9.1 (a) The standard deviation of the four sample differences is given as.68. The standard error is SE (ȳ1  ȳ 2 ) = SE d  = s d n d


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1 CHAPTER 9 Comparison of Paired Samples 9.1 (a) The standard deviation of the four sample differences is given as.68. The standard error is SE (ȳ1  ȳ 2 ) = SE d  = s d n d =.68 4 =.34. (b) H 0 : The mean yields of the two varieties are the same (µ 1 = µ 2 ) H A : The mean yields of the two varieties are different (µ 1 µ 2 ) t s = 1.65/.34 = With df = 3, Table 4 gives t.01 = and t.005 = 5.841; thus,.01 < P <.02. At significance level α =.05, we reject H 0 if P <.05. Since.01 < P <.02, we reject H 0. There is sufficient evidence (.01 < P <.02) to conclude that Variety 2 has a higher mean yield than Variety 1. (c) H 0 : The mean yields of the two varieties are the same (µ 1 = µ 2 ) H A : The mean yields of the two varieties are different (µ 1 µ 2 ) SE (ȳ1  ȳ 2 ) = = t s = 1.65/1.230 = With df = 6, Table 4 gives t.20 =.906 and t.10 = Thus,.20 < P <.40 and we do not reject H 0. There is insufficient evidence (.20 < P <.40) to conclude that the mean yields of the two varieties are different. (By contrast, the correct test, in part (b), resulted in rejection of H 0.) 9.2 (a) The standard deviation of the nine sample differences is given as The standard error is SE d = s d = 59.3 n d 9 = (b) H 0 : The mean weight gains on the two diets are the same (µ 1 = µ 2 ) H A : The mean weight gains on the two diets are different (µ 1 µ 2 ) t s = 22.9/19.77 = With df = 8, Table 4 gives t.20 =.889 and t.10 = Thus,.20 < P <.40 and we do not reject H 0. There is insufficient evidence (.20 < P <.40) to conclude that the mean weight gains on the two diets are different. (c) 22.9 ± (1.860)(19.77) (13.9,59.7) or < µ d < 59.7 lb. (d) We are 90% confident that the average steer gains somewhere between 59.7 pounds more and 13.9 pounds less when on Diet 1 than when on Diet 2 (in a 140day period).
2 Let 1 denote control and let 2 denote progesterone. H 0 : Progesterone has no effect on camp (µ 1 = µ 2 ) H A : Progesterone has some effect on camp (µ 1 µ 2 ) The standard error is SE (ȳ1  ȳ 2 ) = SE d  = s d n d =.40 4 =.20. The test statistic is t s = ȳ 1  ȳ 2 SE (ȳ1  ȳ 2 ) = d  SE d  = = 3.4. To bracket the Pvalue, we consult Table 4 with df = 41 = 3. Table 4 gives t.025 = and t.02 = Thus, the Pvalue is bracketed as.04 < P <.05. At significance level α =.10, we reject H 0 if P <.10. Since.04 < P <.05, we reject H 0. There is sufficient evidence (.04 < P <.05) to conclude that progesterone decreases camp under these conditions. 9.4 (a) Let 1 denote treated side and 2 denote control side. The standard error is SE (ȳ1  ȳ 2 ) = s d = n d 15 = The critical value t.025 is found from Student's t distribution with df = n d  1 = 151 = 14. From Table 4 we find that t(14).025 = The 95% confidence interval is d  ± t.025 SE d.117 ± (2.145)(.2887) (.50,.74) or .50 < µ 1  µ 2 <.74 C. (b) SE (ȳ1  ȳ 2 ) = = ± (2.048)(.460) (using df = 28) (.83,1.06) or .83 < µ 1  µ 2 < 1.06 C. This interval is wider than the one obtained in part (a). 9.5 Let 1 denote treated side and 2 denote control side. H 0 : The electrical treatment has no effect on collagen shrinkage temperature (µ 1 = µ 2 ) H A : The electrical treatment tends to reduce collagen shrinkage temperature (µ 1 < µ 2 ) We note that ȳ 1 > ȳ 2, so the data do not deviate from H 0 in the direction specified by H A. Thus, P >.50 and we do not reject H 0. There is no evidence (P >.50) that the electrical treatment tends to reduce collagen shrinkage temperature under these conditions. 9.6 The data provide fairly strong evidence (P =.03) that desipramine is more effective than clomipramine in reducing the compulsion to pull one's hair.
