Chapter 6 EXERCISES

  1. 6.91 Why not a 100% confidence interval? The most common confidence levels are 90%, 95%, and 99%. If you were to consider a 100% confidence interval for the mean μ, what would it look like? Explain your reasoning and why such an interval is of no practical use.

  2. 6.92 Telemarketing wages. An advertisement in the student newspaper asks you to consider working for a telemarketing company. The ad states, “Earn between $500 and $1000 per week.” Do you think that the ad is describing a confidence interval? Explain your answer.

  3. 6.93 Exercise and statistics exams. A study examined whether light exercise performed an hour before the final exam in statistics affects how students perform on the exam. The P-value was given as 0.13.

    1. State null and alternative hypotheses that could be used for this study. (Note: There is more than one correct answer.)

    2. Do you reject the null hypothesis? State your conclusion in plain language.

    3. What other facts about the study would you like to know for a proper interpretation of the results?

  4. 6.94 Roulette. A roulette wheel has 18 red slots among its 38 slots. You observe many spins and record the number of times that red occurs. Now you want to use these data to test whether the probability of a red has the value that is correct for a fair roulette wheel. State the hypotheses H0 and Ha that you will test.

  5. 6.95 Food selection by children in school cafeterias. A group of researchers examined whether children’s food selection in a school cafeteria met the standards set by the School Meals Initiative. They measured food selection and food intake of 2049 fourth- through sixth-grade students in 33 schools over a three-day period using digital photography. The following table summarizes some of the food intake measurements:29

    Boys n=852 Girls n=1197
    Food intake Mean St. dev. Mean St. dev.
    Energy (kilojoules) 2448 717 2170 693
    Protein (g) 24.5 7.5 22.1 7.7
    Calcium (mg) 324.1 130.6 265.0 128.9

    Given the large sample sizes, we can assume that the sample standard deviations are the population standard deviations.

    1. Compute 95% confidence intervals for all three intake measures for the boys.

    2. Compute 95% confidence intervals for all three intake measures for the girls.

    3. In the next chapter, we will describe the confidence interval for the difference between two means. For now, let’s compare the boys’ and girls’ confidence intervals for each food intake measure. Do you think these pairs of intervals provide strong evidence against the null hypothesis that the boys and girls consume, on average, the same amount? Explain your answer.

  6. Applet NAEP 6.96 Coverage percent of 95% confidence interval. For this exercise, you will use the Confidence Intervals applet. Set the confidence level at 95% and the number of sample to 10 to simulate 10 confidence intervals. Record the percent hit. Simulate another 10 intervals by resampling. Record the percent hit for your 20 intervals. Repeat the process of simulating 10 additional intervals and recording the results until you have a total of 200 intervals. Create a time plot of your results and write a summary of what you have found.

  7. Applet NAEP 6.97 Coverage percent of 90% confidence interval. Refer to the previous exercise. Do the simulations and report the results for 90% confidence.

  8. NAEP 6.98 Effect of sample size on significance. You are testing the null hypothesis that μ=0 versus the alternative μ>0 using α=0.05. Assume that σ=16. Suppose that x¯=8 and n=10. Calculate the test statistic and its P-value. Repeat assuming the same value of x¯ but with n=20. Do the same for sample sizes of 30, 40, and 50. Plot the values of the test statistic versus the sample size. Do the same for the P-values. Summarize what this demonstration shows about the effect of the sample size on significance testing.

  9. 6.99 Survey response and margin of error. Suppose that a business conducts a marketing survey. As is often done, the survey is conducted by telephone. As it turns out, the business was only able to elicit responses from fewer than 10% of the randomly chosen customers. The low response rate is attributable to many factors, including caller ID screening. Undaunted, the marketing manager was pleased with the sample results because the margin of error was quite small, and thus the manager felt that the business had a good sense of the customers’ perceptions on various issues. Do you think the small margin of error is a good measure of the accuracy of the survey’s results? Explain.

