Statistical Inference from Two Samples: Hypothesis Testing and Confidence Intervals - Prof, Assignments of Mathematical Statistics

Download Statistical Inference from Two Samples: Hypothesis Testing and Confidence Intervals - Prof and more Assignments Mathematical Statistics in PDF only on Docsity! CHAPTER 9 Section 9.1 1. a.       4.5.41.4  YEXEYXE , irrespective of sample sizes. b.           0724. 100 0.2 100 8.1 2222 2 1  nm YVXVYXV  , and the s.d. of 2691.0724.  YX . 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 YX  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 YX  will typically be quite complicated. 2. The test statistic value is n s m s yx z 2 2 2 1    , and H0 will be rejected if either 96.1z or 96.1z . We compute 85.4 33.433 2100 45 1900 45 2200 400,40500,42 22    z . Since 4.85 > 1.96, reject H0 and conclude that the two brands differ with respect to true average tread lives. 3. The test statistic value is   n s m s yx z 2 2 2 1 5000    , and H0 will be rejected at level .01 if 33.2z . We compute   76.1 93.396 700 45 1500 45 2200 5000800,36500,42 22    z , which is not > 2.33, so we don’t reject H0 and conclude that the true average life for radials does not exceed that for economy brand by significantly more than 500. 265 Chapter 9: Inferences Based on Two Samples 4. a. From Exercise 2, the C.I. is       33.849210033.43396.1210096.1 2 2 2 1  n s m s yx  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   95.635295.652570093.396645.15700  . 5. a. Ha says that the average calorie output for sufferers is more than 1 cal/cm2/min below that for non-sufferers.     1414. 10 4. 10 2. 2222 2 1  nm  , so     90.2 1414. 105.264.   z . At level .01, H0 is rejected if 33.2z ; since –2.90 < -2.33, reject H0. b.   0019.90.2 P c.   8212.92.1 1414. 12.1 33.21         d.     15.65 2. 28.133.22. 2 2    nm , so use 66. 1 Chapter 9: Inferences Based on Two Samples a. The hypotheses are H0: 521   and Ha: 521   . At level .001, H0 should be rejected if 08.3z . Since   08.389.2 2272. 58.596.65   z , H0 cannot be rejected in favor of Ha at this level, so the use of the high purity steel cannot be justified. b. 121  o , so   2891.53.2272. 1 08.3        11.   n s m s zYX 2 2 2 1 2/   . Standard error = n s . Substitution yields       22 2 12/ SESEzyx   . Using ,05. 96.12/ z , so        41.2,99.02.03.096.18.35.5 22  . We are 95% confident that the true average blood lead level for male workers is between 0.99 and 2.41 higher than the corresponding average for female workers. 12. The C.I. is     46.277.89104.58.277.858.2 2 2 2 1  n s m s yx  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. 05.21  , d = .04, 05.,01.   , and the test is one-tailed, so    38.49 0016. 645.133.20025.0025. 2   n , so use n = 50. 14. The appropriate hypotheses are H0: 0 vs. Ha: 0 , where 212   . ( 0 is equivalent to 212   , so normal is more than twice schizophrenic) The estimator of  is YX 2̂ , with       nm YVarXVarVar 2 2 2 144ˆ    ,  is the square root of  ̂Var , and ̂ is obtained by replacing each 2i with 2iS . The test statistic is then    ˆ ˆ (since 0o ), and H0 is rejected if .33.2z With   97.35.669.22ˆ  and     9236. 45 03.4 43 3.24 ˆ 22  , 05.1 9236. 97.   z ; Because –1.05 > -2.33, H0 is not rejected. 15. 1 Chapter 9: Inferences Based on Two Samples a. As either m or n increases,  decreases, so   o 21 increases (the numerator is positive), so            oz 21 decreases, so             oz 21 decreases. b. As  decreases, z increases, and since z is the numerator of n , n increases also. 16. nn s n s yx z 2 2. 2 2 2 1     . For n = 100, z = 1.41 and p-value =    1586.41.112  . 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.       1743.17 44.1694. 21.37 99 2 10 6 2 10 5 2 10 6 10 5 22 22       b.       217.21 411.694. 01.24 149 2 15 6 2 10 5 2 15 6 10 5 22 22       c.       1827.18 411.018. 