26. a.

96 | 5 9 97 | 0 0 0 1 1 1 2 3 3 3 4 4 4 97 | 5 5 6 6 6 6 6 6 7 8 8 8 8 8 9 9 9 98 | 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 2 2 2 3 3 4 4 4 4 4 4 4 4 4 4 4 4 98 | 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 8 8 8 8 8 8 8 9 9 99 | 0 0 1 2 4 99 | 5 6

b. The condensed stemplot reduces the number of rows so that the stemplot is not too large to be understandable.

6 – 7 | 79 * 778 8 – 9 | 45678 * 049 10 – 11 | 348 * 234477 12 – 13 | 01234 * 5 14 – 15 | 05 * 4569 16 – 17 | * 049 18 – 19 | * 6 20 – 21 | 1 * 3

Chapter Quick Quiz

1. The class width is 1.00

2. The class boundaries are –0.005 and 0.995

3. No

4. 61 min., 62 min., 62 min., 62 min., 62 min., 67 min., and 69 min.

5. No

6. Bar graph

7. Scatterplot

8. Pareto Chart

9. The distribution of the data set

10. The bars of the histogram start relatively low, increase to a maximum value and then decrease. Also, the histogram is symmetric with the left half being roughly a mirror image of the right half.

Review Exercises

1. Volume (cm3) Frequency

900 – 999 1

1000 – 1099 10

1100 – 1199 4

1200 – 1299 3

1300 – 1399 1

1400 – 1499 1



2. No, the distribution does not appear to be normal because the graph is not symmetric.

3. Although there are differences among the frequencies of the digits, the differences are not too extreme given the relatively small sample size, so the lottery appears to be fair.

4. The sample size is not large enough to reveal the true nature of the distribution of IQ scores for the population from which the sample is obtained.

8 | 7 7 9 9 | 6 6 10 | 1 3 3

5. A time-series graph is best. It suggests that the amounts of carbon monoxide emissions in the United States are increasing.



6. A scatterplot is best. The scatterplot does not suggest that there is a relationship.

7. A Pareto chart is best.


1. Pareto chart.

2. Nominal, because the responses consist of names only. The responses do not measure or count anything, and they cannot be arranged in order according to some quantitative scale.

3. Voluntary response sample. The voluntary response sample is not likely to be representative of the population, because those with special interests or strong feelings about the topic are more likely than others to respond and their views might be very different from those of the general population.

4. By using a vertical scale that does not begin at 0, the graph exaggerates the differences in the numbers of responses. The graph could be modified by starting the vertical scale at 0 instead of 50.

5. The percentage is241

0.376 37.6%641

= = . Because the percentage is based on a sample and not a population

that percentage is a statistic.

6.

Grooming Time (min.) Frequency 0 – 9 2

10 – 19 3 20 – 29 9

30 – 39 4

40 – 49 2



7. Because the frequencies increase to a maximum and then decrease and the left half of the histogram is roughly a mirror image of the right half, the data appear to be from a population with a normal distribution.

8. Stemplot

0 | 0 5 1 | 2 5 5 2 | 0 2 4 5 5 5 7 7 8 3 | 0 0 5 5 4 | 0 5

Chapter 3: Statistics for Describing, Exploring, and Comparing Data 29


Chapter 3: Statistics for Describing, Exploring, and Comparing Data

Section 3-2

1. No. The numbers do not measure or count anything, so the mean would be a meaningless statistic.

2. The term average is not used in statistic. The term mean should be used for the value obtained when data values are added, then the sum is divided by the number of data values.

3. No. The price exactly in between the highest and lowest is the midrange, not the median.

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

5. The mean is 332 302 235 225 100 90 88 84 75 67

159.810

+ + + + + + + + += million.

The median is 90 100

$952

+= million.

There is no mode.

The midrange is 332 67

$199.52

+= million.

Apart from the obvious and trivial fact that the mean annual earnings of all celebrities is less than $332 million, nothing meaningful can be known about the mean of the population.

6. The mean is 54410 51991 51730 51300 51196 51190 51122 51115 51037 50875

10

+ + + + + + + + +

$51,596.6= .


$51,1932

+= .

There is no mode.


$52,642.52

+= .

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

7. The mean is 371 356 393 544 326 520 501

430.17

+ + + + + += hic.

The median is 393 hic. There is no mode.


4352

+= hic.

The safest of these cars appears to be the Hyundai Elantra. Because the measurements appear to vary substantially from a low of 326 hic to a high of 544 hic, it appears that some small cars are considerably safer than others.

8. The mean is 774 649 1210 546 431 612

703.76

+ + + + += hic.


630.52

+= hic.

There is no mode.


820.52

+= hic.

30 Chapter 3: Statistics for Describing, Exploring, and Comparing Data


8. (continued)

All of the measures of center are less than 1000 hic, but that does not indicate that all of the individual booster seats satisfy the requirement. One of the booster seats has a measurement of 1210 hic, which does not satisfy the specified requirement of being less than 1000 hic.

9. The mean is 58 22 27 29 21 10 10 8 7 9 11 9 4 4

$16.414

+ + + + + + + + + + + + += million.

The median is 10 10

102

+= million.

The modes are $4 million, $9 million, and $10 million.


$312

+= million.

The measures of center do not reveal anything about the pattern of the data over time, and that pattern is a key component of a movie’s success. The first amount is highest for the opening day when many Harry Potter fans are most eager to see the movie, the third and fourth values are from the first Friday and the first Saturday, which are the popular weekend days when movie attendance tends to spike.

10. The mean is 78 81 95 73 69 79 92 73 90 97

82.710

+ + + + + + + + += manatees.

The median is 79 81

802

+= manatees.

The mode is 73 manatees.


832

+= manatees.

The measures of center do not reveal anything about the pattern of the data over time, and it is important to monitor the number of manatee deaths caused by collisions with watercraft, so that corrective action might be taken.

11. The mean is 55.99 69.99 48.95 48.92 71.77 59.68

$59.226

+ + + + += .

The median is 55.99 59.68

$57.842

+= .

There is no mode.

The midrange is 48.92 71.77

$60.352

+= .

None of the measures of center are most important here. The most relevant statistic in this case is the minimum value of $48.92, because that is the lowest price for the software. Here, we generally care about the lowest price not the mean price or median price.

12. The mean is 17,688, 241 1 19,628,585 12,407,800 14,765,410

$12,898,007.405

+ + + += .

The median is $14,765,410. There is no mode.


$9,814,2932

+= .

The compensation amount of $1 for Jobs is an outlier because it is very far from all the other values.

13. The mean is 3 6.5 6 5.5 20.5 7.5 12 11.5 17.5

11.05 g/g10

μ+ + + + + + + += .


9.5 g/g2

μ+= .

The mode is 20.5 g/gμ .



13. (continued)

The midrange is 3 20.5

11.75 g/g2

μ+= .

There is not enough information given here to assess the true danger of these drugs, but ingestion of any lead is generally detrimental to good health. All of the decimal values are either 0 or 5, so it appears that the lead concentrations were rounded to the nearest one-half unit of measurement.

14. The mean is 0.56 0.75 0.10 0.95 1.25 0.54 0.88

0.7197

+ + + + + += ppm.

The median is 0.75 ppm. There is no mode.


0.6752

+= ppm.

Fairway has the tuna with the lowest level of mercury, so it has the healthiest tuna. Because of the large range of values, it does not appear that the different stores are getting their tuna from the same supplier.

15. The mean is 4 4 4 4 4 4 4.5 4.5 4.5 4.5 4.5 4.5 6 6 8 9 9 13 13 15

20

+ + + + + + + + + + + + + + + + + + + =

6.5 years.


4.52

+= years.

The modes are 4 years and 4.5 years.


9.52

+= years.

It is common to earn a bachelor’s degree in four years, but the typical college student requires more than four years.

16. The mean is 0.38 0.55 1.54 1.55 0.5 0.6 0.92 0.96 1.00 0.86 1.46

0.93811

+ + + + + + + + + += W/kg.

The median is 0.92 W/kg. There is no mode.


0.9652

+= W/kg.

If purchasing a cell phone with concern about radiation emissions, you might be more interested in the fact that the maximum emission is 1.55 W/kg, which is less than the FCC standard of 1.6 W/kg. You might also be interested in the radiation emission for the particular cell phone you are considering.

17. The mean is ( ) ( ) ( ) ( ) ( ) ( )15 18 32 21 9 32 11 2

14.38

− + − + − + − + − + − + +=− min.

The median is ( ) ( )15 18

16.52

− + −=− .

The mode is –32 min.

The midrange is ( )32 11

10.52

− +=− .

Because the measures of center are all negative values, it appears that the flights tend to arrive early before the scheduled arrival times, so the on-time performance appears to be very good.



18. The mean is ( ) ( ) ( ) ( ) ( )11 3 0 2 3 2 2 5 2 7 2 4 1 8 1 0 5 2

18

+ + + − + + − + − + + − + + + + + + + + − +.

= 1.9 kg.

The median is 1 2

1.52

+= kg.

The mode is –2 kg.

The midrange is ( )5 11

32

− += kg.

No, because the mean weight gain is only 1.9 kg, which is below the 6.8 kg weight gain given in the legend.

19. The mean is 9 23 25 88 12 19 74 77 76 73 78

50.411

+ + + + + + + + + += .

The median is 73. There is no mode.


48.52

+= .

The numbers do not measure or count anything; they are simply replacements for names. The data are at the nominal level of measurement, and it makes no sense to compute the measures of center for these data.

20. The mean is 2 1 1 1 1 1 1 4 1 2 2 1 2 3 3 2 3 1 3 1 3 1 3 2 2

25

+ + + + + + + + + + + + + + + + + + + + + + + +

= 1.9. The median is 2. The mode is 1.

The midrange is 1 4

2.52

+= .

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

21. White drivers’ mean is 73 mi/h. White drivers’ median is 73 mi/h. African American drivers’ mean is 74 mi/h. African American drivers’ median is 74 mi/h. Although the African American drivers have a mean speed greater than the white drivers, the difference is very small, so it appears that drivers of both races appear to speed about the same amount.

22. Collection contractor was Brinks had a mean of $1.55 million, and a median of $1.55 million. Collection contractor was not Brinks had a mean of $1.73 million and a median of $1.65 million. The data do suggest that collections were considerably lower when Brinks was the collection contractor.

23. Obama had a mean of $653.9 and a median of $452. McCain had a mean of $458.5 and a median of $350. The contributions appear to favor Obama because his mean and median are substantially higher. With 66 contributions to Obama and 20 to McCain, Obama collected substantially more in total contributions.

24. Jefferson Valley had a mean of 7.15 min. and a median of 7.2 min. Providence had the same results as Jefferson Valley. Although the measures of center are the same, the Providence times are much more varied than the Jefferson Valley times.

25. The mean is 1.184 the median is 1.235. Yes, it is an outlier because it is a value that is very far away from all the other sample values.

26. The mean is 21 min. and the median is 18.5 min. The mean taxi-out time is important for calculating and scheduling the arrival times.



27. The mean is 15 years and the median is 16 years. Presidents receive Secret Service protection after they leave office, so the mean is helpful in planning for the cost and resources used for that protection.

28. The mean is 101 and the median is 96.5. The mean of 101 does not differ from the population mean of 100 by an amount that is substantial, so it appears that the sample is consistent with the population.

29. ( ) ( ) ( ) ( ) ( ) ( ) ( )27 24.5 34 34.5 13 44.5 2 54.5 4 64.5 1 74.5 1 84.5

35.827 34 13 2 4 1 1

+ + + + + +=

+ + + + + +. This result is quite close

to the mean of 35.9 years found by using the original list of data values.

30. ( ) ( ) ( ) ( ) ( ) ( )24.5 1 34.5 26 44.5 35 54.5 13 64.5 6 74.5 1

44.51 26 35 13 6 1

+ + + + +=

+ + + + +years. This result is not substantially

different from the mean of 44.1 found by using the original list of data values.

31. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )4 54.5 10 64.5 25 74.5 43 84.5 26 94.5 8 104.5 3 114.5 2 124.5

84.74 10 25 43 26 8 3 2

+ + + + + + +=

+ + + + + + +. This result

is close to the mean of 84.4 found using the original list of data values.

32. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 8 7 2 12 5 17 7 22 4 27 6 32 0 37 1

158 2 5 7 4 6 0 1

+ + + + + + +=

+ + + + + + +years. When rounded, this result is the

same mean of 15 years found using the original list of data values.

33. a. ( )5 0.62 0.3 0.4 1.1 0.7 0.6x = − − − − = parts per million

b. n – 1

34. The mean ignoring the presidents who are still alive is 15 years. The mean including the presidents who are still alive is at least 15.2 years. The results do not differ by much.

35. The mean is 39.07, the 10% trimmed mean is 27.677, and the 20% trimmed mean is 27.176. By deleting the outlier of 472.2, the trimmed means are substantially different from the untrimmed mean.

36. The mean of 47 mi/h is not the actual average speed, because more time was spent at the lower speed. The harmonic mean is 45.3 mi/h, and it does represent the true “average” value.

37. The geometric mean is 5 1.017 1.037 1.052 1.051 1.027 1.036711036⋅ ⋅ ⋅ ⋅ = , or 1.0367 when rounded. Single percentage growth rate is 3.67%. The result is not exactly the same as the mean which is 3.68%.

38. The root mean square (RMS) is 114.8 volts, which is very different from the mean of 0 volts.

39. The median is ( )( )27 34 13 2 4 1 1 127 1

230 10 33.97058834

⎛ ⎞+ + + + + + + ⎟⎜ − + ⎟⎜ ⎟⎜ ⎟⎜ ⎟+ =⎜ ⎟⎜ ⎟⎟⎜ ⎟⎜ ⎟⎜⎝ ⎠

years, which is rounded

to 34 years. The value of 33 years is better because it is based on the original data and does not involve interpolation.

Section 3-3

1. The IQ scores of a class of statistics students should have less variation, because those students are a much more homogeneous group with IQ scores that are likely to be closer together.

2. Parts (a), (b), and (d) are true.

3. Variation is a general descriptive term that refers to the amount of dispersion or spread among the data values, but the variance refers specifically to the square of the standard deviation.

4. s, σ , s2, 2σ



5. The range is $332 $67 $265− = million.

The variance is ( ) ( )

( )

2

2 10 350, 292 2,553,60410548

10 9s

−= = square of million dollars.

The standard deviation is 10,548 $102.703s = = million.

Because the data values are 10 highest from the population, nothing meaningful can be known about the standard deviation of the population.

6. The range is $54,410 $50,875 $3535− = .


( )

2

2 10 26,631,884,700 515,9661,088,153.8

10 9s

−= = square dollars.

The standard deviation is 1,088,153.8 $1043.10s = = .

Because the data values are the 10 highest from the population, nothing meaningful can be known about the standard deviation of the population.

7. The range is 544 326 218− = hic.


( )7 1,342,439 9,066,121

7879.87 6

−= hic squared.

The standard deviation is 7879.8 88.8= hic. Although all of the cars are small, the range from 326 hic to 544hic appears to be relatively large, so the head injury measurements are not about the same.

8. The range is 1210 431 779− = hic.


( )

2

2 6 3,342,798 422274,383.5

6 5s

−= = hic squared.

The standard deviation is 74,383.5 272.7s = = hic.

Because the data values are the 10 highest from the population, nothing meaningful can be known about the standard deviation of the population.

9. The range is 58 4 $54− = million.

The variance is ( )

( )14 6487 52441

210.914 13

−= square of million dollars.

The standard deviation is 210.9 $14.5= . An investor would care about the gross from opening day and the rate of decline after that, but the measures of center and variation are less important.

10. The range is 97 69 28− = manatees.


( )

2

2 10 69,303 827101.1

10 9s

−= = manatees squared.

The standard deviation is 101.1 10.1s = = manatees. The measures of variation reveal nothing about the pattern over time.

11. The range is $71.77 $48.92 $22.85− = .

The variance is ( )

( )6 21,535.3844 126,238

99.1416 5

−= dollars squared.

The standard deviation is 99.141 $9.957= . The measures of variation are not very helpful in trying to find the best deal.



12. The range is $19,628,584 $1 $19,628,584− = .


( )

2

2 10 1,070,126,052,084,410 64, 490,03759,583,269,405,325.10

10 9s

−= = dollars

squared.

The standard deviation is 59,583,269,405,325.1 $7,719,020= .

The amount of $1 for Jobs is an outlier, and it has a great effect on the measures of deviation.

13. The range is 20.5 3 17.5 g/gμ− = .

The variance is ( )

( )( )210 1596.75 12,210.25

41.75 g/g10 9

μ−

= .

The standard deviation is 41.75 6.46 g/gμ= .

If the medicines contained no lead, all of the measures would be 0 /μg g , and the measures of variation

would all be 0 as well.

14. The range is 1.25 0.10 1.15− = ppm.


( )

2

2 7 4.42 5.030.134

7 6s

−= = ppm squared.

The standard deviation is 0.134 0.366= ppm. If the tuna sushi contained no mercury, all of the measures would be 0 ppm, and the measures of variation would all be 0 as well.

15. The range is 15 4 11− = years.

The variance is ( )

( )20 1078.5 16,900

12.320 19

−= years2.

The standard deviation is 12.3 3.5= years. No, because 12 years is within 2 standard deviations of the mean.

16. The range is 1.55 0.38 1.17− = W/kg.


( )

2

2 11 11.41 10.320.179

11 10s

−= = (W/kg)2.

The standard deviation is 0.179 0.423= W/kg. No. Same models of cell phones are sold much more than others, so the measures from the different models should be weighted according to their size in the population.

17. The range is ( )11 32 43− − = min.

The variance is ( )

( )8 3244 12,996

231.48 7

−= min. squared.

The standard deviation is 231.4 15.2= min. The standard deviation can never be negative.

18. The range is ( )11 5 16− − = kg.


( )

2

2 18 340 3216.5

18 17s

−= = kg2.

The standard deviation is 16.5 4.1= kg. The weight gain of 6.8 kg is not unusual because it is within 2 standard deviations of the mean. Although a gain of 6.8 kg is not unusual, the mean weight gain of 1.9 kg is not close to the legendary 6.8 kg, so an individual weight gain of 6.8 kg does not support the legend.



19. The range is 88 9 79− = .

The variance is ( )

( )11 38,078 306,916

1017.711 10

−= .

The standard deviation is 1017.7 31.9= . The data are at the nominal level of measurement and it makes no sense to compute the measures of variation for these data.

20. The range is 4 1 3− = .


( )

2

2 25 72 470.9

25 24s

−= = .

The standard deviation is 0.9 0.95= . Because the data are at the nominal level of measurement, these results make no sense.

21. The mean of the White drivers is 73 and the standard deviation is 2.906 the coefficient of variation for the

White drivers is 2.906

100% 4%73

⋅ = . The mean for the African American 74 and the standard deviation is

2.749 the coefficient of variation for the African American drivers is2.749

100% 3.7%74

⋅ = . The variation is

about the same.

22. The mean of the collection contractor was Brinks is 1.55 and the standard deviation is 0.178 the coefficient

of variation is 0.178

100% 11.5%1.55

⋅ = . The mean of the collection contractor was not Brinks is 1.73 and the

standard deviation is 0.2214 the coefficient of variation is 0.2214

100% 12.8%1.73

⋅ = . The variation is about

the same.

23. The mean of Obama contributors is $654 and the standard deviation is $523 the coefficient of variation is $523

100% 80%$654

⋅ = . The mean of McCain contributors is $459 and the standard deviation is $418 the

coefficient of variation is$418

100% 90%$459

⋅ = . The variation among Obama contributors is a little less than

the variation among the McCain contributors.

24. The mean of Jefferson Valley is 7.15 and the standard deviation is 0.477 the coefficient of variation

is0.477

100% 6.7%7.16

⋅ = . The mean of Providence is 7.15 and the standard deviation is 1.822 the coefficient

of variation is1.822

100% 25.5%7.15

⋅ = . The variation among Jefferson Valley waiting times is much less

than among the Providence waiting times.

25. The range is 2.95, the variance is 0.345, and the standard deviation is 0.587.

26. The range is 37 min., the variance is 85.5 min. squared, and the standard deviation is 9.2 min.

27. The range is 36 years, the variance is 94.5 years squared, and the standard deviation is 9.7 years.

28. The range is 42, the variance is 174.5, and the standard deviation is 13.2

29. The standard deviation 2.95

0.7384

= , which is not substantially different from 0.587

30. The standard deviation 37

9.34

= min., which is very close to 9.2 min.



31. The standard deviation 36

94

= years, this is reasonably close to 9.7 years.

32. The standard deviation42

10.54

= , which is not substantially different from 13.2.

33. No. The pulse rate of 99 beats per minute is between the minimum usual value of 54.3 beats per minute and the maximum usual value of 100.7 beats per minute.

34. Yes. The pulse rate of 45 beats per minute is not between the minimum usual value of 46.7 beats per minutes and the maximum usual value of 87.9 beats per minute.

35. Yes. The volume of 11.9 oz. is not between the minimum usual value of 11.97 oz. and the maximum usual value of 12.41 oz.

36. No. The weight of 0.8133 lb. is between the minimum usual value of 0.8127 and the maximum usual value of 0.8355 lb.

37. ( )

( )82 84, 408.5 8,637,721

12.382 81

s−

= = years. This result is not substantially different from the standard

deviation of 11.1 years found from the original list of data values.

38. ( )

( )82 169,980.5 13,315,201

9.782 81

s−

= = years. The result is not substantially different from the standard

deviation of 9 years found from the original list of sample values.

39. ( )

( )121 889,106.69 104,941,584.81

13.5121 120

s−

= = . The result is very close to the standard deviation of 13.4

found from the original list of sample values.

40. ( )

( )33 10,552 246,016

9.833 32

s−

= = years. The result is very close to the standard deviation of 9.7 years

found from the original list of sample values.

41. a. 95%

b. 68%

42. a. 68%

b. 99.7%

43. At least 75% of women have platelet counts within 2 standard deviations of the mean. The minimum is 150 and the maximum is 410.

44. At least 89% of healthy adults have body temperatures within 3 standard deviations of the mean. The minimum is 96.34◦F and the maximum is 100.06◦F.

45. a. ( ) ( ) ( )2 2 2

2 2 4.33 3 4.33 8 4.336.9

3σ

− + − + −= = min2

b. The nine possible samples of two values are the following: [(2 min, 2 min), (2 min, 3 min), (2 min, 8 min), (3 min, 2 min), (3 min, 3 min), (3 min, 8 min), (8 min, 2 min), (8 min, 3 min), (8 min, 8 min)] and they have the following corresponding variances: [0, 0.707, 18, 0.707, 0, 12.5, 18, 12.5, 0] which have the mean of 6.934.

c. The population variances of the nine samples above are [0, 0.3535, 9, 0.3535, 0, 6.25, 9, 6.25, 0]

d. Part (b), because repeated samples result in variances that target the same value (6.9 min.2) as the population variance. Use division by 1n− .

e. No. The mean of the sample variances (6.9 min.2) equals the population variance, but the mean of the sample standard deviations (1.9 min.) does not equal the mean of the population standard deviation (2.6 min.)



46. The mean absolute deviation of the population is 2.4 minutes. With repeated samplings of size 2, the nine different possible samples have mean absolute deviations of 0, 0, 0, 0.5, 0.5, 2.5, 2.5, 3, and 3. With many such samples, the mean of those nine results is 1.3 minutes, showing that the sample mean absolute deviations tend to center about the value of 1.3 minutes instead if the mean absolute deviation of the population, which is 2.4 minutes. The sample mean deviations do not target the mean deviation of the population. This is not good. This indicates that a sample mean absolute deviation is not a good estimator of the mean absolute deviation of a population.

Section 3-4

1. Madison’s height is below the mean. It is 2.28 standard deviations below the mean.

2. 2.00 should be preferred, because it is 2.00 standard deviations above the mean and would correspond to the highest of the five different possible scores.

3. The lowest amount is $5 million, the first quartile Q1 is $47 million, the second quartile Q2 (or median) is $104 million, the third quartile Q3 is $121 million, and the highest gross amount is $380 million.

4. All three values are the same.

5. a. The difference is $3,670,505 $4,939,455 $1,268,950− =−

b. $1, 268,950

0.16$7,775,948

= standard deviations

c. 0.16z =−

d. Usual

6. a. The difference is 1.766

b. 1.766

3.010.587


c. 3.01z =

d. Unusual

7. a. The difference is $1 $1,449,779 $1,449,778− =−

b. $1,449,778

2.75$527,651

= standard deviation

c. 2.75z =−

d. Unusual

8. a. The difference is 15.3 beats per minute

b. 15.3

1.4910.3


c. 1.49z =−

d. Usual

9. Z scores of –2 and 2. A z score of –2 means a score of 2 15 100 70x =− ⋅ + = . A z score of 2 means a score of 2 15 100 130x = ⋅ + =

10. Z scores of –2 and 2. A z score of –2 means a hip breadth of 2 2.5 36.6 31.6x =− ⋅ + = cm. A z score of 2 means a hip breadth of 2 2.5 36.6 41.6x = ⋅ + = cm

11. Two standard deviations from the mean: 1.240 2 0.578 0.084− ⋅ = and 1.240 2 0.578 2.396+ ⋅ =

12. Two standard deviations from the mean: 16215 2 7301 1613− ⋅ = words and 16215 2 7301 30817+ ⋅ = words



13. The tallest man z score is 247 175

10.297

z−

= = the tallest women z score is 236 162

12.336

z−

= = . De-

Fen Yao is relatively taller, because her z score of 12.33, which is greater than the z score of 10.29 for Sultan Kosen. De-Fen Yao is more standard deviations above the mean than Sultan Kosen.

14. With a z score of 45 35.9

0.8211.1

z−

= = , Sandra Bullock was relatively younger than Jeff Bridges, who has

a z score of 60 44.1

1.779.0

z−

= = .

15. The SAT score of 1490 has a z score of1490 1518

0.09325

z−

= =− , and the ACT score of 17 has a z score

of17 21.1

0.854.8

z−

= =− . The z score of –0.09 is a larger number than the z score of –0.85, so the SAT

score of 1490 is relatively better.

16. The male has a higher count because his z score is 4.91 5.072

0.410.395

z−

= =− , which is a higher number

than the z score of 4.32 4.577

0.670.382

z−

= =− for the female.

17. The percentile for 213 sec. is 3

100 1324

⋅ = , so the 13th percentile


100 3324

⋅ = , so the 33rd percentile


100 5024

⋅ = , so the 50th percentile


100 8324

⋅ = , so the 83rd percentile

21. 60

60 2414.4

100P

⋅= = , pick 15th entry which is 251 sec.

22. 1

234 235234.5

2Q

+= = sec.

23. 3

255 255255

2Q

+= = sec.

24. 40

40 249.6

100P

⋅= = , pick 10th entry which is 243 sec.

25. 50

245 250247.5

2P

+= = sec.

26. 75

75 2418

100P

⋅= = , which is entry 255 sec.

27. 25 1 234.5P Q= = sec.

28. 85

85 2420.4

100P

⋅= = , pick the 21st entry which is 260 sec.



29. The five number summary: 1 sec, 8709 sec, 10,074.5 sec, 11,445 sec, 11,844 sec

30. The five number summary: 81 min, 88 min, 94.5 min, 98 min, 106 min

31. The five number summary : 4 min, 14 min, 18 min, 32 min, 63 min

32. The five number summary: 70 mi/h, 72 mi/h, 74 mi/h, 78 mi/h, 79 mi/h

33. It appears that males have lower pulse rates than females

Male Pulse

Female Pulse

34. Although actresses include the oldest age of 80 years, the boxplot for actresses shows that they have ages that are generally lower than those of actors.

Actresses

Actors

35. The weights of regular Coke appear to be generally greater than those of diet Coke, probably due to the sugar in cans of regular Coke.

CKREGWT

CKDIETWT



36. The low lead level group has much more variation and the IQ scores tend to be higher than the IQ scores from the high lead level group.

Low Lead

High Lead

37. Outliers for actresses 60 years, 61 years, 63 years, 70 years, and 80 years. Outliers for actors: 76 years. The modified boxplots show that only one actress has an age that is greater than any actor.

38. Using interpolation, 17 21.6P = . Using figure 3-5, 17 22P = . In this case, the results are close, but in some

other cases the results might be quite different.

Chapter Quick Quiz

1. The mean is 14 minutes

2. The median is 12 minutes

3. The mode is 12 minutes

4. The variance is 2 2(5 min) = 25 min

5. 6 11.4

0.777

z−

= =−

6. Standard deviation, variance, range, mean absolute deviation

7. Sample mean x , population mean μ

8. 2 2, , ,s σ s σ

9. 75%

10. Minimum, first quartile Q1, second quartile Q2 (or median), third quartile Q3, maximum

Review Exercises

1. a. 1550 1642 1538 1497 1571

1559.65

x+ + + +

= = mm

b. The median is 1550 mm

c. There is no mode



1. (continued)

d. The midrange is 1497 1642

1569.52

+= mm

e. The range is 1642 1497 145− = mm

f. ( ) ( ) ( ) ( ) ( )2 2 2 2 21550 1559.6 1642 1559.6 1538 1559.6 1497 1559.6 1571 1559.6

5 1s

− + − + − + − + −=

−

53.37= mm

g. 2 253.37 2849.3s = = mm2

h. 1

25 51.25

100Q

⋅= = , pick second entry (in ordered list) which is 1538 mm

i. 3

75 53.75

100Q

⋅= = , pick the fourth entry (in the ordered list) which is 1571 mm

2. 1642 1559.6

1.5453.37

z−

= = . The eye height is not unusual because its z score is between –2 and 2, so it is

within two standard deviations of the mean.

3. The five number summary: 1497, 1538, 1550, 1571, 1642

Because the boxplot shows a distribution of data that is roughly symmetric, the data could be from a population with a normal distribution, but the data are not necessarily from a population with a normal distribution, because there is no way to determine whether a histogram is roughly a bell shape.

4. The mean is 10053.5. The ZIP codes do not measure or count anything. They are at the nominal level of measurement, so the mean is a meaningless statistic.

5. The male z score is28 26.601

0.265.359

z−

= = . The female z score is29 28.441

0.087.394

z−

= = . The male has

a larger relative BMI because the male has the larger z score.

6. a. The answers may vary but a mean around $8 or $9 is reasonable.

b. A reasonable standard deviation would be around $1 or $2.

7. Based on a minimum age of 23 years and a maximum age of 70 years an estimate of the age standard

deviation would be 70 23

11.754

−= years.

8. A minimum usual sitting height of 914 2 36 842− ⋅ = mm and a maximum sitting height of 914 2 36 986+ ⋅ = mm. The maximum usual height of 986 mm is more relevant for designing overhead bin storage.

9. The minimum value is 963 cm3, the first quartile is 1034.5 cm3, the second quartile (or median) is 1079 cm3, the third quartile is 1188.5 cm3, and the maximum value is 1439 cm3.

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


1. a. Continuous

b. Ratio



2.

Hand Length (mm) Frequency

150 – 159 1

160 – 169 0

170 – 179 2

180 – 189 0

190 – 199 3

200 – 209 1

210 – 219 1

3. Hand length histogram

4.

15 | 8 16 | 17 |3 9 18 | 19 | 5 6 9 20 | 7 21 | 4

5. a. 173 179 207 158 196 195 214 199

190.18

x+ + + + + + +

= = mm

b. The median is 195.5 mm

c. ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

2 2 2 2 2173 190.1 179 190.1 207 190.1 158 190.1 196 190.1

2 2 2195 190.1 214 190.1 199 190.1 2440.88

− + − + − + − + −

+ − + − + − =

2440.88

18.7 mm7

s = = mm

d. 2 218.7 348.7s = = mm2

e. The range is 214 158 56− = mm

6. Yes. The frequencies increase to a maximum, and then they decrease. Also, the frequencies preceding the maximum are roughly a mirror image of those that follow the maximum.

7. No. Even though the sample is large, it is a voluntary response sample, so the responses cannot be considered to be representative of the population of the United States.

8. The vertical scale does not begin at 0, so the differences among different outcomes are exaggerated.

Chapter 4: Probability 45


Chapter 4: Probability Section 4-2

1. 1

( ) 0.000110,000

P A = = , 1 9999

( ) 1 0.999910,000 10,000

P A = − = =

2. The probability of a baby being born a boy is 1

2 or 0.5

3. Part (c).

4. The answers vary, but an answer in the neighborhood of 0.99 is reasonable.

5. 5:2, 7

3, –0.9,

456

123

6. 1

4 or 0.25

7. 1

5 or 0.2

8. 0

9. Unlikely, neither unusually low nor unusually high

10. Unlikely, unusually high

11. Unlikely, unusually low

12. Unlikely, neither unusually low nor unusually high

13. 1

4 or 0.25

14. 0.2

15. 1

2 or 0.5

16. 1

2 or 0.5

17. 1

5 or 0.2

18. 1

36 or 0.0278

19. 0

20. 1

21. 6

1000 or 0.006. The employer would suffer because it would be at a risk by hiring someone who uses

drugs.

22. 90

1000 or 0.09. The person tested would suffer because he or she would be suspected of using drugs when

in reality he or she does not use drugs.

23. 50

1000 or 0.05. This result is not close to the probability of 0.134 for a positive test result.

24. 950

1000 or 0.95. This result is not very close to the probability of 0.866 for a negative test result.

25. 879

945 or 0.93. Yes, the technique appears to be effective.

26. 239

291 or 0.821. Yes, the technique appears to be effective.

27. 304

300,000,000 or 0.00000101. No, the probability of being struck is much greater on an open golf course

during a thunder storm. The golfer should seek shelter.

46 Chapter 4: Probability


28. 428

580 = 0.738; yes

29. a. 1

365

b. Yes

c. He already knew

d. 0

30. 2834

0.67169 1227 2834

=+ +

. No, it is not unlikely, because the responses are from a voluntary response

survey; the results are not likely to be very good.

31. 10,427,000

135,933,000 or 0.0767. No, a crash is not unlikely. Given that car crashes are so common, we should take

precautions such as not driving after drinking and not using a cell phone or texting.

32. 117

1,000,000,000 or 0.000000117. Yes, it is unlikely. The air travel fatality rate is much higher than that of

cars. The comparison isn’t fair because car trips involve much shorter distances than trips by air.

33. 8

0.009858 804

=+

. It is unlikely 34. 141

0.175141 663

=+

. It is unlikely

35. 8

0.00993492 8 306

=+ +

. Yes, it is unlikely. The middle seat lacks an outside view, easy access to the

aisle, and a passenger in the middle seat has passengers on both sides instead of on one side only.

36. 19

0.020219 441 235 103 66 75

=+ + + + +

. Yes, it is unlikely.

37. 3

8 or 0.375

38. 3

8 or 0.375

39. {bb, bg, gb, gg}; 1

2or 0.5

40. {bbbb, bbbg, bbgb, bbgg, bgbb, bgbg, bggb, bggg, gbbg, gbbb, gbgb, gbgg, ggbb, ggbg, gggb, gggg}; 4

16

or 0.25

41. a. brown /brown, brown/blue, blue/brown, blue/blue

b. 1

4

c. 3

4

42. a. 0

b. 0

c. 0.5

d. 0

43. a. 999 : 1

b. 499 : 1

c. The description is not accurate. The odds against winning are 999:1 and the odds in favor are 1:999, not 1:1000

44. a. 18

38 or 0.474

b. 10 : 9

c. $18

d. $ 20



45. a. $16

b. 8 : 1

c. About 9.75 : 1, which becomes 39 : 4

d. $21.50

46. 1

37 or 0.0270

47. Relative risk:

262103 0.93922

1671

= Odds ratio:

262103

261

2103 0.93822

167122

11671

−=

−

The probability of a headache with Nasonex (0.0124) is slightly less than the probability of a headache with the placebo (0.0132), so Nasonex does not appear to pose a risk of headache.

48. 1 49.

1

4

Section 4-3

1. Based on the rule of the complements, the sum of P(A) and its complement must always be 1, so the sum cannot be 0.5

2. A is the event of betting on the pass line and not winning (or losing). 251

( ) 0.507495

P A = =

3. Because it is possible to select someone who is male and a Republican, events M and R are not disjoint. Both events can occur at the same time when someone is randomly selected.

4. It is certain that an event occurs or does not occur.

5. Disjoint

6. Not disjoint

7. Not disjoint

8. Not disjoint

9. Disjoint

10. Disjoint

11. Not disjoint

12. Disjoint

13. 1 0.47 0.53− =

14. 1 0.198 0.802− =

15. ( ) 0.45P D = , where ( )P D is the probability of randomly selecting someone who does not choose a direct

in-person encounter as the most fun way to flirt.

16. ( )P I denotes the probability of screening a driver and finding that he or she is not intoxicated, and

( ) 0.99112P I =

17. 1

18. 44 90 860

0.9941000

+ +=

19. 90 860 6

0.9561000

+ +=

20. 44 6 860

0.911000

+ +=

21. 13

28 or 0.464. That probability is not as high as it should be.



22. 15

28 or 0.536. That probability is not as low as it should be.

23. 16

28 or 0.571 24.

17

28 or 0.607

25. a. 11

0.78614

= or 78.6%

b. 2

0.14314

= or 14.3%

c. The physicians given the labels with concentrations appear to have done much better. The results suggest that labels described as concentrations are much better than labels described as ratios.

26. a. 3

0.21414

= or 21.4%

b. 12

0.85714

= or 85.7%

c. The physicians given the labels with ratios appear to have done much worse. The results suggest that label described as ratios are much worse than labels described as concentrations

Use the following table for Exercises 27–32

Age

18–21 22–29 30–39 40–49 50–59

60 and over

Total

Responded 73 255 245 136 138 202 1049

Refused 11 20 33 16 27 49 156

Total 84 275 278 152 165 251 1205

27. 156

1205 = 0.129. Yes. A high refusal rate results in a sample that is not necessarily representative of the

population, because those who refuse may well constitute a particular group with opinions different from others.

28. 202

1205 = 0.168

29. 1049 84 73 1060

1205 1205 1205 1205+ − = = 0.88

30. 156 251 49 358

1205 1205 1205 1205+ − = = 0.297

31. 1049 275 278 255 245 1102

1205 1205 1205 1205

+ ++ − =

= 0.915

32. 156 84 251 11 49 431

1205 1205 1205 1205

+ ++ − = = 0.358

33. 300

Positive Test Result Negative Test Result Total

Subject Used Marijuana

119 3 122

Subject Did not Use Marijuana

24 154 178

Total 143 157 300



34. 119 3 24

0.487300

+ += 35.

3 154 240.603

300

+ +=

36. 122

300 =0.407. No, the general population probably has a marijuana usage rate less than 0.407, or 40.7%.

37. 27

300 = 0.09. With an error rate of 0.09 or 9%, the test does not appear to be highly accurate.

38. 273

300 = 0.91. Exercise 37 results in the probability of a wrong result and this exercise result in the

probability of a correct result, so these exercises deal with events that are complements.

39. 3

4 or 0.75

40. No. Here is one example: A = event of selecting a male under 30 years of age, B = selecting a female, C = selecting a male over 18 years of age.

41. ( or ) ( ) ( ) 2 ( and )P A B P A P B P A B= + −

42. ( or or )

( ) ( ) ( ) ( and ) ( and ) ( and ) ( and and )

P A B C

P A P B P C P A B P A C P B C P A B C= + + − − − +

43. a. 1 ( ) ( ) ( and )P A P B P A B− − +

b. 1 ( and )P A B−

c. No

Section 4-4

1. The probability that the second selected senator is a Democrat given that the first selected senator was a Republican.

2. R and D are dependent events, because the probability of a Democrat on the second selection is affected by the outcome of the first selection. Because it was stipulated that the second selection must be a different senator, the sampling is done without replacement, so only 99 senators are available for the second selection.

3. False. The events are dependent because the radio and air conditioner are both powered by the same electrical system. If you find that your car’s radio does not work, there is a greater probability that the air conditioner will also not work.

4. Because the selections are based on different numbers, the sampling is done without replacement and the events are dependent. Because the sample size of 1068 is less than 5% of the population size of 28,741,346, the events can be treated as being independent (based on the 5% guideline for cumbersome calculations).

5. a. The events are dependent

b. 1

132 or 0.00758

6. a. Independent

b. 1

4 or 0.25

7. a. Independent

b. 1

12 or 0.0833

8. a. Dependent

b. 1

42 or 0.0238



9. a. Independent

b. 5 5

0.000507222 222

⋅ =

10. a. Independent

b. 1

100 or 0.01

11. a. Dependent

b. 58 1

0.00586100 99

⋅ =

12. a. Dependent

b. 8 7

0.00566100 99

⋅ =

13. a. 90 90

0.00811000 1000

⋅ = . Yes, it is unlikely

b. 90 89

0.008021000 999

⋅ = . Yes, it is unlikely

14. a. 6 6 6

0.0000002161000 1000 1000

⋅ ⋅ = . Yes, it is unlikely

b. 6 5 4

0.000000121000 999 998


15. a. 904 904 904

0.7391000 1000 1000

⋅ ⋅ = . No, it is not unlikely

b. 904 903 902

0.7391000 999 998

⋅ ⋅ = . No , it is not unlikely

16. a. 860 860 860 860

0.5471000 1000 1000 1000

⋅ ⋅ ⋅ = . No, it is not unlikely

b. 860 859 858 857

0.5461000 999 998 997

⋅ ⋅ ⋅ = . No, it is not unlikely

17. 8330 8329 8328

0.8388834 8833 8832

⋅ ⋅ = . No, the entire batch consists of malfunctioning pacemakers.

18. 708 707 706 705

0.583810 809 808 807

⋅ ⋅ ⋅ = . The scheme is not likely to detect the large number if defects. With a

probability of 0.583, it is more likely that the entire batch will be accepted.

19. a. 2

0.02100

=

b. 2 2

0.0004100 100

⋅ =

c. 2 2 2

0.000008100 100 100

⋅ ⋅ =

d. By using one backup drive, the probability of failure is 0.02, and with three independent disk drives, the probability drops to 0.000008. By changing from one drive to three, the likelihood of failure drops from 1 chance in 50 to only 1 chance in 125,000, and that is a very substantial improvement in reliability. BACK UP YOUR DATA.

20. 0.0035 0.0035 0.0000123⋅ = . With one radio there is a 0.0035 probability of a serious problem, but with two independent radios, the probability of a serious problem drops to 0.0000123, which is substantially lower. The flight becomes much safer with two independent radios.



21. a. 1

365 or 0.00274

b. 1 1

0.00000751365 365

⋅ =

c. 1

365 or 0.00274

22. a. 1 1

0.045 5⋅ = b.

1

5 or 0.2

c. 1 1 1 1 1 1 1 1 1

0.0000005125 5 5 5 5 5 5 5 5⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = . Yes, it is unlikely, but perhaps there was a strong need to staff

the department so that the hirings were more likely to occur on the same day.

23.

Positive Test Result Negative Test Result Total

Subject used marijuana True Positive

119

False Negative

3

122

Subject did not use marijuana

False Positive

24

True Negative

154

178

Total 143 157 300

119 118 154 153 154 119 119 1540.828

300 299 300 299 300 299 300 299⋅ + ⋅ + ⋅ + ⋅ = . No, it is not unlikely

24. 24 23 3 2 24 3 3 24

0.00783300 299 300 299 300 299 300 299

⋅ + ⋅ + ⋅ + ⋅ = . Yes, it is unlikely

25. 24 23 22

0.000454300 299 298


26. 154 153 152

0.134300 299 298

⋅ ⋅ = . No it is not unlikely

27. a. 2518 252

0.92518

−=

b. 50

2518 2520.00513

2518

⎛ ⎞− ⎟⎜ =⎟⎜ ⎟⎜⎝ ⎠. Using the 5% guideline for cumbersome calculations

28. a. 1021 61

0.941021

−=

b. 40

1021 610.0851

1021


29. a. 162 161

0.143427 426

⋅ =

b. 10

427 1620.00848

427




30. a. 492 306

0.99492 8 306

+=

+ +

b. 798 797

0.98806 805

⋅ =

c. 25

7980.779

806

⎛ ⎞⎟⎜ =⎟⎜ ⎟⎜⎝ ⎠

31. a. 0.99 0.99 0.99 0.01 0.01 0.99 0.9999⋅ + ⋅ + ⋅ =

b. 0.99 0.99 0.9801⋅ =

c. The series arrangement provides better protection.

32. 365 364 363 341

0.431365 365 365 365

⋅ ⋅ ⋅ ⋅ =…

Section 4-5

1. a. Answers vary, but 0.98 is a reasonable estimate.

b. Answers vary, but 0.999 is a reasonable estimate.

2. A conditional probability is a probability of an event calculated with the knowledge that some other event has occurred.

3. The probability that the polygraph indicates lying given that the subject is actually telling the truth.

4. Confusion of the inverse is to think that the following two probabilities are the same: (1) the probability of a polygraph indication of lying when the subject is telling the truth; (2) the probability of a subject telling the truth when the polygraph indicates lying. Confusion of the inverse is to think that

( | ) ( | )P A B P B A= or to use one of those probabilities in place of the other.

5. At least one of the five children is a boy. 31

32 or 0.969

6. At least one of the five children is a girl. 31

32 or 0.969

7. None of the digits is 0. 4

90.656

10

⎛ ⎞⎟⎜ =⎟⎜ ⎟⎜⎝ ⎠

8. At least one of the digits is 7. 4

91 0.344

10

⎛ ⎞⎟⎜− =⎟⎜ ⎟⎜⎝ ⎠

9. 10

41 0.893

5

⎛ ⎞⎟⎜− =⎟⎜ ⎟⎜⎝ ⎠. The chance of passing is reasonably good

10. 0.92 0.92 0.92 0.08 0.08 0.92 0.9936⋅ + ⋅ + ⋅ = . The probability of having to complete the exam without a working calculator drops from 0.08 to 0.0064 (or 64 chances in 10000), so she does gain a substantial increase in reliability.

11. 0.5 or 50%

12. 1

5 or 0.2

13. ( )51 0.512 0.965− =

14. ( )51 0.545 0.952− = . The system cannot continue indefinitely because eventually there would be no

women to give birth.



15. ( )31 1 0.0423 0.122− − = . Given that the three cars are in the same family, they are not randomly selected

and there is a good chance that the family members have similar driving habits, so the probability might not be accurate.

16. a. ( )51 1 0.124 0.484− − =

b. ( )50.124 0.0000293=

c. The detective is much better than average, or the detective was given five easy cases.

17. ( )41 1 0.67 0.988− − = . It is very possible that the result is not valid because it is based on data from a

voluntary response survey.

18. ( )121 1 0.41 0.998− − = . It is very possible that the result is not valid because it is based on data from a

voluntary response survey.

19. 90

950 or 0.0947. This is the probability of the test making it appear that the subject uses drugs when the

subject is not a drug user.

20. 6

50 or 0.12. The employer would suffer by hiring a job applicant who appears to not use drugs, but the

applicant actually does use drugs.

21. 6

866 or 0.00693. This result is substantially different from the result found in Exercise 20. The

probabilities P(subject uses drugs | negative test result) and P(negative test result | subject uses drugs) are not equal.

22. a. 860

950 or 0.905

b. 860

866 or 0.993

c. The results are different

23. 44

134 or 0.328

24. 860

866 or 0.993

25. a. 1

3 or 0.333

b. 5

10 or 0.5

26. a. 2

3 or 0.667

b. 10

10 or 1

27. 10

20 or 0.5

28. 1

4 or 0.25

29. a. ( )21 0.02 0.9996− =

b. ( )31 0.02 0.999992− =

30. ( )21 0.0035 0.99998775− = . Rounding the result to three significant digits would yield a probability of

1.00, but that would be misleading because it would suggest that it is certain that both radios will work. Yes, the probability is high enough to ensure flight safety.

31. ( )81 1 0.134 0.684− − = . The probability is not low, so further testing of the individual samples will be

necessary for about 68% of the combined samples.



32. ( )51 1 0.005 0.0248− − = . The probability is quite low, indicating that further testing of the individual

samples will be necessary for about 2% of the combined samples.

33. a. 365 364 363 341

0.431365 365 365 365

⋅ ⋅ ⋅ ⋅ =…

b. 1 0.431 0.569− =

34. 1

3 or 0.333

35. a. 0.8 0.01

0.07480.8 0.01 0.1 0.99

⋅=

⋅ + ⋅

b. 0.8

c. The estimate of 75% is dramatically greater than the actual rate of 7.48%. They exhibited confusion of the inverse. A consequence is that they would unnecessarily alarm patients who are benign, and they might start treatments that are not necessary.

Section 4-6

1. The symbol ! is the factorial symbol that represents the product of decreasing whole numbers, as in 4! 4 3 2 1 24= ⋅ ⋅ ⋅ = . Four people can stand in line 24 different ways.

2. Combinations, because order does not count and five numbers are selected (from 1 to 39) without replacement.

3. Because repetition is allowed, numbers are selected with replacement, so neither of the two permutation rules applies. The fundamental counting rule can be used to show that the number of possible outcomes is

10 10 10 10 10,000⋅ ⋅ ⋅ = , so the probability of winning is 1

10,000.

4. Only the fundamental counting rule applies.

5. 1 1 1 1 1

10 10 10 10 10,000⋅ ⋅ ⋅ =

6. 1 1 1 1 1 1 1 1 1 1

10 10 10 10 10 10 10 10 10 1,000,000,000⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =

7. 1 1 1 1 1 1 1 1 1 1 1

9 8 7 6 5 4 3 2 1 9! 362,880⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = =

8. 1 1

7! 5040=

9. The number of combinations is27!

17,383,860(27 12)!12!

=−

. Because that number is so large, it is not

practical to make a different CD for each possible combination.

10. 1 1 1 1 1

52 51 52 51 1326⋅ + ⋅ =

11. 1 1 1 1 1

50 49 48 47 5,527,200⋅ ⋅ ⋅ = . No 5,527,200 is too many possibilities to list.

12. 8!

336(8 3)!

=−

13. 11!

34,6504!4!2!

=

14. 10!

50,4003!3!2!

=



15. 44!

7,059,052(44 6)!6!

=−

. The probability is 1

7,059,052

16. 53!

22,957,480(53 6)!6!

=−


22,957, 480

17. 1 1

4! 24= 18.

1 1

7! 5040=

19. a. 41!

749,398(41 5)!5!

=−


749,398

b. 4

1 1

10,00010=

c. $10,000

20. a. ( )

38!575,757

39 5 !5!=

−. The probability is

1

575,757

b. 3

1 1

100010=

c. $1000

21. a. ( )

12!11,880

12 4 !=

−

b. ( )

12!495

12 4 !4!=

−

c. 1

495

22. a. ( )

16!10,461,394,944,000

16 14 !=

−

b. ( )

16!120

16 14 !14!=

−

c. 1

120

23. The number of possible “combinations” is 50 50 50 125,000⋅ ⋅ = . The fundamental counting rule can be used. The different possible codes are ordered sequences of numbers, not combinations, so the name of “combination lock” is not appropriate. Given that “fundamental counting rule lock” is a bit awkward, a better name would be something like “number lock”.

24. 1 1

100 100 100 100 100,000,000=

⋅ ⋅ ⋅. No, there are too many different possibilities

25. 5! 120= ; AMITY; 1

120

26. 6!

3602= ; HARROW;

1

360

27. 5! 5! 5! 5!

265!0! 4!1! 3!2! 2!3!

+ + + =

28. a. 16

1 1

10,000,000,000,000,00010= b.

12

1 1

1,000,000,000,00010=

c. 8

1 1

100,000,00010= . The number of possibilities (100,000,000) is still quite large, so there is no

reason to worry.8

1 1

100,000,00010=



29. ( )

5!10

5 2 !2!=

−

30. a. 1

4 or 0.25 b.

3

16 or 0.188

c. Trick question. There is no finite number of attempts, because you could continue to get the wrong position every time.

31. 4 4 4 64⋅ ⋅ =

32. a. 31

b. 31

1 10.000000000466

2 2,147,483,648

⎛ ⎞⎟⎜ = =⎟⎜ ⎟⎜⎝ ⎠

33. 59 5 39 1

1 1

195,249,054C C=

⋅ 34.

56 5 46 15

1 1

175,711,536C C=

⋅

35. 10 5

2 2

252C= . Yes, if everyone treated is of one gender while everyone in the placebo group is of the

opposite gender, you would not know if different reactions are due to the treatment or gender.

36. 8 2 9 144⋅ ⋅ =

37. 2 3 4 5 6 726 26 36 26 36 26 36 26 36 26 36 26 36 26 36 2,095,681,645,538+ ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ =

38. a. 5 2 10C =

b. 2

( 1)

2n

n nC

−=

c. 4! 24=

d. ( 1)!n−

39. 12 ways: {25p, 1n 20p, 2n 15p, 3n 10p, 4n 5p, 5n, 1d 15p, 1d 1n 10p, 1d 2n 5p, 1d 3n, 2d 5p,2d 1n} (Note: 25p represents 25 pennies, etc.)

40. The probability is 0. If 9 of the letters are in the correct envelopes, the 10th letter must also be in the correct envelope, so it is impossible for the 10th letter to go into the wrong envelope.

