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.