6.12 Margin of error and the confidence interval. A study of stress on the campus of your university reported a mean stress level of 78 (on a 0 to 100 scale with a higher score indicating more stress) with a margin of error of 5 for 95% confidence. The study was based on a random sample of 64 undergraduates.

6.13 Changing the sample size. Consider the setting of the previous exercise. Suppose that the sample mean is again 78 and the population standard deviation is 20. Make a diagram similar to Figure 6.5 (page 351) that illustrates the effect of sample size on the width of a 95% interval. Use the following sample sizes: 9, 25, 81, and 100. Summarize what the diagram shows.

6.14 Changing the confidence level. Consider the setting of the previous two exercises. Suppose that the sample mean is still 78, the sample size is 64, and the population standard deviation is 20. Make a diagram similar to Figure 6.6 (page 353) that illustrates the effect of the confidence level on the width of the interval. Use 80%, 90%, 95%, and 99%. Summarize what the diagram shows.

6.15 Confidence interval mistakes and misunderstandings. Suppose that 500 randomly selected alumni of the University of Okoboji were asked to rate the university’s academic advising services on a 1 to 10 scale. The sample mean was found to be 8.6. Assume that the population standard deviation is known to be σ = 2.2.

6.16 More confidence interval mistakes and misunderstandings. Suppose that 100 randomly selected members of the Karaoke Channel were asked how much time they typically spend on the site during the week.⁶ The sample mean was found to be 3.8 hours. Assume that the population standard deviation is known to be σ = 2.9.

6.17 The state of stress in the United States. Since 2007, the American Psychological Association has supported an annual nationwide survey to examine stress across the United States.⁷ This year, a total of 720 millennials (18- to 33-year-olds) were asked to indicate their average stress level (on a 10-point scale) during the past month. The mean score was 5.5. Assume that the population standard deviation is 2.8.

6.18 Inference based on integer values. Refer to Exercise 6.17. The data for this study are integer values between 1 and 10. Explain why the confidence interval based on the Normal distribution should be a good approximation.

6.19 Mean TRAP in young women. For many important processes that occur in the body, direct measurement of characteristics of the process is not possible. In many cases, however, we can measure a biomarker, a biochemical substance that is relatively easy to measure and is associated with the process of interest. Bone turnover is the net effect of two processes: the breaking down of old bone, called resorption, and the building of new bone, called formation. One biochemical measure of bone resorption is tartrate-resistant acid phosphatase (TRAP), which can be measured in blood. In a study of bone turnover in young women, serum TRAP was measured in 31 subjects.⁸ The mean was 13.2 units per liter (U/l). Assume that the standard deviation is known to be 6.5 U/l. Give the margin of error and find a 95% confidence interval for the mean TRAP amount in young women represented by this sample.

6.20 Mean OC in young women. Refer to the previous exercise. A biomarker for bone formation measured in the same study was osteocalcin (OC), measured in the blood. For the 31 subjects in the study, the mean was 33.4 nanograms per milliliter (ng/ml). Assume that the standard deviation is known to be 19.6 ng/ml. Report the 95% confidence interval.

6.21 Populations sampled and margins of error. Consider the following two scenarios. (A) Take a simple random sample of 200 freshman students at your college or university. (B) Take a simple random sample of 200 students at your college or university. For each of these samples, you will record the amount spent on textbooks used for classes during the fall semester. Which sample should have the smaller margin of error? Explain your answer.

6.22 Average starting salary. The National Association of Colleges and Employers (NACE) Spring Salary Survey shows that the current class of college graduates received an average starting-salary offer of $48,127.⁹ Your institution collected an SRS (n = 300) of its recent graduates and obtained a 95% confidence interval of ($46,382, $48,008). What can we conclude about the difference between the average starting salary of recent graduates at your institution and the overall NACE average? Write a short summary.

6.23 Consumption of sweet snacks. A recent study reported that the U.S. per capita consumption of sweet snacks among healthy weight children aged 12 to 19 years is 251.2 kilocalories per day (kcal/d).¹⁰ This was based on 24-hour dietary recall records of n = 2265 adolescents.

6.24 Total sleep time of college students. In Example 5.4 (page 293), the total sleep time per night among college students was approximately Normally distributed with mean μ = 6.78 hours and standard deviation σ = 1.24 hours. You initially plan to take an SRS of size n = 175 and compute the average total sleep time.

6.25 Determining sample size. Refer to the previous exercise. You really want to use a sample size such that about 95% of the averages fall within ±5 minutes of the true mean μ = 6.78.

