Negative chi-square difference PreviousNext
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 Cristina Sturaro posted on Monday, June 16, 2008 - 6:02 am
Hi Bengt and Linda,

I'm testing for gender differences an autoregressive cross-lagged model using a multiple group model (boys vs. girls). First I let the parameters free for boys and girls. Then I rerun the model with the parameter estimates equal between boys and girls. As a result, the chi-square of the model with free estimates has a higher chi-square (and of course fewer Df) than the one with the fixed parameters, resulting in a negative chi-square difference value.

Is a negative chi-square difference acceptable? And if yes, shall I report the value with the negative sign?

Thanks.
 Linda K. Muthen posted on Monday, June 16, 2008 - 9:46 am
If you are using MLM or MLR, this can happen and the results are not meaningful.
 Cristina Sturaro posted on Wednesday, June 18, 2008 - 12:45 am
Hi Linda,

thank you for the fast reply.

Could you please clarify me something?

What did you mean by "the results are not meaningful"? Does it mean that the chi-square test is not valid? Or that the differences I've found between boys and girls in the free estimated model are not significant and are the results of the reduced sample sizes of model male and female? Or maybe something else?

Thank you
 Linda K. Muthen posted on Wednesday, June 18, 2008 - 8:49 am
Yes, the test is not working correctly so it should not be interpreted.
 uclaalice86 posted on Wednesday, September 17, 2008 - 10:20 pm
Hi,
I am running into the same issue as mentioned above while using MLR. If the chi-sq test cannot be interpreted, what are alternatives to test fixing just one parameter in my model?
thanks!
 Linda K. Muthen posted on Thursday, September 18, 2008 - 6:33 am
You can use MODEL TEST. See the user's guide for further information.
 Ellen Laupper posted on Friday, December 11, 2015 - 4:36 am
Hi

I have the same negative chi-square difference problem using ML.



Thank you
 Ashley Baker posted on Tuesday, October 11, 2016 - 5:29 pm
I have a similar situation regarding Christina(the first poster in this thread) for my multi-group path analysis comparisons, using MLR, and getting a negative chi-square diff test.

I understand how to use model test for comparing one path at a time , but how would I go about comparing my unconstrained model to my partially constrained model using the model test command? Looking at the user guide, I am still not clear about this.

For example:
Unconstrained
t2a ON t2d;
t2a ON t2p;
t2a ON t2pm;
t2d ON t2s;
t2d ON t2p;
t2s ON t2p;
t2p ON t2m;

Constrained Model
t2a ON t2d;
t2aON t2p(1);
t2a ON t2pm;
t2d ON t2s(2);
t2d ON t2p(3);
t2s ON t2p;
t2p ON t2m;

Thank you!
 Bengt O. Muthen posted on Tuesday, October 11, 2016 - 5:41 pm
You give parameter labels for your unconstrained model in each group. Then you say like:

Model Test:

0 = a1-a2;
0 = b1-b2;
0 = c1-c2;

where 1 refers to group 1 and 2 refers to group 2 and a, b, c refer to 3 different parameters.
 Ashley Baker posted on Tuesday, October 11, 2016 - 6:14 pm
Hi Bengt,

Thanks for your quick response. My apologies, I am still not quite understanding. This is what I did:
MODEL: !unconstrained
t2a ON t2d;
t2a ON t2p(a1);
t2a ON t2pm;
t2d ON t2s(b1);
t2d ON t2p;
t2s ON t2p(c1);
t2p ON t2m;

MODEL: !partial
t2a ON t2d;
t2aON t2p(a2);
t2a ON t2pm;
t2d ON t2s(b2);
t2d ON t2p(c2);
t2s ON t2p;
t2p ON t2m;

MODEL TEST:
0=a1-a2;
0=b1-b2;
0=c1-c2;
And I received this warning: "The following parameter label is ambiguous. Check that the corresponding parameter has not been changed. Parameter label: A1". Thanks again for the help!
 Bengt O. Muthen posted on Wednesday, October 12, 2016 - 9:56 am
You have to give parameter labels in each of your groups. So use this style:

Model male:

- labeling

Model female:

- labeling
 Ashley Baker posted on Wednesday, October 12, 2016 - 1:02 pm
Hi Bengt,

Thank you for your response and that worked for me.