84.7 149 2 15 6 2 10 2 2 15 6 10 2 22 22       1 Chapter 9: Inferences Based on Two Samples d.       2605.26 098.395. 84.12 2311 2 24 6 2 12 5 2 24 6 12 5 22 22       18. With H0: 021   vs. Ha: 021   , we will reject H0 if  valuep .       68.6 45 2 5 240. 2 6 164. 2 5 240. 6 164. 22 22     , and the test statistic 17.6 1265. 78.95.2173.22 5 240. 6 164. 22    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 H0 and conclude that there is a difference in the densities of the two brick types. 19. For the given hypotheses, the test statistic 20.1007.3 6.3103.1297.115 6 38.5 6 03.5 22      t , and the d.f. is       96.9 5 8241.4 5 2168.4 8241.42168.4 22 2     , so use d.f. = 9. We will reject H0 if ;764.29,01.  tt since –1.20 > -2.764, we don’t reject H0. 20. We want a 95% confidence interval for 21   . 262.29,025. t , so the interval is    80.6,40.20007.3262.26.13  . 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 H0: 021   vs. Ha: 021   , and the test statistic 46.23329. 82. 07569.0351125. 53.271.1      t . Using d.f.       1.15 9 07569. 7 0351125. 07569.0351125. 22 2     , or 15, the rejection region is 1 Chapter 9: Inferences Based on Two Samples b. 0.5 1.5 2.5 Comparative Box Plot for High Quality and Poor Quality Fabric Quality Poor Quality High extensibility (%) The comparative boxplot does not suggest a difference between average extensibility for the two types of fabrics. c. We test 0: 210  H vs. 0: 21  aH . With degrees of freedom   5.10 00017906. 0433265. 2  , which we round down to 10, and using significance level .05 (not specified in the problem), we reject H0 if 228.210,025. tt . The test statistic is   38. 0433265. 08.   t , which is not 228.2 in absolute value, so we cannot reject H0. There is insufficient evidence to claim that the true average extensibility differs for the two types of fabrics. 24. a. 95% upper confidence bound: x + t.05,65-1SE = 13.4 + 1.671(2.05) = 16.83 seconds b. Let μ1 and μ2 represent the true average time spent by blackbirds at the experimental and natural locations, respectively. We wish to test H0: μ1 – μ2 = 0 v. Ha: μ1 – μ2 > 0. The relevant test statistic is 22 76.105.2 7.94.13   t = 1.37, with estimated df = 49 76.1 64 05.2 )76.105.2( 44 222   ≈ 112.9. Rounding to t = 1.4 and df = 120, the tabulated P-value is very roughly .082. Hence, at the 5% significance level, we fail to reject the null hypothesis. The true average time spent by blackbirds at the experimental location is not statistically significantly higher than at the natural location. 1 Chapter 9: Inferences Based on Two Samples c. 95% CI for silvereyes’ average time – blackbirds’ average time at the natural location: (38.4 – 9.7) ± (2.00) 22 06.576.1  = (17.96 sec, 39.44 sec). The t-value 2.00 is based on estimated df = 55. 25. We calculate the degrees of freedom       95.53 3027 2 31 8.7 2 28 5.5 2 31 8.7 28 5.5 22 22     , or about 54 (normally we would round down to 53, but this number is very close to 54 – of course for this large number of df, using either 53 or 54 won’t make much difference in the critical t value) so the desired confidence interval is   31 8.7 28 5.5 2268.13.885.91   131.6,269.931.22.3  . Because 0 does not lie inside this interval, we can be reasonably certain that the true difference 21   is not 0 and, therefore, that the two population means are not equal. For a 95% interval, the t value increases to about 2.01 or so, which results in the interval 506.32.3  . Since this interval does contain 0, we can no longer conclude that the means are different if we use a 95% confidence interval. 