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