Testing slope variance PreviousNext
Mplus Discussion > Growth Modeling of Longitudinal Data >
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 Oxnard Montalvo posted on Tuesday, April 29, 2014 - 10:49 pm
Hi,
When assessing the fit of a GMM, is it possible to test whether a slope or intercept variance should be free, or held equal to zero, using a chi-square difference test on the difference in the loglikelihoods?

I.e. if the less restricted model M1 is (with likelihood L1), k=3;

VARIABLE: NAMES are t1 t2 t3 t4 t5;
CATEGORICAL = t1 t2 t3 t4 t5;
CLASSES = c(3);
ANALYSIS: TYPE = MIXTURE;
ALGORITHM = integration;
STARTS 200;
MODEL:
%OVERALL%
i s | t1@0 t2@1 t3@2 t4@3 t5@4;
S20;

And the more restricted model M2 (likelihood M2)is the same except;
MODEL:
%OVERALL%
i s | t1@0 t2@1 t3@2 t4@3 t5@4;
I-S@0

Then can we use a chi-square difference test on L1-L2 with 2df?
 Linda K. Muthen posted on Wednesday, April 30, 2014 - 11:12 pm
I would not do a difference test in this way with a mixture model. I would use MODEL TEST to test whether coefficients are equal. I would not test if a variance is zero. This is on the border of the permissible parameter space.
 Oxnard Montalvo posted on Saturday, May 03, 2014 - 12:52 am
Could you please clarify what exactly you recommend testing using MODEL TEST?
 Linda K. Muthen posted on Saturday, May 03, 2014 - 5:43 pm
Holding the parameters equal may change the classes from where they are not equal.
 Oxnard Montalvo posted on Wednesday, May 07, 2014 - 3:17 am
I see.
So to determine whether or not I should set the slope variance to zero (as in an LCGA), or let it be free, would the following be the correct use of MODEL TEST?

VARIABLE: NAMES are t1 t2 t3 t4 t5;
CATEGORICAL = t1 t2 t3 t4 t5;
CLASSES = c(3);
ANALYSIS: TYPE = MIXTURE;
ALGORITHM = integration;
STARTS 200;
MODEL:
%OVERALL%
i s | t1@0 t2@1 t3@2 t4@3 t5@4;

MODEL TEST: s = 0;

?
 Linda K. Muthen posted on Wednesday, May 07, 2014 - 7:50 pm
What you specify in MODEL TEST is no different from what is given in the results section in the third column.

Note that testing a variance against zero may not be correct as zero in on the boundary of the admissible parameter space.

You can use BIC. Look at the BIC with variance of zero versus BIC when the variance is free.
 Oxnard Montalvo posted on Wednesday, May 07, 2014 - 10:40 pm
Ok thankyou, I'm very grateful for the advice.
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