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)<sigma^2(e)<SSW/(chisq,0.975,80)]
	[10.328/106.6286<sigma^2(e)<10.328/57.15317]

1f) The ratio of the variance among fields to the variance of plots 
	within fields is 7.31.

	sigma^2(a)/sigma^2(e) = 0.944/0.129

1g) The 95% confidence interval for the ratio of variances 
       is (3.745,16.597).  
       You can conclude from this interval that the variance of among 
       fields is larger than the variance of plots within fields since 
       the confidence interval includes only numbers greater than 1.
 
	F(0) = (92.102/19)/(10.328/80) = 37.548
	F(0.025,19,80) = 1.904
	F(0.975,19,80) = 0.447
	(1/5)*[(37.548/1.904)-1] = 3.745
	(1/5)*[(37.548/0.447)-1] = 16.597

1h) The intraclass correlation is 0.88.
	
	0.944/(0.944 + 0.129)

1i) F(0) = MSA/MSW = 37.548 
	df = 19,80
	p-value = 5.140921e-32 < 0.0001
	There is significant evidence of field-to-field variation in 
	susceptibility to insect damage.


2)                         Sum of
      Source    DF         Squares     Mean Square    F Value    Pr > 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