I have a copy of the instructions from your web site for chi-square difference testing using the Satorra-Bentler Scaled Chi-Square. I would like to perform such tests with nested models in the context of CFA. I am currently using EQS version 6.0. With this version, the robust estimation option provides the SB Scaled Chi-Square along with the robust CFI and the robust RMSEA measures of fit. This is my question. Is the MLM procedure in M-Plus 2 (maximum likelihood parameter estimates with robust standard errors and a mean-adjusted chi-square test statistic) the same estimation procedure as the robust procedure in EQS 6.0?
If this is so, then I can follow the instructions in the paper on your WEB site to calculate the correct values for the SB Scaled Chi Square for comparing nested models.
So, a traditional chisquare diff test results favours the Comparison Model, with chi-square-diff = 88.00 (df = 3). OK A SBS scaled diff test also favours the comparison Model, but following the calculations on http://www.statmodel.com/chidiff.html, the test-value is 675.14. It's so huge I really hesitate to report it. Am I doing something wrong?
Thanks for all help, -- Wim Beyers Belgium
bmuthen posted on Monday, September 30, 2002 - 9:42 am
It looks like you have done the calculations correctly according to our web site. The test value of 675 is certainly much larger than 88, but the p values are both very small and therefore similar so this could all be ok.
Two other thoughts:
You might have run into a local optimum in one of your 4 runs.
You can run this in Mplus to get the scaling correction factors directly to see if they agree.
Kaja LeWinn posted on Thursday, June 23, 2005 - 3:05 pm
Hello, We are trying to use the Satorra-Benter scaled chi-squared test using the instructions on this site. We have found that when we run our comparison model in MLM and ML we get different degrees of freedom (the same model, using two different estimation techniques gives us two different dfs). This does not happen in the nested model. We are confused by this. One characteristic of our model that might be of significance is that we are looking for measurement invariance using the grouping command. We were hoping this could be explained to us, as well as how we should approach the equation under point 3 of the online instructions (i.e. which degrees of freedom to we choose, those from the ML or MLM model). Thanks for your help, Kaja
I need to test out differences between nested models, and wonder if I can use the Satorra-Bentler scaled chi square in this particular case:
I have a model with a set of predictors that are all regarded as metric (although a couple of them are dichotomies). Some of them could be analysed as latent variables, but I have chosen to start out using simple sumscores. It is therefore simply a path analysis that I am doing. I have two outcome variables, the most dependent one is a dichotomy. The other one, which is also regarded as a mediator, is an ordered categorical variable. Since the sample is based on clusters, I am using the cluster option in order to adjust for the design effect.
I understand that the S-B formula is based on ML and MLM estimators. The programme, however, does not permit ML and MLM estimators with this analysis. It only provides WLSMV estimation.
Can my commands be changed in such a way that it allows for ML and MLM estimation? Or is there an alternative procedure that can be used for testing the difference between models?
In Chapter 15 under the ESTIMATOR option, there is a table that shows the estimators available for TYPE=COMPLEX and TYPE=TWOLEVEL. I'm not sure which you want to use for your clustered data. With the WLS estimator, difference testing is done in the usual way. With WLSM and MLR, you need to use a scaling correction factor which is given in the output and how to use it is shown on the website. With WLSMV, use the DIFFTEST option. See a description in the Mplus User's Guide. When you do not obtain chi-square, you can use -2 times the loglikelihood difference.
I'm performing a CFA with type=complex as independency assumption is violated in my data set [gathered within families].
I used the Satorra-Bentler scales chi-square corrections for MLR as described on the Mplus website.
My restricted model has an unscaled chi-square of 19.696 with df=13, and scaling correction factor of 1.123.
My less restricted model has an unscaled chi-sqaure of 10.704 with df=12, and scaling correction factor of 1.213.
Calculating the diff test scaling correction: (13*1.123-12*1.213)/1=.043
the corrected chi-sq diff test would then be: (19.696-10.704)/.043=209.116
I do not think this is correct. Can someone help me out?
bmuthen posted on Friday, February 03, 2006 - 9:47 am
It looks like you have done this correctly. The asymptotics of this correction does not always work out in small samples as has been noted by the authors, although note that the p value will be zero for both uncorrected and corrected chi-square difference testing. In Mplus Version 4 you will also have access to a Wald test which avoids these problems.
