Structure within factor levels
We again consider the usual model for a single factor,
yij = µ |
+ |
(explained by X) βi |
+ |
(unexplained) εij |
where β1 = 0. If there are several levels for the factor, there is often some internal structure within these levels. On the previous two pages, one factor level corresponded to a control treatments. We now extend this to experiments where the treatments can be grouped into two sets.
For example, in experiment about different crop varieties, it is often known that some varieties are relatively similar to each other — perhaps they were developed in the same region or are hybrids with similar 'parents'. Irrigation systems being compared may include several variations on drip-feeding and also different schedules for delivering water through sprays.
Special cases of the general model
In experiments where the factor levels can be meaningfully grouped into groups A and B, questions about the factor levels might be:
Each of these questions can be examined using constraints on the model parameters:
The questions are answered using analysis of variance based on the sums of squares explained by the constraints.
Yields of wheat
The diagram below shows the yields of wheat in an experiment involving five varieties that are grouped into two A-varieties and two B-varieties.
Use the checkboxes to impose the three sets of constraint described above.
(Note that dividing the constraint on the right by 6 makes it clearer that the average β-value for the A varieties is the same as the average value for the B varieties.)
Analysis of variance
Each linear constraint on the model parameters increases the residual sum of squares. This corresponds to an explained sum of squares (explained by going in the opposite direction and releasing the constraint) that has 1 degree of freedom.
The constraints can therefore be described by explained sums of squares in an analysis of variance table. These explained sums of squares can be used to answer questions about differences between the groups and between the levels within each group.
The following analysis of variance table initially shows a combined sum of squares explained by the five treatments.
Click Split varieties to separate the treatment sum of squares into three components, corresponding to the three sets of constraints that were described above.
Note that the constraint of all B-varieties being equal corresponds to two linear constraints since there are three of the B-varieties — count the number of equals signs in the constraint:
β3 = β4 = β5
This constraint therefore has two degrees of freedom.
From the p-values, we would conclude that there is only weak evidence of any difference between the three B-varieties. However there is very strong evidence of a difference between the two A-varieties. It is also almost certain that the average yield from the A-varieties differes from that of the B-varieties.