Chapter Quick Quiz

1. 0 (not an option)

2. 10 3

0.710

−=

3. 1 (all days contain the letter y)

4. 0.2 0.2 0.04⋅ =

5. Answers vary, but an answer such as 0.01 or lower is reasonable

6. 288 224 512

0.61201 126 288 224 839

+= =

+ + +

7. 224 288 201 713

0.85839 839

+ += =

8. 126

0.15839

=

9. 126 125

0.0224839 839

⋅ =

10. 126 126

0.36126 224 350

= =+

Review Exercises

1. 392 58

0.4381028

+=

2. 392

0.41392 564

=+

3. 58

0.80658 14

=+



4. It appears that you have a substantially better chance of avoiding prison if you enter a guilty plea.

5. 392 58 392 564 392

0.9861028 1028 1028

+ ++ − =

6. 450 449

0.1911028 1027

⋅ =

7. 72 71

0.004841028 1027

⋅ =

8. 72 578 14

0.6191028 1028 1028

+ − =

9. 392

0.3811028

=

10. 14

0.01361028

=

11. Answers vary, but DuPont data show that about 8% of cars are red, so any estimate between 0.01 and 0.2 would be reasonable.

12. a. 1 0.35 .65− =

b. ( )40.35 0.015=

c. Yes, because the probability is so small

13. a. 1

365

b. 31

365

c. Answers vary, but it is probably small, such as 0.02

d. Yes

14. 10

2131 1 0.0211

100,000

⎛ ⎞⎟⎜− − =⎟⎜ ⎟⎜ ⎟⎝ ⎠. No

15. 42 6

1 1

5,245,786C=

16. 39 5

1 1

575,757C=

17. 1 1 1 1

10 10 10 1000⋅ ⋅ =

18. 12 3 1320P = . The probability is1

1320


1. a. The mean of –8.9 years is not close to the value of 0 years that would be expected with no gender discrepancy.

b. The median of –13.5 years is not close to the value of 0 years that would be expected with no gender discrepancy.

c. ( )( ) ( )( ) ( )( )2 2 2

20 8.9 15 8.9 15 8.910.6

11s

− − − + − − − + + − − −= =

…years.

d. ( )22 10.6 113.2s = = years2

e. 1 15Q =− years

f. 3 5Q =− years

g. The boxplot suggests that the data have a distribution that is skewed.



2. a. 100 77.5

1.9411.6

z−

= = . No, the pulse rate of 100 beats per minute is within 2 standard deviations away

from the mean, so it is not unusual.

b. 50 77.5

2.3711.6

z−

= =− . Yes, the pulse rate of 50 beats per minutes is more than 2 standard deviations

away from the mean so it is unusual.

c. Yes, because the probability of 1

256 (or 0.0039) is so small.

d. No, because the probability of 1

8 (or 0.125) is not very small.

3. a. 2346

.46 46%5100

= =

b. 0.46 46%=

c. Stratified sample

4. The graph is misleading because the vertical scale does not start at 0. The vertical scale starts at the frequency of 500 instead of 0, so the difference between the two response rates is exaggerated. The graph incorrectly makes it appear that “no” responses occurred 60 times more often than the number of “yes” responses, but comparisons of the actual frequencies shows that the “no” responses occurred about four times more often than the number of “yes” responses.

5. a. A convenience sample

b. If the students at the college are mostly from a surrounding region that includes a large proportion of one ethnic group, the results will not reflect the general population of the United States.

c. 0.35 0.4 0.75+ =

d. ( )21 0.6 0.64− =

6. The straight-line pattern of the points suggests that there is a correlation between chest size and weight.

7. a. 39 5

1 1

575,757C=

b. 1

19

c. 39 5 19 1

1 1

10,939,383C C=

⋅

Chapter 5: Discrete Probability Distributions 59


Chapter 5: Discrete Probability Distributions Section 5-2

1. The random variable is x, which is the number of girls in three births. The possible values of x are 0, 1, 2, and 3. The values of the random variable x are numerical.

2. The random variable is discrete because the number of possible values is 4, and 4 is a finite number. The random variable is discrete if it has a finite number of values or a countable number of values.

3. Table 5-7 does describe a probability distribution because the three requirements are satisfied. First, the variable x is a numerical random variable and its values are associated with probabilities. Second, Σ ( ) 0.125 0.375 0.375 0.125 1P x = + + + = as required. Third, each of the probabilities is between 0 and 1

inclusive, as required.

4. a. Yes, because 0.0208 0.05≤

b. No, because 0.089 0.05>

5. a. Continuous random variable

b. Discrete random variable

c. Not a random variable

d. Discrete random variable

e. Continuous random variable

f. Discrete random variable

6. a. Not a random variable

b. Continuous random variable

c. Discrete random variable

d. Discrete random variable

e. Not a random variable

f. Discrete random variable 7. Probability distribution with

(0 0.0625) (1 0.25) (2 0.375) (3 0.25) (4 0.0625) 2μ= ⋅ + ⋅ + ⋅ + ⋅ + ⋅ =

2 2 2 2 2(0 2) 0.0625 (1 2) 0.25 (2 2) 0.375 (3 2) 0.25 (4 2) 0.0625 1σ = − ⋅ + − ⋅ + − ⋅ + − ⋅ + − ⋅ =

8. Probability distribution with (0 0.659) (1 0.287) (2 0.05) (3 0.004) (4 0.001) (5 0) 0.4μ= ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ =

2 2 2 2 2(0 0.4) 0.659 (1 0.4) 0.287 (2 0.4) 0.05 ... (4 0.4) 0.001 (5 0.4) 0

0.6

σ = − ⋅ + − ⋅ + − ⋅ + + − ⋅ + − ⋅

=

9. Not a probability distribution because the sum of the probabilities is 0.601, which is not 1 as required. Also, Ted clearly needs a new approach.

10. Not a probability distribution because the responses are not values of a numerical random variable.

11. Probability distribution with (0 0.041) (1 0.2) (2 0.367) (3 0.299) (4 0.092) 2.2μ= ⋅ + ⋅ + ⋅ + ⋅ + ⋅ =

2 2 2 2 2(0 2.2) 0.041 (1 2.2) 0.2 (2 2.2) 0.367 (3 2.2) 0.299 (4 2.2) 0.092 1σ = − ⋅ + − ⋅ + − ⋅ + − ⋅ + − ⋅ =

12. Probability distribution with (0 0.0) (1 0.003) (2 0.025) (3 0.111) (4 0.279) (5 0.373) (6 0.208) 4.6μ= ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ =

2 2 2 2 2(0 4.6) 0.0 (1 4.6) 0.003 (2 4.6) 0.025 (3 4.6) 0.111 ... (6 4.6) 0.208

1

σ = − ⋅ + − ⋅ + − ⋅ + − ⋅ + + − ⋅

=

13. Not a probability distribution because the responses are not values of a numerical random variable. Also, sum of the probabilities is 1.18 instead of 1 as required.

14. Not a probability distribution because the sum of the probabilities is 0.967 instead of 1 as required. The discrepancy between 0.967 and 1 is too large to attribute to rounding errors.

15. (0 0.001) (1 0.01) (2 0.044) (9 0.01) (10 0.001) 5μ= ⋅ + ⋅ + ⋅ + + ⋅ + ⋅ =…

2 2 2 2 2(0 5) 0.001 (1 5) 0.01 (2 5) 0.044 (9 5) 1.60.01 (10 5) 0.001σ = − ⋅ + − ⋅ + − ⋅ + + − ⋅ ⋅ =+ −…

60 Chapter 5: Discrete Probability Distributions


16. Lower limit: 2 5 2(1.6) 1.8μ σ− = − = girls; Upper limit: 2 5 2(1.6) 8.2μ σ+ = + = girls Yes, 1 girl is an unusually low number of girls, because 1 girl is outside the range of usual values.

17. a. ( 8) 0.044P X = =

b. ( 8) 0.044 0.01 0.001 0.055P X ≥ = + + =

c. The result from part (b)

d. No, because the probability of 8 or more girls is 0.055, which is not very low (less than or equal to 0.05)

18. a. ( 1) 0.01P X = =

b. ( 1) 0.001 0.01 0.011P X ≤ = + =


d. Yes, because the probability of 0.011 is very low (less than or equal to 0.05)

19. (0 0.377) (1 0.399) (2 0.176) (3 0.041) (4 0.005) (5 0) (6 0) 0.9μ= ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ = 2 2 2 2 2(0 0.9) 0.377 (1 0.9) 0.399 ... (4 0.9) 0.005 (5 0.9) 0 (6 0.9) 0

0.9

σ = − ⋅ + − ⋅ + + − ⋅ + − ⋅ + − ⋅

=

20. Lower limit: 2 0.9 2(0.9) 0.9μ σ− = − =− ;Upper limit: 2 0.9 2(0.9) 2.7μ σ+ = + = Yes; 3 is above the range of usual values, so 3 is an unusually high number of failures among 6 cars tested.

21. a. ( 3) 0.041P X = =

b. ( 3) 0.041 0.005 0 0 0.046P X ≥ = + + + =

c. The probability from part (b)

d. Yes, because the probability of three or more failures is 0.046 which is very low (less than or equal to 0.05)

22. a. ( 1) 0.399P X = =

b. ( 1) 0.377 0.399 0.776P X ≤ = + =


d. No, because the probability of 0.776 is not very low (less than or equal to 0.05)

23. a. 10 10 10 1000⋅ ⋅ =

b. 1

1000

c. $500 $1 $499− =

d. 1

$1 1 $500 $0.50 501000

− ⋅ + ⋅ =− =− cents

e. The $1 bet on the pass line in craps is better because its expected value of –1.4 cents is much greater than the expected value of –50 cents for the Texas Pick 3 lottery.

24. a. 10 10 10 0 10,000⋅ ⋅ ⋅ =

b. 1

10,000

c. $5000 $1 $4999− =

d. 1

$1 1 $5000 $0.50 5010,000

− ⋅ + ⋅ =− =− cents

e. Because both bets have the same expected value of –50 cents, neither bet is better than the other.



25. a. 5 33

$0.26 $30 $5 0.3938 38

− + ⋅ − ⋅ =−

b. The bet on the number 27 is better because its expected value of –26 cents is greater than the expected value of –39 cents for the other bet.

26. a. 0.01 1 5 10 750,000 1,000,000

$131,477.5426 26 26 26 26 26

+ + + + + + =…

b. 2 2 21 1 1(0.01 131, 477.54) (1 131, 477.54) (1,000,000 131, 477.54) $253,584.47

26 26 26− ⋅ + − ⋅ + + − ⋅ =…

c. Lower limit: $131,477.54 2 $253,584.47 $375,691.40− ⋅ =−

Upper limit: $131,477.54 2 $253,584.47 $638,646.48+ ⋅ =

d. Yes, because the values are above the range of usual values given in part (c)

Section 5-3

1. The given calculation assumes that the first two adults include Wal-Mart and the last three adults do not include Wal-Mart, but there are other arrangements consisting of two adults who include Wal-Mart and three who do not. The probabilities corresponding to those other arrangements should also be included in the result.

2. The format of Formula 5-5 requires that the probability p and the variable x refer to the same outcome. If p is the probability of an adult including Wal-Mart, then x should count the number of people who include Wal-Mart.

3. Because the 30 selections are made without replacement, they are dependent, not independent. Based on the 5% guideline for cumbersome calculations, the 30 selections can be treated as being independent. (The 30 selections constitute 3% of the population of 1000 responses, and 3% is not more than 5% of the population.) The probability can be found by using the binomial probability formula.

4. The 0+ indicates that the probability is a very small positive value. (The actual value is 0.00000296.) The notation of 0+ does not indicate that the event is impossible; it indicates that the event is possible, but very unlikely.

5. Not binomial. Each of the weights has more than two possible outcomes.

6. Binomial

7. Binomial

8. Not binomial. Each of the responses has more than two possible outcomes.

9. Not binomial. Because the senators are selected without replacement, the selections are not independent. (The 5% guideline for cumbersome calculations cannot be applied because the 40 selected senators constitute 40% of the population of 100 senators, and that exceeds 5%.)

10. Not binomial. Because the senators are selected without replacement, they are not independent. . (The 5% guideline for cumbersome calculations cannot be applied because the 10 selected senators constitute 10% of the population of 100 senators, and that exceeds 5%.). Also, the numbers of terms have more than two possible outcomes.

11. Binomial. Although the events are not independent, they can be treated as being independent by applying the 5% guideline. The sample size of 380 is no more than 5% of the population of all smartphone users.

12. Binomial. Although the events are not independent, they can be treated as being independent by applying the 5% guideline. The sample size of 427 is not more than 5% of the population of all women.



13. a. 4 4 1

0.1285 5 5⋅ ⋅ =

b. {WWC, WCW, CWW}; 0.128 for each

c. 0.128 3 0.384⋅ =

14. a. 1 1 9 9

0.008110 10 10 10

⋅ ⋅ ⋅ =

b. {MMXX, MXMX, MXXM, XXMM, XMXM, XMMX}; each has a probability of 0.0081

c. 0.0081 6 0.0468⋅ =

15. 3 25 3 0.2 0.8 0.051C ⋅ ⋅ =

16. 3 2 4 1 5 05 3 5 4 5 50.2 0.8 0.2 0.8 0.2 0.8 0.057C C C⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ =

17. 3 2 4 1 5 05 3 5 4 5 50.2 0.8 0.2 0.8 0.2 0.8 0.057C C C⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ =

18. 0 5 1 4 2 35 0 5 1 5 20.2 0.8 0.2 0.8 0.2 0.8 0.943C C C⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ =

19. 0 55 5 0.2 0.8 0.328C ⋅ ⋅ =

20. 5 05 5 0.2 0.8 0.00032 0+C ⋅ ⋅ = =

21. 3 58 3 0.45 0.55 0.257C ⋅ ⋅ =

22. 6 1016 6 0.45 0.55 0.168C ⋅ ⋅ =

23. 16 420 16 0.45 0.55 0.00125C ⋅ ⋅ =

24. 9 211 9 0.45 0.55 0.0126C ⋅ ⋅ =

25. ( 2) 0.033 0.132 0.297 0.356 0.178 0.996P X ≥ = + + + + = ; yes

26. ( 5) 0.000 0.004 0.033 0.132 0.297 0.356 0.822P X ≤ = + + + + + =

27. ( 2) 0.000 0.004 0.033 0.037P X ≤ = + + = ; yes, because the probability of 2 or fewer peas with green

pods is small (less than or equal to 0.05).

28. ( 5) 0.356 0.178 0.534P X ≥ = + = ; no, because the probability of 0.534 is not small (less than or equal to

0.05)

29. a. 5 16 5 0.20 0.80 0.002C ⋅ ⋅ = (Tech: 0.00154)

b. 6 06 6 0.20 0.80 0+C ⋅ ⋅ = 0+ (Tech: 0.000064)

c. 0.002 0 0.002+ = (Tech: 0.00160)

d. Yes, the small probability from part (c) suggests that 5 is an unusually high number.

30. a. 2 57 2 0.80 0.20 0.004C ⋅ ⋅ = (Tech: 0.00430)

b. 1 67 1 0.80 0.20 0+C ⋅ ⋅ = 0+ (Tech: 0.000358)

c. 0.004 0 0.004+ = (Tech: 0.00467)

d. Yes, the small probability from part (c) suggests that 2 is unusually low



31. a. 0 55 0 0.20 0.80 0.328C ⋅ ⋅ =

b. 1 45 1 0.20 0.80 0.410C ⋅ ⋅ = 0.410

c. 0.328 0.410 0.738+ = (Tech: 0.737)

d. No, the probability from part (c) is not small, so 1 is not an unusually low number

32. a. 8 08 8 0.90 0.10 0.430C ⋅ ⋅ =

b. 7 18 7 0.90 0.10 0.383C ⋅ ⋅ =

c. 0.43 0.383 0.813+ =

d. No, the probability from part (c) is not small, so 7 is not unusually high

33. 12 820 12 0.48 0.52 0.101C ⋅ ⋅ = . No, because the probability of exactly 12 is 0.101, the probability of 12 or

more is greater than 0.101, so the probability of getting 12 or more is not very small, so 12 us not unusually high

34. 6 1824 6 0.25 0.75 0.185C ⋅ ⋅ = . The probability is not very small, so it is not unlikely

35. 10 212 10 0.805 0.195 0.287C ⋅ ⋅ = . No, because the flights all originate from New York, they are not randomly

selected flights, so the 80.5% on-time rate might not apply

36. 20 1030 20 0.72 0.28 0.125C ⋅ ⋅ = . No, because the probability of exactly 20 is 0.125, the probability of 20 or

fewer having those concerns is greater than 0.125, so the probability of getting 20 or fewer is not very small, so 20 is not unusually low.

37. a. 0 1212 0 0.45 0.55 0.000766C ⋅ ⋅ =

b. 1 0.000766 0.999− =

c. 0 12 1 1112 0 12 10.45 0.55 0.45 0.55 0.00829C C⋅ ⋅ + ⋅ ⋅ =

d. Yes, the very low probability of 0.00829 would suggest that the 45 share value is wrong

38. 20 5 21 021 20 21 21(1 0.0995) 0.0995 (1 0.0995) 0.0995 0.368C C⋅ − ⋅ + ⋅ − ⋅ = . With 21 booked passengers, there is

a probability of 0.368 that more than 19 passengers will show up, and that the flight will be overbooked. It does not seem wise to schedule in such a way that the flights will be overbooked about 37% of the time.

39. a. 13 114 13 0.5 0.5 0.000854C ⋅ ⋅ =

b. 14 014 14 0.5 0.5 0.000061C ⋅ ⋅ =

c. 14 0 13 114 14 14 130.5 0.5 0.5 0.5 0.000916C C⋅ ⋅ + ⋅ ⋅ =

d. Yes. The probability of getting 13 girls or a result of 14 girls is 0.000916, so chance does not appear to be a reasonable explanation for the 13 girls. Because 13 is an unusually high number of girls, it appears that the probability of a girl is higher with the XSORT method, and it appears that the XSORT method is effective.

40. a. 8 1220 8 0.5 0.5 0.12C ⋅ ⋅ =

b. No. If the success rate is equal to 50% , it is likely (with probability 0.252) that we get 8 successes or a result that is more extreme(fewer than 8 successes). This indicates that with a 50% success rate, the occurrence of 8 successes in 20 challenges could be reasonably explained by chance.

41. 0 2424 01 0.006 0.994 0.134C− ⋅ ⋅ = . It is not unlikely for such a combined sample to test positive.

42. 0 2416 01 0.00114 0.99886 0.0181C− ⋅ ⋅ = . It is unlikely for such a combined sample to test positive.



43. 1 39 0 4040 1 40 00.03 0.97 0.03 0.97 0.662C C⋅ ⋅ + ⋅ ⋅ = . The probability shows that about 2/3 of all shipments

will be accepted. With about 1/3 of the shipments rejected, the supplier would be wise to improve quality.

44. 2 28 1 29 0 30

30 2 30 1 30 00.02 0.98 0.02 0.98 0.02 0.98 0.978C C C⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ = . About 98% of all shipments will be

accepted. Almost all shipments will be accepted, and only 2% of the shipments will be rejected.

45. ( )4( 5) 0.06 1 0.06 0.0468P X = = − =

46. 12! 18 18 2

0.004855!4!3! 38 38 38

⋅ ⋅ ⋅ =

47. a. 6! 43! (6 43)!

(4) 0.000969(6 4)!4! (43 6 4)!(6 4)! (6 43 6)!6!

P+

= ⋅ ÷ =− − + − + −

b. 6! 43! (6 43)!

(6) 0.0000000715(6 6)!6! (43 6 6)!(6 6)! (6 43 6)!6!

P+

= ⋅ ÷ =− − + − + −

c. 6! 43! (6 43)!

(0) 0.436(6 0)!0! (43 6 0)!(6 0)! (6 43 6)!6!

P+

= ⋅ ÷ =− − + − + −

Section 5-4

1. n = 270, p = 0.07, q = 0.93

2. 270 0.07 0.93 4.2⋅ ⋅ = people. Yes, both expressions will yield the same result because they are equivalent. They are equivalent because 1q p= −

3. Variance is 150 0.933 0.067 9.4⋅ ⋅ = executives2

4. The mean of 140.0 executives is expressed with μ . The mean is calculated for the population of all groups

of 150 executives, not just one sample group. Because the mean is calculated for a population, it is a parameter.

5. 60 0.2 12μ np= = ⋅ = correct guesses and (1 ) 60 0.2 0.8 3.1σ np p= − = ⋅ ⋅ = correct guesses.

Minimum: 12 2(3.1) 5.8− = correct guesses, maximum: 12 2(3.1) 18.2+ = correct guesses.

6. 14 0.5 7μ np= = ⋅ = girls and (1 ) 14 0.5 0.5 1.9σ np p= − = ⋅ ⋅ = girls. Minimum: 7 2(1.9) 3.2− = ,

maximum 7 2(1.9) 10.8+ = girls

7. 1013 0.66 668.6μ np= = ⋅ = worriers and (1 ) 1013 0.66 0.34 15.1σ np p= − = ⋅ ⋅ = worriers. Minimum:

668.6 2(15.1) 638.4− = worriers, maximum: 668.6 2(15.1) 698.8+ = worriers

8. 94 0.064 6μ np= = ⋅ = subjects with headaches and (1 ) 94 0.064 0.936 2.4σ np p= − = ⋅ ⋅ = subjects

with headaches. Minimum: 6 2(2.4) 1.2− = subjects with headaches, maximum: 6 2(2.4) 10.8+ =

subjects with headaches.

9. a. 291 0.5 145.5μ np= = ⋅ = boys and (1 ) 291 0.5 0.5 8.5σ np p= − = ⋅ ⋅ = boys

b. Yes. Using the range rule of thumb, the minimum value is 145.5 2(8.5) 128.5− = boys and the

maximum value is 145.5 2(8.5) 162.5+ = boys. Because 239 boys is above the range of usual values,

it is unusually high. Because 239 boys is unusually high, it does appear that the YSORT method of gender selection is effective.



10. a. 580 0.25 145μ np= = ⋅ = and (1 ) 580 0.25 0.75 10.4σ np p= − = ⋅ ⋅ =

b. No, it is within the range of usual values of 145 2(10.4) 124.2− = and145 2(10.4) 165.8+ = . It does

not provide strong evidence against Mendel’s theory.

11. a. 100 0.20 20μ np= = ⋅ = and (1 ) 100 0.2 0.8 4σ np p= − = ⋅ ⋅ =

b. No, because 25 orange M&Ms is within the range of usual values of 20 2(4) 12− = and 20 2(4) 28+ = . The claimed rate of 20% does not necessarily appear to be wrong,

because that rate will usually result in 12 to 28 orange M&Ms (among 100), and the observed number of orange M&Ms is within that range.

12. a. 100 0.14 14μ np= = ⋅ = and (1 ) 100 0.14 0.86 3.5σ np p= − = ⋅ ⋅ =

b. No, because 8 yellow M&Ms is within the range of usual values of 14 2(3.5) 7− = and

14 2(3.5) 21+ = . The claimed rate of 14% does not necessarily appear to be wrong, because that rate

will usually result in 7 to 21 yellow M&Ms (among 100), and the observed number of yellow M&Ms is within that range.

13. a. 420,095 0.00034 142.8μ np= = ⋅ = and (1 ) 420,095 0.00034 0.999666 11.9σ np p= − = ⋅ ⋅ =

b. No, 135 is not unusually low or high because it is within the range of usual values 142.8 2(11.9) 119− = and 142.8 2(11.9) 166.6+ =

c. Based on the given results, cell phones do not pose a health hazard that increases the likelihood of cancer of the brain or nervous system.

14. a. 280 0.5 140μ np= = ⋅ = and (1 ) 280 0.5 0.5 8.4σ np p= − = ⋅ ⋅ =

b. The result of 123 correct identifications is just outside the range of usual values of 140 2(8.4) 123.2− = and140 2(8.4) 156.8+ = , but this indicates that 123 is unusually low. If the

touch therapists really had an ability to select the correct hand, they would have made more than 156.8 correct identifications. Therefore, they do not appear to have that ability.

15. a. 2600 0.06 156μ np= = ⋅ = and (1 ) 2600 0.06 0.94 12.1σ np p= − = ⋅ ⋅ =

b. The minimum usual frequency is 156 2(12.1) 131.8− = and the maximum is 156 2(12.1) 180.2+ = .

The occurrence of r 178 times is not unusually low or high because it is within the range of usual values.

16. a. 2600 0.127 330.2μ np= = ⋅ = and (1 ) 2600 0.127 0.813 17σ np p= − = ⋅ ⋅ =

b. The minimum usual frequency is 330.2 2(17) 296.2− = and the maximum is 330.2 2(17) 364.2+ = .

The occurrence of e 290 times is unusually because it is below the range of usual values.

17. a. 370 0.2 74μ np= = ⋅ = and (1 ) 370 0.2 0.8 7.7σ np p= − = ⋅ ⋅ =

b. The minimum usual number is 74 2(7.7) 58.6− = and the maximum is 74 2(7.7) 89.4+ = . The value

of 90 is unusually high because it is above the range of usual values.

18. a. 1

50 1.338

μ np= = ⋅ = and 1 37

(1 ) 50 1.138 38

σ np p= − = ⋅ ⋅ =

b. The minimum usual value is 1.3 2(1.1) 0.9− =− and the maximum is 1.3 2(1.1) 3.5+ =

The result of 0 wins is not unusually low because 0 wins is within the range of usual values.



19. a. 1

30 0.0821918365

μ np= = ⋅ = and 1 364

(1 ) 30 0.2862981365 365

σ np p= − = ⋅ ⋅ =

b. The minimum usual value is 0.0821918 2(0.2862981) 0.4904044− =− and the maximum is

0.0821918 2(0.2862981) 0.654788+ = . The result of 2 students born on the 4th of July would be

unusually high, because 2 is above the range of usual values.

20. a. 1

2600 0.000013195,249,054

μ np= = ⋅ = and

1 195,249,053(1 ) 2600 0.003649

195,249,054 195,249,054σ np p= − = ⋅ ⋅ =

b. The minimum usual values is 0.000013 2(0.003649) 0.007285− =− and the maximum is

0.000013 2(0.003649) 0.007311+ = . It is unusual to buy a ticket each week for 50 years and win at

least once, because 1 win (or more) is outside the range of usual values.

21. From the range of usual values we get 60μ= and 6σ = . Using the formulas for the mean and the

standard deviation we get 2

1 0.4σ

pμ

= − = which leads to 150μ

np

= = and 0.6q =

22. 170.

23. The probability of selecting a girl out of 40 is given by 10 30 12

40 12

( ) n nC CP X n

C−⋅

= = the following table list

the probabilities of selecting the number of girls from 0 to 10

Number of girls (X = n) Probability P(X = n)

0 0.0154815

1 0.0977782

2 0.2420012

3 0.3073032

4 0.2200011

5 0.0918266

6 0.0223190

7 0.0030608

8 0.0002207

9 0.0000073

10 0.0000001

The mean is [ ( )]μ x P x= ⋅∑

0 0.0154815 1 0.0977782 2 0.2420012 3 0.3073032 4 0.2200011 5 0.0918266

6 0.0223190 7 0.0030608 8 0.0002207 9 0.0000073 10 0.0000001 3

μ= ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ +⋅ + ⋅ + ⋅ + ⋅ + ⋅ =

The standard deviation is 2 2[ ( )]σ x P x μ= ⋅ −∑

2 2 2 2 2 2

2 2 2 2 2 3

0 0.0154815 1 0.0977782 2 0.2420012 3 0.3073032 4 0.2200011 5 0.0918266

6 0.0223190 7 0.0030608 8 0.0002207 9 0.0000073 10 0.0000001 3

1.3

σ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ +

=⋅ + ⋅ + ⋅ + ⋅ + ⋅ −

=



Section 5-5

1. 535

0.929576

μ= = , which is the mean number of hits per region. 2x = , because we want the probability

that a randomly selected region had exactly 2 hits, and 2.71828e = which is a constant used in all applications of Formula 5 – 9.

2. The mean is 194

4.246

μ= = , the standard deviation is 4.2 2.1σ = = and the variance is 2 4.2σ =

3. With 50n = , the first requirement of 100n ≥ is not satisfied. With 50n = and 0.001p = the second

requirement of 10np ≤ is satisfied. Because both requirements are not satisfied, we should not use the

Poisson distribution as an approximation to the binomial.

4. With 100n = and 0.001p = the two requirements are satisfied. For 101 wins, the Poisson approximation

gives a small positive probability, but the actual probability of 101 wins is 0 since it is impossible to get 101 wins in 100 tries.

5. 0 8.58.5

(0) 0.0002030!

eP

−⋅= = ; Yes it is unlikely.

6. 6 8.58.5

(6) 0.1076!

eP

−⋅= = ; No it is not unlikely

7. 10 8.58.5

(10) 0.1110!

eP


8. 12 8.58.5

(12) 0.060412!

eP


9. a. 268

6.541

μ= =

b. 0 6.56.5

1 0.9980!

e−⋅− =

c. Yes. Based on the result in part (b), we are quite sure (with probability 0.998) that there is at least one earthquake measuring 6.0 or higher on the Richter scale, so there is a very low probability (0.002) that there will be no such earthquake in a year.

10. a. 5469

133.441

μ= =

b. 133 133.4133.4

(133) 0.0346133!

eP

−⋅= =

c. No. Although the probability of exactly 133 earthquakes measured at 6.0 or higher on the Richter scale is quite small (0.0346), the number 133 is so close to the mean of 133.4 that this year would be quite ordinary, and it would not be unusual.

11. a. 22713

62.2365

μ= =

b. 50 62.262.2

(50) 0.015550!

eP

⋅= =



12. 196

9.820

μ= =

a. 0 9.89.8

(0) 0.4970!

eP

−⋅= =

b. 1 9.89.8

(1) 0.3481!

eP

−⋅= =

c. 2 9.89.8

(2) 0.1222!

eP

−⋅= =

d. 3 9.89.8

(3) 0.02843!

eP

−⋅= =

e. 4 9.89.8

(4) 0.004974!

eP

−⋅= = . The expected frequencies of 139, 97, 34, 8 , and 1.4 compare

reasonably well to the actual frequencies, so the Poisson distribution does provide good results.

13. a. 2 0.9290.929

(2) 0.172!

eP

−⋅= =

b. The expected number of regions with exactly 2 hits is 98.2

c. The expected number of regions with 2 hits is close to 93, which is the actual number of regions with 2 hits.

14. a. 12429 0.000011 0.1367μ= ⋅ =

b. 0 0.1370.137

(0) 0.8720!

eP

−⋅= = and

1 0.1370.137(1) 0.119

1!

eP

−⋅= = . So the probability of 0 or 1 is

0.872 0.119 0.991+ =

c. 1 0.991 0.009− =

d. No, the probability of more than one case is extremely small, so the probability of getting as many as four cases is even smaller.

15. a. 26 30.430.4

(26) 0.055826!

eP

−⋅= = . The expected value is 34 0.0558 1.9⋅ = cookies. The expected

number of cookies is very close to the actual number of cookies with 26 chocolate chips which is 2.

b. 30 30.430.4

(30) 0.072430!

eP


number of cookies is very different from the actual number of cookies with 26 chocolate chips which is 6.

16. a. 18 19.619.6

(18) 0.087518!

eP


number of cookies is not very close to the actual number of cookies with 18 chocolate chips which is 5.

b. 21 19.619.6

(21) 0.082621!

eP


number of cookies is very close to the actual number of cookies with 21 chocolate chips which is 3.

17. a. No. With 12n = and 16p = the requirement of 100n ≥ is not satisfied, so the Poisson distribution

is not a good approximation to the binomial distribution.

b. No. The Poisson distribution approximation to the binomial distribution yields 3 22

(3) 0.183!

eP

−⋅= = and the binomial distribution yields

3 9

12 3

1 5(3) 0.197

6 6P C

⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜= ⋅ ⋅ =⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠. The

Poisson approximation of 0.18 is too far from the correct result of 0.197.



Chapter Quick Quiz

1. Yes

2. 1

100 205⋅ =

3. 100 0.2 0.8 4σ = ⋅ ⋅ =

4. The range of usual values has a minimum value of 200 2 10 180− ⋅ = and a maximum value of 200 2 10 220+ ⋅ = . Therefore, 232 girls in 400 is an unusually high number of girls since it is outside the range of usual values.

5. The range of usual values has a minimum value of 200 2 10 180− ⋅ = and a maximum value of 200 2 10 220+ ⋅ = . Therefore, 185 girls in 400 is not an unusually high number of girls since it is inside the range of usual values.

6. Yes. The sum of the probabilities is 0.999 and it can be considered to be 1.

7. 0+ indicates that the probability is a very small positive number. It does not indicate that it is impossible for none of the five flights to arrive on time.

8. ( 3) 0.198 0.409 0.338 0.945P x ≥ = + + =

9. 0 0 1 0.006 2 0.048 3 0.198 4 0.409 5 0.338 4.022μ= ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ = and

2 2 2 2 2 2 20 0 1 0.006 2 0.048 3 0.198 4 0.409 5 0.338 4.022 0.893σ = ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ − = The range of usual values is from 2.236 to 5.808. Since zero is outside the range of usual values it is an unusually low number.

10. 0 0 1 0.006 2 0.048 3 0.198 4 0.409 5 0.338 4.022μ= ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ = and

2 2 2 2 2 2 20 0 1 0.006 2 0.048 3 0.198 4 0.409 5 0.338 4.022 0.893σ = ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ − = The range of usual values is from 2.236 to 5.808. Since 5 is inside the range of usual values it is not an unusually high number.

Review Exercise

1. 0 66 0( 0) 0.4 0.6 0.0467P X C= = ⋅ ⋅ = 2. 4 2

6 4( 4) 0.4 0.6 0.138P X C= = ⋅ ⋅ =

3. 600 0.4 240μ= ⋅ = and 600 0.4 0.6 12σ = ⋅ ⋅ = . The range of usual values has a minimum of

240 2 12 216− ⋅ = and a maximum value of 240 2 12 264+ ⋅ = . The result of 200 with brown eyes is unusually low.

4. The probability of 239 or fewer ( )0.484 is relevant for determining whether 239 is an unusually low

number. Because that probability is not very small, it appears that 239 is not an unusually low number of people with brown eyes.

5. Yes. The three requirements are satisfied. There is a numerical random variable x and its values are associated with corresponding probabilities. The sum of the probabilities is 1.001, so the sum is 1 when we allow for a small discrepancy due to rounding. Also, each of the probability values is between 0 and 1 inclusive.

6. 0 0.674 1 0.28 2 0.044 3 0.003 4 0 0.4μ= ⋅ + ⋅ + ⋅ + ⋅ + ⋅ = and

2 2 2 2 2 20 0.674 1 0.28 2 0.044 3 0.003 4 0 0.4 0.6σ = ⋅ + ⋅ + ⋅ + ⋅ + ⋅ − = The range of usual values has a minimum of 0.4 2 0.6 0.8− ⋅ =− and a maximum value of 0.4 2 0.6 1.6+ ⋅ = . Yes, 3 is an unusually high number of males with tinnitus among four randomly selected males.

7. The sum of the probabilities is 0.902 which is not 1as required. Because the three requirements are not satisfied, the given information does not describe a probability distribution.



8. 1 1 1 1 1

$75 $300 $75,000 $500,000 $1,000,000 $315,0755 5 5 5 5⋅ + ⋅ + ⋅ + ⋅ + ⋅ = . Because the offer is well

below her expected value, she should continue the game (although the guaranteed prize of $193000 had considerable appeal).

9. a. 1 1 1

$1,000,000 $100,000 $25,000900,000,000 110,000,000 110,000,000

1 1$5000 $2500 $0.012

36,667,000 27,500,000

⋅ + ⋅ + ⋅

+ ⋅ + ⋅ =

b. $0.012 minus the cost of the postage stamp. Since the expected value of winning is much smaller than the cost of a postage stamp, it is not worth entering the contest.

10. a. 18

0.630

μ= =

b. 0 0.60.6

(0) 0.5490!

eP

−⋅= =

c. 30 0.549 16.5⋅ = days

d. The expected number of days is 16.5, and that is reasonably close to the actual number of days which is 18.


1. a. The mean is 22.2 24.8 24.2 26.9 23.8

24.45

x+ + + +

= = hours

b. The median is 24.2 hours

c. The range is 26.9 22.2 4.7− = hours

d. The standard deviation is 2 2 2 2 2(22.2 24.4) (24.8 24.4) (24.2 24.4) (26.9 24.4) (23.8 24.4)

1.75

s− + − + − + − + −

= =

e. The variance is 2.9 hours2

f. The minimum is 24.4 2 1.7 21− ⋅ = hours and the maximum is 24.4 2 1.7 27.8+ ⋅ = hours.

g. No, because none of the times are outside the range of the usual values

h. Ratio

i. Continuous

j. The given times come from countries with very different population sizes, so it does not make sense to treat the given times equally. Calculations of statistics should take the different population sizes into account. Also, the sample is very small, and there is no indication that the sample is random.

2. a. 1 1 1 1 1

0.000110 10 10 10 10,000

⋅ ⋅ ⋅ = =

b.

x P(x)

–$1 0.9999

$4999 0.0001

c. 365 0.0001 0.0365⋅ =

d. 1 0.03650.0365

(1) 0.03521!

eP

−⋅= =

e. $1 0.9999 $4999 0.0001 $0.50− ⋅ + ⋅ =− or –50 cents.



3. a. 121 51

0.282611

+=

b. 121

0.303121 279

=+

c. 51

0.24251 160

=+

d. 51

0.29751 121

=+

e. 121 51 121 51

0.0792611 611

+ +⋅ =

f. 121 279 121 51 121

0.738611 611 611

+ ++ − =

g.

121611

0.703172611

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜⎝ ⎠=

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜⎝ ⎠

4. Because the vertical scale begins at 60 instead of 0, the difference between the two amounts is exaggerated. The graph makes it appear that men’s earnings are roughly twice those of women, but men earn roughly 1.2 times the earnings of women.

5. a. Frequency distribution or frequency table

b. Probability distribution

c. 0 9 1 7 2 12 3 10 4 10 5 11 6 8 7 8 8 14 9 11

4.79 7 12 10 10 11 8 8 14 11

x⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅

= =+ + + + + + + + +

This value is a statistic

d. 0 0.1 1 0.1 2 0.1 3 0.1 4 0.1 5 0.1 6 0.1 7 0.1 8 0.1 9 0.1 4.5μ= ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ = . This value is

a parameter

e. The random generation of 1000 digits should have a mean close to 4.5 from part (d). The mean of 4.5 is the mean for the population of all random digits; so samples will have means that tend to center about 4.5

6. a. 4 1216 4 0.1 0.9 0.0514C ⋅ ⋅ =

b. 0 1616 01 0.1 0.9 0.815C− ⋅ ⋅ =

c. This is a voluntary response sample. This suggests that the results might not be valid, because those with a strong interest in the topic are more likely to respond.

Chapter 6: Normal Probability Distributions 73


Chapter 6: Normal Probability Distributions Section 6-2

1. The word “normal” has a special meaning in statistics. It refers to a specific bell-shaped distribution that can be described by Formula 6-1.

2.

3. The mean and standard deviation have values of 0μ= and 1σ =

4. The notation αz represents the z score that has an area of α to its right.

5. ( )0.2 5 1.25 0.75− =

6. ( )0.2 0.75 0 0.15− =

7. ( )0.2 3 1 0.40− =

8. ( )0.2 4.5 1.5 0.60− =

9. ( )0.44 0.6700P z < =

10. ( )1.04 0.8508P z >− =

11. ( ) ( ) ( )0.84 1.28 1.28 0.84 0.8997 0.2005 0.6992P z P z P z− < < = < − <− = − = (Tech: 0.6993)

12. ( ) ( ) ( )1.07 0.67 0.67 1.07 0.7486 0.1423 0.6063P z P z P z− < < = < − <− = − =

13. 1.23z =

14. 0.51z =−

15. 1.45z =−

16. 0.82z =

17. ( )2.04 0.0207P z <− =

18. ( )0.19 0.4247P z <− =

19. ( )2.33 0.9901P z < =

20. ( )1.96 0.9750P z < =

21. ( )0.82 1 0.7939 0.2061P z > = − =

22. ( )1.82 1 0.9656 0.0344P z > = − =

23. ( )1.50 1 0.0668 0.9332P z >− = − =

24. ( )0.84 1 0.2005 0.7995P z >− = − =

25. ( ) ( ) ( )0.25 1.25 1.25 0.25 0.8944 0.5987 0.2957P z P z P z< < = < − < = − = (Tech: 0.2956)

26. ( ) ( ) ( )1.23 2.37 2.37 1.23 0.9911 0.8907 0.1004P z P z P z< < = < − < = − = 0.1004 (Tech: 0.1005)

27. ( ) ( ) ( )2.75 2.00 2.00 2.75 0.0228 0.0030 0.0198P z P z P z− < <− = <− − <− = − =

28. ( ) ( ) ( )1.93 0.45 0.45 1.93 0.3264 0.0268 0.2996P z P z P z− < <− = <− − <− = − =

29. ( ) ( ) ( )2.20 2.50 2.50 2.20 0.9938 0.0139 0.9799P z P z P z− < < = < − <− = − =

30. ( ) ( ) ( )0.62 1.78 1.78 0.62 0.9625 0.2676 0.6949P z P z P z− < < = < − <− = − = (Tech: 0.6948)

74 Chapter 6: Normal Probability Distributions


31. ( ) ( ) ( )2.11 4.00 4.00 2.11 0.9999 0.0174 0.9825P z P z P z− < < = < − <− = − = (Tech: 0.9827)

32. ( ) ( ) ( )3.90 2.00 2.00 3.90 0.9772 0.0001 0.9771P z P z P z− < < = < − <− = − = 0. (Tech: 0.0772)

33. ( )3.65 0.9999P z < =

34. ( )3.80 0.9999P z >− = 0.9999

35. ( )0 0.5000P z < =

36. ( )0 0.5000P z > =

37. 90 1.28P =

38. 5 1.645P =−

39. 2.5 1.96P =− and 97.5 1.96P =

40. 0.5 2.575P =− and 99.5 2.575P =

41. 0.025 1.96z =

42. 0.01 2.33z =

43. 0.05 1.645z =

44. 0.03 1.88z =

45. ( ) ( ) ( )1 1 1 1 0.8413 0.1587 0.6826 68.26%P z P z P z− < < = < − <− = − = = (Tech: 68.27%)

46. ( ) ( ) ( )2 2 2 2 0.9772 0.0228 0.9544 95.44%P z P z P z− < < = < − <− = − = = (Tech: 95.45%)

47. ( ) ( ) ( )3 3 3 3 0.9987 0.0013 0.9974 99.74%P z P z P z− < < = < − <− = − = = (Tech: 99.73%)

48. ( ) ( ) ( )3.5 3.5 3.5 3.5 0.9999 0.0001 0.9998 99.98%P z P z P z− < < = < − <− = − = = (Tech: 99.95%)

49. a. ( ) ( ) ( )1 1 1 1 0.8413 0.1587 0.6826 68.26%P z P z P z− < < = < − <− = − = = (Tech: 68.27%)

b. ( ) ( ) ( )2 or 2 2 2 0.0228 0.0228 0.0456 4.56%P z z P z P z<− > = <− + > = + = =

c. ( ) ( ) ( )1.96 1.96 1.96 1.96 0.975 0.020 0.9500 95%P z P z P z− < < = < − <− = − = =

d. ( ) ( ) ( )2 2 2 2 0.9772 0.0228 0.9544 95.44%P z P z P z− < < = < − <− = − = = (Tech: 95.45%)

e. ( ) ( )3 1 3 1 0.9987 0.0013 0.13%P z P z> = − < = − = =

50. a. 2.5μ= min. and 5

1.412

σ = = min.

b. The probability is 13

or 0.5774, and it is very different from the probability of 0.6826 that would be

obtained by incorrectly using the standard normal distribution. The distribution does affect the results very much.

Section 6-3

1. a. 0μ= and 1σ =

b. The z scores are numbers without units of measurements

2. a. The area equals the maximum probability value of 1.

b. The median is the middle value and for normally distributed scores that is also the mean, which is 100.

c. The mode is also 100.

d. The variance is the square of the standard deviation which is 225.

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

4. No. Randomly generated digits have a uniform distribution, but not a normal distribution. The probability

of a digit less than 3 is 3

0.310

=



5. 118

118 1001.2

15xz =

−= = which has an area of 0.8849 to the left of it

6. 90

91 1000.6

15xz =

−= =− which has an area of 0.7257 to the right of it

7. 133

133 1002.2

15xz =

−= = which has an area of 0.9861 to the left of it. 79

110 1001.4

15xz =

−= =− which has

an area of 0.0808 to the left of it. The area between the two scores is 0.9861 0.0808 0.9053− = .

8. 124

124 1001.6

15xz =


112 1000.8

15xz =

−= = which has

an area of 0.7881 to the left of it. The area between the two scores is 0.9452 0.7881 0.1571− = .

9. 2.44z = , which means 2.44 15 100 136x = ⋅ + =

10. 1z = , which mean 1 15 100 115x = ⋅ + =

11. 2.07z =− , which means 2.07 15 100 69x =− ⋅ + =

12. 1.33z = , which means 1.33 15 100 120x = ⋅ + =

13. 85

85 1001

15xz =

−= =− , which has an area of 0.1587 to the left of it

14. 70

70 1002

15xz =

−= =− , which has an area of 0.9772 to the right of it

15. 90

90 1000.67

15xz =

−= =− which has an area of 0.2514 to the left of it. 110

110 1000.67

15xz =

−= = which

has an area of 0.7486 to the left of it. The area between the two scores is 0.7486 0.2514 0.4972− = . (Tech: 0.4950)

16. 120 1.33xz = = which has an area of 0.9082 to the left of it.

110

110 1000.67

15xz =

−= = which has an area of 0.7486 to the left of it. The area between the two scores is

0.9082 0.7486 0.1596− = (Tech: 0.1613)

17. 1.27z = which means the score is 1.27 15 100 119x = ⋅ + =

18. 0.67z =− which means the score is 0.67 15 100 90x =− ⋅ + =

19. 0.67z = which means the score is 0.67 15 100 110x = ⋅ + =

20. 2.07z = which means the minimum score is 2.07 15 100 131x = ⋅ + =

21. a. 78

78 63.85.46

2.6xz =

−= = which has an area of 0.9999 to the left of it.

62

62 63.80.69

2.6xz =

−= =− which has an area of 0.2451 to the left of it. Therefore, the percentage of

qualified women is 0.9999 0.2451 0.7548− = or 75.48%. (Tech 95.56%.) Yes, about 25% of women are not qualified because of their heights.

b. 78

78 69.53.54

2.4xz =


62 69.53.13

2.4xz =

−= =−

which has an area of 0.0009 to the left of it. Therefore, the percentage of men is 0.9999 0.0009 0.9990− = or 99.90%. (Tech: 99.89%.) No, only about 0.1% of men are not qualified because of their heights.



21. (continued)

c. The z score with 2% to the left of it is –2.04 which corresponds to a height of 2.04 2.6 63.8 58.5x =− ⋅ + = in. The z score with 2% to the right of it is 2.04 which corresponds to a height of 2.04 2.6 63.8 69.1x = ⋅ + = in.

d. The z score with 1% to the left of it is –2.33 which corresponds to a height of 2.33 2.4 69.5 63.9x =− ⋅ + = in. The z score with 1% to the right of it is 2.33 which corresponds to a

height of 2.33 2.4 69.5 75.1x = ⋅ + = in.

22. a. 64

64 63.80.08

2.6xz =

−= = and 77

77 63.85.08

2.6xz =

−= = . The area between the two z scores is

0.9999 0.5319 0.4680− = or 46.80%. (Tech: 46.93%.)

b. 64

64 69.52.29

2.4xz =

−= =− and 77

77 69.53.13

2.4xz =

−= = . The area between the two z scores is

0.9991 0.0110 0.9881− = or 98.81%.

c. The z score with 3% to the left of it for women is –1.88 which corresponds to a height of 1.88 2.6 63.8 58.9− ⋅ + = in. The z score with 3% to the right of it for men is 1.875 or 1.88 which

corresponds to a height of 1.88 2.4 69.5 74⋅ + = in

23. a. The height minimum is 4 12 8 56⋅ + = in. and the height maximum is 6 12 3 75⋅ + = in. The z score for

women for the minimum is 56 63.8

32.6

−=− , and the z score for women for the maximum is

75 63.84.31

2.6

−= . The area between the z scores is 0.9999 0.0013 0.9986− = or 99.86%

b. The z score for men for the minimum is 56 69.5

5.632.4

−=− , and the z score for men for the maximum

is 56 69.5

2.292.4

−= . The area between the z scores is 0.9890 0.0001 0.9898− = or 98.98%. (Tech:

98.90%.)

c. The z score with 5% for women to the left of it is –1.65 which corresponds to a height of 1.65 2.6 63.8 59.5− ⋅ + = in. the z score with 5% of men to the right of it is 1.625 which corresponds to

a height of 1.625 2.4 69.5 73.4⋅ + = in.

24. a. The z score for the minimum height is 51.6 69.5

7.462.4

−=− which has an area of 0.0001 or 0.01% to

the left of it meaning that practically no man can fit without bending. (Tech: 0.00%.)

b. The z score for the minimum height is 51.6 63.8

4.692.6

−=− which has an area of 0.0001 or 0.01% to

the left of it meaning that practically no women can fit without bending. (Tech: 0.00%.)

c. The door design is very inadequate, but the jet is relatively small and seats only six people. A much higher door would require such major changes in the design and cost of the jet, that the greater height is not practical.

d. The z score for 60% is 0.25 which corresponds to a height 0.25 2.4 69.5 70.1⋅ + = in for men.

25. a. The z score for 174 lb. is 174 182.9

0.2240.8

−=− which has an area of 0.4129 to the left of it. (Tech:

0.4137.)

b. 3500

25140

= people c. 3500

19.14182.9

= , so 19 people



25. (continued)

d. The mean weight is increasing over time, so safety limits must be periodically updated to avoid an unsafe condition

26. a. The z score that has 95% of the area to the left of it is 1.67 which corresponds to a height of 1.67 1.2 21.4 23.4⋅ + = in. If there is clearance for 95% of males, there will certainly be clearance for all women in the bottom 5%

b. Men’s z score is 23.5 21.4

1.751.2

−= and that has an area of 0.9599 or 99.95%. Women’s z score is

23.5 19.63.55

1.1

−= and that has an area of 0.9999 or 99.99%. (Tech 99.98%.) The table will fit almost

everyone except about 4% of the men with the largest sitting knee heights

27. a. The z score for a 308 day pregnancy is 308 268

2.6715

−= which corresponds to a probability of 0.0038

or 0.38%. Either a very rare event occurred or the husband is not the father.

b. The z score corresponding to 3% is –1.87 which corresponds to a pregnancy of 1.87 15 268 240− ⋅ + = days

28. a. The z score for a temperature of 100.6 is 100.6 98.2

3.870.62

−= which corresponds to a an area of

1 0.9999 0.0001 0.01%− = = to the right of it.; yes

b. The z score for a probability of 5% is 1.65 which corresponds to a temperature of 1.65 0.62 98.2 99.22⋅ + = degrees.

29. a. The z score for an earthquake of magnitude 2 is 2 1.184

1.390.587

−= which is 0.9177 or 91.77% of

earthquakes. (Tech: 99.78%.)

b. The z score is 4 1.184

4.800.587

−= which is 0.0001 or 0.01% of earthquakes. (Tech: 0.00%.)

c. The z score for 95% of earthquakes is 1.645 or 1.65 which corresponds to an earthquake magnitude of 1.645 0.587 1.184 2.15⋅ + = , so not all earthquakes about the 95th percentile will cause items to shake.

30. The z score for a probability of 99% is 2.33 which corresponds to a hip breadth of 2.59 0.00436 0.78386 0.7951⋅ + = in.

31. The z score for 1P is –2.33 which corresponds to a count of 2.33 2.6 24 17.9− ⋅ + = chocolate chips. (Tech:

18 chocolate chips.) The z score for 99P is 2.33 which corresponds to a count of 2.33 2.6 24 30.1⋅ + =

chocolate chips. (Tech: 10.0 chocolate chips.) The values can be used to identify cookies with an unusually low number of chocolate chips or an unusually high number of chocolate chips, so those numbers can be used to monitor the production process to ensure that the numbers of chocolate chips stay within reasonable limits.

32. a. The minimum weight has a z score of 5.64 5.67

0.50.06

−=− which has a corresponding probability of

0.3085 and the maximum weight has a z score of 5.7 5.67

0.50.06

−= which has a corresponding

probability of 0.6915. Therefore, the percentage of quarters rejected is ( )1 0.6915 0.3085 0.6170− − = . (Tech: 61.71%.) That percentage is too high because most quarters

will be rejected.

b. The z score for a probability of the top 2.5% and the bottom 2.5% is 2 and –2 respectively. Therefore the weight minimum is 2 0.06 5.67 5.5− ⋅ + = g and the weight maximum is 2 0.06 5.67 5.79⋅ + = g



33. a. The mean is 67.25 beats per minute and the standard deviation is 10.335 beats per minute. The histogram for the data confirms that the distribution is roughly normal.