3 9.7 SE d  = s d n d = 3 =.57. The confidence interval is 10.9 ± (2.052)(.57) or (9.7,12.1) With the outliers deleted, the mean of the remaining 26 differences is 11.0 and the standard deviation is 2.1. SE d = s d 2.1 = =.41. The confidence interval is 11.0 ± (2.060)(.41) or (10.1,11.8). This interval is n d 26 more narrow than the previous interval that was based on all of the data, including the outliers, but the difference is not great. 9.9 There is no single correct answer. Any data set with Y 1 and Y 2 varying, but d not varying, is correct; for example: Y 1 Y 2 d See Section III of this Manual (a) 34 Yield, Variety Yield, Variety 1 Yes, the upward trend indicates that the pairing was effective.
4 132 (b) Weight gain, Diet Weight gain, Diet 1 The upward trend here is rather weak, which indicates that the pairing was not especially effective. (c) Shrinkage Temp, Control Shrinkage Temp, Treated Yes, the upward trend indicates that the pairing was effective (a) B s = 6. Looking under n d = 9 in Table 7, we see that there is no entry less than or equal to 6. Therefore, P >.20. (b) B s = 7. Looking under n d = 9 in Table 7, we see that the only column with a critical value less than or equal to 7 is the column headed.20 (for a nondirectional alternative), and the next column is headed.10. Therefore,.10 < P <.20. (c) B s = 8. Looking under n d = 9 in Table 7, we see that the rightmost column with a critical value less than or equal to 8 is the column headed.05 (for a nondirectional alternative), and the next column is headed.02. Therefore,.02 < P <.05. (d) B s = 9. Looking under n d = 9 in Table 7, we see that the rightmost column with a critical value less than or equal to 9 is the column headed.01 (for a nondirectional alternative), and the next column is headed.002. Therefore,.002 < P <.01.
5 9.15 (a) P >.20 (b).10 < P <.20 (c).02 < P <.05 (d).002 < P <.01 (e) P <.001 (f) P < Let p denote the probability that oral conjugated estrogen will decrease PAI1 level. H 0 : Oral conjugated estrogen has no effect on PAI1 level (p =.5) H A : Oral conjugated estrogen has an effect on PAI1 level (p.5) N + = 8, N  = 22, B s = 22. With n d = 30, 22 falls under the.02 heading (for a nondirectional alternative) in Table 7. Thus,.01 < P <.02 and we reject H 0. There is sufficient evidence (.01 < P <.02) to conclude that oral conjugated estrogen tends to decrease PAI1 level For the sign test, the hypotheses can be stated as H 0 : p=.5 H A : p>.5 where p denotes the probability that the rat in the enriched environment will have the larger cortex. The hypotheses may be stated informally as H 0 : Weight of the cerebral cortex is not affected by environment H A : Environmental enrichment increases cortex weight There were 12 pairs. Of these, there were 10 pairs in which the relative cortex weight was greater for the "enriched" rat than for his "impoverished" littermate; thus N + = 10 and N  = 2. To check the directionality of the data, we note that N + > N  Thus, the data so deviate from H 0 in the direction specified by H A. The value of the test statistic is B s = larger of N + and N  = 10. Looking in Table 7, under n d = 12 for a directional alternative, we see that the rightmost column with a critical value less than or equal to 10 is the column headed.025 and the next column is headed.01. Therefore,.01 < P <.025. At significance level α =.05, we reject H 0 if P <.05. Since P <.025, we reject H 0. There is sufficient evidence (.01 < P <.025) to conclude that environmental enrichment increases cortex weight.
