Because the results are deemed to be not significant then the next step is to test for the main effects of the treatment.
We now define a new variable equal to the sum of the observations for each animal. To test for the main treatment effect, consider the following linear combination of the observations for each dog; that is, the sum of all the data points collected for animal j receiving treatment i.
\(Z_{ij} = Y_{ij1}+Y_{ij2}+\dots + Y_{ijt}\)
This is going to be a random variable and a scalar quantity. We could then define the mean as:
\(E(Z_{ij}) = \mu_{Z_i} \)
Consider testing the following hypothesis that all of these means are equal to one another against the alternative that at least two of them are different, or:
\(H_0\colon \mathbf{\mu}_{Z_1} =\mathbf{\mu}_{Z_2} = \dots = \mathbf{\mu}_{Z_a} \)
Using SAS
ANOVA on the data Zij is carried out using the following MANOVA statement in the SAS program as shown below:
Download the SAS Program: dog.sas
h=treat sets the hypothesis test about treatments.
Then we set m = p1+p2+p3+p4 to define the random variable Z as in the above.
Now, we must make sure that we are looking at the correct part of the output! We have defined a new variable MVAR in this case, a single variable which indicates that we are summing these four.
Results for Wilks Lambda:
| Statistic | Value | F Value | Num DF | Den DF | Pr > F |
|---|---|---|---|---|---|
| Wilks' Lambda | 0.63985247 | 6.00 | 3 | 32 | 0.0023 |
| Pillai's Trace | 0.3601453 | 6.00 | 3 | 32 | 0.0023 |
| Hotelling-Lawley Trace | 0.56286025 | 6.00 | 3 | 32 | 0.0023 |
| Roy's Greatest Root | 0.56286025 | 6.00 | 3 | 32 | 0.0023 |
This indicates that there is a significant main effect of treatment. That is that the mean response of our four-time variables differs significantly among treatments.
Conclusion
Treatments have a significant effect on the average coronary sinus potassium over the first four time points following occlusion \( \left( \Lambda = 0.640; F = 6.00; d. f. = 3, 32; p = 0.0023 \right) \).
In comparing this result with the results obtained from the split-plot ANOVA, we find that they are identical. The F-value, p-value and degrees of freedom are all identical. This is not an accident! This is mathematical equality.