chapter 1

Chapter 9: Inferences Based on Two Samples22. Let µ1= the true average strength for wire-brushing preparation and let µ2= the averagestrength for hand-chisel preparation. Since we are concerned about any possible differencebetween the two means, a two-sided test is appropriate. We test H µ − µ 0 vs.Ha: µ1− µ2≠022 21.58 4.01( + )12 122 221.58 4.01( ) ( )0:1 2=. We need the degrees of freedom to find the rejection region:2.3964ν === 14.332.0039 + .1632125+11 11reject H o if t ≥ t .= 2. 145 . The test statistic is025,1419.20 − 23.13 − 3.93t == = −3.159, which is ≤ −2. 14521.244221.58 4.01( + )1212, which we round down to 14, so we, so we reject H o andconclude that there does appear to be a difference between the two population averagestrengths.23.a. Normal plotsNormal Probability Plot for High Quality FabricNormal Probability Plot for Poor Quality FabricProbability.999.99.95.80.50.20.05.01.001Probability.999.99.95.80.50.20.05.01.001Av erage: 1. 50833StD ev : 0. 444206N: 240.81.3H:1.82.3Anders on-D arling N ormality Tes tA-Squared: 0. 396P-Value: 0.344Av erage: 1. 58750St Dev : 0.530330N: 241.0 1.5 2.0 2.5P:Anderson-D arling N ormality Tes tA-Squared: -10.670P-Value: 1. 000Using Minitab to generate normal probability plots, we see that both plots illustratesufficient linearity. Therefore, it is plausible that both samples have been selected fromnormal population distributions.268