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Hi Bengt and Linda, I'm testing for gender differences an autoregressive crosslagged 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 chisquare of the model with free estimates has a higher chisquare (and of course fewer Df) than the one with the fixed parameters, resulting in a negative chisquare difference value. Is a negative chisquare difference acceptable? And if yes, shall I report the value with the negative sign? Thanks. 


If you are using MLM or MLR, this can happen and the results are not meaningful. 


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 chisquare 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 


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 chisq test cannot be interpreted, what are alternatives to test fixing just one parameter in my model? thanks! 


You can use MODEL TEST. See the user's guide for further information. 


Hi I have the same negative chisquare difference problem using ML. Thank you 


I have a similar situation regarding Christina(the first poster in this thread) for my multigroup path analysis comparisons, using MLR, and getting a negative chisquare 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! 


You give parameter labels for your unconstrained model in each group. Then you say like: Model Test: 0 = a1a2; 0 = b1b2; 0 = c1c2; where 1 refers to group 1 and 2 refers to group 2 and a, b, c refer to 3 different parameters. 


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=a1a2; 0=b1b2; 0=c1c2; 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! 


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 chisquare 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 chisquare 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=a1a2; 0=b1b2; 0=c1c2; Thank you again for your clarification. 


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! 

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