My apologies for my questions (applied researcher, here), but I am unclear on how the syntax below is comparing the partially constrained model to the unconstrained model? Before I ran the constrained model and unconstrained models in separate runs, and then calculated out chi-square differences using the loglikelihood information, but had fit indices for both. Using model test I only receive fit indices for the constrained model. Are these two approaches the same, except for that the model test command helps to correct for the negative chi-square difference?


USEVARIABLES = t2m t2p t2s t2d t2a t2tran t2pm;
GROUPING IS t2tran (0=PED 1=ADULT);
MISSING ARE ALL (-999);

ANALYSIS: TYPE IS general;
ESTIMATOR = MLR;

MODEL:
t2a ON t2d;
t2a ON t2p;
t2a ON t2pm;
t2d ON t2s;
t2d ON t2p;
t2s ON t2p(;
t2p ON t2m;

MODEL PED:
t2a ON t2d;
t2a ON t2p(a1);
t2a ON t2pm;
t2d ON t2s(b1);
t2d ON t2p;
t2s ON t2p(c1);
t2p ON t2m;

MODEL ADULT:
t2a ON t2d;
t2aON t2p(a2);
t2a ON t2pm;
t2d ON t2s(b2);
t2d ON t2p(c2);
t2s ON t2p;
t2p ON t2m;

MODEL TEST:
0=a1-a2;
0=b1-b2;
0=c1-c2;

Thank you again for your clarification.
 Bengt O. Muthen posted on Wednesday, October 12, 2016 - 4:12 pm
It looks like you set it up right. The two approaches provide the same testing, just two different statistical approaches - they are the same in large samples. You are right that the Model Test approach doesn't give the overall fit for the constrained model.
 Ashley Baker posted on Wednesday, October 12, 2016 - 8:58 pm
Thank you for all the help!
 Alaine Garmendia posted on Thursday, June 21, 2018 - 8:42 am
Hi,
I am conducting cross-lagged relationships between two variables (A and B).
I would like to compare four models: the stability model, the standard causality model (from A to B), the reverse causality model (from B to A) and the reciprocal causality (from A to B and from B to A).
I am conducting the analysis with MLR estimator and I calculated the Chi Square difference with the Satorra and Bentler formula. However, the difference is negative and thus, I cannot calculate the p value. Does this mean that there is not a significant difference between the models? I did the same analysis with ML estimator and the p values are not significant. I guess this means that model fit does not really improve from one to another.
Is ML trustworhty or I should stick to MLR?
Thank you very much
 Bengt O. Muthen posted on Thursday, June 21, 2018 - 3:42 pm
Q1: First make sure the models are nested. See our new NESTED option in Version 8.1. You can read about it in the Version 8.1 Mplus Language Addendum at

http://www.statmodel.com/ugexcerpts.shtml

Q2: If the MLR SEs aren't too different from those of ML, perhaps the degree of non-normality that you have doesn't have much effect and you can use ML.
 Alaine Garmendia posted on Friday, June 29, 2018 - 1:36 am
Thank you very much for your responses.
Related to the nested option, there is not such an option for version 8?
I am using the same variables in the four models, the only difference between the models is that I add more regression paths from one to another (cross lagged relationships) and I ask Mplus to calculate those parameters. Would those models be nested?
For example, in the simplest model, in the stability model I only add the auto-regressive Betas (A Time 2 on A time 1) and B time 2 on B time 1). Should I add the cross lagged relationships between variables and constrain them to be 0? For example, B time 2 On A time 1=0).
My doubt is whether I should include the same paths in the four models in order to have nested models and constrain them to be 0 when I do not want to estimate them.
Thank you very much again
 Bengt O. Muthen posted on Saturday, June 30, 2018 - 6:52 am
Q1: The NESTED option is new and introduced for version 8.1.

Q2: Yes.

Q3: Not needed; they are still nested. You can check that you get the same result both ways.

You can also use Model Test. That is, include the cross-lagged effects (A2 on B1 etc) and test if they are significant.
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