Inferences About Population Variances

Chapter 11 Inferences About Population VariancesLearning Objectives1. 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) s2/ 2 has a chi-square distribution and be able to use this result to develop a confidence interval estimate of  2.4. Be able to compute p-values using the chi-square distribution.5. Know how to test hypotheses involving  2.6. Understand the role of statistical inference in developing conclusions about two population variances.2 2 7. Know that the sampling distribution of s1/ s 2 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 Chapter 11Solutions:1. a. 11.070 b. 27.488 c. 9.591 d. 23.209 e. 9.3902. s2 = 252 2 a. With 19 degrees of freedom .05 = 30.144 and .95 = 10.11719(25) 19(25)  2  30.144 10.11715.76   2  46.952 2 b. With 19 degrees of freedom .025 = 32.852 and .975 = 8.90719(25) 19(25)  2  32.852 8.90714.46   2  53.33 c. 3.8    7.32 2 2 (n 1) s (16  1)(9.5) 3.  2   27.08  0 50Degrees of Freedom = (16 - 1) = 15Using  2 table, p-value is between .025 and .05Actual p-value = .0281 p-value  .05, reject H0 Critical value approach 2 .05 = 24.9962 Reject H0 if   24.99627.08 > 24.996, reject H011 - 2 Inferences About Population Variances4. a. n = 18 s2 = .362 .05 = 27.5872 .95 = 8.672 (17 degrees of freedom)17(.36) 17(.36)  2  27.587 8.672.22   2  .71 b. .47    .84(x  x )2 5. a. s2 i  31.07 n 1 s 31.07  5.572 2 b. .025 = 16.013 .975 = 1.690(8 1)(31.07) (8  1)(31.07)  2  16.013 1.69013.6   2  128.7 c. 3.7    11.3(x  x )2 6. a. s2 i  176.96 n 1 s 176.96  13.302 2 b. .025 = 11.143 .975 = 0.484(5 1)(176.96) (5  1)(176.96)  2  11.143 .48463.5   2  1462.58.0    38.2(x  x )2 7. a. s2 i  2.62 n 1 s 2.62  1.6211 - 3 Chapter 112 2 b. .025 = 16.013 .095 = 1.690(8 1)(2.62) (8  1)(2.62)  2  16.013 1.6901.14   2  10.85 c. 1.07    3.29(x  x )2 .0929 8. a. s2 i   .00845 n 1 12  1 b. s .00845  .092 c. 11 degrees of freedom2 2 .025 = 21.920 .975 = 3.816(12 1).00845 (12  1).00845  2  21.920 3.816.0042   2  .0244.065   .1562 9. H0:  .00042 Ha:  .00042 2 (n 1) s (30  1)(.0005)  2   36.25  0 .0004Degrees of freedom = n - 1 = 29Using  2 table, p-value is greater than .10Actual p-value = .1664 p-value > .05, do not reject H0. The product specification does not appear to be violated.10.  2 .75)2 = .56252 H0:  .56252 Ha:  .5625(n 1) s2 (30  1)(.95) 2  2    46.53  2 .5625Degrees of freedom = n - 1 = 2911 - 4 Inferences About Population VariancesUsing  2 table, p-value is between .01 and .025Actual p-value = .0208 p-value  .05, reject H0. The standard deviation for television sets is greater than the standard deviation for VCR’s.2 11. H0:  = .0092162 Ha:   .009216(n 1) s2 (20  1)(.114) 2  2    26.79  2 .009216Degrees of freedom = n - 1 = 19Using  2 table, area in upper tail is greater than .10Two-tail p-value is greater than .20Actual p-value = .2191 p-value > .05, do not reject H0. Cannot conclude the variance in interest rates has changed.