chapter 1

CHAPTER 15Section 15.11. We test : µ 100H vs. H : µ ≠ 1000=a. The test statistic is s + = sum of the ranksassociated with the positive values of ( x −100), and we reject H o at significance level .05s+. (from Table A.13, n = 12, with α / 2 = . 02612( 13)+≤ − 64 = 78 − 64 = 142x ( −100)if ≥ 64value of . 025), or if s .iiix ranks105.6 5.6 7*90.9 -9.1 1291.2 -8.8 1196.9 -3.1 396.5 -3.5 591.3 -8.7 10100.1 0.1 1*105 5 6*99.6 -0.4 2107.7 7.7 9*103.3 3.3 4*92.4 -7.6 8, which is close to the desiredS + = 27, and since 27 is neither ≥ 64 nor ≤ 14 , we do not reject H o . There is not enoughevidence to suggest that the mean is something other than 100.H vs. H : µ > 25 . With n = 5 and α ≈ . 03 , reject H o if ≥ 152. We test0: µ = 25as+.From the table below we arrive at s + =1+5+2+3 = 11, which is not ≥ 15 , so do not reject H o .It is still plausible that the mean = 25.x ( x − 25)ranksii25.8 0.8 1*36.6 11.6 5*26.3 1.3 2*21.8 -3.2 427.2 2.2 3*457