90858075706560555045

9

8

7

6

5

4

3

2

1

0

PULSE

Freq

uen

cy

b. The z score for the bottom 2.5% is –1.95 which corresponds to a pulse of 1.95 10.335 67.25 47− ⋅ + =

beats per minute, and the z score for the top 2.5% is 1.95 which corresponds to a pulse of 1.95 10.335 67.25 87.5⋅ + = beats per minute.

34. a. The mean is 0.78386 lb. and the standard deviation is 0.00436 lb. The histogram confirms that the distribution of weights is roughly normal.

0.7940.7900.7860.7820.7780.774

9

8

7

6

5

4

3

2

1

0

Diet Pepsi Weights

Freq

uen

cy

b. The z score for the bottom 0.5% is –2.59 which has a corresponding weight of,

2.59 0.00436 0.78386 0.7726− ⋅ + = lb., the z score for the top 0.5% is 2.59 which has a corresponding weight of 2.59 0.00436 0.78386 0.7951⋅ + = lb.

35. a. The new mean is equal to the old one plus the new points which is 75. The standard deviation is unchanged at 10 (since we added the same amount to each student.)

b. No, the conversion should also account for variation.

c. The z score for the bottom 70% is 0.52 which has a corresponding score of 0.52 10 40 45.2⋅ + = , and the z score for the top 10% is 1.28 which has a corresponding score of 1.28 10 40 52.8⋅ + =

d. Using a scheme like the one in part (c), because variation is included in the curving process.

36. a. 30

30 242.31

2.6xz =

−= = which has a percentage of 0.9896, 20

20 241.54

2.6xz =

−= =− which has a

percentage of 0.0618. Therefore, the percentage between 20 and 30 chocolate chips is 0.9896 0.0618 0.9278− = or 92.78%. (Tech: 92.75%)

b. 30.5

30.5 242.5

2.6xz =

−= = which has a percentage of 0.9938 and 19.5

19.5 241.73

2.6xz =

−= =− which has

a percentage of 0.0418. Therefore, the percentage between 19.5 and 30.5 is 0.9938 0.0418 0.9520− = or 95.20%.

c. The use of the continuity correction changes the result by a relatively small but not insignificant amount.



37. The z score for Q1 is –0.67, and the z score for Q3 is 0.67. The IQR is ( )0.67 0.67 1.34− − = .

1.5 2.01IQR⋅ = , so 1 1.5 0.67 2.01 2.68Q IQR− ⋅ =− − =− and 3 1.5 0.67 2.01 2.68Q IQR+ ⋅ = + =

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

38. a. The z score for the 95th percentile is 1.645. This corresponds to an SAT score of 1.645 312 1511 2024.24⋅ + = or 2024. The z score of 1.645 corresponds to an ACT score of 1.645 5.1 21.1 29.4895⋅ + = or 29.5.

b. The score of 2100 corresponds to a z score of 2100 1511

1.89312

−= which corresponds to an ACT score

of 1.89 5.1 21.1 30.73⋅ + = or 30.7.

Section 6-4

1. a. The sample mean will tend to center about the population parameter of 5.67 g.

b. The sample mean will tend to have a distribution that is approximately normal.

c. The sample proportions will tend to have a distribution that is approximately normal.

2. a. Without replacement

b. (1) When selecting a relatively small sample from a large population, it makes no significant difference whether we sample with replacement or without replacement. (2) Sampling with replacement results in independent events that are unaffected by previous outcomes, and independent events are easier to analyze and they result in simpler calculations and formulas.

3. Sample mean, sample variance, sample proportion

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

5. No. The sample is not a simple random sample from the population of all college Statistics students. It is very possible that the students at Broward College do not accurately reflect the behavior of all college Statistics students.

6. a. Normal

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

4.510

+ + + + + + + + +=

c. 5

0.510

= which is the proportion of the five odd numbers {1, 3, 5, 7, 9} to the ten digits {0, 1, 2, 3, 4,

5, 6, 7, 8, 9}

7. a. The mean of the population is 4 5 9

63

μ + += = , and the variance is

2 2 22 (4 6) (5 6) (9 6)

4.73

σ − + − + −= =

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

Sample Variance Probability

0 3/9

0.5 2/9

8 2/9

12.5 2/9



7. (continued)

c. The sample variances’ mean is 3 0 2 0.5 2 8 2 12.5

4.79

⋅ + ⋅ + ⋅ + ⋅=

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

8. a. The population standard deviation (using the result from the previous problem) is 4.7 2.160σ = =

b. By taking the square root of the sample variances from the previous problem we get

Sample Standard Deviation Probability

0.000 3/9

0.707 2/9

2.828 2/9

3.536 2/9

c. The mean of the sample standard deviations is 3 0 2 0.707 2 2.828 2 3.536

1.5719

⋅ + ⋅ + ⋅ + ⋅=

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

9. a. The population median is 5

b. The possible sample of size 2 are {(4, 4), (4, 5), (4, 9), (5, 4), (5, 5), (5, 9), (9, 4), (9, 5), (9, 9)} which have the following medians {4, 4.5, 6.5, 4.5, 5, 7, 6.5, 7, 9}

Sample Median Probability

4 1/9

4.5 2/9

5 1/9

6.5 2/9

7 2/9

9 1/9

c. The mean of the sampling distribution of the sampling median is 4 4.5 4.5 5 6.5 6.5 7 7 9

69

+ + + + + + + +=

d. No. The mean of the sampling distribution of the sample medians is 6, and it is not equal to the value of the population median of 5, so the sample medians do not target the value of the population median.

10. a. The proportion of odd numbers is 2/3 (there are two odd numbers from the population of 4, 5, and 9)

b. The possible sample of size 2 are {(4, 4), (4, 5), (4, 9), (5, 4), (5, 5), (5, 9), (9, 4), (9, 5), (9, 9)} which have the following proportion of odd numbers {0, 0.5, 0.5, 0.5, 1, 1, 0.5, 1, 1}

Sample Proportion Probability

0 1/9

0.5 4/9

1 4/9

c. The mean of the sampling distribution of sample proportions is 0 0.5 0.5 0.5 0.5 1 1 1 1 2

0.679 3

+ + + + + + + += =



10. (continued)

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

11. a. The possible samples of size 2 are {(56, 56), (56, 49), (56, 58), (56, 46), (49, 56), (49, 49), (49, 58), (49, 46), (58, 56), (58, 49), (58, 58), (58, 46), (46, 56), (46, 49), (46, 58), (46, 46)}

Sample Mean Age Probability

46 1/16

47.5 2/16

49 1/16

51 2/16

52 2/16

52.5 2/16

53.5 2/16

56 1/16

57 2/16

58 1/16

b. The mean of the population is 56 49 58 46

52.254

+ + += and the mean of the sample means is

46 47.5 47.5 49 51 51 52 52 52.5 52.5 53.5 53.5 56 57 57 5852.25

16

+ + + + + + + + + + + + + + +=

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

12. a. The possible samples of size 2 are {(56, 56), (56, 49), (56, 58), (56, 46), (49, 56), (49, 49), (49, 58), (49, 46), (58, 56), (58, 49), (58, 58), (58, 46), (46, 56), (46, 49), (46, 58), (46, 46)}

Sample Median Age Probability

46 1/16

47.5 2/16

49 1/16

51 2/16

52 2/16

52.5 2/16

53.5 2/16

56 1/16

57 2/16

58 1/16

b. The median of the population is 49 56

52.52

+= and the median of the sample medians is

52 52.552.25

2

+= . The two values are not equal.

c. The sample medians do not target the population median of 52.5, so the sample medians do not make good estimators of the population medians



13. a. The possible samples of size 2 are {(56, 56), (56, 49), (56, 58), (56, 46), (49, 56), (49, 49), (49, 58), (49, 46), (58, 56), (58, 49), (58, 58), (58, 46), (46, 56), (46, 49), (46, 58), (46, 46)} which have the following ranges and associated probabilities

Sample Range Probability

0 4/16

2 2/16

3 2/16

7 2/16

9 2/16

10 2/16

12 2/16

b. The range of the population is 58 46 12− = , the mean of the sample ranges is 4 0 2 2 2 3 2 7 2 9 2 10 2 12

5.37516

⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅= . The values are not equal.

c. The sample ranges do not target the population range of 12, so sample ranges do not make good estimators of the population range.

14. a. The possible samples of size 2 are {(56, 56), (56, 49), (56, 58), (56, 46), (49, 56), (49, 49), (49, 58), (49, 46), (58, 56), (58, 49), (58, 58), (58, 46), (46, 56), (46, 49), (46, 58), (46, 46)} which have the following variances and associated probabilities

Sample Variance Probability

0 4/16

2 2/16

4.5 2/16

24.5 2/16

40.5 2/16

50 2/16

72 2/16

b. The variance of the population is 2 2 2 2(56 52.25) (49 52.25) (58 52.25) (46 52.25)

24.18754

− + − + − + −=

The mean of the sample variances is 4 0 2 2 2 4.5 2 24.5 2 40.5 2 50 2 72

24.187516

⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅=

The two values are equal

c. The sample variances do target the population variance , so sample variances do make good estimators of the population variance.

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

Proportion of Girls Probability

0 0.25

1 / 2 0.5

2/2 0.25

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



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

Proportion of Girls Probability

0 1/3

1/3 3/8

2/3 3/8

3/3 1/8

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

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

a.

Proportion Correct Probability

0

2

4 4 16

5 5 25⋅ =

1

2

1 4 82

5 5 25

⎛ ⎞⎟⎜⋅ ⋅ =⎟⎜ ⎟⎜⎝ ⎠

2

2

1 1 1

5 5 25

⎛ ⎞⎟⎜ ⋅ =⎟⎜ ⎟⎜⎝ ⎠

b. The mean is 16 0 8 0.5 1 1

0.225

⋅ + ⋅ + ⋅=

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

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

Proportion of Defective Probability

0 9 / 25

0.5 12 / 25

1 4 / 25

b. The mean is 9 0 12 0.5 4 1

0.425

⋅ + ⋅ + ⋅=

c. Yes. The population proportion is 0.4 (2 out of 5) and the mean of the sampling proportions is also 0.4. Yes, the mean of the sampling distribution of proportions is always equal to the population proportion.

19. 1 1

(0) 0.252(2 2 0)!(2 0)! 4

P = = =− ⋅ ⋅

, 1

(0.5) 0.52(2 2 0.5)!(2 0.5)!

P = =− ⋅ ⋅

,

1(1) 0.25

2(2 2 1)!(2 1)!P = =

− ⋅ ⋅. The formula yields values which describes the sampling distribution of

the sample proportions. The formula is just a different way of presenting the same information in the table that describes the sampling distribution.



20. Sample values of the mean absolute deviation (MAD) do not usually target the value of the population MAD, so a MAD statistic is not good for estimating a population MAD. If the population of {4, 5, 9} from Example 5 is used, the sample MAD values of 0, 0.5, 2, and 2.5 have corresponding probabilities of 3/9, 2/9, 2/9, and 2/9. For these values, the population MAD is 2, but the sample MAD values have a mean of 1.1, so the mean of the sample MAD values is not equal to the population MAD.

Section 6-5

1. Because the sample size is greater than 30, the sampling distribution of the mean ages can be approximated

by a normal distribution with mean μ and standard deviation40

σ.

2. No. Because the original population is normally distributed, the sample means will be normally distributed for any sample size, not just those greater than 30.

3. 60.5xμ = cm and it represents the mean of the population consisting of all sample means.

6.61.1

36xσ = = cm, and it represents the standard deviation of the population consisting of all sample

means.

4. Because the digits are equally likely to occur, they have a uniform distribution. Because the sample means are based on samples of size 3 drawn from a population that does not have a normal distribution, we should not treat the sample means as having a normal distribution.

5. a. 222.7

222.7 205.52

8.6xz =

−= = , which has a probability of 0.9772.

b. 207

207 205.51.22

8.6

49

xz =

−= = , which has a probability of 0.8888. (Tech: 0.8889.)

6. a. 196.9

196.9 205.51

8.6xz =

−= =− , which has a probability of 0.1587.

b. 205

205 205.50.35

8.6

36

xz =

−= =− , which has a probability of 0.3632. (Tech: 0. 3636.)

7. a. 218.4

218.4 205.51.5

8.6xz =

−= = , which has a probability of 1 0.9332 0.0668− = to the right of it

b. 204

204 205.50.52

8.6

9

xz =

−= =− which has a probability of 1 0.3015 0.6985− = to the right of it. (Tech:

0.6996.)

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

8. a. 195

195 205.51.22

8.6xz =

−= =− which has an area of 1 0.1112 0.8888− = to the right of it. (Tech:

0.8889.)

b. 203 205.5

1.458.6

25

z−

= =− which has an area of 1 0.0735 0.9265− = to the right of it. (Tech: 0.9270.)

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



9. a. 231.5

231.5 205.53.02

8.6xz =

−= = and 179.7

179.7 205.53

8.6xz =

−= =− which has a probability of

0.9987 0.0013 0.9974− = between them. (Tech: 0.9973.)

b. 206

206 205.50.37

8.6

40

xz =

−= = and 204

204 205.51.1

8.6

40

xz =



10. a. 200

200 205.50.64

8.6xz =

−= =− and 180

180 205.52.97

8.6xz =

−= =− which have a probability of


b. 206

206 205.50.41

8.6

50

xz =

−= = and 198

198 205.56.17

8.6

50

xz =



11. 195.3

195.3 182.91.22

40.8

16

xz =

−= = which has a probability of 1 0.8888 0.1112− = to the right of it. The

elevator appears to be relatively safe because there is a very small chance that it will be overloaded with 16 male passengers. (Tech: 0.1121.)

12. 195.3

195.3 1742.09

40.8

16

xz =

−= = which has a probability of 1 0.9817 0.0183− = to the right of it. The elevator

appears to be relatively safe because there is a very small chance of overloading. Using the outdated mean that is too low has the effect of making the elevator appear to be much safer than it actually is. (Tech: 0.0183.)

13. a. 25

25 22.652.94

0.8xz =

−= = and 21

21 22.652.06

0.8xz =

−= =− , so 0.9984 0.0197 0.9787 97.87%− = =

of women can fit into the hats. (Tech: 0.9788.)

b. The z scores for the smallest 2.5% and the largest 2.5% head circumferences are –1.96 and 1.96 respectively. This corresponds to head circumferences of ( )0.8 1.96 22.65 21.08⋅ − + = and

0.8 1.96 22.65 24.22⋅ + =

c. 23

23 22.653.5

0.8

64

xz =

−= = and 22

22 22.656.5

0.8

64

xz =


0.9998 0.0000 0.9998 99.98%− = = between them. No, the hats must fit individual women, not the mean from 64 women. If all hats are made to fit head circumferences between 22 in. and 23 in., the hats will not fit about half those women.

14. a. 22

22 18.23.8

1xz =

−= = which has a probability of 0.9999. So the percentage is 99.99%

b. 18.5

18.5 18.21.8

1

36

xz =

−= = which has a probability of 0.9641. No, when considering the diameters of

manholes, we should use a design based on individual men, not samples of 36 men.



15. a. The mean weight of passengers is 3500

14025

= lb.

b. 140

140 182.95.26

40.8

25

xz =

−= =− which has a probability of 0.99999 (or 1.0000) to the right of it. (Tech:

0.0.99999993.)

c. 175

175 182.90.87

40.8

20

xz =

−= =− which has a probability of 0.8078 to the right of it. (Tech: 0.8067.)

d. Given that there is a 0.8078 probability of exceeding the 3500 lb. limit when the water taxi is loaded with 20 random men, the new capacity of 20 passengers does not appear to be safe enough because the probability of overloading is too high.

16. a. 0.8535

0.8535 0.85650.06

0.0518xz =

−= =− which has a probability of 1 0.4761 0.5239− = to the right of it.

(Tech: 0.5231.)

b. 0.8535

0.8535 0.85651.25

0.0518

465

xz =

−= =− which has a probability of 1 0.1056 0.8944− = to the right of it.

(Tech: 0.8941.) Instead of filling each bag with exactly 465 M&Ms, the company probably fills the bags so that the weight is as stated. In any event, the company appears to be doing a good job of filling the bags.

17. a. 167

167 182.90.39

40.8xz =


0.6516.)

b. 167

167 182.91.35

40.8

12

xz =

−= =− which has a probability of 1 0.0885 0.9115− = to the right of it

c. There is a high probability that the gondola will be overloaded if it is occupied by 12 more people, so it appears that the number of allowed passengers should be reduced.

18. a. The z score for 1% is –2.33 which corresponds to a pulse rate of 2.33 11.6 77.5 50.5− ⋅ + = beats per minute. The z score for 99% is 2.33 which corresponds to a pulse rate of 2.33 11.6 77.5 104.5⋅ + = beats per minute.

b. 85

85 77.53.23

11.6

25

xz =

−= = and 70

70 77.53.23

11.6

25

xz =


0.9994 0.0006 0.9988− = between them

c. Instead of the mean pulse rate from the patients in a day , the cutoff values should be based on individual patients, so it would be better to use the pulse rates of 50.5 beats per minute and 104.5 beats per minute.



19. a. 211

211 1651.01

45.6xz =

−= = and 140

140 1650.55

45.6xz =



b. 211

211 1656.05

45.6

36

xz =

−= = and 140

140 1653.29

45.6

36

xz =



c. Part (a) because the ejection seats will be occupied by individual women, not groups of women.

20. a. 140

140 182.97.44

40.8

50

xz =


1.0000 when rounded to four decimal places.)

b. 140

174 182.90.82

40.8

14

xz =


0.7928.)

21. a. 72

72 69.51.04

2.4xz =

−= = which has a probability of 0.8508. (Tech: 0.8512.)

b. 72

72 69.514.76

2.4

100

xz =

−= = which has a probability of 0.9999. (Tech: 1.0000 when rounded to four

decimal places.)

c. The probability of Part (a) is more relevant because it shows that 85.08% of male passengers will not need to bend. The result from part (b) gives us useful information about the comfort and safety of individual male passengers.

d. Because men are generally taller than women, a design that accommodates a suitable proportion of men will necessarily accommodate a greater proportion of women.

22. 167.6

167.6 182.92.28

40.8

37

xz =

−= =− which has a probability of 1 0.0113 0.9887− = to the right of it. There is

a 0.9887 probability that the aircraft is overloaded. Because that probability is so high, the pilot should take action, such as removing excess fuel and/or requiring that some passengers disembark and take a later flight.

23. a. Yes. The sampling is without replacement and the sample size of 50 is greater than 5% of the finite

population size of 275. 16 275 50

2.0504584275 150x

σ −= =

−

b. 105

105 95.54.63

2.0504584xz =

−= = and 95.5

95 95.50.24

2.0504584xz =


1 0.4053 0.5947− = . (Tech: 0.5963.)

24. a. Yes. The sampling is without replacement (because each sample of 16 elevator passengers consists of 16 different people) and the sample size of 16 is greater than 5% of the finite population size of 300.

40 300 169.7459365

300 116xσ −

= =−

b. 3000

187.516

= lb.



24. (continued)

c. 187.5

187.5 1771.08

9.7459365xz =

−= = which has a probability of 1 0.8599 0.1401− = . The probability is not as

low as it should be, since it would be overloaded 14% of the time. (Tech: 0.1407.)

d. The z score for 0.999 is 3.1 which means that we need to solve for n in the equation

40 163.1

16 1

n

n

−=

−which has a solution of 14n = passengers.

25. a. 4 5 9

63

μ + += = ,

2 2 2(4 6) (5 6) (9 6)2.160246899

3σ − + − + −= =

b. The possible samples of size 2 are:{ (4, 5), (4, 9), (5, 4), (5, 9), (9, 4), (9, 5)} which have the following means {4.5, 6.5, 4.5, 7, 6.5, 7} respectively.

c. 4.5 6.5 4.5 7 6.5 7

66x

μ + + + + += = and

2 2 2 2 2 2(4.5 6) (6.5 6) (4.5 6) (7 6) (6.5 6) (7 6)1.08012345

6xσ − + − + − + − + − + −

= =

d. It is clear that 6x

μ μ= = . 2.160246899 3 2

1.080123453 12x

σ σ−= = =

−

Section 6-6

1. The histogram should be approximately bell-shaped, and the normal quantile plot should have points that approximate a straight line pattern.

2. Either the points are not reasonably close to a straight line pattern, or there is some systematic pattern that is not a straight line pattern.

3. We must verify that the sample is from a population having a normal distribution. We can check for normality using a histogram, identifying the number of outliers, and constructing a normal quantile plot.

4. Because the histogram is roughly bell-shaped, conclude that the data are from a population having a normal distribution.

5. Not normal. The points show a systematic pattern that is not a straight line pattern.

6. Normal. The points are reasonably close to a straight line pattern, and there is no other pattern that is not a straight line pattern.

7. Normal. The points are reasonably close to a straight line pattern, and there is no other pattern that is not a straight line pattern.

8. Not normal. The points are not reasonably close to a straight line pattern, and there appears to be a pattern that is not a straight line pattern.



9. Not normal

1209060300-30-60

14

12

10

8

6

4

2

0

Arrival DelayFr

equ

ency

10. Normal

192184176168160

12

10

8

6

4

2

0

Height

Freq

uen

cy

11. Normal

150140130120110100

7

6

5

4

3

2

1

0

Systolic

Freq

uen

cy

12. Not normal

320240160800

40

30

20

10

0

No Exposure

Freq

uen

cy



13. Not normal

14. Normal

15. Normal

16. Not normal



17. Normal. The points have coordinates (–131, –1.28), (134, –0.52), (139, 0), (143, 0.52), (145, 1.28)

18. Not normal. The points have coordinates (13, –1.38), (14, –0.67), (15, –0.21), (15, 0.21), (31, 0.67), (37, 1.38)

19. Not normal. The points have coordinates (1034, –1.53), (1051, –0.89), (1067, –0.49), (1070, –0.16), (1079, 0.16), (1079, 0.49), (1173, 0.89), (1272, 1.53)



20. Normal. The points have coordinates (0.825, –1.59), (0.825, –0.97), (0.855, –0.59), (0.864, –0.28), (0.869, 0), (0.886, 0.28), (0.887, 0.59), (0.912, 0.97), (0.942, 1.59)

21. a. Yes.

b. Yes.

c. No.

22. a. The magnitudes are from a normally distributed

b. The original measurements have a lognormal distribution

c. We can reverse the process of taking values be an exponent of 10. The normal quantile plot indicates that the original values are not from a population with a normal distribution.

23. The original values are not from a normally distributed population.

800006000040000200000-20000-40000

0.99

0.95

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.05

0.01

Net Worth

Pro

babi

lity



23. (continued)

After taking the logarithm of each value, the values appear to be from a normally distributed population.

54321

0.99

0.95

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.05

0.01

Log(Net Worth)

Pro

babi

lity

The original values are from a population with a lognormal distribution.

Section 6-7

1. The Minitab display shows that the region representing 235 wins is a rectangle. The result of 0.0068 is an approximation, but the result of 0.0066 is better because it is based on an exact calculation. The approximation differs from the exact result by a very small amount.

2. The continuity correction is used to compensate for the fact that a continuous distribution (normal) is used to approximate a discrete distribution (binomial). The discrete number of 13 is represented by the interval from 12.5 to 13.5.

3. 1

0.25

p = = , 4

0.85

q = = , 25 0.2 5μ= ⋅ = , 25 0.2 0.8 2σ = ⋅ ⋅ = . The value of 5 for the mean shows

that for many people who make random guesses for the 25 questions, the mean number of correct answers is 5. For many people who make random guesses, the standard deviation of 2 is a measure of how much the numbers of correct responses vary.

4. Yes. The circumstances correspond to 25 independent trials of a binomial experiment in which the probability of success is 0.2. Also, with 25n = , 0.2p = , 0.8q = , the requirements of 5np ≥ and

5nq ≥ are both satisfied.

5. The requirements are satisfied with a mean of 13 0.4 5.2⋅ = and the standard deviation of

13 0.4 0.6 1.766⋅ ⋅ = . Therefore, 2.5

2.5 5.21.53

13 0.4 0.6xz =

−= =−

⋅ ⋅ which has a probability of 0.0630. (Tech:

0.0632.)

6. The requirement of 5nq ≥ is not satisfied. Normal approximation should not be used.

7. The requirement of 5nq ≥ is not satisfied. Normal approximation should not be used.

8. The requirements are satisfied with a mean of 10 and a standard deviation of 25 0.4 0.6 2.449⋅ ⋅ = .

Therefore, 9.5

9.5 100.20

25 0.4 0.6xz =

−= =−

⋅ ⋅ which has a probability of 1 0.4207 0.5793− = to the right of it.

(Tech: 0.5809.)

9. 100 0.22 22, 100 0.22 0.78 4.1425μ σ= ⋅ = = ⋅ ⋅ =

19.5

19.5 220.60

100 0.22 0.78xz =

−= =−

⋅ ⋅ which has a probability of 0.2743. (Tech: 0.2731.)



10. 24.5

24.5 220.60

100 0.22 0.78xz =

−= =

⋅ ⋅ which has a probability of 1 0.7257 0.2743− = to the right of it. (Tech:

0.2731.)

11. 22.5

22.5 220.12

100 0.22 0.78xz =

−= =

⋅ ⋅ and 23.5

23.5 220.36

100 0.22 0.78xz =

−= =

⋅ ⋅ which have a probability of


12. 18.5

18.5 220.84

100 0.22 0.78xz =

−= =−

⋅ ⋅ and 19.5

19.5 220.60

100 0.22 0.78xz =

−= =−


0.2743 0.2005 0.0738− = . (Tech: 0.0738.)

13. 611 0.3 183.3, 611 0.3 0.7 11.3274μ σ= ⋅ = = ⋅ ⋅ =

a. 172.5

172.5 183.30.95

611 0.3 0.7xz =

−= =−

⋅ ⋅ and 171.5

171.5 183.31.04

611 0.3 0.7xz =

−= =−


0.1711 0.1492 0.0219− = between them. (Tech using normal approximation: 0.0214; Tech using binomial: 0.0217)

b. 172.5

172.5 183.30.95

11.3274xz =

−= =− which has a probability of 0.1711. The result of 172 overturned calls

is not unusually low. (Tech using normal approximation: 0.1702; Tech using binomial: 0.1703.)

c. The result from part (b) is useful. We want the probability of getting a result that is at least as extreme as the one obtained.

d. If the 30% rate is correct, there is a good chance (17.11%) of getting 172 or fewer calls overturned, so there is not strong evidence against the 30% rate.

14. 611 0.33 201.63, 611 0.33 0.67 11.6229μ σ= ⋅ = = ⋅ ⋅ =

a. 172.5

172.5 201.632.51

611 0.33 0.67xz =

−= =−

⋅ ⋅and 171.5

171.5 201.632.59

611 0.33 0.67xz =

−= =−

⋅ ⋅which have a probability of


b. 172.5

172.5 201.632.51

611 0.33 0.67xz =

−= =−

⋅ ⋅which has a probability of 0.006. The result of 172 overturned

calls is unusually low. (Tech using normal approximation: 0.0061; Tech using binomial: 0.0056.)


d. If the 33% rate is correct, there is a very small chance (0.6%) of getting 172 or fewer calls overturned, so there is not strong evidence against the 33% rate.

15. 580 0.75 435, 580 0.75 0.25 10.4283μ σ= ⋅ = = ⋅ ⋅ =

a. 428.5

428.5 4350.62

580 0.75 0.25xz =

−= =−

⋅ ⋅and 427.5

427.5 4350.72

580 0.75 0.25xz =

−= =−

⋅ ⋅which have a probability

of 0.2676 0.2358 0.0318− = between them . (Tech using normal approximation: 0.0305; Tech using binomial: 0.0301.)

b. 428.5

428.5 4350.62

580 0.75 0.25xz =

−= =−

⋅ ⋅which has a probability of 0.2676. The result of 428 peas with

green pods is not unusually low. (Tech using normal approximation: 0.2665; Tech using binomial: 0.2650.)




15. (continued)

d. No. Assuming that Mendel’s probability of 3/4 is correct, there is a good chance (26.76%) of getting the results that were obtained. The obtained results do not provide strong evidence against the claim that the probability of a pea having a green pod is 3/4

16. 1004 0.25 251, 1004 0.25 0.75 13.7204μ σ= ⋅ = = ⋅ ⋅ =

a. 290.5

290.5 2512.88

1004 0.25 0.75xz =

−= =

⋅ ⋅which has a probability of 1.0000 0.9980 0.0020− = to the right of

it. (Tech using normal approximation: 0.0020; Tech using binomial: 0.0023.)

b. Because the probability of getting 291 or more with the value of 25% is so small, the result of 291 is unusually high.

c. The results do suggest that the rate is greater than 25%.

17. 945 0.5 472.5, 945 0.5 0.5 15.3704μ σ= ⋅ = = ⋅ ⋅ =

a. 879.5

879.5 472.526.48

945 0.5 0.5xz =

−= =

⋅ ⋅ and 878.5

878.5 472.526.41

945 0.5 0.5xz =

−= =


0.0000 or 0+ (a very small positive probability that is extremely close to 0) between them.

b. 878.5

878.5 472.526.41

945 0.5 0.5xz =

−= =

⋅ ⋅which has a probability of 0.0001. (Tech: 0.0000 or 0+, which is a

very small positive probability that is extremely close to 0). If boys and girls are equally likely, 879 girls in 945 births is unusually high.

c. The result from part (b) is more relevant, because we want the probability of a result that is at least as extreme as the one obtained.

d. Yes. It is very highly unlikely that we would get 879 girls in 945 births by chance. Given that the 945 couples were treated with the XSORT method, it appears that this method is effective in increasing the likelihood that a baby will be a girl.

18. 523 0.85 444.55, 523 0.85 0.15 8.1659μ σ= ⋅ = = ⋅ ⋅ =

517.5

517.5 444.558.93

523 0.85 0.15xz =

−= =

⋅ ⋅ which has a probability of 0.0001 to the right of it. (Tech using normal

approximation: 0.0000 or 0+; Tech using binomial: 0.0000 or 0+.) It appears that many adult males say that they wash their hands in a public restroom when they actually do not.

19. 1002 0.61 611.22, 1002 0.61 0.39 15.4394μ σ= ⋅ = = ⋅ ⋅ =

700.5

700.5 611.225.78

1002 0.61 0.39xz =

−= =

⋅ ⋅which has a probability of 0.0001 to the right of it. (Tech 0.0000.) The

result suggests that the surveyed people did not respond accurately.

20. 420,095 0.00034 142.83, 420,095 0.000344 0.999656 11.9492μ σ= ⋅ = = ⋅ ⋅ =

135.5

135.5 142.830.61

420,095 0.000344 0.999656xz =

−= =−

⋅ ⋅ which has a probability of 0.2709. (Tech using normal

approximation: 0.2697; Tech using binomial: 0.2726.) Media reports appear to be wrong.

21. The probability of six or fewer should be computed. 50 0.2 10, 50 0.2 0.8 2.8284μ σ= ⋅ = = ⋅ ⋅ =

6.5

6.5 101.24

50 0.2 0.8xz =

−= =−

⋅ ⋅which has a probability of 0.1075. (Tech using normal approximation:

0.1080; Tech using binomial: 0.1034.) Because that probability is not very small, the evidence against the rate of 20% is not very strong.



22. The probability of three or fewer should be computed. 50 0.2 10, 50 0.2 0.8 2.8284μ σ= ⋅ = = ⋅ ⋅ =

3.5

3.5 102.30

2.8284xz =

−= =− which has a probability of 0.0107. (Tech using normal approximation: 0.0108;

Tech using binomial: 0.0057.) Because that probability is very small, the evidence against the rate of 20% is very strong. It appears that the rate of smoking among statistics students is lower than the 20% rate for the general population.

23. The probability of 170 or fewer should be computed. 1000 0.20 200, 1000 0.2 0.8 12.6491μ σ= ⋅ = = ⋅ ⋅ =

170.5

170.5 2002.33

1000 0.2 0.8xz =

−= =−

⋅ ⋅which has a probability of 0.0099. (Tech using normal approximation:

0.0098; Tech using binomial: 0.0089.) Because the probability of 170 or fewer is so small with the assumed 20% rate, it appears that the rate is actually less than 20%.

24. The probability of 175 or more should be computed.

250 0.67 167.5, 250 0.67 0.33 7.4347μ σ= ⋅ = = ⋅ ⋅ =

174.5

174.5 167.50.94

250 0.67 0.33xz =

−= =

⋅ ⋅which has a probability of 1.0000 0.8264 0.1736− = to the right of it.

(Tech using normal approximation: 0.1732; Tech using binomial: 0.1734.) If the internet access rate is 67%, there is a relatively high probability of 17.32% of getting 175 or more households with internet access when 250 households are surveyed. It does not appear that the 67% rate is too low.

25. a. In order to make a profit Marc will need to win over $1000. With 35:1 odds a $5 bet wins $175. Therefore, Marc needs 6 winning bets in order to make a profit.

1 1 37200 5.2632, 200 2.2638

38 38 38μ σ= ⋅ = = ⋅ ⋅ =

5.5

5.5 5.26320.10

2.2638xz =

−= = which has a probability of 1.0000 0.5398 0.4602− = to the right of it.

(Tech using normal approximation: 0.4583; tech using binomial: 0.4307)

b. Since the odds of winning are 1:1 Marc would need 101 wins or more to make a

profit.244 244 251

200 98.5859, 200 7.0704495 495 495

μ σ= ⋅ = = ⋅ ⋅ =

100.5

100.5 98.58590.27

7.0704xz =

−= = which has a probability of 1.0000 0.6064 0.3936− =

(Tech using normal approximation: 0.3933; tech using binomial: 0.3932)

c. The roulette game provides a better likelihood of making a profit.

26. The z score that corresponds to a 0.95 probability is 1.645. This means that we have to solve the equation

1.645 0.9005 0.0995 0.9005 213n n⋅ ⋅ ⋅ + ⋅ = for n. This has a solution of 229 reservations. (Tech: 230.)

Chapter Quick Quiz

1. 0μ= and 1σ =



2.

3. 98 2.05P = (Tech: 2.05375)

4. ( ) ( )1 1 1 1 0.1587 0.8413P z P z>− = − <− = − =

5. ( ) ( ) ( )1.37 2.42 2.42 1.37 0.9922 0.9147 0.0775P z P z P z< < = < − < = − = (Tech: 0.0076)

6. 4.2

4.2 4.5770.99

0.382xz =

−= =− which have a probability of 0.1611. (Tech: 0.1618.)

7. 5.4

5.4 4.5772.15

0.382xz =

−= = which has a probability of 1.0000 0.9842 0.0158− = . (Tech: 0.0156.)

8. The z score for 80 0.84P = which corresponds to a red blood count of 0.84 0.382 4.577 4.898⋅ + =

9. 4.444 4.577

1.740.382

25

z−

= =− which has a probability of 0.0409

10. 4.2

4.2 4.5770.99

0.382xz =

−= =− and 5.4

5.4 4.5772.15

0.382xz =

−= = which have a probability of

0.9842 0.1611 0.8231− = or 82.31%. (Tech: 82.26%)

Review Exercises

1. a. The probability to the left of a z score of 2.93 is 0.9983

b. The probability to the right of a z score of –1.53 is 1.0000 0.0630 0.9370− =

c. The probability between z scores –1.07 and 2.07 is 0.9808 0.1423 0.8385− =

d. The z score for 30 0.52P =−

e. 0.27

0.27 01.08

1

16

xz =

−= = which has a probability of 1 0.8599 0.1401− =

2. a. 1605

1605 15161.41

63xz =

−= = which has a probability of 1 0.9207 0.0793− = or 7.93%. (Tech: 7.89%.)

b. The z score for the lowest 1% is –2.33 which corresponds to a standing eye height of 2.33 63 1516 1369.2− ⋅ + = mm. (Tech: 1369.4.)

3. a. 1500

1500 16342.03

66xz =

−= =− which has a probability of 1 0.0212 0.9788− = or 97.88%

b. The z score for the lowest 95% is 1.645 which corresponds to a standing eye height of 1.645 66 1634 1742.6⋅ + = mm



4. a. Normal distribution

b. 21.1xμ =

c. 5.1

0.5780x

σ = =

5. a. An unbiased estimator is a statistic that targets the value of the population parameter in the sense that the sampling distribution of the statistic has a mean that is equal to the mean of the corresponding parameter.

b. Mean, variance and proportion

c. True

6. a. 72

72 69.51.04

2.4xz =

−= = which has a probability of 0.8508 or 85.08%. (Tech: 85.12.) With about

15% of all men needing to bend, the design does not appear to be adequate, but the Mark VI monorail appears to be working quite well in practice.

b. The z score for 99% is 2.33 which corresponds to a doorway height of 2.33 2.4 69.5 75.1⋅ + =

7. a. 175

175 182.90.19

40.9xz =

−= =− which has a probability of 1 0.4247 0.5753− = . (Tech: 0.5766.)

b. 175

175 182.92.82

40.9

213

xz =

−= =− which has a probability of 1 0.0024 0.9976− = . Yes, if the plane is full

of male passengers, it is highly likely that it is overweight.

8. a. No. A histogram is far from bell shaped.

32000240001600080000

12

10

8

6

4

2

0

Salary

Freq

uen

cy

b. No. The sample has a size of 26 which does not satisfy the condition at least 30, and the values do not appear to be from a population having a normal distribution.

9. 3 3 1

1064 798, 1064 14.12444 4 4

μ σ= ⋅ = = ⋅ ⋅ =

787.5

787.5 7980.74

3 11064

4 4

zz =

−= =−

⋅ ⋅ which has a probability of 0.2296. (Tech using normal approximation:

0.2286; Tech using binomial: 0.2278.) The occurrence of 787 offspring plants with long stem is not unusually low because its probability is not small. The results are consistent with Mendel’s claimed proportion of 3/4



10. 64 0.8 51.2, 64 0.8 0.2 3.2μ σ= ⋅ = = ⋅ ⋅ =

a. 49.5

49.5 51.20.53

3.2xz =

−= =− which has a probability of 1 0.2981 0.7019− = to the right of it. (Tech

using normal approximation: 0.7024; Tech using binomial: 0.7100)

b. 49.5 0.53xz = =− and 50.5

50.5 51.20.22

3.2xz =




1. a. 14,500,000 145,000,000 14,000,000 5,000,000 3,500,000

$10,300,0005

x+ + + +

= =

b. The median is $14,000,000

c. 2 2 2 2(14,500 10,300) (14,500 10,300) ... (5000 10,300) (3500 10,300)

5 1$5552.027 (in thousands of dollars) which is $5,552,027.

s− + − + + − + −

=−

=

d. 2 30,825,003,810,000s = square dollars

e. 14,500,000

14,500,000 10,300,0000.76

5,552,027xz =

−= =

f. Ratio

g. Discrete

h. No, the starting players are likely to be the best players who receive the highest salaries.

2. a. A is the event of selecting someone who does not have the belief that college is not a good investment. NOTE: This is not the same as selecting someone who believes that college is a good investment.

b. ( ) 1 0.1 0.9P A = − =

c. 0.1 0.1 0.1 0.001P = ⋅ ⋅ =

d. The sample is a voluntary response sample. This suggests that the 10% rate might not be very accurate, because people with strong feelings or interest about the topic are more likely to respond.

3. a. 2500

2500 33691.53

567xz =

−= =− which has a probability of 0.0630. (Tech: 0.0627)

b. The z score for the bottom 10% is –1.28, which correspond to the weight 1.28 567 3369 2642.24− ⋅ + = g. (Tech: 2642 g.)

c. 1500

1500 33693.3

567xz =

−= =− which has a probability of 0.0005

d. 3400

3400 33690.27

567

25

xz =

−= = which has a probability of 1 0.6064 0.3936− = . (Tech: 0.3923) to the

right of it.



4. a. The vertical scale does not start at 0, so differences are somewhat distorted. By using a scale ranging from 1 to 29 for frequencies that range 2 to 14, the graph is flattened, so differences are not shown as they should be.

b. The graph depicts a distribution that is not exactly normal, but it is approximately normal because it is roughly bell shaped.

c. Minimum: 42 years; maximum: 70 years. Using the range rule of thumb, the standard deviation is

estimated to be 70 42

74

−= years. The estimate of 7 years is very close to the actual standard

deviation of 6.6 years, so the range rule of thumb works quite well here.

5. a. ( 3) 0.1 0.1 0.1 0.001P X = = ⋅ ⋅ =

b. ( 1) 1 ( 0) 1 (0.9 0.9 0.9) 0.271P X P X≥ = − = = − ⋅ ⋅ =

c. The requirement that 5np ≥ is not satisfied, indicating that the normal approximation would result in

errors that are too large.

d. 50 0.1 5μ= ⋅ =

e. 50 0.1 0.9 2.1213σ = ⋅ ⋅ =

f. No, 8 is within two standard deviations of the mean and is within the range of values that could easily occur by chance.

Chapter 7: Estimates and Sample Sizes 101


Chapter 7: Estimates and Sample Sizes Section 7-2

1. The confidence level (such as 95%) was not provided.

2. When using 26% to estimate the value of the population percentage, the maximum likely difference between 26% and the true population percentage is three percentage points, so the interval from 23% to 29% is likely to contain the true population percentage.

3. p = 0.26 is the sample proportion; q = 0.74 (found from evaluating 1 – p ); n = 1910 is the sample size;

0.03E = is the margin of error; p is the population proportion, which is unknown. The value of α is 0.05.

4. The 95% confidence interval will be wider than the 80% confidence interval. A confidence interval must be wider in order for us to be more confident that it captures the true value of the population proportion. (Think of estimating the age of a classmate. You might be 90% confident that she is between 20 and 30, but you might be 99.9% confident that she is between 10 and 40.)

5. 1.28

6. 2.575 (Tech: 2.576)

7. 1.645

8. 2.05

9. 0.186 0.0641

0.061,2

E−

= = so 0.125 ± 0.061

10. 0.335 0.165

0.085,2

E−

= = so 0.250 ± 0.085

11. 0.0268 < p < 0.133 12. 0.183 < p < 0.357

13. a. 531

ˆ 0.5301002

p = =

b. ( )( )531 471

1002 1002/ 2

ˆ ˆ1.96 0.0309

1002αpq

E zn

= = =

c. ˆ ˆ 0.530 0.0309 0.530 0.0309 0.499 0.561p E p p E p p− < < − ⇒ − < < − ⇒ < <

d. We have 95% confidence that the interval from 0.499 to 0.561 actually does contain the true value of the population proportion.

14. a. 490

ˆ 0.610806

p = =

b. ( )( )490 316

806 806/ 2

ˆ ˆ2.58 0.0443

806αpq

E zn

= = =

c.

ˆ ˆ

0.610 0.0443 0.610 0.0443

0.566 0.654

p E p p E

p

p

− < < −

− < < −

< <


102 Chapter 7: Estimates and Sample Sizes


15. a. 1083

ˆ 0.4302518

p = =

b. ( )( )1083 1435

2518 2518/ 2

ˆ ˆ1.65 0.0162

2518αpq

E zn

= = =

c.

ˆ ˆ

0.430 0.0162 0.430 0.0162

0.414 0.446

p E p p E

p

p

− < < −− < < −

< <


16. a. 543

ˆ 0.5401005

p = =

b. ( )( )543 462

1005 1005/ 2

ˆ ˆ1.28 0.0201

1005αpq

E zn

= = =

c.

ˆ ˆ

0.540 0.0201 0.540 0.0201

0.520 0.560

p E p p E

p

p

− < < −− < < −

< <


17. a. 879

ˆ 0.930945

p = =

b. ( )( )879 66

945 945/ 2

ˆ ˆ 879ˆ 1.96

945 9450.914 0.946

αpq

p zn

p

± = ±

< <

c. Yes. The true proportion of girls with the XSORT method is substantially greater than the proportion of (about) 0.5 that is expected when no method of gender selection is used.

18. a. 239

ˆ 0.821291

p = =

b. ( )( )239 52

291 291/ 2

ˆ ˆ 239ˆ 2.56

291 9450.763 0.879

αpq

p zn

p

± = ±

< <

c. Yes. The true proportion of boys with the YSORT method is substantially greater than the proportion of (about) 0.5 that is expected when no method of gender selection is used.

19. a. 0.5

b. 123

ˆ 0.439280

p = =

c. ( )( )123 157

280 280/ 2

ˆ ˆ 123ˆ 2.56

280 2800.363 0.515 or 36.3% 51.5%

αpq

p zn

p p

± = ±

< < < <

d. If the touch therapists really had an ability to select the correct hand by sensing an energy field, their success rate would be significantly greater than 0.5, but the sample success rate of 0.439 and the confidence interval suggest that they do not have the ability to select the correct hand by sensing an energy field.



20. a. ( )( )152 428

580 580/ 2 0.025

ˆ ˆ 152ˆ

580 5800.226 0.298 or 22.6% 29.8%

αpq

p z zn

p p

± = ±

< < < <

b. No, the confidence interval includes 0.25, so the true percentage could easily equal 25%.

21. a. ( )427 0.29 124=

b. ( )( )

/ 2 0.025

0.29 0.71ˆ ˆˆ 0.29

4270.247 0.333 or 24.7% 33.3%

αpq

p z zn

p p

± = ±

< < < <

c. Yes. Because all values of the confidence interval are less than 0.5, the confidence interval shows that the percentage of women who purchase books online is very likely less than 50%.

d. No. The confidence interval shows that it is possible that the percentage of women who purchase books online could be less than 25%.

e. Nothing.

22. If the subjects chose to respond to the posted question, the sample is a voluntary response sample, so the confidence interval could be very misleading.

( )( )

/ 2 0.025

0.208 0.792ˆ ˆˆ 0.208

1440.142 0.274 or 14.2% 27.4%

αpq

p z znp p

± = ±

< < < <

(using x = 30: 14.2% < p < 27.5%).

23. a. ( )514 0.459 236=

b. ( )( )

/ 2 0.025

0.459 0.541ˆ ˆˆ 0.459

5140.402 0.516

αpq

p z zn

p

± = ±

< <

(using x = 236: 0.403 < p < 0.516).

c. ( )( )

/ 2 0.10

0.459 0.541ˆ ˆˆ 0.459

5140.431 0.487

αpq

p z zn

p

± = ±

< <

d. The 95% confidence interval is wider than the 80% confidence interval. A confidence interval must be wider in order to be more confident that it captures the true value of the population proportion. (See Exercise 4.)

24. a. ( )514 0.90 463=

b. ( )( )

/ 2 0.005

0.90 0.10ˆ ˆˆ 0.90

5140.866 0.934

αpq

p z zn

p

± = ±

< <

(using x = 463: 0.867 < p < 0.935).

c. ( )( )

/ 2 0.10

0.90 0.10ˆ ˆˆ 0.90

5140.883 0.917

αpq

p z zn

p

± = ±

< <

(using x = 463: 0.884 < p < 0.918).

d. The 95% confidence interval is wider than the 80% confidence interval. A confidence interval must be wider in order to be more confident that it captures the true value of the population proportion. (See Exercise 4.)



25. No, the confidence interval limits contain the value of 0.13, so the claimed rate of 13% could be the true percentage for the population of brown M&Ms.

( )( )

/ 2 0.01

0.08 0.92ˆ ˆˆ 0.08

1000.0168 0.143

αpq

p z zn

p

± = ±

< <

. (Tech: 0.0169 0.143)p< <

26. a. ( )( )

/ 2 0.01

0.70 0.30ˆ ˆˆ 0.70

10020.666 0.734

αpq

p z zn

p

± = ±

< <

(Tech: 0.666 < p < 0.733)

b. No. Because 0.61 is not included in the confidence interval, it does not appear that the responses are consistent with the actual voter turnout.

27. a. ( )( )

/ 2 0.05

0.000321 0.999679ˆ ˆˆ 0.000321

420,095

0.0276% 0.0366%

αpq

p z zn

p

± = ±

< <

(using x = 135: 0.0276% < p < 0.0367%).

b. No, because 0.0340% is included in the confidence interval.

28. a. ( )3005 0.817 2455=

b. ( )( )

/ 2 0.005

0.817 0.183ˆ ˆˆ 0.817

30050.805 0.829 or 80.5% 82.9%

αpq

p z znp p

± = ±

< < < <

c. Nothing.

29. [ ] [ ] ( )2 2

/ 2

2 2

ˆ ˆ 1.645 0.25752

0.03αz pq

nE

= = =

30. [ ] [ ] ( )2 2

/ 2

2 2

ˆ ˆ 1.28 0.25256

0.04αz pq

nE

= = = (Tech: 257)

31. [ ] [ ] ( )( )2 2

/ 2

2 2

ˆ ˆ 2.575 0.15 0.85339

0.05αz pq

nE

= = =

32. [ ] [ ] ( )( )2 2

/ 2

2 2

ˆ ˆ 2.33 0.15 0.85770

0.03αz pq

nE

= = = (Tech: 767)

33. a. [ ] [ ] ( )2 2

/ 2

2 2

ˆ ˆ 1.96 0.251537

0.025αz pq

nE

= = =

b. [ ] [ ] ( )( )2 2

/ 2

2 2

ˆ ˆ 1.96 0.38 0.621449

0.025αz pq

nE

= = =

34. a. [ ] [ ] ( )2 2

/ 2

2 2

ˆ ˆ 2.575 0.2516,577

0.01αz pq

nE

= = = (Tech: 16,588)

b. [ ] [ ] ( )( )2 2

/ 2

2 2

ˆ ˆ 2.575 0.90 0.105968

0.01αz pq

nE

= = = (Tech: 5972)

c. Yes. Using the additional survey information from part (b) dramatically reduces the sample size.



35. a. [ ] [ ] ( )2 2

/ 2

2 2

ˆ ˆ 1.645 0.25271

0.05αz pq

nE

= = =

b. [ ] [ ] ( )( )2 2

/ 2

2 2

ˆ ˆ 1.645 0.85 0.15139

0.05αz pq

nE

= = = (Tech: 138)

c. No. A sample of students at the nearest college is a convenience sample, not a simple random sample, so it is very possible that the results would not be representative of the population of adults.

36. a. [ ] [ ] ( )2 2

/ 2

2 2

ˆ ˆ 1.28 0.25456

0.03αz pq

nE

= = = (Tech: 457)

b. [ ] [ ] ( )( )2 2

/ 2

2 2

ˆ ˆ 1.28 0.84 0.16245

0.03αz pq

nE

= = = (Tech: 246)

c. No. Flights between New York and San Francisco might not be representative of the population of all Southwest flights.

37. Greater height does not appear to be an advantage for presidential candidates. If greater height is an advantage, then taller candidates should win substantially more than 50% of the elections, but the confidence interval shows that the percentage of elections won by taller candidates is likely to be anywhere between 36.2% and 69.7%.

18

ˆ 0.52934

p = = . ( )( )18 16

34 34/ 2

ˆ ˆ 18ˆ 1.96

34 340.362 0.697 or 36.2% 69.7%.

αpq

p zn

p p

± = ±

< < < <

38. No, the confidence interval is based on sample data consisting of flights from New York (JFK) to Los Angeles, and arrival delays for that route might be very different from arrival delays for the population that includes all routes.

44

ˆ 0.91748

p = = . ( )( )44 4

48 48/ 2

ˆ ˆ 44ˆ 1.645

48 480.851 0.980 or 85.1% 98.0%.

αpq

p zn

p p

± = ±

< < < <

39. a. [ ]

[ ] ( )( )( )[ ]

( )( )[ ] ( )

2 2

/ 2

2 22 2/ 2

ˆ ˆ 200 0.5 0.5 1.96178

ˆ ˆ 1 0.5 0.5 1.96 200 1 0.025

α

α

Npq zn

pq z N E= = =

+ − + −

b. [ ]

[ ] ( )( )( )[ ]

( )( )[ ] ( )

2 2

/ 2

2 22 2/ 2

ˆ ˆ 200 0.38 0.62 1.96176

ˆ ˆ 1 0.38 0.62 1.96 200 1 0.025

α

α

Npq zn

pq z N E= = =

+ − + −

40. ( )( )3 5

8 8/ 2

ˆ ˆ 3ˆ 1.96

8 80.0395 0.710;

αpq

p zn

p

± = ±

< <

no

41. The upper confidence interval limit is greater than 100%. Given that the percentage cannot exceed 100%, change the upper limit to 100%.

( )( )44 4

48 48/ 2

ˆ ˆ 44ˆ 2.575

48 480.814 1.019 or 81.4% 101.9%.

αpq

p zn

p p

± = ±

< < < <



42. a. The requirement of at least 5 successes and at least 5 failures is not satisfied, so the normal distribution cannot be used.

b. 3

0.07540

=

43. Because we have 95% confidence that p is greater than 0.831, we can safely conclude that more than 75% of adults know what Twitter is.

( )( )0.85 0.15ˆ ˆ 44

ˆ 1.64548 1007

0.831

αpq

p zn

p

+ = +

>

(Tech: p > 0.832).

Section 7-3

1. a. sec 233.4 sec 256.65 secμ< <

b. The best point estimate of μ is 256.65 233.4

245.025 sec2

x+

= = . The margin of error is

256.65 233.4 11.625 sec.

2E

−= =

2. a. df = 39

b. 2.023

c. In general, the number of degrees of freedom for a collection of sample data is the number of sample values that can vary after certain restrictions have been imposed on all data values.

3. We have 95% confidence that the limits of 233.4 sec and 256.65 sec contain the true value of the mean of the population of all duration times.

