Dr. Hackney STA Solutions pg 177

Dr. Hackney STA Solutions pg 177 - i s only through = 2 i ....

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Second Edition 11-5 b. From Exercise 5.8(a) We know that s 2 = 1 k - 1 k X i =1 y i · - ¯ ¯ y ) 2 = 1 2 k ( k - 1) X i,i 0 y i · - ¯ y i 0 · ) 2 . Then 1 k ( k - 1) X i,i 0 t 2 ii 0 = 1 2 k ( k - 1) X i,i 0 y i · - ¯ y i 0 · ) 2 s 2 p /n = k X i =1 y i · - ¯ ¯ y ) 2 ( k - 1) s 2 p /n = i n y i · - ¯ ¯ y ) 2 / ( k - 1) s 2 p , which is distributed as F k - 1 ,N - k under H 0 : θ 1 = ··· = θ k . Note that X i,i 0 t 2 ii 0 = k X i =1 k X i 0 =1 t 2 ii 0 , therefore t 2 ii 0 and t 2 i 0 i are both included, which is why the divisor is k ( k - 1), not k ( k - 1) 2 = ( k 2 ) . Also, to use the result of Example 5.9(a), we treated each mean ¯ Y i · as an observation, with overall mean ¯ ¯ Y . This is true for equal sample sizes. 11.13 a. L ( θ | y ) = ± 1 2 πσ 2 ² Nk/ 2 e - 1 2 k i =1 n i j =1 ( y ij - θ i ) 2 2 . Note that k X i =1 n i X j =1 ( y ij - θ i ) 2 = k X i =1 n i X j =1 ( y ij - ¯ y i · ) 2 + k X i =1 n i y i · - θ i ) 2 = SSW + k X i =1 n i y i · - θ i ) 2 , and the LRT statistic is λ = (ˆ τ 2 / ˆ τ 2 0 ) Nk/ 2 where ˆ τ 2 = SSW and ˆ τ 2 0 = SSW + X i n i y i · - ¯ y ·· ) 2 = SSW + SSB. Thus λ < k if and only if SSB/SSW is large, which is equivalent to the F test. b. The error probabilities of the test are a function of the
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Unformatted text preview: i s only through = 2 i . The distribution of F is that of a ratio of chi squared random variables, with the numerator being noncentral (dependent on ). Thus the Type II error is given by P ( F &gt; k | ) = P 2 k-1 ( ) / ( k-1) 2 N-k / ( N-k ) &gt; k P 2 k-1 (0) / ( k-1) 2 N-k / ( N-k ) &gt; k = , where the inequality follows from the fact that the noncentral chi squared is stochastically increasing in the noncentrality parameter....
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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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