26. Let 1 the true average potential drop for alloy connections and let 2 the true average potential drop for EC connections. Since we are interested in whether the potential drop is higher for alloy connections, an upper tailed test is appropriate. We test 0: 210  H vs. 0: 21  aH . Using the SAS output provided, the test statistic, when assuming unequal variances, is t = 3.6362, the corresponding df is 37.5, and the p-value for our upper tailed test would be ½ (two-tailed p-value) =   0004.0008.2 1  . Our p- value of .0004 is less than the significance level of .01, so we reject H0. We have sufficient evidence to claim that the true average potential drop for alloy connections is higher than that for EC connections. 27. The approximate degrees of freedom for this estimate are       83.8 175.101 59.893 75 2 8 3.8 2 6 3.11 2 8 3.8 6 3.11 22 22     , which we round down to 8, so 306.28,025. t and the desired interval is    4674.5306.29.18306.24.213.40 8 3.8 6 3.11 22   5.31,3.6607.129.18  . Because 0 is not contained in this interval, there is strong evidence that 21   is not 0; i.e., we can conclude that the population means are not 1 Chapter 9: Inferences Based on Two Samples equal. Calculating a confidence interval for 12   would change only the order of subtraction of the sample means, but the standard error calculation would give the same result as before. Therefore, the 95% interval estimate of 12   would be ( -31.5, -6.3), just the negatives of the endpoints of the original interval. Since 0 is not in this interval, we reach exactly the same conclusion as before; the population means are not equal. 28. We will test the hypotheses: 10: 210  H vs. 10: 21  aH . The test statistic is     08.2 17.2 5.410 5 44.4 10 75.2 22     yx t The degrees of freedom       559.5 95.3 08.22 49 2 5 44.4 2 10 75.2 2 5 44.4 10 75.2 22 22     , and the p-value from table A.8 is approx .045, which is < .10 so we reject H0 and conclude that the true average lean angle for older females is more than 10 degrees smaller than that of younger females. 29. Let 1 the true average compression strength for strawberry drink and let 2 the true average compression strength for cola. A lower tailed test is appropriate. We test 0: 210  H vs. 0: 21  aH . The test statistic is 10.2 154.29 14    t .       3.25 8114.77 36.1971 14 15 14 4.29 4.44 22 2    , so use df=25. The p-value 023.)10.2(  tP . This p-value indicates strong support for the alternative hypothesis. The data does suggest that the extra carbonation of cola results in a higher average compression strength. 30. a. We desire a 99% confidence interval. First we calculate the degrees of freedom:       24.37 2626 2 26 3.4 2 26 2.2 2 26 3.4 26 2.2 22 22     , which we would round down to 37, except that there is no df = 37 row in Table A.5. Using 36 degrees of freedom (a more conservative choice), 719.236,005. t , and the 99% C.I. is 2 Chapter 9: Inferences Based on Two Samples 33. Let μ1 and μ2 represent the true mean body mass decrease for the vegan diet and the control diet, respectively. We wish to test the hypotheses H0: μ1 – μ2 ≤ 1 v. Ha: μ1 – μ2 > 1. The relevant test statistic is 32 8.2 32 2.3 1)8.38.5( 22   t = 1.33, with estimated df = 60 using the formula. Rounding to t = 1.3, Table A.8 gives a one-sided P-value of .098 (a computer will give the more accurate P-value of .094). Since our P-value > α = .05, we fail to reject H0 at the 5% level. We do not have statistically significant evidence that the true average weight loss for the vegan diet exceeds that for the control diet by more than 1 kg. 34. a. Following the usual format for most confidence intervals: statistic  (critical value)critical value) (critical value)standard error), a pooled variance confidence interval for the difference between two means is   nmpnm styx 11 2,2/   . b. The sample means and standard deviations of the two samples are 90.13x , 225.11 s , 20.12y , 010.12 s . The pooled variance estimate is  2 ps     2222 2 1 010.1244 14 225.1 244 14 2 1 2 1                                 s nm n s nm m 260.1 , so 1227.1ps . With df = m+n-1 = 6 for this interval, 447.26,025. t and the desired interval is      4 1 4 11227.1447.220.1290.13   64.3,24.943.17.1  . This interval contains 0, so it does not support the conclusion that the two population means are different. c. Using the two-sample t interval discussed earlier, we use the CI as follows: First, we need to calculate the degrees of freedom.       