6 We have n d = 12. The null distribution is a binomial distribution with n = 12 and p =.5. Since B s = 10 and H A is directional, we need to calculate the probability of 10, 11, or 12 plus (+) signs. We apply the binomial formula n C j p j (1  p) nj, as follows: j = 10, n  j = 2: j = 11, n  j = 1: j = 12, n  j = 0: (66)(.5 10 )(.5 2 ) = (12)(.5 11 )(.5 1 ) = (1)(.5 12 )(.5 0 ) = The Pvalue is the sum of these probabilities: P = = Let p denote the probability that a patient will have fewer minor seizures with valproate then with placebo. H 0 : Valproate is not effective against minor seizures (p =.5) H A : Valproate is effective against minor seizures (p >.5) N + = 14, N  = 5, B s = 14; the data deviate from H 0 in the direction specified by H A. Eliminating the pair with d = 0, we refer to Table 7 with n d = 19. The only entry of 14 falls under the.05 heading (for a directional alternative). Thus,.025 < P <.05 and we reject H 0. There is sufficient evidence (.025 < P <.05) to conclude that valproate is effective against minor seizures We need to find the probability of 14 or more successes for a binomial distribution with n = 19 and p =.5. For the normal approximation to the binomial, the mean is np = (19)(.5) = 9.5 and the SD is z = np(1  p) = (19)(.5)(.5) = P = = = 1.84; Table 3 gives We need to find the probability of 22 or more successes, or of 8 or fewer successes, for a binomial distribution with n = 30 and p =.5. For the normal approximation to the binomial, the mean is np = (30)(.5) = 15 and the SD is np(1  p) = (30)(.5)(.5) = We find the probability of 22 or more successes and double this probability (since the normal curve is symmetric). z = = 2.37; Table 3 gives P = 2( ) = Let p denote the probability that the Northern member of a pair will dominate in more episodes than the Carolina.
7 H 0 : Dominance is balanced between the subspecies (p =.5) H A : One of the subspecies tends to dominate the other (p.5) N + = 8, N  = 0, B s = 8. Looking under n d = 8 in Table 7, we see that the rightmost column with a critical value less than or equal to 8 is the column headed.01 (for a nondirectional alternative), and the next column is headed.002. Therefore,.002 < P <.01. There is sufficient evidence (.002 < P <.01) to conclude the Carolina subspecies tends to dominate the Northern P = 2(.5 8 ) = (a) The null distribution is a binomial distribution with n = 7 and p =.5. Since B s =7 and H A is nondirectional, we need to calculate the probability of 7 successes or of 0 successes. The probability of 7 successes is.5 7 = Likewise, the probability of 0 successes is.5 7 = Thus, P = 2( ) = (b) With n d = 7, the smallest possible pvalue is ; thus P cannot be less than (a) (i) P = (2)[(105)(.5 13 )(.5 2 ) + (15)(.5 14 )(.5) + (1)(.5 15 )] = (ii) P = (2)[(15)(.5 14 )(.5) + (1)(.5 15 )] = (iii) P = (2)(.5 15 ) = (b) If B s = 14, then P = <.002; if B s = 13, then P = >.002. Thus, the critical value 14 corresponds to a Pvalue that is as close to.002 as possible without exceeding it. (c) The entry for.005 would be 14, because <.005 but > Let p denote the probability that hunger rating is higher when taking mcpp than when taking the placebo. H 0 : p =.5 H A : p.5 N + = 3, N  = 5, B s = 5. Looking under n d = 8 in Table 7, we see that the leftmost column has an entry of 7, so the Pvalue is greater than.20. There is insufficient evidence (P >.20) to conclude that hunger ratings differ on the two treatments P = (2)[(56)(.5 5 )(. 5 3 ) + (28)(.5 6 )(. 5 2 ) + (8)(.5 7 )(. 5 1 ) + (1)(.5 8 )] = (a) P >.20 (b).10 < P <.20 (c).02 < P <.05 (d).01 < P < (a) P >.20 (b).05 < P <.10 (c).002 < P <.01 (d).002 < P < H 0 : Hunger rating is not affected by treatment (mcpp vs. placebo) H A : Treatment does affect hunger rating 135
8 136 The absolute values of the differences are 5, 7, 28, 47, 80, 7, 8, and 20. The ranks of the absolute differences are 1, 2.5, 6, 7, 8, 2.5, 4, and 5. The signed ranks are 1, 2.5, 6, 7, 8, 2.5, 4, and 5. Thus, W+ = = 9 and W = = 27. W s = 27 and n d = 8; reading Table 8 we find Pvalue >.20 and H 0 is not rejected. There is insufficient evidence (P >.20) to conclude that treatment has an effect H 0 : Weight change is not affected by treatment (mcpp vs. placebo) H A : Treatment does affect weight change The absolute values of the differences are 1.1, 1.6, 2.1, 0.3, 0.6, 2.2, 0.9, 0.7, and 0.4. The ranks of the absolute differences are 6, 7, 8, 1, 3, 9, 5, 4, and 2. The signed ranks are 6, 7, 8, 1, 3, 9, 5, 4, and 2. Thus, W+ = = 15 and W = = 30. W s = 30 and n d = 9; reading Table 8 we find Pvalue >.20 and H 0 is not rejected. There is insufficient evidence (P >.20) to conclude that treatment has an