MORE REVIEW QUESTIONS FOR EXAM 3

Question 1 is the "full problem" that should have appeared in its entirety
in the Course Packet on p. 17-15. Question 3 is a replacement problem for
the one on p. 17-16, as you already did that question as part of HW 7!

1. Scientists conducted a survey of stream segments in western
Washington to assess the effect of three different levels of commercial
timber harvest (none, moderate, intensive) on instream salmon spawning
habitat. One of the responses measured on each stream segment (the
sampling unit) was the percent of stream area comprised of pools. (Pools
are related to needed habitat for salmon spawning.) Data and
the summary statistics are as follows:

NONE: 43, 73, 36, 46, 42. n = 5, xbar=48.00, s^2=208.50.
MODERATE: 39, 62, 20, 44. n = 4, xbar=41.25, s^2=298.25
INTENSIVE: 37, 14, 21, 30, 07. n = 5, xbar=21.80, s^2=144.7.

1a. At the .05 level of significance [and assuming normality and equal
variances], test to see whether the three levels of timber harvest yield
the same mean pool fraction of stream area. Include the complete analysis
of variance (ANOVA) table. You may do this either by hand (it is indeed
do-able by hand) or by using statistical software like SPSS. NOTE: in
writing up your "conclusions" statement, AVOID the words "accept",
"reject", and "hypothesis"; rather, express conclusions in terms of the
original research question.

1b. Are the three levels of timber harvest a "fixed effects" model or a
"random effects" model? Explain your answer.

1c. If we were to use a sample size of n=6 for each group (total sample
size = 18), how far apart would the largest and smallest population
means have to be in order to reject the null hypothesis with 90% power
and level of significance = .05?

1d. Now, at the alpha = 5 percent level of significance, do a parametric
(normality based) multiple comparison of means using *both* the Tukey and
the Student-Newman Keuls methods of multiple comparisons. Do the two
methods yield the same results?--comment upon your answer.

1e. Compute a 95% confidence interval for the INTENSIVE mean.

1f. Compute a 95% confidence interval for the mean that you get by
pooling the two means from the NONE and the MODERATE groups.

3. The following data are from an experiment to look at
weight gain [in gms] of lab mice under 3 different diets which have
low, medium, and high amounts of protein and other nutrients. It was
also felt that male mice might respond to the diets differently than
female mice, so sex of the animal was also noted.
FEMALES MALES

LOW 40.8 49.2
40.5 43.8
43.9 37.6

MED 51.5 50.9
50.0 57.9
51.9 59.9

HIGH 62.7 74.0
56.4 72.1
64.7 73.9

MEAN

The Sums of Squares for the different sources of variation for the
above data are as follows:

SOURCE OF VARIATION SS
Diet 1831.9
Sex 179.9
Diet x Sex Interaction 82.4
Error 161.0
---------------------- -----
TOTAL 2255.2

a. At the .10 level of significance, test for the presence of interaction
between Diet and Sex. Why did the result of the test turn out the way it
did? Use a PLOT of the means [plot can be done by hand] to explain why.

b. At the .05 level of significance, test for the overall effect of Sex on
the mean weight gain. Why did the result of the test turn out the way it
did?--refer to your plot from [a] to answer this.

c. At the .05 level of significane, test for the overall effect of the
three Diets on the mean weight gain. Why did the result of the test turn
out the way it did?--refer to your plot from [a] to answer this.

d. Compute a 95% confidence interval for the average weight
gain on the LOW diet (regardless of Sex).

e. Compute a 95% confidence interval for the *difference between the male
and female* mean weight gain for the HIGH diet. Again, that is, HIGH
DIET only, (male - female) difference regarding mean weight gain.