chapter 1

Chapter 9: Inferences Based on Two Samples2 259. We test H : σ = σ vs.( 2.75)( 4.44)0 1 22Ha2 2: σ ≠ σ12. The calculated test statistic isf = = .384 . With numerator d.f. = m – 1 = 10 – 1 = 9, and denominator d.f. = n –21 = 5 – 1 = 4, we reject H 0 if f ≥ F .= 6. 00 orf ≤ F1F13.6305,9,4. 95,9,4= = =.05,4,9.275. Since .384 is in neither rejection region, we donot reject H 0 and conclude that there is no significant difference between the two standarddeviations.60. With =1σ true standard deviation for not-fused specimens and2=: σ = σ0 1Ha: σ1> σσ true standarddeviation for fused specimens, we test H 2vs. 2. The calculatedtest statistic isf=( 277.3)( 205.9)22= 1.814. With numerator d.f. = m – 1 = 10 – 1 = 9, anddenominator d.f. = n – 1 = 8 – 1 = 7, f = 1.814< 2.72 = F.10,9,7. We can say that the p-value > .10, which is obviously > .01, so we cannot reject H o . There is not sufficientevidence that the standard deviation of the strength distribution for fused specimens is smallerthan that of not-fused specimens.261. Let =2σ variance in weight gain for low-dose treatment, and =1σ variance in weight2 2gain for control condition. We wish to test H σ = vs. 2 2H : σ > σ . The teststatistic isf=( 54)( 32)2122: σ0 1 2sf = , and we reject H o at level .05 if > F . 05,19,22≈ 2. 08s22= 2.85 ≥ 20.8more variability in the low-dose weight gains.f ., so reject H o at level .05. The data does suggest that there is2a1262. H : σ = σ0 1 2will be rejected in favor of a: σ1≠ σ2f ≤ F . 975,47,44≈ .or if f ≥ F .≈ 1. 8 . Because f = 1. 22 , H o is not rejected. The data does not025,47,44suggest a difference in the two variances.H if either 56282