578.5 0686. 3971. 33 2 4 01.1 2 4 225.1 2 4 01.1 4 225.1 22 22     so 571.25,025. t . Then the interval is      74.3,34.7938.571.270.1571.22.129.13 4 01.1 4 225.1 22  . This interval is slightly wider, but it still supports the same conclusion. 2 Chapter 9: Inferences Based on Two Samples 35. There are two changes that must be made to the procedure we currently use. First, the equation used to compute the value of the t test statistic is:     nm s yx t p 11    where sp is defined as in Exercise 34 above. Second, the degrees of freedom = m + n – 2. Assuming equal variances in the situation from Exercise 33, we calculate sp as follows:     544.25.2 16 9 6.2 16 7 22             ps . The value of the test statistic is, then,     2.224.2 10 1 8 1 544.2 55.408.32    t . The degrees of freedom = 16, and the p- value is P ( t < -2.2) = .021. Since .021 > .01, we fail to reject H0. Section 9.3 36. 25.7d , 8628.11Ds 1 Parameter of Interest: D true average difference of breaking load for fabric in unabraded or abraded condition. 2 0:0 DH  3 0: DaH  4 ns d ns d t DD D / 0 /      5 RR: 998.27,01. tt 6 73.1 8/8628.11 025.7   t 7 Fail to reject H0. The data does not indicate a significant mean difference in breaking load for the two fabric load conditions. 2 Chapter 9: Inferences Based on Two Samples 37. a. This exercise calls for paired analysis. First, compute the difference between indoor and outdoor concentrations of hexavalent chromium for each of the 33 houses. These 33 differences are summarized as follows: n = 33, 4239.d , 3868.ds , where d = (indoor value – outdoor value). Then 037.232,025. t , and a 95% confidence interval for the population mean difference between indoor and outdoor concentration is    2868.,5611.13715.4239. 33 3868. 037.24239.        . We can be highly confident, at the 95% confidence level, that the true average concentration of hexavalent chromium outdoors exceeds the true average concentration indoors by between .2868 and .5611 nanograms/m3. b. A 95% prediction interval for the difference in concentration for the 34th house is       3758,.224.113868.037.24239.1 331132,025.  ndstd . This prediction interval means that the indoor concentration may exceed the outdoor concentration by as much as .3758 nanograms/m3 and that the outdoor concentration may exceed the indoor concentration by a much as 1.224 nanograms/m3, for the 34th house. Clearly, this is a wide prediction interval, largely because of the amount of variation in the differences. 38. a. The median of the “Normal” data is 46.80 and the upper and lower quartiles are 45.55 and 49.55, which yields an IQR of 49.55 – 45.55 = 4.00. The median of the “High” data is 90.1 and the upper and lower quartiles are 88.55 and 90.95, which yields an IQR of 90.95 – 88.55 = 2.40. The most significant feature of these boxplots is the fact that their locations (medians) are far apart. Normal:High: 90 80 70 60 50 40 Comparative Boxplots for Normal and High Strength Concrete Mix 2 Chapter 9: Inferences Based on Two Samples Differences P e rc e n t 5.02.50.0-2.5-5.0-7.5 99 95 90 80 70 60 50 40 30 20 10 5 1 Normal Probability Plot of Differences Normal 46. With    5,6, 11 yx ,    14,15, 22 yx ,    0,1, 33 yx , and    20,21, 44 yx , 1d and 0ds (the dI’s are 1, 1, 1, and 1), while s1 = s2 = 8.96, so sp = 8.96 and t = .16. Section 9.4 47. H0 will be rejected if 33.201.  zz . With 150.ˆ1 p , and 300.ˆ 2 p , 263. 800 210 600200 8030 ˆ    p , and 737.ˆ q . The calculated test statistic is     18.4 0359. 