STAT 330: 98-1

Assignment 6 Solutions

1. Chapter 9 Q 54: Let be the population standard deviation for control rats and be the treatment population standard deviation. Then to test against the one sided alternative we compute and get upper tail P-values from F tables with 19 and 22 degrees of freedom. We get F = 2.85 and P just a bit bigger than 0.01 so that we conclude that the treatment population is more variable. (While the question asks a one-tailed question it is not entirely clear to me that the tail was not chosen after seeing the data; if so you should double the P-value. The conclusion is not really changed.)
2. Chapter 9 Q 56: The quantity has an F distribution so that

The right hand inequality holds if and only if

Similarly, the first inequality can be rearranged to be

so that the range between these two limits is a level confidence interval for . For the data we have and . The critical point while to find the lower tail critical point we use . The interval is then 0.00549/(0.0258*9.28) to (9.28)(0.00549)/0.0258. Note that some people may have put on top.

3. Chapter 10 Q 10:

1. . Use and the fact that summing over j just multiplies by J to get .
2. . The variance of any average of J independent quantities each with variance is just so we get .
3. which, using (a) and the rule in (b) is .
4. We have . Expand out the square and use the fact that to see that

Take expected values and put in the results of (b) and (c) to get

5. Under the second of these terms is 0 so while under the alternative the second term is positive so that .

4. Chapter 10 Q 42: For we have , , and all the other so that the confidence interval for is (where we use the level for a 95% confidence interval). For the other intervals it is the values of the which change. They are (1,0,-1), (0,1,-1) and (0.5,0.5,-1) respectively. Only the contrast is judged significantly different from 0. (Note the use of not ; these are 95% confidence intervals.)
5. Chapter 10 Q 44: We have and . Subtracting we get so the the SSE for the y's is times the SSE for the x's. Similarly we have so that the new SSTr is times the old SSTr. The factors then cancel out in the formula for the F statistic so that the new F statistic is exactly equal to the old F statistic.
6. Chapter 10 Q46:

1. The two samples are now for and for . The two sample variances are and . Then the pooled estimate of is

which is just

which is in turn just the MSE.

2. The pooled statistic squared is just

Now write as to see that

Now examine the Treatment Sum of Squares. First note that . Use this to see that the Treatment Sum of Squares is given by

which simplifies to

This last is the numerator of .

Richard Lockhart
Fri Mar 6 10:16:01 PST 1998