HW3_sol - 1. 8‐4 a) 95% CI for μ , n = 10, σ = 20 x =...

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Unformatted text preview: 1. 8‐4 a) 95% CI for μ , n = 10, σ = 20 x = 1000, z = 1.96 x − zσ / n ≤ μ ≤ x + zσ / n 1000 − 1.96(20 / 10 ) ≤ μ ≤ 1000 + 1.96(20 / 10 ) 987.6 ≤ μ ≤ 1012.4 b) .95% CI for μ , n = 25, σ = 20 x = 1000, z = 1.96 x − zσ / n ≤ μ ≤ x + zσ / n 1000 − 1.96(20 / 25 ) ≤ μ ≤ 1000 + 1.96(20 / 25 ) 992.2 ≤ μ ≤ 1007.8 c) 99% CI for μ , n = 10, σ = 20 x = 1000, z = 2.58 x − zσ / n ≤ μ ≤ x + zσ / n 1000 − 2.58(20 / 10 ) ≤ μ ≤ 1000 + 2.58(20 / 10 ) 983.7 ≤ μ ≤ 1016.3 d) 99% CI for μ , n = 25, σ = 20 x = 1000, z = 2.58 x − zσ / n ≤ μ ≤ x + zσ / n 1000 − 2.58(20 / 25 ) ≤ μ ≤ 1000 + 2.58(20 / 25 ) 989.7 ≤ μ ≤ 1010.3 e) When n is larger, the CI is narrower. The higher the confidence level, the wider the CI. 2. 8.7 a) Find n for the length of the 95% CI to be 40. Za/2 = 1.96 1/2 length = (1.96)(20) / n = 20 39.2 = 20 n ⎛ 39.2 ⎞ n=⎜ ⎟ = 3.84 ⎝ 20 ⎠ 2 Therefore, n = 4. b) Find n for the length of the 99% CI to be 40. Za/2 = 2.58 1/2 length = (2.58)(20) / n = 20 51.6 = 20 n ⎛ 51.6 ⎞ n=⎜ ⎟ = 6.66 ⎝ 20 ⎠ 2 Therefore, n = 7. 3. 8‐8 b) Interval (1): 3124.9 ≤ μ ≤ 3215.7 was calculated with 95% Confidence because it has a smaller half‐length, and therefore a smaller confidence interval. The 99% confidence level will make the interval larger. Interval (1): 3124.9 ≤ μ ≤ 3215.7 and Interval (2): 3110.5 ≤ μ ≤ 3230.1 Interval (1): half‐length =90.8/2=45.4 and Interval (2): half‐length =119.6/2=59.8 a) x1 = 3124.9 + 45.4 = 3170.3 x 2 = 3110.5 + 59.8 = 3170.3 The sample means are the same. 4. 8‐14 a) 95% Two‐sided CI on the true mean life of a 75‐watt light bulb For α = 0.05, zα/2 = z0.025 = 1.96 , and ⎯x = 1014, σ =25 , n=20 x − z 0.025 ⎜ ⎜ ⎛σ ⎞ ⎛σ ⎞ ⎟ ≤ μ ≤ x + z 0.025 ⎜ ⎟ ⎟ ⎜ ⎟ ⎝ n⎠ ⎝ n⎠ ⎛ 25 ⎞ ⎛ 25 ⎞ ⎟ ≤ μ ≤ 1014 + 1.96⎜ 1014 − 1.96⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 20 ⎠ ⎝ 20 ⎠ 1003 ≤ μ ≤ 1025 b) 95% One‐sided CI on the true mean piston ring diameter For α = 0.05, zα = z0.05 =1.65 and ⎯ x = 1014, σ =25 , n=20 x − z0.05 σ n ≤μ ⎛ 25 ⎞ 1014 − 1.65⎜ ⎟≤μ ⎝ 20 ⎠ 1005 ≤ μ The lower bound of the one sided confidence interval is lower than the lower bound of the two‐ sided confidence interval even though the level of significance is the same. This is because all of the Type I probability (or α) is in the left tail (or in the lower bound). 