Message/Author 

S. Hoult posted on Friday, December 28, 2001  3:52 pm



I am attempting to obtain chisquare difference tests of nested models using the SB scaled chisquare. I see the formulas for calculating this difference test, but want to clarify what is meant by comparison model. Is this (or can this be) be an independence model? In this case, then, would I specify the idependence model (where the covariances of all ksi=0 and the error variances for all observed vars=0) for each subgroup and use the ML and MLM chisquare results from this as my comparison model for each nested model? Thank you for any assistance. 

bmuthen posted on Wednesday, January 02, 2002  12:00 am



The comparison model is an "H1" model, that is a model that is less restrictive than the model you are considering. 

Anonymous posted on Friday, February 18, 2005  6:51 pm



I apologize for the seemingly silly question, but ... I am trying to compute a chisquare difference test using the SatorraBentler scaled chisquare. Step 4 requires the use of the "regular chisquare values" T0 and T1. Where are these values on the output? Then, once step 4 is calculated, is the resulting number the significance of the chisquare difference test? Thank you for your help. 

bmuthen posted on Friday, February 18, 2005  11:10 pm



Regular chisquare values are those using the ML estimator. The calculations give a chisquare value and its df  you have to look up the significance. 

Anonymous posted on Saturday, February 19, 2005  12:16 am



Okay. If I may followup ... I am using the MLR estimator for a TYPE=COMPLEX MEANSTRUCTURE MISSING analysis. If I rerun with the ML estimator, I get an error message that it defaults to MLR due to the COMPLEX command. If the COMPLEX command is removed, I get an error message that it cannot run with my CLUSTER variable. If I remove the CLUSTER variable, is it still correct to compare the ML chisquare with the MLR chisquare? Thank you! 

bmuthen posted on Saturday, February 19, 2005  5:13 am



Stay with the MLR estimator. The MLR chisquare multiplied by the scaling correction factor is the ML chisquare. 

Anonymous posted on Wednesday, February 23, 2005  10:07 pm



I am trying to compute a chisquare difference test using the SatorraBentler scaled chisquare. You said that the calculations give a chisquare value and its df  you have to look up the significance. I would like to make sure that is the TRd value a chisquare value? There is no df value in your step 4 and how do we get the df value then to look up the significance? Thank you! 

bmuthen posted on Saturday, February 26, 2005  10:50 pm



Yes, Trd is the chi square value. The df is the usual difference in df's for the h0 and h1 models you are considering. 

Anonymous posted on Wednesday, March 02, 2005  1:05 am



I would like to know how to cite your instruction of "Chisquare difference testing using the SB scaled chisquare" on this website? http://www.statmodel.com/chidiff.html Thank you for your help 


Something like the following should work: Muthen, L. and Muthen, B. (2005) Chisquare difference testing using the SB scaled chisquare. Note on Mplus website, www.statmodel.com. 

Anonymous posted on Friday, March 04, 2005  1:47 am



I am trying to compute a chisquare difference test using the SatorraBentler scaled chisquare. Now, I got TR0>TR1, which is expected. but I have problem with T0<T1 that makes negative TRd. I really would like to have your advice because I am stuck here. I used Normal Theory Weighted Least Squares ChiSquare in ML estimation for the T values and SB chisquare in DWLS estimation for the TR values. Thank you so much for all your responses. 

bmuthen posted on Friday, March 04, 2005  12:16 pm



Are you saying that your regular ML chisquare for the H0 model fits better than that for the H1 model? If so, it doesn't seem that H0 is nested within (is a special case of ) H1. Or am I misunderstanding something? 

Anonymous posted on Friday, March 04, 2005  11:20 pm



Yes, the regular ML chisquare for the H0 model fits better than that for the H1 model. But why the SB chisquare for the H1 model fits better than that for the H0 model? My problem is T0<T1 H0: 152.90(TR0) 500.40(T0) 128(df0) H1: 149.77(TR1) 500.44(T1) 125(df1) Then, TRd became negative C0: 3.27 C1: 3.34 cd: 0.41 TRd: 0.10 Thanks very much for your time. 

bmuthen posted on Saturday, March 05, 2005  12:21 am



If you are saying that you have an ML analysis with chisquare (cs) = 500.40 with 128 df for an H0 model that is nested within an H1 model with cs=500.44 and 125, then I would guess that the cs difference is due to numerical imprecision and that the extra restrictions are completely harmless. But to be able to tell for sure, send you input, output, and data to the Mplus office at support@statmodel.com and give your license number. 

Anonymous posted on Saturday, March 05, 2005  11:09 pm



Thanks Bengt. Does your instruction for computing chisquare difference test using the SatorraBentler scaled chisquare apply to LISREL package? Because I am using LISREL for this project and will learn how to use Mplus after this one. Thank you so much for your responses. 


If you are using the correct chisquare, then it should apply. 


Bengt, I too have occasionally run into situations where the model is nested and this same problem with the cs difference arises. I see your note above regarding numerical imprecision, i.e., that the additional constraints are harmless. My interpretation has been similar to yours with regard to the constraints, but I've not articulated the numerical imprecision hypothesis that you note. I find it logical, etc., However, I'm wondering if there is a citation that might discuss the numerical precision of the cs, etc., so that I could more forcefully make this argument in publications? Best, Adam 

bmuthen posted on Monday, March 07, 2005  4:42 pm



I think you want to avoid the T0, T1 reversal simply by changing the convergence criterion, making it stricter until T1 is no larger than T0  that's the imprecision that I had in mind. 


