Model fit without chi square PreviousNext
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 Gareth posted on Wednesday, December 02, 2009 - 5:46 am
My model has latent variable interactions, therefore no chi square is available in the output.

I have no particular competing/alternative model with which to calculate 2 x loglikelihood difference.

What alternatives are available for reporting model fit in this situation? Could I compare this model to one without the latent variable interactions?
 Linda K. Muthen posted on Wednesday, December 02, 2009 - 9:46 am
I would get a good fitting model without the interaction and then add the interaction hoping it does not worsen model fit which you cannot assess.
 Gareth posted on Friday, December 04, 2009 - 9:33 am
I could compare a model with two LV interactions with one that has one LV interaction, using the -2*LL test with correction factor for MLR.

Are the degrees of freedom for this test the difference in the number of parameters (in my example, 1 less free parameter), so that 3.84 is the cut off at p < .05?

Does it matter if the chi square is negative -3.29, which presumably arises because the comparison model is more, rather than less, complex?
 Linda K. Muthen posted on Friday, December 04, 2009 - 11:05 am
The degrees of freedom is the difference in the number of parameters. And, yes, 3.84 is the chi-square cutoff for one degree of freedom.

You should subtract the chi-square of the least restrictive model from the chi-square of the most restrictive model. This should be positive.
 newuser posted on Wednesday, January 06, 2010 - 7:09 pm
Hi, Dr. Muthen;
I am trying to compare an interaction model to one without the interaction. I am wondering
1. Can I just take the ¨C2*difference between loglikelihood to get the chi-square, or should I do the TRd? From the earlier postings, it seems ¨C2*LL can be used, then when should the TRd be used?
2. Is the difference between the degree of freedom from the two models¡¯ ¡°Number of Free Parameters¡± under ¡°info criteria¡±?
3. Only the loglikelihood of the two ¡°H0¡± should be compared, not the ¡°H1¡±?
4. If the two models can be expressed as Y on X, W, W*X is compared to Y on X, W. I have trouble understanding why introducing the interaction term to the Y on X, W model make the two models nested, because after the interaction is introduced, the models are not identical anymore.
5. the output for the main-effect-only- model don't have a Scaling correction factor reported, is there a way to request one?
Thanks for your help in advance! I tried to post a similar message earlier, but don¡¯t see it, so trying it again.
 Linda K. Muthen posted on Thursday, January 07, 2010 - 8:40 am
The following two models are nested:

y ON x w wx;
y ON x w wx@0;

The difference in degrees of freedom is the difference in the number of parameters. The test is -2 times the H0 loglikelihood difference.

You should obtain a scaling correction factor if you are using a recent version of Mplus and the estimator requires one. Please send the output and your license number if you think there should be a scaling correction factor and there is not one.
 newuser posted on Thursday, January 07, 2010 - 9:34 am
Dear Dr. Muthen, thank you so much for your reply, just to make sure, in getting the "-2 times the H0 loglikelihood difference"you mentioned, do I need to divide that # by the cd (obtained through scaling correction factor and df) to get a TRd? Or do I just use the raw difference from the two likilihoods? Thanks gain!!
 Linda K. Muthen posted on Thursday, January 07, 2010 - 10:05 am
It depends on the estimator you are using. Which estimator is it?
 newuser posted on Thursday, January 07, 2010 - 10:38 am
I did not specify any estimator in my model, I think it is using the default, ML (I have a SEM latent model)?
 Linda K. Muthen posted on Thursday, January 07, 2010 - 1:43 pm
ML does not require a scaling correction factor. The results will show the estimator used.
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