Hw 8 solutions 1a) The degrees of freedom for the among-fields sum of squares is 20 – 1 = 19. 1b) The degrees of freedom for plots within fields is (5-1)*20 = 80. 1c) The estimated variance component corresponding among-field variation is 0.944. sigma^2(a) = (MSA – MSW)/r sigma^2(a) = [(92.102/19) – (10.328/80)]/5 = 0.944 1d) The estimated variance component corresponding to variation of plots within field is 0.129. sigma^2(e) = MSW = 10.328/80 = 0.129. 1e) The 95% confidence interval for the variance of plots within fields is (0.097, 0.181). [SSW/(chisq,0.025,80) F Model 11 14.02125000 1.27465909 6.81 0.0012 Error 12 2.24500000 0.18708333 Corrected Total 23 16.26625000 Source DF Type I SS Mean Square F Value Pr > F temp 2 7.83250000 3.91625000 20.93 0.0001 cooler(temp) 9 6.18875000 0.68763889 3.68 0.0196 Source DF Type III SS Mean Square F Value Pr > F temp 2 7.83250000 3.91625000 20.93 0.0001 cooler(temp) 9 6.18875000 0.68763889 3.68 0.0196 Source Type III Expected Mean Square temp Var(Error) + 2 Var(cooler(temp)) + Q(temp) cooler(temp) Var(Error) + 2 Var(cooler(temp)) 2a) Source DF Model 11 (3*4)-1 or 2+9 Error 12 (3*4)*(2-1) or 23-11 Corrected Total 23 (3*4*2)-1 temp 2 3-1 cooler(temp) 9 3*(4-1) 2b) The number is n=2. Recall that n is the number of observations per experimental unit. 2c) F = MStemp/MScooler(temp) = 3.916/0.688 = 5.695 df = 2, 9 p-value = 0.0252 There is significant evidence of a difference between temperature means. 2d) The standard error of the treatment mean is 0.293. sqrt(MScooler(temp)/2*4) = sqrt(0.688/8) 2e) The 95% confidence interval for the mean of the 34 degree treatment is (5.389,6.713). 6.05 +/- t(0.975,9)*SE(trt mean) 6.05 +/- 2.262157*0.293 2f) The 95% confidence interval for the difference between the 40 degree treatment mean and the 34 degree treatment mean is (0.087,1.963). estimate: 7.075 – 6.05 = 1.025 SE: sqrt[(MScooler(temp)/2*4)*((1^2)+(-1^2))] = sqrt[(0.688/8)*2] = 0.415 CI: 1.025 +/- t(0.975,9)*SE(trt diff) 1.025 +/- 2.262157*0.415 2g)t = 3.23 df = 9 p-value = 0.0104 There is significant evidence that the mean for 40 degrees was greater than the mean for 46 degrees. Difference: 7.075 - 5.7375 = 1.338 SE: same as 2f = 0.415 t = diff/se = 1.338/0.415 2h) The variance component that corresponds to variation of cuts of beef within coolers is 0.187. This is just the mean square for cuts of beef within coolers. SAS calls this MSE. 2i) The variance component that corresponds to variation of coolers within temperature treatment groups is 0.25. (MScooler(temp) – MSE)/2 = (0.688-0.187)/2