4. When we say that the confidence interval methods of this section are robust against departures from normality, we mean that these methods work reasonably well with distributions that are not normal, provided that departures from normality are not too extreme. The given dotplot does appear to satisfy the loose normality requirement. Also, there are 40 dots, so the sample size of 40 satisfies the condition of n > 30.

5. Neither the normal nor the Student t distribution applies.

6. / 2αt = 1.729

7. / 2αt = 2.708

8. / 2αz = 2.575 (Tech: 2.576)

9. Because the sample size is greater than 30, the confidence interval yields a reasonable estimate of μ , even

though the data appear to be from a population that is not normally distributed.

/ 2

5.0139.808 2.403

508.104 km 11.512 km

(Tech: 8.103 km 513 km)

αs

x tn

μμ

± = ± ⋅

< << <

10. / 2

0.3660.719 1.943

70.450 ppm 0.988 ppm

αs

x tn

μ

± = ± ⋅

< < (If the original values are used, the upper limit is 0.987 ppm.)



11. The $1 salary of Jobs is an outlier that is very far away from the other values, and that outlier has a dramatic effect on the confidence interval.

/ 2

7719.0512898 2.776

53315.1 thousand dollars 22480.9 thousand dollars

(Tech: 3313.5 thousand dollars 6 22,482.5 thousand dollars)

αs

x tn

μμ

± = ± ⋅

< << <

12. The confidence interval is an estimate of the population mean and it does not apply to individual sample values.

/ 2

2.5523.95 2.71

4022.86 chocolate chips 25.04 chocolate chips

αs

x tn

μ

± = ± ⋅

< <

13. Because the confidence interval does not contain 98.6°F, it appears that the mean body temperature is not 98.6°F, as is commonly believed.

/ 2

0.6298.2 1.98

106

98.08 F 98.32 F

αs

x tn

μ

± = ± ⋅

< <

14. Because the confidence interval does not include 0 or negative values, it does appear that the weight loss program is effective with a positive loss of weight. Because the amount of weight lost is relatively small, the weight loss program does not appear to be very practical.

/ 2

4.82.1 1.68

400.8 lb 3.4 lb

αs

x tn

μ

± = ± ⋅

< <

15. Because the confidence interval includes the value of 0, it is very possible that the mean of the changes in LDL cholesterol is equal to 0, suggesting that the garlic treatment did not affect LDL cholesterol levels. It does not appear that garlic is effective in reducing LDL cholesterol.

/ 2

210.4 2.4

496.8 mg/dL 7.6 mg/dL

αs

x tn

μ

± = ± ⋅

− < <

16. The confidence interval includes the mean of 102.8 min that was measured before the treatment, so the mean could be the same after the treatment. This result suggests that the zoplicone treatment has no effect.

/ 2

42.398.9 2.6

1671.4 min 126.4 min

αs

x tn

μ

± = ± ⋅

< <

17. The data appear to have a distribution that is far from normal, so the confidence interval might not be a good estimate of the population mean. The population is likely to be the list of box office receipts for each day of the movie’s release. Because the values are from the first 14 days of release, the sample values are not a simple random sample, and they are likely to be the largest of all such values, so the confidence interval is not a good estimate of the population mean.

/ 2

14.516.4 3.01

144.7 million dollars 28.1 million dollars

αs

x tn

μ

± = ± ⋅

< <



18. The confidence interval does not contain the value of 4 years. The data appear to have a distribution that is far from normal, so the confidence interval might not be a good estimate of the population mean.

/ 2

3.516.5 1.73

205.1 years 7.9 years

αs

x tnμ

± ± ⋅

< <

19. The sample data meet the loose requirement of having a normal distribution. Because the confidence interval is entirely below the standard of 1.6 W/kg, it appears that the mean amount of cell phone radiation is less than the FCC standard, but there could be individual cell phones that exceed the standard.

/ 2

0.4230.938 1.81

110.707 W/kg 1.169 W/kg

αs

x tn

μ

± = ± ⋅

< <

20. The sample data meet the loose requirement of having a normal distribution

/ 2

7.6699833.6 2.62


αs

x tn

μ

± = ± ⋅

< <

21. The sample data meet the loose requirement of having a normal distribution. We cannot conclude that the population mean is less than 7 g/gμ , because the confidence interval shows that the mean might be greater

than that level.

/ 2

6.4611.05 2.26

106.43 g/g 15.67 g/g

αs

x tn

μ μ μ

± = ± ⋅

< <

22. The sample data meet the loose requirement of having a normal distribution. The values are typical because they are between 950 cm3 and 1800 cm3.

/ 2

3 3

117.441130.2 3.25

10

1009.5 cm 1250.9 cm

αs

x tn

μ

± = ± ⋅

< <

23. Although final conclusions about means of populations should not be based on the overlapping of confidence intervals, the confidence intervals do overlap, so it appears that both populations could have the same mean, and there is not clear evidence of discrimination based on age.

CI for ages of unsuccessful applicants CI for ages of successful applicants

/ 2

7.246.96 2.07


αs

x tn

μ

± = ± ⋅

< < / 2

5.0344.5 2.05


αs

x tn

μ

± = ± ⋅

< <

24. Although final conclusions about means of populations should not be based on the overlapping of confidence intervals, the confidence intervals do overlap, so it appears that both populations could have the same mean, and there is not clear evidence that skull breadths changed from 4000 b.c. to 150 a.d.

CI for 4000 b.c. CI for 150 a.d.

/ 2

4.6128.7 2.201

12125.8 mm 131.6 mm

αs

x tn

μ

± = ± ⋅

< <

5.0155133.33 2.201

12130.1 mm 136.5 mmμ

± ⋅

< <



25. The sample size is 2 2

/ 2 1.645 1568

3αz σ

nE

⎡ ⎤ ⎡ ⎤⋅⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

, and it does appear to be very reasonable.

26. The required sample size is 2 2

/ 2 2.58 0.79104

0.2αz σ

nE

⎡ ⎤ ⎡ ⎤⋅⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

. Limiting the sample to students at your

college would result in a convenience sample that might not be representative of the population of all college students, so it does not make sense to collect the entire sample at your college.


/ 2 2.33 2157405

250αz σ

nE

⎡ ⎤ ⎡ ⎤⋅⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

(Tech: 403). It is not likely that you would

find that many two-year-old used Corvettes in your region.


/ 2 1.96 210753

15αz σ

nE

⎡ ⎤ ⎡ ⎤⋅⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

. A major obstacle to getting a good estimate

of the population mean is that it would be very difficult to actually measure times spent on Facebook, so you must rely on reported times that can be very inaccurate.

29. Use 2400 600

4504

σ −= = to get a sample size of

2 2

/ 2 2.33 450110

100αz σ

nE

⎡ ⎤ ⎡ ⎤⋅⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

. The margin of error

of 100 points seems too high to provide a good estimate of the mean SAT score.

30. Use 45,000 0

11,2504

σ −= = to get a sample size of

2 2

/ 2 2.575 1125083,919

100αz σ

nE

⎡ ⎤ ⎡ ⎤⋅⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

(Tech: 83,973). The sample size seems too large to be practical.

31. With the range rule of thumb, use 90 46

114

σ −= = to get a required sample size of

2 2

/ 2 1.96 11117

2αz σ

nE

⎡ ⎤ ⎡ ⎤⋅⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

.

With s = 10.3, the required sample size is 2 2

/ 2 1.96 10.3102

2αz σ

nE

⎡ ⎤ ⎡ ⎤⋅⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

. The better estimate of s is the

standard deviation of the sample, so the correct sample size is likely to be closer to 102 than 117.

32. With the range rule of thumb, use 2.95 0

0.73754

σ −= = to get a required sample size of

2 2

/ 2 1.96 0.737553

0.2αz σ

nE

⎡ ⎤ ⎡ ⎤⋅⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

. With s = 0.587, the required sample size is 34. The better estimate of s

is the standard deviation of the sample, so the sample size of 34 is the better result.

33. / 2

0.58731.1842 2.8

500.963 1.407

(Tech: 0.962 1.407)

αs

x tn

μμ

± = ± ⋅

< << <

34. / 2

119.5172.5 2.023

40134.3 ng/mL 210.7 ng/mL

ασ

x zn

μ

± = ± ⋅

< <

35. / 2

5.0139.808 2.33

508.156 km 11.46 km

(Tech: 8.159 km 11.457 km)

ασ

x zn

μμ

± = ± ⋅

< << <



36. / 2

0.3660.719 1.645

70.491 ppm 0.947 ppm

ασ

x zn

μ

± = ± ⋅

< < (If the original values are used, the upper limit is 0.946 ppm.)

37. / 2

7717.812898 1.96

56133.05 thousand dollars 19662.95 thousand dollars

(Tech: 6131.9 thousand dollars 19,663.3 thousand dollars)

ασ

x zn

μμ

± = ± ⋅

< << <

38. / 2

2.5523.95 2.576

4022.91 chocolate chips 24.99 chocolate chips

ασ

x zn

μ

± = ± ⋅

< <

39. The sample data do not appear to meet the loose requirement of having a normal distribution. The effect of the outlier on the confidence interval is very substantial. Outliers should be discarded if they are known to be errors. If an outlier is a correct value, it might be very helpful to see its effects by constructing the confidence interval with and without the outlier included.

/ 2

7.185111.375 2.26

1024.54 m 106.04 m

(Tech: 24.55 m 106.05 m)

αs

x tn

μμ

± = ± ⋅

− < <− < <

40. The second confidence interval is narrower, indicating that we have a more accurate estimate when the relatively large sample is from a relatively small finite population.

Large population: / 2

0.05180.8565 2.26

1000.8462 g 0.8668 g

αs

x tn

μ

± = ± ⋅

< <

Finite population: / 2

0.0518 465 1000.8565 2.26

1 100 11000.8474 g 0.8656 g

αs N n

x tnn

μ

− −± = ± ⋅

− −< <

41. The confidence interval based on the first sample value is much wider than the confidence interval based on all 10 sample values.

9.68 3.0

26.0 m 32.0 m

x

μ±

− < <

Section 7-4

1. ( ) ( )2 22916.591 mg/dL 2252.1149 mg/dL 30.3 mg/dL 47.5 mg/dLσ σ< < ⇒ < < . We have 95%

confidence that the limits of 30.3 mg/dL and 47.5 mg/dL contain the true value of the standard deviation of the LDL cholesterol levels of all women.

2. The format implies that s = 15.7, but s is given as 14.3. In general, a confidence interval for σ does not have s at the center.

3. The original sample values can be identified, but the dotplot shows that the sample appears to be from a population having a uniform distribution, not a normal distribution as required. Because the normality requirement is not satisfied, the confidence interval estimate of s should not be constructed using the methods of this section.



4. The normality requirement for a confidence interval estimate of σ has a much stricter normality requirement than the loose normality requirement for a confidence interval estimate of μ . Departures from

normality have a much greater effect on confidence interval estimates of σ than on confidence interval estimates of μ .

5. df = 24. 2Lχ = 9.886 and 2

Rχ = 45.559.

( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

25 1 0.24 25 1 0.24

45.559 9.8860.17 mg 0.37 mg

R L

n s n sσ

χ χ

σ

σ

− −< <

− −< <

< <

6. df = 19. 2Lχ = 6.844 and 2

Rχ = 38.582.

( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

20 1 0.04111 20 1 0.04111

38.582 6.8440.02885 g 0.06850 g

R L

n s n sσ

χ χ

σ

σ

− −< <

− −< <

< <

7. df = 39. 2Lχ = 24.433 (Tech: 23.654) and 2

Rχ = 59.342 (Tech: 58.120).

( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

40 1 65.2 40 1 65.2;df 40

59.342 24.43352.9 82.4 (Tech: 53.4 83.7)

R L

n s n sσ

χ χ

σ

σ σ

− −< <

− −< < =

< < < <

8. df = 49. 2Lχ = 32.357 (Tech: 31.555) and 2

Rχ = 71.420 (Tech: 70.222).

( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

50 1 0.587 50 1 0.587;df 50

71.420 32.3570.486 0.722 (Tech: 0.490 0.731)

R L

n s n sσ

χ χ

σ

σ σ

− −< <

− −< < =

< < < <

9. ( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

106 1 0.62 106 1 0.62;df 100

124.342 77.929

0.579 F 0.720 F (Tech: 0.557 F 6 s 6 0.700 F)

R L

n s n sσ

χ χ

σ

σ

− −< <

− −< < =

< <

10. ( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

40 1 2.55 40 1 2.55;df 40

55.758 26.5092.13 chocolate chips 3.09 chocolate chips

(Tech: 2.16 chocolate chips 3.14 chocolate chips)

R L

n s n sσ

χ χ

σ

σσ

− −< <

− −< < =

< << <



11. The confidence interval shows that the standard deviation is not likely to be less than 30 mL, so the variation is too high instead of being at an acceptable level below 30 mL. (Such one-sided claims should be tested using the formal methods presented in Chapter 8.)

( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

24 1 42.8 24 1 42.8

44.181 9.26030.9 mL 67.45 mL

R L

n s n sσ

χ χ

σ

σ

− −< <

− −< <

< <

12. a. ( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

40 1 10.3 40 1 10.3;df 40

53.672 13.7877.9 beats per minute 14.1 beats per minute

(Tech: 7.9 beats per minute 14.4 beats per minute)

R L

n s n sσ

χ χ

σ

σσ

− −< <

− −< < =

< << <

b. ( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

40 1 11.6 40 1 11.6

53.672 13.7878.9 beats per minute 15.9 beats per minute

(Tech: 9.0 beats per minute 16.2 beats per minute)

R L

n s n sσ

χ χ

σ

σσ

− −< <

− −< <

< << <

c. The confidence intervals are not dramatically different, so it appears that the populations of pulse rates of men and women have about the same standard deviation.

13. ( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

7 1 0.36576 7 1 0.36576

12.592 1.6350.252 ppm 0.701 ppm

R L

n s n sσ

χ χ

σ

σ

− −< <

− −< <

< <

0

14. Because traffic conditions vary considerably at different times during the day, the confidence interval is an estimate of the standard deviation of the population of speeds at 3:30 on a weekday, not other times.

( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

12 1 4.075 12 1 4.075

19.675 4.5752.9 mi/h 6.9 mi/h

R L

n s n sσ

χ χ

σ

σ

− −< <

− −< <

< <



15. CI for ages of unsuccessful applicants:

( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

25 1 7.22 25 1 7.22

45.559 9.8865.2 years 11.5 years

R L

n s n sσ

χ χ

σ

σ

− −< <

− −< <

< <

CI for ages of successful applicants:

( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

29 1 5.026 29 1 5.026

50.993 12.4613.7 years 7.5 years

R L

n s n sσ

χ χ

σ

σ

− −< <

− −< <

< <

Although final conclusions about means of populations should not be based on the overlapping of confidence intervals, the confidence intervals do overlap, so it appears that the two populations have standard deviations that are not dramatically different.

16. a. ( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

10 1 0.4767 10 1 0.4767

19.023 2.7000.33 min 0.87 min

R L

n s n sσ

χ χ

σ

σ

− −< <

− −< <

< <

b. ( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

10 1 1.8216 10 1 1.8216

19.023 2.7001.25 min 3.33 min

R L

n s n sσ

χ χ

σ

σ

− −< <

− −< <

< <

c. The variation appears to be significantly lower with a single line. The single line appears to be better.

17. ( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

37 1 0.0165 37 1 0.0165; df=40

63.691 22.1640.01239 g 0.02111 g

(Tech: 0.01291 g 0.02255 g)

R L

n s n sσ

χ χ

σ

σσ

− −< <

− −< <

< << <

18. ( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

37 1 6.5613 37 1 6.5613; df=40

63.691 22.1645.2 years 8.9 years

R L

n s n sσ

χ χ

σ

σ

− −< <

− −< <

< <



19. 33,218 is too large. There aren’t 33,218 statistics professors in the population, and even if there were, that sample size is too large to be practical.

20. The sample size of 48 is very practical, although the sample should be selected from the population of all McDonald’s restaurants with drive-up windows.

21. The sample size is 768. Because the population does not have a normal distribution, the computed minimum sample size is not likely to be correct.

22. The sample size is 1336. The population of incomes does not have a normal distribution, so the computed sample size is not likely to be correct.

23. 2 2

2/ 2

1 12 1 1.645 2 105 1 82.072

2 2L αχ z k⎡ ⎤ ⎡ ⎤= − + − = − + ⋅ − =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ and

2 22

/ 2

1 12 1 1.645 2 105 1 129.635

2 2R αχ z k⎡ ⎤ ⎡ ⎤= + − = + ⋅ − =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

(Tech using / 2 1.644853626αz = : 2Lχ = 82.073 and 2

Rχ = 129.632). The approximate values are quite close

to the actual critical values.

Chapter Quick Quiz

1. 40% 3.1% 40% 3.1%

36.9% 43.1%

p

p

− < < +< <

2. 0.511 0.449

ˆ 0.4802

p+

= =

3. We have 95% confidence that the limits of 0.449 and 0.511 contain the true value of the proportion of females in the population of medical school students.

4. z = 1.645

5. [ ] [ ] ( )2 2

/ 2

2 2

ˆ ˆ 1.645 0.25752

0.03αz pq

nE

= = =

6. 2 2

/ 2 2.575 15373

2αz σ

nE

⎡ ⎤ ⎡ ⎤⋅⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

(Tech: 374)

7. The sample must be a simple random sample and there is a loose requirement that the sample values appear to be from a normally distributed population.

8. The degrees of freedom is the number of sample values that can vary after restrictions have been imposed on all of the values. For the sample data in Exercise 7, df = 5.

9. t = 2.571

10. 2Lχ = 0.831 and 2

Rχ = 12.833

Review Exercises

1. a. 284

ˆ 0.510 51.0%557

p = = =

b. ( )( )284 273

557 557/ 2

ˆ ˆ 284ˆ 1.96

557 55746.8% 55.1%

αpq

p zn

p

± = ±

< <

c. No, the confidence interval shows that the population percentage might be 50% or less, so we cannot safely conclude that the majority of adults say that they are underpaid.



2. [ ] [ ] ( )2 2

/ 2

2 2

ˆ ˆ 2.575 0.254145

0.02αz pq

nE

= = = (Tech: 4147)

3. 2 2

/ 2 2.33 16155

3αz σ

nE

⎡ ⎤ ⎡ ⎤⋅⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

(Tech: 154)

4. a. Student t distribution

b. Normal distribution

c. The distribution is not normal, Student t, or chi-square.

d. 2χ (chi-square distribution)

e. Normal distribution

5. a. [ ] [ ] ( )2 2

/ 2

2 2

ˆ ˆ 2.33 0.25543

0.05αz pq

nE

= = = (Tech: 542)

b. 2 2

/ 2 2.33 337247

50αz σ

nE

⎡ ⎤ ⎡ ⎤⋅⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

(Tech: 246)

c. 543

6. Because the entire confidence interval is above 50%, we can safely conclude that the majority of adults consume alcoholic beverages.

( )( )/ 2

0.64 0.36ˆ ˆˆ 0.64 1.65

101161.5% 66.5%

αpq

p zn

p

± = ±

< <

7. / 2

259.7754143 2.201

1222.1 sec 308.1 sec

αs

x tn

μ

± = ± ⋅

− < <

8. Because women and men have some notable physiological differences, the confidence interval does not necessarily serve as an estimate of the mean white blood cell count of men.

/ 2

2.287.15 1.685

406.54 7.76

αs

x tn

μ

± = ± ⋅

< <

9. There is 95% confidence that the limits of 37.5 g and 47.9 g contain the true mean deceleration measurement for all small cars.

/ 2

5.642.7 2.447

737.5 g 47.9 g

αs

x tn

μ

± = ± ⋅

< <

10. ( ) ( )

( ) ( )

2 2

2 2

2 2

1 1

7 1 5.6 7 1 5.6

14.449 1.2373.6 g 12.3 g

R L

n s n sσ

χ χ

σ

σ

− −< <

− −< <

< <


1. 5.5x = ; median = 5.0; s = 3.8



2. The range of usual values is from ( )5.5 2 3.8 2.1− =− to ( )5.5 2 3.8 3.8+ = (or from 0 to 13.1).

3. Ratio level of measurement; discrete data. 4.

2 2

/ 2 1.96 5.833

2αz σ

nE

⎡ ⎤ ⎡ ⎤⋅⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

campuses

5. The population should include only colleges of the same type as the sample, so the population consists of all large urban campuses with residence halls.

/ 2

5.85.5 2.02

403.6 7.4

αs

x tn

μ

± = ± ⋅

< <

6. The graphs suggest that the population has a distribution that is skewed (to the right) instead of being normal. The histogram shows that some taxi-out times can be very long, and that can occur with heavy traffic, but little or no traffic cannot make the taxi-out time very low. There is a minimum time required, regardless of traffic conditions. Construction of a confidence interval estimate of a population standard deviation has a strict requirement that the sample data are from a normally distributed population, and the graphs show that this strict normality requirement is not satisfied.

7. a. ( )( )

/ 2

0.59 0.31ˆ ˆˆ 0.59 1.96

10030.560 0.620

αpq

p zn

p

± = ±

< <

(or 0.560 < p < 0.621 if using x = 592)

b. Because the survey was about shaking hands and because it was sponsored by a supplier of hand sanitizer products, the sponsor could potentially benefit from the results, so there might be some pressure to obtain results favorable to the sponsor.

c. [ ] [ ] ( )2 2

/ 2

2 2

ˆ ˆ 1.96 0.251083

0.025αz pq

nE

= = =

8. There does not appear to be a correlation between HDL and LDL cholesterol levels.

9. a. 185 175

1.119

z−

= = and ( )1.11 13.35%P z > = (Tech: 13.32%).

Yes, losing about 13% of the market would be a big loss.

b. 5th percentile: 175 1.645 9 160.2 mmx μ z σ= + ⋅ = − ⋅ =

95th percentile: 175 1.645 9 189.8 mmx μ z σ= + ⋅ = + ⋅ =

10. a. There are 310 possible tickets so the probability of winning by purchasing one ticket is 1

.1000

b. 1 999

1 .1000 1000

− = c. 10

9990.990.

1000

⎛ ⎞⎟⎜ =⎟⎜ ⎟⎜⎝ ⎠

Chapter 8: Hypothesis Testing 117


Chapter 8: Hypothesis Testing Section 8-2

1. Rejection of the aspirin claim is more serious because the aspirin is a drug treatment. The wrong aspirin dosage can cause adverse reactions. M&Ms do not have those same adverse reactions. It would be wise to use a smaller significance level for testing the aspirin claim.

2. Estimates and hypothesis tests are both methods of inferential statistics, but they have different objectives. We could use the sample weights to construct a confidence interval estimate of the mean weight of all M&Ms, but hypothesis testing is used to test some claim made about the mean weight of all M&Ms.

3. a. H0: 98.6 Fμ= °

b. H1: 98.6 Fμ≠ °

c. Reject the null hypothesis or fail to reject the null hypothesis.

d. No. In this case, the original claim becomes the null hypothesis. For the claim that the mean body temperature is equal to 98.6°F, we can either reject that claim or fail to reject it, but we cannot state that there is sufficient evidence to support that claim.

4. The P-value of 0.001 is preferred because it corresponds to the sample evidence that most strongly supports the alternative hypothesis that the XSORT method is effective.

5. a. 0.20p =

b. H0: 0.20p = and H1: 0.20p ≠

6. a. 0.5p >

b. H0: 0.5p = and H1: 0.5p >

7. a. 76μ≤

b. H0: 76μ= and H1: 76μ<

8. a. 50σ ≥

b. H0: 50σ = and H1: 50σ >

9. There is not sufficient evidence to warrant rejection of the claim that 20% of adults smoke.

10. There is sufficient evidence to support the claim that when parents use the XSORT method of gender selection, the proportion of baby girls is greater than 0.5.

11. There is not sufficient evidence to warrant rejection of the claim that the mean pulse rate of adult females is 76 or lower.

12. There is sufficient evidence to reject the claim that pulse rates of adult females have a standard deviation of at least 50.

13. ( )( )0.75 0.25

1021

ˆ 0.89 0.7510.33

pqn

p pz

− −= = = (or z = 10.35 if using x = 909)

14. ( )( )0.48 0.52

1002

ˆ 0.48 0.501.27

pqn

p pz

− −= = =− (or z = –1.26 if using x = 481)

15. ( ) ( )2 2

22 2

1 40 1 2.288.110

5

n sχ

σ− −

= = =

16. 7.15 8

2.3582.28 40

x μt

s n

− −= = =−

118 Chapter 8: Hypothesis Testing


17. P-value ( )2 0.0228.P z= > = Critical value: 1.645z = .

18. P-value ( )2 0.0228.P z= <− = Critical value: 1.645z =− .

19. P-value ( )2 1.75 0.0802.P z= ⋅ <− = (Tech: 0.0801). Critical values: 1.96, 1.96z z=− = .

20. P-value ( )2 1.50 0.1336.P z= ⋅ > = . Critical values: 1.96, 1.96z z=− = .

21. P-value ( )2 1.23 0.2186.P z= ⋅ <− = (Tech: 0.2187). Critical values: 1.96, 1.96z z=− = .

22. P-value ( )2 2.50 0.0124.P z= ⋅ > = Critical values: 1.96, 1.96z z=− = .

23. P-value ( )3.00 0.0013.P z= <− = Critical value: 1.645z =− .

24. P-value ( )2.88 0.0020.P z= > = Critical value: 1.645z = .

25. a. Reject H0.

b. There is sufficient evidence to support the claim that the percentage of blue M&Ms is greater than 5%.

26. a. Fail to reject H0.

b. There is not sufficient evidence to support the claim that fewer than 20% of M&M candies are green.


b. There is not sufficient evidence to warrant rejection of the claim that women have heights with a mean equal to 160.00 cm.

28. a. Reject H0.

b. There is sufficient evidence to warrant rejection of the claim that women have heights with a standard deviation equal to 5.00 cm.

29. a. H0: 0.5p = and H1: 0.5p >

b. 0.01α=

c. Normal distribution.

d. Right-tailed.

e. 1.00z =

f. P-value ( )1.00 0.1587.P z= > =

g. 2.33z =

h. 0.01

30. a. H0: 0.5p = and H1: 0.5p ≠

b. 0.05α=

c. Normal distribution.

d. Two-tailed.

e. 1.00z =

f. P-value ( )2 1.00 0.3174.P z= ⋅ > =

(Tech: 0.3173)

g. 1.96, 1.96z z=− =

h. 0.05

31. Type I error: In reality 0.1p = , but we reject the claim that 0.1p = . Type II error: In reality 0.1p ≠ , but

we fail to reject the claim that 0.1p = .

32. Type I error: In reality 0.001p = , but we reject the claim that 0.001p = . Type II error: In reality

0.001p ≠ , but we fail to reject the claim that 0.001p = .

33. Type I error: In reality 0.5p = , but we support the claim that 0.5p > . Type II error: In reality 0.5p > ,

but we fail to support that conclusion.

34. Type I error: In reality 0.9p = , but we support the claim that 0.9p < . Type II error: In reality 0.9p < ,

but we fail to support that conclusion.



35. The power of 0.96 shows that there is a 96% chance of rejecting the null hypothesis of 0.08p = when the

true proportion is actually 0.18. That is, if the proportion of Chantix users who experience abdominal pain is actually 0.18, then there is a 96% chance of supporting the claim that the proportion of Chantix users who experience abdominal pain is greater than 0.08.

36. a. From 0.5p = , ( )( )0.5 0.5

ˆ 0.5 1.645 0.602812564

p = + =

From 0.65p = , ( )( )0.65 0.35

64

0.6028125 .650.791z

−= =− ; Power = ( )0.791 0.7852.P z >− = (Tech: 0.7857)

b. Assuming that 0.5p = , as in the null hypothesis, the critical value of 1.645z = corresponds to

ˆ 0.6028125p = , so any sample proportion greater than 0.6028125 causes us to reject the null

hypothesis, as shown in the shaded critical region of the top graph. If p is actually 0.65, then the null hypothesis of 0.5p = is false, and the actual probability of rejecting the null hypothesis is found by

finding the area greater than ˆ 0.6028125p = in the bottom graph, which is the shaded area. That is, the

shaded area in the bottom graph represents the probability of rejecting the false null hypothesis.

37. From 0.5p = , ( )( )0.5 0.5

ˆ 0.5 1.645pn

= + , from 0.55p = , ( )( )0.55 0.45

ˆ 0.55 0.842pn

= − ; Since

( )( )0.842 0.8000P z >− = , so:

( )( ) ( )( )

2

0.5 0.5 0.55 0.450.5 1.645 0.55 0.842

0.5 1.645 0.25 0.55 0.842 0.2475

0.05 1.645 0.25 0.842 0.2475

1.645 0.25 0.842 0.2475617

0.05

n n

n n

n

n

+ = −

+ = −

= +

⎛ ⎞+ ⎟⎜ ⎟⎜= =⎟⎜ ⎟⎟⎜⎝ ⎠

Section 8-3

1. The P-value method and the critical value method always yield the same conclusion. The confidence interval method might or might not yield the same conclusion obtained by using the other two methods.

2. 411

ˆ 0.410.1003

p = = The symbol p is used to represent a sample proportion.

3. P-value = 0.00000000550. Because the P-value is so low, we have sufficient evidence to support the claim that 0.5p < .

4. a. The symbol p represents the population proportion, but the P-value is a probability of getting sample results that are at least as extreme as those obtained (assuming that the null hypothesis is true).

b. If the P-value is very low (such as less than or equal to 0.05), “the null must go” means that we should reject the null hypothesis.

c. The statement that “if the P is high, the null will fly” suggests that with a high P-value, the null hypothesis has been proved or is supported, but we should never make such a conclusion.

5. a. Left-tailed.

b. 1.94z =−

c. P-value = 0.0260 (rounded)

d. H0: 0.1p = . Reject the null hypothesis.

e. There is sufficient evidence to support the claim that less than 10% of treated subjects experience headaches.



6. a. Two-tailed.

b. 1.45z =

c. P-value = 0.146

d. H0: 0.35p = . Fail to reject the null hypothesis.

e. There is not sufficient evidence to warrant rejection of the claim that 35% of homes have guns in them.

7. a. Two-tailed.

b. 0.82z =−

c. P-value = 0.4106

d. H0: 0.35p = . Fail to reject the null hypothesis.

e. There is not sufficient evidence to warrant rejection of the claim that 35% of adults have heard of the Sony Reader.

8. a. Left-tailed.

b. 2.53z =−

c. P-value = 0.0057

d. H0: 0.5p = . Reject the null hypothesis.

e. There is sufficient evidence to support the claim that fewer than half of adults say that public speaking is the activity that they dread most.

9. H0: 0.25p = . H1: 0.25p ≠ . Test statistic: ( )( )

152580

0.25 0.75

580

0.250.67z

−= = . Critical values: 2.575z =±

(Tech: 2.576± ). P-value ( )2 0.67 0.5028P z= ⋅ > = (Tech: 0.5021). Fail to reject H0. There is not

sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow. MINITAB Test of p = 0.25 vs p not = 0.25 Sample X N Sample p 95% CI Z-Value P-Value 1 152 580 0.262069 (0.226280, 0.297858) 0.67 0.502

10. H0: 0.13p = . H1: 0.13p ≠ . Test statistic: ( )( )0.13 0.87

100

0.08 0.131.49z

−= =− . Critical values: 1.96z =± .

P-value ( )2 1.49 0.1362P z= ⋅ > = (Tech: 0.1371). Fail to reject H0. There is not sufficient evidence to

warrant rejection of the claim that 13% of M&Ms are brown. MINITAB Test of p = 0.13 vs p not = 0.13 Sample X N Sample p 95% CI Z-Value P-Value 1 8 100 0.080000 (0.026828, 0.133172) -1.49 0.137

11. H0: 0.5p = . H1: 0.5p > . Test statistic: ( )( )

5311002

0.5 0.5

1002

0.51.90z

−= = . Critical value: 1.645z = . P-value

( )1.90 0.0287P z= > = (Tech: 0.0290). Reject H0. There is sufficient evidence to support the claim that

the majority of adults feel vulnerable to identify theft. MINITAB Test of p = 0.5 vs p > 0.5 Sample X N Sample p Z-Value P-Value 1 531 1002 0.529940 1.90 0.029




492806

0.5 0.5

806

0.56.27z


( )6.27 0.0001P z= > = (Tech: 0.000000000182). Reject H0. There is sufficient evidence to support the

claim that the majority of adults prefer window seats when they fly. MINITAB Test of p = 0.5 vs p > 0.5 Sample X N Sample p Z-Value P-Value 1 492 806 0.610422 6.27 0.000


879945

0.5 0.5

945

0.526.45z



the XSORT method is effective in increasing the likelihood that a baby will be a girl. MINITAB Test of p = 0.5 vs p > 0.5 Sample X N Sample p Z-Value P-Value 1 879 945 0.930159 26.45 0.000


239291

0.5 0.5

291

0.510.96z



the YSORT method is effective in increasing the likelihood that a baby will be a boy. MINITAB Test of p = 0.5 vs p > 0.5 Sample X N Sample p Z-Value P-Value 1 239 291 0.821306 10.96 0.000


123280

0.5 0.5

280

0.52.03z

−= =− . Critical values: 1.645z =± . P-value

( )2 2.03 0.0424P z= ⋅ <− = (Tech: 0.0422). Reject H0. There is sufficient evidence to warrant rejection of

the claim that touch therapists use a method equivalent to random guesses. However, their success rate of 123/280 (or 43.9%) indicates that they performed worse than random guesses, so they do not appear to be effective.

MINITAB Test of p = 0.5 vs p not = 0.5 Sample X N Sample p 95% CI Z-Value P-Value 1 123 280 0.439286 (0.381154, 0.497417) -2.03 0.042


123280

0.5 0.5

280

0.52.03z

−= =− . Critical values: 2.575z =± (Tech:

2.576± ). P-value ( )2 2.03 0.0424P z= ⋅ <− = (Tech: 0.0422). Reject H0. There is not sufficient evidence

to warrant rejection of the claim that touch therapists use a method equivalent to random guesses. However, their success rate of 123/280 (or 43.9%) indicates that they performed worse than random guesses, so they do not appear to be effective.




17. H0: 13p = . H1: 1

3p < . Test statistic: ( )( )1 2

3 3

172 1611 3

611

2.72z−

= =− . Critical value: 2.33z =− . P-value

( )2.72 0.0033P z= <− = . Reject H0. There is sufficient evidence to support the claim that fewer than 1/3

of the challenges are successful. Players don’t appear to be very good at recognizing referee errors. MINITAB Test of p = 0.3333 vs p < 0.3333 Sample X N Sample p Z-Value P-Value 1 172 611 0.281506 -2.72 0.003


308611

0.43 0.57

601

0.433.70z

−= = . Critical values: 1.645z =± .

P-value ( )2 3.70 0.0002P z= ⋅ > = . Reject H0. There is sufficient evidence to warrant rejection of the

claim that the percentage who believe that they voted for the winning candidate is equal to 43%. There appears to be a substantial discrepancy between how people said that they voted and how they actually did vote.

MINITAB Test of p = 0.43 vs p not = 0.43 Sample X N Sample p 95% CI Z-Value P-Value 1 308 611 0.504092 (0.464447, 0.543736) 3.70 0.000

19. H0: 0.000340p = . 0.000340p ≠ . Test statistic: ( )( )

135420,095

0.000340 0.99966

420,095

0.0003400.66z

−= =− . Critical values:

2.81z =± . P-value ( )2 0.66 0.5092P z= ⋅ <− = (Tech: 0.5122). Fail to reject H0. There is not sufficient

evidence to support the claim that the rate is different from 0.0340%. Cell phone users should not be concerned about cancer of the brain or nervous system.



8561007

0.75 0.25

1007

0.757.33z



more than 75% of adults know what Twitter is. MINITAB Test of p = 0.75 vs p > 0.75 Sample X N Sample p Z-Value P-Value 1 856 1007 0.850050 7.33 0.000


235414

0.5 0.5

414

0.52.75z

−= = . Critical values: 1.96z =± . P-value

( )2 2.75 0.0060P z= ⋅ > = (Tech: 0.0059). Reject H0. There is sufficient evidence to warrant rejection of

the claim that the coin toss is fair in the sense that neither team has an advantage by winning it. The coin toss rule does not appear to be fair.





3971

0.5 0.5

71

0.50.83z


( )2 0.83 0.2033P z= ⋅ > = (Tech: 0.2031). Fail to reject H0. There is not sufficient evidence to support the

claim that among smokers who try to quit with nicotine patch therapy, the majority are smoking a year after the treatment. The results show that about half of those who use nicotine patch therapy are successful in quitting smoking.

MINITAB Test of p = 0.5 vs p > 0.5 Sample X N Sample p Z-Value P-Value 1 39 71 0.549296 0.83 0.203

23. H0: 0.5p = . H1: 0.5p < . Test statistic: ( )( )

152380

0.5 0.5

380

0.53.90z

−= =− . Critical value: 2.33z =− . P-value

( )3.90 0.0001P z= <− = (Tech: 0.0000484). Reject H0. There is sufficient evidence to support the claim

that fewer than half of smartphone users identify the smartphone as the only thing they could not live without. Because only smartphone users were surveyed, the results do not apply to the general population.

MINITAB Test of p = 0.5 vs p < 0.5 Sample X N Sample p Z-Value P-Value 1 152 380 0.400000 -3.90 0.000

24. H0: 0.5p = . H1: 0.5p < . Test statistic: ( )( )

606212000

0.5 0.5

12000

0.51.13z

−= = . Critical value: 1.645z =− . P-value

( )1.13 0.8708P z= < = (Tech: 0.8712). Fail to reject H0. There is not sufficient evidence to support the

claim that less than 0.5 of the deaths occur the week before Thanksgiving. Based on these results, there is no indication that people can temporarily postpone their death to survive Thanksgiving.

MINITAB Test of p = 0.5 vs p < 0.5 Sample X N Sample p Z-Value P-Value 1 6062 12000 0.505167 1.13 0.871

25. H0: 0.25p = . H1: 0.25p > . Test statistic: ( )( )0.25 0.75

427

0.29 0.251.91z

−= = or z = 1.93 (using x = 124). Critical

value: 1.645z = (assuming a 0.05 significance level). P-value ( )1.92 0.0281P z= > = (using ˆ 0.29p = )

or 0.0268 (using x = 124) (Tech P-value = 0.0269). Reject H0. There is sufficient evidence to support the claim that more than 25% of women purchase books online.


26. H0: 0.5p = . H1: 0.5p < . Test statistic: ( )( )0.5 0.5

514

0.459 0.51.86z

−= =− or z = –1.85 (using x = 236). Critical

value: 2.33z =− . P-value ( )1.86 0.0314P z= <− = (using ˆ 0.459p = ) or 0.0322 (using x = 236) (Tech

P-value = 0.0320). Fail to reject H0. There is not sufficient evidence to support the claim that less than half of all human resource professionals say that body piercings are big grooming red flags.





514

0.90 0.757.85 z


value: 2.33z = . P-value ( )7.85 0.0001 P z= > = (Tech: 0.0000). Reject H0. There is sufficient evidence

to support the claim that more than 3/4 of all human resource professionals say that the appearance of a job applicant is most important for a good first impression.


28. H0: 0.61p = . H1: 0.61p ≠ . Test statistic: ( )( )0.61 0.39

1002

0.70 0.615.84z


values: 1.96z =± (assuming a 0.05 significance level). P-value ( )2 5.81 0.0002P z= ⋅ > = (Tech:

0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim that the percentage of all voters who say that they voted is equal to 61%. The results suggest that either survey respondents are not being truthful or they have an incorrect perception of reality.



870

0.39 0.79129.09z

−= =− or z = –29.11 (using x = 339).

Critical value: 2.33z =− . P-value ( )2 29.09 0.0001P z= ⋅ <− = (Tech: 0.0000). Reject H0. There is

sufficient evidence to support the claim that the percentage of selected Americans of Mexican ancestry is less than 79.1%, so the jury selection process appears to be unfair.



703

0.61 0.55.83z

−= = or z = 5.85 (using x = 429). Critical value:

1.645z = . P-value ( )5.83 0.0001P z= > = (Tech: 0.0000). Reject H0. There is sufficient evidence to

support the claim that most workers get their jobs through networking. MINITAB Test of p = 0.5 vs p > 0.5 Sample X N Sample p Z-Value P-Value 1 429 703 0.610242 5.85 0.000


25,000

0.77 0.757.30z



more than 75% of television sets in use were tuned to the Super Bowl. MINITAB Test of p = 0.75 vs p > 0.75 Sample X N Sample p Z-Value P-Value 1 19250 25000 0.770000 7.30 0.000




1500

0.47 0.52.32z


( )2.32 0.0102P z= <− = (Tech: 0.0101). Fail to reject H0. There is not sufficient evidence to support the

claim that fewer than half of all households have a high-definition television. Because the use of high-definition televisions is growing rapidly, these results are not likely to be valid today.


33. Among 100 M&Ms, 19 are green. H0: 0.16p = . H1: 0.16p ≠ . Test statistic: ( )( )0.16 0.84

100

0.19 0.160.82z

−= = .

Critical values: 1.96z =± . P-value ( )2 0.82 0.4122P z= ⋅ > = (Tech: 0.4132). Fail to reject H0. There is

not sufficient evidence to warrant rejection of the claim that 16% of plain M&M candies are green. MINITAB Test of p = 0.16 vs p not = 0.16 Sample X N Sample p 95% CI Z-Value P-Value 1 19 100 0.190000 (0.113110, 0.266890) 0.82 0.413

34. Among 48 flights, 44 are on time. H0: 0.795p = . H1: 0.795p ≠ . Test statistic: ( )( )

4448

0.795 0.205

48

0.7952.09z

−= = .

Critical values: 1.96z =± . P-value ( )2 2.09 0.0366 P z= ⋅ > = (Tech: 0.0368). Reject H0. There is

sufficient evidence to warrant rejection of the claim that 79.5% of flights are on time. With 91.7% of the 48 flights arriving on time, American Airlines appears to have a better on-time performance.


35. H0: 0.5p = . H1: 0.5p > . Using the binomial probability distribution with an assumed proportion of

0.5p = , the probability of 7 or more heads is 0.0352, so the P-value is 0.0352. Reject H0. There is

sufficient evidence to support the claim that the coin favors heads.

36. a. H0: 0.10p = . H1: 0.1p ≠ . Test statistic: ( )( )0.1 0.9

1000

0.119 0.12.00z

−= = . Critical values: 1.96z =± . Reject

H0. There is sufficient evidence to warrant rejection of the claim that the proportion of zeros is 0.1.

b. H0: 0.10p = . H1: 0.1p ≠ . Test statistic: ( )( )0.1 0.9

1000

0.119 0.12.00z

−= = . P-value

( )2 2.00 0.0456P z= ⋅ > = (Tech: 0.0452). There is sufficient evidence to warrant rejection of the

claim that the proportion of zeros is 0.1.

c. 0.0989 0.139p< < ; because 0.1 is contained within the confidence interval, fail to reject

H0: 0.10p = . There is not sufficient evidence to warrant rejection of the claim that the proportion of

zeros is 0.1. MINITAB Sample X N Sample p 95% CI 1 119 1000 0.119000 (0.098932, 0.139068)

d. The traditional and P-value methods both lead to rejection of the claim, but the confidence interval method does not lead to rejection of the claim.



37. a. From 0.40p = , ( )( )0.4 0.6

ˆ 0.4 1.645 0.28650

p = − =

From 0.25p = , ( )( )0.25 0.75

50

0.286 0.250.588z

−= = ; Power = ( )0.588 0.7224.P z < = (Tech: 0.7219)

b. 1 – 0.7224 = 0.2776 (Tech: 0.2781)

c. The power of 0.7224 shows that there is a reasonably good chance of making the correct decision of rejecting the false null hypothesis. It would be better if the power were even higher, such as greater than 0.8 or 0.9.

Section 8-4

1. The requirements are (1) the sample must be a simple random sample, and (2) either or both of these conditions must be satisfied: The population is normally distributed or 30n > . There is not enough information given to determine whether the sample is a simple random sample. Because the sample size is not greater than 30, we must check for normality, but the value of 583 sec appears to be an outlier, and a normal quantile plot or histogram suggests that the sample does not appear to be from a normally distributed population.

6005004003002001000

5

4

3

2

1

0

Freq

uen

cy

6005004003002001000-100-200-300

0.99

0.95

0.9

0.8

0.7

0.6

0.50.4

0.3

0.2

0.1

0.05

0.01

Pro

babi

lity

Probability Plot

2. df denotes the number of degrees of freedom. For the sample of 12 times, df 12 1 11= − = .

3. A t test is a hypothesis test that uses the Student t distribution, such as the method of testing a claim about a population mean as presented in this section. The t test methods are much more likely to be used than the z test methods because the t test does not require a known value of σ , and realistic hypothesis tests of claims about μ typically involve a population with an unknown value of σ .

4. Use a 90% confidence level. The given confidence interval does contain the value of 90 sec, so it is possible that the value of μ is equal to 90 sec or some lower value, so there is not sufficient evidence to

support the claim that the mean is greater than 90 sec.

5. P-value < 0.005 (Tech: 0.0013).

6. 0.025 < P-value < 0.05 (Tech: 0.0480).

7. 0.02 < P-value < 0.05 (Tech: 0.0365).

8. 0.01 < P-value < 0.02 (Tech: 0.0183)

9. H0: 24μ= . H1: 24μ< . Test statistic: 7.323t =− . Critical value: 1.685t =− . P-value < 0.005. (The

display shows that the P-value is 0.00000000387325.) Reject H0. There is sufficient evidence to support the claim that Chips Ahoy reduced-fat cookies have a mean number of chocolate chips that is less than 24 (but this does not provide conclusive evidence of reduced fat).

10. H0: 10μ= km. H1: 10μ≠ km. Test statistic: 0.27t =− . Critical values: 2.678t =± (approximately). P-

value > 0.20. (Minitab shows a P-value of 0.790.) Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the earthquakes are from a population with a mean depth equal to 10 km.



11. H0: 33μ= years. H1: 33μ≠ years. Test statistic: 35.9 33

2.36711.1 82

t−

= = . Critical values 2.639t =±

(approximately). P-value > 0.02 (Tech: 0.0204). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the mean age of actresses when they win Oscars is 33 years.

MINITAB Test of mu = 33 vs not = 33 N Mean StDev SE Mean 95% CI T P 82 35.90 11.10 1.23 (33.46, 38.34) 2.37 0.020

12. H0: 1.800μ= lb. H1: 1.800μ> lb. Test statistic: 1.911 1.800

0.8211.065 62

t−

= = . Critical value: 1.671t =

(approximately). P-value > 0.10 (Tech: 0.2075). Fail to reject H0. There is not sufficient evidence to support the claim that the mean weight of discarded plastic from the population of households is greater than 1.800 lb.

MINITAB Test of mu = 1.8 vs > 1.8 N Mean StDev SE Mean T P 62 1.911 1.065 0.135 0.82 0.208

13. H0: 0.8535μ= g. H1: 0.8535μ≠ g. Test statistic: 0.8635 0.8535

0.7650.0570 19

t−

= = . Critical values:

2.101t =± . P-value > 0.20 (Tech: 0.4543). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the mean weight of all green M&Ms is equal to 0.8535 g. The green M&Ms do appear to have weights consistent with the package label.

MINITAB Test of mu = 0.8535 vs not = 0.8535 N Mean StDev SE Mean 95% CI T P 19 0.8635 0.0570 0.0131 (0.8360, 0.8910) 0.76 0.454

14. H0: 98.6μ≠ °F. H1: 98.6μ≠ °F. Test statistic: 98.2 98.6

6.6420.62 106

t−

= =− . Critical values: 1.984t =±

(approximately). P-value < 0.01 (Tech: 0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim that the mean body temperature of the population is equal to 98.6°F. There is sufficient evidence to conclude that the common belief is wrong.

MINITAB Test of mu = 98.6 vs not = 98.6 N Mean StDev SE Mean 95% CI T P 106 98.2000 0.6200 0.0602 (98.0806, 98.3194) -6.64 0.000

15. H0: 0μ= lb. H1: 0μ> lb. Test statistic: 3 0

3.8724.9 40

t−

= = . Critical value: 2.426t = .

P-value < 0.005 (Tech: 0.0002). Reject H0. There is sufficient evidence to support the claim that the mean weight loss is greater than 0. Although the diet appears to have statistical significance, it does not appear to have practical significance, because the mean weight loss of only 3.0 lb does not seem to be worth the effort and cost.

MINITAB Test of mu = 0 vs > 0 N Mean StDev SE Mean T P 40 3.000 4.900 0.775 3.87 0.000



16. H0: 12.0μ= min. H1: 12.0μ< min. Test statistic: 10.5 12.0

0.337430.8 48

t−

= =− . Critical value: 2.412t =−

(approximately). P-value > 0.10 (Tech: 0.3687). Fail to reject H0. There is not sufficient evidence to support the claim that the mean departure delay time for all such flights is less than 12.0 min. A flight operations manager is not justified in reporting that the mean departure time is less than 12.0 min.

MINITAB Test of mu = 12 vs < 12 N Mean StDev SE Mean T P 48 10.50 30.80 4.45 -0.34 0.369

17. H0: 0μ= . H1: 0μ> . Test statistic: 0.4 0

0.13321.0 49

t−

= = . Critical value: 1.676t = (approximately,

assuming a 0.05 significance level). P-value > 0.10 (Tech: 0.4472). Fail to reject H0. There is not sufficient evidence to support the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. The results suggest that the garlic treatment is not effective in reducing LDL cholesterol levels.


18. H0: 102.8μ= min. H1: 102.8μ< min. Test statistic: 98.9 102.8

0.36942.3 16

t−

= =− . Critical value:

1.753t =− (assuming a 0.05 significance level). P-value > 0.10 (Tech: 0.3587). Fail to reject H0. There is not sufficient evidence to support the claim that after treatment with Zopiclone, subjects have a mean wake time less than 102.8 min. This result suggests that the Zoplicone treatment is not effective.

MINITAB Test of mu = 102.8 vs < 102.8 N Mean StDev SE Mean T P 16 98.9 42.3 10.6 -0.37 0.359

19. H0: 4μ= years. H1: 4μ> years. Test statistic: 3.189t = . Critical value: 2.539t = .

P-value < 0.005 (Tech: 0.0024). Reject H0. There is sufficient evidence to support the claim that the mean time required to earn a bachelor’s degree is greater than 4.0 years. Because 30n ≤ and the data do not appear to be from a normally distributed population, the requirement that “the population is normally distributed or 30n > ” is not satisfied, so the conclusion from the hypothesis test might not be valid. However, some of the sample values are equal to 4 years and others are greater than 4 years, so the claim does appear to be justified.

141210864

12

10

8

6

4

2

0

Years of College

Freq

uen

cy

MINITAB Test of mu = 4 vs > 4 Variable N Mean StDev SE Mean T P CollegeYears 20 6.500 3.506 0.784 3.19 0.002



20. The sample data meet the loose requirement of having a normal distribution. H0: 30μ= years.

H1: 30μ> years. Test statistic: 1.818t = . Critical value: 1.761t = . P-value < 0.05 (Tech: 0.0453). Reject

H0. There is sufficient evidence to support the claim that the mean age of all race car drivers is greater than 30 years.

MINITAB Test of mu = 30 vs > 30 Variable N Mean StDev SE Mean T P Ages 15 33.60 7.67 1.98 1.82 0.045

21. The sample data meet the loose requirement of having a normal distribution. H0: 14μ= mg/g.

H1: 14μ< mg/g. Test statistic: 1.444t =− . Critical value: 1.833t =− . P-value > 0.05 (Tech: 0.0913).

Fail to reject H0. There is not sufficient evidence to support the claim that the mean lead concentration for all such medicines is less than 14 mg/g.

MINITAB Test of mu = 14 vs < 14 Variable N Mean StDev SE Mean T P Lead 10 11.05 6.46 2.04 -1.44 0.091

22. The sample data meet the loose requirement of having a normal distribution. H0: 1100μ= cm3.

H1: 1100μ≠ cm3. Test statistic : 0.813t = . Critical values: 3.250t =± . P-value > 0.20 (Tech: 0.4371).

Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the population of brain volumes has a mean equal to 1100.0 cm3.

MINITAB Test of mu = 1100 vs not = 1100 Variable N Mean StDev SE Mean 95% CI T P Volume 10 1130.2 117.4 37.1 (1046.2, 1214.2) 0.81 0.437

23. The sample data meet the loose requirement of having a normal distribution. H0: 63.8μ= in.

H1: 63.8μ> in. Test statistic: 23.824t = . Critical value: 2.821t = . P-value < 0.005 (Tech: 0.0000).

Reject H0. There is sufficient evidence to support the claim that supermodels have heights with a mean that is greater than the mean height of 63.8 in. for women in the general population. We can conclude that supermodels are taller than typical women.

MINITAB Test of mu = 63.8 vs > 63.8 Variable N Mean StDev SE Mean T P Heights 10 69.825 0.800 0.253 23.82 0.000

24. The sample data meet the loose requirement of having a normal distribution. H0: 65μ= mi/h. H1:

65μ< mi/h. Test statistic: 3.684t =− . Critical value: 1.796t =− . P-value < 0.005 (Tech: 0.0018).

