Modeling and Multivariate Methods - SAS

Appendix A Statistical Details 663
The Factor Models
LSM(B1) – LSM(B2) = 2β 1
Note that this shows that a test on the β 1 parameter is equivalent to testing that the least squares means are
the same. But because β 1 is not estimable, the test is not testable, meaning there are no degrees of freedom
for it.
Now, construct the test for the least squares means across the A levels.
LSM(A1) = ( 1 ⁄ 2) ( μ+ α 1
+ β 1
+ αβ 11
+ μ + α 1
– β 1
– αβ 11
)
= μ + α 1
LSM(A2) = ( 1 ⁄ 2) ( μ+ α 2
+ β 1
+ αβ 21
+ μ + α 2
– β 1
– αβ 21
)
= μ + α 2
LSM(A3) = ( 1 ⁄ 2) ( μ– α 1
– α 2
+ β 1
– αβ 11
– αβ 21
+
μ– α 1
– α 2
– β 1
+ αβ 11
+ αβ 21
)
= μ – α 1
– α 2
LSM(A1) – LSM(A3) =
LSM(A2) – LSM(A3) =
Neither of these turn out to be estimable, but there is another comparison that is estimable; namely
comparing the two A columns that have no missing cells.
LSM(A1) – LSM(A2) =
This combination is indeed tested by JMP using a test with 1 degree of freedom, although there are two
parameters in the effect.
The estimability can be verified by taking its inner product with the singularity combination, and checking
that it is zero:
Table A.9 Verification
2α 1
+ α 2
2α 2
+ α 1
α 1
– α 2
singularity
combination
parameters combination to be tested
m 1 0
a 1 –1 1
a 2 –1 –1
b 1 –1 0
ab 11 1 0
ab 21 1 0