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sta108_handout6

# sta108_handout6 - Handout 6 F-test for lack of t(requires...

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Handout 6 F-test for lack of fit (requires replication): Growth data (reexpressed): j = 1 j = 2 j = 3 j = 4 j = 5 X 1 = 5 X 2 = 10 X 3 = 15 X 4 = 20 X 5 = 25 i = 1 Y 11 = 68 . 5 Y 12 = 79 . 4 Y 13 = 85 . 3 Y 14 = 88 . 0 Y 15 = 90 . 4 i = 2 Y 21 = 70 . 0 Y 22 = 77 . 6 Y 24 = 88 . 7 Y 25 = 89 . 6 i = 3 Y 34 = 86 . 3 Mean ¯ Y j ¯ Y 1 = 69 . 25 ¯ Y 2 = 78 . 5 ¯ Y 3 = 85 . 3 ¯ Y 4 = 87 . 67 ¯ Y 5 = 90 . 3 n 1 = 2 n 2 = 2 n 3 = 1 n 4 = 3 n 5 = 2 Model: Y ij = μ j + ε ij , i = 1 , ..., n j , j = 1 , ..., c, where { ε ij } are i.i.d. N (0 , σ 2 ). So E ( Y ij ) = μ j and V ar ( Y ij ) = σ 2 . The total number of observations is n = n 1 + · · · + n c . Here, n = 10 and c = 5. We want to test, at a level of significance α = 0 . 05 , H 0 : μ j = β 0 + β 1 X j for all j , against H 1 : μ j ’s do not lie on a straight line. Full model: Y ij = μ j + ε ij . Reduced model: Y ij = β 0 + β 1 X j + ε ij . For the full model, the estimate of μ j is ˆ μ j = ( Y 1 j + Y 2 j + · · · + Y n j j ) /n j = ¯ Y j . So we have SSE full = X 1 j c X 1 i n j ( Y ij - ¯ Y j ) 2 = 6 . 11 [also called SSPE ] , df ( SSPE ) = df ( SSE full ) = n - c = 5 . For the reduced model, the fit is : ˆ Y = 66 . 57 + 1 . 02 X . We then have SSE red = X 1 j c X 1 i n j ( Y ij - ( b 0 + b

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sta108_handout6 - Handout 6 F-test for lack of t(requires...

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