Reject H0. There is sufficient evidence to support the claim that the sample is from a population with a mean that is less than the speed limit of 65 mi/h.

MINITAB Test of mu = 65 vs < 65 Variable N Mean StDev SE Mean T P HWY 12 60.67 4.08 1.18 -3.68 0.002

25. The sample data meet the loose requirement of having a normal distribution. H0: 1.00μ= . H1: 1.00μ> .

Test statistic: 2.218t = . Critical value: 1.676t = (approximately). P-value < 0.025 (Tech: 0.0156). Reject H0. There is sufficient evidence to support the claim that the population of earthquakes has a mean magnitude greater than 1.00.

MINITAB Test of mu = 1 vs > 1 Variable N Mean StDev SE Mean Bound T P MAG 5 0 1.1842 0.5873 0.0831 1.0449 2.22 0.016



26. The sample data meet the loose requirement of having a normal distribution. H0: 120μ= mm Hg. H1:

120μ< mm Hg. Test statistic: 0.424t =− . Critical value: 1.685t =− . P-value > 0.10 (Tech: 0.3370).

Fail to reject H0. There is not sufficient evidence to support the claim that the female population has a mean systolic blood pressure level less than 120.0 mm Hg.

MINITAB Test of mu = 120 vs < 120 Variable N Mean StDev SE Mean T P SYS 40 118.50 22.39 3.54 -0.42 0.337

27. The sample data meet the loose requirement of having a normal distribution. H0: 83μ= kg. H1: 83μ< kg.

Test statistic: 5.524t =− . Critical value: 2.453t =− . P-value < 0.005 (Tech: 0.0000). Reject H0. There is sufficient evidence to support the claim that male college students have a mean weight that is less than the 83 kg mean weight of males in the general population.

MINITAB Test of mu = 83 vs < 83 Variable N Mean StDev SE Mean T P WTSEP 32 72.72 10.53 1.86 -5.52 0.000

28. The sample data meet the loose requirement of having a normal distribution. H0: 120μ= V. H1:

120μ≠ V. Test statistic: 96.358t = . Critical values: 2.708t =± . P-value < 0.01 (Tech: 0.0000). Reject

H0. There is sufficient evidence to warrant rejection of the claim that the mean voltage amount is 120 volts. MINITAB Test of mu = 120 vs not = 120 Variable N Mean StDev SE Mean 95% CI T P Home 40 123.663 0.240 0.038 (123.586, 123.739) 96.36 0.000

29. H0: 24μ= . H1: 24μ< . Test statistic: 19.6 24

7.323.8 40

z−

= =− . Critical value: 1.645z =− .

P-value ( )7.32 0.0001P z= <− = (Tech: 0.0000). Reject H0. There is sufficient evidence to support the

claim that Chips Ahoy reduced-fat cookies have a mean number of chocolate chips that is less than 24 (but this does not provide conclusive evidence of reduced fat).

30. H0: 10μ= km. H1: 10μ≠ km. Test statistic: 9.810 10

0.275.01 50

z−

= =− . Critical values: 2.575z =± . P-

value ( )2 0.27 0.7872P z= ⋅ <− = (Tech: 0.7886). Fail to reject H0. There is not sufficient evidence to

warrant rejection of the claim that the earthquakes are from a population with a mean depth equal to 10 km.

31. H0: 33μ= years. H1: 33μ≠ years. Test statistic: 35.9 33

2.3711.1 82

z−

= = . Critical values: 2.575z =± . P-

value ( )2 2.37 0.0178P z= ⋅ > = (Tech: 0.0180). Fail to reject H0. There is not sufficient evidence to

warrant rejection of the claim that the mean age of actresses when they win Oscars is 33 years.

32. H0: 1.800μ= lb. H1: 1.800μ> lb. Test statistic: 1.911 1.800

0.821.065 62

z−

= = . Critical value: 1.645z = . P-

value ( )0.82 0.2061P z= > = (Tech: 0.2059). Fail to reject H0. There is not sufficient evidence to support

the claim that the mean weight of discarded plastic from the population of households is greater than 1.800 lb.

33. ( )1.645 8 149 3

1.65052478 149 1

A⋅ +

= =⋅ −

. The approximation yields a critical value of

( )21.6505247 /149149 1 1.655t e= − = , which is the same as the result from STATDISK or a TI-83/84 Plus

calculator.



34. Using the normal distribution makes you more likely to reject the null hypothesis because the critical z values are not as extreme as the corresponding critical t values.

35. a. The power of 0.4274 shows that there is a 42.74% chance of supporting the claim that 1μ< W/kg

when the true mean is actually 0.80 W/kg. This value of power is not very high, and it shows that the hypothesis test is not very effective in recognizing that the mean is less than 1.00 W/kg when the actual mean is 0.80 W/kg.

b. β = 0.5726. The probability of a type II error is 0.5726. That is, there is a 0.5726 probability of

making the mistake of not supporting the claim that 1μ< W/kg when in reality the population mean is

0.80 W/kg.

Section 8-5

1. a. The mean waiting time remains the same.

b. The variation among waiting times is lowered.

c. Because customers all have waiting times that are roughly the same, they experience less stress and are generally more satisfied. Customer satisfaction is improved.

d. The single line is better because it results in lower variation among waiting times, so a hypothesis test of a claim of a lower standard deviation is a good way to verify that the variation is lower with a single waiting line.

2. a. The normality requirement for a hypothesis test of a claim about a standard deviation is much more strict, meaning that the distribution of the population must be much closer to a normal distribution.

b. With only 10 sample values, a histogram doesn’t really give us a good picture of the distribution, so a normal quantile plot would be better. Also, we should determine that there are no outliers.

3. Use a 90% confidence interval. The conclusion based on the 90% confidence interval will be the same as the conclusion from a hypothesis test using the P-value method or the critical value method.

4. a. H0: 1.8σ = min. H1: 1.8σ < min.

b. ( ) ( )2 2

22 2

1 10 1 0.50.694

1.8

n sχ

σ− −

= = =

c. Reject H0, the null hypothesis.

d. There is sufficient evidence to support the claim that the standard deviation of waiting times of all customers is less than 1.8 min.

e. The change to a single waiting line is effective because the variation among waiting times is less than it was with multiple lines.

5. H0: 0.15σ = oz. H1: 0.15σ < oz. Test statistic: ( ) 2

22

36 1 0.1118.822

0.15χ

−= = . Critical value of 2χ is

between 18.493 and 26.509, so it is estimated to be 22.501 (Tech: 22.465). P-value < 0.05 (Tech: 0.0116). Reject H0. There is sufficient evidence to support the claim that the population of volumes has a standard deviation less than 0.15 oz.

MINITAB Method Chi-Square DF P-Value Standard 18.82 35 0.012

6. H0: 0.15σ = oz. H1: 0.15σ < oz. Test statistic: ( ) 2

22

36 1 0.0912.600

0.15χ

−= = . Critical value of 2χ is

between 18.493 and 26.509, so it is estimated to be 22.501 (Tech: 22.465). P-value < 0.05 (Tech: 0.0002). Reject H0. There is sufficient evidence to support the claim that the population of volumes has a standard deviation less than 0.15 oz.




7. H0: 0.0230σ = g. H1: 0.0230σ < g. Test statistic: ( ) 2

22

37 1 0.0164818.483

0.0230χ

−= = . Critical value of 2χ

is between 18.493 and 26.509, so it is estimated to be 22.501 (Tech: 23.269). P-value < 0.05 (Tech: 0.0069). Reject H0. There is sufficient evidence to support the claim that the population of weights has a standard deviation less than the specification of 0.0230 g.


8. H0: 0.0230σ = g. H1: 0.0230σ > g. Test statistic: ( ) 2

22

35 1 0.0391098.260

0.0230χ

−= = . Critical value of 2χ

is between 43.773 and 55.758. P-value < 0.005 (Tech: 0.0000). Reject H0. There is sufficient evidence to support the claim that pre-1983 pennies have a standard deviation greater than 0.0230 g. Weights of pre-1983 pennies appear to vary more than those of post-1983 pennies.


9. The data appear to be from a normally distributed population. H0: 10σ = bpm. H1: 10σ ≠ bpm. Test

statistic: ( ) 2

22

40 1 10.341.375

10χ

−= = . Critical value of 2χ = 24.433 and 2χ = 59.342 (approximately).

P-value > 0.20 (Tech: 0.7347). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that pulse rates of men have a standard deviation equal to 10 beats per minute.


10. The data appear to be from a normally distributed population. H0: 10σ = bpm. H1: 10σ ≠ bpm. Test

statistic: ( ) 2

22

40 1 11.652.478

10χ

−= = . Critical values of 2χ = 24.433 and 2χ = 59.342 (approximately).

P-value > 0.10 (Tech: 0.1463). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that pulse rates of women have a standard deviation equal to 10 beats per minute.


11. H0: 3.2σ = mg. H1: 3.2σ ≠ mg. Test statistic: ( ) 2

22

25 1 3.732.086

3.2χ

−= = . Critical values: 2χ = 12.401

and 2χ = 39.364. P-value > 0.20 (Tech: 0.2498). Fail to reject H0. There is not sufficient evidence to

support the claim that filtered 100-mm cigarettes have tar amounts with a standard deviation different from 3.2 mg. There is not enough evidence to conclude that filters have an effect.


12. H0: 28.866σ = cents. H1: 28.866σ ≠ cents. Test statistic: ( ) 2

22

100 1 33.5133.337

28.866χ

−= = . Critical values:

2χ = 67.328 and 2χ = 140.169 (approximately). P-value > 0.02 (Tech: 0.0244). Fail to reject H0. There is

not sufficient evidence to warrant rejection of the claim that the standard deviation is 28.866 cents. Because the amounts from 0 cents to 99 cents are all equally likely, the requirement of a normal distribution is violated, so the results are highly questionable.



12. (continued)


13. The data appear to be from a normally distributed population. H0: 22.5σ = years. H1: 22.5σ < years. Test

statistic: ( ) 2

22

15 1 7.671.627

22.5χ

−= = . Critical value: 2χ = 4.660. P-value < 0.005 (Tech: 0.0000). Reject

H0. There is sufficient evidence to support the claim that the standard deviation of ages of all race car drivers is less than 22.5 years.

MINITAB Variable Method Chi-Square DF P-Value Ages Standard 1.63 14.00 0.000

14. The data appear to be from a normally distributed population. H0: 5σ = mi/h. H1: 5σ ≠ mi/h. Test

statistic: ( ) 2

22

12 1 4.087.307

5.0χ

−= = . Critical values of 2χ = 3.816 and 2χ = 21.920. P-value > 0.20

(Tech: 0.4525). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the standard deviation of speeds is equal to 5.0 mi/h.

MINITAB Variable Method Chi-Square DF P-Value HWY Standard 7.31 11.00 0.453

15. The data appear to be from a normally distributed population. H0: 32.2σ = ft. H1: 32.2σ > ft. Test

statistic: ( ) 2

22

12 1 52.429.176

32.2χ

−= = . Critical value: 2χ = 19.675. P-value = 0.0021. Reject H0. There is

sufficient evidence to support the claim that the new production method has errors with a standard deviation greater than 32.2 ft. The variation appears to be greater than in the past, so the new method appears to be worse, because there will be more altimeters that have larger errors. The company should take immediate action to reduce the variation.

MINITAB Variable Method Chi-Square DF P-Value Errors Standard 29.18 11.00 0.002

16. The data appear to be from a normally distributed population. H0: 15σ = . H1: 15σ < . Test statistic:

( ) 22

2

12 1 9.504.416

15χ

−= = . Critical value: 2χ = 4.575. P-value < 0.05 (Tech: 0.0439). Reject H0. There

is sufficient evidence to support the claim that IQ scores of professional pilots have a standard deviation less than 15.

MINITAB Variable Method Chi-Square DF P-Value IQ Standard 4.42 11.00 0.044

17. The data appear to be from a normally distributed population. H0: 0.15σ = oz. H1: 0.15σ < oz. Test

statistic: ( ) 2

22

36 1 0.080910.173

0.15χ

−= = . Critical value of 2χ is between 18.493 and 26.509, so it is

estimated to be 22.501 (Tech: 22.465). P-value < 0.01 (Tech: 0.0000). Reject H0. There is sufficient evidence to support the claim that the population of volumes has a standard deviation less than 0.15 oz.

MINITAB Variable Method Chi-Square DF P-Value CKDTVOL Standard 10.17 35.00 0.000



18. The data appear to be from a normally distributed population. H0: 0.0230σ = g. H1: 0.0230σ ≠ g. Test

statistic: ( ) 2

22

35 1 0.0391156.155

0.0230χ

−= = . Critical values of 2χ = 13.787 and 2χ = 53.672

(approximately). P-value < 0.01 (Tech: 0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim that the population of weights has a standard deviation equal to 0.0230 g.

MINITAB Variable Method Chi-Square DF P-Value Wheat Standard 156.16 34.00 0.000

19. Critical ( )2

2 11.645 2 35 1 22.189

2χ = − + ⋅ − = , which is reasonably close to the value of 22.465 obtained

from STATDISK and Minitab.

20. Critical

3

2 2 235 1 1.645 22.642

9 35 9 35χ

⎛ ⎞⎛ ⎞⎟⎜ ⎟⎜ ⎟⎟⎜ ⎜= − + − =⎟⎟⎜ ⎜ ⎟⎟⎜ ⎟⎜⋅ ⋅ ⎟⎜ ⎝ ⎠⎝ ⎠, which is very close to the value of 22.465

obtained from STATDISK and Minitab.

Chapter Quick Quiz

1. H0: 0μ= sec. H1: 0μ≠ sec.

2. a. Two-tailed.

b. Student t.


b. There is not sufficient evidence to warrant rejection of the claim that the sample is from a population with a mean equal to 0 sec.

4. There is a loose requirement of a normally distributed population in the sense that the test works reasonably well if the departure from normality is not too extreme.

5. a. H0: 0.5p = . H1: 0.5p > .

b. ( )( )0.5 0.5

511

0.64 0.56.33z

−= =

c. P-value = 0.0000000001263996. There is sufficient evidence to support the claim that the majority of adults are in favor of the death penalty for a person convicted of murder.

6. ( )2 2.00 0.0456P z= ⋅ <− = (Tech: 0.0455)

7. The only true statement is the one given in part (a).

8. No. All critical values of x2 are greater than zero.

9. True. 10. False.

Review Exercises

1. a. False.

b. True.

c. False.

d. False.

e. False.



2. H0: 23p = . H1: 2

3p ≠ . Test statistic: ( )( )2 1

3 3

657 21010 3

1010

1.09z−

= =− . Critical values: 2.575z =± (Tech: 2.576± ).

P-value ( )2 1.09 0.2758P z= ⋅ <− = (Tech: 0.2756). Fail to reject H0. There is not sufficient evidence to

warrant rejection of the claim that 2/3 of adults are satisfied with the amount of leisure time that they have. MINITAB Test of p = 0.666667 vs p not = 0.666667 Sample X N Sample p 95% CI Z-Value P-Value 1 657 1010 0.650495 (0.621089, 0.679901) -1.09 0.27


678737

0.75 0.25

737

0.7510.65z

−= = or 10.66z = (if using ˆ 0.92p = ).

Critical value: 2.33z =± . P-value = 0.0001 (Tech: 0.0000). Reject H0. There is sufficient evidence to support the claim that more than 75% of us do not open unfamiliar e-mail and instant-message links. Given that the results are based on a voluntary response sample, the results are not necessarily valid.


4. H0: 3369μ= g. H1: 3369μ< g. Test statistic: 3245 567

19.962446 81

t−

= =− . Critical value: 2.328t =−

(approximately). P-value < 0.005 (Tech: 0.0000). Reject H0. There is sufficient evidence to support the claim that the mean birth weight of Chinese babies is less than the mean birth weight of 3369 g for Caucasian babies.

5. H0: 567σ = g. H1: 567σ ≠ g. Test statistic: ( ) 2

22

81 1 46654.038

567χ

−= = . Critical values of 2χ = 51.172

and 2χ = 116.321. P-value is between 0.02 and 0.05 (Tech: 0.0229). Fail to reject H0. There is not

sufficient evidence to warrant rejection of the claim that the standard deviation of birth weights of Chinese babies is equal to 567 g.


6. H0: 1.5μ= mg/m3. H1: 1.5μ> mg/m3. Test statistic : 0.049t = . Critical value: 2.015t = .

P-value > 0.10 (Tech: 0.4814). Fail to reject H0. There is not sufficient evidence to support the claim that the sample is from a population with a mean greater than the EPA standard of 1.5 mg/m3. Because the sample value of 5.40 mg/m3 appears to be an outlier and because a normal quantile plot suggests that the sample data are not from a normally distributed population, the requirements of the hypothesis test are not satisfied, and the results of the hypothesis test are therefore questionable.

6420-2-4

0.99

0.95

0.9

0.8

0.7

0.60.50.40.3

0.2

0.1

0.05

0.01

Air Lead

Pro

babi

lity

Probability Plot of Air Lead



6. (continued)

MINITAB Test of mu = 1.5 vs > 1.5 Variable N Mean StDev SE Mean T P Air Lead 6 1.538 1.914 0.781 0.05 0.481

7. H0: 25μ= . H1: 25μ≠ . Test statistic: 24.2 25

0.56714.1 100

t−

= =− . Critical values: 1.984t =±

(approximately). P-value > 0.20 (Tech: 0.5717). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the sample is selected from a population with a mean equal to 25.

MINITAB Test of mu = 25 vs not = 25 N Mean StDev SE Mean 95% CI T P 100 24.20 14.10 1.41 (21.40, 27.00) -0.57 0.572

8. a. A type I error is the mistake of rejecting a null hypothesis when it is actually true. A type II error is the mistake of failing to reject a null hypothesis when in reality it is false.

b. Type I error: Reject the null hypothesis that the mean of the population is equal to 25 when in reality, the mean is actually equal to 25. Type II error: Fail to reject the null hypothesis that the population mean is equal to 25 when in reality, the mean is actually different from 25.

9. The 2χ test has a reasonably strict requirement that the sample data must be randomly selected from a

population with a normal distribution, but the numbers are selected in such a way that they are all equally likely, so the population has a uniform distribution instead of the required normal distribution. Because the requirements are not all satisfied, the 2χ 2 test should not be used.

10. The sample data meet the loose requirement of having a normal distribution. H0: 1000μ= HIC.

H1: 1000μ< HIC. Test statistic: 10.177t =− . Critical value: 3.747t =− . P-value < 0.005 (Tech:

0.0003). Reject H0. There is sufficient evidence to support the claim that the population mean is less than 1000 HIC. The results suggest that the population mean is less than 1000 HIC, so they appear to satisfy the specified requirement.

MINITAB Test of mu = 1000 vs < 1000 Variable N Mean StDev SE Mean T P Booster 5 653.8 76.1 34.0 -10.18 0.000


1. a. 53.3 wordsx =

b. Median = 52.0 words

c. s = 15.7 words

d. s2 = 245.1 words2

e. Range = 45 words

2. a. Ratio. b. Discrete.

c. The sample is a simple random sample if it was selected in such a way that all possible samples of the same size have the same chance of being selected.

3. 42.1 words 64.5 wordsμ< <

MINITAB Variable N Mean StDev SE Mean 95% CI X 10 53.30 15.66 4.95 (42.10, 64.50)



4. H0: 48.0μ= words. H1: 48.0μ> words. Test statistic: 53.3 48.0

1.07015.7 10

t−


P-value > 0.10 (Tech: 0.1561). Fail to reject H0. There is not sufficient evidence to support the claim that the mean number of words on a page is greater than 48.0. There is not enough evidence to support the claim that there are more than 70,000 words in the dictionary.


5. a. 38.8 36.0

2;1.4

z−

= = ( )2 2.28%.P z > =

b. 98th percentile: 36.0 2.054 1.4 38.9 in.x μ z σ= + ⋅ = + ⋅ =

c. 37.0 36.0

1.43;1.4 4

z−

= = ( )1.43 92.36%.P z < = (Tech: 0.9234)

6. a. ( )30.125 0.00195= . It is unlikely because the probability of the event occurring is so small.

b. ( )( )0.097 0.125 0.0121= c. ( )51 0.875 0.487− =

7. No. The distribution is very skewed. A normal distribution would be approximately bell-shaped, but the displayed distribution is very far from being bell-shaped.

8. Because the vertical scale starts at 7000 and not at 0, the difference between the number of males and the number of females is exaggerated, so the graph is deceptive by creating the wrong impression that there are many more male graduates than female graduates.

9. a. ( )0.372 1003 373=

b. 34.2% 40.2%p< <

MINITAB Sample X N Sample p 95% CI 1 373 1003 0.371884 (0.341974, 0.401795)

c. Yes. With test statistic 8.11z =− and with a P-value close to 0, there is sufficient evidence to support the claim that less than 50% of adults answer “yes.” MINITAB Test of p = 0.5 vs p < 0.5 Sample X N Sample p Z-Value P-Value 1 373 1003 0.371884 -8.11 0.000

d. The required sample size depends on the confidence level and the sample proportion, not the population size.


1003

0.372 0.58.11z


( )8.11 0.0001P z= <− = (Tech: 0.0000). Reject H0. There is sufficient evidence to support the claim that

fewer than 50% of Americans say that they have a gun in their home.

Chapter 9: Inferences from Two Samples 139


Chapter 9: Inferences from Two Samples Section 9-2

1. The samples are simple random samples that are independent. For each of the two groups, the number of successes is at least 5 and the number of failures is at least 5. (Depending on what we call a success, the four numbers are 33, 115, 201,229 and 200,745 and all of those numbers are at least 5.) The requirements are satisfied.

2. 1 201,229n = , 1

33ˆ 0.000163992

201,299p = = , 1ˆ 1 0.000163992 0.999836q = − = , 2n =200,745,

2

115ˆ 0.000572866

200,745p = = , 2ˆ 1 0.000572866 0.999427q = − = ,

33 1150.000368183

201,299 200,745p

+= =

+, and 1 0.000368183 0.999632q = − = .

3. a. H0: 1 2p p= . H1: 1 2p p< .

b. If the P-value is less than 0.001 we should reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the rate of polio is less for children given the Salk vaccine than it is for children given a placebo.

4. a. 0.90, or 90%

b. Because the confidence interval limits do not contain 0, there appears to be a significant difference between the two proportions. Because the confidence interval consists of negative values only, it appears that the first proportion is less than the second proportion. There is sufficient evidence to support the claim that the rate of polio is less for children given the Salk vaccine than it is for children given a placebo.

c. The P-value method and the critical value method are equivalent in the sense that they will always lead to the same conclusion, but the confidence interval method is not equivalent to them.

5. Test statistic: 12.39z =− (rounded). The P-value of 3.137085E–35 is 0.0000 when rounded to four decimal places. There is sufficient evidence to warrant rejection of the claim that the vaccine has no effect.

6. Test statistic: 2.17z = . P-value: 0.030. Because the P-value is greater than the significance level of 0.01, conclude that there is not sufficient evidence to warrant rejection of the claim that for those saying that monitoring e-mail is seriously unethical, the proportion of workers is the same as the proportion of managers.

For Exercises 7 – 18, assume that the data fit the requirements for the statistical methods for two proportions unless otherwise indicated.

7. a. H0: 1 2p p= . H1: 1 2p p> . Test statistic: 6.44z = . Critical value: 2.33z = . P-value: 0.0001 (Tech:

0.0000). Reject H0. There is sufficient evidence to support the claim that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25.

MINITAB Difference = p (1) - p (2) Test for difference = 0 (vs > 0): Z = 6.44 P-Value = 0.000

b. 98% CI: 0.117 1 2p p< − < 0.240. Because the confidence interval limits do not include 0, it appears

that the two proportions are not equal. Because the confidence interval limits include only positive values, it appears that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25.

MINITAB Difference = p (1) - p (2) 98% CI for difference: (0.116836, 0.240360)

140 Chapter 9: Inferences from Two Samples


7. (continued)

c. The results suggest that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25, but the results cannot be used to verify the cause of that difference.

8. a. H0: 1 2p p= . H1: 1 2p p< . Test statistic: 1.66z =− . Critical value: 2.33z =− . P-value: 0.0485

(Tech: 0.0484). Fail to reject H0. There is not sufficient evidence to support the claim that the rate of dementia among those who use ginkgo is less than the rate of dementia among those who use a placebo. There is not sufficient evidence to support the claim that ginkgo is effective in preventing dementia.

MINITAB Difference = p (1) - p (2) Test for difference = 0 (vs < 0): Z = -1.66 P-Value = 0.048

b. 98% CI: –0.0542 1 2p p< − < 0.00909 (Tech: –0.0541 1 2p p< − < 0.00904). Because the confidence

interval limits include 0, there does not appear to be a significant difference between dementia rates for those treated with ginkgo and those given a placebo. There is not sufficient evidence to support the claim that the rate of dementia among those who use ginkgo is less than the rate of dementia among those who use a placebo. There is not sufficient evidence to support the claim that ginkgo is effective in preventing dementia.

MINITAB Difference = p (1) - p (2) 98% CI for difference: (-0.0541115, 0.00904103)

c. The sample results suggest that ginkgo is not effective in preventing dementia.

9. a. H0: 1 2p p= . H1: 1 2p p> . Test statistic: 6.11z = . Critical value: 1.64z = 5. P-value: 0.0001 (Tech:

0.0000). Reject H0. There is sufficient evidence to support the claim that the fatality rate is higher for those not wearing seat belts.

MINITAB Test for difference = 0 (vs > 0): Z = 6.11 P-Value = 0.000

b. 90% CI: 0.00556 1 2p p< − < 0.0122. Because the confidence interval limits do not include 0, it

appears that the two fatality rates are not equal. Because the confidence interval limits include only positive values, it appears that the fatality rate is higher for those not wearing seat belts.


c. The results suggest that the use of seat belts is associated with lower fatality rates than not using seat belts.

10. a. H0: 1 2p p= . H1: 1 2p p≠ . Test statistic: 18.26z = . Critical values: 2.575z =± (Tech: 2.576± ). P-

value: 0.0002 (Tech: 0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim that the survival rates are the same for day and night.

MINITAB Difference = p (1) - p (2) Test for difference = 0 (vs not = 0): Z = 18.26 P-Value = 0.000

b. 99% CI: 0.0441 1 2p p< − < 0.0579. Because the confidence interval limits do not contain 0, there

appears to be a significant difference between the two proportions. There is sufficient evidence to warrant rejection of the claim that the survival rates are the same for day and night.


c. The data suggest that for in-hospital patients who suffer cardiac arrest, the survival rate is not the same for day and night. It appears that the survival rate is higher for in-hospital patients who suffer cardiac arrest during the day.



11. a. H0: 1 2p p= . H1: 1 2p p≠ . Test statistic: 0.57z = . Critical values: 1.96z =± . P-value: 0.5686

(Tech: 0.5720). Fail to reject H0. There is not sufficient evidence to support the claim that echinacea treatment has an effect.


b. 95% CI: –0.0798 1 2p p< − < 0.149. Because the confidence interval limits do contain 0, there is not

a significant difference between the two proportions. There is not sufficient evidence to support the claim that echinacea treatment has an effect.


c. Echinacea does not appear to have a significant effect on the infection rate. Because it does not appear to have an effect, it should not be recommended.

12. a. H0: 1 2p p= . H1: 1 2p p< . Test statistic: 2.44z =− . Critical value: 2.33z =− . P-value: 0.0074.

Reject H0. There is sufficient evidence to support the claim that the incidence of malaria is lower for infants who use the bednets.


b. 98% CI: 0.0950 1 2p p< − < –0.00118 (Tech: –0.0950 1 2p p< − < –0.00125). Because the

confidence interval does not include 0 and it includes only negative values, it appears that the rate of malaria is lower for infants who use the bednets.

MINITAB Difference = p (1) - p (2) 98% CI for difference: (-0.0949568, -0.00125315)

c. The bednets appear to be effective.

13. a. H0: 1 2p p= . H1: 1 2p p≠ . Test statistic: 0.40z = . Critical values: 1.96z =± . P-value: 0.6892

(Tech: 0.6859). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that men and women have equal success in challenging calls.


b. 95% CI: –0.0318 1 2p p< − < 0.0484. Because the confidence interval limits contain 0, there is not a

significant difference between the two proportions. There is not sufficient evidence to warrant rejection of the claim that men and women have equal success in challenging calls.


c. It appears that men and women have equal success in challenging calls.

14. a. H0: 1 2p p= . H1: 1 2p p≠ . Test statistic: 1.91z = . Critical values: 1.96z =± . P-value: 0.0562 (Tech:

0.0567). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that New York City police and Los Angeles police have the same proportion of hits.


b. 95% CI:–0.000455 1 2p p< − < 0.130 (Tech: –0.000454 1 2p p< − < 0.130). Because the confidence

interval limits contain 0, there does not appear to be a significant difference between the two proportions. There is not sufficient evidence to warrant rejection of the claim that New York City police and Los Angeles police have the same proportion of hits.



14. (continued)


c. There does not appear to be a difference between the hit rates of New York City police and Los Angeles police.

15. a. H0: 1 2p p= . H1: 1 2p p> . Test statistic: 9.97z = . Critical value: 2.33z = . P-value: 0.0001 (Tech:

0.0000). Reject H0. There is sufficient evidence to support the claim that the cure rate with oxygen treatment is higher than the cure rate for those given a placebo. It appears that the oxygen treatment is effective.


b. 98% CI: 0.467 1 2p p< − < 0.687. Because the confidence interval limits do not include 0, it appears

that the two cure rates are not equal. Because the confidence interval limits include only positive values, it appears that the cure rate with oxygen treatment is higher than the cure rate for those given a placebo. It appears that the oxygen treatment is effective.


c. The results suggest that the oxygen treatment is effective in curing cluster headaches.

16. a. H0: 1 2p p= . H1: 1 2p p< . Test statistic: 1.85z =− . Critical value: 1.64z =− 5. P-value: 0.0322

(Tech: 0.0324). Reject H0. There is sufficient evidence to support the claim that when given a single large bill, a smaller proportion of women in China spend some or all of the money when compared to the proportion of women in China given the same amount in smaller bills.


b. 90% CI: –0.201 1 2p p< − < –0.0127. Because the confidence interval does not include 0 and it

includes only negative values, it appears that the first proportion is less than the second proportion. There is sufficient evidence to support the claim that when given a single large bill, a smaller proportion of women in China spend some or all of the money when compared to the proportion of women in China given the same amount in smaller bills.

MINITAB Difference = p (1) - p (2) 90% CI for difference: (-0.200605, -0.0127280)

c. Because the P-value is 0.0322 (Tech: 0.0324), the difference is significant at the 0.05 significance level, but not at the 0.01 significance level. The conclusion does change.

17. a. H0: 1 2p p= . H1: 1 2p p< . Test statistic: 1.17z =− . Critical value: 2.33z =− . P-value: 0.1210

(Tech: 0.1214). Fail to reject H0. There is not sufficient evidence to support the claim that the rate of left-handedness among males is less than that among females.


b. 98% CI: –0.0849 1 2p p< − < 0.0265 (Tech: –0.0848 1 2p p< − < 0.0264). Because the confidence

interval limits include 0, there does not appear to be a significant difference between the rate of left-handedness among males and the rate among females. There is not sufficient evidence to support the claim that the rate of left-handedness among males is less than that among females.



17. (continued)


c. The rate of left-handedness among males does not appear to be less than the rate of left-handedness among females.

18. a. H0: 1 2p p= . H1: 1 2p p≠ . Test statistic: 2.30z = . Critical values: 2.575z =± (Tech: 2.576± ). P-

value: 0.0214 (Tech: 0.0213). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the rate of those who finish is the same for men and women.


b. 99% CI: –0.000409 1 2p p< − < 0.00553 (Tech: –0.000409 1 2p p< − < 0.00554). Because the

confidence interval limits contain 0, there does not appear to be a significant difference between the two proportions. There is not sufficient evidence to warrant rejection of the claim that the rate of those who finish is the same for men and women.


c. It appears that men and women finish the New York City marathon at the same rate.

19. a. 0.0227 1 2p p< − < 0.217; because the confidence interval limits do not contain 0, it appears that

1 2p p= can be rejected.


b. 0.491 1p< <0.629; 0.371 2p< <0.509; because the confidence intervals do overlap, it appears that

1 2p p= cannot be rejected.

MINITAB Sample X N Sample p 95% CI 1 112 200 0.560000 (0.488250, 0.629944) 2 88 200 0.440000 (0.370056, 0.511750

c. H0: 1 2p p= . H1: 1 2p p≠ . Test statistic: 2.40z = . P-value: 0.0164. Critical values: 1.96z =± .

Reject H0. There is sufficient evidence to reject 1 2p p= .


d. Reject 1 2p p= . Least effective: Using the overlap between the individual confidence intervals.

20. Hypothesis test: With a test statistic of 1.96z =− 15, P-value = 0.05 (Tech: 0.0498), reject 1 2p p= .

Confidence interval: –0.422 1 2p p< − < 0.0180, which suggests that we should not reject 1 2p p=

(because 0 is included). The hypothesis test and confidence interval lead to different conclusions about the equality of 1 2p p= .

MINITAB Difference = p (1) - p (2) 95% CI for difference: (-0.422046, 0.0180456) Test for difference = 0 (vs not = 0): Z = -1.96 P-Value = 0.050

21. 2 2

/ 22 2

1.6453383

2 2 0.02αznE

= = =⋅

(Tech: 3382)



Section 9-3

1. Independent: b, d, e

2. –17.32 cm 1 2μ μ< − < –11.61 cm

3. Because the confidence interval does not contain 0, it appears that there is a significant difference between the mean height of women and the mean height of men. Based on the confidence interval, it appears that the mean height of men is greater than the mean height of women.

4. a. Yes.

b. 90%

5. a. H0: 1 2μ μ= . H1: 1 2μ μ≠ . Test statistic: 2.979t =− . Critical values: 2.032t =± (Tech: 2.002± ). P-

value < 0.01 (Tech: 0.0042). Reject H0. There is sufficient evidence to warrant rejection of the claim that the samples are from populations with the same mean. Color does appear to have an effect on creativity scores. Blue appears to be associated with higher creativity scores.

MINITAB Difference = mu (1) - mu (2) T-Test of difference = 0 (vs not =): T-Value = -2.98 P-Value = 0.004 DF = 58

b. 95% CI: –0.98 1 2μ μ< − < –0.18 (Tech: –0.97 1 2μ μ< − < –0.19)

MINITAB Difference = mu (1) - mu (2) 95% CI for difference: (-0.970, -0.190)

6. a. H0: 1 2μ μ= . H1: 1 2μ μ≠ . Test statistic: 2.647t = . Critical values: 2.032t =± (Tech: 1.995± ). P-

value < 0.02 (Tech: 0.0101). Reject H0. There is sufficient evidence to warrant rejection of the claim that the samples are from populations with the same mean. Color does appear to have an effect on word recall scores. Red appears to be associated with higher word memory recall scores.

MINITAB Difference = mu (1) - mu (2) T-Test of difference = 0 (vs not =): T-Value = 2.65 P-Value = 0.010 DF = 68

b. 95% CI: 0.83 1 2μ μ< − < 6.33 (Tech: 0.88 1 2μ μ< − < 6.28)

MINITAB Difference = mu (1) - mu (2) 95% CI for difference: (0.88, 6.28)

7. a. H0: 1 2μ μ= . H1: 1 2μ μ> . Test statistic: 0.132t = . Critical value: 1.729t = . P-value > 0.10 (Tech:

0.4480). Fail to reject H0. There is not sufficient evidence to support the claim that the magnets are effective in reducing pain. It is valid to argue that the magnets might appear to be effective if the sample sizes are larger.

MINITAB Difference = mu (1) - mu (2) T-Test of difference = 0 (vs >): T-Value = 0.13 P-Value = 0.448 DF = 33

b. 90% CI: –0.61 1 2μ μ< − < 0.71 (Tech: –0.59 1 2μ μ< − < 0.69)

MINITAB Difference = mu (1) - mu (2) 90% CI for difference: (-0.592, 0.692)

8. a. H0: 1 2μ μ= . H1: 1 2μ μ< . Test statistic: 0.676t =− . Critical value: 2.345t =− (Tech: –2.337). P-

value > 0.10 (Tech: 0.2499). Fail to reject H0. There is not sufficient evidence to support the claim that the mean number of words spoken in a day by men is less than that for women.

MINITAB Difference = mu (1) - mu (2) T-Test of difference = 0 (vs <): T-Value = -0.68 P-Value = 0.250 DF = 364



8. (continued)

b. 98% CI: –2443.6 words 1 2μ μ< − < 1350.6 words (Tech: –2436.8 words 1 2μ μ< − < 1343.8 words)

MINITAB Difference = mu (1) - mu (2) 98% CI for difference: (-2437, 1344)

9. a. The sample data meet the loose requirement of having a normal distribution. H0: 1 2μ μ= . H1: 1 2μ μ> .

Test statistic: 0.852t = . Critical value: 2.426t = (Tech: 2.676). P-value > 0.10 (Tech: 0.2054). Fail to reject H0. There is not sufficient evidence to support the claim that men have a higher mean body temperature than women.


b. 98% CI: –0.54 F 1 2μ μ< − < 1.02 F (Tech: –0.51 F 1 2μ μ< − < 0.99 F)



value > 0.10 (Tech: 0.1316). Fail to reject H0. There is not sufficient evidence to support the claim that men and women have different mean body temperatures.


b. 99% CI: –0.19°F 1 2μ μ< − < 0.61°F (Tech: –0.17°F 1 2μ μ< − < 0.59°F)



value < 0.005 (Tech: 0.0004). Reject H0. There is sufficient evidence to support the claim that the mean maximal skull breadth in 4000 b.c. is less than the mean in a.d. 150.


b. 98% CI: –8.13 mm 1 2μ μ< − < –1.47 mm (Tech: –8.04 mm 1 2μ μ< − < –1.56 mm)


12. a. H0: 1 2μ μ= . H1: 1 2μ μ≠ . Test statistic: 0.941t =− . Critical value: 2.201t =± (Tech: 2.080).

P-value > 0.20 (Tech: 0.3573). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that Flight 1 and Flight 3 have the same mean arrival delay time.


b. 95% CI: –18.1 min 1 2μ μ< − < 7.3 min (Tech: –17.4 min 1 2μ μ< − < 6.6 min)





value < 0.005 (Tech: 0.0014). Reject H0. There is sufficient evidence to support the claim that students taking the nonproctored test get a higher mean than those taking the proctored test.


b. 98% CI: –25.54 1 2μ μ< − < –3.10 (Tech: –25.27 1 2μ μ< − < –3.37)


14. a. H0: 1 2μ μ= . H1: 1 2μ μ≠ . Test statistic: 0.770t =− . Critical values: 2.756t =± (Tech: 2.666± ). P-

value > 0.20 (Tech: 0.4443). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the samples are from populations with the same mean.


b. 99% CI: –18.17 1 2μ μ< − < 10.23 (Tech: –17.71 1 2μ μ< − < 9.77)



value > 0.20 (Tech: 0.2066). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that males and females have the same mean BMI.


b. 95% CI: –1.08 1 2μ μ< − < 4.76 (Tech: –1.04 1 2μ μ< − < 4.72)


16. a. H0: 1 2μ μ= . H1: 1 2μ μ> . Test statistic: 2.282t = . Critical value: 1.725t = (Tech: 2.004).

P-value < 0.05 (Tech: 0.0132). Reject H0. There is sufficient evidence to support the claim that the mean IQ score of people with low lead levels is higher than the mean IQ score of people with high lead levels.


b. 90% CI: 1.5 1 2μ μ< − < 10.5 (Tech: 1.6 1 2μ μ< − < 10.4)


17. a. H0: 1 2μ μ= . H1: 1 2μ μ> . Test statistic: 0.089t = . Critical value: 1.725t = (Tech: 2.029).

P-value > 0.10 (Tech: 0.4648.) Fail to reject H0. There is not sufficient evidence to support the claim that the mean IQ score of people with medium lead levels is higher than the mean IQ score of people with high lead levels.

MINITAB Variable N Mean StDev LOW LEAD 22 87.23 14.29 Difference = mu (1) - mu (2) T-Test of difference = 0 (vs >): T-Value = 0.09 P-Value = 0.464 DF = 35



17. (continued)

b. 90% CI: –5.9 1 2μ μ< − < 6.6 (Tech: –5.8 1 2μ μ< − < 6.4)

MINITAB Difference = mu (1) - mu (2) Estimate for difference: 0.33 90% CI for difference: (-5.80, 6.45)


Test statistic: 12.533t = . Critical value: 2.821t = (Tech: 2.411). P-value < 0.005 (Tech: 0.0000). Reject H0. There is sufficient evidence to support the claim that supermodels have heights with a mean that is greater than the mean height of women in the general population. We can conclude that supermodels are taller than typical women.

MINITAB Variable N Mean StDev Model Height 10 69.825 0.800 Difference = mu (1) - mu (2) T-Test of difference = 0 (vs >): T-Value = 12.53 P-Value = 0.000 DF = 45

b. 98% CI: 4.7 in 1 2μ μ< − < 7.4 in. (Tech: 4.9 in. 1 2μ μ< − < 7.2 in.)



value > 0.025 (Tech: 0.0442). Fail to reject H0. There is not sufficient evidence to support the claim that the mean longevity for popes is less than the mean for British monarchs after coronation.

MINITAB Difference = mu (Popes) - mu (Kings and Queens) T-Test of difference = 0 (vs <): T-Value = -1.81 P-Value = 0.045 DF = 16

b. 98% CI: –23.6 years 1 2μ μ< − < 4.4 years (Tech: –23.2 years 1 2μ μ< − < 4.0 years)

MINITAB Difference = mu (Popes) - mu (Kings and Queens) 98% CI for difference: (-23.28, 4.10)

20. a. H0: 1 2μ μ= . H1: 1 2μ μ> . Test statistic: 3.265t = . Critical value: 1.796t = (Tech: 1.746t = ). P-

value < 0.005 (Tech: 0.0024). Reject H0. There is sufficient evidence to support the claim that the mean amount of strontium-90 from Pennsylvania residents is greater than the mean from New York residents.

MINITAB Difference = mu (Pennsylvania) - mu (New York) T-Test of difference = 0 (vs >): T-Value = 3.27 P-Value = 0.003 DF = 15

b. 90% CI: 5.0 mBq 1 2μ μ< − < 17.3 mBq (Tech: 5.2 mBq 1 2μ μ< − < 17.1 mBq)

MINITAB Difference = mu (Pennsylvania) - mu (New York) 90% CI for difference: (5.17, 17.16)

21. H0: 1 2μ μ= . H1: 1 2μ μ≠ . Test statistic: 32.773t = . Critical values: 2.023t =± (Tech: 1.994± ). P-value

< 0.01 (Tech: 0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim that the two populations have equal means. The difference is highly significant, even though the samples are relatively small.

MINITAB Difference = mu (Pre-1964 Quarters) - mu (Post-1964 Quarters) T-Test of difference = 0 (vs not =): T-Value = 32.77 P-Value = 0.000 DF = 70



22. –9.1 years 1 2μ μ< − < 5.4 years (Tech: –9.0 years 1 2μ μ< − < 5.3 years). Because the confidence

interval includes 0, there is not a significant difference between the two population means. It appears that the sample of men and the sample of women are from populations with the same mean.

MINITAB N Mean StDev MALE AGE 40 36.4 1 6.5 FEMALE AGE 40 38.3 15.6 Difference = mu (MALE AGE) - mu (FEMALE AGE) 95% CI for difference: (-9.00, 5.30)

23. 0.03795 lb 1 2μ μ< − < 0.04254 lb (Tech: 0.03786 lb 1 2μ μ< − < 0.04263 lb). Because the confidence

interval does not include 0, there appears to be a significant difference between the two population means. It appears that the cola in cans of regular Pepsi weighs more than the cola in cans of Diet Pepsi, and that is probably due to the sugar in regular Pepsi that is not in Diet Pepsi.

MINITAB N Mean StDev PPREGWT 36 0.82410 0.00570 PPDIETWT 36 0.78386 0.00436 Difference = mu (PPREGWT) - mu (PPDIETWT) 95% CI for difference: (0.03786, 0.04263)

24. H0: 1 2μ μ= . H1: 1 2μ μ≠ . Test statistic: 22.095t = . Critical values: 2.023t =± (Tech: 2.003± ).

P-value < 0.01 (Tech: 0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim that the two populations have equal means. The difference is due to the sugar in regular Coke that is not in diet Coke.

MINITAB N Mean StDev CKREGWT 36 0.81682 0.00751 CKDIETWT 36 0.78479 0.00439 Difference = mu (CKREGWT) - mu (CKDIETWT) T-Test of difference = 0 (vs not =): T-Value = 22.10 P-Value = 0.000 DF = 56


Test statistic: 1.046t = . Critical value: 2.381t = (Tech: 2.382). P-value > 0.10 (Tech: 0.1496). Fail to reject H0. There is not sufficient evidence to support the claim that men have a higher mean body temperature than women.

MINITAB Difference = mu (1) - mu (2) T-Test of difference = 0 (vs >): T-Value = 1.05 P-Value = 0.150 DF = 68 Both use Pooled StDev = 0.6986

b. 98% CI: –0.31°F 1 2μ μ< − < 0.79°F. The test statistic became larger, the P-value became smaller,

and the confidence interval became narrower, so pooling had the effect of attributing more significance to the results.

MINITAB Difference = mu (1) - mu (2) 98% CI for difference: (-0.307, 0.787) Both use Pooled StDev = 0.6986

26. a. H0: 1 2μ μ= . H1: 1 2μ μ< . Test statistic: 0.682t =− . Critical value: 2.336t =− . P-value > 0.10

(Tech: 0.2477). Fail to reject H0. There is not sufficient evidence to support the claim that the mean number of words spoken in a day by men is less than that for women.

MINITAB Difference = mu (1) - mu (2) T-Test of difference = 0 (vs <): T-Value = -0.68 P-Value = 0.248 DF = 394 Both use Pooled StDev = 7954.1009



26. (continued)

b. 98% CI: –2417.4 words 1 2μ μ< − < 1324.4 words (Tech: –2417.2 words 1 2μ μ< − < 1324.3 words).

The test statistic became larger, the P-value became smaller, and the confidence interval became narrower, so pooling had the effect of attributin

MINITAB 98% CI for difference: (-2417, 1324) Both use Pooled StDev = 7954.1009

27. H0: 1 2μ μ= . H1: 1 2μ μ≠ . Test statistic: 15.322t = . Critical values: 2.080t =± . P-value < 0.01 (Tech:

0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim that the two populations have the same mean.

( ) ( )1 2

2 2

0.049 0.000

22 22p p

μ μt

s s

− − −=

+

; ( ) ( )

( ) ( )

2 22 22 1 0.015 22 1 0

0.00012522 1 22 1ps

− + −= =

− + −

28. df = 77.3502249. Using “df = smaller of 1 1n − and 2 1n − ” is a more conservative estimate of the number

of degrees of freedom (than the estimate obtained with Formula 9-1) in the sense that the confidence interval is wider, so the difference between the sample means needs to be more extreme to be considered a significant difference.

29. a. H0: 1 2μ μ= . H1: 1 2μ μ< . Test statistic: 3.002t =− . Critical value based on 68.9927614 degrees of

freedom: 2.381t =− (Tech: –2.382). P-value < 0.005 (Tech: 0.0019). Reject H0. There is sufficient evidence to support the claim that students taking the nonproctored test get a higher mean than those taking the proctored test.

b. –25.68 1 2μ μ< − < –2.96 (Tech: –25.69 1 2μ μ< − < –2.95)

Section 9-4

1. Parts (c) and (e) are true.

2. 2.4d = mi/gal and 1.1ds = mi/gal. dμ represents the mean of the differences from the population of

paired data.

3. The test statistic will remain the same. The confidence interval limits will be expressed in the equivalent values of km/L.

4. The first confidence interval shows that we have 95% confidence that the limits of 1.0 mi/gal and 3.8 mi/gal contain the mean of the population of differences, but the second confidence interval shows that we have 95% confidence that the limits of –7.8 mi/gal and 12.6 mi/gal contain the difference between the two population means. Because the first confidence interval does not include 0 mi/gal and consists of positive values only, it appears that the old ratings are higher than the new ratings. Because the second confidence interval does include 0 mi/gal, there does not appear to be a significant different between the mean of the old ratings and the mean of the new ratings.

5. H0: 0dμ = cm. H1: 0dμ > cm. Test statistic: 0.036t = (rounded). Critical value: 1.692t = .

P-value > 0.10 (Tech: 0.4859). Fail to reject H0. There is not sufficient evidence to support the claim that for the population of heights of presidents and their main opponents; the differences have a mean greater than 0 cm (with presidents tending to be taller than their opponents).

6. –2.7 cm dμ< < 2.8 cm. The confidence interval includes 0 cm, so it is very possible that the mean of the

differences is equal to 0 cm, indicating that there is no significant difference between heights of presidents and heights of their opponents.



7. a. 11.6d =− years b. 17.2ds = years

c. Test statistic 11.6 0

1.50817.2 5

d

d

d μt

s n

− − −= = =−

d. H0: 0dμ = . H1: 0dμ ≠ . Critical values: 2.776t =±

8. a. 0.35 Fd =− b. 0.30 Fds =

c. Test statistic: 0.35 0

2.3330.30 4

d

d

d μt

s n

− − −= = =−

d. H0: 0dμ = . H1: 0dμ ≠ . Critical values: 3.182t =±

9. H0: 0dμ = . H1: 0dμ ≠ . Test statistic: 11.6 0

1.50717.21 5

t− −

= =− . Critical values: 2.776t =± .

P-value > 0.20 (Tech: 0.2063). Fail to reject H0. There is not sufficient evidence to support the claim that there is a difference between the ages of actresses and actors when they win Oscars.

MINITAB Paired T for Actress - Actor T-Test of mean difference = 0 (vs not = 0): T-Value = -1.51 P-Value = 0.206


2.3330.30 4

t− −


P-value > 0.10 (Tech: 0.1018). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that there is no difference between body temperatures measured at 8 A.M. and at 12 A.M.

MINITAB Paired T for 8A.M. - 12A.M. T-Test of mean difference = 0 (vs not = 0): T-Value = -2.33 P-Value = 0.102

11. 1.0 min dμ< < 12.0 min. Because the confidence interval includes only positive values and does not

include 0 min, it appears that the taxi-out times are greater than the corresponding taxi-in times, so there is sufficient evidence to support the claim of the flight operations manager that for flight delays, more of the blame is attributable to taxi-out times at JFK than taxi-in times at LAX.

MINITAB Paired T for Out - In 90% CI for mean difference: (0.99, 12.01)

6.5d = min ; df 12 1 11= − =

/ 2

10.631.796 5.5

12d

αs

E tn

= ⋅ = ⋅ = min

12. –66.8 cm3 dμ< < 49.8 cm3 (Tech: –66.7 cm3 dμ< < 49.7 cm3). Because the confidence interval includes

0 cm3, the mean of the differences could be equal to 0 cm3, so there does not appear to be a significant difference.

MINITAB Paired T for First Born - Second Born 99% CI for mean difference: (-66.7, 49.7)

8.5d =− cm3 ; df 10 1 9= − =

/ 2

56.73.250 58.3

10d

αs

E tn

= ⋅ = ⋅ = cm3

13. H0: 0dμ = . H1: 0dμ > . Test statistic: 7279 0

2.5796913 6

t−


P-value < 0.025 (Tech: 0.0247). Reject H0. There is sufficient evidence to support the claim that among couples, males speak more words in a day than females.

MINITAB Paired T for Male - Female T-Test of mean difference = 0 (vs > 0): T-Value = 2.58 P-Value = 0.025




17.3399.31 4

t− −


P-value > 0.01 (Tech: 0.0001). Reject H0. There is sufficient evidence to support the claim of a difference in measurements between the two arms. The right and left arms should yield the same measurements, but the given data show that this is not happening.

MINITAB Paired T for Right arm - Left arm T-Test of mean difference = 0 (vs not = 0): T-Value = -17.34 P-Value = 0.000

15. –6.5 dμ< < –0.2. Because the confidence interval does not include 0, it appears that there is sufficient

evidence to warrant rejection of the claim that when the 13th day of a month falls on a Friday, the numbers of hospital admissions from motor vehicle crashes are not affected. Hospital admissions do appear to be affected.

MINITAB Paired T for Friday the 6th - Friday the 13th 95% CI for mean difference: (-6.49, -0.17)

3.33d =− cm3 ; df 6 1 5= − =

/ 2

3.012.571 3.2

6d

αs

E tn

= ⋅ = ⋅ = cm3

16. –4.2 in. dμ< < 2.2 in. Because the confidence interval limits contain 0, there is not sufficient evidence to

support a claim that there is a difference between self-reported heights and measured heights. We might believe that males would tend to exaggerate their heights, but the given data do not provide enough evidence to support that belief.

MINITAB Paired T for Reported - Measured 99% CI for mean difference: (-4.16, 2.16)

1.0d =− in. ; df 12 1 11= − =

/ 2

3.523.106 3.2

12d

αs

E tn

= ⋅ = ⋅ = in.

17. H0: 0dμ = . H1: 0dμ < . Test statistic: 1.57 0

1.0804.60 10

t− −

= =− . Critical value: 1.833t =− .

