Solutions to Homework 2, Stat 402, Spring 2004

1a.	The experimental units are the litters.

b.	The observational units are animals.

c.	The number of replications per treatment is 2.

d.	An alternative approach would be to do a RCBD by assigning one animal per 
litter at random to treatment A and then another randomly selected animal per 
litter to treatment B.

Below is a possible method of assigning treatments:

Litter#    ID#    R.N.    Treatment
  1         1      8          B
  1         2      7          -
  1         3      0          A

  2         4      2          A
  2         5      3          B

  3         6      6          -
  3         7      5          -
  3         8      7          B
  3         9      5,0        A

  4        10      4          -
  4        11      1          A
  4        12      4,9        B


The assignments of treatments were based on the rule that the smallest random 
number in the litter group received treatment A and the largest random number 
received treatment B.  If a random number was duplicated within a litter group, 
a second digit was drawn to replace the duplicate digit.

2a.	c = 7.

b.	c = 8.

c.	c is the mean.

d.	c is the median.

3a.	Calculate the overall mean = [(72*8)+(68*8)+(55*7)]/23 = 65.43.  

Calculate SSB = (8*(72-65.43)^2)+(8*(68-65.43)^2)+(7*(55-65.43)^2) = 1159.65.

Calculate SSE = ((8-1)*312)+((8-1)*310)+((7-1)*321) = 6280.


	The rest of the ANOVA calculations are now straight forward:

	Source      df   SS         MS        F      p-value
	Treatment   2    1159.65    579.83    1.85   > 0.1
	Error       20   6280       314
	Total       22   7439.65

b.	There is no evidence of differences among the phosphorous content means of 
the three varieties.  (F = 1.85, w/2 and 20 df, p-value > 0.1)

c.	No, this is not a violation of the assumption of equal variances.  The 
sample variances are estimates of the true variances.  The observed difference 
could be easily explained by chance.  As a rough rule, we generally don't become 
too concerned about non-constant variance unless one variance is 4 times larger 
than another.

d.	t(0.975,20) = 2.086    SE = (314/7)^0.5 = 6.7
	CI: 55+/-2.086*6.7   (41 to 69)
	The 95% confidence interval for the mean phosphorous content of leaves 
from variety C trees is 41 to 69.

4.	143 replicates would be needed in each treatment group to have a test with 
a power of 80%.  Below is the input necessary to do the calculations.

data one;
  do r=2 to 150;
    alpha=0.05;
      t=2;
      variance=9;
      sumtausq=0.5;


	The sumtausq is the only one not given directly in the problem.  Recall 
that tau represents a treatment effect.  A treatment effect is simply a 
treatment mean minus the average of all treatment means.  It is fairly easy to 
see, that in the case of two treatments, the treatment effects must be + and - 
1/2 the distance between the two treatments.  More formally, suppose the mean 
for the control group is X.  We want to be able to detect a difference of 1 
unit, so the mean of the new treatment is either X + 1 or X - 1.  Thus, the 
overall mean of these two groups would be either (X+X+1)/2=X+1/2 or (X+X-1)/2=X-
1/2.  Either way, the two treatment effects are 1/2 and -1/2.  Squaring and 
summing these results in 0.25 + 0.25 = 0.5 which is the sumtausq.