effect H 0 : HLA compatibility has no effect on graft survival time H A : Survival time tends to be greater when compatibility score is close The differences tend to be positive, which is consistent with H A. The absolute values of the differences are 12, 6, 42+, 67, 5, 5, 6, 20, 11, 18+, and 1. The ranks of the absolute differences are 7, 4.5, 10, 11, 2.5, 2.5, 4.5, 9, 6, 8, and 1. The signed ranks are 7, 4.5, 10, 11, 2.5, 2.5, 4.5, 9, 6, 8, and 1. Thus, W+ = = 60.5 and W = = 5.5. W s = 60.5 and n d = 11; reading Table 8 we find.005 < Pvalue <.01 and H 0 is rejected. There is strong evidence (.005 < Pvalue <.01) to conclude that survival time tends to be greater when compatibility score is close H 0 : Alcoholism has no effect on brain density H A : Alcoholism reduces brain density The differences tend to be negative, which is consistent with H A. The absolute values of the differences are 1.2, 1.7,.5, 4.7, 3.3,.4, 2.7, 1.8,.1,.3, and 1.4. The ranks of the absolute differences are 5, 7, 4, 11, 10, 3, 9, 8, 1, 2, and 6. The signed ranks are 5, 7, 4, 11, 10, 3, 9, 8, 1, 2, and 6. Thus, W+ = = 5 and W = = 61. W s = 61 and n d = 11; reading Table 8 we find.001 < Pvalue <.005 and H 0 is rejected. There is strong evidence (.001 < Pvalue <.005) to conclude that alcoholism is associated with reduced brain density. This was an observational study, so drawing a causeeffect inference is risky. We should stop short of saying that alcoholism reduces brain density.
9 9.34 Let 1 denote 5 weeks and 2 denote baseline. (a) Let N denote no coffee. H 0 : Mean cholesterol does not change in the "no coffee" condition (µ N,1 = µ N,2 ) H A : Mean cholesterol does change in the "no coffee" condition (µ N,1 µ N,2 ) 137 SE = 27/ 25 = t s = 35/5.40 = With df = 24, Table 4 gives t.0005 = We reject H 0. There is sufficient evidence (P <.001) to conclude that mean cholesterol is reduced in the "no coffee" condition. (b) Let U denote usual coffee. H 0 : Mean cholesterol does not change in the "usual coffee" condition (µ U,1 = µ U,2 ) H A : Mean cholesterol does change in the "usual coffee" condition (µ U,1 µ U,2 ) SE = 56/ 8 = t s = 26/19.8 = With df = 7, Table 4 gives t.20 = and t.10 = We do not reject H 0. There is insufficient evidence (.20 < P <.40) to conclude that mean cholesterol is changed in the "usual coffee" condition. (c) Let N denote no coffee, U denote usual coffee, and d denote the change from baseline. H 0 : Mean cholesterol is not affected by discontinuing coffee (µ N,d = µ U,d ) H A : Mean cholesterol is affected by discontinuing coffee (µ N,d µ U,d ) SE = = t s = (3526)/20.52 = Formal (7.1) gives df = 8.1 and the conservative df value is min{24,7} = 7, whereas the liberal df value is n 1 + n 22 = 31. Using df = 8, Table 4 gives t.01 = and t.005 = 3.355, which implies that.01 < P <.02. (Using df = 30, the closest value to 31 in Table 4, we get t.005 = and t.0005 = 3.646, which implies that.001 < P <.01.) We reject H 0. (d) There is sufficient evidence (.01 < P <.02) to conclude that mean cholesterol is reduced by discontinuing coffee Food Intake (cal) Premenstrual Postmenstrual
10 No. This result suggests that the population mean difference between the right eye and the left eye is less than 1.1 mg/dl, but positive and negative differences cancel each other out when such a mean is calculated. The result says nothing at all about the typical or average magnitude of the difference between the two eyes No. "Accurate" prediction would mean that the individual differences (d's) are small. To judge whether this is the case, one would need the individual values of the d's; using these, one could see whether most of the magnitudes ( d 's) are small (a) ȳ 1  ȳ 2 = d  = 1, s d = 1.2. SE (ȳ1  ȳ 2 ) = 1.2/ 15 = ± (2.145)(.3098) (df = 14) (1.66,.34) or < µ 1  µ 2 < (b) SE (ȳ1  ȳ 2 ) = = ± (2.145)(.8595) (using df = 14) (2.84,0.84) or < µ 1  µ 2 < This interval is much wider than the one constructed in part (a) H 0 : The before and after means are the same (µ 1 = µ 2 ) H A : The before and after means are different (µ 1 µ 2 ) SE (ȳ1  ȳ 2 ) = 1.2/ 15 = t s = 1/.3098 = With df = 14, Table 4 gives t.005 = and t.0005 = 4.140; thus,.001 < P <.01. We reject H 0 ; there is strong evidence (.001 < P <.01) of a before and after difference (a) Let p denote the probability that a before count is higher than the corresponding after count. H 0 : p =.5 H A : p.5 N + = 2, N  = 10, B s = 10. Looking under n d = 12 in Table 7, we see that.02 < P <.05. There is sufficient evidence (.02 < P <.05) to conclude that the after count tends to be higher than the before count. (b) P = (2)[(66)(.5 10 )( 5 2 ) + (12)(.5 11 )( 5 1 ) + (1)(.5 12 )] =.0386.
