Message/Author 

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? 


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? 


The degrees of freedom is the difference in the number of parameters. And, yes, 3.84 is the chisquare cutoff for one degree of freedom. You should subtract the chisquare of the least restrictive model from the chisquare 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 chisquare, 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 maineffectonly 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. 


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!! 


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)? 


ML does not require a scaling correction factor. The results will show the estimator used. 

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