Also, to be clear -- if the test is not working -- is that attributable to some data characteristics or user error? I'm confident I'm doing the test correctly (as is comes out positive for about 50% of my comparisons). Thanks!
bmuthen posted on Monday, February 20, 2006 - 6:30 pm
The asymptotics of the test fails to kick in - this has been observed among the creators of the test (Satorra-Bentler) in several applications. No user error, and no fault of the data (apart from perhaps not having a large enough n). An alternative is to use a Wald test which is part of the upcoming Mplus Version 4 (see new announcement on the home page). This test is robust to the same type of violations as MLR/MLM.
Would the Wald result here be similar to what EQS calls the Lagrange Multiplier test?
For example if I were to take my model and constrain a covariance to one - would the Lagrange value associated with freeing that parameter give me a test of the difference between a comparison model and a model with that constraint (given that is the only constraint I'm testing?
Thanks again - great comments.
bmuthen posted on Tuesday, February 21, 2006 - 3:16 pm
No, Wald testing pertains to restrictions on a given H0 model, whereas LM tests (= Mod Ind) pertains to relaxing restrictions on a given H0 model.
Yes. The Wald test, however, would make it unnecessary for you to run that second run with your covariance restricted.
jennybr posted on Monday, February 27, 2006 - 6:56 pm
HI, I want to compute a chi-square difference test, however, my two df's are the same. Does this mean that my models are not nested?
I have run a multi-group (4 groups) CFA using LISREL. The output indicates that the factor loadings are equivalent across the four groups - only one set of loadings in the output. However, a chi-square difference test (Santorra-Bentler Chi-square) is significant - contrained model minus unconstrained model. Does this sometimes happen? If so, what is the explanation and how does one proceed?
This would indicate that constraining the factor loadings to be equal across groups significantly worsens the fit of the model. There must be some factor loadings that are not invariant across groups. You can read about testing for measurement invariance in Chapter 13 of the Mplus User's Guide which is on the website. It is at the end of the multiple group discussion.
I am searching for the formula for calculating degrees of freedom when using WLSMV, but cannot find the correct info in the latest User's Guide or in the Technical Appendices. I am using MPlus version 3.0 for this CFA. Can I interpret the degrees of freedom as accurate in the output?
It is formula 110 in Technical Appendix 4 on the website. The only interpretable value for WLSMV is the p-value for chi-square. The degrees of freedom are not calculated in the regular way. Difference testing of nested models can be carried out using the DIFFTEST option.
I know that when using the Satorra-Bentler chi-square (MLM), you must calculate a corrected chi-square difference test. It appears from the discussion above that this SAME procedure should be used for the Yuan-Bentler T2 test statistic (MLR). Is this true, or when using MLR (Yuan-Bentler T2 test statistic) is it possible to just use a regular difference test?
I would like to use the Chi-square difference test but I am unsure if my two models are nested. I would like to compare a CFA with conflict as 1 scale (items 1,2,3,4,5,6,7,8) and a CFA with conflict as 2 scales( task conflict items 3,6,7 and relationship conflict items 1,2,5). The last model has 1 df less. Could I consider these models nested? If so could I use the Satorra-Bentler Scaled Difference Test?
Thanks for your quick reply. I think I understand. So if I would make a second order factor (conflict) of the two first order factors (task conflict and relationship conflict)and compare this to a CFA of one factor (conflict) the models would be nested and thus I can use the S-B Scaled Difference Test?
A second-order factor model with two indicators is not identified. To be nested, each model would need to include the same set of dependent variables. I don't think if identified, your suggestion has that.
I ran a multilevel CFA with continous variables. I would like to compare two models: (1) 2-factors within and 1 factor between vs. (2) 2-factors within and 2 factors between. As much as I understand Model 2 is not nested within model 1, so I can't use the S-B Scaled Difference Test. Regarding my chi-square value and the fit indices I cannot decide which model fits better, because they are nearly the same. My question: Is there another way to test if they are different? If not, can I conclude both models are equivalent although I couldn't really prove it? Thanks very much! Janine
BIC could be helpful to balance parsimony against improving the loglikelihood. If the fit information is about the same, I would decide based on the interpretation and usefulness of the model; I wouldn't be so concerned about testing. If the model with 2 between factors doesn't add interpretational value, parsimony speaks for the 1-factor model on between.