11 The scatterplot shows a positive relationship between before and after counts. The pairing removes the variability between cats from the analysis and is, therefore, effective. After (Y ) Before (Y ) Let 1 denote central and 2 denote top. ȳ 1  ȳ 2 = d  = 2.533, s d = SE (ȳ1  ȳ 2 ) =.41312/ 6 = ± (2.015)(.1687) (df = 5) (2.19,2.87) or 2.19 < µ 1  µ 2 < 2.87 percent The standard error is SE (ȳ1  ȳ 2 ) = 1.86/ 15 = ± (1.761)(.48) (df = 14) (1.35,3.05) or 1.35 < µ 1  µ 2 < 3.05 species It must be reasonable to regard the 15 differences as a random sample from a normal population. We must trust the researchers that their sampling method was random. The normality condition can be verified with a normal probability plot. The plot below is fairly linear (although the plateaus show that there are several differences that have the same value), which supports the normality condition Difference nscores
12 The null and alternative hypotheses are H 0 : The average number of species is the same in pools as in riffles (µ 1 =µ 2 ) H A : The average numbers of species in pools and in riffles differ (µ 1 µ 2 ) The standard error is SE (ȳ1  ȳ 2 ) = SE d  = s d n d = =.48. The test statistic is t s = ȳ 1  ȳ 2 SE (ȳ1  ȳ 2 ) = d  SE d  = = To bracket the Pvalue, we consult Table 4 with df = 151 = 14. Table 4 gives t.0005 = Thus, the Pvalue for the nondirectional test is bracketed as P <.001. At significance level α =.10, we reject H 0 if P <.10. Since P <.001, we reject H 0. There is sufficient evidence (P <.001) to conclude that the average number of species in pools is greater than in riffles (a) Let p denote the probability that there are more species in a pool than in its adjacent riffle. H 0 : The two habitats support equal levels of diversity (p =.5) H A : The two habitats do not support equal levels of diversity (p.5) N + = 12, N  = 1, B s = 12. Eliminating the two pairs with d = 0, we refer to Table 7 with n d = 13. The rightmost column with a critical value of 12 is the column headed.01 for a nondirectional alternative (i.e., for a twotailed test), and the next column is headed.002. Therefore,.002 < P <.01. There is sufficient evidence (.002 < P <.01) to conclude that species diversity is greater in pools than in riffles. (b) P = (2)[(13)(.5 12 )(.5 1 ) ] = H 0 : Pools and riffles support equal levels of diversity H A : Pools and riffles support different levels of diversity The absolute values of the differences are 3, 3, 4, 3, 4, 5, 1, 1, 1, 4, 1, 4, and 1. The ranks of the absolute differences are 7, 7, 10.5, 7, 10.5, 13, 3, 3, 3, 10.5, 3, 10.5 and 3. The signed ranks are 7, 7, 10.5, 7, 10.5, 13, 3, 3, 3, 10.5, 3, 10.5 and 3. Thus, W+ = = 88 and W = 3. W s = 88 and n d = 13; reading Table 8 we find.001 < Pvalue <.002 and H 0 is rejected. There is strong evidence (.001 < Pvalue <.002) to conclude that the diversity levels differ between pools and riffles There are several ties in the data, which means that the Pvalue from the Wilcoxon test is only approximate.
