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### Transcript of Solution Manual - Probability and Statistics 6th Ed. by J. L. Devore Chapter 9

261 CHAPTER 9 Section 9.1 1. a. ( ) ( ) ( ) 4 . 5 . 4 1 . 4 Y E X E Y X E , irrespective of sample sizes. b. ( ) ( ) ( ) ( ) ( )0724 .1000 . 21008 . 12 2 2221 + + + n mY V X V Y X V , and the s.d. of 2691 . 0724 . Y X . c. A normal curve with mean and s.d. as given in a and b (because m = n = 100, the CLT implies that both X and Y have approximately normal distributions, so Y X does also). The shape is not necessarily that of a normal curve when m = n = 10, because the CLT cannot be invoked. So if the two lifetime population distributions are not normal, the distribution of Y X will typically be quite complicated. 2. The test statistic value is nsmsy xz2221+ , and Ho will be rejected if either 96 . 1 z or 96 . 1 z . We compute 85 . 433 . 4332100451900452200400 , 40 500 , 4222 + z . Since 4.85 > 1.96, reject Ho and conclude that the two brands differ with respect to true average tread lives. 3. The test statistic value is ( )nsmsy xz22215000+ , and Ho will be rejected at level .01 if 33 . 2 z . We compute ( )76 . 193 . 3967004515004522005000 800 , 36 500 , 4322 + z , which is not > 2.33, so we dont reject Ho and conclude that the true average life for radials does not exceed that for economy brand by more than 500. Chapter 9: Inferences Based on Two Samples 262 4. a. From Exercise 2, the C.I. is ( ) ( ) ( ) 33 . 849 2100 33 . 433 96 . 1 2100 96 . 12221t t + t nsmsy x ( ) 33 . 2949 , 67 . 1250 . In the context of this problem situation, the interval is moderately wide (a consequence of the standard deviations being large), so the information about 1 and 2 is not as precise as might be desirable. b. From Exercise 3, the upper bound is ( ) 95 . 6352 95 . 652 5700 93 . 396 645 . 1 5700 + + . 5. a. Ha says that the average calorie output for sufferers is more than 1 cal/cm2/min below that for nonsufferers. ( ) ( )1414 .1016 .1004 .2 2 2221 + +n m , so ( ) ( )90 . 21414 .1 05 . 2 64 . z . At level .01, Ho is rejected if 33 . 2 z ; since 2.90 < -2.33, reject Ho. b. ( ) 0019 . 90 . 2 P c. ( ) 8212 . 92 . 11414 .1 2 . 133 . 2 1 ,`

.| + d. ( )( )15 . 652 .28 . 1 33 . 2 2 .22+ n m , so use 66. Chapter 9: Inferences Based on Two Samples 263 6. a. Ho should be rejected if 33 . 2 z . Since ( )33 . 2 53 . 33296 . 14056 . 287 . 16 12 . 18 + z , Ho should be rejected at level .01. b. ( ) ( ) 3085 . 50 .3539 .0 133 . 2 1 ,`

.| c. ( )06 . 37 0529 .96 . 11169 .28 . 1 645 . 11 96 . 14056 . 22 + + nn n, so use n = 38. d. Since n = 32 is not a large sample, it would no longer be appropriate to use the large sample test. A small sample t procedure should be used (section 9.2), and the appropriate conclusion would follow. 7. 1 Parameter of interest: 2 1 the true difference of means for males and females on the Boredom Proneness Rating. Let 1 mens average and 2 womens average. 2 Ho: 02 1 3 Ha: 02 1 > 4 ( ) ( )nsmsy xnsmsy xz o222122210+ + 5 RR: 645 . 1 z 6 ( )83 . 114868 . 49783 . 426 . 9 40 . 102 2+ oz 7 Reject Ho. The data indicates the Boredom Proneness Rating is higher for males than for females. Chapter 9: Inferences Based on Two Samples 264 8. a. 1 Parameter of interest: 2 1 the true difference of mean tensile strength of the 1064 grade and the 1078 grade wire rod. Let 1 1064 grade average and 2 1078 grade average. 2 Ho: 102 1 3 Ha: 102 1 < 4 ( ) ( ) ( )nsmsy xnsmsy xz o2221222110+ + 5 RR: < value p 6 ( ) ( )57 . 28210 .61290 . 21293 . 110 6 . 123 6 . 1072 2 + z 7 For a lower-tailed test, the p-value = ( ) 0 57 . 28 , which is less than any , so reject Ho. There is very compelling evidence that the mean tensile strength of the 1078 grade exceeds that of the 1064 grade by more than 10. b. The requested information can be provided by a 95% confidence interval for 2 1 : ( ) ( ) ( ) ( ) 588 . 5 , 412 . 6 210 . 96 . 1 6 96 . 12221 t + t nsmsy x . 9. a. point estimate 2 . 6 7 . 13 9 . 19 y x . It appears that there could be a difference. b. Ho: 02 1 ,Ha: 02 1 , ( )14 . 144 . 52 . 6608 . 15601 . 397 . 13 9 . 192 2 + z , and the p-value = 2[P(z > 1.14)] = 2( .1271) = .2542. The p value is larger than any reasonable , so we do not reject H0. There is no significant difference. c. No. With a normal distribution, we would expect most of the data to be within 2 standard deviations of the mean, and the distribution should be symmetric. 2 sds above the mean is 98.1, but the distribution stops at zero on the left. The distribution is positively skewed. d. We will calculate a 95% confidence interval for , the true average length of stays for patients given the treatment. ( ) 8 . 21 , 0 . 10 9 . 9 9 . 19601 . 3996 . 1 9 . 19 t t Chapter 9: Inferences Based on Two Samples 265 10. a. The hypotheses are Ho: 52 1 and Ha: 52 1 > . At level .001, Ho should be rejected if 08 . 3 z . Since ( )08 . 3 89 . 22272 .5 8 . 59 6 . 65< z , Ho cannot be rejected in favor of Ha at this level, so the use of the high purity steel cannot be justified. b. 12 1 o , so ( ) 2891 . 53 .2272 .108 . 3 ,`

