ch8 - Chapter 8 IN FERENCES CONCERNING VARIANCES 8.1 (a)...

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Unformatted text preview: Chapter 8 IN FERENCES CONCERNING VARIANCES 8.1 (a) The sample variance 32 is given by 2 1 s =n_1Z(xi—:E)2 = 17.6 Thus, the sample standard deviation is \/ 17.6 = 4.195. (b) The minimum observation is 21. The maximum observation is 32. thus, the range is 11. Since the sample size is 6, the expected length of the range is 2.5340. Thus, we estimate a by 11/2.534 = 4.341. _ 8.2 The range of the data is 8140 — 7840 = 300. Since the sample size is 6, the estimate is 300/ 2.534 = 118.39. The sample standard deviation is 104.50. 8.3 (a) The sample standard deviation is 1.787. (b) The range of the data is 68.4 — 61.8 = 6.6. The sample size is 10, so the esti— mate of the standard deviation is 6.6/ 3.07 8 = 2.144. The relative difference 253 254 8.4 8.5 8.6 Chapter 8 INF ERENCES CONCERNING VARIANCES is 2.144 — 1.787 1.787 - 100 = 20.0 percent. The sample variance is 10.28571. The sample size is 8. The >801 /2 with 7 degrees of freedom is 20.278 and ximfl with 7 degrees of freedom is .989. Thus, if the data are from a normal population, a 99 percent confidence interval for 02 is 7(10.28571) < 02 20.278 7(10.28571) < .989 01' 3.55 < 02 < 72.80. The sample variance is .025. Since the sample size is 5, the x201 /2 with 4 degrees of freedom is 14.860 and x101 /2 with 4 degrees of freedom is .207. Thus, if the data are from a normal population, a 99 percent confidence interval for a2 is 4(.025) 2 < a 4(.025) 14.860 .207 < 01‘ .0067 < 02 < .483, and the 99 percent confidence interval for the standard deviation is .082 < 0' < .695 The sample variance is 1.787 and the sample size is 10. The flog/2 with 9 degrees of freedom is 21.666 and xim/g with 9 degrees of freedom is 2.088. If the data are from a normal population, a 98 percent confidence interval for a is 9 - (1.787)2 21.666 9- (1.787)? < a < 2.088 8.7 8.8 8.9 255 or 1.152 < 0 < 3.710 Assuming a normal population, we use the statistic 2 (n _ llsz X “ —"—2 ~ 00 Since the alternative is a > 600, we reject the null hypothesis 0 = 600 if X2 > X205 with 5 degrees of freedom or if X2 > 11.076. In this case, 3 = 648 so the test statistic 2 M _ 6002 =5.832. X Thus, we cannot reject the null hypothesis at the .05 level of significance. If the data are from a normal population, we can use the statistic 2: (71—1)le2 X 03 The null hypothesis is 0 = .010 and the alternative is a $5 .010. Since the sample size is 12, we reject the null hypothesis if X2 > )801/2 with 11 degrees of freedom or if X2 < xim/Z with 11 degrees of freedom. Thus, we reject the null hypothesis 0 =_ .010 if X2 > 26757 or if X2 < 2.603. In this case, 5 = .0086 and the test statistic 2 = 11-(.0086)2 (.010? = 8.136 X so we cannot reject the null hypothesis at the .01 level of significance. Since the data are from a normal population, we can use the statistic 2 (n — USQ X — 2 - 00 The null hypothesis is a = 15.0 and the alternative is a > 15.0. 'Since the sample 256 8.10 8.11 Chapter 8 INFERENCES CONCERNING VARIANCES size is 71, we reject the null hypothesis if X2 > X205 with'70 degrees of freedom Thus, we reject the null hypothesis 0 = 15.0 if X2 > 90.531. In this case, 8 = 19.3 minutes and the test statistic 2 2 70119.3)2 (15 (D2 = 115.886 X so we reject the null hypothesis 0 = 15.0 in favor of the alternative 0 > 15.0, at the .05 level of significance. The sample variance is .92 = .00011 and the sample size is 15. The null hypothesis is a = .015 and the alternative is a 74 .015. Thus, since the data are from a normal population, we can use the statistic 2 (n —1)S2 0% X We reject the null hypothesis when X2 > x201 /2 with 14 degrees of freedom or when X2 < ximfl with 14 degrees of freedom. Thus, we reject the null hypothesis when X2 > 31.319 or when X2 < 4.075. In this case 2 z 14- (.00011) (.015? = 6.884 X so we cannot reject the null hypothesis at the .01 level of significance. The sample standard deviation is s = 1.32 and the sample size is 10. If the data are from a normal population, we can use the statistic 2 = (n— 1)S2 x 08 The null hypothesis is a = 1.20 and the alternative is a > 1.20. Thus, we reject the null hypothesis when X2 > xfim with 9 degrees of freedom or when X2 > 16.919. 