Review exercises are short and straightforward exercises that help you solidify the basic ideas and skills in each part of this book. We have provided “hints’’ that indicate where you can find the relevant material for the odd-
IV.1. We’ve been hacked. An October 2014 Gallup Poll asked a random sample of 1017 adults if they or another household member had information from a credit card used at a store stolen by computer hackers in the previous year. Of these adults, 275 said Yes. We can act as if the sample were an SRS. Give a 95% confidence interval for the proportion of all adults who had information from a credit card used at a store by themselves or another household member stolen by computer hackers in the year prior to October 2014. (Hint: See pages 495–499.)
IV.1 0.243 to 0.298.
IV.2. Do you drink? In a July 2014 Gallup Poll a random sample of 1013 adults were asked whether they drank alcoholic beverages (liquor, wine, or beer) or completely abstained from drinking any alcohol. Among respondents, 365 said that they completely abstain from drinking alcohol. Assume that the sample was an SRS. Give a 95% confidence interval for the proportion of all adults who completely abstain from drinking alcoholic beverages.
IV.3. We’ve been hacked. Exercise IV.1 concerns a random sample of 1017 adults. Suppose that (unknown to the pollsters) exactly 25% of all adults had information from a credit card used at a store by themselves or another household member stolen by computer hackers in the year prior to October 2014. Imagine that we take very many SRSs of size 1017 from this population and record the percentage in each sample who claim to have had information from a credit card used at a store by themselves or another household member stolen by computer hackers in the year prior to October 2014. Where would the middle 95% of all values of this percentage lie? (Hint: See pages 495–497.)
IV.3 0.223 to 0.277.
IV.4. Do you drink? Exercise IV.2 concerns a random sample of 1013 adults. Suppose that, in the population of all adults, exactly 35% would say that they completely abstain from drinking alcoholic beverages. Imagine that we take a very large number of SRSs of size 1013. For each sample, we record the proportion ˆp of the sample who would say that they totally abstain from alcoholic beverages.
(a) What is the sampling distribution that describes the values ˆp would take in our samples?
(b) Use this distribution and the 68–
IV.5. Honesty in the media. A Gallup Poll conducted from December 5 to 8, 2013, asked a random sample of 1031 adults to rate the honesty and ethical standards of people in a variety of professions. Among the respondents, 206 rated the honesty and ethical standards of TV reporters as very high or high. Assume that the sample was an SRS. Give a 95% confidence interval for the proportion of all adults who would rate the honesty and ethical standards of TV reporters as very high or high. (Hint: See pages 498–499.)
IV.5 0.175 to 0.224.
IV.6. Roulette. A roulette wheel has 18 red slots among its 38 slots. You observe many spins and record the number of times the ball falls in a red slot. Now you want to use these data to test whether the probability p of the ball falling in a red slot has the value that is correct for a fair roulette wheel. State the hypotheses H0 and Ha that you will test.
IV.7. Why not? Table 11.1 (page 244) records the percentage of residents aged 65 or older in each of the 50 states. You can check that this percentage is 14% or higher in 12 of the states. So the sample proportion of states with at least 14% of elderly residents is ˆp=12/50=0.24. Explain why it does not make sense to go on to calculate a 95% confidence interval for the population proportion p. (Hint: See pages 494–495.)
IV.7 We have information about all states—the whole population of interest—so we already know the value of the population proportion. Thus, there is no need to estimate the population proportion using a confidence interval.
IV.8. Helping welfare mothers. A study compares two groups of mothers with young children who were on welfare two years ago. One group attended a voluntary training program that was offered free of charge at a local vocational school and was advertised in the local news media. The other group did not choose to attend the training program. The study finds a significant difference (P<0.01) between the proportions of the mothers in the two groups who are still on welfare. The difference is not only significant but quite large. The report says that with 95% confidence the percentage of the nonattending group still on welfare is 21% ± 4% higher than that of the group who attended the program. You are on the staff of a member of Congress who is interested in the plight of welfare mothers and who asks you about the report.
(a) Explain in simple language what “a significant difference (P < 0.01)’’ means.
(b) Explain clearly and briefly what “95% confidence’’ means.
(c) This study is not good evidence that requiring job training of all welfare mothers would greatly reduce the percentage who remain on welfare. Explain this to the member of Congress.
IV.9. Beating the system. Some doctors think that health plan rules restrict their ability to treat their patients effectively, so they bend the rules to help patients get reimbursed by their health plans. Here’s a sentence from a study on this topic: “Physicians who agree with the statement ‘Today it is necessary to game the system to provide high-
IV.9 (a) If there were no difference between the two groups of doctors, results like these would be very rare (occur less than 1 in 1000 times). (b) The two proportions are given for comparison and are very different.
