midterm2solu - Midterm 2 Solution Problem 1. (a):...

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Unformatted text preview: Midterm 2 Solution Problem 1. (a): Advantage: eliminate between-block variation. Disadvantage: loss of degree of freedom. (b): ¯ y ·· = 17 . 2+20 . 8+14 . 6 3 = 17 . 533 SS Block = a ∑ (¯ y · j- ¯ y .. ) 2 = 3(11 . 6667- 17 . 5333) 2 +3(18 . 6667- 17 . 5333) 2 + 3(28- 17 . 5333) 2 + 3(20 . 6667- 17 . 5333) 2 + 3(8 . 6667- 17 . 5333) 2 = 701 . 07 SS Trt = b ∑ (¯ y i ·- ¯ y ·· ) 2 = 5(17 . 2- 17 . 5333) 2 +5(20 . 8- 17 . 5333) 2 +5(14 . 6- 17 . 5333) 2 = 96 . 93 SS E = SST- SS Block- SS Trt = 41 . 73 H : τ 1 = τ 2 = τ 3 = 0 vs H 1 : at least one is not Use F-test to test H : F = MS Trt MS E = SS Trt / ( a- 1) SS E / ( a- 1)( b- 1) = 96 . 93 / 2 41 . 73 / 8 = 9 . 29. F . 05 (2 , 8) = 4 . 46, which is smaller than F . Therefore H is rejected. (c): CD = q . 05 ( a, ( a- 1)( b- 1)) √ (2) q MS E ( 1 b + 1 b ) = 4 . 04 · q 41 . 73 8 1 5 = 4 . 13 ¯ y 2 ·- ¯ y 1 · = 3 . 6 < CD ¯ y 1 ·- ¯ y 3 · = 2 . 6 < CD ¯ y 2 ·- ¯ y 3 · = 6 . 2 > CD Therefore catalyst A is not different from catalyst B, and catalyst A is not...
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This note was uploaded on 12/09/2011 for the course STAT 514 taught by Professor Staff during the Fall '08 term at Purdue University.

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midterm2solu - Midterm 2 Solution Problem 1. (a):...

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