hwk5_sol

# hwk5_sol - ISOM 111 L6-L7 Fall 2009 1 Homework 5 Solutions...

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ISOM 111 L6-L7, Fall 2009 1 Homework 5 Solutions I . Problems 7.40 (p.281) of the textbook; and (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 p ± z 0 . 025 r ˆ p (1 - ˆ p ) n = [0 . 41 ± 1 . 96 r 0 . 41(1 - 0 . 41) 1000 ] = [0 . 3795 , 0 . 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 first-year 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 = » 0 . 4405(1 - 0 . 4405) · z 0 . 025 0 . 02 · 2 = 2367 . II . Problem 8.48 (p.318) of the textbook (let the significance level α be 0.01). Solution : (a) We’re testing a “not equal to” alternative when σ is known. The observed test statistic z = ¯ x - μ 0 σ/ n = 3 . 3 - 4 0 . 71 / 800 ≈ - 27 . 89 . So the p -value p = P ( | Z | ≥ | z | ) = P ( | Z | ≥ 27 . 89) < P ( | Z | ≥ 3 . 09) 0 . 002 . Since the p -value is smaller than α = 0 . 01, we reject the null hypothesis at 1% significance level. And since the sample mean ¯ x = 3 . 3 < 4, we estimate that the true population mean is smaller than 4. (b) Similarly as in Part (a), the observed test statistic z = ¯ x - μ 0 σ/ n = 4 . 3 - 4 0 . 66 / 500 10 . 16 ,

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ISOM 111 L6-L7, Fall 2009 2 and the p -value p = P ( | Z | ≥ | z | ) = P ( | Z | ≥ 10 . 16) < P ( | Z | ≥ 3 . 09) 0 . 002 . This is smaller than α = 0 . 01, so we reject the null hypothesis at 1% significance level. Finally, since the sample mean ¯ x = 4 . 3 > 4, we estimate that the true population mean is bigger than 4.
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