Probit or logit PreviousNext
Mplus Discussion > Categorical Data Modeling >
 Courtney Bagge posted on Tuesday, August 07, 2007 - 3:35 pm
Hi there,

I am trying to run a logistic regression model using type = complex missing h1. I wanted to first compare the models with and without taking into account my complex sampling design. First, I ran this model using type = missing h1, estimator = ML and obtained a parameter estimate for one of my variables (var1) of .264 and Est./S.E. of 1.68. Next, I ran this model using type= complex missing h1, estimator = wlmsv (default) and parameterization = theta. I obtained a parameter estimate of .334 and a Est./S.E. of 2.989.

I thought that I would have the same parameter estimate, but different standard errors when comparing the two analyses. When I am using type = missing h1 I cannot specify type = ml (to specify a logistic regression). Is the parameter estimate when using type = complex a probit instead of a logit? If so, is this why the parameter estimates differ? If I am indeed obtaining a probit is there anyway to specify a logit and if not is it possible to convert a probit into an OR?

***I am also running a multinomial logistic regression with a 4-level nominal outcome using type=complex missing h1. Are the parameters in this case probits or logits?

Thanks in advance!
 Linda K. Muthen posted on Tuesday, August 14, 2007 - 2:27 pm
If you want to compare with and without complex survey features, you need to keep the estimator constant. Maximum likelihood give logistic regression as the default. Weighted least squares gives probit regresison. Use MLR for both analyses.

Multinomial logistic regression gives logits.
 patrick sturgis posted on Wednesday, April 02, 2008 - 7:40 am
Dear Linda/Bengt

I am estimating a growth curve model with binary outcome, clustering and weights. From the output, documentation and discussion, it seems to be the case that I cannot get any measures of model fit for my model. Is this correct?

Second, it is possible to test for differences in model fit for nested models when using MLR by using the approach set out on the mplus website. However, having read this, I am not sure how to interpret the result. The material on the website says:

Compute the chi-square difference test (TRd) as follows:

TRd = -2*(L0 - L1)/cd
= -2*(-2606 + 2583)/2.014 = 22.840

but what is one supposed to do with the 22.840? Is this a chi square value? If so, what are the degrees of freedom?


 Linda K. Muthen posted on Wednesday, April 02, 2008 - 8:08 am
There are many situations where you get fit statistics and many when you don't. If you don't, they are not available for that model. If you want to know why in your particular case you don't get fit statsitics, send your output and license number to

The value is a chi-square value. The degrees of freedom is the difference in the number of free parameters.
Back to top
Add Your Message Here
Username: Posting Information:
This is a private posting area. Only registered users and moderators may post messages here.
Options: Enable HTML code in message
Automatically activate URLs in message