Modeling and Multivariate Methods - SAS

130 Fitting Standard Least Squares Models Chapter 3
Examples with Statistical Details
Sometimes you can investigate the functional contribution of a covariate. For example, some transformation
of the covariate might fit better. If you happen to have data where there are exact duplicate observations for
the regressor effects, it is possible to partition the total error into two components. One component
estimates error from the data where all the x values are the same. The other estimates error that can contain
effects for unspecified functional forms of covariates, or interactions of nominal effects. This is the basis for
a lack of fit test. If the lack of fit error is significant, then the fit model platform warns that there is some
effect in your data not explained by your model. Note that there is no significant lack of fit error in this
example, as seen by the large probability value of 0.7507.
The covariate, x, has a substitution effect with respect to Drug. It accounts for much of the variation in the
response previously accounted for by the Drug variable. Thus, even though the model is fit with much less
error, the Drug effect is no longer significant. The effect previously observed in the main effects model now
appears explainable to some extent in terms of the values of the covariate.
The least squares means are now different from the ordinary mean because they are adjusted for the effect of
x, the covariate, on the response, y. Now the least squares means are the predicted values that you expect for
each of the three values of Drug, given that the covariate, x, is held at some constant value. The constant
value is chosen for convenience to be the mean of the covariate, which is 10.7333.
So, the prediction equation gives the least squares means as follows:
fit equation:
-2.696 - 1.185*Drug[a - f] - 1.0761*Drug[d - f] + 0.98718*x
for a: -2.696 - 1.185*(1) -1.0761*(0) + 0.98718*(10.7333) = 6.71
for d: -2.696 - 1.185*(0) -1.0761*(1) + 0.98718*(10.7333) = 6.82
for f: -2.696 - 1.185*(-1) -1.0761*(-1) + 0.98718*(10.7333) = 10.16
Figure 3.55 shows a leverage plot for each effect. Because the covariate is significant, the leverage values for
Drug are dispersed somewhat from their least squares means.
Figure 3.55 Comparison of Leverage Plots for Drug Test Data