IV.10. Smoking in the United States. A July 2015 nationwide random survey of 1009 adults asked whether they had smoked cigarettes in the last week. Among the respondents, 192 said they had. Assume that the sample was an SRS.
(a) Give a 95% confidence interval for the proportion of all American adults who smoked in the week preceding the survey.
(b) Write a short paragraph for a news report based on the survey results.
IV.11. When shall we call you? As you might guess, telephone sample surveys get better response rates during the evening than during the weekday daytime. One study called 2304 randomly chosen telephone numbers on weekday mornings. Of these, 1313 calls were answered and only 207 resulted in interviews. Of 2454 calls on weekday evenings, 1840 were answered and 712 interviews resulted. Give two 95% confidence intervals, for the proportions of all calls that are answered on weekday mornings and on weekday evenings. Are you confident that the proportion is higher in the evening? (Hint: See pages 498–499, 554.)
IV.11 For weekday mornings, 0.550 to 0.590; for weekday evenings, 0.733 to 0.767. The entire confidence interval for the evening proportion is considerably higher than the confidence interval for the morning proportion.
IV.12. Smoking in the United States. Does the survey of Exercise IV.10 provide good evidence that fewer than one-
(a) State the hypotheses to be tested.
(b) If your null hypothesis is true, what is the sampling distribution of the sample proportion ˆp? Sketch this distribution.
(c) Mark the actual value of ˆp on the curve. In your opinion, does it appear surprising enough to give good evidence against the null hypothesis?
IV.13. When shall we call you? Suppose that we know that 57% of all calls made by sample surveys on weekday mornings are answered. We make 2454 calls to randomly chosen numbers during weekday evenings. Of these, 1840 are answered. Is this good evidence that the proportion of answered calls is higher in the evening?
IV.13 (a)H0 : p=0.57 and Ha : p>0.57, where p is the proportion of answered calls in the evening. (b) If H0 is true, then the proportion ˆp from an SRS of 2454 would have (approximately) a Normal distribution with mean 0.57 and standard deviation 0.0100. (c) The observed sample proportion ˆp = 18402454 =0.7498 lies so far out in the high tail of the sampling distribution that it is very strong evidence against the null hypothesis.
IV.14. Not significant. The study cited in Exercise IV.9 looked at the factors that may affect whether doctors bend medical plan rules. Perhaps doctors who fear being prosecuted will bend the rules less often. The study report said, “Notably, greater worry about prosecution for fraud did not affect physicians’ use of these tactics (P = .34).’’ Explain why the result (P = 0.34) supports the conclusion that doctors’ fears about potential prosecution did not affect behavior.
IV.15. Going to church. Opinion polls show that about 40% of Americans say they attended religious services in the last week. This result has stayed stable for decades. Studies of what people actually do, as opposed to what they say they do, suggest that actual church attendance is much lower. One study calculated 95% confidence intervals based on what a sample of Catholics said and then based on a sample of actual behavior. In Chicago, for example, the 95% confidence interval from the opinion poll said that between 45.7% and 51.3% of Catholics attended mass weekly. The 95% confidence interval from actual counts said that between 25.7% and 28.9% attended mass weekly.
(b) The poll in Chicago found that 48.5% of Catholics claimed to attend mass weekly. Why don’t we just say that “48.5% of all Catholics in Chicago claim to attend mass’’ instead of giving the interval 45.7% to 51.3%? (Hint: See pages 554.)
The following exercises are based on the optional sections of Chapters 21 and 22.
IV.15 (a) People are either reluctant to admit that they don’t attend regularly or believe that they attend more regularly than they do. (b) Sample results vary from population truth; while we hope the proportion in our sample is close to the population proportion, we cannot assume the two are exactly equal. Thus, we give a range of plausible values for the population proportion via a confidence interval. (c) Both intervals are based on methods that work (include the true population proportion) about 95% of the time.
IV.16. We’ve been hacked. An October 2014 Gallup Poll asked a random sample of 1017 adults if they or another household member had information from a credit card used at a store stolen by computer hackers in the previous year. Of these adults, 275 said Yes. Assume that the sample was an SRS. Give 90% and 99% confidence intervals for the proportion of all adults who had information from a credit card used at a store by themselves or another household member stolen by computer hackers in the year prior to October 2014. Explain briefly what important fact about confidence intervals is illustrated by comparing these two intervals and the 95% confidence interval from Exercise IV.1.
