is correct based on a significance level of 5% [computer help #24].Hint: Make sure that you lay out all the steps involved—as in Section 1.6.1—and include a short sentence summarizing your conclusion; that is, who do you think is correct, the journalist or the student?10 1.10Consider the housing market represented by the sale prices in the HOMES1 data file.As suggested in Section 1.3, calculate the probability of finding an affordable home (less than ) in this housing market. Assume that the population of sale prices () is normal, with mean and standard deviation .As suggested in Section 1.5, calculate a 90% confidence interval for the population mean in this housing market. Recall that the sample mean , the sample standard deviation , and the sample size . Check your answer using statistical software [computer help #23].Practice the mechanics of hypothesis tests by conducting the following tests using a significance level of 5%.: versus : ;: versus : ;: versus : ;: versus : .As suggested in Section 1.7, calculate a 90% prediction interval for an individual sale price in this market.
11 10.11Consider the COUNTRIES data file from Problem 7. Calculate a 95% prediction interval for the variable . Discuss why this interval is so much wider than the confidence interval calculated in Problem 7 part (c).Hint: Calculate by hand (using the fact that the sample mean of is 69.787, the sample standard deviation is 9.2504, and the 97.5th percentile of the t‐distribution with 54 degrees of freedom is approximately 2.005) and check your answer using statistical software (if possible—see the discussion of the“ones trick” in Section 1.7).
12 10.12This problem is adapted from one in Frees (1995). The HOSP data file contains data on charges for patients at a Wisconsin hospital in 1989, as analyzed by Frees (1994). Managers wish to estimate health care costs and to measure how reliable their estimates are. Suppose that a risk manager for a large corporation is trying to understand the cost of one aspect of health care, hospital costs for a small, homogeneous group of claims, the charges (in thousands of dollars) for female patients aged 30–49 who were admitted to the hospital for circulatory disorders.Calculate a 95% confidence interval for the population mean, . Use the following in your calculation: the sample mean, , is 2.9554, the sample standard deviation, , is 1.48104, and the 97.5th percentile of the t‐distribution with 32 degrees of freedom is 2.037. Check your answer using statistical software [computer help #23].Also calculate a 95% prediction interval for an individual claim, . Does this interval seem reasonable given the range of values in the data?Transform the data by taking the reciprocal of the claim values (i.e., ). Calculate a 95% confidence interval for the population mean of the reciprocal‐transformed claims. Use the following sample statistics: the sample mean of is 0.3956 and the sample standard deviation of is 0.12764. Check your answer using statistical software [computer help #23].Back‐transform the endpoints of the interval you just calculated into the original units of (thousands of dollars).Do the same for a 95% prediction interval—that is, calculate the reciprocal‐transformed interval and back‐transform to the original units. Does this interval seem reasonable given the range of values in the data? If so, why did transforming the data help here?
13 10.13The following questions allow you to practice important concepts from Chapter 1 without having to use a computer.In the construction of confidence intervals, will an increase in the sample size lead to a wider or narrower interval (if all other quantities are unchanged)?Suppose that a 95% confidence interval for the population mean, , turns out to be . Give a definition of what it means to be “95% confident” here.A government department is supposed to respond to requests for information within 5 days of receiving the request. Studies show a mean time to respond of 5.28 days and a standard deviation of 0.40 day for a sample of requests. Construct a 90% confidence interval for the mean time to respond. Then do an appropriate hypothesis test at significance level 5% to determine if the mean time to respond exceeds 5 days. (You may find some of the following information useful in answering these questions: The 90th percentile of the t‐distribution with 8 degrees of freedom is 1.397; the 95th percentile of the t‐distribution with 8 degrees of freedom is 1.860.)Students have claimed that the average number of classes missed per student during a quarter is 2. College professors dispute this claim and believe that the average is more than this. They sample students and find that the sample mean is 2.3 and the sample standard deviation is 0.6. State the null and alternative hypotheses that the professors wish to test. Then calculate the test statistic for this test and, using a 5% significance level, determine who appears to be correct, the students or the professors. (You may find some of the following information useful: The 95th percentile of the t‐distribution with 15 degrees of freedom is 1.753; the 97.5th percentile of the t‐distribution with 15 degrees of freedom is 2.131.)Consider the following computer output:Sample size, 150Mean2.94Standard deviation0.50Suppose that we desire a two‐tail test of the null hypothesis that the population mean is equal to 3 versus the alternative hypothesis that the population mean is not equal to 3. Find upper and lower limits for the p‐value for the test. (You may find some of the following information useful: The 90th percentile of the t‐distribution with 149 degrees of freedom is 1.287; the 95th percentile of the t‐distribution with 149 degrees of freedom is 1.655.)A developer would like to see if the average sale price of condominiums in a particular locality has changed in the last 12 months. A study conducted 12 months ago indicated that the average sale price of condominiums in this locality was $. Data on recent sales were as follows:Sample size, 28MeanStandard deviationWrite down the null and alternative hypotheses for this problem. Then specify the rejection region for conducting a two‐tail test at significance level 5%. Based on the computer output, would you reject or fail to reject the null hypothesis for this test? (You may find some of the following information useful: The 95th percentile of the t‐distribution with 27 degrees of freedom is 1.703; the 97.5th percentile of the t‐distribution with 27 degrees of freedom is 2.052.)In a hypothesis test, is it true that the smaller the p‐value, the less likely you are to reject the null hypothesis? Explain.
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