OPMT600 CH8 HW.docx - OPMT 600 Chapter 8 Homework N = 60 =...

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OPMT 600 Chapter 8 Homework N = 60 ẋ = 80 σ = 15 3A) (76.2 , 83.8) (1 - ∞) = 0.95 ∞ = 0.05 (∞ / 2) = 0.025 Z a/2 = Z 0.025 = 1.96 (use table) ± Z a/2 (σ / √n) = 80 ± 1.96 (15 / √60) = 80 ± 1.96 (15 / 7.7460) = 80 ± 1.96 (1.9365) = 80 ± 3.7955 = (80 – 3.7955, 80+ 3.7955) = (76.2045 , 83.7955) = (76.2, 83.8) 3B) (77.3, 82.7) ± Z a/2 (σ /√n) = 80 ± 1.96 (15/√120) = 80 ± 1.96 (15 / 10.9545) = 80 ± 1.96 (1.3693) = 80 ± 2.6838 = (80 – 2.6838, 80 + 2.6838) = (77.3162 , 82.6838) = (77.3, 82.7) 3C) The length of the interval decreases as the sample size increases (i.e. if the sample size is larger, then the spread of the interval is smaller). N = ??? = (160+152)/2 = 156 σ = 15 95% = (152, 160) 4) Sample Size = 54 (1 - ∞) = 0.95 ∞ = 0.05 (∞ / 2) = 0.025 Z a/2 = Z 0.025 = 1.96 (use table) ± Z a/2 (σ / √n) 156 + 1.96 (15 / √n) = 160 1.96 (15 / √n) = 4 29.40 = 4* √n √n = 7.35 N = 54.02 12A) t 12, 0.025 = 2.179 12B) t 50, 0.05 = 1.676 12C) t 30, 0.01 = 2.457 12D) t 25, 0.90 = 0.127 12E) t 45, 0.95 = 0.064
N = 65 ẋ = 19.5 σ = 5.2 15) 90% Confidence Interval = (18.4, 20.6) and 95% Confidence Interval = (18.2 , 20.76) (1 - ∞) = 0.90 ∞ = 0.10 (∞ / 2) = 0.05 Z a/2 = Z 0.005 = 1.645 (use table) ± Z a/2 (σ / √n) = 19.5 ± 1.645 (5.2 / √65) = 19.5 ± 1.645 (5.2 / 8.0623) = 19.5 ± 1.645 (0.64498) = 19.5 ± 1.0610 = (19.5 – 1.0610, 19.5+ 1.0610) = (18.439, 20.561) = (18.4, 20.6) (1 - ∞) = 0.95 ∞ = 0.05

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