9.4 - Approach 2: MANOVA

When taking a multivariate approach, we collect the observations over time from the same dog, dog j receiving treatment i into a vector:

\(\mathbf{Y}_{ij} = \left(\begin{array}{c}Y_{ij1}\\ Y_{ij2} \\ \vdots\\ Y_{ijt}\end{array}\right)\)

We treat the data collected at different points in time as if it were data from different variables. Basically, we have a vector of observations for dog j receiving treatment i and each entry corresponds to data collected at a particular point in time.

The usual assumptions are made for a one-way MANOVA. In this case:

  1. Dogs receiving treatment i have common mean vector \(\mu_{i}\)
  2. All dogs have common variance-covariance matrix \(\Sigma\)
  3. Data from different dogs are independently sampled
  4. Data are multivariate normally distributed

The Analysis

Step 1: Use a MANOVA to test for overall differences between the mean vectors of the four different observations and the treatments.

Using SAS

We will use the Dog SAS program to perform this multivariate analysis.

Download the SAS program: dog.sas


We use the glm procedure to analyze these data. In this case, we look at a one-way manova.  We only really have one classification variable - treatment.

The model statement includes the variables of interest on the left-hand side of the equal sign.  In this case, they are p1, p2, p3, and p4, (the potassium levels at four different points in time).  We put the explanatory variable, treatment, on the right-hand side of the equal sign.

The first manova statement tests the hypothesis that the mean vector of observations over time does not depend on treatment. The print option asks for the error of sums of squares and cross products matrix as well as the partial correlations.

The second manova statement tests for the main effects of treatment. We'll return to this later.

The third manova statement tests for the interaction between treatment and time. We'll also return to this later.

Right now, the result that we want to focus on is the Wilks Lambda of 0.484, and the corresponding F-approximation of 2.02 with 12, 77 d.f. A p-value of 0.0332 indicates that we can reject the null hypothesis that there is no treatment effect.

Our Conclusion at this point: There are significant differences between at least one pair of treatments in at least one measurement of time \(\left( \Lambda = 0.485; F = 2.02; d.f. = 12, 77; p = 0.0332 \right) \).

Next Steps... Section

If we find that there is a significant difference, then with repeated measures data we tend to focus on a couple of additional questions:

  • First Question

    Is there a significant treatment by time interaction? Or, in other words, does the effect of treatment depend on the observation time? Previously in the ANOVA analysis, this question was evaluated by looking at the F-value, 2.06. This was reported as a significant result. If we find that this is a significant interaction, the next thing we need to address is, what is the nature of that interaction?

  • Alternative Question

    If we do not find a significant interaction, then we can collapse the data and determine if the average sinus potassium level over time differs significantly among treatments.  Here, we are looking at the main effects of treatment.

Let's proceed...