Wald test vs. DIFFTEST PreviousNext
Mplus Discussion > Structural Equation Modeling >
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 Ulrich Schroeders posted on Wednesday, July 28, 2010 - 1:12 pm
Dear Drs. Muthén,

I have a model with three latent variables and want to test whether constraining the correlation between two of them (and constraining the remaining correlations to equality), deteriorates model fit. Mplus provides two testing methods:

a) Chi-square difference test using DIFFTEST option (X^2 = 5.6, df = 2, p = .06)
b) Wald test of parameter constraints (X^2 = 6.3, df = 2, p = .04)

Shouldn't the results be exactly the same? Is there a problem with putting the correlation to the upper boundary (such as described in a Psych Meth paper from Stoel et al., 2006)? I'm using WLSMV estimator.

Thanks in advance,
Ulrich
 Linda K. Muthen posted on Wednesday, July 28, 2010 - 4:30 pm
The results are asymptotically equivalent but may differ in a given sample. Yes, a correlation of one is on the boundary.
 Ulrich Schroeders posted on Wednesday, July 28, 2010 - 5:14 pm
Dear Linda,

thanks. A short follow-up question: Are there cases where one test is preferable to the other?

Cordially,
Ulrich
 Linda K. Muthen posted on Wednesday, July 28, 2010 - 9:03 pm
I think there is a literature showing that one tests works better than the other in various situations. I don't think the situations are that well-defined. You can take a look at this literature for more information. I'm not all that familiar with it.
 Joe King posted on Thursday, December 15, 2011 - 10:42 pm
so if i do two models, and have a constraint and then release the constraint on the second one, and use the scaled likelihood test (i am using MLR) wont it change the covariance matrix on more than just those variables that it were constrained then released, thus making the MODEL TEST command a more reliable method to determine equality of parameter estimates?
 Linda K. Muthen posted on Friday, December 16, 2011 - 12:09 am
Difference testing and MODEL TEST are asymptotically equivalent.
 Joe King posted on Saturday, December 17, 2011 - 1:09 am
Dr. Muthen, i am very appreciative of your responses, and i do hope this next question isnt a bother, its on another subject. The model i am running creates a factor using a group of binary variables, then we are looking at how those factors are influenced by predictors, its 3 factors that are created, most of our predictors are also binary, does this mean we shouldnt look at the STDYX but only the STD or raw metrics?
 Linda K. Muthen posted on Saturday, December 17, 2011 - 1:09 pm
Factors are continuous. If you have f2 ON f1, STD and StdYX are the same because standardization is done using latent variables. There are no observed variables involved in the standardization.
 Joe King posted on Saturday, December 17, 2011 - 6:36 pm
well my question is more focused on the predictors to the factors so if we have v1 on f1, with v1 being a binary predictor and we are wanting to assess the relationship between it and the factor, those are different between StdYX and STD, so which one is best to use?
 Linda K. Muthen posted on Saturday, December 17, 2011 - 8:40 pm
For the regression of f1 on a binary predictor, use StdY. For the regression of a factor indicator on a factor, use StdYX.
 Joe King posted on Saturday, December 17, 2011 - 8:56 pm
so factor indicator you mean the variables which create the factor (which are all binary) as opposed to the predictor which we are interested in seeing how it is related to our factors (they are all in the same model).
 Linda K. Muthen posted on Saturday, December 17, 2011 - 11:06 pm
Yes.
 Kätlin Peets posted on Tuesday, October 29, 2013 - 11:47 am
What is the correct way of reporting the Wald test statistic? Is it (for instance) Wald chi-square (1) = XXX, p = .04. My reviewers also ask me to report the sample size. Would this be correct then?: Wald chi-square (1, N = XXX) = XXX, p = .04?
 Linda K. Muthen posted on Tuesday, October 29, 2013 - 3:10 pm
As long as it is clear what you are reporting, I don't know of any rules for this.
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