HW Solutions Stat 80

HW Solutions Stat 80 - A 1 1 1 2 2 2 3 3 3 c B 1 2 3 1 2 3...

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AB Freq. X 1 X 2 X 3 X 4 X 1 X 3 X 1 X 4 X 2 X 3 X 2 X 4 1 1 3 1 1 0 1 01000 1 2 4 1 1 0 0 10100 13 4 1 1 0 1 1 1 100 2 1 4 1 0 1 1 00010 2 2 2 1 0 1 0 10001 23 4 1 0 1 1 1 1 31 4 1 1 110 10 32 4 1 1 101 0 1 33 4 1 1 11 11111 c. X β entries: µ .. + α 1 + β 1 +( αβ ) 11 12 µ .. + α 1 + β 2 αβ ) 12 µ .. + α 1 β 1 β 2 ( αβ ) 11 ( αβ ) 12 = µ .. + α 1 + β 3 αβ ) 13 21 µ .. + α 2 + β 1 αβ ) 21 22 µ .. + α 2 + β 2 αβ ) 22 µ .. + α 2 β 1 β 2 ( αβ ) 21 ( αβ ) 22 = µ .. + α 2 + β 3 αβ ) 23 µ .. α 1 α 2 + β 1 ( αβ ) 11 ( αβ ) 21 = µ .. + α 3 + β 1 αβ ) 31 µ .. α 1 α 2 + β 2 ( αβ ) 12 ( αβ ) 22 = µ .. + α 3 + β 2 αβ ) 32 µ .. α 1 α 2 β 1 β 2 αβ ) 11 αβ ) 12 αβ ) 21 αβ ) 22 = µ .. + α 3 + β 3 αβ ) 33 d. Y ijk = µ .. + α 1 X ijk 1 + α 2 X ijk 2 + β 1 X ijk 3 + β 2 X ijk 4 + ± ijk e. Full model : ˆ Y =7 . 18704 3 . 28426 X 1 + . 63796 X 2 2 . 53426 X 3 + . 73796 X 4 +1 . 16481 X 1 X 3 . 04074 X 1 X 4 + . 15926 X 2 X 3 + . 33704 X 2 X 4 , SSE ( F )=1 . 5767 Reduced model : ˆ Y . 12711 3 . 33483 X 1 + . 62861 X 2 2 . 58483 X 3 + . 72861 X 4 , ( R )=29 . 6474 H 0 : all ( αβ ) ij equal zero, H a : not all ( αβ ) ij equal zero. F = (28 . 0707 / 4) ÷ (1 . 5767 / 24) = 106 . 82, F ( . 95; 4 , 24)=2 . 78. If F 2 . 78 conclude H 0 , otherwise H a . Conclude H a . P -value = 0+ f. ¯ Y 11 . =2 . 5333, ¯ Y 12 . =4 . 6000, ¯
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This note was uploaded on 12/21/2011 for the course STA 2014 taught by Professor Davehatley during the Fall '11 term at UNF.

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