Dr. Hackney STA Solutions pg 180

Dr. Hackney STA Solutions pg 180 - 11-8 Solutions Manual...

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Unformatted text preview: 11-8 Solutions Manual for Statistical Inference Factor out the terms with yn and do some algebra on the middle term to get f (y1 , y2 , . . . , yn ) = yn i i -1 e-yn n-1 1 (1 ) yi-1 1 + yi-1 1 n-1 j=i (1 1 n-1 j=1 (1 1 -1 + yj ) 1 i -1 i=2 n-1 1 (i ) n-1 j=i (1 + yj ) i=1 (1 + yi ) + yj ) . We see that Yn is independent of the other Yi (and has a gamma distribution), but there does not seem to be any other obvious conclusion to draw from this density. b. The Yi are related to the F distribution in the ANOVA. For example, as long as the sum of the i are integers, Yi = Xi+1 i j=1 Xj = 2Xi+1 2 i j=1 Xj = 2 i+1 2 i j=1 F j i+1 , i j=1 j . Note that the F density makes sense even if the i are not integers. 11.21 a. Grand mean y Total sum of squares = = i=1 j=1 3 5 188.54 15 3 5 = 12.57 (yij - y ) 2 = 1295.01. Within SS = 1 5 1 (yij - yi ) 2 5 5 = 1 (y1j - 3.508) + 1 3 2 (y2j - 9.274) + 1 2 (y3j - 24.926) 2 = Between SS = = ANOVA table: 1.089 + 2.189 + 63.459 = 66.74 5 1 (yij - yi ) 2 5(82.120 + 10.864 + 152.671) = 245.65 5 = 1228.25. Source Treatment Within Total df 2 12 14 SS 1228.25 66.74 1294.99 MS 614.125 5.562 F 110.42 Note that the total SS here is different from above round off error is to blame. Also, F2,12 = 110.42 is highly significant. b. Completing the proof of (11.2.4), we have k ni k ni (yij - y ) i=1 j=1 2 = i=1 j=1 ((yij - yi ) + (i - y )) y 2 ...
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