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MBS-ProblemSolutions-Ch11

# MBS-ProblemSolutions-Ch11 - Chapter 11 Inferences About...

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Chapter 11 Inferences About Population Variances Learning Objectives 1. Understand the importance of variance in a decision-making situation. 2 Understand the role of statistical inference in developing conclusions about the variance of a single population. 3. Know that the sampling distribution of ( n - 1) s 2 / σ 2 has a chi-square distribution and be able to use this result to develop a confidence interval estimate of σ 2 . 4. Be able to compute p -values using the chi-square distribution. 5. Know how to test hypotheses involving σ 2 . 6. Understand the role of statistical inference in developing conclusions about two population variances. 7. Know that the sampling distribution of 2 2 1 2 / s s has an F distribution and be able to use this result to test hypotheses involving two population variances. 8. Be able to compute p -values using the F distribution. 11 - 1

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Chapter 11 Solutions: 1. a. 11.070 b. 27.488 c. 9.591 d. 23.209 e. 9.390 2. s 2 = 25 a. With 19 degrees of freedom 2 .05 χ = 30.144 and 2 .95 χ = 10.117 2 19(25) 19(25) 30.144 10.117 σ 15.76 σ 2 46.95 b. With 19 degrees of freedom 2 .025 χ = 32.852 and 2 .975 χ = 8.907 2 19(25) 19(25) 32.852 8.907 σ 14.46 σ 2 53.33 c. 3.8 σ 7.3 3. 2 2 2 2 0 ( 1) (16 1)(9.5) 27.08 50 n s χ σ - - = = = Degrees of Freedom = (16 - 1) = 15 Using 2 χ table, p -value is between .025 and .05 Using Excel, p -value = CHIDIST(27.08,15) = .0281 p -value .05, reject H 0 Critical value approach 2 .05 χ = 24.996 Reject H 0 if 2 χ 24.996 27.08 > 24.996, reject H 0 4. a. n = 18 11 - 2
Inferences About Population Variances s 2 = .36 2 .05 χ = 27.587 2 .95 χ = 8.672 (17 degrees of freedom) 2 17(.36) 17(.36) 27.587 8.672 σ .22 σ 2 .71 b. .47 σ .84 5. a. 2 2 ( ) 31.07 1 i x x s n Σ - = = - 31.07 5.57 s = = b. 2 .025 χ = 16.013 2 .975 χ = 1.690 2 (8 1)(31.07) (8 1)(31.07) 16.013 1.690 σ - - 13.6 σ 2 128.7 c. 3.7 σ 11.3 6. a. 4.41 .2205 12 i x x n = = = 2 2 ( ) 47.9500 1 i x x s n Σ - = = - 47.9500 6.92 s = = b. 2 .025 χ = 32.852 2 .975 χ = 8.907 2 (20 1)(47.95) (20 1)(47.95) 32.852 8.907 σ - - 27.73 σ 2 102.28 5.27 σ 10.11 The 95% confidence interval for the population standard deviation ranges from a quarterly standard deviation of 5.27% to 10.11%. 7. a. 2 2 ( ) 57.7230 1 i x x s n Σ - = = - 11 - 3

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Chapter 11 57.7230 7.60 s = = b. With 11 df, 2 .025 χ = 21.920 2 .975 χ = 3.816 2 (12 1)(57.7230) (12 1)(57.7230) 21.920 3.816 σ - - 28.9668 σ 2 166.3923 c. 5.38 σ 12.90 8. a. 2 2 ( ) .0929 .00845 1 12 1 i x x s n Σ - = = = - - b. .00845 .092 s = = c. 11 degrees of freedom 2 .025 χ = 21.920 2 .975 χ = 3.816 2 (12 1).00845 (12 1).00845 21.920 3.816 σ - - .0042 σ 2 .0244 .065 σ .156 9. H 0 : σ 2 .0004 H a : σ 2 .0004 2 2 2 0 ( 1) (30 1)(.0005) 36.25 .0004 n s χ σ - - = = = Degrees of freedom = n - 1 = 29 Using 2 χ table, p -value is greater than .10 Using Excel, p -value = CHIDIST(36.25,29) = .1664 p -value > .05, do not reject H 0 . The product specification does not appear to be violated.
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MBS-ProblemSolutions-Ch11 - Chapter 11 Inferences About...

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