chapter 1

Chapter 9: Inferences Based on Two Samples81. We wish to test H 0 :2Unpooled:With H o : − µ 0ν=1 2=22 2.79 1.52( + )14 122.79221.52214 12µ1= µ versus H a : µ1≠ µ2µ vs. H a : µ − µ 0 , we will reject H o if − value < α( ) ( )1 2≠= 15.95 ≈16, and the test statistic+13 118.48 −9.36−.96t == = −1.97leads to a p-value of 2[ P(t > 1.97)]2 2.79 1.52+ .486914 12≈ 2 .031 ≈ .Pooled:( ) 062p .The degrees of freedom ν = m = n − 2 = 14 + 12 − 2 = 24 and the pooled variance⎛ 13 ⎞ 2 ⎛ 11 ⎞ 2⎜ ⎟ ⎜ ⎟⎝ 24 ⎠ ⎝ 24 ⎠− .96 − .96== ≈ −2.11 11.181 + .465is (.79) + ( 1.52) = 1. 39701412, so s =1. 181 . The test statistic ist . The p-value = 2[ P( t 24 > 2.1 )] = 2( .023) = .046.With the pooled method, there are more degrees of freedom, and the p-value is smaller thanwith the unpooled method.p82. Because of the nature of the data, we will use a paired t test. We obtain the differences bysubtracting intake value from expenditure value. We are testing the hypotheses H 0 : µ d = 0 vs1.757H a : µ d ? 0. Test statistic t = = 3. 88 with df = n – 1 = 6 leads to a p-value of 2[ P( t >1.19773.88 ) ˜ .004. Using either significance level .05 or .01, we would reject the null hypothesisand conclude that there is a difference between average intake and expenditure. However, atsignificance level .001, we would not reject.83.a. With n denoting the second sample size, the first is m = 3n. We then wish900 40020 2( 2.58)+3nn= , which yields n = 47, m = 141.b. We wish to find the n which minimizes ( z )n which minimizes900 400+400 − n n900 400 −900400 − n2α / 2+400n, or equivalently, the. Taking the derivative with respect to n and−2−2equating to 0 yields ( n ) − 400n= 0 , whence 9n2 4( 400 − n) 25 2 + 3200n− 640,000 = 0n . This yields n = 160, m = 400 – n = 240.= , or290