HW Solutions Stat 101

HW Solutions Stat 101 - Since np = n(3 1)/(4 1) = 2n/3, 2...

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Since n p = n (3 1) / (4 1)=2 n/ 3, σ 2 { ˆ D j } =2 σ 2 (3) / (4 n p )=9 σ 2 / (4 n ) { ˆ D j } = 1 2 q [ . 95; 4 , 8 n/ 3 3] s 9 σ 2 4 n For σ 2 =2 . 0 and { ˆ D j }≤ 1 . 5, so we need to iterate to fnd n so that n 2 q 2 [ . 95; 4 , 8 n/ 3 3] We iteratively fnd n 28. Since design 2 in Table 28.1 has n = 3, we require that design 2 be repeated 10 times. Thus, n = 30, and n b = 40. 28.14. e ijk : ij =1 j =2 j =3 j =4 1 . 1375 . 0875 . 0125 . 0625 2 . 0125 . 0125 . 1625 . 1375 3 . 1375 . 0875 . 0625 . 0125 4 . 0125 . 0125 . 0875 . 0625 r = . 986 28.15. a. ¯ Y .. 1 =1 . 725, ¯ Y .. 2 =1 . 900, ¯ Y .. 3 =2 . 175, ¯ Y .. 4 =2 . 425 b. Source SS df MS Rows (sales volumes) 5 . 98187 3 1 . 99396 Columns (locations) . 12188 3 . 04062 Treatments (prices) 1 . 13688 3 . 37896 Error . 11875 6 . 01979 Total 7 . 35938 15 H 0 : all τ k equal zero ( k =1 , ..., 4), H a : not all τ k equal zero. F = . 37896 /. 01979 = 19 . 149, F ( . 95; 3 , 6 )=4 . 76. I± F 4.76 conclude H 0 , otherwise H a . Conclude H a . P -value = . 002 c. ˆ L 1 = ¯ Y .. 1 ¯ Y .. 2 = . 175, ˆ L 2 = ¯ Y .. 1 ¯ Y .. 3 = . 450, ˆ L 3 = ¯ Y .. 1 ¯ Y .. 4 = . 700, ˆ L 4 = ¯ Y .. 2 ¯ Y .. 3 = . 275, ˆ L 5 = ¯ Y .. 2 ¯ Y .. 4 = . 525 , ˆ L 6 = ¯ Y .. 3 ¯ Y .. 4 = . 250, s { ˆ L i } = . 09947 ( i =1 , ..., 6), q ( . 90; 4 , 6)=4 . 07, T
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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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