  10. 6.100 Reporting margins of error. An AP News article from July 17, 2014, reported Commerce Department estimates of changes in the construction industry:30

    Construction fell 9.3 percent last month to a seasonally adjusted annual rate of 893,000 homes.

    If we turn to the original Commerce Department report (released the same day), it states: 31

    Privately owned housing starts in June were at a seasonally adjusted annual rate of 893,000. This is 9.3 percent (10.3%) below the revised May estimate of 985,000.

    1. The 10.3% figure is the margin of error based on a 90% level of confidence. Given that fact, what is the 90% confidence interval for the percent change in housing starts from May to June?

    2. Explain why a credible media report should state:

      The Commerce Department has no evidence that privately owned housing starts rose or fell in June from the previous month.

  11. 6.101 CEO pay. A study of the pay of corporate chief executive officers (CEOs) examined the increase in cash compensation of the CEOs of 104 companies, adjusted for inflation, in a recent year. The mean increase in real compensation was x¯=6.9%, and the standard deviation of the increases was s=55%. Is this good evidence that the mean real compensation μ of all CEOs increased that year? The hypotheses are

    H0:   μ=0 (no increase)Ha:   μ>0 (an increase)

    Because the sample size is large, the sample s is close to the population σ, so take σ=55%.

    1. Sketch the Normal curve for the sampling distribution of x¯ when H0 is true. Shade the area that represents the P-value for the observed outcome x¯=6.9%.

    2. Calculate the P-value.

    3. Is the result significant at the α=0.05 level? Do you think the study gives strong evidence that the mean compensation of all CEOs went up?

  12. 6.102 Meaning of “statistically significant.” When asked to explain the meaning of “statistically significant at the α=0.01 level,” a student says, “This means there is only probability 0.01 that the null hypothesis is true.” Is this an essentially correct explanation of statistical significance? Explain your answer.

  13. 6.103 More on the meaning of “statistically significant.” Another student, when asked why statistical significance appears so often in research reports, says, “Because saying that results are significant tells us that they cannot easily be explained by chance variation alone.” Do you think that this statement is essentially correct? Explain your answer.

  14. 6.104 What’s wrong? For each of the following statements, explain what is wrong and why.

    1. The z statistic for a two-sided test is −3.18. The researcher concludes that the null hypothesis is not rejected for α=0.05 because −3.18<1.96.

    2. The P-value for a two-sided test with z=1.67 is 2P(z≤1.67).

    3. The z statistic for a test of H0:   μ=50 with x¯=45, n=25 and σ=10 is z=(50−45)/2.

    4. The probability of a Type I error is equal to 1 minus the power of the test.

  15. NAEP 6.105 Simulation study of the confidence interval. Use a computer to generate n=15 observations from a Normal distribution with mean 20 and standard deviation 5: N(20, 5). Find the 95% confidence interval for μ. Repeat this process 100 times and then count the number of times that the confidence interval includes the value μ=20. Explain your results.

  16. NAEP 6.106 Simulation study of a test of significance. Use a computer to generate n=15 observations from a Normal distribution with mean 20 and standard deviation 5: N(20, 5). Test the null hypothesis that μ=20 using a two-sided significance test. Repeat this process 100 times and then count the number of times that you reject H0. Explain your results.

  17. NAEP 6.107 Another simulation study of a test of significance. Use the same procedure for generating data as in the previous exercise. Now test the null hypothesis that μ=24. Explain your results.

  18. NAEP 6.108 Simulation study of power. Explain why the result from the previous exercise approximates the power of detecting a difference of four units (H0: μ=24 versus μ=20) in this setting.

