chapter 1

Chapter 15: Distribution-Free Procedures3. We test H0: µ = 7. 39 vs. Ha: µ ≠ 7. 39 , so a two tailed test is appropriate. With n =14 and α / 2 = . 025 , Table A.13 indicates that H o should be rejected if either( x − 7.39 ’s are -.37, -.04, -.05, -.22, -.11, .38, -.30, -.17, .06, -.44,s+≥ 84or≤ 21 . The i).01, -.29, -.07, and -.25, from which the ranks of the three positive differences are 1, 4, and 13.Since s = 18 ≤ 21 , H o is rejected at level .05.+4. The appropriate test is : µ 30H o if( )H vs. H : µ < 30 . With n = 15, and α = . 100=15 16s+≤ − 83 = 37 .2xi( x i− 30)ranks xi( x i− 30)ranks30.6 0.6 3* 31.9 1.9 5*30.1 0.1 1* 53.2 23.2 15*15.6 -14.4 12 12.5 -17.5 1326.7 -3.3 7 23.2 -6.8 1127.1 -2.9 6 8.8 -21.2 1425.4 -4.6 8 24.9 -5.1 1035 5 9* 30.2 0.2 2*30.8 0.8 4*a, rejectS + = 39, which is not ≤ 37 , so H o cannot be rejected. There is not enough evidence to provethat diagnostic time is less than 30 minutes at the 10% significance level.5. The data is paired, and we wish to test : 0H µ vs. H : µ ≠ 0 . With n = 12 and0 D=α = .05 , H o should be rejected if either s ≥ 64 or if ≤ 14++as .d i -.3 2.8 3.9 .6 1.2 -1.1 2.9 1.8 .5 2.3 .9 2.5rank 1 10* 12* 3* 6* 5 11* 7* 2* 8* 4* 9*s = 72 , and 72 ≥ 64 , so H o is rejected at level .05. In fact for α = . 01+is c = 71, so even at level .01 H o would be rejected.D, the critical value458