Scripting Guide - SAS

166 Data Structures Chapter 7
Matrices
Next, add a column of 1s to the design matrix for the intercept term. You can do this by concatenating J
and Design Nom(), as follows:
x = J(9,1,1) || designNom(factor);
[1 1 0,
1 0 1,
1 -1 -1,
1 1 0,
1 0 1,
1 -1 -1,
1 1 0,
1 0 1,
1 -1 -1]
Now, to solve the normal equation, you need to construct a matrix M with partitions:
X′X X′y
y′X y′y
You can construct matrix M in one step by concatenating the pieces, as follows:
M=(x`*x || x`*y)
|/
(y`*x || y`*y);
[ 9 0 0 27,
0 6 3 2,
0 3 6 1,
27 2 1 93]
Now, sweep M over all the columns in X′X for the full fit model, and over the first column only for the
intercept-only model:
FullFit=sweep(M,[1,2,3]); // full fit model
InterceptOnly=sweep(M,[1]); // model with intercept only
Recall that some of the standard ANOVA results are calculated by comparing the results of these two
models. This example focuses on the full fit model, which produces this swept matrix:
[0.111111111111111 0 0 3,
0 0.222222222222222 -0.11111111111111 0.333333333333333,
0 -0.11111111111111 0.222222222222222 0,
-3 -0.33333333333333 0 11.33333333333333]
Examine the model coefficients from the upper right partition of the matrix. The lower left partition is the
same, except that the signs are reversed: 3, 0.333, 0. The results can be interpreted as follows:
• The coefficient for the intercept term is 3.
• The coefficient for the first level of the factor is 0.333.
• The coefficient for the second level is 0.
• Because of the use of Design Nom(), the coefficient for the third level is the difference, –0.333.
• The lower right partition of the matrix holds the sum of squares, 11.333.