ANSWERS TO REVIEW 3
 
1e. Intensive mean is 21.80. The value for "s" in the CI is the sqrt of
the MSE from the original 1-way ANOVA. So s = sqrt(209.78). The t-value is
based upon 11 df; this is the "within-groups" df that the MSE is based on.
Recall that the MSE is the pooled, within-groups estimate of the variance.
t-value for the 95% CI is 2.201. The CI is:
 
21.80 +/- 2.201 * sqrt(209.78/5) = [7.54, 36.06]
 
Interpretation: we are 95% confident that the true mean percentage of
stream area that is pools for the "Intensive" group is contained in the
above interval.
 
1f. Pooled mean for {None, Moderate} is 45.0 and involves 9 data points.
The value for "s" in the CI is the sqrt of the MSE from the original 1-way
ANOVA. So s = sqrt(209.78). The t-value is based upon 11 df; this is the
"within-groups" df that the MSE is based on. Recall that the MSE is the
pooled, within-groups estimate of the variance. t-value for the 95% CI is
2.201. The CI is:
 
45.0 +/- 2.201 * sqrt(209.78/9) = [34.37, 55.62]
Interpretation: we are 95% confident that the above interval captures the
true mean for the {None,Moderate} groups combined, with respect to percent
of stream area that is pools.
 
2. ANOVA Table:
 
SOURCE OF VARIATION df SS  MS F P-value
Total 19 3125.88
Between Groups  3 2588.212 862.7373 25.673  P<.0005
Within Groups 16  537.668 33.6042
 
Since all the samples sizes are equal (5), the denominator of each
"q-observed" value will be the same: sqrt(MSE/5) = sqrt(33.6042/5) =
2.5925. Numerator of each q-observed is the observed difference between
the means of the two groups being compared. The missing q-observed values
are:
 
1 vs. 2: |53.533 - 83.2|/2.5925 = 11.44
1 vs. 3: |53.533 - 78.722|/2.5925 = 9.716
1 vs. 4: |53.533 - 74.667|/2.5925 = 8.152
4 vs. 3: |74.667 - 78.722|/2.5925 = 1.564
 
Tukey .05 critical value = 4.046 (k=4, dfe=16)
SNK .05 critical values = 4.046 (p=4), 3.649 (p=3), 2.998 (p=2).
 
For the Tukey method, any q-observed values that meet or exceed 4.046 are
deemed "significant" and the two involved means declared statistically
different. This yields the following in terms of means: Group 1 at the
lower end all by itself; Groups {2,3,4} cannot be separated.
 
For the SNK method, we have to use 3 different critical values, depending
upon whether the two means being compared span a range of 4, 3, or 2
means. As it happens, the SNK method yields exactly the same results as
the Tukey method above.
 
3. ANOVA table is as follows:
 
SRCE VARIATION  SS df MS F P-value
Protein Level 7. 2 3.5 2.25 .10<P<.25
Alkaloid 6. 1 6.0 3.86 .05<P<.10
Interaction 39. 2 19.5 12.53  P<.0005
Error 28. 18 1.55
TOTAL 80 23
 
3a. df for Interaction test are 2,18. "F-crit" at alpha = .10 = 2.62
Interaction highly significant as P<.0005. This is seen in the
"interaction bar charts". At the Low and Medium levels of protein, the
presence of alkaloid leads to an increase in growth; but at the High level
of protein, the presence of alkaloid seems to be "too much" for an insect
larva, and there is actually a decrease in growth. This is an example of
what Zar terms "reverse interaction".
 
3b. df for Protein test are 2, 18. "F-crit" at alpha = .05 = 3.55. This is
a nonsignificant result; hence we fail to reject the null hypothesis that
the three Protein means are equal [note the actual observed means are
5.50, 5.25, 4.25]. This does NOT, however, imply that "nothing is going
on", as the interaction effect between the two factors is IGNORED when
doing a main effects test on the Protein Levels.
 
3c. df for Alkaloid test are 1,18. "F-crit" at alpha = .05 = 4.41. This is
a barely nonsignificant result [P-value via computer is .065]; hence we
fail to reject the null hypothesis that the two Alkaloid means are equal
[note the actual observed means are 4.5, 5.5]. This does NOT, however,
imply that "nothing is going on", as the interaction effect between the
two factors is IGNORED when doing a main effects test on the Alkaloid
levels.
 
3d. Pooled mean for {Low+Medium} Protein is: 5.375, involving 16 data
points. The value for "s" in the CI is the sqrt of the MSE from the
original 2-way ANOVA. So s = sqrt(1.55). The t-value is based upon 18 df;
this is the "within-cells" df that the MSE is based on. Recall that the
MSE is the pooled, within-groups estimate of the variance. t-value for the
95% CI is 2.101. The CI is:
 
5.375 +/- 2.101 * sqrt(1.55/16) = [4.72, 6.03].
 
4. ANOVA Table.
 
SRCE VARIATION SS df MS F P-value
Varieties 188.54 3 62.85 144.36 <.005
Blocks 19.79 5 3.96
Residual 6.53 15 .4353
 
Critical value for the F-test on Varieties has 3,15 df; at the .05 level
the tabled value is 3.29. We conclude that there is a difference somewhere
among the Variety means.
 
Since the sample sizes are equal, the denominator for any Tukey or SNK
multiple comparison will be sqrt(.4353/6) = .2694.
 
The Tukey tabled critical value at the .05 level is [k=4, dfe=15] 4.076.
So any two means that are at least 4.076*.2694=1.10 apart will be declared
statistically different. But note that all four of the Variety means are
at least this far apart, so we may declare all Variety measn statistically
different. The SNK tabled critical values are [p=4] 4.076, [p=3]3.674,
[p=2] 3.014. But any two means declared different via the Tukey HSD
procedure will also be declared different via the SNK procedure. Hence the
SNK procedure will also come to the same conclusion: that all 4 Variety
means are statistically different.
 
5. Error df for Plan A is 20. Chart on p. App208 gives phi-value of about
1.95. Using formula for delta [Notes p. 13-5], obtain delta = 2.76.
Error df for Plan B is 20. Chart on p. App207 gives phi-value of about
2.1 [actually slightly less]. Obtain delta = 2.42. So Plan B will allow
one to detect the "finer" Minimum Detectable Effect and is therefore
deemed the Plan of choice.