257 In this case 3 = 1.32 so 9 - (1.32)2 2 = ~————— : 10.89 X (1.20)2 Thus, we cannot reject the null hypothesis at the .05 level of significance. 8.12 The sample standard deviation 5 = 1.8 and the sample size is 31. The null hypoth— esis is a = 2.0 and the alternative is a aé 2.0 . Since the data are from a normal population, we can use the statistic 71—152 X24 2) . (70 We reject the null hypothesis when X2 > xin/Z with 30 degrees of freedom or when X2 < xffiom with 30 degrees of freedom. Thus, we reject the null hypothesis when X2 > 31.319 or when X2 < 4.075. In this case 2 30 - (1.8)2 = ———~—— = 12. 84 X (2.0)2 8 so we cannot reject the null hypothesis at the .01 level of significance. 8.13 Since Exercise 7.67 states that the two samples can be assumed to be from normal populations, we can use the statistic _S%4 F—g which has an F distribution with n M —— 1 and mm — 1 degrees of freedom. The null hypothesis is of = 0% and the alternative hypothesis is of 75 0%. The sample sizes are n M = 8 and nm = 10 . Thus, we reject the null hypothesis when F > Fog/2(7, 9) or when F > 5.61. In this case is?” = (1.81)2 and 8371 = (1.48)2 so 1.81 2 F ~ _ 1.496. 258 8.14 8.15 8.16 Chapter 8 INFERENCES CONCERNING VARIANCES Thus, we cannot reject the null hypothesis at the .02 level of significance. Since the two samples can be assumed to be from normal populations, we can use the statistic F _ a _ S2 m which has an F distribution with 71M — 1 and mm — 1 degrees of freedom. The 2 null hypothesis is 01 = 0% and the alternative hypothesis is of 74 0%. The sample sizes are rm 2 10 and nm = 10. Thus, we reject the null hypothesis when F > Fog/2(9,9) = 5.35. In this case 3?” = 29.111 and 53,, = 11.333 so F_ 29.111 — = 2.57. 11.333 We cannot reject the null hypothesis at the .02 level of significance. The null hypothesis 0% = 0% and the alternative hypothesis is of < 0%. Therefore, we use the statistic S2 F = J— 512 which has an F distribution with m — 1 = 20 and n1 — 1 = 14 degrees of freedom. The null hypothesis of = of < a; if F > F01(20,14) = 3.51. In this case 4.2 2 F: —— =242 (2.7) a; will be rejected favor of the alternative hypothesis so we cannot reject the null hypothesis at the .01 level of significance. This analysis assumes that the two samples come from normal populations and that the samples are independent. The null hypothesis of = a; and the alternative hypothesis is a; > 0%. Since the 259 two samples come from normal populations, we can use the statistic _53 F__ 512 which has an F distribution with 712 — 1 = 29 and n1 — 1 = 39 degrees of freedom. The null hypothesis of = a; will be rejected favor of the alternative hypothesis 0% > 012 if F > F05(29,39) = 1.76. In this case 18.7 2 F4 _ 1.51 so we cannot reject the null hypothesis at the .05 level of significance. 8.17 (a) The two samples are not from normal populations so we cannot directly use a two-sample t test. Also, the population variances may be unequal. (b) If we take the logarithm of each observation, then the transformed data are samples from two normal populations. On this scale, we can then apply the F test for the equality of variances. Specifically, we would test the null hypothesis of = 03 versus the alternative hypothesis of 79 0% using the logarithms of the original data. 8.18 The sample size is 5 and fies/2 with 4 degrees of freedom is 11.143 and X105” with 4 degrees of freedom is .484. The sample standard deviation is 5.7. Thus, the 95 percent confidence interval for a is 4 - (5.7)2 < a < 4 - (5.7)2 11.143 .484 01' 3.42 < a < 16.39 8.19 For alloy 1 the sample size is 58 and X205” with 57 degrees of freedom is 79.8 and ximfl with 57 degrees of freedom is 38.0. The sample standard deviation is 1.80. 260 8.20 8.21 Chapter 8 INF ERENCES CONCERNING VARIANCES Thus, the 95 percent confidence interval for 01 is‘ 57 - (1.80)? . 79.8 57- (1.80)2 < 01 < 38.0 01' 1.52 < 01 < 2.20 For alloy 2, the sample size is 27 and the xim/g with 26 degrees of freedom is 13.844 and Kola/2 with 26 degrees of freedom is 41.923. The sample standard deviation is 2.42. Thus, the 95 percent confidence interval for 02 is 26- (2.42)2 < a < 26- (2.42)2 41.923 2 13.844 or 1.91 < 02 < 3.32. The sample standard deviation is 4.9 and s2 = (4.9)2 = 24.01. The sample size is 25, the X?05/2 with 24 degrees of freedom is 39.364 and x§__05/2 with 24 degrees of freedom is 12.401. Thus, assuming the data are from a normal population, the 95 percent confidence interval for a is /24 - (4.9)2 < a < [24 - (4.9)2 39.364 12.401 3.83 < 0’ < 6.82. 