.| 11. ( )nsmsz Y X22212 / + t . Standard error = ns. Substitution yields ( ) ( ) ( )2221 2 / SE SE z y x + t . Using , 05 . 96 . 12 / z , so ( ) ( ) ( ) ( ) 41 . 2 , 99 . 0 2 . 0 3 . 0 96 . 1 8 . 3 5 . 52 2 + t . Because we selected , 05 . we can state that when using this method with repeated sampling, the interval calculated will bracket the true difference 95% of the time. The interval is fairly narrow, indicating precision of the estimate. 12. The C.I. is ( ) ( ) 46 . 2 77 . 8 9104 . 58 . 2 77 . 8 58 . 22221t t + t nsmsy x ( ) 31 . 6 , 23 . 11 . With 99% confidence we may say that the true difference between the average 7-day and 28-day strengths is between -11.23 and -6.31 N/mm2. 13. 05 .2 1 , d = .04, 05 . , 01 . , and the test is one-tailed, so ( )( )38 . 490016 .645 . 1 33 . 2 0025 . 0025 .2+ + n , so use n = 50. 14. The appropriate hypotheses are Ho: 0 vs. Ha: 0 < , where 2 12 . ( 0 < is equivalent to 2 12 < , so normal is more than twice schizo) The estimator of is Y X 2 , with ( ) ( ) ( )n mY Var X Var Var222144 + + , is the square root of ( ) Var , and is obtained by replacing each 2i with 2iS . The test statistic is then (since 0 o ), and Ho is rejected if . 33 . 2 z With ( ) 97 . 35 . 6 69 . 2 2 and ( ) ( )9236 .4503 . 4433 . 2 42 2 + , 05 . 19236 .97 . z ; Because 1.05 > -2.33, Ho is not rejected. Chapter 9: Inferences Based on Two Samples 266 15. a. As either m or n increases, decreases, so o 2 1 increases (the numerator is positive), so ,`

.| oz2 1decreases, so ,`

.| oz2 1 decreases. b. As decreases, z increases, and since z is the numerator of n , n increases also. 16. nnsnsy xz22 .2221+ . For n = 100, z = 1.41 and p-value = ( ) [ ] 1586 . 41 . 1 1 2 . For n = 400, z = 2.83 and p-value = .0046. From a practical point of view, the closeness of x and y suggests that there is essentially no difference between true average fracture toughness for type I and type I steels. The very small difference in sample averages has been magnified by the large sample sizes statistical rather than practical significance. The p-value by itself would not have conveyed this message. Section 9.2 17. a. ( )( ) ( )17 43 . 1744 . 1 694 .21 . 379 92106210521061052 22 2 +++ b. ( )( ) ( )21 7 . 21411 . 694 .01 . 2414 92156210521561052 22 2 +++ c. ( )( ) ( )18 27 . 18411 . 018 .84 . 714 92156210221561022 22 2 +++ d. ( )( ) ( )26 05 . 26098 . 395 .84 . 1223 112246212522461252 22 2 +++ Chapter 9: Inferences Based on Two Samples 267 18. With Ho: 02 1 vs. Ha: 02 1 , we will reject Ho if < value p . ( )( ) ( )6 8 . 64 525240 .26164 .25240 .6164 .2 22 2 ++ , and the test statistic 17 . 61265 .78 . 95 . 21 73 . 225240 .6164 .2 2 + t leads to a p-value of 2[ P(t > 6.17)] < 2(.0005) =.001, which is less than most reasonable s ' , so we reject Ho and conclude that there is a difference in the densities of the two brick types. 19. For the given hypotheses, the test statistic 20 . 1007 . 36 . 3 10 3 . 129 7 . 115638 . 5603 . 52 2 ++ t , and the d.f. is ( )( ) ( )96 . 958241 . 452168 . 48241 . 4 2168 . 42 22++ , so use d.f. = 9. We will reject Ho if ; 764 . 29 , 01 . t t since 1.20 > -2.764, we dont reject Ho. 20. We want a 95% confidence interval for 2 1 . 262 . 29 , 025 . t , so the interval is ( ) ( ) 20 . 3 , 40 . 10 007 . 3 262 . 2 6 . 3 t . Because the interval is so wide, it does not appear that precise information is available. 21. Let 1 the true average gap detection threshold for normal subjects, and 2 the corresponding value for CTS subjects. The relevant hypotheses are Ho: 02 1 vs. Ha: 02 1 < , and the test statistic 46 . 23329 .82 .07569 . 0351125 .53 . 2 71 . 1 + t . Using d.f. ( )( ) ( )1 . 15907569 .70351125 .07569 . 0351125 .2 22++ , or 15, the rejection region is 602 . 215 , 01 . t t . Since 2.46 is not 602 . 2 , we fail to reject Ho. We have insufficient evidence to claim that the true average gap detection threshold for CTS subjects exceeds that for normal subjects. Chapter 9: Inferences Based on Two Samples 268 22. Let 1 the true average strength for wire-brushing preparation and let