This answer is done in terms of PROC MIXED, but applies to any modeling procedure using the GLM parameterization for CLASS effects
The T-tests produced by the SOLUTION option may be testing different hypotheses than the Type 3 F-tests. To check the hypothesis tested with the F-test, add the /E3 option to the MODEL statement. Here's an example.
The DATA step below generates data for a simple model:
options ls=64;
/* generate sample data. */
data check;
seed = 213845740;
do v1 = 1 to 2;
do n = 1 to 20;
x1 = ranuni(seed);
y = v1 + x1 + rannor(seed);
output;
end;
end;
run;
This PROC MIXED code estimates the model for this data:
proc mixed data=check;
class v1;
model y = v1 x1 v1*x1 / s e3;
lsmeans v1 / pdiff e;
run;
Here are the results of the /S option for the model above:
Solution for Fixed Effects
Standard
Effect v1 Estimate Error DF t Value Pr > |t|
Intercept 2.3605 0.3681 36 6.41 <.0001
v1 1 -1.7722 0.5554 36 -3.19 0.0029
v1 2 0 . . . .
x1 0.3744 0.7303 36 0.51 0.6113
x1*v1 1 1.3046 1.0224 36 1.28 0.2101
x1*v1 2 0 . . . .
The estimates here need a little interpretation because of the paramaterization that PROC MIXED (and PROC GLM) use. The CLASS statement sets up an overparamterized model, using 1 dummy variablefor each level of a class effect. This parameterization causes one level of the class variable to be set to zero. We choose to make that level the last, or highest ordered. The other parameters for the CLASS variables are differences between that level and the last level. For instance, the -1.7722 above for V1 level 1 is really the difference between level 1 and level 2 of V1. So, the test on V1 level 1 is a test of the difference between those two levels of V1. The INTERCEPT value is the actual estimate for the last level of V1, making the actual estimate for V1 level 1 equal to 2.3605 - 1.7722.
A similar interpretation can be made for the interaction term. The estimate for X1 is actually the estimate for X1*V1 level 2, while the estimate for X1*V1 is a test of the difference in the slopes of X1 across the two levels of V1.
Now take a look at the results of the Tests of Fixed Effects section:
Type 3 Tests of Fixed Effects
Num Den
Effect DF DF F Value Pr > F
v1 1 36 10.18 0.00
x1 1 36 4.03 0.05
x1*v1 1 36 1.63 0.21
These results can be hard to understand if you are not familiar with Type 3 tests. To help understand the tests, the /E3 option shows exactly how each test is formed. For V1, the /E3 option shows:
Type 3 Coefficients
for v1
Effect v1 Row1
Intercept
v1 1 1
v1 2 -1
x1
x1*v1 1
x1*v1 2
The coefficients on V1, the 1 and -1, show that the Type 3 test is testing a difference in the levels of V1. This is the same test you get in the solution output for V1.
For X1, the /E3 option shows:
Type 3 Coefficients
for x1
Effect v1 Row1
Intercept
v1 1
v1 2
x1 1
x1*v1 1 0.5
x1*v1 2 0.5
which implies that the Type 3 test is testing that the average slope of X1 across the two levels of V1 is not equal to 0. We do not get a similar test in the solution option. Remember, the test on X1 in the solution option is a test that the slope on X1 for the second level of V1 is not zero.
For X1*V1, the /E3 option shows:
Type 3 Coefficients
for x1*v1
Effect v1 Row1
Intercept
v1 1
v1 2
x1
x1*v1 1 1
x1*v1 2 -1
which is a test of the equality of the slopes on X1 across the two levels of V1. This is the same test we get in the solution output.
Here are the results of the LSMEANS statement with the PDIFF option:
Differences of Least Squares Means
Standard
Effect v1 _v1 Estimate Error DF t Value Pr > |t|
v1 1 2 -1.1531 0.2721 36 -4.24 0.0001
This result is different from the test on V1 in either the solution output or Type 3 tests output. The /E option on the LSMEANS statement shows why:
Coefficients for v1
Least Squares Means
Effect v1 Row1 Row2
Intercept 1 1
v1 1 1
v1 2 1
x1 0.4745 0.4745
x1*v1 1 0.4745
x1*v1 2 0.4745
In the Type 3 tests, we test the difference in the two levels of V1 assuming that the value of X1 is 0. In the LSMEANS result, we test the difference in the two levels of V1 at the mean value of the covariate, X1, of .4745. Since the slopes of X1 are different across the two levels of V1, you will get a different test depending on the value of X1 used. This difference causes some statisticians to mean-center their covariates before running an analysis. If the mean value of the covariate is 0, then these test results will agree.
All of the above hypothesis tests are interesting, but they do test different things. By understanding how the model is parameterized and using the /E3 and /E options, you can see exactly what tests are performed.