I'm doing a chi-square difference test for MLR. I get a negative cd-value (=-.016). I think this is due to the complex model (M1: df=1085 M2: df=1086, M1: Scale Cor=1.070 M2:1.069). It follows that also the scaled chi-square difference is negative (-240,639). Is this result interpretable?
How should I interpret it?
Do I have to use the modulus of cd? Or does it mean, that the model with df=1086 fits better??
I have the same problems with MLR-testing as described above (negative chi-square). In this situation, is it o.k. to use the Wald-test in combination with MLR and COMPLEX/CLUSTER in order to test some simple 1df constraints?
Dear Drs. Muthen, I am trying the formula for positive SB-difference statistic values at invariance testing across groups (3). since the method requires estimating a M10 model, with the M0 output as starting values, I do not know which starting values to use in case of several groups. can you help me further, please?
I am trying to run a fully unconstrained multiple group CFA model to compare to my constrained model, but I am not able to figure out the syntax for this. For the constrained model, I am using the syntax below. What do I need to add to make it unconstrained? Thank you!
Title: CFA; low.hi; current; King DATA: FILE IS low.hi.current.csv; VARIABLE: NAMES ARE b1-b5 c1-c2 c3-c7 d1-d5 g; CATEGORICAL ARE b1-b5 c1-c2 c3-c7 d1-d5; GROUPING IS g (1 = low 2 = high); MODEL: f1 BY b1-b5; f2 BY c1-c2; f3 BY c3-c7; f4 BY d1-d5;
Thank you for your prompt response! I realize this is probably a really easy question, but the syntax I posted above - is this for a fully unconstrained model and then syntax on slides 169 and 170 for a fully constrained model?
The syntax above is for a model with factor loadings and intercepts constrained to be equal across groups. This is the Mplus default. See Chapter 14 of the user's guide for a full description of multiple group analysis in Mplus.
On Slides 169 and 170, the first syntax is for a fully constrained model. The second is for an unconstrained model. And the third is for a partially constrained model.
Thanks again. We got a bit closer, but now are running into this problem -
When the fully unconstrained model is run, we're getting error messages like this:
THE MODEL ESTIMATION TERMINATED NORMALLY
THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL. PROBLEM INVOLVING PARAMETER 81.
THE CONDITION NUMBER IS -0.244D-17.
We're not able to get fit statistics because the parameter estimates couldn't be computed. When I check the output, I can find out that, in this model, parameter 81 is d5. So then I delete the line of code freeing d5 the model will run. I'm not sure what the condition number is yet. Can you tell us what the condition number indicates?
Ian Clara posted on Tuesday, December 07, 2010 - 6:39 am
Good morning. I am trying to conduct a chi-square difference test using the WLSMV estimation and type=complex. I have used the difftest option and for the two models that I want to test it says that they are not nested (although there is a single path that is different -- set to zero in one model and freely estimated in the other). I wanted to try to conduct the chi-square difference test by hand, but I can't determine how to get the required output in Mplus v5. How can I obtain the scaling correction factors, or the log likelihood?
Difference testing using WLSMV cannot be done by hand. If you use WLSM, you will obtain a scaling correction factor.
Fred Mueller posted on Wednesday, February 22, 2012 - 6:08 am
Dear Linda and Bengt,
I would like to compare two nested models and I am using MLR as an estimator. Instead of using the (Santorra-Bentler Scaled) Chi-Square difference test, I would like to use the small difference in fit test by MacCallum, Browne, and Cai (2006, Psych Methods). For this test, I also have to indicate the Chi-Square of both models. How do I have to proceed? Can I just multiply the (regularI Chi-Square with the scaling correction factor to get the correct Chi-Square or is it more complicated?
i am comparing 2 measurement models for 18 items using MLR. It would be great if you could help me with some answers:
The first model is a one-factor model where all items are explaining by the same factor and the second model is a three factor model where each factor explains 6 items.
