Dr. Hackney STA Solutions pg 156

# Dr. Hackney STA Solutions pg 156 - 9-14 Solutions Manual...

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Unformatted text preview: 9-14 Solutions Manual for Statistical Inference 9.52 a. The LRT of H0 : = 0 versus H1 : = 0 is based on the statistic (x) = sup,=0 L (, 0 | x) . sup,(0,) L(, 2 | x) In the denominator, 2 = (xi - x)2 /n and = x are the MLEs, while in the numerator, ^ ^ 2 0 and are the MLEs. Thus ^ 2 20 -n/2 - (x) = e (xi -)2 x 2 2 0 (xi -)2 x 2 2 = -n/2 - (2^ 2 ) e 2 0 2 ^ -n/2 e - (xi -)2 x 2 2 0 e-n/2 , and, writing 2 = [(n - 1)/n]s2 , the LRT rejects H0 if ^ 2 0 n-1 2 n s -n/2 e - (n-1)s2 2 2 0 < k , (n-1)s2 , 2 0 where k is chosen to give a size test. If we denote t = then T 2 under H0 , n-1 and the test can be written: reject H0 if tn/2 e-t/2 < k . Thus, a 1 - confidence set is 2 : tn/2 e-t/2 k = 2 : (n - 1)s2 2 n/2 e- (n-1)s2 2 /2 k . Note that the function tn/2 e-t/2 is unimodal (it is the kernel of a gamma density) so it follows that the confidence set is of the form 2 : tn/2 e-t/2 k = = 2 : a t b (n - 1)s2 b 2 (n - 1)s2 (n - 1)s2 2 : 2 , b b = 2 : a where a and b satisfy an/2 e-a/2 = bn/2 e-b/2 (since they are points on the curve tn/2 e-t/2 ). Since n = n+2 - 1, a and b also satisfy 2 2 n+2 2 1 a((n+2)/2)-1 e-a/2 = 2(n+2)/2 n+2 2 1 b((n+2)/2)-1 e-b/2 , 2(n+2)/2 or, fn+2 (a) = fn+2 (b). b. The constants a and b must satisfy fn-1 (b)b2 = fn-1 (a)a2 . But since b((n-1)/2)-1 b2 = b((n+3)/2)-1 , after adjusting constants, this is equivalent to fn+3 (b) = fn+3 (a). Thus, the values of a and b that give the minimum length interval must satisfy this along with the probability constraint. The confidence interval, say I(s2 ) will be unbiased if (Definition 9.3.7) c. P2 2 I(S ) P2 2 I(S ) = 1 - . Some algebra will establish P2 2 I(S ) 2 2 2 = P2 = P2 2 (n - 1)S (n - 1)S 2 2 b a 2 2 2 2 n-1 2 n-1 b a = 2 2 bc fn-1 (t) dt, ac ...
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## This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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