Chapter 11 Inferences About Population Variances Learning Objectives Understand the importance of variance in a decision-making situation Understand the role of statistical inference in developing conclusions about the variance of a single population Know that the sampling distribution of (n - 1) s2/σ has a chi-square distribution and be able to use this result to develop a confidence interval estimate of σ Be able to compute p-values using the chi-square distribution Know how to test hypotheses involving σ Understand the role of statistical inference in developing conclusions about two population variances Know that the sampling distribution of s12 / s22 has an F distribution and be able to use this result to test hypotheses involving two population variances Be able to compute p-values using the F distribution Solutions: 11 - © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 11 a 11.070 b 27.488 c 9.591 d 23.209 e 9.390 s2 = 25 2 a With 19 degrees of freedom χ.05 = 30.144 and χ.95 = 10.117 19(25) 19(25) ≤σ2 ≤ 30.144 10.117 15.76 ≤ σ ≤ 46.95 2 b With 19 degrees of freedom χ.025 = 32.852 and χ.975 = 8.907 19(25) 19(25) ≤σ2 ≤ 32.852 8.907 14.46 ≤ σ ≤ 53.33 c 3.8 ≤ σ ≤ 7.3 χ2 = (n − 1) s (16 − 1)(9.5)2 = = 27.08 σ 02 50 Degrees of Freedom = (16 - 1) = 15 Using χ table, p-value is between 025 and 05 Exact p-value using χ =27.08 is 0281 p-value ≤ 05, reject H0 Critical value approach χ.05 = 24.996 Reject H0 if χ ≥ 24.996 27.08 > 24.996, reject H0 a n = 18 11 - © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Inferences About Population Variances s2 = 36 χ.05 = 27.587 χ.95 = 8.672 (17 degrees of freedom) 17(.36) 17(.36) ≤σ2 ≤ 27.587 8.672 22 ≤ σ ≤ 71 b .47 ≤ σ ≤ 84 a s2 = Σ( xi − x )2 = 31.07 n −1 s = 31.07 = 5.57 b χ.025 = 16.013 χ.975 = 1.690 (8 − 1)(31.07) (8 − 1)(31.07) ≤σ2 ≤ 16.013 1.690 13.6 ≤ σ ≤ 128.7 c 3.7 ≤ σ ≤ 11.3 a x= ∑ xi 4.41 = = 2205 n 12 s2 = Σ( xi − x )2 = 47.9500 n −1 s = 47.9500 = 6.92 b χ.025 = 32.852 χ.975 = 8.907 (20 − 1)(47.95) (20 − 1)(47.95) ≤σ2 ≤ 32.852 8.907 27.73 ≤ σ ≤ 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% a s2 = Σ( xi − x )2 = 57.7230 n −1 11 - © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 11 s = 57.7230 = 7.60 b With 11 df, χ.025 = 21.920 χ.975 = 3.816 (12 − 1)(57.7230) (12 − 1)(57.7230) ≤σ2 ≤ 21.920 3.816 28.9668 ≤ σ ≤ 166.3923 c 5.38 ≤ σ ≤ 12.90 a x = 78 s b ∑ (x − x) = i n −1 = 5.2230 = 4748 12 − s = 4748 = 6891 c 11 degrees of freedom χ.025 = 21.920 χ.975 = 3.816 (12 − 1).4748 (12 − 1).4748 ≤σ2 ≤ 21.920 3.816 2383 ≤ σ ≤ 1.3687 4882 ≤ σ ≤ 1.1699 H0: σ ≤ 0004 Ha: σ > 0004 χ2 = (n − 1) s (30 − 1)(.0005) = = 36.25 σ 02 0004 Degrees of freedom = n - = 29 Using χ table, p-value is greater than 10 Exact p-value using χ = 36.25 is 1664 p-value > 05, not reject H0 The product specification does not appear to be violated 10 σ = (18.2)2 = 331.24 H0: σ ≤ 331.24 Ha: σ > 331.24 11 - © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Inferences About Population Variances χ2 = (n − 1) s (36 − 1)(22.2) = = 52.075 σ 02 (18.2) Degrees of freedom = n - = 35 Using χ table, p-value is between 025 and 05 Exact p-value corresponding to χ = 