Fit indices when assessing latent var... PreviousNext
Mplus Discussion > Structural Equation Modeling >
 Chris Stride posted on Wednesday, October 19, 2005 - 7:45 am
I am currently trying to fit a SEM which contains interaction terms between continuous latent variables. I have upgraded from v2 to v3.13 and have hence successfully used the next XWITH command to compute my interaction term.

However, when I run the analyses, the only fit indices in the output are the AIC and BIC. No chi-squared stat, CFI, TLI, RMSEA.

From this I have 2 questions...

i) Is it possible to get these indices for this type of analyses?

ii) And a sort of related question - can a SEM which contains such an interaction effect between factors be considered as nested w.r.t.o. the model that contains just the main effects? And if so, is there a way of testing for whether the model is significantly improved by fitting the interaction?

 Linda K. Muthen posted on Wednesday, October 19, 2005 - 9:05 am
i) No, this is not possible. These fit statistics are available only for models where means, variances, and covariances are sufficient statistics for model estimation. They are printed whenever they are available.

ii) I think these models would be considered nested. It is the case that -2 times the loglikelihood differenece is distributed as chi-square and can be used to test the difference between two nested models.
 Xiaojing Yan posted on Tuesday, March 20, 2007 - 8:40 am
Dear All,

I am currently doing a similar analysis and have two broad questions:

1. Is it possible to get the effect size of interaction between two continuous latent variables? It seems for RANDOM model the R-sq is not available. How can I solve this? What's the problem if I use the R-sq change from means interaction?

2. May I have references on interaction between continuous latent variables?
 Xiaojing Yan posted on Tuesday, March 20, 2007 - 8:46 am
Many thanks.
 Bengt O. Muthen posted on Tuesday, March 20, 2007 - 4:54 pm
1. To me, effect size concerns a standardized mean difference between two conditions (treatment-control, male-female, etc). But it sounds like you use this term for R-square. R-square when predictors involve products of latent variables is less straightforward because the variance and covariance related to a product of two variables is involved. R-square for the means would not be the same. I would simply see if the influence of the product term is significant.

2. Our User's Guide refers to Klein-Moosbrugger in Psychometrika 2000
 Xiaojing Yan posted on Wednesday, March 21, 2007 - 11:23 am
Many thanks.
 Heiko Schimmelpfennig posted on Saturday, September 08, 2007 - 1:04 pm
Dear Bengt,

is it just as difficult to calculate R-square if the predictors involve a quadratic or cubic latent variable?

Thanks in advance.
 Linda K. Muthen posted on Tuesday, September 18, 2007 - 4:48 am
Yes, this is more involved also.
 Heiko Schimmelpfennig posted on Monday, October 29, 2007 - 5:42 am
Thank you for your reply.

Nevertheless, is it possible to calculate R-square by hand in these cases with the help of the Mplus-Output? Maybe do you have any references?
 Linda K. Muthen posted on Monday, October 29, 2007 - 6:54 am
I don't know. You might want to check the Aiken and West book on regression analysis with observed variable interactions or contact Andreas Klein.
 Steve posted on Friday, July 12, 2013 - 4:14 am
Dear Linda,

I have read the above discussion (and recognizing that it is a bit dated now), I am wondering if it is now possible somehow to estimate model fit statistics (chi-squared stat, CFI, TLI, RMSEA) when including an interaction term in the model. I have run my analysis and this information is not included in the output. I would like to do a chi-square difference test after adding in the interaction variable. Mplus is specifically referenced as being able to do this in a recent paper (Graves, Sarkis & Zhu, 2013, p. 86, Journal of Environmental Psychology), and I would like to replicate this process if you could please explain how to get this output.

Many thanks.
 Linda K. Muthen posted on Friday, July 12, 2013 - 10:18 am
They probably did difference testing using -2 times the loglikelihood difference which is distributed as chi-square. This is the same as the z-test of the interaction.
 Steve posted on Friday, July 12, 2013 - 1:07 pm
Dear Linda,

Thank you very much for this information. Have a nice weekend.
 Miguel Villodas posted on Wednesday, October 09, 2013 - 10:09 am

I am a bit confused about how to compare an SEM model with an interaction to a "nested" model without an interaction using the log likelihood value. Specifically, I only get an Ho loglikelihood value in my output for the model with the interaction and do not get a scaling factor for my model without the interactions. Could you please clarify which of these I should compare?
 Linda K. Muthen posted on Wednesday, October 09, 2013 - 10:16 am
You should get an H0 loglikelihood and scaling correction factor in both outputs. This is what you compare. It is not really necessary, however, because the z-test of the interaction gives the same information.
 Miguel Villodas posted on Thursday, October 10, 2013 - 5:45 am
Thank you Linda.
 Sabrina Thornton posted on Monday, November 25, 2013 - 8:30 am
Hi Linda,

Please can you elaborate on your post on October 09, 2013?

"You should get an H0 loglikelihood and scaling correction factor in both outputs. This is what you compare. It is not really necessary, however, because the z-test of the interaction gives the same information."

Does that mean that the significant level of the interaction term (Est./S.E.) would provide enough information for reporting the interaction effect? If not, can you let me know where I can find "the z-test of the interaction" you referred to?

I am comparing two nested models using H0 log likelihood. I specified a model with only direct effects and the other with the direct effects and the interaction term:

Null model
H0 value -4103.707
free parameters 70

Interaction model (the interaction term is significant, p < 0.05)
H0 value -4103.948
free parameters 71

-2*(-4103.948-(-4103.707))=0.482, which is not significant at the 0.05 level for 1 df. Does that mean that the model is not significantly different from the null model, i.e. the fit is not significantly worsen when adding the interaction term, and therefore the interaction model can be accepted?
 Linda K. Muthen posted on Monday, November 25, 2013 - 9:22 am
The z-test is in the third column of the results. It is the parameter estimate divided by its standard error.
 Sabrina Thornton posted on Monday, November 25, 2013 - 9:33 am
Hi Linda,

Thanks. So does it mean that the z-test for the interaction term would be sufficient when reporting the interaction effect? Does the log likelihood value comparison between two nested models (without vs. with interaction term) necessary when the z-test is significant? if so, can you advise as to whether my reasoning in the previous post is appropriate? Thanks.
 Linda K. Muthen posted on Monday, November 25, 2013 - 11:22 am
The loglikelihood test with one degree of freedom is the same as the z-test. You don't need both.
 Chong M. Chow posted on Thursday, April 30, 2015 - 6:36 pm
Dear Drs. Muthen,

For a latent interaction model, I have learned that it is possible to compute a chi-square statistic by comparing the log-likelihood of interaction model versus independence model. May I know if it is possible to hand compute other fit indices (e.g., TLI, CFI) based on this particular chi-square?
 Bengt O. Muthen posted on Friday, May 01, 2015 - 1:42 pm
I don't think it is known how well or how poorly CFI would work in this context.
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