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Please let me know whether I'm using the formulas correctly, found in "Difference Testing Using ChiSquare," http://www.statmodel.com/chidiff.shtml). cd = (d0 * c0  d1*c1)/(d0  d1) = (951 x 5.750  950 x 5.722)/(951  950) TRd = (T0*c0  T1*c1)/cd = (2728.083 x 5.750  2717.094 x 5.722)/cd COMPARISON MODEL Loglikelihood H0 Value 154318.940 H0 Scaling Correction Factor 5.722 for MLR H1 Value 151645.790 H1 Scaling Correction Factor 2.887 for MLR ... ChiSquare Test of Model Fit Value 2717.094* Degrees of Freedom 950 PValue 0.0000 Scaling Correction Factor 1.968 for MLR NESTED MODEL: Loglikelihood H0 Value 154323.099 H0 Scaling Correction Factor 5.750 for MLR H1 Value 151645.790 H1 Scaling Correction Factor 2.887 for MLR ... ChiSquare Test of Model Fit Value 2728.083* Degrees of Freedom 951 PValue 0.0000 Scaling Correction Factor 1.963 for MLR 


Yes, you are doing this correctly. 


Thanks for your confirmation. Let me ask a followup question. TRd was found to be 4.304958... Since the material says, "For MLM and MLR the products T0*c0 and T1*c1 are the same as the corresponding ML chisquare values," am I supposed to use 3.841 as critical value to determine whether the calculated SB scaled chisquare difference is significant at the level of .05 or not? That is, is the difference (4.304958...) significant since TRd > 3.841? 


I forgot asking another question. The equality constraint was imposed on a single parameter (which measures the effect of child maltreatment on violent offenses) for two ethnic groups, whites and Asian Americans. In the comparison model, the coefficient was found to be .031 (SE = .027) for whites, whereas it was .654 (SE = 1.430). As you can see, both coefficients are not significant, although the SB scaled chisquare difference is larger than 3.841. As I supposed to say the coefficient is significantly different between whites and Asian Americans even though the coefficient was found to be not significant in each ethnic group? 


1st post: Right. 2nd post: Each coefficient being significantly different from zero or not is not the same as testing that they are the same. Typically, if you use the independentsample z test of equality using your SEs, you get the same thing as the chi2. 


Hi Linda and Bengt, Step 1 on the Mplus website (http://www.statmodel.com/chidiff.shtml) for Difference Testing Using the Loglikelihood is: 1. Estimate the nested and comparison models using MLR. The printout gives loglikelihood values L0 and L1 for the H0 and H1 models, respectively, as well as scaling correction factors c0 and c1 for the H0 and H1 models, respectively. Does this refer to H0 and H1 values given for the SAME model (i.e., in the same output file); or for DIFFERENT models (estimated in separate runs, with separate output files)? I ask because while I have seen BOTH H0 and H1 values in some output files, I only see H0 for in a model I estimated using a NBI dependent variable, as seen below. There is no H1 value offered. Can I still utilize the steps on the website to compare the fit of this model with that of another nested model, using the H0 values only (the ones provided for each distinct modelbecause I did not get H0 and H1 values together in one output file). Thanks. MODEL FIT INFORMATION Number of Free Parameters 14 Loglikelihood H0 Value 1189.806 H0 Scaling Correction Factor for MLR 1.1902 


To do difference testing you need to run two analyses. The first is the least restrictive model. It is referred to as H1 in the writeup. The nested model is referred to as H0 in the write up. In both cases, the H0 values are taken from the output to use in the computations. 

EFried posted on Tuesday, April 02, 2013  1:55 pm



When comparing 2 models using the MLR estimator, each model provides 3 scaling correction factors and 2 loglikelihoods. I don't find it specified which one to use for model comparison (http://www.statmodel.com/chidiff.shtml). Thank you 


The one for the H0 model  what is posted above. 

ri ri posted on Friday, August 29, 2014  3:21 am



I would like to compare two Mediation models: model 1: xmy1y2y3 xmy1y4 model 2: xmy1y4 xmy2y3 y3 and y4 are binary. Both models seem to work in Terms of Mediation. SO I Need to prove one model is better than the other. Can I follow the instruction of doing the chi square test for WLSMV? I did not find any description regarding H0 and H1 models, scale correction info in Output. 