13 9.49 The null and alternative hypotheses are H 0 : Caffeine has no effect on RER (µ 1 =µ 2 ) H A : Caffeine has some effect on RER (µ 1 µ 2 ) We proceed to calculate the differences, the standard error of the mean difference, and the test statistic. Subject Placebo Caffeine Difference Mean 7.33 SD 5.59 The standard error is SE (ȳ1  ȳ 2 ) = SE d  = s d n d = = The test statistic is t s = ȳ 1  ȳ 2 SE (ȳ1  ȳ 2 ) = d  SE d  = = To bracket the Pvalue, we consult Table 4 with df = 91 = 8. Table 4 gives t.005 = and t.0005 = Thus, the Pvalue for the nondirectional test is bracketed as.001 < P <.01. At significance level α =.05, we reject H 0 if P <.05. Since P <.01, we reject H 0. To determine the directionality of departure from H 0, we note that d  > 0; that is, ȳ 1 > ȳ 2. There is sufficient evidence (.001 < P <.01) to conclude that caffeine tends to decrease RER under these conditions. 141
14 RER (%) Placebo Caffeine 9.51 Let p denote the probability that RER for a subject is higher after taking placebo than after taking caffeine. H 0 : RER is not affected by caffeine (p =.5) H A : RER is affected by caffeine (p.5) N + = 9, N  = 0, B s = 9. Looking under n d = 9 in Table 7, we see that the rightmost column with a critical value less than or equal to 9 is the column headed.01 (for a nondirectional alternative), and the next column is headed.002. Therefore,.002 < P <.01. There is sufficient evidence (.002 < P <.01) to conclude that caffeine tends to decrease RER under these conditions H 0 : Mean CP is the same in regenerating and in normal tissue (µ 1 = µ 2 ) H A : Mean CP is different in regenerating and in normal tissue (µ 1 µ 2 ) SE (ȳ1  ȳ 2 ) = 4.89/ 8 = t s = 4.64/1.727 = With df = 7, Table 4 gives t.02 = and t.01 = Thus,.02 < P <.04 and we reject H 0. There is sufficient evidence (.02 < P <.04) to conclude that mean CP is different in regenerating and in normal tissue (a) Let 1 denote control and 2 denote benzamil. H 0 : Benzamil does not impair healing (µ 1 = µ 2 ) H A : Benzamil impairs healing (µ 1 > µ 2 ) ȳ 1  ȳ 2 = d  =.09706; s d = SE (ȳ1  ȳ 2 ) =.14768/ 17 = t s =.09706/ = P =.0077, so we reject H 0. There is sufficient evidence (P =.0077) to conclude that benzamil impairs healing.
15 (b) Let p denote the probability that the control limb heals more than the benzamil limb. H 0 : Benzamil does not impair healing (p =.5) H A : Benzamil impairs healing (p >.5) N + = 11, N  = 4, B s = 1. The two animals with d = 0 are eliminated. P =.059, so we do not reject H 0. There is insufficient evidence (P =.059) to conclude that benzamil impairs healing. [Remark: Unlike the t test in part (a), the sign test does not take account of the fact that the negative d's are smaller in magnitude than the positive d's. This illustrates the inferior power of the sign test.] (c) (.021,.173) or.021 < µ 1  µ 2 <.173 mm 2. (d) Benzamil Control Yes, the upward trend indicates that the pairing was effective Summary statistics are as follows: Experimental group Control group Rest (1) Work (2) Difference (3) Rest (4) Work (5) Difference (6) Mean SD (a) Column (1) versus column (4) H 0 : Mean ventilation at rest is the same in the two conditions (µ 1 = µ 4 ) H A : Mean ventilation at rest is different in the two conditions (µ 1 µ 4 ) t s = 2.757, df = 13.97, P =.015. We reject H 0. There is sufficient evidence (P =.015) to conclude that mean ventilation at rest is higher in the "to be hypnotized" condition than in the "control" condition. (b) (i) Column (1) versus column (2): H 0 : Hypnotic suggestion does not change mean ventilation (µ 1 = µ 2 ) H A : Hypnotic suggestion increases mean ventilation (µ 1 < µ 2 ) t s = , df = 7, P = We reject H 0. There is sufficient evidence (P =.0087) to conclude that hypnotic suggestion increases mean ventilation.
