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 ... Chi-Square Test of Model Fit Value 2717.094* Degrees of Freedom 950 P-Value 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 ... Chi-Square Test of Model Fit Value 2728.083* Degrees of Freedom 951 P-Value 0.0000 Scaling Correction Factor 1.963 for MLR
Thanks for your confirmation. Let me ask a follow-up 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 chi-square values," am I supposed to use 3.841 as critical value to determine whether the calculated S-B scaled chi-square 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 S-B scaled chi-square 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?
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 independent-sample z test of equality using your SEs, you get the same thing as the chi2.
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 model--because 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).
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 z-test 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?
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 chi-square test of model fit.
Thus in the following output for the nested model, I would use 1.5862 as the scaling correction factor.
H0 Value -15770.576 H0 Scaling Correction Factor 1.5862 for MLR H1 Value -15733.949 H1 Scaling Correction Factor 1.5549 for MLR ............................. .....Chi-Square Test of Model Fit
Value 50.133* Degrees of Freedom 5 P-Value 0.0000 Scaling Correction Factor 1.4612 for MLR
If you use chi-square for the difference testing, you should use the scaling correction factor under chi-square. 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 chi-square 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 email@example.com and we can tell you which number to use.
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 chi-square p-value. I follow the instruction given in your website on difference testing on chi-square, to compute the scaled difference in chi-square. 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 p-value right? like the ML estimator's chi-square result for model fit test?
is there any information how the Satorra-Bentler scaled chi-square difference test is influenced by high Nīs (e.g. n > 30 000)? Is it influenced at all? I am investigating measurement invariance with large subpopulation-samples and complex survey data (students nested in teachers) and therefore I use the SB-chi-square to test nested models.