Dr. Hackney STA Solutions pg 167

# Dr. Hackney STA Solutions pg 167 - 10-10 Solutions Manual...

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10-10 Solutions Manual for Statistical Inference c. Under H 0 , ˆ p 1 - ˆ p 2 1 n 1 + 1 n 2 p (1 - p ) n(0 , 1) and both ˆ p 1 and ˆ p 2 are consistent, so ˆ p 1 (1 - ˆ p 1 ) p (1 - p ) and ˆ p 2 (1 - ˆ p 2 ) p (1 - p ) in probability. Therefore, by Slutsky’s Theorem, ˆ p 1 - ˆ p 2 ˆ p 1 (1 - ˆ p 1 ) n 1 + ˆ p 2 (1 - ˆ p 2 ) n 2 n(0 , 1) , and ( T ** ) 2 χ 2 1 . It is easy to see that T ** = T in general. d. The estimator (1 /n 1 + 1 /n 2 p (1 - ˆ p ) is the MLE of Var(ˆ p 1 - ˆ p 2 ) under H 0 , while the estimator ˆ p 1 (1 - ˆ p 1 ) /n 1 + ˆ p 2 (1 - ˆ p 2 ) /n 1 is the MLE of Var(ˆ p 1 - ˆ p 2 ) under H 1 . One might argue that in hypothesis testing, the first one should be used, since under H 0 , it provides a better estimator of variance. If interest is in finding the confidence interval, however, we are making inference under both H 0 and H 1 , and the second one is preferred. e. We have ˆ p 1 = 34 / 40, ˆ p 2 = 19 / 35, ˆ p = (34 + 19) / (40 + 35) = 53 / 75, and T = 8 . 495. Since χ 2 1 ,. 05 = 3 . 84, we can reject H 0 at α = . 05.
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