mont4e_sm_ch09_supplemental

mont4e_sm_ch09_supplemental - Supplemental Exercises 9-100...

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9-48 Supplemental Exercises 9-100 α =0.01 a.) n=25 9783 . 0 ) 02 . 2 ( ) 31 . 0 33 . 2 ( 25 / 16 86 85 01 . 0 = Φ = Φ = + Φ = z β n=100 9554 . 0 ) 70 . 1 ( ) 63 . 0 33 . 2 ( 100 / 16 86 85 01 . 0 = Φ = Φ = + Φ = z n=400 8599 . 0 ) 08 . 1 ( ) 25 . 1 33 . 2 ( 400 / 16 86 85 01 . 0 = Φ = Φ = + Φ = z n=2500 2119 . 0 ) 80 . 0 ( ) 13 . 3 33 . 2 ( 2500 / 16 86 85 01 . 0 = Φ = Φ = + Φ = z
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9-49 b) n=25 31 . 0 25 / 16 85 86 0 = = z P -value: 3783 . 0 6217 . 0 1 ) 31 . 0 ( 1 = = Φ n=100 63 . 0 100 / 16 85 86 0 = = z P -value: 2643 . 0 7357 . 0 1 ) 63 . 0 ( 1 = = Φ n=400 25 . 1 400 / 16 85 86 0 = = z P -value: 1056 . 0 8944 . 0 1 ) 25 . 1 ( 1 = = Φ n=2500 13 . 3 2500 / 16 85 86 0 = = z P -value: 0009 . 0 9991 . 0 1 ) 13 . 3 ( 1 = = Φ The data would be statistically significant when n=2500 at α =0.01 9-101 Sample Mean = ± p Sample Variance = ± ( ± ) pp n 1 Sample Size, n Sampling Distribution Sample Mean Sample Variance a. 50 Normal p () 1 50 b. 80 Normal p 1 80 c. 100 Normal p 1 100 d) As the sample size increases, the variance of the sampling distribution decreases. 9-102 n Test statistic P-value conclusion a. 50 12 . 0 50 / ) 10 . 0 1 ( 10 . 0 10 . 0 095 . 0 0 = = z 0.4522 Do not reject H 0 b. 100 15 . 0 100 / ) 10 . 0 1 ( 10 . 0 10 . 0 095 . 0 0 = = z 0.4404 Do not reject H 0 c. 500 37 . 0 500 / ) 10 . 0 1 ( 10 . 0 10 . 0 095 . 0 0 = = z 0.3557 Do not reject H 0 d. 1000 53 . 0 1000 / ) 10 . 0 1 ( 10 . 0 10 . 0 095 . 0 0 = = z 0.2981 Do not reject H 0 e. The P-value decreases as the sample size increases. 9-103. σ = 12, δ = 205 200 = 5, α 2 0 025 = ., z 0.025 = 1.96, a) n = 20: 564 . 0 ) 163 . 0 ( 12 20 5 96 . 1 = Φ = Φ = β
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9-50 b) n = 50: 161 . 0 839 . 0 1 ) 986 . 0 ( 1 ) 986 . 0 ( 12 50 5 96 . 1 = = Φ = Φ = Φ = β c) n = 100: 0116 . 0 9884 . 0 1 ) 207 . 2 ( 1 ) 207 . 2 ( 12 100 5 96 . 1 = = Φ = Φ = Φ = d) , which is the probability of a Type II error, decreases as the sample size increases because the variance of the sample mean decreases. Consequently, the probability of observing a sample mean in the acceptance region centered about the incorrect value of 200 ml/h decreases with larger n. 9-104 σ = 14, δ = 205 200 = 5, α 2 0 025 = ., z 0.025 = 1.96, a) n = 20: 6406 . 0 ) 362 . 0 ( 14 20 5 96 . 1 = Φ = Φ = b) n = 50: 2877 . 0 7123 . 0 1 ) 565 . 0 ( 1 ) 565 . 0 ( 14 50 5 96 . 1 = = Φ = Φ = Φ = c) n = 100: 0537 . 0 9463 . 0 1 ) 611 . 1 ( 1 ) 611 . 1 ( 14 100 5 96 . 1 = = Φ = Φ = Φ = d) The probability of a Type II error increases with an increase in the standard deviation. 9-105 σ = 8, δ = 204 200 = 4, α 2 0 025 = z 0.025 = 1.96. a) n = 20: β= =− = ΦΦ 196 42 0 8 028 .( . ) 1 Φ (0.28) = 1 0.61026 = 0.38974 Therefore, power = 1 β = 0.61026 b) n = 50: 45 0 8 258 . ) 1 Φ (2.58) = 1 0.99506 = 0.00494 Therefore, power = 1 β = 0.995 c) n = 100: 4 100 8 304 . ) 1 Φ (3.04) = 1 0.99882 = 0.00118 Therefore, power = 1 β = 0.9988 d) As sample size increases, and all other values are held constant, the power increases because the variance of the sample mean decreases. Consequently, the probability of a Type II error decreases, which implies the power increases. 9-106 a) α =0.05 n=100 3632 . 0 ) 35 . 0 ( ) 0 . 2 65 . 1 ( 100 / ) 5 . 0 ( 5 . 0
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mont4e_sm_ch09_supplemental - Supplemental Exercises 9-100...

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