chapter 1

Chapter 15: Distribution-Free Procedures15. Let µ1and µ2denote true average cotanine levels in unexposed and exposed infants,respectively. The hypotheses of interest are H µ − µ = 25 vs.Ha: µ1− µ2< −25( 7 + 8 + 1) − 71 41w ≤ 7 =0:1 2−. With m = 7, n = 8, H o will be rejected at level .05 if. Before ranking, -25 is subtracted from each x I (i.e. 25 isadded to each), giving 33, 36, 37, 39, 45, 68, and 136. The corresponding ranks in thecombined set of 15 observations are 1, 3, 4, 5, 6, 8, and 12, from which w = 1 + 3 + … + 12 =39. Because 39 ≤ 41, H o is rejected. The true average level for exposed infants appears toexceed that for unexposed infants by more than 25 (note that H o would not be rejected usinglevel .01).16.a.X rank Y rank0.43 2 1.47 91.17 8 0.8 70.37 1 1.58 110.47 3 1.53 100.68 6 4.33 160.58 5 4.23 150.5 4 3.25 142.75 12 3.22 13We verify that w = sum of the ranks of the x’s = 41.b. We are testing µ − µ 0H vs. H µ − µ 00:1 2=a:1 2