(1)Is it right to say that these models are nested as the 3-factor-model with the correlations between the factors set to 1 would equal the 1-factor-model?
(2) When i compute the S-B scaled Chi square difference test, the result is negative. Therefore i tried computing the strictly positive S-B scaled Chi square difference test. Thus i tried to estimate the more strict M0 model first which would be the 3-factor-model with between-factor-correlations set to one. However, the estimation did not converge. What am i doing wrong? Heres the input:
MODEL: factor1 by f10-f16; factor2 by f20-f26; factor3 by f30-f36; factor1 WITH factor2 @1; factor2 WITH factor3 @1; factor3 WITH factor1 @1;
Yes, but from the examples in the Webnote describing the strictly positive S-B Test i thought in order to compare the models i would need the same notation in the syntax (i.e. 3 correlations set to one instead of just "factor1 by f10-f36") so that i can save the start values for the M10 Model and then free the correlations between the factors.
If you are using the MLR estimator with categorical data you should use the unscaled likelihood ratio test. The S-B is designed to be used for the case when you are treating the variables as continuous.
Strictly speaking there is a bit of a problem in using LRT for this purpose (overfactoring) see
Hi I have been working with ESEM models with MLR estimator. To compare nested models I have been using S-B x^2 and it has worked fine until I got a negative estimate when comparing two models. I found your web note on how to compute the strictly positive S-B x^2 but when I tried to run the M0 model with the svalues command to get the starting values for the M10 model I got the following warning text in Mplus:
*** WARNING in MODEL command The SVALUES option in the OUTPUT command is not available with the use of EFA factors (ESEM). Request for SVALUES will be ignored.
You can just use the final results in the M0 output and put these values (manually) as starting value for M10 run. The SVALUES option is just a convenience feature that does this for you but you can do it manually as well.
The strictly positive S-B chi-square is not very easy to do with ESEM. You have to work with the unrotated model. We may eventually update the web note to include a step by step computation for the strictly positive S-B for ESEM. Consider using Model Test as a simpler alternative.
Send the outputs from the m0 and m1 analyses along with your attempt at the m10 model. Include your license number.
Sabrina posted on Saturday, April 27, 2013 - 3:04 pm
Hello. I computed the chi-square difference test for 2 nested models and it was negative so I followed the instructions in WebNote 12, but I received the following error message and can't figure out what to do next?
THE MODEL ESTIMATION TERMINATED NORMALLY
THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL. PROBLEM INVOLVING PARAMETER 131.
THE CONDITION NUMBER IS -0.494D-04. THE ROBUST CHI-SQUARE COULD NOT BE COMPUTED.
I have a longitudinal cross-lagged path model with four outcomes over three assessment points. Three of the outcomes are continuous and the fourth is a count, so I am using MLR. I also am using the knownclass method to compare invariance across males and females. My analysis command looks like this:
I want to test whether two parameters in the model are equal to each other (although they are invariant across sex), so I constrained them to be equal and compared this to a model where they were free. However, I get a negative scaled chi-square difference value. I read webnote #12 and tried to estimate model 10, but when I pasted the start values from model 0 in, the new model does not converge. I'm sure I'm doing something wrong but am not sure what.
Use MODEL TEST. Label the parameters in the MODEL command.
0 - p1 -p2;
jtw posted on Friday, September 20, 2013 - 5:52 am
I am comparing the fit of nested models (i.e., bifactor, correlated traits, higher-order, uni-dimensional) using difftest. I have clustered data and obtain different conclusions when type=COMPLEX is used as compared to when I do not adjust for clustering. Specifically, the bifactor model is preferred when I do not account for clustering, whereas the correlated traits model seems to be preferred when I account for clustering.
It may be helpful to know that I obtain all chi-square difference test statistics when not adjusting for clustering. However, I do not obtain the chi-square the difference test statistic for the correlated traits model when adjusting for clustering (because the chi-square value in this case is actually lower than the value for the bifactor model; i.e., the models are not technically nested). Also, it is worth noting that there are no real substantive differences with respect to any of the factor loadings in either case; so, the only issue at hand is which model is preferred.
