chapter 1

50. Let = . 05Chapter 9: Inferences Based on Two Samplespˆ1qˆ1 pˆ2qˆ2α . A 95% confidence interval is ( pˆ p ) ± z ( + )126( − )224 171 126 140( )( ) ( )( )1− ˆ2 α / 2⎛395 395 266 266⎞224395 266± 1.96 ⎜ + ⎟ = .0934 ± .0774395 266=⎝⎠mn(.0160,.1708)= .51.H : p = p0 will be rejected in favor of H a: p1≠ p2if either z ≥1. 645 or.011z ≤ −1.645 . With p ˆ1= . 193 , and p ˆ 2= . 182 , p ˆ = . 188 , = = 1. 48.00742Since 1.48 is not ≥ 1. 645a.1 2z ., H o is not rejected and we conclude that no difference exists.b. Using formula (9.7) with p 1 = .2, p 2 = .18, α = . 1 , β = . 1 , and 1. 645n =2( 1.645 .5(.38)( 1.62)+ 1.28 .16 + .1476) = 6582.0004z ,α / 2=52. Let p 1 = true proportion of irradiated bulbs that are marketable; p 2 = true proportion ofuntreated bulbs that are marketable; The hypotheses are H p − p 0 vs.H0: p1− p2>0. The test statistic is119272p ˆ 2= = .661, ˆ = = . 756180360The p-value = 1−Φ( 4.2) ≈ 0beneficial.1 1( + )mn0:1 2=pˆ1− pˆ2153z = . With p . 850pq ˆ ˆˆ1 =180=.850 − .661p , z == = 4.2 .1 1(.756)(.244)( + )180180.189.045, and, so reject H o at any reasonable level. Radiation appears to be53.a. A 95% large sample confidence interval formula for ln ( θ ) isln( θ ˆ)± zα/ 2m − x n − y+mx nygives the confidence interval for θ itself.. Taking the antilogs of the upper and lower bounds18911,034b. ˆ = = 1. 818θ , ln ( θ ) = . 59810411,03710,845ˆ10,933, and the standard deviation is+ = .1213 , so the CI for ( θ )( 11,034)( 189) ( 11,037)( 104). 598 1.96 (.1213) = (.360,.836)the CI for θ to be ( 1 .43,2.31).279ln is± . Then taking the antilogs of the two bounds gives