6.26 Inference based on skewed data. The mean OC for the 31 subjects in Exercise 6.20 was 33.4 ng/ml. In our calculations, we assumed that the standard deviation was known to be 19.6 ng/ml. Use the 68–95–99.7 rule from Chapter 1 (page 57) to find the approximate bounds on the values of OC that would include these percents of the population. If the assumed standard deviation is correct, this distribution may be highly skewed. Why? (Hint: The measured values for a variable such as this are all positive.) Do you think that this skewness will invalidate the use of the Normal confidence interval in this case? Explain your answer.

6.27 Average hours per week listening to the radio. The Student Monitor surveys 1200 undergraduates from four-year colleges and universities throughout the United States semiannually to understand trends among college students.¹¹ Recently, the Student Monitor reported that the average amount of time listening to the radio per week was 11.5 hours. Of the 1200 students surveyed, 83% said that they listened to the radio, so this collection of listening times has around 204 (17% × 1200) zeros. Assume that the standard deviation is 8.3 hours.

6.28 Average minutes per week listening to the radio. Refer to the previous exercise.

6.29 Outlook on life. Since 2008, the Gallup-Healthways Well-Being Index tracks how people feel about their daily lives. In 2014, 54.1% of the respondents were classified as “thriving.” This classification is based on how a respondent rates his or her current and future lives. This is the highest percent of respondents in this category since the index started. Material provided with the results noted:

Results are based on telephone interviews . . . with a random sample of 176,903 adults, living in all 50 U.S. states and the District of Columbia. For results based on the total sample of national adults, the margin of sampling error is ±1 percentage points at the 95% confidence level.¹²

The poll uses a complex multistage sample design, but the sample percent has approximately a Normal sampling distribution.

6.30 Fuel efficiency. Computers in some vehicles calculate various quantities related to performance. One of these is the fuel efficiency, or gas mileage, usually expressed as miles per gallon (mpg). For one vehicle equipped in this way, the miles per gallon were recorded each time the gas tank was filled, and the computer was then reset.¹³ Here are the mpg values for a random sample of 20 of these records:

Suppose that the standard deviation is known to be σ = 3.5 mpg.

6.31 Fuel efficiency in metric units. In the previous exercise, you found an estimate with a margin of error for the average miles per gallon. Convert your estimate and margin of error to the metric units kilometers per liter (kpl). To change mpg to kpl, use the fact that 1 mile = 1.609 kilometers and 1 gallon = 3.785 liters.

6.32 How many “hits”? The Confidence Interval applet lets you simulate large numbers of confidence intervals quickly. Select 95% confidence and then sample 50 intervals. Record the number of intervals that cover the true value (this appears in the “Hit” box in the applet). Press the “Reset” button and repeat 30 times. Make a stemplot of the results and find the mean. Describe the results. If you repeated this experiment very many times, what would you expect the average number of hits to be?

6.33 Required sample size for specified margin of error. A new bone study is being planned that will measure the biomarker TRAP described in Exercise 6.19. Using the value of σ given there, 6.5 U/l, find the sample size required to provide an estimate of the mean TRAP with a margin of error of 1.5 U/l for 95% confidence.

6.34 Adjusting required sample size for dropouts. Refer to the previous exercise. In similar previous studies, about 20% of the subjects drop out before the study is completed. Adjust your sample size requirement so that you will have enough subjects at the end of the study to meet the margin of error criterion.

6.35 Radio poll. A national public radio (NPR) station invites listeners to enter a dispute about a proposed “pay as you throw” waste collection program. The station asks listeners to call in and state how much each 10 gallon bag of trash should cost. A total of 179 listeners call in. The station calculates the 95% confidence interval for the average fee to be $0.53 to $1.39. Is this result trustworthy? Explain your answer.

6.36 Accuracy of a laboratory scale. To assess the accuracy of a laboratory scale, a standard weight known to weigh 10 grams is weighed repeatedly. The scale readings are Normally distributed with unknown mean (this mean is 10 grams if the scale has no bias). The standard deviation of the scale readings is known to be 0.0002 gram.

6.37 More than one confidence interval. As we prepare to take a sample and compute a 95% confidence interval, we know that the probability that the interval we compute will cover the parameter is 0.95. That’s the meaning of 95% confidence. If we plan to use several such intervals, however, our confidence that all of them will give correct results is less than 95%. Suppose that we plan to take independent samples each month for five months and report a 95% confidence interval for each set of data.