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Mplus Discussion > Structural Equation Modeling >
 Andy Cohen posted on Wednesday, July 18, 2007 - 1:28 pm
I am trying to compare a model of partial mediation to one of no mediation and am having difficulty interpreting the model fit stats (log likelihood, AIC, etc.) that I am getting.

My base model, which is a count dependent variable, models y on x1-x9 and has a log likelihood of -559. I've then tried to model a path model where x1 is modelled on x2, thus testing whether x1 mediates the path between x2 and y. When I add this statement to the model statement, I get log likelihood of -1492, on the surface suggesting a much poorer fitting model. But I don't think so. Any variant of the mediation test (i.e. different relationships between x1 and the other 8 variables) yield similar fit statistics (to the -1492) and the directions are intuitive and align other preliminary exploratory analyses. So, can you help me understand how to reconcile the larger (in absolute value) fit statistics with the more complex model?

 Bengt O. Muthen posted on Wednesday, July 18, 2007 - 7:01 pm
I think you are running into the following. The likelihood is computed conditional on the covariates. In the first model you have 9 and in the second you have 8. This means that the likelihood metrics are not comparable. I think you can get around this by letting x1 be taken off the covariate list also in the first model. This is done by referring to an x1 parameter, such as its mean - it then turn into a "y variable" in Mplus parlance, and is no longer conditioned on in the likelihood.
 Andy Cohen posted on Thursday, July 19, 2007 - 5:00 am
Thanks for the response. I'm not sure I understand, but let me try. My first, base, model, is

y on x1 x2...x9;

My second model is

y on x1 x2...x9;
x1 on x2;

Are you suggesting that I make my first model look like this?:

y on x1 x2...x9;

Or did I misunderstand?

 Linda K. Muthen posted on Thursday, July 19, 2007 - 9:32 am
 Ahmad Siddiquei posted on Wednesday, September 19, 2018 - 10:49 pm
Hi Dr. Muthen,

How can I compare two models using log-likelihood difference test, with 1-1-2 SEM? Is there any online calculator to do it?

One model one without direct path from Independent to Dependant variables and one with direct path from Independent to Dependant variables.

 Bengt O. Muthen posted on Thursday, September 20, 2018 - 5:43 pm
If the interest is in only 1 parameter (the direct effect), you can simply look at the Est/SE that the program provides for that parameter.
 Ahmad Siddiquei posted on Tuesday, September 25, 2018 - 10:31 pm
Hi Dr Muthen,

Following my earlier question, the direct path from independent variable to dependent variable is insignificant, indicating full-mediation.
But it says:

The chi-square value for MLM, MLMV, MLR, cannot be used for chi-square difference testing in the regular way. Do I need to conduct a separate analysis for that?
 Bengt O. Muthen posted on Wednesday, September 26, 2018 - 6:29 pm
See our left-margin special web page:

 Ahmad Siddiquei posted on Wednesday, September 26, 2018 - 8:51 pm
I did Satorra-Bentler chi-square difference testing and it worked. Thank you.

One last question is about the following statement:

For MLM and MLR the products T0*c0 and T1*c1 are the same as the corresponding ML chi-square values.

As it says, I am using the chi-square value (302.07 and 292.12) of both nested and comparison model given in the model outputs, rather than using the product values (T0*c0 and T1*c1) i.e. multiplying chi-square values with respective scaling correction factor to compute TRd.

Is that correct?
 Bengt O. Muthen posted on Friday, September 28, 2018 - 11:08 am
 Ahmad Siddiquei posted on Friday, September 28, 2018 - 3:10 pm
Thanks. You are amazing!
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