# hw7_soln7 - Problem 7.33 Let μ 1 denote the average...

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Problem 7.22 In this problem, we have 507 = n and 28008 . 507 / 142 = p . For a two-sided 99% confidence interval, we use 576 . 2 * = z : n z n z n p p n z z p 2 2 2 2 *) ( 4 *) ( ) 1 ( 2 *) ( 1 * + + ± + - 507 576 . 2 ) 507 ( 4 576 . 2 507 ) 28008 . 1 ( 28008 . ) 507 ( 2 576 . 2 2 2 2 2 1 576 . 2 28008 . + + ± + = - ) 334 . , 231 (. 013078 . 1 ) 0201 (. 576 . 2 28662 . ± = Problem 7.24 Here, we have 487 = n and 072 . % 2 . 7 = = p . For a 99% upper confidence bound, we use the value 33 . 2 * = z : n z n z n p p n z z p 2 2 2 2 *) ( 4 *) ( ) 1 ( 2 *) ( 1 * + + + + - 487 33 . 2 ) 487 ( 4 33 . 2 487 ) 072 . 1 ( 072 . ) 487 ( 2 33 . 2 2 2 2 2 1 33 . 2 072 . + + + + = - = + = .104 011 . 1 ) 012 (. 33 . 2 0776 . upper bound is .104. We are 99% confident that at most 10.4% of all births by nonsmoking women in this metropolitan area result in low birth weight. Problem 7.29 For 90% confidence, the associated z value is 1.645. Since nothing is known about the likely values of π we use .25, the largest possible value of π (1- π ), in the sample size formula: n = (.25) 2 05 . 645 . 1 = 270.6. To be conservative, we round this value up to the next highest integer and use n = 271.

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Unformatted text preview: Problem 7.33 Let μ 1 denote the average toughness for the high-purity steel and let μ 2 denote the average toughness for the commercial purity steel. Then, a lower 95% confidence bound for μ 1-μ 2 is given by: ( 1 x- 2 x ) - z 2 2 2 1 2 1 n s n s + = (65.6-59.2) - (1.645) 32 ) 1 . 1 ( 32 ) 4 . 1 ( 2 2 + = 6.4 - .518 = 5.882. Because this lower interval bound exceeds 5, it gives a reliable indication that the difference between the population toughness levels does exceed 5. Problem 7.34 We need the following equation to be true: ( 29 n n B 2 2 2 1 value critical z σ + = So, ( 29 n n 2 2 2 2 96 . 1 5 . + = Solving for n: 123 ≈ n A sample of 123 batteries of each type should be taken....
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hw7_soln7 - Problem 7.33 Let μ 1 denote the average...

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