# 9.6 - Step 3: Test for the main effects of treatments

9.6 - Step 3: Test for the main effects of treatments

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:

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:

MANOVA Test Criteria and Exact F Statistics for
the Hypothesis of No Overall treat Effect
on the Variables Defined by the M Matrix Transformation
H = Type III SSCP Matrix for treat
E = Error SSCP Matrix

S=1 M=0.5 N=15

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.

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