chapter 1

Chapter 15: Distribution-Free Procedures11. With X identified with pine (corresponding to the smaller sample size) and Y with oak, wewish to test H0: µ1− µ2= 0 vs. Ha: µ1− µ2≠ 0 . From Table A.14 with m = 6 and n= 8, H o is rejected in favor of H a at level .05 if either w ≥ 61 or if w ≤ 90 − 61 = 29 (theactual α is 2(.021) = .042). The X ranks are 3 (for .73), 4 (for .98), 5 (for 1.20), 7 (for 1.33),8 (for 1.40), and 10 (for 1.52), so w = 37. Since 37 is neither ≥ 61 nor ≤ 29 , H o cannot berejected.12. The hypotheses of interest are µ − µ 1H vs. H µ − µ 10:1 2=a:1 2>, where 1(X) refersto the original process and 2 (Y) to the new process. Thus 1 must be subtracted from each x Ibefore pooling and ranking. At level .05, H o should be rejected in favor of H a if w ≥ 84 .x – 1 3.5 4.1 4.4 4.7 5.3 5.6 7.5 7.6rank 1 4 5 6 8 10 15 16y 3.8 4.0 4.9 5.5 5.7 5.8 6.0 7.0rank 2 3 7 9 11 12 13 14Since w = 65, H o is not rejected.13. Here m = n = 10 > 8, so we use the large-sample test statistic from p. 663.H0: µ1− µ2= 0 will be rejected at level .01 in favor of Ha: µ1− µ2≠ 0 if eitherz ≥ 2.58 or z ≤ −2. 58 . Identifying X with orange juice, the X ranks are 7, 8, 9, 10, 11,16, 17, 18, 19, and 20, so w = 135. With( m + n + 1)mn12=≥ 2.58 nor −2. 58( )m m + n +1 = 1052and135 −105175 = 13.22 , z = = 2. 2713.22p − value ≈ 2 1− Φ 2.27 = .0232. Because 2.27 is neither≤ , H o is not rejected. ( ( )) .14.x 8.2 9.5 9.5 9.7 10.0 14.5 15.2 16.1 17.6 21.5rank 7 9 9 11 12.5 16 17 18 19 20y 4.2 5.2 5.8 6.4 7.0 7.3 9.5 10.0 11.5 11.5rank 1 2 3 4 5 6 9 12.5 14.5 14.5The denominator of z must now be computed according to (15.6). With 3τ , [ ( )( ) ( )( ) ( )( ) ] 21= 2 2175 .0219 2 3 4 1 2 3 1 2 3 174.3σ = −+ + = , so138.5 −105z == 2.54 . Because 2.54 is neither ≥ 2. 58 nor ≤ −2. 58174.21rejected.461τ , 21=τ ,2=, H o is not