P-value > 0.10 (Tech: 0.1540). Fail to reject H0. There is not sufficient evidence to support the claim that Harry Potter and the Half-Blood Prince did better at the box office. After a few years, the gross amounts from both movies can be identified, and the conclusion can then be judged objectively without using a hypothesis test.

MINITAB Paired T for Phoenix - Prince T-Test of mean difference = 0 (vs < 0): T-Value = -1.08 P-Value = 0.154

18. H0: 0dμ = . H1: 0dμ > . Test statistic: 18.58 0

6.37110.10 12

t−


P-value < 0.005 (Tech: 0.00003). Reject H0. There is sufficient evidence to support the claim that Captopril is effective in lowering systolic blood pressure.

MINITAB Paired T for Before - After T-Test of mean difference = 0 (vs < 0): T-Value = 6.37 P-Value = 1.000

19. 0.69 dμ< < 5.56. Because the confidence interval limits do not contain 0 and they consist of positive

values only, it appears that the “before” measurements are greater than the “after” measurements, so hypnotism does appear to be effective in reducing pain.

MINITAB Paired T for Before - After 95% CI for mean difference: (0.69, 5.56)

3.13d = ; df 8 1 7= − =

/ 2

2.912.365 2.43

8d

αs

E tn

= ⋅ = ⋅ =



20. –7.3°F dμ< < 6.3°F. Because the confidence interval limits do contain 0°F, there is not a significant

difference between the actual high temperatures and those that were forecast five days earlier. This suggests that the forecast temperatures are reasonably accurate.

MINITAB Paired T for Actual High - Forecast High 99% CI for mean difference: (-7.28, 6.28)

0.5d =− °F ; df 8 1 7= − =

/ 2

5.483.500 6.8

8d

αs

E tn

= = = °F

21. H0: 0dμ = . H1: 0dμ ≠ . Test statistic: 5.553t =− . Critical values: 1.990t =± . P-value < 0.01 (Tech:

0.0000). Reject H0. There is sufficient evidence to support the claim that there is a difference between the ages of actresses and actors when they win Oscars.

MINITAB Paired T for Actresses - Actors T-Test of mean difference = 0 (vs not = 0): T-Value = -5.55 P-Value = 0.000

22. H0: 0dμ = . H1: 0dμ ≠ . Test statistic: 0.124t = . Critical values: 2.028t =± . P-value > 0.20 (Tech:

0.9023). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that there is no difference between body temperatures measured at 8 a.m. and at 12 a.m.

MINITAB Paired T for 8 AM - 12 AM T-Test of mean difference = 0 (vs not = 0): T-Value = 0.12 P-Value = 0.902

23. H0: 0dμ = . H1: 0dμ < . Test statistic: 1.560t =− . Critical value of t is between –1.671 and –1.676

(Tech: –1.673). P-value > 0.05 (Tech: 0.0622). Fail to reject H0. There is not sufficient evidence to support the claim that among couples, males speak fewer words in a day than females.

MINITAB Paired T for M1 - F1 95% upper bound for mean difference: 135 T-Test of mean difference = 0 (vs < 0): T-Value = -1.56 P-Value = 0.062

24. H0: 0dμ = sec. H1: 0dμ > sec. Test statistic: 0.938t = . Critical value: 1.694t = . P-value > 0.10 (Tech:

0.1776). Fail to reject H0. There is not sufficient evidence to support the claim that the mean of the differences is greater than 0 sec. There is not sufficient evidence to support the claim that more time is devoted to showing tobacco than alcohol. For animated children’s movies, no time should be spent showing the use of tobacco or alcohol.

MINITAB Paired T for Tobacco Use (sec) - Alcohol Use (sec) T-Test of mean difference = 0 (vs > 0): T-Value = 0.94 P-Value = 0.176

25. H0: 6.8dμ = kg. H1: 6.8dμ ≠ kg. Test statistic: 11.833t =− . Critical values: 1.994t =± (Tech:

1.997± ). P-value < 0.01 (Tech: 0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim that 6.8dμ = kg. It appears that the “Freshman 15” is a myth, and college freshman might gain some

weight, but they do not gain as much as 15 pounds. MINITAB Paired T for WTAPR - WTSEP T-Test of mean difference = 6.8 (vs not = 6.8): T-Value = -11.83 P-Value = 0.000

Section 9-5

1. a. No.

b. No.

c. The two samples have the same standard deviation (or variance).



2. a. 2 21 6.60 43.56s = = cm2 and 2 2

2 6.02 36.2404s = = cm2

b. H0: 2 21 2σ σ=

c. 2122

43.561.2120

36.2404

sF

s= = =

d. There is not sufficient evidence to support the claim that heights of men and heights of women have different variances.

3. The F test is very sensitive to departures from normality, which means that it works poorly by leading to wrong conclusions when either or both of the populations has a distribution that is not normal. The F test is not robust against sampling methods that do not produce simple random samples. For example, conclusions based on voluntary response samples could easily be wrong.

4. No. Unlike some other tests which have a requirement that samples must be from normally distributed populations or the samples must have more than 30 values, the F test has a requirement that the samples must be from normally distributed populations, regardless of how large the samples are.

5. H0: 1 2σ σ= . H1: 1 2σ σ≠ . Test statistic: 1.7341F = . Upper critical F value is between 1.8752 and 2.0739

(Tech: 1.9611). P-value: 0.1081. Fail to reject H0. There is not sufficient evidence to support the claim that weights of regular Coke and weights of regular Pepsi have different standard deviations.

6. H0: 1 2σ σ= . H1: 1 2σ σ> . Test statistic: 1.0110F = . Critical F value is less than 1.3519 (Tech: 1.2848).

P-value: 0.4745. Fail to reject H0. There is not sufficient evidence to support the claim that ages of student cars vary more than the ages of faculty cars.

7. H0: 1 2σ σ= . H1: 1 2σ σ≠ . Test statistic: 2

2

5.901.1592

5.48F = = . Upper critical F value is between 1.8752

and 2.0739 (Tech: 1.9678). P-value: 0.6656. Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the samples are from populations with the same standard deviation. The background color does not appear to have an effect on the variation of word recall scores.

MINITAB Test for Equal Variances F-Test (Normal Distribution) Test statistic = 1.16, p-value = 0.666

8. H0: 1 2σ σ= . H1: 1 2σ σ> . Test statistic: 2

2

8632.51.3979

7301.2F = = . Critical F value is between 1.0000 and

1.3519 (Tech: 1.2642). P-value: 0.0094. Reject H0. There is sufficient evidence to support the claim that the numbers of words spoken in a day by men vary more than the numbers of words spoken in a day by women.



2

2.29.3364


2.0960 (Tech: 2.0842). P-value: 0.0000. Reject H0. There is sufficient evidence to support the claim that the treatment group has errors that vary more than the errors of the placebo group.

MINITAB F-Test (Normal Distribution) Test statistic = 9.34, p-value = 0.000




2

0.891.8184


2.0772 (Tech: 1.9983). P-value: 0.0774. Fail to reject H0. There is not sufficient evidence to support the claim that men have body temperatures that vary more than the body temperatures of women.



2

1.42.1267


2.2341 (Tech: 2.1682). P-value: 0.0543. Fail to reject H0. There is not sufficient evidence to support the claim that those given a sham treatment (similar to a placebo) have pain reductions that vary more than the pain reductions for those treated with magnets.



2

5.351.0876


and 2.1540 (Tech: 2.1010). P-value: 0.8226. Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the variation of maximal skull breadths in 4000 b.c. is the same as the variation in a.d. 150.



2

10.63834.1648


2.8536 (Tech: 2.8179). P-value: 0.0130. Reject H0. There is sufficient evidence to support the claim that amounts of strontium-90 from Pennsylvania residents vary more than amounts from New York residents.



2

18.60284.3103


and 2.5699 (Tech: 2.5308). P-value: 0.0023. Reject H0. There is sufficient evidence to warrant rejection of the claim that both populations of longevity times have the same variation.

MINITAB Test for Equal Variances: Kings/Queens, Popes F-Test (Normal Distribution) Test statistic = 4.31, p-value = 0.002


2

6.064651.0073

6.04264F = = . Upper critical F value: 4.0260. P-value:

0.9915. Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that females and males have heights with the same amount of variation.





2

22.66271.7619

17.0734F = = . Critical F value: 3.1789. P-value:

0.2058. Fail to reject H0. There is not sufficient evidence to support the claim that males have weights with more variation than females.



2

20.68831.2397


1.8409 (Tech: 1.7045). P-value: 0.2527. Fail to reject H0. There is not sufficient evidence to support the claim that males have weights with more variation than females.



2

0.05761.3213

0.0501F = = . Upper critical F value: 2.5411 (Tech:

2.5412). P-value: 0.5399. Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same amount of variation.


19. a. No solution provided.

b. 1 4c = , 2 0c =

c. Critical value = ( )log 0.05 / 2

540

log40 40

=⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜⎝ ⎠+

.

d. Fail to reject 2 21 2σ σ= .

20. Test statistic: 2.055t = . Critical values: 2.023t =± (Tech: 1.996± ). P-value > 0.05 (Tech: 0.0438). Using Table A-3 with df = the smaller of 1 1n − and 2 1n − , fail to reject 2 2

1 2σ σ= . Using technology with

df found from Formula 9-1, reject 2 21 2σ σ= .

MINITAB Difference = mu (pre) - mu (post) T-Test of difference = 0 (vs not =): T-Value = 2.05 P-Value = 0.044 DF = 67

21. 0.2727LF = , 2.8365RF =

Chapter Quick Quiz

1. H0: 1 2p p= . H1: 1 2p p≠ .

2. 347 305

0.875386 359

p+

= =+

3. ( )2 2.04 0.0414P z⋅ > = 0.0414

4. 0.00172 1 2p p< − < 0.0970

5. Because the data consist of matched pairs, they are dependent.

6. H0: 0dμ = . H1: 0dμ > .

7. There is not sufficient evidence to support the claim that front repair costs are greater than the corresponding rear repair costs.

8. F distribution



9. False. 10. True.

Review Exercises

1. H0: 1 2p p= . H1: 1 2p p> . Test statistic: 3.12z = . Critical value: 2.33z = . P-value: 0.0009. Reject H0.

There is sufficient evidence to support a claim that the proportion of successes with surgery is greater than the proportion of successes with splinting. When treating carpal tunnel syndrome, surgery should generally be recommended instead of splinting.


2. 98% CI: 0.0581 1 2p p< − < 0.332 (Tech: 0.0583 1 2p p< − < 0.331). The confidence interval limits do

not contain 0; the interval consists of positive values only. This suggests that the success rate with surgery is greater than the success rate with splints.


3. H0: 1 2p p= . H1: 1 2p p< . Test statistic: 1.91z =− . Critical value: 1.645z =− . P-value: 0.0281 (Tech:

0.0280). Reject H0. There is sufficient evidence to support the claim that the fatality rate of occupants is lower for those in cars equipped with airbags.


4. H0: 0dμ = . H1: 0dμ > . Test statistic: 4.712t = . Critical value: 3.143t = . P-value < 0.005 (Tech:

0.0016). Reject H0. There is sufficient evidence to support the claim that flights scheduled 1 day in advance cost more than flights scheduled 30 days in advance. Save money by scheduling flights 30 days in advance.

MINITAB Paired T for Flight scheduled one day in adv - Flight scheduled 30 days in adv T-Test of mean difference = 0 (vs > 0): T-Value = 4.71 P-Value = 0.002

5. H0: 0dμ = . H1: 0dμ ≠ . Test statistic: 0.574t =− . Critical values: 2.365t =± . P-value > 0.20 (Tech:

0.5840). Fail to reject H0. There is not sufficient evidence to support the claim that there is a difference between self-reported heights and measured heights of females aged 12–16.

MINITAB Paired T for Reported Height - Measured Height T-Test of mean difference = 0 (vs not = 0): T-Value = -0.57 P-Value = 0.584

6. H0: 1 2μ μ= . H1: 1 2μ μ> . Test statistic: 2.879t = . Critical value: 2.429t = (Tech: 2.376). P-value <

0.005 (Tech: 0.0026). Reject H0. There is sufficient evidence to support the claim that “stress decreases the amount recalled.”


7. 98% CI: 1.3 1 2μ μ< − < 14.7 (Tech: 1.4 1 2μ μ< − < 14.6). The confidence interval limits do not contain

0; the interval consists of positive values only. This suggests that the numbers of details recalled are lower for those in the stress population.




8. H0: 1 2p p= . H1: 1 2p p≠ . Test statistic: 4.20z =− . Critical values: 2.575z =± . P-value: 0.0002 (Tech:

0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim that the acceptance rate is the same with or without blinding. Without blinding, reviewers know the names and institutions of the abstract authors, and they might be influenced by that knowledge.

MINITAB Difference = p (1) - p (2) Test for difference = 0 (vs not = 0): Z = -4.20 P-Value = 0.000

9. H0: 1 2μ μ= . H1: 1 2μ μ≠ . Test statistic: 0.679t = . Critical values: 2.014t =± approximately (Tech:

1.985± ). P-value > 0.20 (Tech: 0.4988). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim of no difference between the mean LDL cholesterol levels of subjects treated with raw garlic and subjects given placebos. Both groups appear to be about the same.


10. H0: 1 2σ σ= . H1: 1 2σ σ≠ . Test statistic: 1.1480F = . Upper critical F value is between 1.6668 and 1.8752

(Tech: 1.7799). P-value: 0.6372. Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the two populations have LDL levels with the same standard deviation.

MINITAB F-Test (Normal Distribution) Test statistic = 1.15, p-value = 0.637


1. a. Because the sample data are matched with each column consisting of heights from the same family, the data are dependent.

b. Mean: 63.81 in.; median: 63.70 in.; mode: 62.2 in.; range: 8.80 in.; standard deviation: 2.73 in.; variance: 7.43 in2

c. Ratio

2. There does not appear to be a correlation or association between the heights of mothers and the heights of their daughters.

69686766656463626160

69

68

67

66

65

64

63

62

61

60

Heights of Mothers (in.)

Hei

ghts

of

Dau

ghte

rs (

in.)

3. 61.86 in. μ< < 65.76 in. We have 95% confidence that the limits of 61.86 in. and 65.76 in. actually

contain the true value of the mean height of all adult daughters.

MINITAB Variable N Mean StDev SE Mean 95% CI Daughters 10 63.810 2.726 0.862 (61.860, 65.760)



4. H0: 0dμ = . H1: 0dμ ≠ . Test statistic: 0.283t = . Critical values: 2.262t =± . P-value > 0.20 (Tech:

0.7834). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim of no significant difference between the heights of mothers and the heights of their daughters.

MINITAB Paired T for Heights of Mothers (in.) - Heights of Daughters (in.) T-Test of mean difference = 0 (vs not = 0): T-Value = 0.28 P-Value = 0.783

5. Because the points lie reasonably close to a straight-line pattern and there is no other pattern that is not a straight-line pattern and there are no outliers, the sample data appear to be from a population with a normal distribution.

6. 0.109 1p< < 0.150. Because the entire range of values in the confidence interval lies below 0.20, the

results do justify the statement that “fewer than 20% of Americans choose their computer and/or Internet access when identifying what they miss most when electrical power is lost.”


7. No. Because the Internet users chose to respond, we have a voluntary response sample, so the results are not necessarily valid.

8. [ ] [ ] ( )2 2

/ 2

2 2

ˆ ˆ 2.17 0.252944

0.02αz pq

nE

= = = . The survey should not be conducted using only local phone

numbers. Such a convenience sample could easily lead to results that are dramatically different from results that would be obtained by randomly selecting respondents from the entire population, not just those having local phone numbers.

9. a. 152.1 162.0

1.5;6.6

z−

= =− ( )1.5 0.9332.P z >− =

b. 152.1 162.0

3;6.6 4

z−

= =− ( )1.5 0.0.9987.P z >− =

c. 80th percentile: 162.0 0.842 6.6 167.6 cm.x μ z σ= + ⋅ = + ⋅ =

10. No. Because the states have different population sizes, the mean cannot be found by adding the 50 state means and dividing the total by 50. The mean income for the U.S. population can be found by using a weighted mean that incorporates the population size of each state.

Chapter 10: Correlation and Regression 159


Chapter 10: Correlation and Regression Section 10-2

1. r represents the value of the linear correlation computed by using the paired sample data. ρ represents the

value of the linear correlation coefficient that would be computed by using all of the paired data in the population. The value of r is estimated to be 0 (because there is no correlation between sunspot numbers and the Dow Jones Industrial Average).

2. No. The value of 0r = suggests that there is no linear relationship, but there might be some other relationship that is not linear in the sense that the pattern of points in the scatterplot is not a straight-line pattern.

3. The headline is not justified because it states that increased salt consumption is the cause of higher blood pressure levels, but the presence of a correlation between two variables does not necessarily imply that one is the cause of the other. Correlation does not imply causality. A correct headline would be this: “Study Shows That Increased Salt Consumption Is Associated with Higher Blood Pressure.”

4. Table A-6 shows that the critical values of r are 0.312± (assuming a 0.05 significance level), so there is sufficient evidence to support a claim of a linear correlation between the before and after weights. The value of r does not indicate that the diet is effective in reducing weight. While the diet might be effective in reducing weight, there could be a linear correlation if the diet has no effect so that the before and after weights are about the same, or there could be a linear correlation if the diet causes people to gain weight.

5. H0: 0ρ= . H1: 0;ρ≠ Yes. With 0.687r = and critical values of 0.312± , there is sufficient evidence to

support the claim that there is a linear correlation between the durations of eruptions and the time intervals to the next eruptions.

6. H0: 0ρ= . H1: 0;ρ≠ No. With 0.091r = and critical values of 0.312± , there is not sufficient evidence

to support the claim that there is a linear correlation between the durations of eruptions and the heights of eruptions.

7. H0: 0ρ= . H1: 0;ρ≠ No. With 0.149r = and a P-value of 0.681 (or critical values of 0.632± ), there is

not sufficient evidence to support the claim that there is a linear correlation between the heights of fathers and the heights of their sons.

8. H0: 0ρ= . H1: 0;ρ≠ Yes. With 0.765r = and critical values of 0.497± , there is sufficient evidence to

support the claim that there is a linear correlation between calories and sugar in a gram of cereal.

9. a.

160 Chapter 10: Correlation and Regression


9. (continued)

b. H0: 0ρ= . H1: 0;ρ≠ 0.816.r = Critical values: 0.602.r =± P-value = 0.002. There is sufficient

evidence to support the claim of a linear correlation between the two variables. MINITAB Pearson correlation of x and y = 0.816 P-Value = 0.002

c. The scatterplot reveals a distinct pattern that is not a straight line pattern.

10. a.

15.012.510.07.55.0

13

12

11

10

9

8

7

6

5

x

y

b. H0: 0ρ= . H1: 0;ρ≠ 0.816.r = Critical values: 0.602.r =± P-value = 0.002. There is sufficient

evidence to support the claim of a linear correlation between the two variables. MINITAB Pearson correlation of x and y = 0.816 P-Value = 0.002

c. The scatterplot reveals a perfect straight-line pattern, except for the presence of one outlier.

11. a. There appears to be a linear correlation.

b. H0: 0ρ= . H1: 0;ρ≠ 0.906.r = Critical values: 0.632r =± (for a 0.05 significance level). There is

a linear correlation. MINITAB Pearson correlation of x and y = 0.906 P-Value = 0.000

c. H0: 0ρ= . H1: 0;ρ≠ 0r = . Critical values: 0.666r =± (for a 0.05 significance level). There does

not appear to be a linear correlation. MINITAB Pearson correlation of x and y = 0.000 P-Value = 1.000

d. The effect from a single pair of values can be very substantial, and it can change the conclusion.

12. a. There does not appear to be a linear correlation.

b. There does not appear to be a linear correlation.

c. H0: 0ρ= . H1: 0;ρ≠ 0r = . Critical values: 0.950r =± (for a 0.05 significance level). There does

not appear to be a linear correlation. The same results are obtained with the four points in the upper right corner.

MINITAB Pearson correlation of x and y = 0.000 P-Value = 1.000



12. (continued)

d. H0: 0ρ= . H1: 0;ρ≠ 0.985r = . Critical values: 0.707r =± (for a 0.05 significance level). There is

a linear correlation. MINITAB Pearson correlation of x and y = 0.985 P-Value = 0.000

e. There are two different populations that should be considered separately. It is misleading to use the combined data from women and men and conclude that there is a relationship between x and y.

13. H0: 0ρ= . H1: 0;ρ≠ 0.959r =− . Critical values: 0.878r =± . P-value = 0.010. There is sufficient

evidence to support the claim that there is a linear correlation between weights of lemon imports from Mexico and U.S. car fatality rates. The results do not suggest any cause-effect relationship between the two variables.

MINITAB Pearson correlation of Lemon Imports and Fatality Rate = -0.959 P-Value = 0.010

550500450400350300250200

16.0

15.8

15.6

15.4

15.2

15.0

Lemon Imports

Fata

lity

Rat

e

Scatterplot of Fatality Rate vs Lemon Imports

14. H0: 0ρ= . H1: 0;ρ≠ 0.543r = . Critical values: 0.707r =± . P-value = 0.164. There is not sufficient

evidence to support the claim that there is a linear correlation between PSAT scores and SAT scores. Because the data are from a voluntary response sample, the results are very questionable.

MINITAB Pearson correlation of PSAT and SAT = 0.543 P-Value = 0.164

210200190180170160150140

2400

2300

2200

2100

2000

1900

PSAT

SAT

Scatterplot of SAT vs PSAT



15. H0: 0ρ= . H1: 0;ρ≠ 0.561.r = Critical values: 0.632r =± . P-value = 0.091. There is not sufficient

evidence to support the claim that there is a linear correlation between enrollment and burglaries. The results do not change if the actual enrollments are listed as 32,000, 31,000, 53,000, etc.

MINITAB Pearson correlation of Enrollment and Burglaries = 0.561 P-Value = 0.091

555045403530

160

140

120

100

80

60

40

20

0

Enrollment

Bu

rgla

ries

Scatterplot of Burglaries vs Enrollment

16. H0: 0ρ= . H1: 0;ρ≠ 0.997.r =− Critical values: 0.754.r =± P-value = 0.000. There is sufficient

evidence to support the claim that there is a linear correlation between altitude and outside air temperature. The results do not change if the altitudes are converted to meters and the temperatures are converted to the Celsius scale.

MINITAB Pearson correlation of Alt and Temp = -0.997 P-Value = 0.000

35302520151050

50

25

0

-25

-50

Alt

Tem

p

Scatterplot of Temp vs Alt

17. H0: 0ρ= . H1: 0;ρ≠ 0.864.r = Critical values: 0.666.r =± P-value = 0.003. There is sufficient

evidence to support the claim that there is a linear correlation between court incomes and justice salaries. The correlation does not imply that court incomes directly affect justice salaries, but it does appear that justices might profit by levying larger fines, or perhaps justices with higher salaries impose larger fines.

MINITAB Pearson correlation of Income and Salary = 0.864 P-Value = 0.003

16001400120010008006004002000

100

90

80

70

60

50

40

30

20

10

Income

Sala

ry

Scatterplot of Salary vs Income




evidence to support the claim that there is a linear correlation between the opening bids suggested by the auctioneer and the final winning bids.

MINITAB Pearson correlation of Open and Win = 0.947 P-Value = 0.015

1600140012001000800600400200

700

600

500

400

300

200

100

Open

Win

Scatterplot of Win vs Open


evidence to support the claim that there is a linear correlation between amounts of redshift and distances to clusters of galaxies. Because the linear correlation coefficient is 1.000, it appears that the distances can be directly computed from the amounts of redshift.

MINITAB Pearson correlation of Red and Dist = 1.000 P-Value = 0.000

0.080.070.060.050.040.030.020.01

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

Red

Dis

t

Scatterplot of Dist vs Red


evidence to support the claim that there is a linear correlation between weights and prices. The results do not necessarily apply to other populations of diamonds, such as those with different color and clarity ratings.

MINITAB Pearson correlation of Weight and Price = 0.968 P-Value = 0.002

1.00.90.80.70.60.50.40.3

6000

5000

4000

3000

2000

1000

0

Weight

Pri

ce

Scatterplot of Price vs Weight




evidence to support the claim of a linear correlation between the overhead width of a seal in a photograph and the weight of a seal.

MINITAB Pearson correlation of Width and Weight = 0.948 P-Value = 0.004

10.09.59.08.58.07.57.0

260

240

220

200

180

160

140

120

100

Width

Wgh

t

Scatterplot of Weight vs Width

22. H0: 0ρ= . H1: 0;ρ≠ 0.283.r =− Critical values: 0.754.r =± P-value = 0.539. There is not sufficient

evidence to support the claim of a linear correlation between the repair costs from full-front crashes and full-rear crashes.

MINITAB Pearson correlation of Front and Rear = -0.283 P-Value = 0.539

45004000350030002500200015001000

3500

3000

2500

2000

1500

1000

Front

Rea

r

Scatterplot of Rear vs Front

23. H0: 0ρ= . H1: 0;ρ≠ 0.867.r = Critical values: 0.878.r =± P-value = 0.057. There is not sufficient

evidence to support the claim of a linear correlation between the systolic blood pressure measurements of the right and left arm.

MINITAB Pearson correlation of Right Arm and Left Arm = 0.867 P-Value = 0.057

10510095908580

180

170

160

150

140

Right Arm

Left

Arm

Scatterplot of Left Arm vs Right Arm




evidence to support the claim of a linear correlation between number of cricket chirps and temperature. MINITAB Pearson correlation of Chirps and Temp = 0.874 P-Value = 0.005

12001150110010501000950900850

95

90

85

80

75

70

Chirps

Tem

p(°F

)

Scatterplot of Temp(°F) vs Chirps


evidence to support the claim that there is a linear correlation between prices of regular gas and prices of premium gas. Because there does not appear to be a linear correlation between prices of regular and premium gas, knowing the price of regular gas is not very helpful in getting a good sense for the price of premium gas.

MINITAB Pearson correlation of Reg and Prem = 0.197 P-Value = 0.640

2.8752.8502.8252.8002.7752.750

3.09

3.08

3.07

3.06

3.05

3.04

3.03

3.02

3.01

3.00

Reg

Pre

m

Scatterplot of Prem vs Reg


evidence to support the claim that there is a linear correlation between prices of regular gas and prices of mid-grade gas. Because there does not appear to be a linear correlation between prices of regular and mid-grade gas, knowing the price of regular gas is not very helpful in getting a good sense for the price of mid-grade gas.

MINITAB Pearson correlation of Reg and Mid = 0.399 P-Value = 0.327

2.8752.8502.8252.8002.7752.750

3.00

2.95

2.90

2.85

2.80

Reg

Mid

Scatterplot of Mid vs Reg




evidence to support the claim that there is a linear correlation between diameters and circumferences. A scatterplot confirms that there is a linear association between diameters and volumes.

MINITAB Pearson correlation of Diam and Circ = 1.000 P-Value = 0.000

252015105

80

70

60

50

40

30

20

10

Diam

Cir

cum

Scatterplot of Circum vs Diam


evidence to support the claim that there is a linear correlation between diameters and volumes. Although the results suggest that there is a linear correlation between diameters and volumes, the scatterplot suggests that there is a very strong correlation that is not linear.

MINITAB Pearson correlation of Diam and Vol = 0.978 P-Value = 0.000

252015105

8000

7000

6000

5000

4000

3000

2000

1000

0

Diam

Vol

Scatterplot of Vol vs Diam

29. H0: 0ρ= . H1: 0;ρ≠ 0.063.r =− Critical values: 0.444.r =± P-value = 0.791. There is not sufficient

evidence to support the claim of a linear correlation between IQ and brain volume. MINITAB Pearson correlation of IQ and VOL = -0.063 P-Value = 0.791

1301201101009080

1500

1400

1300

1200

1100

1000

900

IQ

VO

L

Scatterplot of VOL vs IQ



30. H0: 0ρ= . H1: 0;ρ≠ 0.917.r = Critical values: 0.279r =± (approximately) (Tech: 0.285± ). P-value

= 0.000. There is sufficient evidence to support the claim of a linear correlation between departure delay times and arrival delay times.

MINITAB Pearson correlation of Dep Delay and Arr Delay = 0.917 P-Value = 0.000

1501251007550250

125

100

75

50

25

0

-25

-50

Dep Delay

Arr

Del

ay

Scatterplot of Arr Delay vs Dep Delay

31. H0: 0ρ= . H1: 0;ρ≠ 0.319.r = Critical values: 0.254r =± (approximately) (Tech: ±0.263). P-value =

0.017. There is sufficient evidence to support the claim of a linear correlation between the numbers of words spoken by men and women who are in couple relationships.

MINITAB Pearson correlation of M1 and F1 = 0.319 P-Value = 0.017

400003500030000250002000015000100005000

50000

40000

30000

20000

10000

0

F1

M1

Scatterplot of M1 vs F1


evidence to support the claim of a linear correlation between magnitudes of earthquakes and their depths. MINITAB Pearson correlation of MAG and DEPTH = 0.027 P-Value = 0.852

20151050

3.0

2.5

2.0

1.5

1.0

0.5

0.0

DEPTH

MA

G

Scatterplot of MAG vs DEPTH



33. a. 0.911r =

MINITAB Pearson correlation of y and x = 0.911 P-Value = 0.031

b. 0.787r =

MINITAB Pearson correlation of y and x^2 = 0.787 P-Value = 0.114

c. 0.9999r = (largest)

MINITAB Pearson correlation of y and LOG(x) = 1.000 P-Value = 0.000

d. 0.976r =

MINITAB Pearson correlation of y and SQRT(x) = 0.976 P-Value = 0.005

e. 0.948r =−

MINITAB Pearson correlation of y and 1/x = -0.948 P-Value = 0.014

34. 2 2

2.4850.445

2 2.485 27 2

tr

t n=± =± =±

+ − + −

Section 10-3

1. The symbol y represents the predicted pulse rate. The predictor variable represents height. The response

variable represents pulse rate.

2. The regression line has the property that the sum of squares of the residuals is the lowest possible sum (where a residual is the difference between an observed value of y and a predicted value of y).

3. If r is positive, the regression line has a positive slope and rises from left to right. If r is negative, the slope of the regression line is negative and it falls from left to right.

4. The first equation represents the regression line that best fits sample data, whereas the second equation represents the regression line that best fits all paired data in a population.

5. The regression line fits the points well, so the best predicted time for an interval after the eruption is ( )ˆ 47.4 0.180 120 69 min.y = + =

6. The regression line does not fit the points well, so the best predicted height is 127.2 ft.y =

7. The regression line does not fit the points well, so the best predicted height is 68.0 in.y =

8. The regression line fits the points well, so the best predicted value is ( )ˆ 3.46 1.01 0.40y = +

3.86 calories.=

9. ˆ 3.00 0.500y x= + . The data have a pattern that is not a straight line.

MINITAB Predictor Coef SE Coef T P Constant 3.001 1.125 2.67 0.026x 0.5000 0.1180 4.24 0.002

15.012.510.07.55.0

10

9

8

7

6

5

4

3

x

y

Scatterplot of y vs x



10. ˆ 3.00 0.500y x= + . There is an outlier.

MINITAB Predictor Coef SE Coef T P Constant 3.002 1.124 2.67 0.026x 0.4997 0.1179 4.24 0.002

15.012.510.07.55.0

13

12

11

10

9

8

7

6

5

4

x

y

Scatterplot of y vs x

11. a. ˆ 0.264 0.906y x= +

MINITAB Predictor Coef SE Coef T P Constant 0.2642 0.5649 0.47 0.653 x 0.9057 0.1499 6.04 0.000

b. ˆ 2 0y x= + (or ˆ 2y = )

MINITAB Predictor Coef SE Coef T P Constant 2.0000 0.8165 2.45 0.044 x -0.0000 0.3780 -0.00 1.000

c. The results are very different, indicating that one point can dramatically affect the regression equation.

12. a. ˆ 0.0846 0.985y x= +


b. ˆ 1.5 0y x= + (or ˆ 1.5y = )


c. ˆ 9.5 0y x= + (or ˆ 9.5y = )


d. The results are very different, indicating that combinations of clusters can produce results that differ dramatically from results within each cluster alone.



13. ˆ 16.5 0.00282y x= − ; The regression line fits the points well, so the best predicted value is

( )ˆ 16.5 0.00282 500 15.1y = − = fatalities per 100,000 population.

MINITAB Predictor Coef SE Coef T P Constant 16.4909 0.1880 87.70 0.000 Lemon -0.00282 0.0004815 -5.86 0.010

550500450400350300250200

16.0

15.8

15.6

15.4

15.2

15.0

Lemon Imports

Fata

lity

Rat

e

Scatterplot of Fatality Rate vs Lemon Imports

550500450400350300250200

0.20

0.15

0.10

0.05

0.00

-0.05

-0.10

Lemon Imports

RES

I1

Scatterplot of Residuals vs Lemon Imports

0.30.20.10.0-0.1-0.2-0.3

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity

Probability Plot of ResidualsNormal

14. ˆ 1314 4.56y x= + ; The regression line does not fit the points well, so the best predicted value is 2153.y =

The result is not close to the actual reported value of 2400. Because the data are from a voluntary response sample, the results have questionable validity.

MINITAB Predictor Coef SE Coef T P Constant 1313.7 532.5 2.47 0.049 PSAT 4.562 2.878 1.59 0.164

210200190180170160150140

2400

2300

2200

2100

2000

1900

PSAT

SAT

Scatterplot of SAT vs PSAT



14. (continued)

210200190180170160150140

200

100

0

-100

-200

PSAT

Res

idu

als

Scatterplot of Residuals vs PSAT

4003002001000-100-200-300-400

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity


15. ˆ 36.8 3.47y x=− + ; The regression line does not fit the points well, so the best predicted value is 87.7y =

burglaries. The predicted value is not close to the actual value of 329 burglaries.

MINITAB Predictor Coef SE Coef T P Constant -36.77 66.50 -0.55 0.595 Enrollment 3.467 1.807 1.92 0.091

555045403530

160

140

120

100

80

60

40

20

0

Enrollment

Bu

rgla

ries

Scatterplot of Burglaries vs Enrollment

555045403530

50

25

0

-25

-50

-75

Enrollment

Res

idu

als

Scatterplot of Residuals vs Enrollment

100500-50-100

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity




16. ˆ 72.5 3.68y x= − ; The best predicted value is ( )ˆ 72.5 3.68 6.327 49.2°F.y = − = The predicted value is

close to the actual value of 48°F. MINITAB Predictor Coef SE Coef T P Constant 72.498 3.017 24.03 0.000 Altitude -3.6843 0.1326 -27.78 0.000

35302520151050

75

50

25

0

-25

-50

Altitude

Tem

p(°F

)

Scatterplot of Temp(°F) vs Altitude

35302520151050

4

3

2

1

0

-1

-2

-3

-4

-5

Altitude

Res

idu

als

Scatterplot of Residuals vs Altitude

1050-5-10

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity


17. ˆ 27.7 0.0373y x= + ; The best predicted value is ( )ˆ 27.7 0.0373 83.941 30.8,y = + = which represents

$30,800. The predicted value is not very close to the actual salary of $26,088. The possible outliers might explain the inaccuracy.

MINITAB Predictor Coef SE Coef T P Constant 27.701 5.519 5.02 0.002 Income 0.03728 0.008201 4.55 0.003

16001400120010008006004002000

100

90

80

70

60

50

40

30

20

10

Court Income

Just

ice

Sala

ry

Scatterplot of Justice Salary vs Court Income



17. (continued)

16001400120010008006004002000

25

20

15

10

5

0

-5

-10

-15

Court Income

Res

idu

als

Scatterplot of Residuals vs Court Income

3020100-10-20-30

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity


18. ˆ 4.62 0.429y x=− + ; The best predicted value is ( )ˆ 4.62 0.429 300 $124.y =− + = The predicted value is

not very close to the actual winning bid of $250. The one influential outlier would account for this inaccuracy.

MINITAB Predictor Coef SE Coef T P Constant -4.62 65.17 -0.07 0.948 Opening Bid 0.4291 0.0841 5.10 0.015

1600140012001000800600400200

700

600

500

400

300

200

100

Opening Bid

Win

nin

g B

id

Scatterplot of Winning Bid vs Opening Bid

1600140012001000800600400200

100

50

0

-50

-100

Opening Bid

Res

idu

als

Scatterplot of Residuals vs Opening Bid

2001000-100-200

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity




19. ˆ 0.00440 14.0y x=− + ; The best predicted value is ( )ˆ 0.00440 14.0 0.0126 0.172y =− + = billion light-

years. The predicted value is very close to the actual distance of 0.18 light-years. MINITAB Predictor Coef SE Coef T P Constant -0.004396 0.00125 -3.51 0.025 Redshift 13.9999 0.0278 503.40 0.000

0.080.070.060.050.040.030.020.01

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

Redshift

Dis

tan

ce

Scatterplot of Distance vs Redshift

0.080.070.060.050.040.030.020.01

0.0015

0.0010

0.0005

0.0000

-0.0005

-0.0010

-0.0015

-0.0020

Redshift

Res

idu

als

Scatterplot of Residuals vs Redshift

0.0030.0020.0010.000-0.001-0.002-0.003

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity


20. ˆ 2010 7180y x=− + ; best predicted value is ( )ˆ 2010 7180 1.5 $8760.y =− + = (Tech: $8759). The

predicted value is far from the actual price of $16,097. The weight of 1.50 carats is well beyond the scope of the available sample weights, so the extrapolation might be off by a considerable amount.

MINITAB Predictor Coef SE Coef T P Constant -2007.0 571.8 -3.51 0.025 Weight 7177.0 935.8 7.67 0.002

1.00.90.80.70.60.50.40.3

6000

5000

4000

3000

2000

1000

0

Weight

Pri

ce

Scatterplot of Price vs Weight



20. (continued)

1.00.90.80.70.60.50.40.3

500

250

0

-250

-500

-750

Weight

Res

idu

als

Scatterplot of Residuals vs Weight

10005000-500-1000

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity


21. ˆ 157 40.2y x=− + ; The best predicted weight is ( )ˆ 157 40.2 2 76.6 kg.y =− + =− (Tech: –76.5 kg). That

prediction is a negative weight that cannot be correct. The overhead width of 2 cm is well beyond the scope of the available sample widths, so the extrapolation might be off by a considerable amount.

MINITAB Predictor Coef SE Coef T P Constant -156.88 57.41 -2.73 0.052 Width 40.182 6.712 5.99 0.004

10.09.59.08.58.07.57.0

260

240

220

200

180

160

140

120

100

Width

Wei

ght

Scatterplot of Weight vs Width

10.09.59.08.58.07.57.0

15

10

5

0

-5

-10

-15

-20

Width

Res

idu

als

Scatterplot of Residuals vs Width

403020100-10-20-30-40

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity




22. ˆ 2060 0.186y x= − ; The regression line does not fit the data well, so the best predicted cost is $1615.y =

The predicted cost of $1615 is very different from the actual cost of $982. MINITAB Predictor Coef SE Coef T P Constant 2062.6 782.0 2.64 0.046 Front -0.1856 0.2818 -0.66 0.539

45004000350030002500200015001000

3500

3000

2500

2000

1500

1000

Front

Rea

r

Scatterplot of Rear vs Front

45004000350030002500200015001000

1500

1000

500

0

-500

-1000

Front

Res

idu

als

Scatterplot of Residuals vs Front

200010000-1000-2000

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity


23. ˆ 43.6 1.31y x= + ; The regression line does not fit the data well, so the best predicted value is

163.2 mm Hg.y =

MINITAB Predictor Coef SE Coef T P Constant 43.56 39.93 1.09 0.355 Right Arm 1.3147 0.4361 3.01 0.057

10510095908580

180

170

160

150

140

Right Arm

Left

Arm

Scatterplot of Left Arm vs Right Arm



23. (continued)

10510095908580

15

10

5

0

-5

-10

Right Arm

Res

idu

als

Scatterplot of Residuals vs Right Arm

20100-10-20

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity


24. ˆ 27.6 0.0523y x= + ; best predicted value is ( )ˆ 27.6 0.0523 3000y = + = 185°F (Tech: 184°F). The value of

3000 chirps in 1 minute is well beyond the scope of the available sample data, so the extrapolation might be off by a considerable amount.

MINITAB Predictor Coef SE Coef T P Constant 27.63 12.17 2.27 0.064 Chirps 0.05227 0.01188 4.40 0.005

12001150110010501000950900850

95

90

85

80

75

70

Chirps

Tem

p(°F

)

Scatterplot of Temp(°F) vs Chirps

12001150110010501000950900850

5.0

2.5

0.0

-2.5

-5.0

-7.5

Chirps

Res

idu

als

Scatterplot of Residuals vs Chirps

1050-5-10

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity




25. ˆ 2.57 0.172y x= + ; The regression line does not fit the data well, so the best predicted value is $3.05.y =

The predicted price is not very close to the actual price of $2.93. MINITAB Predictor Coef SE Coef T P Constant 2.5662 0.9732 2.64 0.039 Regular 0.1718 0.3491 0.49 0.640

2.8752.8502.8252.8002.7752.750

3.09

3.08

3.07

3.06

3.05

3.04

3.03

3.02

3.01

3.00

Regular

Pre

miu

m

Scatterplot of Premium vs Regular

2.8752.8502.8252.8002.7752.750

0.050

0.025

0.000

-0.025

-0.050

Regular

Res

idu

als

Scatterplot of Residuals vs Regular

0.080.060.040.020.00-0.02-0.04-0.06-0.08

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity



$2.91.y = The predicted price is not too far from the actual price.

MINITAB Predictor Coef SE Coef T P Constant 0.640 2.125 0.30 0.773 Regular 0.8129 0.7621 1.07 0.327

2.8752.8502.8252.8002.7752.750

3.00

2.95

2.90

2.85

2.80

Regular

Mid

-Gra

de

Scatterplot of Mid-Grade vs Regular



26. (continued)

2.8752.8502.8252.8002.7752.750

0.10

0.05

0.00

-0.05

-0.10

-0.15

Regular

Res

idu

als

Scatterplot of Residuals vs Regular

0.150.100.050.00-0.05-0.10-0.15

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity


27. ˆ 0.00396 3.14y x=− + ; The best predicted value is ( )ˆ 0.00396 3.14 1.50 4.7 cm.y =− + = Even though

the diameter of 1.50 cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference.

MINITAB Predictor Coef SE Coef T P Constant -0.00396 0.01883 -0.21 0.840 Diameter 3.14274 0.00129 2443.98 0.000

252015105

80

70

60

50

40

30

20

10

Diameter

Cir

cum

fere

nce

Scatterplot of Circumference vs Diameter

252015105

0.04

0.03

0.02

0.01

0.00

-0.01

-0.02

-0.03

-0.04

-0.05

Diameter

Res

idu

als

Scatterplot of Residuals vs Diameter

0.0500.0250.000-0.025-0.050-0.075

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity




28. ˆ 2010 347y x=− + ; The best predicted value is ( ) 3ˆ 2010 347 1.50 1489.5 cmy =− + =− (Tech: –1489.8

cm3). The predicted value is negative and is far from the actual volume of 1.8 cm3. The diameter of 1.50 cm is beyond the scope of the sample diameters, and the predicted value is way wrong. The scatterplot and residual plot suggest that a nonlinear model would yield better results.

MINITAB Predictor Coef SE Coef T P Constant -2010.7 441.0 -4.56 0.004 Diameter 347.30 30.11 11.53 0.000

252015105

8000

7000

6000

5000

4000

3000

2000

1000

0

-1000

Diameter

Vol

um

e

Scatterplot of Volume vs Diameter

252015105

1000

500

0

-500

-1000

Diameter

Res

idu

als

Scatterplot of Residuals vs Diameter

150010005000-500-1000-1500

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity


29. ˆ 109 0.00670y x= − ; The regression line does not fit the data well, so the best predicted IQ score is

101y = .

MINITAB Predictor Coef SE Coef T P Constant 108.55 28.17 3.85 0.001 VOLUME -0.00670 0.02487 -0.27 0.791

150014001300120011001000900

130

120

110

100

90

80

VOLUME

IQ

Scatterplot of IQ vs VOLUME



29. (continued)

150014001300120011001000900

30

20

10

0

-10

-20

VOLUME

Res

idu

als

Scatterplot of Residuals vs VOLUME

3020100-10-20-30

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity


30. ˆ 18.4 904y x=− + ; best predicted arrival delay time is ( )ˆ 18.4 904 0 18.4 minutes.y =− + =− That is, if a

flight has no departure delay, we can predict that the flight will arrive 18.4 minutes early. MINITAB Predictor Coef SE Coef T P Constant -18.366 1.870 -9.82 0.000 Dep Delay 0.90356 0.05801 15.58 0.000

1501251007550250

125

100

75

50

25

0

-25

-50

Dep Delay

Arr

Del

ay

Scatterplot of Arr Delay vs Dep Delay

1501251007550250

30

20

10

0

-10

-20

-30

Dep Delay

Res

idu

als

Scatterplot of Residuals vs Dep Delay

3020100-10-20-30

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity




31. ˆ 13, 400 0.302y x= + ; The best predicted value is ( )ˆ 13, 400 0.302 10,000 16,400y = + = (Tech: 16,458).

MINITAB Predictor Coef SE Coef T P Constant 13439 2239 6.00 0.000 M1 0.3019 0.1222 2.47 0.017

50000400003000020000100000

40000

35000

30000

25000

20000

15000

10000

5000

M1

F1

Scatterplot of F1 vs M1

50000400003000020000100000

20000

15000

10000

5000

0

-5000

-10000

-15000

M1

Res

idu

als

Scatterplot of Residuals vs M1

20000100000-10000-20000

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity



9.81 kmy = .

MINITAB Predictor Coef SE Coef T P Constant 9.535 1.625 5.87 0.000 MAG 0.231 1.232 0.19 0.852

3.02.52.01.51.00.50.0

20

15

10

5

0

MAG

DEP

TH

Scatterplot of DEPTH vs MAG



32. (continued)

3.02.52.01.51.00.50.0

10

5

0

-5

-10

MAG

Res

idu

als

Scatterplot of Residuals vs MAG

1050-5-10-15

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity


33. With 1 0β = , the regression line is horizontal so that different values of x result in the same y value, and

there is no correlation between x and y.

34. a. 61.8

b. The sum of squares of the residuals is 101.3, which is larger than 61.8.

ˆ 125.4 1.727y x= + y y− ( )2y y− ˆ 120 2.00y x= + y y− ( )2

y y−

176.6919 1.3919 1.937386 179.4 4.1 16.81

176.6919 –1.1081 1.227886 179.4 1.6 2.56

179.6278 –5.7722 33.31829 182.8 –2.6 6.76

180.3186 5.0186 25.18635 183.6 8.3 68.89

173.0652 0.3652 0.133371 175.2 2.5 6.25 Sum 61.80328 Sum 101.27

Section 10-4

1. The value of 17.5436 cmes = is the standard error of estimate, which is a measure of the differences

between the observed weights and the weights predicted from the regression equation. It is a measure of the variation of the sample points about the regression line.

2. We have 95% confidence that the limits of 50.7 kg and 123.0 kg contain the value of the weight for a male with a height of 180 cm. The major advantage of using a prediction interval is that it provides us with a range of likely weights, so we have a sense of how accurate the predicted weight is likely to be. The terminology of prediction interval is used for an interval estimate of a variable, whereas the terminology of confidence interval is used for an interval estimate of a population parameter.

3. The coefficient of determination is ( )22 0.356 0.127.r = = We know that 12.7% of the variation in weight

is explained by the linear correlation between height and weight, and 87.3% of the variation in weight is explained by other factors and/or random variation.

4. For the paired weights, 0es = because there is an exact conversion formula. For a textbook that weighs

4.5 lb, the predicted weight is 4.5

2.04 kg2.205

= , and there is no prediction interval because the conversion

yields an exact result.

5. ( )22 0.933 0.870.r = = 87.0% of the variation in waist size is explained by the linear correlation between

weight and waist size, and 13.0% of the variation in waist size is explained by other factors and/or random variation.



6. ( )22 0.963 0.927.r = = 92.7% of the variation in weight is explained by the linear correlation between

chest size and weight, and 7.3% of the variation in weight is explained by other factors and/or random variation.

7. ( )22 0.793 0.629.r = − = 62.9% of the variation in highway fuel consumption is explained by the linear

correlation between weight and highway fuel consumption, and 37.1% of the variation in highway fuel consumption is explained by other factors and/or random variation.

8. ( )22 0.751 0.564.r = = 56.4% of the variation in household size is explained by the linear correlation

between weight of discarded plastic and household size, and 43.6% of the variation in household size is explained by other factors and/or random variation.

9. 0.842.r = Critical values: 0.312r =± (assuming a 0.05 significance level). P-value = 0.000. There is sufficient evidence to support a claim of a linear correlation between foot length and height.

10. ( )20.842 0.709.r = = 70.9% of the variation in height is explained by the linear correlation between foot

length and height.

11. ( )ˆ 64.1 4.29 29.0 =189 cmy = +

12. 177 cm 200 cmy< < . We have 95% confidence that the limits of 177 cm and 200 cm contain the height

of someone with a foot length of 29.0 cm.

13. 160 cm 183 cmy< <

( )

( )( ) ( )

( )( )

( ) ( )

2 2

0/ 2 2 22

ˆ 64.1 4.29 25 171.35

40 25 25.681 11 2.024 5.50571 1 11.299

40Σ Σ 40 26530.92 1027.2α e

y

n x xE t s

n n x x

= + =

− −= + + = + + =

− −

14. 156 cm 186 cmy< < (Tech: 156 cm 187 cmy< < )

( )

( )( ) ( )

( )( )

( ) ( )

2 2

0/ 2 2 22

ˆ 64.1 4.29 25 171.35

40 25 25.681 11 2.712 5.50571 1 15.139

40Σ Σ 40 26530.92 1027.2α e

y

n x xE t s

n n x x

= + =

− −= + + = + + =

− −

15. 149 cm 168 cmy< <

( )

( )( ) ( )

( )( )

( ) ( )

2 2

0/ 2 2 22

ˆ 64.1 4.29 22 158.48

40 22 25.681 11 1.686 5.50571 1 9.797

40Σ Σ 40 26530.92 1027.2α e

y

n x xE t s

n n x x

= + =

− −= + + = + + =

− −

16. 164 cm 187 cmy< <

( )

( )( ) ( )

( )( )

( ) ( )

2 2

0/ 2 2 22

ˆ 64.1 4.29 26 175.64

40 26 25.681 11 2.204 5.50571 1 12.289

40Σ Σ 40 26530.92 1027.2α e

y

n x xE t s

n n x x

= + =

− −= + + = + + =

− −



17. a. 10,626.59

b. 68.83577

c. 38.0°F 60.4°Fy< <

MINITAB Analysis of Variance Source DF SS MS F P Regression 1 10627 10627 771.88 0.000 Residual Error 5 69 14 Total 6 10695 Predicted Values for New Observations Obs Fit SE Fit 95% CI 95% PI 1 49.19 2.31 (43.26, 55.12) (37.96, 60.42)

18. a. 3210.364

b. 1087.191

c. $10,400 $105,000y< <

MINITAB Analysis of Variance Source DF SS MS F P Regression 1 3210.4 3210.4 20.67 0.003 Residual Error 7 1087.2 155.3 Total 8 4297.6 Predicted Values for New Observations Obs Fit SE Fit 99% CI 99% PI 1 57.53 5.08 (39.75, 75.31) (10.43, 104.63)

19. a. 0.466276 b. 0.000007359976

c. 0.168 billion light-years 0.176 billion light-yearsy< <

MINITAB Analysis of Variance Source DF SS MS F P Regression 1 0.46628 0.46628 253411.69 0.000 Residual Error 4 0.00001 0.00000 Total 5 0.46628 Predicted Values for New Observations Obs Fit SE Fit 90% CI 90% PI 1 0.17200 0.00095 (0.16997, 0.17403) (0.16847, 0.17554)

20. a. 16,139,685

b. 1,097,655

c. $2051 $5419y< <

MINITAB Analysis of Variance Source DF SS MS F P Regression 1 16139685 16139685 58.82 0.002 Residual Error 4 1097655 274414 Total 5 17237340 Predicted Values for New Observations Obs Fit SE Fit 95% CI 95% PI 1 3735 306 (2886, 4583) (2051, 5419)



21. 058.9 103β< < ; 12.46 3.98β< <

CI for 0β

( )( )2 2

0 0 0

2 2

0 / 2 1160.7Σ240

1 1 29.0280.93; 2.024 5.94376 22.06

40 33933Σα e x

n

b E β b E

xb E t s

n x

− < < +

= = + = + =−−

CI for 1β

( )

1 1 1

1 / 2 2 2

2

5.943763.2186; 2.024 0.757

1160.7Σ 33933Σ 40

eα

b E β b E

sb E t

xx

n

− < < +

= = ⋅ = ⋅ =

−−

22. 172 cm 176 cmy< <

( )

( )( ) ( )

( )( )

( ) ( )

2 2

0/ 2 2 22

ˆ ˆ

ˆ 80.93 3.2186 29 174.269

40 29 29.021 12.024 5.94376 1.902

40Σ Σ 40 33933 1160.7α e

y E y y E

y

n x xE t s

n n x x

− < < +

= + =

− −= + = + =

− −

Section 10-5

1. The response variable is weight and the predictor variables are length and chest size.

2. No, it is not better to use the regression equation with the three predictor variables of length, chest size, and neck size. The adjusted 2R value of 0.925 is just a little less than 0.933, so in this case it is better to use two predictor variables instead of three.

