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Chapter 6 Lab Problems 6.20 Clothing for runners. Your company sells exercise clothing and equipment on the Internet. To design the clothing, you collect data on the physical characteristics of your different types of customers. Here are the weights (in kilograms) for a sample of 24 male runners. Assume that these runners can be viewed as a random sample of your potential male customers. Suppose also that the standard deviation of the population is known to be σ = 4.5 kg. (a) What is , the standard deviation of ? (b) Give a 95% confidence interval for μ, the mean of the population from which the sample is drawn. (c) Will the interval contain the weights of approximately 95% of similar runners? Explain your answer. 6.21 Pounds versus kilograms. Suppose that the weights of the runners in Exercise 6.20 were recorded in pounds rather than kilograms. Use your answers to Exercise 6.20, and the fact that 1 kilogram equals 2.2 pounds, to answer these questions. (a) What is the mean weight of these runners? (b) What is the standard deviation of the mean weight? (c) Give a 95% confidence interval for the mean weight of the population of runners that these runners represent. 6.22 99% versus 95% confidence interval. Find a 99% confidence interval for the mean weight μ of the population of male runners in Exercise 6.20. Is the 99% confidence interval wider or narrower than the 95% interval found in Exercise 6.20? Explain in plain language why this is true. 6.57 Compare student loan debt for different groups of students. One purpose of the National Student Loan Survey is to compare the debt of different subgroups of 8 students. For example, the 525 borrowers who last attended a private four-year college had a mean debt of $21,200, while those who last attended a public four-year college had a mean debt of $17,100. The difference of $4100 is fairly large, but we know that these numbers are estimates of the true means. If we took a different sample, we would get different estimates. Can we conclude from these data that the private four-year students have greater debt than public four-year borrowers? We answer this question by computing the probability of obtaining a difference as large or larger than the observed $4100 assuming that, in fact, there is no difference in the true means. This probability is 0.17. What do you conclude? Illustrate the probability result with a sketch and write a short paragraph explaining your answer 6.69 Corn yield. The mean yield of corn in the United States is about 135 bushels per acre. A survey of 50 farmers this year gives a sample mean yield of = 138.4 bushels per acre. We want to know whether this is good evidence that the national mean this year is not 135 bushels per acre. Assume that the farmers surveyed are an SRS from the population of all commercial corn growers and that the standard deviation of the yield in this population is σ = 10 bushels per acre. Report the value of the test statistic , give a sketch illustrating the P-value, and report the P-value for the test of Are you convinced that the population mean is not 135 bushels per acre? Is your conclusion correct if the distribution of corn yields is somewhat non-Normal? Why? 6.93 Student loan poll. A local television station announces a question for a call-in opinion poll on the six o’clock news and then gives the response on the eleven o’clock news. Today’s question concerns a proposed increase in funds for student loans. Of the 2372 calls received, 1921 oppose the increase. The station, following standard statistical practice, makes a confidence statement: “81% of the Channel 13 Pulse Poll sample oppose the increase. We can be 95% confident that the proportion of all viewers who oppose the increase is within 1.6% of the sample result.” Is the station’s conclusion justified? Explain your answer.