IV.17. Do you drink? In a July 2014 Gallup Poll a random sample of 1013 adults were asked whether they drank alcoholic beverages (liquor, wine, or beer) or completely abstained from drinking any alcohol. Among respondents, 365 said that they completely abstain from drinking alcohol. Assume that the sample was an SRS. Is this good evidence that more than one-
The following exercises are based on the optional material in Chapters 21, 22, and 24.
IV.17 Our hypotheses are H0 : p=13 and Ha : p>13, where p is the proportion of American adults who say they completely abstain from alcohol. If the null hypothesis is true, then the proportion from an SRS of 1013 adults would have (approximately) a Normal distribution with p = 1/3 and standard deviation 0.0148. Sample proportion: 0.360; Standard score: 1.8; P-value: 0.0359 using Table B. Conclusion: we do have enough evidence to conclude that more than one-third of American adults say they completely abstain from alcohol.
IV.18. Honesty in the media. A Gallup Poll conducted from December 5 to 8, 2013, asked a random sample of 1031 adults to rate the honesty and ethical standards of people in a variety of professions. Among the respondents, 206 rated the honesty and ethical standards of TV reporters as very high or high. Assume that the sample was an SRS. Give a 90% confidence interval for the proportion of all adults who rate the honesty and ethical standards of TV reporters as very high or high. For what purpose might a 90% confidence interval be less useful than a 95% confidence interval? For what purpose might a 90% interval be more useful?
IV.19. A poll of voters. You are the polling consultant to a member of Congress. An SRS of 500 registered voters finds that 37% name “economic problems’’ as the most important issue facing the nation. Give a 90% confidence interval for the proportion of all voters who hold this opinion. Then explain carefully to the member of Congress what your conclusion reveals about voters’ opinions. (Hint: See pages 502–505.)
IV.19 0.335 to 0.406. We are 90% confident the proportion of all voters who think economic problems are the most important issue facing the nation is between 0.335 to 0.406.
IV.20. Smoking in the United States. Carry out the significance test called for in Exercise IV.12 in all detail. Show the five steps of the test (hypotheses, sampling distribution, data, P-value, conclusion) clearly.
IV.21. When shall we call you? Carry out the significance test called for in Exercise IV.13 in all detail. Show the five steps of the test (hypotheses, sampling distribution, data, P-value, conclusion) clearly. (Hint: See pages 528–530.)
IV.21 Our hypotheses are H0 : p=0.57 and Ha : p>0.57, where p is the proportion of answered calls in the evening. If the null hypothesis is true, then the proportion ˆp from an SRS of 2454 would have (approximately) a Normal distribution with mean 0.57 and standard deviation 0.0100. Sample proportion: 0.7498; Standard score: 17.99; P-value: P < 0.0003 based on Table B. Conclusion: we have very strong (overwhelming) evidence that more than 57% of evening calls are answered.
IV.22. CEO pay. A study of 104 corporations found that the pay of their chief executive officers had increased an average of ˉx=6.9 % per year in real terms. The standard deviation of the percentage increases was s = 17.4%.
(a) The 104 individual percentage increases have a right-
(b) Give a 95% confidence interval for the mean percentage increase in pay for all corporate CEOs.
(c) What must we know about the 104 corporations studied to justify the inference you did in part (b)?
exIV-24
IV.23. Water quality. An environmentalist group collects a liter of water from each of 45 random locations along a stream and measures the amount of dissolved oxygen in each specimen. The mean is 4.62 milligrams (mg) and the standard deviation is 0.92 mg. Is this strong evidence that the stream has a mean oxygen content of less than 5 mg per liter? (Hint: See pages 531–535.)
IV.23 We test H0 : μ=5mg/liter and Ha : μ<5mg/liter. If the null hypothesis were true, the sample mean ˉx of an SRS of 45 measurements would have (approximately) a normal distribution with mean 5 mg/liter and approximate standard deviation 0.1371. Standard score: −2.77; P-value: 0.0028. Conclusion: we have strong evidence that the stream has a mean oxygen content of less than 5 mg/liter.
IV.24. Pleasant smells. Do pleasant odors help work go faster? Twenty-
−7.37 | −3.14 | 4.10 | −4.40 | 19.47 | −10.80 | −0.87 |
8.70 | 2.94 | −17.24 | 14.30 | −24.57 | 16.17 | −7.84 |
8.60 | −10.77 | 24.97 | −4.47 | 11.90 | −6.26 | 6.67 |
(a) We hope to show that work is faster on the average with the scented mask. State null and alternative hypotheses in terms of the mean difference in times μ for the population of all adults.