Any recommendations of which results should be reported in this case? Thank you in advance for your time.
I understand nested models require two things: 1) more restrictions (higher degrees of freedom) AND 2) worse fit (which can be examined via Tech 5 output).
When a model is more restricted yet has slightly better fit, MPlus does not execute the DIFFTEST and returns the error indicating that the models are not nested. In this specific situation, is the conclusion to be drawn: 1) the more restricted model (e.g., correlated traits) is preferred over the less restricted model (e.g., bifactor model); or 2) no firm conclusions should be drawn because the DIFFTEST is not executed. I'm thinking the answer is #1. Am I right? However, if it is #2, what can be done?
You cannot compare the chi-square values of the estimators that end in V, WLSMV and MLMV. You would need to compare their fitting functions which you will find in TECH5 at the bottom of the first column. The lower the fitting function the better.
The conditions that you mention are necessary but not sufficient conditions for nesting. Please send the outputs and your license number to firstname.lastname@example.org for further information.
I have a longitudinal cross-lagged path model with six outcomes over three assessment points. All outcomes are continuous and non normally distributed, so I used MLR. I am also using the grouping method to compare invariance across males and females, therefore doing Satorra-Bentler scaled chi-square difference testing. Of the 141 parameters tested for invariance, 19 resulted in negative values. I read the articles and web notes about computing the strictly positive Satorra-Bentler chi-square difference test and I have a few questions:
1- When I use the SVALUES option to get all the final parameter estimates from model M0, estimates regarding the structural links (the "ONs") are not provided. Is that normal? Does this mean we do not have to include these estimates in the model M10?
2- I read that we must halt iterations in model M10 (i.e., the number of iterations have to be equal to 0) and that we have to increase convergence in order to do so. I did that and looked at the TECH5 part of the output to verify that iterations are null. I was wondering which type of iterations to look at? Is it the Quasi-Newton type? These seem to be at 0 whereas the other ones (for instance, EM algorithm iterations) never are at 0.
Kaisa Perko posted on Monday, January 12, 2015 - 1:25 am
Hello I'm a novice in SEM and would highly appreciate any expert advice. I'm struggling with a discrepancy between the chi-square test and a path coefficient.
I'm comparing two nested models with the scaled Satorra-Bentler chi-square difference test. In the restricted model, two paths between latent variables (from one IV to two DVs) are fixed at zero. In the unrestricted model, these paths are freely estimated.(For information, there are also other IVs in the model).
The scaled chi-square test favors the restricted model (nonsignicant result). However, in the unrestricted model, one of the freed paths is significant (p<.05) . Thus, the chi-square test seems to reject a significant path. Accordingly, the conclusions are different depending on whether I estimate all the hypothesized paths at once, or step by step using nested models. I wonder if I'm making a mistake if I follow the chi-square test and accordingly consider the restricted model as the final model?
These tests can give different results, particularly when the sample isn't large. As a check, you also want to use the Wald test of both coefficients being zero - you do that using the Model Test command.
Kaisa Perko posted on Tuesday, January 13, 2015 - 4:34 am
Thank you very much for responding.
The sample size is 549. Wald test yielded a significant result (p=0.0365) for the path which is significant in the unrestricted model, and a nonsignificant result for the path which is nonsignificant in the unrestricted model.
As the discrepancy with the scaled chi-square remains, may I ask your opinion, would I stay on safer ground if I 1) follow the scaled chi-square test which suggests the restricted model as the final one or 2) adopt a "all at once" approach (all paths freely estimated) and ignore the chi-square test? My aim is to evaluate whether one of the IVs is redundant to the other IVs, and the conclusions differ between alternatives 1 and 2.
You should test both paths jointly when doing the Wald Model Test. If you only do one at a time you are not doing anything differently than the printed z test (chi-2 = z*z).
Kaisa Perko posted on Wednesday, January 14, 2015 - 5:29 am
Of course, thank you. Testing both paths jointly yielded a nonsignificant result in accordance with the scaled chi-square difference test. I additionally found out that also the chi-square difference test suggests the unconstrained model if I fix and free the single path instead both of the paths. So the original discrepancy appears more understandable now, and I need to rethink my analysis strategy against this background.