Model test vs Chi-square difference PreviousNext
Mplus Discussion > Structural Equation Modeling >
 Jeremy Miles posted on Thursday, July 05, 2007 - 11:21 am
I'm testing some nested models, using either the Wald test (using model test: ), or using the Satorra method described here: .

However, I'm getting quite different results. With the Wald test, I get a chi-square difference of about 12 (with 4 df), with the chi-square difference test, I get 32.

(I just swap "model test:" for "model constraint:", to make sure the models are the same).

Is this because the constraint is fixing a correlation to 1.00, and therefore I'm having a boundary problem? Or is it something else? Which (if either) of these results should I trust?


 Linda K. Muthen posted on Thursday, July 05, 2007 - 11:56 am
I can't answer this without seeing exactly what you are doing. You should get the same results using the Wald test versus testing two nested models. I have seen examples where these are the same. So perhaps there are some defaults that change. If you send the inputs, data, outputs, and your license number to, I can take a look at it.
 Kihan Kim posted on Wednesday, September 22, 2010 - 9:42 am
I was going to run Wald chi-square difference test with one degree of freedom by constrainting one path (i.e., cc on itv) to 0.

I used the following syntax, but I get the same chi-square value as the model without the MODEL TEST command.

Is there anything wrong in my syntax?

dv3 on dc cc;

dc on itv ts pd cent form12;

cc on itv(p1) ts pd cent form12;

dc with cc;

itv with ts cent form12 pd;
ts with cent form12 pd;
cent with form12 pd;
form12 with pd;

Model Test: p1 = 0;
 Linda K. Muthen posted on Wednesday, September 22, 2010 - 11:18 am
The test of model fit does not change when MODEL TEST is used. What you are testing in MODEL FIT is the same test that is done using the z-test from the regular output.

The following statement

cc on itv(p1) ts pd cent form12;

should be

cc on itv(p1)
ts pd cent form12;

I think if you look at your output you will find everything after (p1) is ignored.
 Marie-Helene Veronneau posted on Friday, April 05, 2013 - 9:44 am
I would like to know if I can use the MODEL TEST command to compare the strength of two different path coefficients predicting the same outcome. I am worried that if the two predictors have a different scale of measurement, then constraining them to equality may not be meaningful, if the constraint applies to unstandardized path coefficients. But if the test allows to compare the strength of the standardized coefficients for the two paths, then this strategy seems OK.
Thanks for your help.
 Linda K. Muthen posted on Monday, April 08, 2013 - 7:01 am
You would need to compare the standardized coefficients in this case. You can use MODEL CONSTRAINT to define the standardized coefficients and then test their difference.
 Timothy Allen posted on Thursday, February 14, 2019 - 2:48 pm

I am running a three group path analysis with observed variables (though I specify the variance of one x-variable that has some missingness to ensure all cases are included in the analysis).

I want to see if one of my paths varies across groups. Comparing a model that constrains the path to one that freely estimates it, I get a significant chi-square difference test result.

When I use MODEL TEST however (e.g., 0 = a1-a3; 0 = a1-a2;), the wald test is not significant.

Is it possible for these two approaches to yield contradicting results? If so, which approach should be favored? (And, if chi-square, how one would probe a significant difference to determine which of the three groups differs from the other two?)
 Bengt O. Muthen posted on Thursday, February 14, 2019 - 5:16 pm
Q1: Yes, for instance with small N.

Q2: Not clear - I wouldn't trust either one if they disagree. You may want to get more opinions on this from SEMNET.
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