hw5sol - 9.3 a. l1= y1+ y2+ y3-3y4 a1=1, a2=1,a3=1, a4= -3...

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9.3 a. l 1 = y 1 + y 2 + y 3 -3 y 4 a 1 =1, a 2 =1,a 3 =1, a 4 = -3 l 2 = y 1 + y 2 - 2 y 3 + 0 y 4 a 1 =1, a 2 =1,a 3 = -2, a 4 = 0 = i 14a i = 1+1+1-3=0 = i 14b i = 1+1-2+0=0 Thus l 1 & l 2 are linear contrast. b. n 1 = n 2 = n 3 = n 4 = n = ( i 14 a i b i /n i )= 1n (1*1 +1*1+1*(-2)+(-3)*0) = 0 Thus l 1 & l 2 are orthogonal 9.4 a. l 1 = y 1 + y 2 + y 3 -3 y 4 a 1 =1, a 2 =1,a 3 =1, a 4 = -3 l 2 = y 1 + y 2 - 2 y 3 + 0 y 4 a 1 =1, a 2 =1,a 3 = -2, a 4 = 0 = i 14a i = 1+1+1-3=0 = i 14b i = 1+1-2+0=0 Thus l 1 & l 2 are linear contrast. b. n 1 = 5, n 2 =6 , n 3 =4, n 4 = 8 = ( i 14 a i b i /n i )= * 1 15 + * 1 16 + *(- ) 1 2 4 + - * 3 08 =- 215 Thus l 1 & l 2 are not orthogonal 9.27 a. assume y 1 is the control group. LSD ij = t 0.025 *( + ) s^2w 1ni 1nj , s 2 w=49.668 y 1 =1.8 (placebo), y 2 =2.5(100mg), y 3 =3.9(1000mg), y 4 =5.7(500mg), y 5 =5.7(2000mg) | y 5 - y 1 | = 3.9 >|LSD 44,45 |=2.94 | y 4 - y 1 | = 3.9 >|LSD 45,45 |=2.92 | y 3 - y 1 | = 1.1 <|LSD 43,45 |=2.96 ---stop: y 4 and y 5 is significantly different from
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hw5sol - 9.3 a. l1= y1+ y2+ y3-3y4 a1=1, a2=1,a3=1, a4= -3...

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