chapter 1

Chapter 9: Inferences Based on Two Samples91.a. Let µ1and µ2denote the true average weights for operations 1 and 2, respectively. Therelevant hypotheses are H µ − µ 0 vs. H µ − µ 0 . The value of thetest statistic is( 1402.24 −1419.63)22( 10.97) ( 9.96)0:1 2=a:1 2≠−17.39−17.39t ==== −6.43.4.011363+3.30672 7.318083+30 30The d.f.2( 7.318083)2( 4.011363) ( 3.30672)ν == 57.5 , so use df = 57. 2. 00029+292t ,. 025,57≈so we can reject H o at level .05. The data indicates that there is a significant differencebetween the true mean weights of the packages for the two operations.b. H0: µ1= 1400 will be tested against : µ >a 11400x −1400with test statistic ts1mH using a one-sample t test= . With degrees of freedom = 29, we reject H o if1402.24 −14002.24t > t . 05,29= 1.699 . The test statistic value is == = 1. 110.972.00t .Because 1.1 < 1.699, H o is not rejected. True average weight does not appear to exceed1400.30λ1λ − = + 2 and ˆλ 1 = Xm n, ˆλ = Y mX2, λˆ =mX − YZ = . With x = 1. 616 and = 2. 55792. Var( X Y )λ ˆ λˆ+m n( ( −5.3)) . 0006++nYn, givingy , z = -5.3 and p-value =2 Φ < , so we would certainly reject H : λ = λ0 1 2in favor ofH : λ ≠ λ .a12λˆˆ1λ293. λˆ1= x = 1. 62 , λˆ2= y = 2. 56 , + = 1. 77m n− . 94 ± 1.96 1.77 = −.94± .35 = −1.29,−.59( )( ) ( ), and the confidence interval is293