(b) Using a calculator, find the mean and standard deviation of the 21 observations. Did the subjects work faster with the scented mask? Is the mean improvement big enough to be important?
(c) Make a stemplot of the data (round to the nearest whole second). Are there outliers or other problems that might hinder inference?
(d) Test the hypotheses you stated in part (a). Is the improvement statistically significant?
exIV-25
IV.25. Sharks. Great white sharks are big and hungry. Here are the lengths in feet of 44 great whites:
18.7 | 12.3 | 18.6 | 16.4 | 15.7 | 18.3 | 14.6 | 15.8 | 14.9 | 17.6 | 12.1 |
16.4 | 16.7 | 17.8 | 16.2 | 12.6 | 17.8 | 13.8 | 12.2 | 15.2 | 14.7 | 12.4 |
13.2 | 15.8 | 14.3 | 16.6 | 9.4 | 18.2 | 13.2 | 13.6 | 15.3 | 16.1 | 13.5 |
19.1 | 16.2 | 22.8 | 16.8 | 13.6 | 13.2 | 15.7 | 19.7 | 18.7 | 13.2 | 16.8 |
IV.25 (a) The distribution looks reasonably symmetric; other than the low (9.4 feet) and high (22.8 feet) outliers, it appears to be fairly Normal. The mean is ˉx=15.59 feet and s = 2.550 feet. (b) 14.96 to 16.22 feet (c) We need to know what population we are examining: Were these all full-grown sharks? Were they all male? (That is, is μ the mean adult male shark length or something else?) Also, can these numbers be considered an SRS from this population?
IV.26. Pleasant smells. Return to the data in Exercise IV.24. Give a 95% confidence interval for the mean improvement in time to solve a maze when wearing a mask with a floral scent. Are you confident that the scent does improve mean working time?
IV.27. Sharks. Return to the data in Exercise IV.25. Is there good evidence that the mean length of sharks in the population that these sharks represent is greater than 15 feet? (Hint: See pages 531–535.)
IV.27 We test H0 : μ=15 feet and Ha : μ>15 feet. If the null hypothesis were true, the sample mean ˉx of an SRS of 44 sharks would have (approximately) a Normal distribution with mean 15 feet and approximate standard deviation 0.3844 feet. Standard score: 1.53; P-value: 0.0668. This is significant at the 10% level, but not at the 5% level. We have some weak evidence that mean shark length is greater than 15 feet.
IV.28. Simpson’s paradox. If we compare average 2015 SAT mathematics scores, we find that female college-
IV.29. Unhappy HMO patients. A study of complaints by HMO members compared those who filed complaints about medical treatment and those who filed nonmedical complaints with an SRS of members who did not complain that year. Here are the data on the number who stayed and the number who voluntarily left the HMO:
No complaint |
Medical complaint |
Nonmedical complaint |
|
Stayed | 721 | 173 | 412 |
Left | 22 | 26 | 28 |
IV.29 (a) Row and column totals: Stayed 1306; Left 76; No Complaint 743; Medical Complaint 199; Nonmedical Complaint 440. (b) No complaint: 22/743 = 2.96%, Medical complaint: 26/199 = 13.07%, and Nonmedical Complaint: 28/440 = 6.36%. (c) 702.14, 188.06, 415.80, 40.86, 10.94, 24.20 All expected counts are greater than 5, so the chi-square test is safe. (d) We test
H0 : There is no relationship between a member complaining and leaving the HMO.
Ha : There is some relationship between a member complaining and leaving the HMO.
The degrees of freedom are df = 2; this X2 value is so large (and the degrees of freedom are so small) that checking a table should not be necessary: this result is significant at α = 0.001 (and much smaller). We have strong evidence that there is a relationship between a member complaining and leaving the HMO.
IV.30. Treating ulcers. Gastric freezing was once a recommended treatment for stomach ulcers. Use of gastric freezing stopped after experiments showed it had no effect. One randomized comparative experiment found that 28 of the 82 gastric-
(a) Outline the design of this experiment.
(b) Make a two-
(c) Write a brief summary that includes the test result and also percentages that compare the success of the two treatments.
IV.31. When shall we call you? In Exercise IV.11, we learned of a study that dialed telephone numbers at random during two periods of the day. Of 2304 numbers called on weekday mornings, 1313 answered. Of 2454 calls on weekday evenings, 1840 were answered.
IV.31 (a) Morning calls: 1313/2304 = 57.0%; Evening calls: 1840/2454 = 75.0%. (b) The percent of calls answered in the evening versus morning are very different. (c) X2 = 172.080 with df = 1, which is highly significant. There is a relationship between the time at which a call was placed and whether it was answered.