3. The unadjusted 2R increases (or remains the same) as more variables are included, but the adjusted 2R is adjusted for the number of variables and sample size. The unadjusted 2R incorrectly suggests that the best multiple regression equation is obtained by including all of the available variables, but by taking into account the sample size and number of predictor variables, the adjusted 2R is much more helpful in weeding out variables that should not be included.

4. 92.8% of the variation in weights of bears can be explained by the variables of length and chest size, so 7.2% of the variation in weights can be explained by other factors and/or random variation.

5. LDL = 47.4 + 0.085 WT + 0.497 SYS.

6. a. 0.149

b. 9.8%, or 0.098

c. 4.9%, or 0.049

7. No. The P-value of 0.149 is not very low, and the values of 2R (0.098) and adjusted 2R (0.049) are not high. Although the multiple regression equation fits the sample data best, it is not a good fit.

8. Predicted ( ) ( )LDL 47.4 0.085 59.3 0.497 122 113 mg/dL.= + + = This result is not likely to be a good

predicted value because the multiple regression equation is not a good model (based on the results from Exercise 7).

9. HWY (highway fuel consumption) because it has the best combination of small P-value (0.000) and highest adjusted 2R (0.920).



10. WT (weight) and HWY (highway fuel consumption) because they have the best combination of small P-value (0.000) and highest adjusted 2R (0.935).

11. CITY = –3.15 + 0.819 HWY. That equation has a low P-value of 0.000 and its adjusted 2R value of 0.920 isn’t very much less than the values of 0.928 and 0.935 that use two predictor variables, so in this case it is better to use the one predictor variable instead of two.

12. Predicted city fuel consumption is ( )CITY 3.15 0.819 36 26.3 mi/gal=− + = (based on the result from

Exercise 11). The predicted value is a good estimate, but it might not be very accurate because the sample consists of only 21 cars.

13. The best regression equation is 1 2ˆ 0.127 0.0878 0.0250y x x= + − , where 1x represents tar and 2x

represents carbon monoxide. It is best because it has the highest adjusted 2R value of 0.927 and the lowest P-value of 0.000. It is a good regression equation for predicting nicotine content because it has a high value of adjusted 2R and a low P-value.

MINITAB Predictor Coef SE Coef T P Constant 0.08000 0.06611 1.21 0.239 100 Tar 0.063333 0.004832 13.11 0.000 S = 0.0869783 R-Sq = 88.2% R-Sq(adj) = 87.7% Predictor Coef SE Coef T P Constant 0.3281 0.1378 2.38 0.026 100 CO 0.039721 0.008967 4.43 0.000 S = 0.185937 R-Sq = 46.0% R-Sq(adj) = 43.7% Predictor Coef SE Coef T P Constant 0.12714 0.05230 2.43 0.024 100 Tar 0.087797 0.007062 12.43 0.000 100 CO -0.025004 0.006130 -4.08 0.000 S = 0.0671065 R-Sq = 93.3% R-Sq(adj) = 92.7%

14. The best regression equation is 1 2ˆ 0.251 0.101 0.0454y x x= + − , where 1x represents tar and 2x represents

carbon monoxide. It is best because it has the highest adjusted 2R value of 0.908 and the lowest P-value of 0.000. It is a good regression equation for predicting nicotine content because it has a high value of adjusted 2R and a low P-value.

MINITAB Predictor Coef SE Coef T P Constant 0.13884 0.08874 1.56 0.131 Menth Tar 0.056746 0.006609 8.59 0.000 S = 0.120760 R-Sq = 76.2% R-Sq(adj) = 75.2% Predictor Coef SE Coef T P Constant 0.3851 0.1559 2.47 0.021 Menth CO 0.03255 0.01005 3.24 0.004 S = 0.205218 R-Sq = 31.3% R-Sq(adj) = 28.3% Predictor Coef SE Coef T P Constant 0.25073 0.05699 4.40 0.000 Menth Tar 0.100692 0.008053 12.50 0.000 Menth CO -0.045432 0.007206 -6.30 0.000 S = 0.0737007 R-Sq = 91.5% R-Sq(adj) = 90.8%



15. The best regression equation is 1ˆ 109 0.00670y x= − , where 1x represents volume. It is best because it has

the highest adjusted 2R value of –0.0513 and the lowest P-value of 0.791. The three regression equations all have adjusted values of 2R that are very close to 0, so none of them are good for predicting IQ. It does not appear that people with larger brains have higher IQ scores.

MINITAB Predictor Coef SE Coef T P Constant 108.55 28.17 3.85 0.001 VOL -0.00670 0.02487 -0.27 0.791 S = 13.5455 R-Sq = 0.4% R-Sq(adj) = 0.0% Predictor Coef SE Coef T P Constant 101.14 12.46 8.11 0.000 WT -0.0018 0.1554 -0.01 0.991 S = 13.5728 R-Sq = 0.0% R-Sq(adj) = 0.0% Predictor Coef SE Coef T P Constant 108.26 29.72 3.64 0.002 VOL -0.00694 0.02616 -0.27 0.794 WT 0.0072 0.1631 0.04 0.965 S = 13.9375 R-Sq = 0.4% R-Sq(adj) = 0.0%

16. The best regression equation is 1 2ˆ 10.0 0.567 0.532y x x=− + + , where 1x represents verbal IQ score and

2x represents performance IQ score. It is best because it has the highest adjusted 2R value of 0.999 and the

lowest P-value of 0.000. Because the adjusted 2R is so close to 1, it is likely that predicted values will be very accurate.

MINITAB Predictor Coef S E Coef T P Constant 11.504 4.091 2.81 0.006 IQV 0.94024 0.04790 19.63 0.000 S = 7.03711 R-Sq = 76.4% R-Sq(adj) = 76.2% Predictor Coef SE Coef T P Constant 10.465 3.548 2.95 0.004 IQP 0.80620 0.03515 22.94 0.000 S = 6.22165 R-Sq = 81.6% R-Sq(adj) = 81.4% Predictor Coef SE Coef T P Constant -10.0200 0.3285 -30.50 0.000 IQV 0.566561 0.004261 132.95 0.000 IQP 0.532216 0.003537 150.49 0.000 S = 0.508785 R-Sq = 99.9% R-Sq(adj) = 99.9%

17. For H0: 1β = 0, the test statistic is 0.7072

5.4860.1289

t = = , the P-value is 0.000, and the critical values are

2.110t =± , so reject H0 and conclude that the regression coefficient of 1 0.707b = should be kept. For H0:

2β = 0, the test statistic is 0.1636

1.2920.1266

t = = , the P-value is 0.213, and the critical values are 2.110t =± ,

so fail to reject H0 and conclude that the regression coefficient of 2 0.164b = should be omitted. It appears

that the regression equation should include the height of the mother as a predictor variable, but the height of the father should be omitted.



18. 10.435 0.979β< < ; 20.104 0.431β− < < . The confidence interval for 2β includes 0, suggesting that the

father’s height be eliminated as a predictor variable.

CI for 1β

1 1 1

1 / 2 1 1 1 / 2 1

1

1

0.7072 2.11 0.1289 0.7072 2.11 0.1289

0.435 0.979

α α

b E β b E

b t s β b t s

ββ

− < < +

− < < +

− ⋅ < < + ⋅

< <

CI for 2β

2 2 2

2 / 2 2 2 2 / 2 2

2

2

0.1636 2.11 0.1266 0.1636 2.11 0.1266

0.104 0.431

α α

b E β b E

b t s β b t s

ββ

− < < +

− < < +

− ⋅ < < + ⋅

− < <

19. 1 2ˆ 3.06 82.4 2.91y x x= + + , where 1x represents sex and 2x represents age.

Female: ( ) ( )ˆ 3.06 82.4 0 2.91 20 61 lby = + + = ; male: ( ) ( )ˆ 3.06 82.4 1 2.91 20 144 lby = + + = . The sex of

the bear does appear to have an effect on its weight. The regression equation indicates that the predicted weight of a male bear is about 82 lb more than the predicted weight of a female bear with other characteristics being the same.

MINITAB Predictor Coef SE Coef T P Constant 3.06 22.46 0.14 0.892 SEX 82.38 20.80 3.96 0.000 AGE 2.9053 0.2974 9.77 0.000

Section 10-6

1. Since the area of a square is the square of its side, the best model is 2y x= ; quadratic; 2 1R =

2. Quadratic is best because it has the highest 2R value, but this is not a good model because the value of 2R is so low. Using the models discussed in this section, it appears that we cannot make accurate predictions of the numbers of points scored in future Super Bowl games. Common sense suggests that no such model could be found.

3. 10.3% of the variation in Super Bowl points can be explained by the quadratic model that relates the variable of year and the variable of points scored. Because such a small percentage of the variation is explained by the model, the model is not very useful.

4. Instead of showing a pattern that approximates the graph of the quadratic equation, the points are scattered about with no obvious pattern. The points do not fit the graph of the quadratic equation well, so the value of

2 0.103R = is very low.



5. Quadratic: 24.88 0.0214 300d t t=− + +

Model 2R Linear 0.962 Quadratic 1.000 Logarithmic 0.831 Exponential 0.933 Power 0.783

y = -4.8786x2 + 0.0214x + 299.96

R2 = 1

6. Power: 335y x=


y = 35.001x3

R2 = 1

7. Exponential: ( )100 1.03xy = The value of 2R is slightly higher for the exponential model.


y = 100(1.03)x

R2 = 1

8. Logarithmic: 0.00476 4.34 lny x= +


y = 4.3426Ln(x) + 0.0048

R2 = 1



9. Power: 0.94565.7y x−= . Prediction for the 22nd day: ( ) 0.94565.7 22y

−= = $3.5 million, which isn’t very

close to the actual amount of $2.2 million. The model does not take into account the fact that movies do better on weekend days.


y = 65.7x-0.945

R2 = 0.8421

10. Quadratic: 2323 1365 45,084y x x=− + + . (with 1975 coded as 1). Projected value for 2020:

( ) ( )2323 10 1365 10 45,084 26,434y =− + + = (Tech: 26,454).


y = -322.77x2 + 1364.7x + 45084

R2 = 0.652

11. Logarithmic: 3.22 0.293lny x= +


y = 0.2933Ln(x) + 3.2178

R2 = 0.9972



12. Power: 0.8630.313y x−=


y = 0.3133x-0.8631

R2 = 0.9985

13. Exponential: ( )10 2xy =


y = 10(2)x

R2 = 1

14. Exponential: ( )115 0.938x

y = . (Result is based on 1980 coded as 1.) With 2 0.253R = , this best model is

not a good model. There is a cyclical pattern that does not fit any of the five models included in this section.


y = 115.00(0.938)x

R2 = 0.2535



15. Quadratic: 2125 439 3438y x x= − + . The projected value for 2010 is ( ) ( )2125 21 439 21 3438y = − +

49,344= (Tech: 49,312), which is dramatically greater than the actual value of 11,655.


y = 124.93x2 - 438.96x + 3437.7

R2 = 0.995

16. Quadratic: 231.4 1233 280y x x=− + + . The projected value for 2010 is

( ) ( )231.4 21 1233 21 280y =− + + = 12,326 (Tech: 12,345), which is not dramatically different from the

actual value of 11,655.


y = -31.378x2 + 1233.4x + 280.2

R2 = 0.8992

17. a. Exponential: ( )23 12 xy −= [or ( )0.629961 1.587401

xy = for an initial value of 1 that doubles every 1.5

years].

b. Exponential: ( )1.36558 1.42774x

y = , where 1971 is coded as 1.


y = 1.3656(1.42774)x

R2 = 0.9899

c. Moore’s law does appear to be working reasonably well. With 2 0.990R = , the model appears to be very good.



18. a. 6641.8

b. 73.2

c. The quadratic sum of squares of residuals (73.2) is less than the sum of squares of residuals from the linear model (6641.8).

Chapter Quick Quiz

1. 0.878r =±

2. Based on the critical values of 0.878± (assuming a 0.05 significance level), conclude that there is not sufficient evidence to support the claim of a linear correlation between systolic and diastolic readings.

3. The best predicted diastolic reading is 90.6, which is the mean of the five sample diastolic readings.

4. The best predicted diastolic reading is ( )ˆ 1.99 0.698 125 85.3y =− + = , which is found by substituting 125

for x in the regression equation.

5. 2 0.342r =

6. False; there could be another relationship.

7. False, correlation does not imply causation.

8. 1r =

9. Because r must be between –1 and 1 inclusive, the value of 3.335 is the result of an error in the calculations.

10. 1r =−

Review Exercises

1. a. 0.926r = . Critical values: 0.707r =± (assuming a 0.05 significance level). P-value = 0.001. There is sufficient evidence to support the claim that there is a linear correlation between duration and interval-after time.

MINITAB Pearson correlation of After and Duration = 0.926 P-Value = 0.001

b. ( )22 0.926 0.857 85.7%r = = =

c. ˆ 34.8 0.234y x= +

MINITAB Predictor Coef SE Coef T P Constant 34.770 8.732 3.98 0.007 Duration 0.23406 0.03908 5.99 0.001

280260240220200180160140120100

100

90

80

70

60

Duration

Aft

er

Scatterplot of After vs Duration



1. (continued)

280260240220200180160140120100

5.0

2.5

0.0

-2.5

-5.0

-7.5

Duration

Res

idu

als

Scatterplot of Residuals vs Duration

1050-5-10

0.99

0.95

0.9

0.8

0.70.60.50.40.3

0.2

0.1

0.05

0.01

Residuals

Pro

babi

lity


d. ( )ˆ 34.8 0.234 200 81.6 miny = + =

2. a. The scatterplot suggests that there is not sufficient sample evidence to support the claim of a linear correlation between heights of eruptions and interval-after times.

150140130120110

100

90

80

70

60

Height

Aft

er

Scatterplot of After vs Height

b. 0.269.r = Critical values: 0.707r =± (assuming a 0.05 significance level). P-value = 0.519. There is not sufficient evidence to support the claim that there is a linear correlation between height and interval-after time.

MINITAB Pearson correlation of Height and After = 0.269 P-Value = 0.519

c. ˆ 54.3 0.246y x= +

MINITAB Predictor Coef SE Coef T P Constant 54.27 46.53 1.17 0.288 Height 0.2465 0.3597 0.69 0.519

d. ( )ˆ 54.3 0.246 100 78.9 miny = + =



3. a. The scatterplot suggests that there is not sufficient sample evidence to support the claim of a linear correlation between duration and height.

280260240220200180160140120100

150

140

130

120

110

Duration

Hei

ght

Scatterplot of Height vs Duration

b. 0.389.r = Critical values: 0.707r =± (assuming a 0.05 significance level). P-value = 0.340. There is

not sufficient evidence to support the claim that there is a linear correlation between duration and height.

MINITAB Pearson correlation of Height and Duration = 0.389 P-Value = 0.340

c. ˆ 105 0.108y x= +

MINITAB Predictor Coef SE Coef T P Constant 105.19 23.22 4.53 0.004 Duration 0.1076 0.1039 1.04 0.340

d. The regression line does not fit the points well, so the best predicted height 128.8 ft.y =

4. 0.450.r = Critical values: 0.632r =± (assuming a 0.05 significance level). P-value = 0.192. There is not sufficient evidence to support the claim that there is a linear correlation between time and height. Although there is no linear correlation between time and height, the scatterplot shows a very distinct pattern revealing that time and height are associated by some function that is not linear.

MINITAB Pearson correlation of Height(m) and Time(sec) = 0.450 P-Value = 0.192

2.01.51.00.50.0

5

4

3

2

1

0

Time(sec)

Hei

ght(

m)

Scatterplot of Height(m) vs Time(sec)

5. AFTER = 50.1 + 0.242 Duration – 0.178 BEFORE, or 1 2ˆ 50.1 0.242 0.178y x x= + − . 2R = 0.872; adjusted 2R = 0.820; P-value = 0.006. With high values of 2R and adjusted 2R and a small P-value of 0.006, it

appears that the regression equation can be used to predict the time interval after an eruption given the duration of the eruption and the time interval before that eruption.

MINITAB Predictor Coef SE Coef T P Constant 50.09 22.07 2.27 0.072 Duration 0.24179 0.04177 5.79 0.002 Before - 0.1779 0.2336 -0.76 0.481 S = 5.15785 R-Sq = 87.2% R-Sq(adj) = 82.0%




1. 3.3 lbx = , 5.7 lbs =

2. The highest weight before the diet is 212 lb, which converts to 212 179.4

1.5521.0

z−

= = . The highest weight

is not unusual because its z score of 1.55 shows that it is within 2 standard deviations of the mean.

3. 0 : 0dH μ = . 0 : 0dH μ > . Test statistic: 1.613t = . Critical value: 1.895t = . P-value > 0.05 (Tech:

0.075). Fail to reject H0. There is not sufficient evidence to support the claim that the diet is effective. MINITAB Paired T for Before - After 95% lower bound for mean difference: -0.57 T-Test of mean difference = 0 (vs > 0): T-Value = 1.61 P-Value = 0.075

4. 161.8 lb 197.0 lb.μ< < We have 95% confidence that the interval limits of 161.8 lb and 197.0 lb contain

the true value of the mean of the population of all subjects before the diet. MINITAB Variable N Mean StDev SE Mean 95% CI Before 8 179.38 21.04 7.44 (161.79, 196.96)

5. a. 0.965.r = Critical values: 0.707r =± (assuming a 0.05 significance level). P-value = 0.000. There is sufficient evidence to support the claim that there is a linear correlation between before and after weights. MINITAB Pearson correlation of Before and After = 0.965 P-Value = 0.000

b. 1r = c. 1r =

d. The effectiveness of the diet is determined by the amounts of weight lost, but the linear correlation coefficient is not sensitive to different amounts of weight loss. Correlation is not a suitable tool for testing the effectiveness of the diet.

6. a. 3500 3420

0.162;495

z−

= = ( )0.162 43.64%.P z > = (Tech: 43.58%)

b. 10th percentile: 3420 1.28 495 2786.4 gx μ z σ= + ⋅ = − ⋅ = (Tech: 2785.6 g)

c. 2450 3420

1.96;495

z−

= =− ( )1.96 0.0250.P z <− =

4390 3420

1.96;495

z−

= = ( )1.96 0.0250.P z > =

0.0250 0.0250 0.0500 5.00%.+ = = Yes, many of the babies do require special treatment.

7. a. H0: p = 0.5. H1: p > 0.5. Test statistic: 3.84z = . Critical value: 1.645z = . P-value: 0.0001. Reject H0. There is sufficient evidence to support the claim that the majority of us say that honesty is always the best policy.

MINITAB Test of p = 0.5 vs p > 0.5 95% Lower Sample X N Sample p Bound Z-Value P-Value 1 269 456 0.589912 0.552026 3.84 0.000

b. The sample is a voluntary response (or self-selected) sample. This type of sample suggests that the results given in part (a) are not necessarily valid.



8. a. Nominal

b. Ratio

c. Discrete

d. 304

0.575529

=

e. Parameter

9. a. 2

3040.330

529

⎛ ⎞⎟⎜ =⎟⎜ ⎟⎜⎝ ⎠

b. 304 156 460

0.870529 529

+= =

c. 514

0.972529

=

d. 39

0.0737 7.37%529

= =

10.

OtherMormonJewishCatholicProtestant

300

250

200

150

100

50

0

Members of Congress

Nu

mbe

r

Chapter 11: Goodness-of-Fit and Contingency Tables 199


Chapter 11: Goodness-of-Fit and Contingency Tables Section 11-2

1. The test is to determine whether the observed frequency counts agree with the claimed uniform distribution so that frequencies for the different days are equally likely.

2. 1005

7E = , or 143.571, for each of the seven days of the week. For Sunday, O = 523 and E = 143.571.

3. Because the given frequencies differ substantially from frequencies that are all about the same, the 2χ test

statistic should be large and the P-value should be small.

4. df = 6. Critical value: 2χ = 12.592.

5. Test statistic: 2χ = 1934.979. Critical value: 2χ = 12.592. P-value = 0.000. There is sufficient evidence to

warrant rejection of the claim that the days of the week are selected with a uniform distribution with all days having the same chance of being selected.

6. Test statistic: 2χ = 6.6. Critical value: 2χ = 16.919. P-value = 0.679. There is not sufficient evidence to

support the claim that the sample is from a population of heights in which the last digits do not occur with the same frequency.

7. Critical value: 2χ = 16.919. P-value > 0.10 (Tech: 0.516). There is not sufficient evidence to warrant

rejection of the claim that the observed outcomes agree with the expected frequencies. The slot machine appears to be functioning as expected.

8. Test statistic: 2χ = 4.600. Critical value: 2χ = 7.815. P-value > 0.10 (Tech: P-value = 0.204). There is not

sufficient evidence to warrant rejection of the claim that the tires selected by the students are equally likely. It appears that students do not have the ability to select the same tire.

Tire O E O E− ( )2O E− ( )2

O E

E

−

Left Front 11 40 ⋅0.25 = 10 1 1 0.1

Right Front 15 40 ⋅0.25 = 10 5 25 2.5

Left Rear 8 40 ⋅0.25 = 10 –2 4 0.4

Right Rear 6 40 ⋅0.25 = 10 –4 16 1.6

Sum 4.6

9. Test statistic: 2χ = 10.375. Critical value: 2χ = 19.675. P-value > 0.10 (Tech: 0.497). There is not

sufficient evidence to warrant rejection of the claim that homicides in New York City are equally likely for each of the 12 months. There is not sufficient evidence to support the police commissioner’s claim that homicides occur more often in the summer when the weather is better.

MINITAB N DF Chi-Sq P-Value 512 11 10.375 0.497

10. Test statistic: 2χ = 93.072. Critical value: 2χ = 19.675. P-value < 0.005 (Tech: 0.000). There is sufficient

evidence to warrant rejection of the claim that American born major league baseball players are born in different months with the same frequency. The sample data appear to support Gladwell’s claim.


200 Chapter 11: Goodness-of-Fit and Contingency Tables


11. Test statistic: 2χ = 5.860. Critical value: 2χ = 11.071. P-value > 0.10 (Tech: P-value = 0.320). There is

not sufficient evidence to support the claim that the outcomes are not equally likely. The outcomes appear to be equally likely, so the loaded die does not appear to behave differently from a fair die.



evidence to warrant rejection of the claim that births occur on the days of the week with equal frequency. Because many births are induced or involve Caesarean section, they are scheduled for days other than Saturday or Sunday, so those two days have smaller numbers of births.



sufficient evidence to warrant rejection of the claim that the likelihood of winning is the same for the different post positions. Based on these results, post position should not be considered when betting on the Kentucky Derby race.


14. Test statistic: 2χ = 8.021. Critical value: 2χ = 16.919. P-value > 0.10 (Tech: 0.532). There is not sufficient

evidence to warrant rejection of the claim that the digits are selected in a way that they are equally likely.



evidence to warrant rejection of the claim that the different days of the week have the same frequencies of police calls. The highest numbers of calls appear to fall on Friday and Saturday, and these are weekend days with disproportionately more partying and drinking.



evidence to warrant rejection of the claim that the different days of the week have the same frequencies of police calls. Because March has 31 days, three of the days of the week occur more often than the other days of the week, so the comparison does not make sense with the given data.



evidence to warrant rejection of the claim that the actual numbers of games fit the distribution indicated by the proportions listed in the given table.



17. (continued)

Games Played O E O E− ( )2O E− ( )2

O E

E

−

4 20 103 ⋅0.125 = 12.875 7.125 50.76563 3.942961

5 23 103 ⋅0.2500 = 25.75 –2.75 7.5625 0.293689

6 23 103 ⋅0.3125 = 32.1875 –9.1875 84.41016 2.622451

7 37 103 ⋅0.3125 = 32.1875 4.8125 23.16016 0.719539

Sum 7.578641


evidence to warrant rejection of the claim that the actual eliminations agree with the expected numbers. The leadoff singers do appear to be at a disadvantage because 20 of them were eliminated compared to the expected value of 12.9 eliminations, but that result is not significant in the context of the available sample data.


19. Test statistic: 2χ = 6.682. Critical value: 2χ = 11.071 (assuming a 0.05 significance level). P-value > 0.10

(Tech: 0.245). There is not sufficient evidence to warrant rejection of the claim that the color distribution is as claimed.

Color O E O E− ( )2O E− ( )2

O E

E

−

Red 13 100 ⋅0.13 = 13 0 0 0

Orange 25 100 ⋅0.20 = 20 5 25 1.25

Yellow 8 100 ⋅0.14 = 14 –6 36 2.571429

Brown 8 100 ⋅0.13 = 13 –5 25 1.923077

Blue 27 100 ⋅0.24 = 24 3 9 0.375

Green 19 100 ⋅0.16 = 16 3 9 0.5625

Sum 6.682005


evidence to warrant rejection of the claim that the actual frequencies fit a Poisson distribution.


21. Test statistic: 2χ = 3650.251. Critical value: 2χ = 20.090. P-value < 0.005 (Tech: 0.000). There is

sufficient evidence to warrant rejection of the claim that the leading digits are from a population with a distribution that conforms to Benford’s law. It does appear that the checks are the result of fraud (although the results cannot confirm that fraud is the cause of the discrepancy between the observed results and the expected results).



sufficient evidence to warrant rejection of the claim that the leading digits are from a population with a distribution that conforms to Benford’s law. The author’s check amounts appear to be legitimate.



22. (continued)



evidence to warrant rejection of the claim that the leading digits are from a population with a distribution that conforms to Benford’s law. The tax entries do appear to be legitimate.



sufficient evidence to warrant rejection of the claim that the leading digits are from a population with a distribution that conforms to Benford’s law.


25. a. 6, 13, 15, 6

b. 155.41 162

1;6.595

z−

= =− ( )1 0.1587P z <− = ;

162.005 162

0;6.595

z−

= = ( )1 0 0.5000 0.1587 0.3413P z− < < = − = ;

168.601 162

1;6.595

z−

= = ( )0 1 0.8413 0.5000 0.3413P z< < = − = :

215 280

1;65

z−

= =− ( )1 0.1587P z > =

(Tech: 0.1587, 0.3413, 0.3414, 0.1586)

c. 40 0.1587 6.348⋅ = , 40 0.3413 13.652⋅ = , , 40 0.3413 13.652⋅ = , 40 0.1587 6.348⋅ =

(Tech: 6.348, 13.652, 13.656, 6.344)

d. Test statistic: 2χ = 0.202 (Tech: 0.201). Critical value: 2χ = 11.345. P-value > 0.10 (Tech: 0.977).

There is not sufficient evidence to warrant rejection of the claim that heights were randomly selected from a normally distributed population. The test suggests that the data are from a normally distributed population.

Height O E O E− ( )2O E− ( )2

O E

E

−

Less than 155.410 6 6.348 –0.348 0.121104 0.019078

155.410–162.005 13 13.652 –0.652 0.425104 0.031139

162.005–168.601 15 13.652 1.348 1.817104 0.133102

Greater than 168.601 6 6.348 –0.348 0.121104 0.019078

Sum 0.202395 MINITAB N DF Chi-Sq P-Value 40 3 0.202395 0.977



Section 11-3

1. Because the P-value of 0.216 is not small (such as 0.05 or lower), fail to reject the null hypothesis of independence between the treatment and whether the subject stops smoking. This suggests that the choice of treatment doesn’t appear to make much of a difference.

2. In this context, the word contingency refers to a dependency of one variable on another, and we use a test of independence between the row variable and the column variable to determine whether one variable appears to be contingent on the other. We use the terminology of two-way table because the frequency counts are arranged in a table with two variables: the row variable and the column variable.

3. ( )( )df 3 1 2 1 2= − − = and the critical value is 2χ = 5.991.

4. The test is right-tailed. The test statistic is based on differences between observed frequencies and the frequencies expected with the assumption of independence between the row and column variables. Only large values of the test statistic correspond to substantial differences between the observed and expected values, and such large values are located in the right tail of the distribution.


evidence to warrant rejection of the claim that the form of the 100-Yuan gift is independent of whether the money was spent. There is not sufficient evidence to support the claim of a denomination effect.


evidence to warrant rejection of the claim that success is independent of the type of treatment. The results suggest that the surgery treatment is better.


evidence to warrant rejection of the claim that whether a subject lies is independent of the polygraph test indication. The results suggest that polygraphs are effective in distinguishing between truths and lies, but there are many false positives and false negatives, so they are not highly reliable.

MINITAB Expected counts are printed below observed counts No (Did Yes Not Lie) (Lied) Total 1 15 42 57 27.34 29.66 2 32 9 41 19.66 21.34 Total 47 51 98 Chi-Sq = 25.571, DF = 1, P-Value = 0.000


sufficient evidence to support the claim that the results are discriminatory.

MINITAB Expected counts are printed below observed counts Passed Failed Total 1 17 16 33 12.81 20.19 2 9 25 34 13.19 20.81 Total 26 41 67 Chi-Sq = 4.423, DF = 1, P-Value = 0.035




evidence to warrant rejection of the claim that the sentence is independent of the plea. The results encourage pleas for guilty defendants.

MINITAB Expected counts are printed below observed counts Guilty Not Guilty Plea Plea Total 1 392 58 450 418.48 31.52 2 564 14 578 537.52 40.48 Total 956 72 1028 Chi-Sq = 42.557, DF = 1, P-Value = 0.000


evidence to warrant rejection of the claim that deaths on shifts are independent of whether Gilbert was working. The results favor the guilt of Gilbert.

MINITAB Expected counts are printed below observed counts With Death Without Death Total 1 40 217 257 11.59 245.41 2 34 1350 1384 62.41 1321.59 Total 74 1567 1641 Chi-Sq = 86.481, DF = 1, P-Value = 0.000


evidence to warrant rejection of the claim that the gender of the tennis player is independent of whether the call is overturned.

MINITAB Expected counts are printed below observed counts Yes No Total 1 421 991 1412 416.90 995.10 2 220 539 759 224.10 534.90 Total 641 1530 2171 Chi-Sq = 0.164, DF = 1, P-Value = 0.686


evidence to warrant rejection of the claim that left-handedness is independent of gender. MINITAB Expected counts are printed below observed counts Yes No Total 1 23 217 240 27.79 212.21 2 65 455 520 60.21 459.79 Total 88 672 760 Chi-Sq = 1.364, DF = 1, P-Value = 0.243




evidence to warrant rejection of the claim that the direction of the kick is independent of the direction of the goalkeeper jump. The results do not support the theory that because the kicks are so fast, goalkeepers have no time to react, so the directions of their jumps are independent of the directions of the kicks.

MINITAB Chi-Sq = 14.589, DF = 4, P-Value = 0.006

14. Test statistic: 2χ = 1.358. Critical value: 2χ = 7.815 (assuming a 0.05 significance level). P-value > 0.10

(Tech: 0.715). There is not sufficient evidence to warrant rejection of the claim that the amount of smoking is independent of seat belt use. The theory is not supported by the given data.



evidence to warrant rejection of the claim that getting a cold is independent of the treatment group. The results suggest that echinacea is not effective for preventing colds.


16. Test statistic: 2χ = 9.971. Critical value: 2χ = 9.488 (assuming a 0.05 significance level). P-value < 0.05

(Tech: 0.041). There is sufficient evidence to warrant rejection of the claim that injuries are independent of helmet color. It appears that motorcycle drivers should use yellow or orange helmets.



evidence to warrant rejection of the claim that cooperation of the subject is independent of the age category. The age group of 60 and over appears to be particularly uncooperative.


18. Test statistic: 2χ = 20.054. Critical value: 2χ = 19.675. P-value > 0.05 (Tech: 0.0446). There is sufficient

evidence to warrant rejection of the claim that months of births of baseball players are independent of whether they are born in America. The data do appear to support Gladwell’s claim.



evidence to warrant rejection of the claim that getting an infection is independent of the treatment. The atorvastatin treatment does not appear to have an effect on infections.



evidence to warrant rejection of the claim that left-handedness is independent of parental handedness. It appears that handedness of the parents has an effect on handedness of the offspring, so left-handedness appears to be an inherited trait.



20. (continued)

MINITAB Expected counts are printed below observed counts Yes No Total 1 5360 50928 56288 6067.90 50220.10 2 767 2736 3503 377.63 3125.37 3 741 3667 4408 475.19 3932.81 4 94 289 383 41.29 341.71 Total 6962 57620 64582 Chi-Sq = 784.647, DF = 3, P-Value = 0.000

21. Test statistics: 2χ = 12.1619258 and z = 3.487395274, so that 2 2z χ= . Critical values: 2χ = 3.841 and 2 1.96z =± , so 2 2z χ= (approximately).

MINITAB Expected counts are printed below observed counts Purchased Kept the Gum Money Total 1 27 16 43 18.84 24.16 2 12 34 46 20.16 25.84 Total 39 50 89 Chi-Sq = 12.162, DF = 1, P-Value = 0.000

MINITAB Difference = p (1) - p (2) Estimate for difference: 0.372308 95% CI for difference: (0.178143, 0.566473) Test for difference = 0 (vs not = 0): Z = 3.49 P-Value = 0.000

22. Without Yates’s correction, the test statistic is 2χ = 12.162. With Yates’s correction, the test statistic is 2χ

= 10.717. Yates’s correction decreases the test statistic so that sample data must be more extreme in order to reject the null hypothesis of independence.

Without Yates’s correction

( ) ( ) ( ) ( )2 2 2 2

2 27 18.84 16 24.16 12 20.16 34 25.84 12.162

18.84 24.16 20.16 25.84χ

− − − −= + + + =

With Yates’s Correction

( ) ( ) ( ) ( )2 2 2 2

227 18.84 0.5 16 24.16 0.5 12 20.16 0.5 34 25.84 0.5

18.84 24.16 20.16 25.84

10.717

χ− − − − − − − −

= + + +

=

Chapter Quick Quiz

1. H0: 1 2 3 4 5p p p p p= = = = . H1: At least one of the probabilities is different from the others.

2. O = 23 and 107

21.45

E = = .

3. Right-tailed.

4. df = 4 and the critical value is 2χ = 9.488.

5. There is not sufficient evidence to warrant rejection of the claim that occupation injuries occur with equal frequency on the different days of the week.



6. H0: Response to the question is independent of gender. H1: Response to the question and gender are dependent.

7. Chi-square distribution.

8. Right-tailed.

9. ( )( )df 2 1 3 1 2= − − = and the critical value is 2χ = 5.991.

10. There is not sufficient evidence to warrant rejection of the claim that response is independent of gender.

Review Exercises

1. Test statistic: 2χ = 931.347. Critical value: 2χ = 16.812. P-value: 0.000. There is sufficient evidence to

warrant rejection of the claim that auto fatalities occur on the different days of the week with the same frequency. Because people generally have more free time on weekends and more drinking occurs on weekends, the days of Friday, Saturday, and Sunday appear to have disproportionately more fatalities.


evidence to warrant rejection of the claim that the last digits of 0, 1, 2, . . . , 9 occur with the same frequency. It does appear that the weights were obtained through measurements.



evidence to warrant rejection of the claim that weather-related deaths occur in the different months with the same frequency. The summer months appear to have disproportionately more weather-related deaths, and that is probably due to the fact that vacations and outdoor activities are much greater during those months.


4. Test statistic: 2χ = 10.708. Critical value: 2χ = 3.841. P-value: 0.00107. There is sufficient evidence to

warrant rejection of the claim that wearing a helmet has no effect on whether facial injuries are received. It does appear that a helmet is helpful in preventing facial injuries in a crash.


evidence to warrant rejection of the claim that when flipping or spinning a penny, the outcome is independent of whether the penny was flipped or spun. It appears that the outcome is affected by whether the penny is flipped or spun. If the significance level is changed to 0.01, the critical value changes to 6.635, and we fail to reject the given claim, so the conclusion does change. All expected counts are greater than 5.

Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts Heads Tails Total 1 2048 1992 4040 2007.29 2032.71 2 953 1047 2000 993.71 1006.29 Total 3001 3039 6040 Chi-Sq = 4.955, DF = 1, P-Value = 0.026




evidence to warrant rejection of the claim that home/visitor wins are independent of the sport.

MINITAB Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts Basketball Baseball Hockey Football Total 1 127 53 50 57 287 115.97 58.57 54.47 57.99 2 71 47 43 42 203 82.03 41.43 38.53 41.01 Total 198 100 93 99 490 Chi-Sq = 4.737, DF = 3, P-Value = 0.192


1. H0: p = 0.5. H1: p ≠ 0.5. Test statistic: z = 7.28. Critical values: 1.96z =± . P-value: 0.0002 (Tech: 0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim that among those who die in weather-related deaths, the percentage of males is equal to 50%.

MINITAB Test of p = 0.5 vs p not = 0.5 Exact Sample X N Sample p 95% CI P-Value 1 325 489 0.664622 (0.620854, 0.706384) 0.000

2. 59.0% 65.0%p< < . Because the confidence interval does not include 50% (or “half”), we should reject

the stated claim. MINITAB Sample X N Sample p 95% CI 1 620 1000 0.620000 (0.589098, 0.650193)

3. x = 53.7 years, median = 60.0 years, s = 16.1 years. Because an age of 16 differs from the mean by more than 2 standard deviations, it is an unusual age.

4. 42.2 years 65.2 yearsμ< < . Yes, the confidence interval limits do contain the value of 65.0 years that was

found from a sample of 9269 ICU patients. MINITAB Variable N Mean StDev SE Mean 95% CI AGES 10 53.70 16.09 5.09 (42.19, 65.21)

5. a. r = –0.0458. Critical values: 0.632r =± . P-value = 0.900. There is not sufficient evidence to support the claim that there is a linear correlation between the numbers of boats and the numbers of manatee deaths. MINITAB Pearson correlation of Boats and Manatee Deaths = -0.046 P-Value = 0.900

b. ˆ 96.1 0.137y x= −

MINITAB The regression equation is Manatee Deaths = 96.1 - 0.14 Boats Predictor Coef SE Coef T P Constant 96.14 99.89 0.96 0.364 Boats -0.137 1.053 -0.13 0.900



5. (continued)

c. ( )ˆ 96.1 0.137 84 84.6y = − = manatee deaths (the value of y). The predicted value is not very accurate

because it is not very close to the actual value of 78 manatee deaths.

6. a. 5th percentile: 686 1.645 34 630 mmx μ z σ= + ⋅ = − ⋅ =

b. 650 686

1.0634

z−

= =− and ( )1.06 14.46%P z <− = (Tech: 14.48%). That percentage is too high,

because too many women would not be accommodated.

c. 680 686

0.70634 16

z−

= =− and ( )0.706 76.11%P z >− = (Tech: 0.7599). Groups of 16 women do not

occupy a cockpit; because individual women occupy the cockpit, this result has no effect on the design.

7. a. Statistic.

b. Quantitative.

c. Discrete.

d. The sampling is conducted so that all samples of the same size have the same chance of being selected.

e. The sample is a voluntary response sample (or self-selected sample), and those with strong feelings about the topic are more likely to respond, so it is not a valid sampling plan.

8. a. ( )40.6 0.1296=

b. 1 0.6 0.4− =

Chapter 12: Analysis of Variance 211


Chapter 12: Analysis of Variance Section 12-2

1. a. The chest deceleration measurements are categorized according to the one characteristic of size.

b. The terminology of analysis of variance refers to the method used to test for equality of the three population means. That method is based on two different estimates of a common population variance.

2. As we increase the number of individual tests of significance, we increase the risk of finding a difference by chance alone (instead of a real difference in the means). The risk of a type I error—finding a difference in one of the pairs when no such difference actually exists—is too high. The method of analysis of variance helps us avoid that particular pitfall (rejecting a true null hypothesis) by using one test for equality of several means, instead of several tests that each compare two means at a time.

3. The test statistic is F = 3.288, and the F distribution applies.

4. The P-value is 0.061. Because the P-value is greater than the significance level of 0.05, we fail to reject the null hypothesis of equal means. There is not sufficient evidence to warrant rejection of the claim that the three different categories of car sizes have the same mean chest deceleration in the standard car crash test.

5. Test statistic: F = 0.39. P-value: 0.677. Fail to reject H0: 1 2 3μ μ μ= = . There is not sufficient evidence to

warrant rejection of the claim that the three categories of blood lead level have the same mean verbal IQ score. Exposure to lead does not appear to have an effect on verbal IQ scores.

6. Test statistic: F = 2.3034. P-value: 0.1044. Fail to reject H0: 1 2 3μ μ μ= = . There is not sufficient evidence

to warrant rejection of the claim that the three categories of blood lead level have the same mean full IQ score. Exposure to lead does not appear to have an effect on full IQ scores.

7. Test statistic: F = 11.6102. P-value: 0.000577. Reject H0: 1 2 3μ μ μ= = . There is sufficient evidence to

warrant rejection of the claim that the three size categories have the same mean highway fuel consumption. The size of a car does appear to affect highway fuel consumption.


warrant rejection of the claim that the three size categories have the same mean city fuel consumption. The size of a car does appear to affect city fuel consumption.

9. Test statistic: F = 0.161. P-value: 0.852. Fail to reject H0: 1 2 3μ μ μ= = . There is not sufficient evidence to

warrant rejection of the claim that the three size categories have the same mean head injury measurement. The size of a car does not appear to affect head injuries.

10. Test statistic: F = 0.3476. P-value: 0.7111. Fail to reject H0: 1 2 3μ μ μ= = . There is not sufficient evidence

to warrant rejection of the claim that the three size categories have the same mean pelvis injury measurement. The size of a car does not appear to affect pelvis injuries.

11. Test statistic: F = 27.2488. P-value: 0.000. Reject H0: 1 2 3μ μ μ= = . There is sufficient evidence to warrant

rejection of the claim that the three different miles have the same mean time. These data suggest that the third mile appears to take longer, and a reasonable explanation is that the third lap has a hill.

EXCEL ANOVA

Source of Variation SS df MS F P-value F crit Between Groups 0.103444 2 0.051722 27.24878 3.45E-05 3.885294 Within Groups 0.022778 12 0.001898 Total 0.126222 14

212 Chapter 12: Analysis of Variance



warrant rejection of the claim that the three books have the same mean Flesch Reading Ease score. The data suggest that the books appear to have mean scores that are not all the same.

EXCEL ANOVA

Source of Variation SS df MS F P-value F crit Between Groups 1338.002 2 669.0011 9.469487 0.000562 3.284918 Within Groups 2331.387 33 70.64808 Total 3669.389 35

13. Test statistic: F = 6.1413. P-value: 0.0056. Reject H0: 1 2 3 4μ μ μ μ= = = . There is sufficient evidence to

warrant rejection of the claim that the four treatment categories yield poplar trees with the same mean weight. Although not justified by the results from analysis of variance, the treatment of fertilizer and irrigation appears to be most effective.

EXCEL ANOVA


14. Test statistic: F = 0.3801. P-value: 0.769. Fail to reject H0: 1 2 3 4μ μ μ μ= = = . There is not sufficient

evidence to warrant rejection of the claim that the four treatment categories yield poplar trees with the same mean weight. In the sandy and dry region, there does not appear to be a treatment that is more effective than the others.

EXCEL ANOVA



rejection of the claim that the three different types of cigarettes have the same mean amount of nicotine. Given that the king-size cigarettes have the largest mean of 1.26 mg per cigarette, compared to the other means of 0.87 mg per cigarette and 0.92 mg per cigarette, it appears that the filters do make a difference (although this conclusion is not justified by the results from analysis of variance).

EXCEL ANOVA

Source of Variation SS df MS F P-value F crit Between Groups 2.208267 2 1.104133 18.99312 2.38E-07 3.123907 Within Groups 4.1856 72 0.058133 Total 6.393867 74



16. Test statistic: F = 20.8562. P-value: 0.000. Reject H0: 1 2 3 4μ μ μ μ= = = . There is sufficient evidence to

warrant rejection of the claim that the three samples are from populations with the same mean. It appears that cotinine levels are greater with more exposure to tobacco smoke. (The samples do not appear to be from normally distributed populations, but ANOVA is robust against departures from normality.)

EXCEL ANOVA

Source of Variation SS df MS F P-value Between Groups 518033.017 2 259016.508 20.8562144 1.7912E-08 Within Groups 1453040.85 117 12419.1526 Total 1971073.87 119

560480400320240160800

SMOKER

ETS

NOETS

Data

Dotplot of SMOKER, ETS, NOETS

Each symbol represents up to 2 observations.

17. The Tukey test results show different P-values, but they are not dramatically different. The Tukey results suggest the same conclusions as the Bonferroni test.

18. a. In Exercise 13 we reject the null hypothesis of equal means. The displayed Bonferroni results show that with a P-value of 0.039, there is a significant difference between the mean of the no treatment group (group 1) and the mean of the group treated with both fertilizer and irrigation (group 4).

b. The test statistic is t = –4.007. P-value = 6(0.001018) = 0.00611. Reject the null hypothesis that the mean weight from the irrigation treatment group is equal to the mean from the group treated with both fertilizer and irrigation.

Section 12-3

1. The load values are categorized using two different factors of (1) femur (left or right) and (2) size of car (small, midsize, large).

2. No. To use two individual tests of one-way analysis of variance is to completely ignore the very important feature of the possible effect from an interaction between femur and size. If there is an interaction, it doesn’t make sense to consider the effects of one factor without the other.

3. An interaction between two factors or variables occurs if the effect of one of the factors changes for different categories of the other factor. If there is an interaction effect, we should not proceed with individual tests for effects from the row factor and column factor. If there is an interaction, we should not consider the effects of one factor without considering the effects of the other factor.

4. Yes, the result is a balanced design because each cell has the same number (7) of values.



5. For interaction, the test statistic is F = 1.72 and the P-value is 0.194, so there is not sufficient evidence to conclude that there is an interaction effect. For the row variable of femur (right, left), the test statistic is F = 1.39 and the P-value is 0.246, so there is not sufficient evidence to conclude that whether the femur is right or left has an effect on measured load. For the column variable of size of the car, the test statistic is F = 2.23 and the P-value is 0.122, so there is not sufficient evidence to conclude that the car size category has an effect on the measured load.

6. For interaction, the test statistic is F = 0.34 and the P-value is 0.717, so there is not sufficient evidence to conclude that there is an interaction effect. For the row variable of type (foreign, domestic), the test statistic is F = 5.44 and the P-value is 0.038, so there is sufficient evidence to conclude that the type of car (foreign, domestic) has an effect on measured chest deceleration. For the column variable of size of the car, the test statistic is F = 3.58 and the P-value is 0.060, so there is not sufficient evidence to conclude that the car size category has an effect on the measured chest deceleration.

7. For interaction, the test statistic is F = 1.05 and the P-value is 0.365, so there is not sufficient evidence to conclude that there is an interaction effect. For the row variable of sex, the test statistic is F = 4.58 and the P-value is 0.043, so there is sufficient evidence to conclude that the sex of the subject has an effect on verbal IQ score. For the column variable of blood lead level (LEAD), the test statistic is F = 0.14 and the P-value is 0.871, so there is not sufficient evidence to conclude that blood lead level has an effect on verbal IQ score. It appears that only the sex of the subject has an effect on verbal IQ score.

8. For interaction, the test statistic is F = 41.38 and the P-value is 0.000, so there is sufficient evidence to conclude that there is an interaction effect. The ratings appear to be affected by an interaction between the use of a supplement and the amount of whey. Because there appears to be an interaction effect, we should not proceed with individual tests of the row factor (supplement) and the column factor (amount of whey).

EXCEL

ANOVA

Source of Variation SS df MS F P-value F crit

Sample 0.510417 1 0.510417 17.01389 0.00079419 4.493998

Columns 6.824583 3 2.274861 75.8287 1.12627E-09 3.238872

Interaction 3.724583 3 1.241528 41.38426 9.12956E-08 3.238872

Within 0.48 16 0.03

Total 11.53958 23

9. For interaction, the test statistic is F = 3.7332 and the P-value is 0.0291, so there is sufficient evidence to conclude that there is an interaction effect. The measures of self-esteem appear to be affected by an interaction between the self-esteem of the subject and the self-esteem of the target. Because there appears to be an interaction effect, we should not proceed with individual tests of the row factor (target’s self-esteem) and the column factor (subject’s self-esteem).

EXCEL

ANOVA


Sample 4.5 1 4.5 4.977654 0.029079736 3.986269

Columns 2.861111 2 1.430556 1.582402 0.213176735 3.135918

Interaction 6.75 2 3.375 3.73324 0.029107515 3.135918

Within 59.66667 66 0.90404

Total 73.77778 71



10. For interaction, the test statistic is F = 3.6872 and the P-value is 0.0628, so there is not sufficient evidence to conclude that there is an interaction effect. For the row variable of gender (male, female), the test statistic is F = 10.8629 and the P-value is 0.0022, so there is sufficient evidence to conclude that gender has an effect on pulse rate. For the column variable of age bracket, the test statistic is F = 4.8700 and the P-value is 0.0338, so there is sufficient evidence to conclude that the age bracket has an effect on pulse rate.

EXCEL

ANOVA


Sample 1322.5 1 1322.5 10.86292 0.00221114 4.113165

Columns 592.9 1 592.9 4.870037 0.033790054 4.113165

Interaction 448.9 1 448.9 3.687232 0.062780461 4.113165

Within 4382.8 36 121.7444

Total 6747.1 39

11. a. Test statistics and P-values do not change.

b. Test statistics and P-values do not change.

c. Test statistics and P-values do not change.

d. An outlier can dramatically affect and change test statistics and P-values.

Chapter Quick Quiz

1. H0: 1 2 3μ μ μ= = . Because the displayed P-value of 0.000 is small, reject H0.

2. No. Because we reject the null hypothesis of equal means, it appears that the three different power sources do not produce the same mean voltage level, so we cannot expect electrical appliances to behave the same way when run from the three different power sources.

3. Right-tailed.

4. Test statistic: F = 183.01. In general, larger test statistics result in smaller P-values.

5. The sample voltage measurements are categorized using only one factor: the source of the voltage.

6. Test a null hypothesis that three or more samples are from populations with equal means.

7. With one-way analysis of variance, the different samples are categorized using only one factor, but with two-way analysis of variance, the sample data are categorized into different cells determined by two different factors.

8. For interaction, the test statistic is F = 0.19 and the P-value is 0.832. Fail to reject the null hypothesis of no interaction. There does not appear to be an effect due to an interaction between sex and major.

9. The test statistic is F = 0.78 and the P-value is 0.395. There is not sufficient evidence to support a claim that the length estimates are affected by the sex of the subject.

10. The test statistic is F = 0.13 and the P-value is 0.876. There is not sufficient evidence to support a claim that the length estimates are affected by the subject’s major.

Review Exercises

1. H0: 1 2 3μ μ μ= = . Test statistic: F = 10.10. P-value: 0.001. Reject the null hypothesis. There is sufficient

evidence to warrant rejection of the claim that 4-cylinder cars, 6-cylinder cars, and 8-cylinder cars have the same mean highway fuel consumption amount.



2. For interaction, the test statistic is F = 0.17 and the P-value is 0.915, so there is not sufficient evidence to conclude that there is an interaction effect. For the row variable of site, the test statistic is F = 0.81 and the P-value is 0.374, so there is not sufficient evidence to conclude that the site has an effect on weight. For the column variable of treatment, the test statistic is F = 7.50 and the P-value is 0.001, so there is sufficient evidence to conclude that the treatment has an effect on weight.


rejection of the claim that the three different types of cigarettes have the same mean amount of tar. Given that the king-size cigarettes have the largest mean of 21.1 mg per cigarette, compared to the other means of 12.9 mg per cigarette and 13.2 mg per cigarette, it appears that the filters do make a difference (although this conclusion is not justified by the results from analysis of variance).

EXCEL

ANOVA


Between Groups 1083.707 2 541.8533 42.94364 5.29E-13 3.123907449

Within Groups 908.48 72 12.61778

Total 1992.187 74

4. For interaction, the test statistic is F = 0.8733 and the P-value is 0.3685, so there does not appear to be an effect from an interaction between gender and whether the subject smokes. For gender, the test statistic is F = 0.0178 and the P-value is 0.8960, so gender does not appear to have an effect on body temperature. For smoking, the test statistic is F = 3.0119 and the P-value is 0.1082, so there does not appear to be an effect from smoking on body temperature.

EXCEL

ANOVA


Sample(Gender) 0.005625 1 0.005625 0.017822 0.896012 4.747225

Columns(Smokes) 0.950625 1 0.950625 3.011881 0.108238 4.747225

Interaction 0.275625 1 0.275625 0.873267 0.368476 4.747225

Within 3.7875 12 0.315625

Total 5.019375 15


1. a. 15.5 years, 13.1 years, 22.7 years

b. 9.7 years, 9.0 years, 18.6 years

c. 94.5 years2, 80.3 years2, 346.1 years2

d. Ratio.

2. Test statistic: t = –1.383. Critical values 2.160t =± (assuming a 0.05 significance level). (Tech: P-value = 0.1860.) Fail to reject H0: 1 2μ μ= . There is not sufficient evidence to support the claim that

there is a difference between the means for the two groups.

