Dr. Hackney STA Solutions pg 152

Dr. Hackney STA Solutions pg 152 - 9-10 Solutions Manual...

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Unformatted text preview: 9-10 Solutions Manual for Statistical Inference The log of this is t2 t2 = = t2 /2 as , - /2t - + -t 2 + 2 -(t 2/ ) + 2 1 - t/ 2 so the mgf converges to et /2 , the mgf of a standard normal. b. Since P (2 2 ) = for all , 2Y 2Y, 2 2Y, - 2 z as . 8 In standardizing (9.2.22), the upper bound is nb 2 nb+1 2(Y +a),/2 2 - 2 8 = 8( + a) 8 nb 2 nb+1 [2(Y +a),/2 - 2( + a)] 8( + a) + nb nb+1 2( + a) - 2 8( + a) . While the first quantity in square brackets z/2 , the second one has limit lim 1 nb -2 nb+1 + a nb+1 8( + a) -, so the coverage probability goes to zero. 9.33 a. Since 0 Ca (x) for every x, P ( 0 Ca (X)| = 0) = 1. If > 0, P ( Ca (X)) = P ( max{0, X + a}) = P ( X + a) = P (Z -a) = .95 (since > 0) (Z n(0, 1)) (a = 1.645.) A similar calculation holds for < 0. b. The credible probability is max(0,x+a) min(0,x-a) max(-x,a) 2 1 2 1 1 1 e- 2 t dt e- 2 (-x) d = 2 2 min(-x,-a) = P (min(-x, -a) Z max(-x, a)) . To evaluate this probability we have two cases: (i) (ii) |x| a |x| > a credible probability = P (|Z| a) credible probability = P (-a Z |x|) Thus we see that for a = 1.645, the credible probability is equal to .90 if |x| 1.645 and increases to .95 as |x| . 9.34 a. A 1 - confidence interval for is { : x - 1.96/ n x + 1.96/ n}. We need 2 2(1.96)/ n /4 or n 4(2)(1.96). Thus we need n 64(1.96) 245.9. So n = 246 suffices. b. The length of a 95% confidence interval is 2tn-1,.025 S/ n. Thus we need P S 2tn-1,.025 4 n .9 P S2 2 4t2 n-1,.025 n 16 .9 (n - 1)S 2 (n - 1)n 2 P .9. 2 tn-1,.025 64 2 n-1 ...
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