52.075 is 0317 p-value ≤ 05, reject H0 The standard deviation for Vanguard PRIMECAP fund is greater than the average standard deviation for large cap mutual funds 11 a x= ∑x s2 = ∑(x i n = i 40.28 = 3.3567 12 − x )2 n −1 = 7.5887 = 6899 11 s = 6899 = 8306 b H0: σ = 70 Ha: σ ≠ 70 c χ2 = (n − 1) s (12 − 1)(.6899) = = 10.84 70 σ 02 Degrees of freedom = n - = 11 Using χ table, area in upper tail is greater than 10 Two-tail p-value is greater than 20 Exact two-tailed p-value corresponding to χ = 10.84 is 2(.4568) = 9136 p-value > 05, not reject H0 We cannot conclude the variance in bond yields has changed 12 a s2 = Σ( xi − x )2 = 8106 n −1 b H0: σ = 94 Ha: σ ≠ 94 χ2 = (n − 1) s (12 − 1)(.8106) = = 9.49 σ 02 94 Degrees of freedom = n - = 11 11 - © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 11 Using χ table, area in tail is greater than 10 Two-tail p-value is greater than 20 Exact p-value corresponding to χ = 9.49 is 8465 p-value > 05, cannot reject H0 13 a F.05 = 3.33 b F.025 = 2.76 c F.01 = 4.50 d 14 a F.10 = 1.94 F= s12 5.8 = = 2.4 s22 2.4 Degrees of freedom 15 and 20 Using F table, p-value is between 025 and 05 Exact p-value corresponding to F = 2.4 is 0345 p-value ≤ 05, reject H0 Conclude σ 12 > σ 22 b F.05 = 2.20 Reject H0 if F ≥ 2.20 2.4 ≥ 2.20, reject H0 Conclude σ 12 > σ 22 15 a Larger sample variance is s12 F= s12 8.2 = = 2.05 s22 Degrees of freedom 20 and 25 Using F table, area in tail is between 025 and 05 Two-tail p-value is between 05 and 10 Exact p-value corresponding to F = 2.05 is 0904 p-value > 05, not reject H0 b Since we have a two-tailed test Fα / = F.025 = 2.30 11 - © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Inferences About Population Variances Reject H0 if F ≥ 2.30 2.05 < 2.30, not reject H0 16 For this type of hypothesis test, we place the larger variance in the numerator So the Fidelity variance is given the subscript of H :σ 12 ≤ σ 22 H a :σ 12 > σ 22 F= s12 18.92 = = 1.59 s22 15.0 Degrees of freedom in the numerator and denominator are both 60 Using the F table, p-value is less than 05 Exact p-value corresponding to F = 1.59 is 0375 p-value ≤ 05, reject H0 We conclude that the Fidelity fund has a greater variance than the American Century fund 17 a Population is year old automobiles H :σ 12 ≤ σ 22 H a :σ 12 > σ 22 b F= s12 170 = = 2.89 s22 1002 Degrees of freedom 25 and 24 Using F table, p-value is less than 01 Exact p-value corresponding to F = 2.89 is 0057 p-value ≤ 01, reject H0 Conclude that year old automobiles have a larger variance in annual repair costs compared to year old automobiles This is expected due to the fact that older automobiles are more likely to have some more expensive repairs which lead to greater variance in the annual repair costs 18 We place the larger sample variance in the numerator So, the Merrill Lynch variance is given the subscript of H :σ 12 = σ 22 11 - © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 11 H a :σ 12 ≠ σ 22 F= s12 587 = = 1.44 s22 4892 Degrees of freedom 15 and Using F table, area in tail is greater than 10 Two-tail p-value is greater than 20 Exact p-value corresponding to F = 1.44 is 5906 p-value > 10, not reject H0 We cannot conclude there is a statistically significant difference between the variances for the two companies 19 H :σ 12 = σ 22 H a :σ 12 ≠ σ 22 s12 = 0489 s22 = 0059 F= s12 0489 = = 8.28 s22 0059 Degrees of freedom 24 and 21 Using F table, area in tail is less than 01 Two-tail p-value is less than 02 Exact p-value ≈ p-value ≤ 05, reject H0 The process variances are significantly different Machine offers the best opportunity for process quality improvements Note that the sample means are similar with the mean bag weights of approximately 3.3 grams However, the process variances are significantly different 20 H :σ 12 = σ 22 H a :σ 12 ≠ σ 22 F= s12 11.1 = = 5.29 s22 2.1 Degrees of freedom 25 and 24 Using F table, area in tail is less than 01 11 - © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Inferences About Population Variances Two-tail p-value is less than 02 Exact p-value ≈ p-value ≤ 05, reject H0 The population variances are not equal for seniors and managers 21 Consider the Small cap fund as population and the large cap fund as population H :σ 12 ≤ σ 22 H a :σ 12 > σ 22 F= s12 (13.03) = = 2.15 s22 (8.89) Degrees of freedom 25 and 25 Upper-tail p-value is the area to the right of the test statistic From the F table, the p-value is between 025 and 05 Exact p-value corresponding to F = 2.15 is 0306 p-value ≤ 05, reject H0 The population variance and standard deviation for the small cap growth fund are larger Financial analysts would say the small cap growth fund is riskier 22 a Population - Wet pavement H :σ 12 ≤ σ 22 H a :σ 12 > σ 22 F= s12 322 = = 4.00 s22 16 Degrees of freedom 15 and 15 Using F table, p-value is less than 01 Exact p-value corresponding to F = 4.00 is 0054 p-value ≤ 05, reject H0 Conclude that there is greater variability in stopping distances on wet pavement b Drive carefully on wet pavement because of the uncertainty in stopping distances 23 a s2 = (30) = 900 b 2 χ.05 = 30.144 and χ.95 = 10.117 (19 degrees of freedom) (19)(900) (19)(900) ≤σ2 ≤ 30.144 10.117 11 - © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 11 567 ≤ σ ≤ 1690 c 23.8 ≤ σ ≤ 41.1 24 With 12 degrees of freedom, χ.025 = 23.337 χ.975 = 4.404 (12)(14.95) (12)(14.95) ≤σ2 ≤ 23.337 4.404 114.9 ≤ σ ≤ 609 10.72 ≤ σ ≤ 24.68 25 a x= Σxi = 260.16 n b s2 = Σ( xi − x )2 = 4996.8 n −1 s = 4996.79 = 70.69 c χ.025 = 32.852 χ.975 = 8.907 (19 degrees of freedom) (20 − 1)(4996.8) (20 − 1)(4996.8) ≤σ2 ≤ 32.852 8.907 2890 ≤ σ ≤ 10,659 53.76 ≤ σ ≤ 103.24 26 a H0: σ ≤ 0001 Ha: σ > 0001 χ2 = (n − 1) s (15 − 1)(.014)2 = = 27.44 σ2 0001 Degrees of freedom = n - = 14 Using χ table, p- value is between 01 and 025 Exact p-value corresponding to χ = 27.44 is 0169 p-value ≤ 10, reject H0 Variance exceeds maximum variance requirement b χ.05 = 23.685 11 - 10 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Inferences About Population Variances χ.95 = 6.571 (14 degrees of freedom) (14)(.014) (14)(.014) ≤σ2 ≤ 23.685 6.571 00012 ≤ σ ≤ 00042 27 H0: σ ≤ 02 Ha: σ > 02 χ2 = (n − 1) s (41 − 1)(.16) = = 51.20 σ 02 02 Degrees of freedom = n - = 40 Using χ table, p- value is greater than 10 Exact p-value