5. 8‐23 a) t 0.025,12 = 2.179 d) t 0.0005,15 = 4.073 b) t 0.025, 24 = 2.064 c) t 0.005,13 = 3.012 6. 8‐24 a) t 0.05,14 = 1.761 b) t 0.01,19 = 2.539 c) t 0.001, 24 = 3.467 7. 8‐25 a) Mean = sum = 251.848 = 25.1848 N 10 Variance = = ( stDev) 2 = 1.6052 = 2.5760 b) 95% confidence interval on mean n = 10 x = 25.1848 s = 1.605 t0.025,9 = 2.262 ⎛s⎞ ⎛s⎞ x − t 0.025,9 ⎜ ⎟ ≤ μ ≤ x + t 0.025,9 ⎜ ⎟ ⎝ n⎠ ⎝ n⎠ ⎛ 1.605 ⎞ ⎛ 1.605 ⎞ 25.1848 − 2.262⎜ ⎟ ≤ μ ≤ 25.1848 + 2.262⎜ ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ 24.037 ≤ μ ≤ 26.333 8. 8‐34 The data appear to be normally distributed based on examination of the normal probability plot below. Therefore, there is evidence to support that the solar energy is normally distributed. Probability Plot of Solar Normal - 95% CI 99 Mean StDev N AD P-Value 65.58 4.225 16 0.386 0.349 95 90 80 Percent 70 60 50 40 30 20 10 5 1 50 55 60 65 Solar 70 75 80 95% confidence interval on mean solar energy consumed n = 16 x = 65.58 s = 4.225 t 0.025,15 = 2.131 ⎛s⎞ ⎛s⎞ x − t 0.025,15 ⎜ ⎟ ≤ μ ≤ x + t 0.025,15 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ n⎠ ⎝ n⎠ ⎛ 4.225 ⎞ ⎛ 4.225 ⎞ 65.58 − 2.131⎜ ⎟ ≤ μ ≤ 65.58 + 2.131⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 16 ⎠ ⎝ 16 ⎠ 63.329 ≤ μ ≤ 67.831 9. 8‐39 a) The data appear to be normally distributed based on examination of the normal probability plot below. Therefore, there is evidence to support that the speed‐up of CNN is normally distributed. 99 Probability Plot of Speed Normal - 95% CI Mean StDev N AD P-Value 4.313 0.4328 13 0.233 0.745 95 90 80 Percent 70 60 50 40 30 20 10 5 1 3.0 3.5 4.0 4.5 Speed 5.0 5.5 6.0 b) 95% confidence interval on mean speed‐up n = 13 x = 4.313 s = 0.4328 t 0.025,12 = 2.179 ⎛s⎞ ⎛s⎞ x − t 0.025,12 ⎜ ⎟ ≤ μ ≤ x + t 0.025,12 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ n⎠ ⎝ n⎠ ⎛ 0.4328 ⎞ ⎛ 0.4328 ⎞ 4.313 − 2.179⎜ ⎟ ≤ μ ≤ 4.313 + 2.179⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 13 ⎠ ⎝ 13 ⎠ 4.051 ≤ μ ≤ 4.575 c) 95% lower confidence bound on mean speed‐up n = 13 x = 4.313 s = 0.4328 t 0.05,12 = 1.782 ⎛s⎞ x − t 0.05,12 ⎜ ⎟≤μ ⎜ ⎟ ⎝ n⎠ ⎛ 0.4328 ⎞ ⎟≤μ 4.313 − 1.782⎜ ⎜ ⎟ ⎝ 13 ⎠ 4.099 ≤ μ ...
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This note was uploaded on 09/02/2010 for the course ISYE 2028 taught by Professor Shim during the Summer '07 term at Georgia Tech.

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