chapter 1

Chapter 9: Inferences Based on Two Samples54.a. The “after” success probability is p 1 + p 3 while the “before” probability is p 1 + p 2 , so p 1 +H : p = p0 versusp 3 > p 1 + p 2 becomes p 3 > p 2 ; thus we wish to test3 2H a: p > p .32X1 + X3− X1+ X2X3− X2=( ) ( ) b. The estimator of (p 1 + p 3 ) – (p 1 + p 2 ) isnnc. When H o is true, p 2 = p 3 , sopˆ 2+ pˆ3n. The Z statistic is then⎛ X3− X2 ⎞ p2+ p3Var⎜⎟ =⎝ n ⎠ nX3− X2n X3− X2=pˆ+ pˆX + Xd. The computed value of Z is = 2. 682200 −150n200 + 15032, which is estimated by3., so P = 1 − Φ( 2.68) = . 0037level .01, H o can be rejected but at level .001 H o would not be rejected... At15 + 729p , p ˆ 2= = . 6904042. 550 .690 ± 1.96 .106 = −.14±.21=−.35,.0755. ˆ1 = = . 550, and the 95% C.I. is( ) ( ) ( )− .56. Using p 1 = q 1 = p 2 = q 2 = .5, L ( 1.96 )⎛ .25 .25⎞2.7719= 2 ⎜ + ⎟ = , so L=.1 requires n=769.⎝ n n ⎠ nSection 9.557.a. From Table A.9, column 5, row 8, F 3. 69 .. 01,5,8=b. From column 8, row 5, F 4. 82 .1c.. 95,5,8= = . 207FF ..05,8,5. 01,8,5=280