Review material: If you're rusty in 1-way ANOVA, this is covered in many texts. You should be familiar with both the F test and t-tests for linear contrasts (also called estimates).
The assigned reading focuses on SAS. More conceptual treatments of same material (optional, all on reserve):
Cochran and Cox, section 2.2, Sleuth, Chapter 5; Kuehl, Chapter 5.
review of contrasts: Sleuth Chapter 6; Cochran and Cox, sections 3.4 and esp. 3.5

I expect you'll use the computer for all of these. There are too many to do by hand.

1) Last week, we used hand-calculation to find the sample size for the herbal extract study. In that study, we needed to find n that an alpha=5% test had 90% power to detect a difference of 6.5 when the s.d. is 4.1.
a) use SAS PROC POWER or other computer program to compute the sample size. If you use some other program, please tell us what you used. Last week, you determined n using a shifted-t approximation and hand calculation. You should have gotten n=10. Do you get the same sample size using the computer?

The subsequent parts will change one piece of the problem at a time and examine the consequences for sample size. For each of the following, use the same values as in part a, except for the quantity that is changed
b) use SAS PROC POWER or other computer program to compute the sample size when the required power is 95%. (remember, keep s.d.=4.1, diff=6.5, alpha=5%). What is the consequence of requiring a smaller type II error rate?
c) use SAS PROC POWER or other computer program to compute the sample size when alpha is 1%. (remember to return back to 90% power). What is the consequence of requiring a smaller type I error rate?
d) use SAS PROC POWER or other computer program to compute the sample size when the 'important difference' is 3 (smaller than the 6.5 used in parts a and b). What is the consequence of focusing on a smaller difference?
e) use SAS PROC POWER or other computer program to compute the sample size when s.d. is 7 (larger than the 4.1 used in part a). What is the consequence of a larger variation.

2) Much of the variability in this study is due to variation in cell characteristics. The impact of this variability can be reduced by pairing: a sample of cells is divided into two tubes. One tube is randomly chosen to receive the extract; the other tube in the pair receives the control treatment. The sample size, n, is now the number of cell samples needed = the number of pairs. When the experiment uses a paired design, the s.d. of the difference between replicates is 2.2.

Remember that a paired t-test is equivalent to computing the difference for each sample of cells, computing the s.d. of the differences, and the s.e. of the mean difference from the s.d. of the differences. The mean difference and the s.e. of the mean difference are used in a 'one-sample' test to test H0: mean difference equals zero.

What sample size is needed so that a paired t-test with alpha = 5% has 90% power to detect a true difference of 6.5 when the s.d. (of the differences) = 2.2?

3) Consider a 1-way ANOVA comparing 5 treatments: a control and 4 different seed treatments (A, B, C, and D) to enhance emergence of soybean seeds. Each treatment will be replicated n times in a completely randomized design. The data will be analyzed by one-way ANOVA followed by contrasts. The response is the percent of seeds that emerge. The usual ANOVA assumptions (normality, equal variances, and independence) are reasonable here.

Prior studies suggest that the s.d. = 2.3%. A difference of 1.5% germination is considered important.

a) If the investigators say that the most important contrast is the difference between the control and the average of the four seed treatments, what sample size is needed to get 80% power? Again, assume alpha = 5%.
b) If the investigators tell you that the most important contrast is the difference between treatments C and D, what sample size is needed to get 80% power? Assume alpha=5%.
c) Assume the investigators do not specify any a-priori contrasts. Instead, they are interested in all pairwise differences. Since there are 10 possible differences, you need to adjust for multiple comparisons. You decide to use a Bonferroni correction, i.e. you test each difference using alpha = 0.05 / 10 = 0.005. What sample size is needed to get 80% power for the test of the difference between treatments C and D when alpha = 0.005?