16 144 (ii) Column (4) versus column (5): H 0 : Waking suggestion does not change mean ventilation (µ 4 = µ 5 ) H A : Waking suggestion increases mean ventilation (µ 4 < µ 5 ) Because ȳ 4 > ȳ 5, the data do not deviate from H 0 in the direction specified by H A. Thus, P >.50 and we do not reject H 0. There is no evidence that waking suggestion increases mean ventilation. (iii) Column (3) versus column (6): H 0 : Hypnotic and waking suggestion produce the same mean change in ventilation (µ 3 = µ 6 ) H A : Hypnotic suggestion increases mean ventilation more than does waking suggestion (µ 3 < µ 6 ) t s = , df = 7.5, P = We reject H 0. There is sufficient evidence (P =.0055) to conclude that hypnotic suggestion increases mean ventilation more than does waking suggestion. (c) (i) Sign test for column (1) versus column (2). Let p 1 denote the probability that a person's ventilation after hypnotic suggestion will be higher than that at rest. H 0 : Hypnotic suggestion does not change mean ventilation (p 1 =.5) H A : Hypnotic suggestion increases mean ventilation (p 1 >.5) B s = 8, P = We reject H 0. There is sufficient evidence (P =.0039) to conclude that hypnotic suggestion increases mean ventilation. (ii) Sign test for column (4) versus column (5). Let p 2 denote the probability that a person's ventilation after waking suggestion will be higher than that at rest. H 0 : Waking suggestion does not change mean ventilation (p 2 =.5) H A : Waking suggestion increases mean ventilation (p 2 >.5) N + = 6, N  = 2. Thus, the data do not deviate from H 0 in the direction specified by H A, so P >.50 and we do not reject H 0. There is no evidence that waking suggestion increases mean ventilation. (iii) WilcoxonMannWhitney test for column (3) versus column (6): H 0 : Hypnotic and waking suggestion produce the same mean change in ventilation H A : Hypnotic suggestion increases mean ventilation more than does waking suggestion U s = 63, P = We reject H 0. There is sufficient evidence (P = ) to conclude that hypnotic suggestion increases mean ventilation more than does waking suggestion.
17 145 (d) A normal probability plot of column (3) shows that the data are quite skewed. This could account for two discrepancies: First, to compare column (1) to column (2), we used the differences in column (3); the t test gave P =.0087 whereas the sign test gave P = Second, to compare column (3) to column (6), the t test gave P =.0055 whereas the WilcoxonMannWhitney test gave P = Both of the t tests rest on the questionable condition that the population distribution corresponding to column (3) is normal. The failure of this condition inflates the standard deviation and robs the t test of power, so that the nonparametric tests give stronger conclusions (smaller Pvalues) Ventillation nscores A normal probability plot of column (6) shows that the normality condition appears to be met for these data. 5.5 Ventillation nscores 9.55 (a) By using matched pairs we eliminate the variability that is associated with the variables used to create the pairs (age, sex, etc.). This provides for greater precision and more power in the test. (b) It may be that the pairing variables (age, sex, etc.) are unrelated to blood pressure. If this is the case, then the pairing accomplishes nothing, but it reduces the number of degrees of freedom, and therefore the power, of the test N + = 10, N  = 10, B s = 10. In this case, the data are as evenly balanced as possible, so P = 1. (Table 7 indicates that P >.20.) Thus, we do not reject H 0. There is no evidence that transdermal estradiol has an effect on PAI1 level.
18 A normal probability plot of the data shows that the normality condition is not met. However, a sign test can be conducted. Let p denote the probability that urinary protein excretion will go down after plasmapheresis. H 0 : Plasmapheresis affects urinary protein excretion (p =.5) H A : Plasmapheresis does not affect urinary protein excretion (p.5) N + = 6, N  = 0, B s = 6. From Table 7,.02 < P <.05 (for a twosided test). The exact Pvalue is (2)(.5 6 ) = Thus, there is evidence (P =.03125) to conclude that urinary protein excretion tends to go down after plasmapheresis. Note: Another approach would be to transform the data and then conduct a t test in the transformed scale. For example, taking the reciprocal of each difference yields a fairly symmetric distribution; a t test then gives t s = 5.4 and P =.003.
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