150. 737.263. 300.150. 600 1 200 1      z . Because 33.218.4  , H0 is rejected; the proportion of those who repeat after inducement appears lower than those who repeat after no inducement. 48. a. H0 will be rejected if 96.1z . With 2100. 300 63 ˆ1 p , and 4167. 180 75 ˆ 2 p , 2875.180300 7563 ˆ    p ,     84.4 0427. 2067. 7125.2875. 4167.2100. 180 1 300 1      z . Since 96.184.4  , H0 is rejected. b. 275.p and 0432. , so power =                              0432. 2.0421.96.1 0432. 2.0421.96.1 1      9967.72.254.61  . 2 Chapter 9: Inferences Based on Two Samples 49. 1 Parameter of interest: p1 – p2 = true difference in proportions of those responding to two different survey covers. Let p1 = Plain, p2 = Picture. 2 0: 210  ppH 3 0: 21  ppH a 4  nmqp pp z 11 21 ˆˆ ˆˆ    5 Reject H0 if p-value < .10 6     1910. 213 1 207 1 420 207 420 213 213 109 207 104    z ; p-value = .4247 7 Fail to Reject H0. The data does not indicate that plain cover surveys have a lower response rate. 50. Let 05. . A 95% confidence interval is    n qp m qpzpp 2211 ˆˆˆˆ2/21 ˆˆ            1708,.0160.0774.0934. 266395 96.1 266 140 266 126 395 171 395 224 266 126 395 224        . 51. a. Let p1 and p2 denote the true incidence rates of GI problems for the olestra and control groups, respectively. We wish to test H0: p1 – μ2 = 0 v. Ha: p1 – p2 ≠ 0. The pooled proportion is 563529 )158(.563)176(.529 ˆ   p = .1667, from which the relevant test statistic is z = ]563529)[8333)(.1667(. 158.176. 11    = 0.78. The two-sided P-value is 2P(Z ≥ 0.78) = .433 > α = .05, hence we fail to reject the null hypothesis. The data do not suggest a statistically significant difference between the incidence rates of GI problems between the two groups. b.   39.1210 )05(. )8)(.2(.)85)(.15(.28.12/)65.1)(35(.96.1 2 2   n , so a common sample size of m = n = 1211 would be required. 52. Let p1 = true proportion of irradiated bulbs that are marketable; p2 = true proportion of untreated bulbs that are marketable; The hypotheses are 0: 210  ppH vs. 0: 210  ppH . The test statistic is  nmqp pp z 11 21 ˆˆ ˆˆ    . With 850. 180 153 ˆ1 p , and 2 Chapter 9: Inferences Based on Two Samples 661. 180 119 ˆ 2 p , 756.360 272 ˆ p ,     2.4 045. 189. 244.756. 661.850. 180 1 180 1    z . The p-value =   02.41  , so reject H0 at any reasonable level. Radiation appears to be beneficial. 53. a. A 95% large sample confidence interval formula for  ln is   ny yn mx xm z     2/ ˆln  . Taking the antilogs of the upper and lower bounds gives the confidence interval for  itself. b. 818.1ˆ 037,11 104 034,11 189  ,   598.ˆln  , and the standard deviation is       1213. 104037,11 933,10 189034,11 845,10  , so the CI for  ln is    836,.360.1213.96.1598.  . Then taking the antilogs of the two bounds gives the CI for  to be  31.2,43.1 . We are 95% confident that people who do not take the aspirin treatment are between 1.43 and 2.31 times more likely to suffer a heart attack than those who do. This suggests aspirin therapy may be effective in reducing the risk of a heart attack. 54. a. The “after” success probability is p1 + p3 while the “before” probability is p1 + p2, so p1 + p3 > p1 + p2 becomes p3 > p2; thus we wish to test 230 : ppH  versus 23: ppH a  . b. The estimator of (p1 + p3) – (p1 + p2) is     n XX n XXXX 232131   . c. When H0 is true, p2 = p3, so n pp n XX Var 3223         , which is estimated by n pp 32 ˆˆ  . The Z statistic is then 32 23 32 23 ˆˆ XX XX n pp n XX      . d. The computed value of Z is 68.2 150200 150200    , so   0037.68.21 P . At level .01, H0 can be rejected but at level .001 H0 would not be rejected. 3 Chapter 9: Inferences Based on Two Samples Hence, H0 is not rejected. The data does not suggest a significant difference in the two population variances. 3 Chapter 9: Inferences Based on Two Samples 63.               