PUTTING IT ALL TOGETHER

  1. 6.109 Blood phosphorus level in dialysis patients. Patients with chronic kidney failure may be treated by dialysis, in which a machine removes toxic wastes from the blood, a function normally performed by the kidneys. Kidney failure and dialysis can cause other changes, such as retention of phosphorus, that must be corrected by changes in diet. A study of the nutrition of dialysis patients measured the level of phosphorus in the blood of several patients on six occasions. Here are the data for one patient (in milligrams of phosphorus per deciliter of blood):32

    5.4 5.2 4.5 4.9 5.7 6.3

    The measurements are separated in time and can be considered an SRS of the patient’s blood phosphorus level. Assume that this level varies Normally with σ=0.9 mg/dl. Data set icon for pmgdl.

    1. Give a 95% confidence interval for the mean blood phosphorus level.

    2. The normal range of phosphorus in the blood is considered to be 2.6 to 4.8 mg/dl. Is there strong evidence that this patient has a mean phosphorus level that exceeds 4.8?

  2. 6.110 Cellulose content in alfalfa hay. An agronomist examines the cellulose content of a variety of alfalfa hay. Suppose that the cellulose content in the population has standard deviation σ=8 milligrams per gram (mg/g). A sample of 15 cuttings has mean cellulose content x¯=145 mg/g.

    1. Give a 90% confidence interval for the mean cellulose content in the population.

    2. A previous study claimed that the mean cellulose content was μ=140 mg/g, but the agronomist believes that the mean is higher than that figure. State H0 and Ha and carry out a significance test to see if the new data support this belief.

    3. The statistical procedures used in parts (a) and (b) are valid when several assumptions are met. What are these assumptions?

  3. 6.111 Odor threshold of future wine experts. Many food products contain small quantities of substances that would give an undesirable taste or smell if they are present in large amounts. An example is the “off-odors” caused by sulfur compounds in wine. Oenologists (wine experts) have determined the odor threshold, the lowest concentration of a compound that the human nose can detect. For example, the odor threshold for dimethyl sulfide (DMS) is given in the oenology literature as 25 micrograms per liter of wine (μg/l). Untrained noses may be less sensitive, however. Here are the DMS odor thresholds for 10 beginning students of oenology:

    31 31 43 36 23 34 32 30 20 24

    Assume (this is not realistic) that the standard deviation of the odor threshold for untrained noses is known to be σ=7 μg/1. Data set icon for odor.

    1. Make a stemplot to verify that the distribution is roughly symmetric, with no outliers. (A Normal quantile plot confirms that there are no systematic departures from Normality.)

    2. Give a 95% confidence interval for the mean DMS odor threshold among all beginning oenology students.

    3. Are you convinced that the mean odor threshold for beginning students is higher than the published threshold, 25 μg/l? Carry out a significance test to justify your answer.

  4. 6.112 Where do you buy? Consumers can purchase nonprescription medications at food stores, mass merchandise stores such as Target and Walmart, or pharmacies. About 45% of consumers make such purchases at pharmacies. What accounts for the popularity of pharmacies, which often charge higher prices?

    A study examined consumers’ perceptions of overall performance of the three types of stores, using a long questionnaire that asked about such things as “neat and attractive store,” “knowledgeable staff,” and “assistance in choosing among various types of nonprescription medication.” A performance score was based on 27 such questions. The subjects were 201 people chosen at random from the Indianapolis telephone directory. Here are the means and standard deviations of the performance scores for the sample:33

    Store type x¯ s
    Food stores 18.67 24.95
    Mass merchandisers 32.38 33.37
    Pharmacies 48.60 35.62

    We do not know the population standard deviations, but a sample standard deviation s from so large a sample is usually close to σ. Use s in place of the unknown σ in this exercise.

    1. What population do you think the authors of the study want to draw conclusions about? What population are you certain they can draw conclusions about?

    2. Give 95% confidence intervals for the mean performance for each type of store.

    3. Based on these confidence intervals, are you convinced that consumers think that pharmacies offer higher performance than the other types of stores? (In Chapter 12, we will study a statistical method for comparing the means of several groups.)