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MTWeaver
Calcite | Level 5
Is there a way to get (or calculate) formal test for set of predictors controlling for other model variables in PROC MIXED?

Similar to the TEST statement in PROC REG.

For example, I'd like to test contribution of set of 3 psychological variables after controlling for 3 anthropometirc variables in a mixed model.

Thanks!
5 REPLIES 5
AA1973
Calcite | Level 5
Perhaps you can test the linear contrast that the three parameters of the three psychological variables (assumed to be fixed effects) are equal to the vector (0,0,0). I think you can construct this test with the CONTRAST statement, using only the three psychological variables (you do not need to include all effects that are in the MODEL statement).
It probably looks like this:

CONTRAST '3 psycho var params= vector 0'
psychovar1 1, psychovar2 1, psychovar3 1;

Hope this helps
Dale
Pyrite | Level 9
Instead of testing a set of parameters using the CONTRAST statement as indicated by AA1973, you could construct a likelihood ratio test. The likelihood ratio test requires that you fit two models, one with the three psychological variables plus the anthropometric variables and a second model which has the anthropometric variables only. You must specify METHOD=ML when fitting each of these models.

You can then take the difference in -2LL values returned by the two models and compare that difference to a table of the chi-square distribution with 3 df. If the difference is larger than the tabled value then the set psychological variables is significant at the specified alpha level.
AA1973
Calcite | Level 5
Oh yes, this is another way, also appropriate, to test for the effect of the set of three predictors. The conclusions should be equivalent. If I'm not mistaken, the test with the CONTRAST statement is based on linear model theory, so it should produce an F statistic, while the test based on the difference in -2LL is asymptotic. The p-values for both tests should be very similar. Message was edited by: AA1973
MTWeaver
Calcite | Level 5
Thanks, Dale!

I had tried that, but got results that did not make sense.

The missing piece was that I had forgotten to specify ML, which, as you point out, makes all the difference.

Much chagrined, and grateful...

Michael
MTWeaver
Calcite | Level 5
Thanks - the CONTRAST statement worked - I had only used this with group means (and examples also only use class variables), so did not think it could be applied to continuous variables.

Dangers of ASSUMING! - thanks again!

Michael

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