MINITAB Difference = mu (Presidents) - mu (Monarchs) Estimate for difference: -7.21 95% CI for difference: (-18.33, 3.90) T-Test of difference = 0 (vs not =): T-Value = -1.38 P-Value = 0.187 DF = 15



3. Normal, because the histogram is approximately bell-shaped (or the points in a normal quantile plot are reasonably close to a straight-line pattern with no other pattern that is not a straight-line pattern).

403020100

9

8

7

6

5

4

3

2

1

0

Presidents

Freq

uen

cy

Histogram of Presidents

4. 12.3 years 18.7 yearsμ< <

MINITAB One-Sample T: Presidents Variable N Mean StDev SE Mean 95% CI Presidents 38 15.50 9.72 1.58 (12.30, 18.70)

5. a. H0: 1 2 3μ μ μ= =

b. Because the P-value of 0.051 is greater than the significance level of 0.05, fail to reject the null hypothesis of equal means. There is not sufficient evidence to warrant rejection of the claim that the three means are equal. The three populations do not appear to have means that are significantly different.

6. a. r = 0.918. Critical values: 0.707r =± . P-value = 0.001. There is sufficient evidence to support the claim that there is a linear correlation between September weights and the subsequent April weights. MINITAB Correlations: September, April Pearson correlation of September and April = 0.918 P-Value = 0.001

b. ˆ 9.28 0.823y x= +

MINITAB Regression Analysis: April versus September The regression equation is April = 9.3 + 0.823 September Predictor Coef SE Coef T P Constant 9.28 11.04 0.84 0.433 September 0.8227 0.1450 5.67 0.001

c. ( )ˆ 9.28 0.823 94 86.6 kg,y = + = which is not very close to the actual April weight of 105 kg.

7. a. 345 280

1;65

z−

= = ( )1 0.1587.P z > =

b. 215 280

1;65

z−

= =− ( )1 1 0.8413 0.1587 0.6826P z− < < = − = (Tech: 0.6827)

c. 319 280

3;65 / 25

z−

= = ( )3 0.9987.P z < =

d. 80th percentile: 280 .84 65 334.6x μ z σ= + ⋅ = + ⋅ = (Tech: 334.7)



8. a. 0.20 1000 200⋅ = 200

b. 0.175 0.225p< <

MINITAB Test and CI for One Proportion Sample X N Sample p 95% CI 1 200 1000 0.200000 (0.175621, 0.226159)

c. Yes. The confidence interval shows us that we have 95% confidence that the true population proportion is contained within the limits of 0.175 and 0.225, and 1/4 is not included within that range.

9. a. The distribution should be uniform, with a flat shape. The given histogram agrees (approximately) with the uniform distribution that we expect.

b. No. A normal distribution is approximately bell-shaped, but the given histogram is far from being bell-shaped.

10. Test statistic: 2χ = 10.400. Critical value: 2χ = 16.919 (assuming a 0.05 significance level). P-value >

0.10 (Tech: 0.319). There is not sufficient evidence to warrant rejection of the claim that the digits are selected from a population in which the digits are all equally likely. There does not appear to be a problem with the lottery.

MINITAB Chi-Square Goodness-of-Fit Test for Observed Counts in Variable: C1 N DF Chi-Sq P-Value 200 9 10.4 0.319

Chapter 13: Nonparametric Statistics 219


Chapter 13: Nonparametric Statistics Section 13-2

1. The only requirement for the matched pairs is that they constitute a simple random sample. There is no requirement of a normal distribution or any other specific distribution. The sign test is “distribution free” in the sense that it does not require a normal distribution or any other specific distribution.

2. There are 2 positive signs, 7 negative signs, 1 tie, 9n = , and the test statistic is 2x = (the smaller of 2 and 7).

3. H0: There is no difference between the populations of September weights and April weights. H1: There is a difference between the populations of September weights and April weights. The sample data do not contradict H1 because the numbers of positive signs (2) and negative signs (7) are not exactly the same.

4. The efficiency of 0.63 indicates that with all other things being equal, the sign test requires 100 sample observations to achieve the same results as 63 sample observations analyzed through a parametric test.

5. The test statistic of 1x = is less than or equal to the critical value of 2 (from Table A-7.) There is sufficient evidence to warrant rejection of the claim of no difference. There does appear to be a difference.

6. The test statistic of 5x = is less than or equal to the critical value of 5 (from Table A-7.) There is sufficient evidence to warrant rejection of the claim of no difference. There does appear to be a difference.

7. The test statistic of ( ) 1007

2151 0.522.18

1007 2z

+ −= =− falls in the critical region bounded by 1.96z =− and

1.96. There is sufficient evidence to warrant rejection of the claim of no difference. There does appear to be a difference.


2172 0.510.76

611 2z

+ −= =− falls in the critical region bounded by 1.96z =− and

1.96. There is sufficient evidence to warrant rejection of the claim of no difference. There does appear to be a difference.

9. There are 9 positive signs, 1 negative signs, 0 ties, and 10n = . The test statistic of 1x = is less than or equal to the critical value of 1 (from Table A-7). There is sufficient evidence to warrant rejection of the claim of no difference. There does appear to be a difference.


216 0.55.25

80 2z

+ −= =− falls in the critical region bounded by 1.96z =± . There is

sufficient evidence to warrant rejection of the claim of no difference. There does appear to be a difference.


214 0.50.53

32 2z

+ −= =− does not fall in the critical region bounded by 1.96z =± .

There is not sufficient evidence to warrant rejection of the claim of no difference. There does not appear to be a difference.

MINITAB Sign Test for Median: Ht - HtOpp- Sign test of median = 0.00000 versus not = 0.00000 N Below Equal Above P Median C7 34 14 2 18 0.5966 1.000

12. There are 12 positive signs, 0 negative signs, 0 ties, and 12n = . The test statistic of 0x = is less than or equal to the critical value of 2 (from Table A-7). There is sufficient evidence to warrant rejection of the claim of no difference. There does appear to be a difference.

220 Chapter 13: Nonparametric Statistics



252 0.510.90

291 2z

+ −= =− is in the critical region bounded by 2.575z =± . There

is sufficient evidence to warrant rejection of the claim of no difference. The YSORT method appears to have an effect on the gender of the child. (Because so many more boys were born than would be expected with no effect, it appears that the YSORT method is effective in increasing the likelihood that a baby will be a boy.)


247 0.50.88

104 2z

+ −= =− is not in the critical region bounded by 1.96z =± . There

is not sufficient evidence to warrant rejection of the claim of no difference. It appears that women do not have the ability to predict the sex of their babies.


2123 0.51.97

280 2z

+ −= =− is not in the critical region bounded by 2.575z =± .

There is not sufficient evidence to warrant rejection of the claim that the touch therapists make their selections with a method equivalent to random guesses. The touch therapists do not appear to be effective in selecting the correct hand.


2242 0.56.06

638 2z

+ −= =− is in the critical region bounded by 1.96z =± . There is

sufficient evidence to warrant rejection of the claim that respondents did not have a strong opinion one way or the other. They do appear to favor the opinion that NFL games are not too long. However, the validity of the results is highly questionable because the sample is a voluntary response sample.


212 0.52.37

40 2z


is not sufficient evidence to warrant rejection of the claim that the median is equal to 5.670 g. The quarters appear to be minted according to specifications.

MINITAB Sign Test for Median: Post-1964 Quarters Sign test of median = 5.670 versus not = 5.670 N Below Equal Above P Median Post-1964 Quarters 40 28 0 12 0.0166 5.636


220 0.50.88

47 2z

+ −= =− is not in the critical region bounded by 2.575z =± .

There is not sufficient evidence to warrant rejection of the claim that the median is equal to 1.00.

MINITAB Sign Test for Median: MAGNITUDE Sign test of median = 1.000 versus not = 1.000 N Below Equal Above P Median MAG 50 20 3 27 0.3817 1.235


21 0.55.32

34 2z


sufficient evidence to warrant rejection of the claim that the median amount of Coke is equal to 12 oz. Consumers are not being cheated because they are generally getting more than 12 oz of Coke, not less.



19. (continued)

MINITAB Sign Test for Median: CKREGVOL Sign test of median = 12.00 versus not = 12.00 N Below Equal Above P Median CKREGVOL 36 1 2 33 0.0000 12.20


227 0.52.70

79 2z


sufficient evidence to warrant rejection of the claim that the median age is 30 years.

MINITAB Sign Test for Median: Actresses Sign test of median = 30.00 versus not = 30.00 N Below Equal Above P Median Actresses 82 27 3 52 0.0069 33.00

21. Second approach: The test statistic of ( ) 105

230 0.54.29

105 2z

+ −= =− is in the critical region bounded by

1.645z =− , so the conclusions are the same as in Example 4.

Third approach: The test statistic of ( ) 106

238 0.52.82

106 2z

+ −= =− is in the critical region bounded by

1.645z =− , so the conclusions are the same as in Example 4. The different approaches can lead to very different results; as seen in the test statistics of –4.21, –4.29, and –2.82. The conclusions are the same in this case, but they could be different in other cases.

22. The column entries are *, *, *, *, *, 0, 0, 0.

Section 13-3

1. The only requirements are that the matched pairs be a simple random sample and the population of differences be approximately symmetric. There is no requirement of a normal distribution or any other specific distribution. The Wilcoxon signed-ranks test is “distribution free” in the sense that it does not require a normal distribution or any other specific distribution.

2. a. 1, 1, –4, –3, 0, –1, –5, –8, –3, –1

b. 2.5, 2.5, 7, 5.5, 2.5, 8, 9, 5.5, 2.5

c. 2.5, 2.5, –7, –5.5, –2.5, –8, –9, –5.5, –2.5

d. 5 and 40

e. 5T =

f. Critical value of T is 6.

3. The sign test uses only the signs of the differences, but the Wilcoxon signed-ranks test uses ranks that are affected by the magnitudes of the differences.

4. The efficiency of 0.95 indicates that with all other things being equal, the Wilcoxon signed-ranks test requires 100 sample observations to achieve the same results as 95 sample observations analyzed through a parametric test.

5. Test statistic: 6T = . Critical value: 8T = . Reject the null hypothesis that the population of differences has a median of 0. There is sufficient evidence to warrant rejection of the claim of no difference. There does appear to be a difference.

MINITAB Test of median = 0.000000 versus median not = 0.000000 N for Wilcoxon Estimated N Test Statistic P Median AGE DIFF 10 10 6.0 0.032 -11.50



6. Convert 501.5T = to the test statistic

( )

( )( )

80 80 1501.5

4 5.3680 80 1 2 80 1

24

z

+−

= =−+ ⋅ +

.

Critical values: 1.96z =± . (Tech: P-value = 0.000.) Reject the null hypothesis that the population of differences has a median of 0. There is sufficient evidence to warrant rejection of the claim of no difference. There does appear to be a difference.

MINITAB Test of median = 0.000000 versus median not = 0.000000 N for Wilcoxon Estimated N Test Statistic P Median AGE DIFF 82 80 501.5 0.000 -9.000

7. Convert 247T = to the test statistic

( )

( )( )

32 32 1247

4 0.3232 32 1 2 32 1

24

z

+−

= =−+ ⋅ +

.

Critical values: 1.96z =± . (Tech: P-value = 0.751.) Fail to reject the null hypothesis that the population of differences has a median of 0. There is not sufficient evidence to warrant rejection of the claim of no difference. There does not appear to be a difference.

MINITAB Test of median = 0.000000 versus median not = 0.000000 N for Wilcoxon Estimated N Test Statistic P Median HtDiff 34 32 247.0 0.758 -0.5000

8. Test statistic: 0T = . Critical value: 14T = . Reject the null hypothesis that the population of differences has a median of 0. There is sufficient evidence to warrant rejection of the claim of no difference. There does appear to be a difference.

MINITAB Test of median = 0.000000 versus median not = 0.000000 N for Wilcoxon Estimated N Test Statistic P Median IN-OUT 12 12 0.0 0 .003 -8.500


( )

( )( )

40 40 1196

4 2.8840 40 1 2 40 1

24

z

+−

= =−+ ⋅ +

.

Critical values: 2.57z =± 5. (Tech: P-value = 0.004.) There is sufficient evidence to warrant rejection of the claim that the median is equal to 5.670 g. The quarters do not appear to be minted according to specifications.

MINITAB Test of median = 5.670 versus median not = 5.670 N for Wilcoxon Estimated N Test Statistic P Median Post-1964 Quarters 40 40 196.0 0.004 5.638


( )

( )( )

47 47 1376

4 1.9947 47 1 2 47 1

24

z

+−

= =−+ ⋅ +

.

Critical values: 2.57z =± . (Tech: P-value = 0.047.) There is not sufficient evidence to warrant rejection of the claim that the median is equal to 1.00.



10. (continued)

MINITAB Test of median = 1.000000 versus median not = 1.000000 N for Wilcoxon Estimated N Test Statistic P Median Mag 50 47 376.0 0.047 -0.1400


( )

( )( )

34 34 115.5

4 4.8234 34 1 2 34 1

24

z

+−

= =−+ ⋅ +

.

Critical values: 1.96z =± . (Tech: P-value = 0.000.) There is sufficient evidence to warrant rejection of the claim that the median amount of Coke is equal to 12 oz. Consumers are not being cheated because they are generally getting more than 12 oz of Coke, not less.

MINITAB Test of median = 12.000000 versus median not = 12.000000 N for Wilcoxon Estimated N Test Statistic P Median Volume 36 34 15.5 0.000 -0.2000


( )

( )( )

79 79 1672.5

4 4.4479 79 1 2 79 1

24

z

+−

= =−+ ⋅ +

.

Critical values: 1.96z =± . (Tech: P-value = 0.000.) There is sufficient evidence to warrant rejection of the claim that the median age is 30 years.

MINITAB Test of median = 30.000000 versus median not = 30.000000 N for Wilcoxon Estimated N Test Statistic P Median AGE 82 79 672.5 0.000 -4.000

13. a. Min: 0 and Max: 1 + 2 + … +74 + 75 = 2850

b. 2850

14252

=

c. 2850 850 2000− =

d. ( )1

2

n nk

+−

Section 13-4

1. Yes. The two samples are independent because the flight data are not matched. The samples are simple random samples. Each sample has more than 10 values.

2. 137.5R =

3. H0: Arrival delay times from Flights 19 and 21 have the same median. There are three different possible alternative hypotheses: H1: Arrival delay times from Flights 19 and 21 have different medians. H1: Arrival delay times from Flight 19 have a median greater than the median of arrival delay times from Flight 21. H1: Arrival delay times from Flight 19 have a median less than the median of arrival delay times from Flight 21.

4. The efficiency rating of 0.95 indicates that with all other factors being the same, the Wilcoxon rank-sum test requires 100 sample observations to achieve the same results as 95 observations with the parametric t test of Section 9-3, assuming that the stricter requirements of the parametric t test are satisfied.



5. 1 137.5R = , 2 162.5R = , ( )12 12 12 1

1502Rμ+ +

= = , ( )12 12 12 12 1

17.32112Rσ

⋅ + += = , test statistic:

137.5 1500.72

17.321z

−= =− . Critical values: 1.96z =± . (Tech: P-value = 0.4705.) Fail to reject the null

hypothesis that the populations have the same median. There is not sufficient evidence to warrant rejection of the claim that Flights 19 and 21 have the same median arrival delay time.

6. 1 128R = , 2 172R = R2 = 172, ( )12 12 12 1

1502Rμ+ +

= = , ( )12 12 12 12 1

17.32112Rσ

⋅ + += = , test

statistic: 128 150

1.2717.321

z−

= =− . Critical values: 1.96z =± . (Tech: P-value = 0.2040.) Fail to reject the

null hypothesis that the populations have the same median. There is not sufficient evidence to warrant rejection of the claim that Flights 19 and 21 have the same median taxi-out time.

7. 1 253.5R = , 2 124.5R = , ( )13 13 14 1

1822Rμ+ +

= = , ( )13 14 13 14 1

20.60712Rσ


253.5 1823.47

20.607z

−= = . Critical values: 1.96z =± . (Tech: P-value = 0.0005.) Reject the null hypothesis

that the populations have the same median. There is sufficient evidence to reject the claim that for those treated with 20 mg of atorvastatin and those treated with 80 mg of atorvastatin, changes in LDL cholesterol have the same median. It appears that the dosage amount does have an effect on the change in LDL cholesterol.

8. 1 194.5R = , 2 105.5R = , ( )12 12 12 1

1502Rμ+ +

= = , ( )12 12 12 12 1

17.32112Rσ


194.5 1502.57

17.321z


that the populations have the same median. There is sufficient evidence to reject the claim that the median amount of strontium-90 from Pennsylvania residents is the same as the median from New York residents.

9. 1 501R = , 2 445R = , ( )22 22 21 1

4842Rμ+ +

= = , ( )22 21 22 21 1

41.15812Rσ


501 4840.41

41.158z

−= = . Critical value: 1.645z = . (Tech: P-value = 0.3398.) Fail to reject the null

hypothesis that the populations have the same median. There is not sufficient evidence to support the claim that subjects with medium lead levels have full IQ scores with a higher median than the median full IQ score for subjects with high lead levels. It does not appear that lead level affects full IQ scores.

10. 1 4178R = , 2 772R = , ( )78 78 21 1

39002Rμ+ +

= = , ( )78 21 78 21 1

116.83312Rσ


4178 39002.38

116.833z

−= = . Critical value: 1.645z = . (Tech: P-value = 0.0087.) Reject the null hypothesis

that the populations have the same median. There is sufficient evidence to support the claim that subjects with low lead levels have performance IQ scores with a higher median than the median performance IQ score for subjects with high lead levels. It appears that exposure to lead does have an adverse effect.



11. 1 2420R = , 2 820R = , ( )40 40 40 1

16202Rμ+ +

= = , ( )40 40 40 40 1

103.92312Rσ

⋅ + += = , test

statistic: 2420 1620

7.70103.923

z−

= = . Critical values: 1.96z =± . (Tech: P-value = 0.0000.) Reject the null

hypothesis that the populations have the same median. It appears that the design of quarters changed in 1964.

12. 1 1958R = , 2 670R = , ( )36 36 36 1

13142Rμ+ +

= = , ( )36 36 36 36 1

88.79212Rσ


1958 13147.25

88.792z


that the populations have the same median. It appears that cans of regular Coke have weights different from cans of diet Coke. The cans of regular Coke appear to weigh more because they include more sugar.

13. Using ( )12 12 1

12 11 123.5 86.52

U+

= ⋅ + − = , we get ( )

12 1186.5

2 1.2612 11 12 11 1

12

z

⋅−

= =⋅ ⋅ + +

. The test statistic is

the same value with opposite sign.

14. a.

Rank Rank Sum for Treatment A

1 2 3 4

A A B B 3

A B A B 4

A B B A 5

B B A A 7

B A A B 5

B A B A 6

b. The R values of 3, 4, 5, 6, 7 have probabilities of 1

6,

1

6,

2

6,

1

6, and

1

6, respectively

c. No, none of the probabilities for the values of R would be less than 0.10.

Section 13-5

1. 1 1 10 12.5 5 8 36.5R = + + + + = , 2 3 14 15 2 12.5 6 52.5R = + + + + + = , 3 7 16 10 4 10 47R = + + + + =

Low Lead Level Medium Lead Level High Lead Level

70 (1) 72 (3) 82 (7)

85 (10) 90 (14) 93 (16)

86 (12.5) 92 (15) 85 (10)

76 (5) 71 (2) 75 (4)

84 (8) 86 (12.5) 85 (10)

79 (6)

2. Yes. The samples are independent simple random samples, and each sample has at least five data values.

3. 1 5n = , 2 6n = , 3 5n = , and 5 6 5 16N = + + = .



4. The efficiency rating of 0.95 indicates that with all other factors being the same, the Kruskal-Wallis test requires 100 sample observations to achieve the same results as 95 observations with the parametric one-way analysis of variance test, assuming that the stricter requirements of the parametric test are satisfied.

5. Test statistic: ( )

( )2 2 212 33 22 65

3 15 1 9.980015 15 1 5 5 5

H⎛ ⎞⎟⎜ ⎟= + + − + =⎜ ⎟⎜ ⎟⎜+ ⎝ ⎠

. Critical value: 2 5.991χ = . (Tech:

P-value = 0.0068.) Reject the null hypothesis of equal medians. The data suggest that the different miles present different levels of difficulty.


( )2 2 212 201.5 337.0 128.5

3 36 1 16.948636 36 1 12 12 12

H⎛ ⎞⎟⎜ ⎟= + + − + =⎜ ⎟⎜ ⎟⎜+ ⎝ ⎠

. Critical value:

2 5.991χ = . (Tech: P-value = 0.0002.) Reject the null hypothesis of equal medians. The data suggest that

the books have levels of reading difficulty that are not all the same.


( )2 2 212 86 97 48

3 21 1 4.905421 21 1 7 7 7

H⎛ ⎞⎟⎜ ⎟= + + − + =⎜ ⎟⎜ ⎟⎜+ ⎝ ⎠

. Critical value: 2 5.991χ = . (Tech:

P-value = 0.0861.) Fail to reject the null hypothesis of equal medians. The data do not suggest that larger cars are safer.


( )2 2 212 33.5 85.5 112

3 21 1 11.834921 21 1 7 7 7

H⎛ ⎞⎟⎜ ⎟= + + − + =⎜ ⎟⎜ ⎟⎜+ ⎝ ⎠

. Critical value: 2 5.991χ = .

(Tech: P-value = 0.0027.) Reject the null hypothesis of equal medians. The size of a car does appear to affect highway fuel consumption.


( )2 2 212 5277.5 1112 991.5

3 121 1 8.0115121 121 1 78 22 21

H⎛ ⎞⎟⎜ ⎟= + + − + =⎜ ⎟⎜ ⎟⎜+ ⎝ ⎠

. Critical value:

2 9.210χ = . (Tech: P-value = 0.0182.) Fail to reject the null hypothesis of equal medians. The data do not

suggest that lead exposure has an adverse effect.


( )2 2 212 3629.5 2393.5 1237

3 120 1 59.1546120 120 1 40 40 40

H⎛ ⎞⎟⎜ ⎟= + + − + =⎜ ⎟⎜ ⎟⎜+ ⎝ ⎠

. Critical value:

2 9.210χ = . (Tech: P-value = 0.0000.) Reject the null hypothesis of equal medians. The data suggest that

the amounts of nicotine absorbed by smokers is different from the amounts absorbed by people who don’t smoke.


( )2 2 212 1413.5 650.5 786

3 75 1 27.909875 75 1 25 25 25

H⎛ ⎞⎟⎜ ⎟= + + − + =⎜ ⎟⎜ ⎟⎜+ ⎝ ⎠

. Critical value:

2 5.991χ = . (Tech: P-value: 0.0000.) Reject the null hypothesis of equal medians. There is sufficient

evidence to warrant rejection of the claim that the three different types of cigarettes have the same median amount of nicotine. It appears that the filters do make a difference.


( )2 2 212 82 66.5 82.5

3 21 1 0.614121 21 1 7 7 7

H⎛ ⎞⎟⎜ ⎟= + + − + =⎜ ⎟⎜ ⎟⎜+ ⎝ ⎠

. Critical value: 2 5.991χ = .

(Tech: P-value = 0.7356.) Fail to reject the null hypothesis of equal medians. The data do not suggest that larger cars are safer.



13. Using Σ 16,836T = (see table below) and 25 25 25 75N = + + = , the corrected value of H is

3

27.9098=29.0701

16,8361

75 75−

−

, which is not substantially different from the value found in Exercise 11.

In this case, the large numbers of ties do not appear to have a dramatic effect on the test statistic H.

Nicotine Level Rank t 3t t−

0.2 1.5 2 6

0.6 4.5 2 6

0.7 6.5 2 6

0.8 17.0 19 6840

0.9 28.0 3 24

1.0 33.5 8 504

1.1 48.0 21 9240

1.2 61.0 5 120

1.3 65.5 4 60

1.4 69.0 3 24

1.7 73.5 2 6

SUM 16,836

Section 13-6

1. The methods of Section 10-3 should not be used for predictions. The regression equation is based on a linear correlation between the two variables, but the methods of this section do not require a linear relationship. The methods of this section could suggest that there is a correlation with paired data associated by some nonlinear relationship, so the regression equation would not be a suitable model for making predictions.

2. Data at the nominal level of measurement have no ordering that enables them to be converted to ranks, so data at the nominal level of measurement cannot be used with the methods of rank correlation.

3. r represents the linear correlation coefficient computed from sample paired data; ρ represents the

parameter of the linear correlation coefficient computed from a population of paired data; sr denotes the

rank correlation coefficient computed from sample paired data; sρ represents the rank correlation

coefficient computed from a population of paired data. The subscript s is used so that the rank correlation coefficient can be distinguished from the linear correlation coefficient r. The subscript does not represent the standard deviation s. It is used in recognition of Charles Spearman, who introduced the rank correlation method.

4. The efficiency rating of 0.91 indicates that with all other factors being the same, rank correlation requires 100 pairs of sample observations to achieve the same results as 91 pairs of observations with the parametric test using linear correlation, assuming that the stricter requirements for using linear correlation are met.

5. 1sr = . Critical values are 0.886sr =± (From Table A-9.) Reject the null hypothesis of 0sρ = . There is

sufficient evidence to support a claim of a correlation between distance and time.

6. 1sr =− . Critical values are 0.648sr =± (From Table A-9.) Reject the null hypothesis of 0sρ = . There is

sufficient evidence to support a claim of a correlation between altitude and time.



7. 0.821sr = . Critical values: 0.786sr =± (From Table A-9.) Reject the null hypothesis of 0sρ = . There is

sufficient evidence to support the claim of a correlation between the quality scores and prices. These results do suggest that you get better quality by spending more.

MINITAB Pearson correlation of Rank Price and Rank Quality = 0.821

8. 0.467sr = . Critical values: 0.738sr =± (From Table A-9.) Fail to reject the null hypothesis of 0sρ = .

There is not sufficient evidence to support the claim of a correlation between the quality scores and prices. These results do not suggest that you get better quality by spending more.

MINITAB Pearson correlation of Rank Quality and Rank Price = 0.467

9. 0.929sr =− . Critical values: 0.786sr =± (From Table A-9.) Reject the null hypothesis of 0sρ = . There

is sufficient evidence to support the claim of a correlation between the two judges. Examination of the results shows that the first and third judges appear to have opposite rankings.

MINITAB Pearson correlation of First and Second = -0.929


There is not sufficient evidence to support the claim of a correlation between the two judges. The two judges appear to rank the bands very differently.

MINITAB Pearson correlation of First and Third = 0.607

11. 1sr = . Critical values: 0.886sr =± (From Table A-9.) Reject the null hypothesis of 0sρ = . There is

sufficient evidence to conclude that there is a correlation between overhead widths of seals from photographs and the weights of the seals.

MINITAB Pearson correlation of Rank Width and Rank Weight = 1.000

12. 0.857sr = . Critical values: 0.738sr =± (From Table A-9.) Reject the null hypothesis of 0sρ = . There is

sufficient evidence to conclude that there is a correlation between the number of chirps in 1 min and the temperature.

MINITAB Pearson correlation of Rank Chirps and Rank Temp = 0.857

13. 0.394sr = . Critical values: 1.96

0.31440 1

sr =± =±−

. Reject the null hypothesis of 0sρ = . There is

sufficient evidence to conclude that there is a correlation between the systolic and diastolic blood pressure levels in males.

MINITAB Pearson correlation of RANK SYS and RANK DIAS = 0.394


There is not sufficient evidence to conclude that there is a correlation between brain volumes and IQ scores.

MINITAB Pearson correlation of RANK VOL and RANK IQ = 0.106


0.28648 1

sr =± =±−

. Reject the null hypothesis of 0sρ = . There is

sufficient evidence to conclude that there is a correlation between departure delay times and arrival delay times.

MINITAB Pearson correlation of RANK DEPART and RANK ARRIVE = 0.651




0.28050 1

sr =± =±−

. Fail to reject the null hypothesis of 0sρ = . There

is not sufficient evidence to conclude that there is a correlation between magnitudes and depths of earthquakes.

MINITAB Pearson correlation of RANK MAG and RANK DEPTH = 0.043

17. a. 2

2

2.4470.707

2.447 8 2sr =± =±+ −

is not very close to the values of 0.738sr =± found in Table A-9.

b. 2

2

2.7630.463

2.763 30 2sr =± =±

+ − is quite close to the values of 0.467sr =± found in Table A-9.

Section 13-7

1. No. The runs test can be used to determine whether the sequence of World Series wins by American League teams and National League teams is not random, but the runs test does not show whether the proportion of wins by the American League is significantly greater than 0.5.

2. 1 12n = , 2 8n = , 9G =

3. a. Answers vary, but here is a sequence that leads to rejection of randomness because the number of runs is 2, which is very low: W W W W W W W W W W W W E E E E E E E E

b. Answers vary, but here is a sequence that leads to rejection of randomness because the number of runs is 17, which is very high: W E W E W E W E W E W E W E W E W W W W

4. No. It is very possible that the sequence of data appears to be random, yet the sampling method (such as voluntary response sampling) might be very unsuitable for statistical methods.

5. 1n = 19, 2n =15, 16G = , critical values: 11, 24 (From Table A-10.) Fail to reject randomness. There is

not sufficient evidence to support the claim that we elect Democrats and Republicans in a sequence that is not random. Randomness seems plausible here.

6. 1n = 17, 2n =13, 15G = , critical values: 10, 22 (From Table A-10.) Fail to reject randomness. There is

not sufficient evidence to warrant rejection of the claim that odd and even digits occur in random order.

7. 1n = 20, 2n =10, 16G = , critical values: 9, 20 (From Table A-10.) Fail to reject randomness. There is not

sufficient evidence to reject the claim that the dates before and after July 1 are randomly selected.

8. 1n = 10, 2n =10, 2G = , critical values: 6, 16 (From Table A-10.) Reject randomness. The numbers of

daily newspapers do not appear to be in a random sequence. Because all of the values above the median occur in the beginning and all of the values below the median occur at the end, there appears to be a downward trend in the numbers of daily newspapers.

9. 1n = 24, 2n =21, 17G = , 2 24 21

1 23.424 21Gμ⋅ ⋅

= + =+

, ( )

( ) ( )2

2 24 21 2 24 21 24 213.3007

24 21 24 21 1Gσ

⋅ ⋅ ⋅ ⋅ − −= =

+ + −.

Test statistic: 17 23.4

1.943.3007

z−

= =− . Critical values: 1.96z =± . (Tech: P-value = 0.05252.) Fail to reject

randomness. There is not sufficient evidence to reject randomness. The runs test does not test for disproportionately more occurrences of one of the two categories, so the runs test does not suggest that either conference is superior.



10. 1n = 62, 2n =44, 62G = , 2 62 44

1 52.471762 44Gμ⋅ ⋅

= + =+

, ( )

( ) ( )2

2 62 44 2 62 44 62 444.9741

62 44 62 44 1Gσ

⋅ ⋅ ⋅ ⋅ − −= =

+ + −.

Test statistic: 62 52.4717

1.924.9741

z−

= = . Critical values: 1.96z =± . (Tech: P-value = 0.0554.) Fail to

reject randomness. There is not sufficient evidence to reject randomness.

11. The median is 2453, 1n = 23, 2n =23, 4G = , 2 23 23

1 2423 23Gμ⋅ ⋅

= + =+

,

( )( ) ( )2

2 23 23 2 23 23 23 233.3553

23 23 23 23 1Gσ

⋅ ⋅ ⋅ ⋅ − −= =

+ + −. Test statistic:

4 245.96

3.3553z


1.96z =± . (Tech: P-value = 0.0000.) Reject randomness. The sequence does not appear to be random when considering values above and below the median. There appears to be an upward trend, so the stock market appears to be a profitable investment for the long term, but it has been more volatile in recent years.

12. The mean is 14.250°C, 1n = 26, 2n =24, 8G = , 2 26 24

1 25.9626 24Gμ⋅ ⋅

= + =+

,

( )( ) ( )2

2 26 24 2 26 23 26 243.4936

26 24 26 24 1Gσ

⋅ ⋅ ⋅ ⋅ − −= =

+ + −. Test statistic:

8 25.965.14

3.4935z


1.96z =± . (Tech: P-value = 0.0000.) Reject randomness. The sequence does not appear to be random when considering values above and below the mean. There appears to be an upward trend, so global warming appears to be occurring.

13. a. No solution provided.

b. The 84 sequences yield these results: 2 sequences have 2 runs, 7 sequences have 3 runs, 20 sequences have 4 runs, 25 sequences have 5 runs, 20 sequences have 6 runs, and 10 sequences have 7 runs.

c. With P(2 runs) = 2/84, P(3 runs) = 7/84, P(4 runs) = 20/84, P(5 runs) = 25/84, P(6 runs) = 20/84, and P(7 runs) = 10/84, each of the G values of 3, 4, 5, 6, 7 can easily occur by chance, whereas G = 2 is unlikely because P(2 runs) is less than 0.025. The lower critical value of G is therefore 2, and there is no upper critical value that can be equaled or exceeded.

d. Critical value of G = 2 agrees with Table A-10. The table lists 8 as the upper critical value, but it is impossible to get 8 runs using the given elements.

Chapter Quick Quiz

1. Distribution-free test

2. 57 has rank 1 2 3

23

+ += , 58 has rank 4, and 61 has rank 5.

3. The efficiency rating of 0.91 indicates that with all other factors being the same, rank correlation requires 100 pairs of sample observations to achieve the same results as 91 pairs of observations with the parametric test for linear correlation, assuming that the stricter requirements for using linear correlation are met.

4. The Wilcoxon rank-sum test does not require that the samples be from populations having a normal distribution or any other specific distribution.

5. 4G =

6. Because there are only two runs, all of the values below the mean occur at the beginning and all of the values above the mean occur at the end, or vice versa. This indicates an upward (or downward) trend.

7. Sign test and Wilcoxon signed-ranks test

8. Rank correlation



9. Kruskal-Wallis test

10. Test claims involving matched pairs of data; test claims involving nominal data; test claims about the median of a single population

Review Exercises


244 0.51.65

106 2z

+ −= =− is not less than or equal to the critical value of

1.96z =− . Fail to reject the null hypothesis of 0.5p = . There is not sufficient evidence to warrant

rejection of the claim that in each World Series, the American League team has a 0.5 probability of winning.

2. There are 6 positive signs, 0 negative signs, 0 ties, and 7n = . The test statistic of 0x = is less than or equal to the critical value of 0. There is sufficient evidence to reject the claim of no difference. It appears that there is a difference in cost between flights scheduled 1 day in advance and those scheduled 30 days in advance. Because all of the flights scheduled 30 days in advance cost less than those scheduled 1 day in advance, it is wise to schedule flights 30 days in advance.

3. The test statistic of 0T = is less than or equal to the critical value of 0. There is sufficient evidence to reject the claim that differences between fares for flights scheduled 1 day in advance and those scheduled 30 days in advance have a median equal to 0. Because all of the flights scheduled 1 day in advance have higher fares than those scheduled 30 days in advance, it appears that it is generally less expensive to schedule flights 30 days in advance instead of 1 day in advance.

MINITAB Test of median = 0.000000 versus median not = 0.000000 N for Wilcoxon Estimated N Test Statistic P Median One Day – 30 Days 7 7 28.0 0.022 357.8

4. The sample mean is 54.8 years. 1 19n = , 2 19n = , and the number of runs is 18G = . The critical values

are 13 and 27 (From Table A-10.) Fail to reject the null hypothesis of randomness. There is not sufficient evidence to warrant rejection of the claim that the sequence of ages is random relative to values above and below the mean. The results do not suggest that there is an upward trend or a downward trend.


There is not sufficient evidence to support the claim that there is a correlation between the student ranks and the magazine ranks. When ranking colleges, students and the magazine do not appear to agree.

MINITAB Pearson correlation of Student Ranks and USNEWS Ranks = 0.714


213 0.50.88

32 2z


is not sufficient evidence to warrant rejection of the claim that the population of differences has a median of zero. Based on the sample data, it appears that the predictions are reasonably accurate, because there does not appear to be a difference between the actual high temperatures and the predicted high temperatures.


( )

( )( )

32 32 1230.5

4 0.6232 32 1 2 32 1

24

z

+−

= =−+ ⋅ +

.

Critical values: 1.96z =± . (Tech: P-value = 0.531.) There is not sufficient evidence to warrant rejection of the claim that the population of differences has a median of zero. Based on the sample data, it appears that the predictions are reasonably accurate, because there does not appear to be a difference between the actual high temperatures and the predicted high temperatures.



7. (continued)

MINITAB Test of median = 0.000000 versus median not = 0.000000 N for Wilcoxon Estimated N Test Statistic P Median TEMP 35 32 230.5 0.537 -0.5000


( )2 2 212 91 112.5 174.5

3 27 1 6.630527 27 1 9 9 9

H⎛ ⎞⎟⎜ ⎟= + + − + =⎜ ⎟⎜ ⎟⎜+ ⎝ ⎠

.

Critical value: 2 5.991χ = . (Tech: P-value = 0.0363.) Reject the null hypothesis of equal medians.

Interbreeding of cultures is suggested by the data.

9. 1 60R = , 2 111R = , ( )9 9 9 1

85.52Rμ

+ += = ,

( )9 9 9 9 111.3248

12Rσ⋅ + +

= = , test statistic:

60 85.52.25

11.3248z

−= =− . Critical values: 1.96z =± . (Tech: P-value = 0.0243.) Reject the null hypothesis

that the populations have the same median. Skull breadths from 4000 b.c. appear to have a different median than those from a.d. 150.

10. 0.473sr = . Critical values: 0.587sr =± . Fail to reject the null hypothesis of 0sρ = . There is not

sufficient evidence to support the claim that there is a correlation between weights of plastic and weights of food.

Pearson correlation of Rank Plastic and Rank Rood = 0.473


1. 14.6x = hours, median = 15.0 hours, 1.7s = hours, 2 2.9s = hour2, range = 6.0 hours

2. a. Convenience sample

b. Because the sample is from one class of statistics students, it is not likely to be representative of the population of all fulltime college students.

c. Discrete

d. Ratio

3. The data meet the requirement of being from a normal distribution. H0: 14 hoursμ= . H1: 14 hoursμ> .

Test statistic: 14.55 14

1.4461.701/ 20

t−

= = . Critical value: 1.729t = (assuming a 0.05 significance level). P-

value > 0.05 (Tech: 0.0822). Fail to reject H0. There is not sufficient evidence to support the claim that the mean is greater than 14 hours.

19181716151413121110

0.99

0.95

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.05

0.01



4. The test statistic of 5x = is not less than or equal to the critical value of 4 (from Table A-7.) There is not sufficient evidence to support the claim that the sample is from a population with a median greater than 14 hours.

5. 13.8 hours 15.3 hoursμ< < . We have 95% confidence that the limits of 13.8 hours and 15.3 hours contain

the true value of the population mean.

1.701 1.70114.55 2.101 14.55 2.101

20 20μ− ⋅ < < + ⋅

6. 0.205r = . Critical values: 0.811r =± . P-value = 0.697. There is not sufficient evidence to support the claim of a linear correlation between price and quality score. It appears that you don’t get better quality by paying more.

MINITAB Pearson correlation of Price and Quality = 0.205 P-Value = 0.697

7. 0.543sr =− . Critical values: 0.886sr =± . Fail to reject the null hypothesis of 0sρ = . There is not

sufficient evidence to support the claim that there is a correlation between price and rank. MINITAB Pearson correlation of Rank and Price Rank = -0.543 P-Value = 0.266

8. 0.276 0.343p< < . Because the value of 0.25 is not included in the range of values in the confidence

interval, the result suggests that the percentage of all such telephones that are not functioning is different from 25%.

( )( ) ( )( )229 511 229 511740 740 740 740229 229

1.96 1.96740 740 740 740

p− < < +

9. [ ] ( ) ( )2 2

/ 2

2 2

0.25 1.645 0.251692

0.02αzn

E= = = (Tech: 1691)

10. There must be an error, because the rates of 13.7% and 10.6% are not possible with samples of size 100.

Chapter 14: Statistical Process Control 235


Chapter 14: Statistical Process Control Section 14-2

1. No. If we know that the manufacture of quarters is within statistical control, we know that the three out-of-control criteria are not violated, but we know nothing about whether the specification of 5.670 g is being met. It is possible to be within statistical control by manufacturing quarters with weights that are very far from the desired target of 5.670 g.

2. An x control chart is a plot of sample means and it includes a centerline as well as a line representing an upper control limit and a line representing a lower control limit. An R control chart is a plot of sample ranges and it also includes a centerline as well as a line representing an upper control limit and a line representing a lower control limit. x denotes the mean of all of the 20 sample means, R denotes the mean of the 20 ranges, UCL denotes the value used to locate the upper control limit in a control chart, and LCL denotes the value used to locate the lower control limit in a control chart.

3. To use an x chart without an R chart is to ignore variation, and amounts of variation that are too large will result in too many defective goods or services, even though the mean might appear to be acceptable. To use an R chart without an x chart is to ignore the central tendency, so the goods or services might not vary much, but the process could be drifting so that daily process data do not vary much, but the daily means are steadily increasing or decreasing.

4. The range and mean are both out of statistical control. The control charts show that the elevations are decreasing substantially, so Lake Mead is becoming shallower. The variation is becoming more stable, so the elevations are not varying as much as they did earlier.

5. x = 267.11 lb, R = 54.96 lb, n = 7.

For the R chart: 3LCL 0.076 54.96 4.18 lbD R= = ⋅ = and 4UCL 1.924 54.96 105.74 lbD R= = ⋅ = .

For the x chart: 2LCL 267.11 0.419 54.96 244.08 lbx A R= − = − ⋅ = and

2UCL 267.11 0.419 54.96 290.14 lbx A R= + = + ⋅ = .

6. The run chart does not reveal any patterns suggesting problems that need correcting.

236 Chapter 14: Statistical Process Control


7. The R chart does not violate any of the out-of-control criteria, so the variation of the process appears to be within statistical control

8. The x chart does not violate any of the out-of-control criteria, so the mean of the process appears to be within statistical control.

9. x = 14.250°C, R = 0.414°C, n = 10.

For the R chart: 3LCL 0.223 0.414 0.092 CD R= = ⋅ = and 4UCL 1.777 0.414 0.736 CD R= = ⋅ = .

For the x chart: 2LCL 14.250 0.308 0.414 14.122 Cx A R= − = − ⋅ = and

2UCL 14.250 0.308 0.414 14.377 Cx A R= + = + ⋅ = .

10. The R chart does not violate any of the out-of-control criteria, so the variation of the process appears to be within statistical control.



11. Because there is a pattern of an upward trend and there are points lying beyond the control limits, the x chart shows that the process is out of statistical control.

12. The run chart reveals a very clear pattern of an upward trend, so this graph provides evidence in favor of the theory that we are undergoing global warming.

13. s = 0.0823 g, n = 5. The R chart and the s chart are very similar in their pattern.

3LCL 0 0.0823 0 gB s= = ⋅ = and 4UCL 2.089 0.0823 0.1719 gB s= = ⋅ = .



14. x = 5.6955 g, s = 0.0823 g, n = 5. The two x charts are very similar.

3LCL 5.6955 1.427 0.0823 5.578 gx A s= − = − ⋅ = and

3UCL 5.6955 1.427 0.0823 5.813 gx A s= + = + ⋅ = .

Section 14-3

1. No, the process does not appear to be within statistical control. There is a downward trend, there are at least 8 consecutive points all lying above the centerline, and there are at least 8 consecutive points all lying below the centerline. Because the proportions of defects are decreasing, the manufacturing process is not deteriorating; it is improving.

2. p is the pooled estimate of the proportion of defective items. It is obtained by finding the total number of

defects in all samples combined and dividing that total by the total number of items sampled.

3. LCL denotes the lower control limit. Because the value of –0.000025 is negative and the actual proportion of defects cannot be less than 0, we should replace that value by 0.

4. No. It is very possible that the proportions of defects in repeated samplings behave in a way that makes the process appear to be within statistical control, but the actual proportions of defects could be very high (such as 90%) so that almost all of the dimes fail to meet the manufacturing specifications. Upper and lower control limits of a control chart for the proportion of defects are based on the actual behavior of the process, not the desired behavior.

5. The process appears to be within statistical control.



6. p = 0.282, ( )( )0.282 0.718

LCL 3 0.282 3 0.1470100

pqp

n= − = − =

( )( )0.282 0.718LCL 3 0.282 3 0.4170

100

pqp

n= + = + =

The process appears to be within statistical control. The control chart is the same as the control chart from Exercise 5, except for the values used to locate the centerline and control limits. In this exercise, the proportions of defects are very high. Even though the process is within statistical control, this manufacturing process is yielding far too many defective dimes, so corrective action should be taken to lower the defect rate.

7. p = 0.01407, ( )( )0.01407 0.98593

LCL 3 0.01407 3 0.012953100,000

pqp

n= − = − =

( )( )0.01407 0.98593UCL 3 0.01407 3 0.015187

100,000

pqp

n= + = + =

Because there appears to be a pattern of a downward shift and there are at least 8 consecutive points all lying above the centerline, the process is not within statistical control.

8. p = 0.005176, ( )( )0.005176 0.994824

LCL 3 0.005176 3 0.004495100,000

pqp

n= − = − =

( )( )0.005176 0.994824UCL 3 0.005176 3 0.005857

100,000

pqp

n= + = + =



8. (continued)

There is a pattern of a downward trend and there are at least 8 consecutive points all below the centerline, so the process is not within statistical control. Because there is a downward trend in the rate of violent crimes, we are becoming safer.

9. p = 0.55231, ( )( )0.55231 0.44769

LCL 3 0.55231 3 0.505131000

pqp

n= − = − =

( )( )0.55231 0.44769UCL 3 0.55231 3 0.0.59948

1000

pqp

n= + = + =

The process is out of control because there are points lying beyond the control limits and there are at least 8 points all lying below the centerline. The percentage of voters started to increase in recent years, and it should be much higher than any of the rates shown.

10. p = 0.65361, ( )( )0.65361 0.34639

LCL 3 0.65361 3 0.608471000

pqp

n= − = − =

( )( )0.65361 0.34639UCL 3 0.65361 3 0.69875

1000

pqp

n= + = + =

The process appears to be within statistical control. Ideally, there would be an upward trend due to increasing rates of college enrollments among high school graduates.



11. p = 0.0268, ( )( )0.0268 0.9732

LCL 3 0.0268 3 0.00513500

pqp

n= − = − =

( )( )0.0268 0.9732UCL 3 0.0268 3 0.04847

500

pqp

n= + = + =

There is a pattern of a downward trend and there are at least 8 consecutive points all below the centerline, so the process does not appear to be within statistical control. Because the rate of defects is decreasing, the process is actually improving and we should investigate the cause of that improvement so that it can be continued.

12. p = 0.059335, ( )( )0.059335 0.940665

LCL 3 0.059335 3 0.009218200

pqp

n= − = − =

( )( )0.059335 0.940665UCL 3 0.059335 3 0.10945

200

pqp

n= + = + =

There appears to be a pattern of increasing variation, so the process is not within statistical control. The cause of the increasing variation should be identified and corrected.



13. 10,000 0.00126 12.6np = ⋅ = , ( )( )LCL 3 12.6 3 10,000 0.00126 0.99874 1.9578np npq= − = − =

( )( )UCL 3 12.6 3 10,000 0.00126 0.99874 23.2422np npq= + = + =

Except for the vertical scale, the control chart is identical to the one obtained for Example 1.

Chapter Quick Quiz

1. Process data are data arranged according to some time sequence. They are measurements of a characteristic of goods or services that result from some combination of equipment, people, materials, methods, and conditions.

2. Random variation is due to chance, but assignable variation results from causes that can be identified, such as defective machinery or untrained employees.

3. There is a pattern, trend, or cycle that is obviously not random. There is a point lying outside of the region between the upper and lower control limits. There are at least 8 consecutive points all above or all below the centerline.

4. An R chart uses ranges to monitor variation, but an x chart uses sample means to monitor the center (mean) of a process.

5. No. The R chart has at least 8 consecutive points all lying below the centerline and there are points lying beyond the upper control limit. Also, there is a pattern showing that the ranges have jumped in value for the most recent samples.

6. R = 52.8 ft. In general, a value of R is found by first finding the range for the values within each individual subgroup; the mean of those ranges is the value of R .

7. No. The x chart has a point lying below the lower control limit.

8. x = 3.95 ft. In general, a value of x is found by first finding the mean of the values within each individual

subgroup; the mean of those subgroup means is the value of x .

9. A p chart is a control chart of the proportions of some attribute, such as defective items.

10. Because there is a downward trend, the process is not within statistical control, but the rate of defects is decreasing, so we should investigate and identify the cause of that trend so that it can be continued.

Review Exercises

1. x = 2781.71 kWh, R = 1729.38 kWh, n = 6.

For the R chart: 3LCL 0 1729.38 0 kWhD R= = ⋅ = and

4UCL 2.004 1729.38 3465.678 kWhD R= = ⋅ = .

For the x chart: 2LCL 2781.71 0.483 1729.38 1946.419 kWhx A R= − = − ⋅ = and

2UCL 2781.71 0.483 1729.38 3617.001 kWhx A R= − = − ⋅ = .



2. R = 1729.4 kWh, n = 6. The process variation is within statistical control.

3LCL 0 1729.4 0.0000 kWhD R= = ⋅ = and 4UCL 2.004 1729.4 3465.7176 kWhD R= = ⋅ = .

3. x = 2781.71 kWh, R = 1729.4 kWh, n = 6. The process mean is within statistical control.

2LCL 2781.71 0.483 1729.4 1946.4098 kWhx A R= − = − ⋅ = and

2UCL 2781.71 0.483 1729.4 3617.0102 kWhx A R= + = + ⋅ = .

4. There does not appear to be a pattern suggesting that the process is not within statistical control. There is 1 point that appears to be exceptionally low. (The author’s power company made an error in recording and reporting the energy consumption for that time period.)



5. p = 0.056, ( )( )0.056 0.944

LCL 3 0.056 3 0.01298; use 0100

pqp

n= − = − =−

( )( )0.056 0.944LCL 3 0.056 3 0.12498

100

pqp

n= + = + =

Because there are 8 consecutive points above the centerline and there is an upward trend, the process does not appear to be within statistical control.


1. 0.519 0.581p< < . Because all of the values in the confidence interval estimate of the population

proportion are greater than 0.5, it does appear that the majority of adults believe that it is not appropriate to wear shorts at work.


2. a. 1 0.55 0.45− =

b. ( )50.55 0.0503=

c. ( )51 0.55 0.950− =

3. r = 0.820. Critical values: 0.602r =± . P-value = 0.00202. There is sufficient evidence to support the claim that there is a linear correlation between yields from regular seed and kiln-dried seed. The purpose of the experiment was to determine whether there is a difference in yield from regular seed and kiln-dried seed (or whether kiln-dried seed produces a higher yield), but results from a test of correlation do not provide us with the information we need to address that issue.

MINITAB Pearson correlation of Regular and Kiln-dried = 0.820 P-Value = 0.002

4. H0: 0dμ = . H1: 0dμ < . Test statistic: t = –1.532. Critical value: t = –1.812 (assuming a 0.05 significance

level). P-value < 0.05 (Tech: 0.0783). Fail to reject H0. There is not sufficient evidence to support the claim that kiln-dried seed is better in the sense that it produces a higher mean yield than regular seed. (The sign test can be used to arrive at the same conclusion; the test statistic is x = 3 and the critical value is 1. Also, the Wilcoxon signed-ranks test can be used; the test statistic is T = 13.5 and the critical value is 8.)

Minitab Paired T for Regular - Kiln-dried 95% upper bound for mean difference: 0.200 T-Test of mean difference = 0 (vs < 0): T-Value = -1.53 P-Value = 0.078



5. For the sample of yields from regular seed, x = 20.0 and for the sample of yields from kiln-dried seed, x = 21.0, so there does not appear to be a significant difference. For the sample of yields from regular seed, s = 3.4 and for the sample of yields from kiln-dried seed, s = 4.1, so there does not appear to be a significant difference.

6. p = 0.122, ( )( )0.122 0.878

LCL 3 0.122 3 0.01686; use 050

pqp

n= − = − =−

( )( )0.122 0.878LCL 3 0.122 3 0.26086

50

pqp

n= + = + =

There appears to be a pattern of an upward trend, so the process is not within statistical control.

7.

8. a. 17 15.2

0.72;2.5

z−

= = ( )0.72 23.58%.P z > = With 23.58% of males with head breadths greater than

17 cm, too many males would be excluded.

b. 5th percentile: 15.2 1.645 2.5 11.08 cmx μ z σ= + ⋅ = − ⋅ =

95th percentile: 15.2 1.645 2.5 19.3 cmx μ z σ= + ⋅ = + ⋅ =

9. With a voluntary response sample, the subjects decide themselves whether to be included. With a simple random sample, subjects are selected through some random process in such a way that all samples of the same size have the same chance of being selected. A simple random sample is generally better for use with statistical methods.

10. Sampling method (part c)

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Triola Elem ISM 12e TOC - Palm Beach State College | … does appear to be a potential to create a bias. 6. There does not appear to be a potential to create a bias. 7. There does