HW4_sol - 1. 8‐43 a) 95% upper CI and df = 24 b) 99%...

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Unformatted text preview: 1. 8‐43 a) 95% upper CI and df = 24 b) 99% lower CI and df = 9 c) 90% CI and df = 19 2 2 χ α / 2,df = χ 02.05,19 = 30.14 and χ12−α / 2,df = χ 0.95,19 = 10.12 χ 12−α ,df = χ 02.95, 24 = 13.8484 2 χ α ,df = χ 02.01,9 = 21.6660 2. 8‐52 a) 99% two‐sided confidence interval on σ2 n = 10 s = 1.913 χ 02.005,9 = 23.59 and χ 02.995,9 = 1.73 9(1.913) 2 9(1.913) 2 ≤σ2 ≤ 23.59 1.73 2 1.396 ≤ σ ≤ 19.038 b) 99% lower confidence bound for σ2 2 2 For α = 0.01 and n = 10, χ α , n −1 = χ 0.01,9 = 21.67 9(1.913) 2 ≤σ2 21.67 2 1.5199 ≤ σ c) 90% lower confidence bound for σ2 2 2 For α = 0.1 and n = 10, χ α , n −1 = χ 0.1,9 = 14.68 9(1.913) 2 ≤σ2 14.68 2.2436 ≤ σ 2 1.498 ≤ σ d) The lower confidence bound of the 99% two‐sided interval is less than the one‐sided interval. The lower confidence bound for σ2 in part (c) is greater because the confidence is lower. 3. 8‐58 a) 95% Confidence Interval on the true proportion of helmets showing damage ˆ p= 18 = 0.36 50 ˆ p − zα / 2 n = 50 zα / 2 = 1.96 ˆ ˆ p (1 − p ) n ˆ ˆ p (1 − p ) ˆ ≤ p ≤ p + zα / 2 n 0.36 − 1.96 0.36(0.64) 0.36(0.64) ≤ p ≤ 0.36 + 1.96 50 50 0.227 ≤ p ≤ 0.493 b) n = ⎜ ⎛ zα / 2 ⎞ ⎛ 1.96 ⎞ ⎟ p (1 − p ) = ⎜ ⎟ 0.36(1 − 0.36) = 2212.76 ⎝ 0.02 ⎠ ⎝E⎠ 2 2 n ≅ 2213 c) n = ⎜ ⎛ zα / 2 ⎞ ⎛ 1.96 ⎞ ⎟ p (1 − p ) = ⎜ ⎟ 0.5(1 − 0.5) = 2401 ⎝ 0.02 ⎠ ⎝E⎠ 2 2 4. 8‐40 95% lower bound confidence for the mean wall thickness given x = 4.05 s = 0.08 n = 25 tα,n‐1 = t0.05,24 = 1.711 x − t 0.05, 24 ⎜ ⎜ ⎛s⎞ ⎟≤μ ⎟ ⎝ n⎠ 4.05 − 1.711⎜ ⎜ ⎛ 0.08 ⎞ ⎟≤μ ⎟ ⎝ 25 ⎠ 4.023 ≤ μ There is high confidence that the true mean wall thickness is greater than 4.023 mm. 5. 8‐70 90% prediction interval on wall thickness on the next bottle tested. given x = 4.05 s = 0.08 n = 25 for tα/2,n‐1 = t0.05,24 = 1.711 x − t 0.05, 24 s 1 + 4.05 − 1.711(0.08) 1 + 1 1 ≤ x n +1 ≤ x + t 0.05, 24 s 1 + n n 1 1 ≤ x n +1 ≤ 4.05 − 1.711(0.08) 1 + 25 25 3.91 ≤ x n +1 ≤ 4.19 6. 8‐82 a) 90% tolerance interval on wall thickness measurements that have a 90% CL given x = 4.05 s = 0.08 n = 25 we find k=2.077 x − ks, x + ks 4.05 − 2.077(0.08), 4.05 + 2.077(0.08) (3.88, 4.22) The lower bound of the 90% tolerance interval is much lower than the lower bound on the 95% confidence interval on the population mean (4.023 ≤ μ ≤ ∞) b) 90% lower tolerance bound on bottle wall thickness that has confidence level 90%. given x = 4.05 s = 0.08 n = 25 and k = 1.702 4.05 − 1.702(0.08) 3.91 The lower tolerance bound is of interest if we want to make sure the wall thickness is at least a certain value so that the bottle will not break. x − ks 7. 9‐ 1 a) H 0 : μ = 25, H1 : μ ≠ 25 Yes, because the hypothesis is stated in terms of the parameter of interest, inequality is in the alternative hypothesis, and the value in the null and alternative hypotheses matches. b) H 0 c) H 0 : σ > 10, H1 : σ = 10 No, because the inequality is in the null hypothesis. : x = 50, H1 : x ≠ 50 No, because the hypothesis is stated in terms of the statistic rather than the parameter. d) H 0 : p = 0.1, H1 : p = 0.3 No, the values in the null and alternative hypotheses do not match and both of the hypotheses are equality statements. e) H 0 : s = 30, H1 : s > 30 No, because the hypothesis is stated in terms of the statistic rather than the parameter. ...
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