1. Scientists conducted a survey of stream segments in western
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.
MSTr = 915.65, df=2; MSE=209.78, df=11
F = 915.62/209.78 = 4.365
F0.05(1),2,11 = 3.98; Since 4.365>3.98 we reject the F-test (p=0.0402). Salmon spawning habitat (represented by % pools) is significantly different among levels of timber harvest.
1b. Are the three levels of timber harvest a "fixed effects" model or a
"random effects" model? Explain your answer.
The levels of timber harvest are
fixed effects; if we were to repeat the study we would use the same three
harvest levels.
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?
For n = 6, df would be 18-3=15.
Φ~2.3
δ = sqrt(2*3*2.32*209.78/6)
= 33.21
The largest and smallest means
would need to be 33.21 percent pools apart to reject with 90% power and 0.05 significance.
1d. Now, at the alpha = 5 percent level of significance, do a parametric
(normality based) multiple comparison of means using the Student-Newman
Keuls method of multiple comparisons.
For denominator: Sqrt(MSE/2*(1/5+1/4)) = 6.87
Sqrt(MSE/2*(1/5+1/5))
= 6.477
SNK:
|
|
xbari |
xbari-xbarj: None |
xbari-xbarj: Moderate |
|
Intensive |
21.8 |
26.2 |
19.45 |
|
Moderate |
41.25 |
6.75 |
|
|
None |
48.0 |
|
|
|
|
xbari |
q: None |
q: Moderate |
|
Intensive |
21.8 |
4.04 |
2.83 |
|
Moderate |
41.25 |
0.9825 |
|
|
None |
48.0 |
|
|
q0.05,3
= 3.82, q0.05,2 = 3.113
4.04>3.82, reject. 0.9825, 2.83 both < 3.113, fail to reject
Intensive Moderate None
_______________________
Intensive and None are different,
but neither can be distinguished from Moderate.
There is some kind of statistical error here.
1e. Compute a 95% confidence interval for the INTENSIVE
mean.
xbarintensive = 21.8, sxbar = sqrt(208.78/5)=6.45, df=11,t=2.201
21.8-2.201*6.45< muintensive<21.8+2.201*6.45
7.60< muintensive<36.0
1f. Compute a 95% confidence interval for the mean that
you get by
pooling the two means from the NONE and the MODERATE groups.
xbarnone/moderate =
(4*41.25+5*48)/9=45, sxbar=sqrt(208.78/9)=4.82, df=11,t=2.201
45-2.201*4.82<munone/moderate<45+2.201*4.82
34.39< munone/moderate<55.61
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 41.73 43.8 43.53
43.9 37.6
MED 51.5 50.9
50.0 51.13 57.9 56.23
51.9 59.9
HIGH 62.7 74.0
56.4 61.27 72.1 73.33
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 df MSE
Diet 1831.9 2 915.95
Sex 179.9 1 179.9
Diet x Sex
Interaction 82.4 2 41.2
Error 161.0 12 13.42
---------------------- -----
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.
Fdiet*sex = 41.2/13.42 =
3.071
F0.10,2,12
= 2.81
Reject the F-test:
3.071>2.81. There is a significant
interaction between diet and sex. It
seems that the difference between male and females is greater for the high
protein diet than it is for the low and moderate protein diets. The lines are not parallel.
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.
Fsex = 179.9/13.42 = 13.405
F0.05,1,12
= 4.75, 13.405>4.75, reject the F statistic (p=0.003). There is a significant difference between the
sexes—males have a greater weight gain (as seen by the plot of means).
c. At the .05 level of significance, 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.
Fdiet = 915.95/13.42 = 68.287
F0.05,2,12 = 3.89,
68.287>3.89, reject the F statistic.
There is a significant difference among the diet types—there is a
consistent increase in weight gain as the level of protein in the diet gets
higher, for both sexes.
d. Compute a 95% confidence interval
for the average weight
gain on the LOW diet (regardless of Sex).
xbarlow = 42.63, sxbar = sqrt(13.42/6)
= 1.49, df=12, t=2.179
42.63-2.179*1.49<mulow<42.63+2.179*1.49
39.38<mulw<45.88
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.
xbarmale/high-xbarfemale/high=73.33-61.27=12.06,sxbar1-xbar2=sqrt(13.42*(1/3+1/3))=2.99
df=12, t=2.179
12.06-2.179*2.99<mumale/high-mufemale/high<12.06+2.179*2.99
5.54< mumale/high-mufemale/high<18.57