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Unformatted text preview: ISOM 111 L7L8, Fall 2010 1 Homework 5 Solutions I . In a news story distributed by the Washington Post , Lew Sichelman reports that a substantial fraction of mortgage loans that go into default within the first year of the mortgage were approved on the basis of falsified applications. For instance, loan applicants often exaggerate their income or fail to declare debts. Suppose that a random sample of 1,000 mortgage loans that were defaulted within the first year reveals that 410 of these loans were approved on the basis of falsified applications. (a). Find a point estimate of and a 95 percent confidence interval for p , the proportion of all firstyear defaults that are approved on the basis of falsified applications. (b). Based on your interval, what is a reasonable estimate of the minimum percentage of firstyear defaults that are approved on the basis of falsified applications? (c). Based on your 95% confidence interval found in part (a), find the sample size needed so that the margin of error at 95% confidence will be no more than 2%. Solution : (a) A point estimate of the population proportion p is the sample proportion ˆ p = 410 / 1000 = 0 . 41 . A 95% CI for p is &quot; ˆ p ± z . 025 r ˆ p (1 ˆ p ) n # = &quot; . 41 ± 1 . 96 r . 41(1 . 41) 1000 # = [0 . 3795 , . 4405] . (b) Based on the 95% CI, we are 95% confident that p is at least 0.3795 = 37.95%, hence a reasonable estimate of the minimum percentage of firstyear defaults that are approved on the basis of falsified applications is 37.95%. (c) Again, based on the 95% CI, we are 95% confident that p is at most 0.4405, so the sample size needed so that the margin of error at 95% confidence will be no more than 2% is n = . 4405(1 . 4405) · z . 025 . 02 2 = 2367 . II . Recall from Exercise II of Homework 4 that Bayus (1991) studied the mean numbers of auto dealers visited by early and late replacement buyers. (a). Letting μ be the mean number of dealers visited by early replacement buyers, suppose that we wish to test H : μ ≥ 4 versus H a : μ &lt; 4 . A random sample of 800 early replacement buyers yields a mean number of dealers visited of ¯ x = 3 . 3 . Assuming σ equals .71, calculate the p value and test H versus H a at α = 0 . 01 significance level. Interpret your result. (b). Letting μ be the mean number of dealers visited by late replacement buyers, suppose that we wish to test H : μ ≤ 4 versus H a : μ &gt; 4 . A random sample of 500 late replacement buyers yields a mean number of dealers visited of ¯ x = 4.3. Assuming σ equals .66, calculate the p value and test H versus H a at α = 0 . 01 significance level. Interpret your result. ISOM 111 L7L8, Fall 2010 2 Solution : (a) We’re testing a “smaller than” alternative when σ is known. The observed test statistic z = ¯ x μ σ/ √ n = 3 . 3 4 . 71 / √ 800 ≈  27 . 89 ....
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 Fall '10
 YingYingLi
 Business, Statistics, Standard Deviation, Null hypothesis, Statistical hypothesis testing, ISOM

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