corresponding to χ = 51.20 is 1104 p-value > 05, not reject H0 The population variance does not appear to be exceeding the standard 28 H0: σ ≤ Ha: σ > χ2 = (n − 1) s (22 − 1)(1.5) = = 31.50 σ 02 Degrees of freedom = n - = 21 Using χ table, p-value is between 05 and 10 Exact p-value corresponding to χ = 31.50 is 0657 p-value ≤ 10, reject H0 Conclude that σ > 29 s2 = Σ( xi − x )2 101.56 = = 12.69 n −1 −1 H0: σ = 10 Ha: σ ≠ 10 χ2 = (n − 1) s (9 − 1)(12.69) = = 10.16 σ2 10 11 - 11 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 11 Degrees of freedom = n - = Using χ table, area in tail is greater than 10 Two-tail p- value is greater than 20 Exact p-value corresponding to χ = 10.16 is 5080 p-value > 10, not reject H0 30 a Try n = 15 χ.025 = 26.119 χ.975 = 5.629 (14 degrees of freedom) (14)(64) (14)(64) ≤σ2 ≤ 26.119 5.629 34.3 ≤ σ ≤ 159.2 5.86 ≤ σ ≤ 12.62 ∴ A sample size of 15 was used b n = 25; expect the width of the interval to be smaller χ.05 = 39.364 χ.975 = 12.401 (24 degrees of freedom) (24)(8) (24)(8) ≤σ2 ≤ 39.364 12.401 39.02 ≤ σ ≤ 126.86 6.25 ≤ σ ≤ 11.13 31 H :σ 12 = σ 22 H a :σ 12 ≠ σ 22 Population is women’s scores F= s12 2.46232 = = 1.24 s22 2.21182 Degrees of freedom 19 and 29 11 - 12 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Inferences About Population Variances Using Excel of Minitab, we find the exact two-tail p-value corresponding to F = 1.24 is 5876 p-value > 10, not reject H There is not a statistically significant difference in the variances We cannot conclude that there is a difference in the variability of golf scores for male and female professional golfers 32 H : σ 12 = σ 22 H a : σ 12 ≠ σ 22 Use critical value approach since F tables not have 351 and 72 degrees of freedom F.025 = 1.47 Reject H0 if F ≥ 1.47 F= s12 940 = = 1.39 s22 797 F < 1.47, not reject H0 We are not able to conclude students who complete the course and students who drop out have different variances of grade point averages 33 H : σ 12 = σ 22 H a : σ 12 ≠ σ 22 Population has the larger sample variance F= s12 5.4 = = 2.35 s22 2.3 Degrees of freedom 15 and 15 Using F table, area in tail is between 05 and 10 Two-tail p-value is between 10 and 20 Exact p-value corresponding to F = 2.35 is 1087 p-value > 10, not reject H0 Cannot conclude that there is a difference between the population variances 34 H : σ 12 = σ 22 H a : σ 12 ≠ σ 22 F= s12 25 = = 2.08 s22 12 11 - 13 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 11 Degrees of freedom 30 and 24 Using F table, area in tail is between 025 and 05 Two-tail p-value is between 05 and 10 Exact p-value corresponding to F = 2.08 is 0695 p-value ≤ 10, reject H0 Conclude that the population variances are not equal 11 - 14 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part .. .Chapter 11 a 11. 070 b 27.488 c 9.591 d 23.209 e 9.390 s2 = 25 2 a With 19 degrees of freedom χ.05 = 30.144 and χ.95 = 10 .117 19(25) 19(25) ≤σ2 ≤ 30.144 10 .117 15.76 ≤ σ ≤ 46.95... freedom = n - = 11 11 - © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 11 Using χ table,... 10.16 σ2 10 11 - 11 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 11 Degrees of