(x  x )2 12. a. s2 i  .8106 n 12 b. H0:  .942 Ha:  .942 2 (n 1) s (12  1)(.8106)  2   9.49  0 .94Degrees of freedom = n - 1 = 11Using  2 table, area in upper tail is greater than .10Two-tail p-value is greater than .20Actual p-value = .8457 p-value > .05, cannot reject H0.13. a. F.05 = 3.33 b. F.025 = 2.76 c. F.01 = 4.50 d. F.10  1.9411 - 5 Chapter 112 s1 5.8 14. a. F 2   2.4 s2 2.4Degrees of freedom 15 and 20Using F table, p-value is between .025 and .05Actual p-value = .03342 2 p-value  .05, reject H0. Conclude 1  2 b. F.05 = 2.20Reject H0 if F  2.202 2 2.4  2.20, reject H0. Conclude 1  22 15. a. Larger sample variance is s12 s1 8.2 F 2   2.05 s2 4Degrees of freedom 20 and 25Using F table, area in tail is between .025 and .05Two-tail p-value is between .05 and .10 Actual p-value = .0904 p-value > .05, do not reject H0. b. Since we have a two-tailed testF / 2 F .025  2.30Reject H0 if F  2.302.05 < 2.30, do not reject H0 2 2 16. H0 : 1  22 2 Ha :1  22 2 s1 94 F 2  2  2.63 s2 58 Degrees of freedom 25 and 29Using F table, p-value is less than .01 11 - 6 Inferences About Population VariancesActual p-value = .0067 p-value  .01, reject H0. Conclude adults have a greater variance in online times.17. a. Population 1 is 4 year old automobiles2 2 H0 : 1  22 2 Ha : 1  22 2 s1 170 b. F 2  2  2.89 s2 100Degrees of freedom 25 and 24Using F table, p-value is less than .01 Actual p-value = .0057 p-value  .01, reject H0. Conclude that 4 year old automobiles have a larger variance in annual repair costs compared to 2 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.2 2 18. H0 : 1  22 2 Ha :1  22 2 s1 4.27 F 2  2  3.54 s2 2.27Degrees of freedom 9 and 6Using F table, area in tail is between .05 and .10Two-tail p-value is between .10 and .20 Actual p-value = .1379 p-value > .05, do not reject H0. Cannot conclude any difference between variances of the two industries.2 2 19. H0 : 1  22 2 Ha :1  22 s1  .04892 s2  .005911 - 7 Chapter 112 s1 .0489 F 2   8.28 s2 .0059Degrees of freedom 24 and 21Using F table, area in tail is less than .01Two-tail p-value is less than .02 Actual p-value  0 p-value  .05, reject H0. The process variances are significantly different. Machine 1 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.2 2 20. H0 : 1  22 2 Ha : 1  22 s1 11.1 F 2   5.29 s2 2.1Degrees of freedom 25 and 24Using F table, area in tail is less than .01Two-tail p-value is less than .02 Actual p-value  0 p-value  .05, reject H0. The population variances are not equal for seniors and managers.(x  x )2 21. a. s2  i n 12 sNov = 96642 2 sDec = 19,238 (Population 1 since s is larger)2 2 b. H0 : 1  22 2 Ha : 1  22 s1 19,238 F 2   1.99 s2 9664Degrees of freedom 9 and 9Using F table, area in tail is greater than .1011 - 8 Inferences About Population VariancesTwo-tail p-value is greater than .20 Actual p-value = .3197 p-value > .05, do not reject H0. There is no evidence that the population variances differ.22. a. Population 1 - Wet pavement.2 2 H0 : 1  22 2 Ha : 1  22 2 s1 32 F 2  2  4.00 s2 16Degrees of freedom 15 and 15Using F table, p-value is less than .01Actual p-value = .