chapter 1

Chapter 15: Distribution-Free Procedures6. We wish to test : 5H0µD= vs. Ha: µD> 5 , where µDµblack− µwhiteα ≈ . , H o will be rejected if s+≥ 37 . As given in the table below, s+= 37≥= . With n = 9and 05, whichis 37 , so we can (barely) reject H o at level approximately .05, and we conclude that thegreater illumination does decrease task completion time by more than 5 seconds.d − 5idirankid −5d rank7.62 2.62 3* 16.07 11.07 9*8 3 4* 8.4 3.4 5*9.09 4.09 8* 8.89 3.89 7*6.06 1.06 1* 2.88 -2.12 21.39 -3.61 6iH0µD= vs. Ha: µD> . 20 , where µD= µoutdoor− µindoor. α = . 057. : . 20because n = 33, we can use the large sample test. The test statistic iswe reject H o if z ≥1. 96 .d − . 2idirankid −. 2dirankiZ =sd −. 2i, and( n+1)n+−4, andn( n+1)( 2n+1)24d rank0.22 0.02 2 0.15 -0.05 5.5 0.63 0.43 230.01 -0.19 17 1.37 1.17 32 0.23 0.03 40.38 0.18 16 0.48 0.28 21 0.96 0.76 310.42 0.22 19 0.11 -0.09 8 0.2 0 10.85 0.65 29 0.03 -0.17 15 -0.02 -0.22 180.23 0.03 3 0.83 0.63 28 0.03 -0.17 140.36 0.16 13 1.39 1.19 33 0.87 0.67 300.7 0.5 26 0.68 0.48 25 0.3 0.1 9.50.71 0.51 27 0.3 0.1 9.5 0.31 0.11 110.13 -0.07 7 -0.11 -0.31 22 0.45 0.25 200.15 -0.05 5.5 0.31 0.11 12 -0.26 -0.46 24424 − 280.5 143.5s+= 434 , so z == = 2. 56 . Since 2.56≥ 1. 96 , we reject H o3132.25 55.9665at significance level .05.459