1/ / 1,1,2/2 2 2 2 2 1 2 1 1,1,2/1 nmnm FS S FP . The set of inequalities inside the parentheses is clearly equivalent to 2 1 1,1,2/ 2 2 2 1 2 2 2 1 1,1,2/1 2 2 S FS S FS nmnm      . Substituting the sample values 21s and 2 2s yields the confidence interval for 2 1 2 2   , and taking the square root of each endpoint yields the confidence interval for 1 2   . m = n = 4, so we need 28.93,3,05. F and 108. 28.9 1 3,3,95. F . Then with s1 = .160 and s2 = .074, the C. I. for 2 1 2 2   is (.023, 1.99), and for 1 2   is (.15, 1.41). 64. A 95% upper bound for 1 2   is       10.8 79. 18.359.3 2 2 2 1 9,9,05. 2 2  s Fs . We are confident that the ratio of the standard deviation of triacetate porosity distribution to that of the cotton porosity distribution is at most 8.10. Supplementary Exercises 65. We test 0: 210  H vs. 0: 21  aH . The test statistic is     22.3 524.15 50 241 50 10 41 10 27 757807 222 2 2 1        n s m s yx t . The approximate d.f. is       6.15 9 1.168 9 9.72 241 22 2    , which we round down to 15. The p-value for a two- tailed test is approximately 2P(T > 3.22) = 2( .003) = .006. This small of a p-value gives strong support for the alternative hypothesis. The data indicates a significant difference. Due to the small sample sizes (10 each), we are assuming here that compression strengths for both fixed and floating test platens are normally distributed. And, as always, we are assuming the data were randomly sampled from their respective populations. 3 Chapter 9: Inferences Based on Two Samples 66. a. Although the median of the fertilizer plot is higher than that of the control plots, the fertilizer plot data appears negatively skewed, while the opposite is true for the control plot data. b. A test of 0: 210  H vs. 0: 21  aH yields a t value of -.20, and a two- tailed p-value of .85. (d.f. = 13). We would fail to reject H0; the data does not indicate a significant difference in the means. c. With 95% confidence we can say that the true average difference between the tree density of the fertilizer plots and that of the control plots is somewhere between –144 and 120. Since this interval contains 0, 0 is a plausible value for the difference, which further supports the conclusion based on the p-value. 67. Let p1 = true proportion of returned questionnaires that included no incentive; p2 = true proportion of returned questionnaires that included an incentive. The hypotheses are 0: 210  ppH vs. 0: 210  ppH . The test statistic is  nmqp pp z 11 21 ˆˆ ˆˆ    . 682. 110 75 ˆ1 p , and 673.98 66 ˆ 2 p . At this point we notice that since 21 ˆˆ pp  , the numerator of the z statistic will be > 0, and since we have a lower tailed test, the p-value will be > .5. We fail to reject H0. This data does not suggest that including an incentive increases the likelihood of a response. 3 Fertiliz Control 1000 1100 1200 1300 1400 F er til iz Comparative Boxplot of Tree Density Between Fertilizer Plots and Control Plots Chapter 9: Inferences Based on Two Samples 73. Since we can assume that the distributions from which the samples were taken are normal, we use the two-sample t test. Let 1 denote the true mean headability rating for aluminum killed steel specimens and 2 denote the true mean headability rating for silicon killed steel. Then the hypotheses are 0: 210  H vs. 0: 21  aH . The test statistic is 25.2 086083. 66. 047203.03888. 66.      t . The approximate degrees of freedom       5.57 29 047203. 29 03888. 086083. 22 2    , so we use 57. The two-tailed p-value   028.014.2  , which is less than the specified significance level, so we would reject H0. The data supports the article’s authors’ claim. 