Model fit indices for two-part growth... PreviousNext
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 Scott Weaver posted on Monday, March 13, 2006 - 3:30 pm
Greetings,
I am fitting the binary portion of a semincontinuous growth model (as a preliminary step for building the full semicontinuous model) for the first time. I have a few questions regarding the output:
(1) I notice that Pearson chi-square and likelihood ratio chi-square values are provided. I had previously thought that a chi-sq test of model fit was not available for analyses requiring numerical integration. The Pearson chi-sq test statistical is highly significant (p<.0001) whereas the likelihood chi-sq test is not (p=1.00). Which one should I use?
(1a) Can I use the likelihood ratio chi-sq test for chi-sq difference testing or should I use the Loglikelihood value? How do I incorporate the scaling correction factor when using the LogL for nested model difference testing?

(2) For the residual output - what do you think would be too high of a value for a residual? Basically, I am trying to decide from the output whether the linear model is appropriate or whether I should try other growth forms.

(3) I noticed that the plot of estimated probabilities for being in category 2 was not exactly linear. So is it that the log odds of being in category 2 is constrained to be linear in this model? If I wanted to specify a model of linear growth in propensity, how could I do this?

(4) In a multiple group context, would I use the Known Classes statement?

Thank you in advance!
Scott
 Bengt O. Muthen posted on Tuesday, March 14, 2006 - 5:38 am
1. The Pearson and LRT chi-squares are only for the binary part of the model, not a test of the overall fit of the whole model. When the Pearson and the LRT disagree this much, neither is reliable - this is typically due to many empty cells where it is well-known that the chi-square approximation is very poor.

2. It seems easier to try linear vs quadratic and test if the quadratic growth factor parameters are significant.

3. Linearity is on the logit (log odds) scale.

4. yes.
 Scott Weaver posted on Tuesday, March 14, 2006 - 8:36 am
Thank you for your response. As I am taking the approach of fitting the binary and continuous parts separately to more easily determine the proper growth function prior to looking at the full semicontinuous model.

Regarding the unreliable chi-sq statistics, I do not seem to have any empty cells at the univariate or bivariate level. But I guess at combinations of 3-7 of the variables, there could be missing cells.

How do I use the scale correction factors with the log likelihood for comparing nested models? Can I use the procedures listed at http://www.statmodel.com/chidiff.shtml or do I need to make some adjustment because I am working with log likelihoods and not chi-square values?

Thanks again,
Scott
 Bengt O. Muthen posted on Tuesday, March 14, 2006 - 9:16 am
The unreliable chi-square refers to empty cells in the p-variate table, where p is the total number of binary variables that you use in your model.

Yes, you can use that difference test approach.
 Abhik Bhattacharya posted on Monday, October 09, 2006 - 7:02 pm
Hello,

I am trying to fit this model.

title:
Two-part growth model test for qol and dg-time

Data: File is e:\mpltoprt.dat;
format is 26F12;

Data Twopart:
names = dg1-dg5;
binary = bin1-bin5;
continuous = con1-con5;

Variable:
names are code age racev insured sspre ss2m ss4m
ss9m ewbpre ewb2m ewb4m ewb9m
fwbpre fwb2m fwb4m fwb9m dg1-dg5
prt1m prt2m prt3m prt4m prt5m;
usevariables = dg1-dg5;
categorical = bin1 - bin5;
missing = .;
analysis: type = missing;
estimator = mlr;
model:

intu slpu | bin1@1 bin2@2 bin3@3 bin4@4 bin5@5;
inty slpy | con1@1 con2@2 con3@3 con4@4 con5@5;
slpu@0;
intu with slpy@0;


But getting an error message as:

*** ERROR in Variable command
Unknown variable in CATEGORICAL option:
bin1-bin5

I could not follow why binary variables are not recognized. Could you please help?

Thanks,
Abhik
 Linda K. Muthen posted on Tuesday, October 10, 2006 - 8:38 am
I think your USEVARIABLES statement should be:

USEVARIABLES = bin1-bin5 con1-con5;
 Steven Haas posted on Thursday, March 06, 2008 - 9:00 am
I am trying to compare nested two-part growth models with known latent classes- with and without equality of variances across classes. I am using the MLR estimator.

Would I compare the fit of these models using the Satorra-Bentler scaled chi-square factor or a standard likelihood ratio test?


Thanks
 Linda K. Muthen posted on Thursday, March 06, 2008 - 9:22 am
You need to use the scaling correction factor for MLR in the same way that you would with MLM.
 Steven Haas posted on Thursday, March 06, 2008 - 3:13 pm
Thanks. I have a second question. When choosing between two models I come to different conclusions if I use BIC and the scaled correction chi-square.

Details of the two models.

Both are two-part growth mixture with known classes. One allows variances to be free across classes and the other constrains the varainces to be equal.

Model 1 likelihood -50705 scaling factor .988 df=41 BIC 101785.

Model 2 likelihood -50715 scaling factor 1.094 df=29 BIC 101695.

The Satorra-Bentler is approx 26.00 which is significant at p-value .0106.

SB says model 1 is prefered BIC says model 2.

Any thoughts as to which test is better?
 Linda K. Muthen posted on Friday, March 07, 2008 - 6:19 am
I'm not sure that this is a fair comparison because the number of parameters in the two models is different and BIC has the number of parameters as part of its formula.
 Isaac Washburn posted on Friday, November 09, 2012 - 2:06 pm
I just wanted to check in to make sure that with Mplus 7 fit statistics have not been added for a two-part semicontinuous model. I have reviewers who are asking for them even after I told then mplus does not provide any.
 Linda K. Muthen posted on Friday, November 09, 2012 - 2:25 pm
When we do not provide chi-square and related fit statistics, it is because means, variances, and covariances are not sufficient statistics for model estimation. These fit statistics have not been developed for models where means, variances, and covariances are not sufficient statistics for model estimation. This has not changed.
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