01' The sample standard deviation is 4.9. The sample size is 25. The null hypothesis is 02 = 30.0 and the alternative is 02 < 30.0. If the data are from a normal 8.22 8.23 261 population, we can use the statistic 2 (n — 1)S2 X: 08 We reject the null hypothesis if X2 < X51205 with 24 degrees of freedom or when X2 < 13.484. In this case, 2 = 24 . (4.9)2 X = 19.21. Thus, we cannot reject the null hypothesis at the .05 level of significance. The sample standard deviation is .83 degree Celsius and the sample size is 31. The X202” with 30 degrees of freedom is 50.892 and xim/Q with 30 degrees of freedom is 14.953. Assuming the data are from a normal population, a 98 percent confidence interval for a is 30- (.83)? 50.892 30 - (.83)2 < a 14.953 < 01‘ .637 < 0 < 1.176. The sample size is 101 and )8ch with 100 degrees of freedom is 129.561 and Xf_.05/2 with 100 degrees of freedom is 74.222. The null hypothesis is a2 = .18 and the alternative is 02 75 .18. The sample variance is .13. If the data are from a normal population, we can use the statistic 2 (n —1)S2 X 08 We reject the null hypothesis if X2 > x_205/2 = 129.561 or when X2 < xffiowg 262 8.24 Chapter 8 INFERENCES CONCERNING VARIANCES 74.222 . In this case, 2 (100)(.13) X = 7222. Thus, we reject the null hypothesis 02 = .18, in favor of the alternative 02 7é .18, at the .05 level of significance. The inspector is not making satisfactory measure— ments. The sample variance of the first sample is 109 and the sample size is 9. The sample variance of the second sample is 160.967 and the sample size is 6. Since the two samples can be assumed to be from normal populations, we can use the statistic St F=—S—T2; which has an F distribution with n M — 1 and nm -— 1 degrees of freedom. The null hypothesis of = (7% will be rejected favor of the alternative hypothesis is of 75 0% if F > Fog/2(8, 5) = 10.30. In this case _ 160.967 F” 109 = 1.477 _ so we cannot reject the null hypothesis at the .02 level of significance. 8.25 The variance of the first sample is 7.499 and the sample size is 10. The variance of the second sample is 2.681 and the sample size is 8. The null hypothesis of = a; I and the alternative hypothesis is of 7é 03. Assuming normal populations we use the statistic St F=§ which has an F distribution with Tip; — 1 = 9 and nm — 1 = 7 degrees of freedom. The null hypothesis 01 = 02 will be rejected in favor of the alternative hypothesis 8.26 8.27 263 (71 74 02 if F > F01(9, 7) = 6.72. In this case 7.499 2 ~——— = 2.797 2.681 F so we cannot reject the null hypothesis at the .02 level of significance. The standard deviation of the first sample is 1.80 and the sample size is 58. The standard deviation of the second sample is 2.42 and the sample size is 27. The null hypothesis of = 0% and the alternative hypothesis is of 75 0%. normal—scores plots of these samples fail to contradict the normal assumption so we use the statistic _Si2v1 F—S—TQn which has an F distribution with n M — 1 = 26 and mm —— 1 = 57 degrees of freedom. 2 The null hypothesis 01 = 0% will be rejected in favor of the alternative hypothesis 0% 7ré a; if F > F01(26,57) : 2.11. In this case 242 2 F: — =1. 1 (1.80) 8 so we cannot reject the null hypothesis at the .02 level of significance. (a) No. The distribution which produced these observations is clearly not normal. The dot diagram exhibits a long right hand tail. (b) The natural logarithms look much more normal. They have 1 1 2 61mm.) = 27.2752 2 6111265..) = 67.1147 i=1 i=1 67.1147 — 27.27522/16 15 15 degrees of freedom, xg'025 = 27.448 and and Xa975 = 6.262 so the 95 % so 5 = = 1.172. From the chi—square table7 with 264 confidence interval for 0 is 15(1.172)2 <0 < 15(1.172)2 27.448 6.262 01‘ .87 < a < 1.81 Note however that this result cannot be translated to a statement about the variance on the original scale. ...
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ch8 - Chapter 8 IN FERENCES CONCERNING VARIANCES 8.1 (a)...

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