74. Let 1 denote the true average tear length for Brand A and let 2 denote the true average tear length for Brand B. The relevant hypotheses are 0: 210  H vs. 0: 21  aH . Assuming both populations have normal distributions, the two-sample t test is appropriate. m = 16, 0.74x , s1 = 14.8, n = 14, 0.61y , s2 = 12.5, so the approximate d.f. is       97.27 1315 2 14 5.12 2 16 8.14 2 14 5.12 16 8.14 22 22     , which we round down to 27. The test statistic is 6.2 0.610.74 14 5.12 16 8.14 22    t . From Table A.7, the p-value = P( t > 2.6) = .007. At a significance level of .05, H0 is rejected and we conclude that the average tear length for Brand A is larger than that of Brand B. 75. a. The relevant hypotheses are 0: 210  H vs. 0: 21  aH . Assuming both populations have normal distributions, the two-sample t test is appropriate. m = 11, 1.98x , s1 = 14.2, n = 15, 2.129y , s2 = 39.1. The test statistic is 84.2 252.120 1.31 9207.1013309.18 1.31      t . The approximate degrees of freedom       64.18 14 9207.101 10 3309.18 252.120 22 2    , so we use 18. From Table A.8, the two-tailed p-value   012.006.2  . No, obviously, the results are different. 3 Chapter 9: Inferences Based on Two Samples b. For the hypotheses 25: 210  H vs. 25: 21  aH , the test statistic changes to   556. 252.120 251.31   t . With degrees of freedom 18, the p-value   278.6.  tP . Since the p-value is greater than any sensible choice of  , we fail to reject H0. There is insufficient evidence that the true average strength for males exceeds that for females by more than 25N. 76. a. The relevant hypotheses are 0: 210   H (which is equivalent to saying 021   ) versus 0: 21   aH (which is the same as saying 021   ). The pooled t test is based on d.f. = m + n – 2 = 8 + 9 – 2 = 15. The pooled variance is 2ps 2 2 2 1 2 1 2 1 s nm n s nm m                     22 6.4 298 19 9.4 298 18                 49.22 , so 742.4ps . The test statistic is 0.304.3 742.4 0.110.18** 9 1 8 111        nmps yx t . From Table A.7, the p-value associated with t = 3.0 is 2P( t > 3.0 ) = 2(.004) = .008. At significance level .05, H0 is rejected and we conclude that there is a difference between 1 and  2 , which is equivalent to saying that there is a difference between 1 and 2 . b. No. The mean of a lognormal distribution is   2/ 2   e , where  and  are the parameters of the lognormal distribution (i.e., the mean and standard deviation of ln(x)). So when   21  , then   21  would imply that 21   . However, when   21  , then even if   21  , the two means 1 and 2 (given by the formula above) would not be equal. 77. This is paired data, so the paired t test is employed. The relevant hypotheses are 0:0 dH  vs. 0: daH  , where d denotes the difference between the population average control strength minus the population average heated strength. The observed differences (control – heated) are: -.06, .01, -.02, 0, and -.05. The sample mean and standard deviation of the differences are 024.d and 0305.ds . The test statistic is 8.176.1 024. 5 0305.   t . From Table A.8, with d.f. = 5 – 1 = 4, the lower tailed p- value associated with t = -1.8 is P( t < -1.8) = P( t > 1.8 ) = .073. At significance level .05, H0 should not be rejected. Therefore, this data does not show that the heated average strength exceeds the average strength for the control population. 78. Let 1 denote the true average ratio for young men and 2 denote the true average ratio for elderly men. Assuming both populations from which these samples were taken are normally 3 Chapter 9: Inferences Based on Two Samples distributed, the relevant hypotheses are 0: 210  H vs. 0: 21  aH . The value of the test statistic is       5.7 12 28. 13 22. 71.647.7 22    t . The d.f. = 20 and the p-value is P(t > 7.5) ≈ 0. Since the p-value is 05.  , we reject H0. We have sufficient evidence to claim that the true average ratio for young men exceeds that for elderly men. 79. The normal probability plot below indicates the data for good visibility does not follow a normal distribution, thus a t-test is not appropriate for this small a sample size. (The plot for poor visibility isn’t as bad.) Good P e rc e n t 3210-1 99 95 90 80 70 60 50 40 30 20 10 5 1 80. a. A 95% CI for μ37,dry = 325.73 ± t.025,5(34.97)/ 6 = 325.73 ± 2.571(14.276) = (289.03, 362.43). We are 95% confident that the true average breaking force in a dry medium at 37° is between 289.03 N and 362.43 N. b. The relevant estimated df = 9. A 95% CI for μ37,dry – μ37,wet = (325.73 – 306.09) ± t.025,9 6 97.41 6 97.34 22  = (–30.81,70.09). We are 95% confident that the true average breaking force in a dry medium at 37° is between 30.81 N less and 70.09 N more than the true average breaking force in a wet medium at 37°. c. We wish to test H0: μ37,dry – μ22,dry = 0 v. Ha: μ37,dry – μ22,dry > 0. The relevant test statistic is t = 6 08.39 6 97.34 100)60.17073.325( 22   = 2.58. The estimated df = 9 again, and the approximate P-value is .015. Hence, we reject H0 and conclude that true average force in a dry medium at 37° is indeed more than 100 N greater than the average at 22°. 4 87. 00  , 1021  , d = 1, nn 142.14200  , so          142.14 645.1 n  , giving  .9015, .8264, .0294, and .0000 for n = 25, 100, 2500, and 10,000 respectively. If the si ' referred to true average IQ’s resulting from two different conditions, 121   would have little practical significance, yet very large sample sizes would yield statistical significance in this situation. 88. 0: 210  H is tested against 0: 21  aH using the two-sample t test, rejecting H0 at level .05 if either 131.215,025. tt or if 131.2t . With 20.11x , 68.21 s , 79.9y , 21.32 s , and m = n = 8, sp = 2.96, and t = .95, so H0 is not rejected. In the situation described, the effect of carpeting would be mixed up with any effects due to the different types of hospitals, so no separate assessment could be made. The experiment should have been designed so that a separate assessment could be obtained (e.g., a randomized block design). 89. 210 : ppH  will be rejected at level  in favor of 21: ppH a  if either 645.105. zz . With 10.ˆ 2500 250 1 p , 0668.ˆ 2500 167 2 p , and 0834.ˆ p , 2.4 0079. 0332. z , so H0 is rejected . It appears that a response is more likely for a white name than for a black name. 90. The computed value of Z is 34.1 4634 4634    z . A lower tailed test would be appropriate, so the p-value   05.0901.34.1  , so we would not judge the drug to be effective. 91. a. Let 1 and 2 denote the true average weights for operations 1 and 2, respectively. The relevant hypotheses are 0: 210  H vs. 0: 21  aH . The value of the test statistic is       43.6 318083.7 39.17 30672.3011363.4 39.17 30 96.9 30 97.10 63.141924.1402 22         t . The d.f.       5.57 29 30672.3 29 011363.4 318083.7 22 2    , so use df = 57. 000.257,025. t , so we can reject H0 at level .05. The data indicates that there is a significant difference between the true mean weights of the packages for the two operations. b. 1400: 10 H will be tested against 1400: 1 aH using a one-sample t test with test statistic m s x t 1 1400  . With degrees of freedom = 29, we reject H0 if 699.129,05.  tt . The test statistic value is 1.100.2 24.2140024.1402 30 97.10   t . Because 1.1 < 1.699, H0 is not rejected. True average weight does not appear to exceed 1400. 92. First,          nmnm YXVar 1121   under H0, where λ can be estimated for the variance by the pooled estimate nm YnXm   ̂ . With the obvious point estimates X1̂ , Y2̂ , we have a large-sample test statistic of m Y n X YX nm YX Z             11ˆ 0)(  . With 616.1x and 557.2y , z = -5.3 and p-value =    0006.3.52  , so we would certainly reject 210 :  H in favor of 21:  aH . 93. A large-sample confidence interval for λ1 – λ2 is nm z 212/21 ˆˆ )ˆˆ(     , or n y m x zyx  2/)(  . With 62.1x and 56.2y , the 95% confidence interval for λ1 – λ2 is -.94 ± 1.96(.177) = -.94 ± .35 = (-1.29,-.59).