chapter 1

Chapter 9: Inferences Based on Two Samplesb. For the hypotheses µ − µ = 25H0:1 2− vs. Ha: µ1− µ2< −25− 31.1−( − 25)changes to t == −.556120.252≈ P t < −. 6 = .( ) 278, the test statistic. With degrees of freedom 18, the p-value. Since the p-value is greater than any sensible choice of α , wefail to reject H o . There is insufficient evidence that the true average strength for malesexceeds that for females by more than 25N.76.a. The relevant hypotheses are H µ ∗ − µ∗ 0 (which is equivalent to saying0:1 2=: µ ∗ 1−∗aµ2≠µ − µ 0 ) versus 01 2=1− µ2≠H (which is the same as sayingµ 0 ). The pooled t test is based on d.f. = m + n – 2 = 8 + 9 – 2 = 15. The2pooled variance is s =⎛ 8 −1⎞⎜ ⎟⎝ 8 + 9 − 2 ⎠x * − y *t =1 1+⎛ m − 1 ⎞ 2 ⎛ n −1⎞ 2p ⎜ ⎟s1+ ⎜ ⎟s2⎝ m + n − 2 ⎠ ⎝ m + n − 2 ⎠⎛ 9 − 1 ⎞+ ⎜ ⎟ 4. = 22.49 , so sp= 4. 742⎝8+ 9 − 2 ⎠18.0 −11.0== 3.04 ≈ 3.1 14.742 +( 4.9) 2 ( 6) 2. The test statisticis 0 . From Table A.7, the p-values pmn89associated with t = 3.0 is 2P( t > 3.0 ) = 2(.004) = .008. At significance level .05, H o isrejected and we conclude that there is a difference between µ and µ , which isequivalent to saying that there is a difference between µ1and µ2.∗1∗2b. No. The mean of a lognormal distribution is∗µ∗ + ( σ ) 2 / 2µ = e , where∗µ and∗σ arethe parameters of the lognormal distribution (i.e., the mean and standard deviation of∗ln(x)). So when∗ ∗σ 1= σ 2, then µ ∗ = 1µ would imply that µ 21= µ2. However,∗when∗ ∗σ 1≠ σ 2, then even if µ ∗ = 1µ , the two means µ 21and µ2(given by theformula above) would not be equal.77. This is paired data, so the paired t test is employed. The relevant hypotheses areH : µ 0 vs. : < 0 µ denotes the difference between the population0 d=Haµd, wheredaverage control strength minus the population average heated strength. The observeddifferences (control – heated) are: -.06, .01, -.02, 0, and -.05. The sample mean and standarddeviation of the differences are = −. 024. The test statistic isd and sd= . 0305−.024= = −1.76≈ −1.8t . From Table A.7, with d.f. = 5 – 1 = 4, the lower tailed p-.03055value associated with t = -1.8 is P( t < -1.8) = P( t > 1.8 ) = .073. At significance level .05, H oshould not be rejected. Therefore, this data does not show that the heated average strengthexceeds the average strength for the control population.288