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Dr. Hackney STA Solutions pg 138

# Dr. Hackney STA Solutions pg 138 - Second Edition 8-17...

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Second Edition 8-17 because the cross term is zero. Performing a similar operation on the Y sum yields ˆ σ 2 0 ˆ σ 2 = ( x i - ¯ x ) 2 + ( y i - ¯ y ) 2 + nm n + m x - ¯ y ) 2 ˆ σ 2 = n + m + nm n + m x - ¯ y ) 2 ˆ σ 2 . Because ˆ σ 2 = n + m - 2 n + m S 2 p , large values of ˆ σ 2 0 ˆ σ 2 are equivalent to large values of (¯ x - ¯ y ) 2 S 2 p and large values of | T | . Hence, the LRT is the two-sample t -test. b. T = ¯ X - ¯ Y S 2 p (1 /n + 1 /m ) = ( ¯ X - ¯ Y ) σ 2 (1 /n + 1 /m ) [( n + m - 2) S 2 p 2 ] / ( n + m - 2) . Under H 0 , ( ¯ X - ¯ Y ) n(0 , σ 2 (1 /n +1 /m )). Under the model, ( n - 1) S 2 X 2 and ( m - 1) S 2 Y 2 are independent χ 2 random variables with ( n - 1) and ( m - 1) degrees of freedom. Thus, ( n + m - 2) S 2 p 2 = ( n - 1) S 2 X 2 + ( m - 1) S 2 Y 2 χ 2 n + m - 2 . Furthermore, ¯ X - ¯ Y is independent of S 2 X and S 2 Y , and, hence, S 2 p . So T t n + m - 2 . c. The two-sample t test is UMP unbiased, but the proof is rather involved. See Chapter 5 of Lehmann (1986). d. For these data we have n = 14, ¯ X = 1249 . 86, S 2 X = 591 . 36, m = 9, ¯ Y = 1261 . 33, S 2 Y = 176 . 00 and S 2 p = 433 . 13. Therefore, T = - 1 . 29 and comparing this to a
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