Dr. Hackney STA Solutions pg 156

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

Ask a homework question - tutors are online