Looks like the only difference is that model 2 has a direct effect from m to y2. 

ri ri posted on Saturday, August 30, 2014  4:08 pm



Yes, as far as I know one Needs to do a chi square difference test to compare the two models. In regular way, one just uses the chi square values. But since I have categorical data, I shall do it differently I suppose? I used the difftest command, but could not find the scale correction to calculate the difference with the formula provided at the Website. 


With only one parameter difference you can just look at the ztest for that parameter in the model that is less restrictive. In the general case you use DIFFTEST, first running the less restrictive model and then the more restrictive model. You don't need the scaling correction factors or the computations on the website. DIFFTEST does it for you. 

ri ri posted on Tuesday, September 02, 2014  12:16 am



I tried the DIFFTEST to compare the contrained and uncontrained model, it worked wonderfully! Just I have another methodological question. In the user guide multiple Group Analysis, you wrote an example, Fixing the mean of the variables in Group 2 to Zero. If I compare constrained and unconstrained models, is it necessary to fix the mean to Zero? I also saw some People Center the means of the continous variables in order to minimize multicollinearity. If I compare two path models (such as the above mentioned model comparion), I wonder if mean centering is needed? Thank you very much! 


To compare means across groups, use the model with means zero in all groups versus the model with means zero in one group and free in the others. Centering is not needed. 

Ari J Elliot posted on Wednesday, January 07, 2015  7:44 pm



Hello Drs. Muthen, Regarding chi square difference testing with MLR, please confirm that the H0 scaling correction factor that should be used in the calculation is the one listed under Loglikelihood, NOT the one listed under the MLR chisquare test of model fit. Thus in the following output for the nested model, I would use 1.5862 as the scaling correction factor. Loglikelihood H0 Value 15770.576 H0 Scaling Correction Factor 1.5862 for MLR H1 Value 15733.949 H1 Scaling Correction Factor 1.5549 for MLR ............................. .....ChiSquare Test of Model Fit Value 50.133* Degrees of Freedom 5 PValue 0.0000 Scaling Correction Factor 1.4612 for MLR Thank you! 


If you use chisquare for the difference testing, you should use the scaling correction factor under chisquare. If you use the loglikelihood for the difference testing, you should use the scaling correction factor under loglikelihood. 

Ari J Elliot posted on Thursday, January 08, 2015  11:37 am



Ok thanks. To further clarify, the instructions on the webpage for difference testing using chisquare state: "Be sure to use the correction factor given in the output for the H0 model." Under the chi square I only see one scaling correction factor, whereas for loglikelihood there are correction factors provided for both H0 and H1. Given that correction factors for both the nested and comparison models are used in the calculation I'm not sure to what the reference to the H0 model in the instructions refers. 


You should use the H0 scaling correction factors. If only one is specified, it is the H0. Don't use the one for the baseline model. If you have further questions, send the output and your license number to support@statmodel.com and we can tell you which number to use. 

Cheng posted on Sunday, April 05, 2015  6:37 pm



Dear Muthen, I have a SEM model with these relationships (1) one latent and its three observed variables (2) Five observed variables that involved in path relationships with the latent in (1) (3) A correlation in two of the observed variables in (2). I use MLR estimator and would like to know the chisquare pvalue. I follow the instruction given in your website on difference testing on chisquare, to compute the scaled difference in chisquare. My question is for the H0 (restricted model), which relationships in (1)to(3) I need to constrain? Should I just constrain the path relationships in (2) which is my main interest? Our goal in this test is to get a non significant pvalue right? like the ML estimator's chisquare result for model fit test? 


These are choices the researcher has to make. You may want to discuss it on SEMNET. 

Cheng posted on Monday, April 06, 2015  5:51 pm



Thank you. 

Noa Cohen posted on Tuesday, June 23, 2015  5:26 am



I am trying to calculate the values for a chi square difference test with ML. However, the output does not include the scaling correction factor. What can I do? Thank you. 


With ML you don't need a scaling correction factor. 

Noa Cohen posted on Tuesday, June 23, 2015  8:31 am



Thank you. Other than that I should use the instructions that are stated on the website? because they do not specify ML. 


See pages 486487 of the user's guide. 

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