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Dr. Hackney STA Solutions pg 175

# Dr. Hackney STA Solutions pg 175 - Second Edition 11-3 11.7...

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Second Edition 11-3 11.7 Let U i = ¯ Y i · - θ i . Then k X i =1 n i [( ¯ Y i · - ¯ ¯ Y ) - ( θ i - ¯ θ )] 2 = k X i =1 n i ( U i - ¯ U ) 2 . The U i are clearly n(0 2 /n i ). For K = 2 we have S 2 2 = n 1 ( U 1 - ¯ U ) 2 + n 2 ( U 2 - ¯ U ) 2 = n 1 ± U 1 - n 1 ¯ U 1 + n 2 ¯ U 2 n 1 + n 2 ² 2 + n 2 ± U 2 - n 1 ¯ U 1 + n 2 ¯ U 2 n 1 + n 2 ² 2 = ( U 1 - U 2 ) 2 " n 1 ± n 2 n 1 + n 2 ² 2 + n 2 ± n 1 n 1 + n 2 ² 2 # = ( U 1 - U 2 ) 2 1 n 1 + 1 n 2 . Since U 1 - U 2 n(0 2 (1 /n 1 + 1 /n 2 )), S 2 2 2 χ 2 1 . Let ¯ U k be the weighted mean of k U i s, and note that ¯ U k +1 = ¯ U k + n k +1 N k +1 ( U k +1 - ¯ U k ) , where N k = k j =1 n j . Then S 2 k +1 = k +1 X i =1 n i ( U i - ¯ U k +1 ) 2 = k +1 X i =1 n i ³ ( U i - ¯ U k ) - n k +1 N k +1 ( U k +1 - ¯ U k ) ´ 2 = S 2 k + n k +1 N k N k +1 ( U k +1 - ¯ U k ) 2 , where we have expanded the square, noted that the cross-term (summed up to k ) is zero, and did a boat-load of algebra. Now since U k +1 - ¯ U k n(0 2 (1 /n k +1 + 1 /N k )) = n(0 2 ( N k +1 /n k +1 N k )) , independent of S 2 k , the rest of the argument is the same as in the proof of Theorem 5.3.1(c). 11.8 Under the oneway ANOVA assumptions, Y
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