[1] In March 16, 1998, issue of Fortune magazine, the results of a survey of 2,221 MBA students from across the United States conducted by the Stockholm-based academic consulting firm Universum showed that only 20 percent of MBA students expect to stay at their first job five years or more. Source: Shalley Branch, “MBAs: What Do They Really Want,” Fortune (March 16, 1998), p.167.a) Assuming that a random sample was selected, construct a 98% confidence interval for the proportion of all U.S. MBA students who expect to stay at their first job five years or more.b) Based on the interval from a), can you conclude that there is strong evidence that less than one-fourth of all U.S. MBA students expect to stay? Explain why.[2] An earlier study claims that U.S. adults spend an average of 114 minutes with their families per day. A recently taken sample of 25 adults showed that they spend an average of 109 minutes per day with their families. The sample standard deviation is 11 minutes. Assume that the time spent by adults with their families has an approximate normal distribution. We wish to test whether the mean time spent currently by all adults with their families is less than 114 minutes a day.a) Construct a 95% confidence interval for the mean time spent by all adults with their families.b) Does the sample information support that the mean time spent currently by all adults with their families is less than 114 minutes a day? Explain your conclusion in words.[3] In the case of Casteneda v. Partida, it was found that during a period of 11 years in Hilda County, Texas, 870 people were selected for grand jury duty, and 39% of them were Americans of Mexican ancestry. Among the people eligible for grand jury duty, 79.1% were Americans of Mexican ancestry. We shall use a 0.01 significance level to test the claim that the selection process is biased against Americans of Mexican ancestry.a) Set up the null and alternative hypotheses, and perform the hypothesis test.b) Does the jury selection system appear to be fair?[4] When 40 people used the Weight Watchers diet for one year, their mean weight loss was 3.00 lb. (based on data from Γ’ΒΒComparison of the Atkins, Ornish, Weight Watchers, and Zone Diets for Weight Loss and Heart Disease Reduction,Γ’ΒΒ by Dansinger, et al., Journal of the American Medical Association, Vol. 293, No. 1). Assume that the standard deviation of all such weight changes iss = 4.9 lb. We shall use a 0.01 significance level to test the claim that the mean weight loss is greater than 0.a) Set up the null and alternative hypotheses, and perform the hypothesis test.b) Based on these results, does the diet appear to be effective? Does the diet appear to have practical significance?[5] A local television station has added a consumer spot to its nightly news. The consumer reporter has recently bought sixteen bottles of aspirin from a local drugstore and has counted the aspirins in each bottle. Although the bottles advertised 500 aspirins, the reporter found the following numbers with the mean count 498.8125:499, 498, 496, 501, 493, 495, 497, 502, 496, 502, 499, 501, 500, 498, 501, 503The consumer reporter claims that this is an obvious case of the public being taken advantage of. Using a confidence interval estimate method or a hypothesis testing method, do you think that the reporterΓ’ΒΒs claim is justifiable?
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