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) 2 = 9002 2 b. .05 = 30.144 and .95 = 10.117 (19 degrees of freedom)(19)(900) (19)(900)  2  30.144 10.117567   2  1690 c. 23.8    41.124. With 12 degrees of freedom,2 2 .025 = 23.337 .975 = 4.404(12)(14.95)2 (12)(14.95) 2  2  23.337 4.404114.9   2  60910.72    24.68x 25. a. x i  260.16 n11 - 9 Chapter 11(x  x )2 b. s2 i  4996.8 n 1 s 4996.79  70.692 2 c. .025 = 32.852 .975 = 8.907 (19 degrees of freedom)(20 1)(4996.8) (20  1)(4996.8)  2  32.852 8.9072890   2  10,65953.76    103.242 26. a. H0:  .00012 Ha:  .0001(n 1) s2 (15  1)(.014) 2  2    27.44  2 .0001Degrees of freedom = n - 1 = 14Using  2 table, p- value is between .01 and .025Actual p-value = .0169 p-value  .10, reject H0. Variance exceeds maximum variance requirement.2 b. .05 = 23.685 2 .95 = 6.571 (14 degrees of freedom)(14)(.014)2 (14)(.014) 2  2  23.685 6.571.00012   2  .000422 27. H0:  .022 Ha:  .022 2 2 (n 1) s (41  1)(.16)  2   51.20  0 .02Degrees of freedom = n - 1 = 40Using  2 table, p- value is greater than .10Actual p-value = .110411 - 10 Inferences About Population Variances p-value > .05, do not reject H0. The population variance does not appear to be exceeding the standard.2 28. H0:  2 Ha:  2 2 (n 1) s (22  1)(1.5)  2   31.50  0 1Degrees of freedom = n - 1 = 21Using  2 table, p-value is between .05 and .10Actual p-value = .06572 p-value  .10, reject H0. Conclude that  > 1.(x  x )2 101.56 29. s2 i   12.69 n 1 9  12 H0:  = 102 Ha:   10(n 1) s2 (9  1)(12.69)  2    10.16  2 10Degrees of freedom = n - 1 = 8Using  2 table, area in tail is greater than .10Two-tail p- value is greater than .20Actual p-value = .5086 p-value > .10, do not reject H030. a. Try n = 152 .025 = 26.119 2 .975 = 5.629 (14 degrees of freedom)(14)(64) (14)(64)  2  26.119 5.629 34.3   2  159.25.86    12.6211 - 11 Chapter 11A sample size of 15 was used. b. n = 25; expect the width of the interval to be smaller.2 .05 = 39.3642 .975 = 12.401 (24 degrees of freedom)(24)(8)2 (24)(8) 2  2  39.364 12.40139.02   2  126.866.25    11.132 2 31. H0 : 1  22 2 Ha : 1  2Population 1 is NASDAQ2 2 s1 15.8 F 2  2  4.00 s2 7.9Degrees of freedom 9 and 9Using F table, area in tail is between .025 and .05Two-tail p-value is between .05 and .10Actual p-value = .0510 p-value > .05, do not reject H0. Cannot conclude population variances differ. But with such a low p- value, a larger sample is recommended. 2 2 32. H0 : 1  22 2 Ha : 1  2Use critical value approach since F tables do not have 351 and 72 degrees of freedom.F.025 = 1.47Reject H0 if F  1.472 2 s1 .940 F 2  2  1.39 s2 .797F < 1.47, do 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.11 - 12 Inferences About Population Variances2 2 33. H0 : 1  22 2 Ha : 1  2Population 1 has the larger sample variance.2 s1 5.4 F 2   2.35 s2 2.3Degrees of freedom 15 and 15Using F table, area in tail is between .05 and .10Two-tail p-value is between .10 and .20Actual p-value = .1091 p-value > .10, do not reject H0. Cannot conclude that there is a difference between the population variances.2 2 34. H0 : 1  22 2 Ha : 1  22 s1 25 F 2   2.08 s2 12Degrees of freedom 30 and 24Using F table, area in tail is between .025 and .05Two-tail p-value is between .05 and .10Actual p-value = .0689 p-value  .10, reject H0. Conclude that the population variances are not equal.11 - 13

Inferences About Population Variances

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