**STAT 330: 98-1
**

Assignment 6 Solutions

- 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.) - Chapter 9 Q 56: The quantity has an
*F*distribution so thatThe 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.

- Chapter 10 Q 10:
- . Use and the fact that summing over
*j*just multiplies by*J*to get . - .
The variance of any average of
*J*independent quantities each with variance is just so we get . - which, using (a) and the rule in (b) is .
- 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

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

- . Use and the fact that summing over
- 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.)
- 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. - Chapter 10 Q46:
- 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*. - 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 .

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

Fri Mar 6 10:16:01 PST 1998