Bengt, Perhaps on the West coast you heard the head slapping "Duh" that I performed here in DC in response to your post. Increasing the convergence criterion did indeed resolve the issue, with a resulting cs that was not significant, much as we both suspected. For those following the discussion, it may also be necessary to increase the iterations in order to reach the convergence criterion. Thanks Bengt. Best, Adam 

Anonymous posted on Monday, March 14, 2005  10:55 pm



Hi Bengt, I have got T0>T1 and TR0>TR1. However, d0*c0d1*c1<0 that makes cd<0. Thus, TRd becomes negative. Could I have your suggestions on this problem? Thank you. 

BMuthen posted on Tuesday, March 15, 2005  12:44 am



This has been discussed in the literature by Bentler and Satorra. It is a failure of the asymptotic approximation. There is nothing you can do. You cannot use the test in that case. 

Anonymous posted on Tuesday, March 15, 2005  2:42 am



Bengt, You said that the regular chisquare values are those using the ML estimator. I would like to make sure which one should we use: Normal Theory Weighted Least Squares ChiSquare Or Minimum Fit Function ChiSquare? Thank you for your responses. 

bmuthen posted on Tuesday, March 15, 2005  4:24 am



Normal theory. 

Anonymous posted on Tuesday, June 14, 2005  4:46 pm



Just to clarrify my understanding of the DIFFTEST option. If I follow the example on pg 278 of the users manual I am testing a less restrictive model (H1)versus a more restrictive model (H0)and a nonsigificant chisquare tells me that the restrictive model does not improve the fit of the model whereas a signficant chisquare tells me that the less restrictive model better fits the data. So if I were looking to compare models and I thought the more restrictive model to be more interpretable, I would like to see a nonsigificant chisquare? 

Anonymous posted on Tuesday, June 14, 2005  11:51 pm



Hi Bengt, Back to the problem that I had in Chisquare difference test using SB chisquare. I encountered T0>T1 and TR0>TR1. However, d0*c0d1*c1<0 makes cd<0. You have said that this has been discussed in the literature by Bentler and Satorra. It is a failure of the asymptotic approximation. Do you have the reference of Bentler and Satorra's discussion on this problem? Thank you 


Re 10:46 You want the more restrictive model not to worsen the fit significantly for it to be the accepted model. 

Barbara posted on Wednesday, October 05, 2005  6:21 am



When TRd is negative and this test can not be used to compare models, how should we compare models in this case? Thank you 

bmuthen posted on Saturday, October 08, 2005  5:53 pm



I don't think there is a good answer to that. You can accept the rough approximation of treating the variables as normal using the ML estimator and do a usual chisquare difference test. But this may not be very accurate given strong nonnormality. 

Chan Wai Yen posted on Thursday, December 15, 2005  12:41 am



Hi Bengt, Regarding the SB scaled ChiSq test; I am using TYPE = COMPLEX MGROUP to test for factor and structural invariance. You recommended in one of your post to use the ChiSq from MLR, and multiply it by the correction factor produced in the output to get the ML ChiSq. May I clarify with you regarding the df. Should I also use the df from MLR, or should I run another analysis with ML and use that df? Thanks. 


You can use the degrees of freedom from MLR. 


I have a question concerning Multigroup Analysis and scaled ChiScare testing: I am trying to test in how far my fit improves after relaxing some correlations and allowing them to differ across groups. If I follow the formulas given on the mplushomepage for ChiSquare difference testing (calculating MLM and then ML ChiSquares and dividing them) I obtain very different correction factors that in the MPLUSoutput when running MLM. How do I calculate the correct Difference Test in this case? From Chan Wai Yen's posting above I would assume that I multiply the MLMChisQ with the Scaling correction factor to obtain my MLChisqvalues and then use those in the same way as usually, but I am not entirely sure whether I am right about that. I hope my problem is understandable and ...Thank you. 


What you say in the second paragraph is correct. If you need further help, please send your outputs, computations, and license number to support@statmodel.com. 

Nina Zuna posted on Wednesday, September 27, 2006  2:31 pm



Dear Drs. Muthen, I have satisfactorily conducted the MLR difference test utilizing your website instructions to test for difference in measurement model vs. 2nd order model. My change in chisq. was ChgYBX2=5.859, p=.053. 1. Is the YB chi sq. (and change YB chi sq) also impacted by sample size (n=566), thereby having the same impact as traditional chisq (usu. signif with large Ns)? When I compared the CFI, TLI, RMSEA, and SRMR from MLR meas. model with MLR 2nd order model I noticed very little change.001 less in CFI and SRMR with 2nd order model; no change was observed in TLI and RMSEAwere exactly the same. 2. Given the borderline signif, am I still safe to use other fit statistics to determine if the 2nd order model explains the data as well as the lower order meas. model? Since my readings have indicated that rescaled chisqs. are not distributed like chisqs. I wasn't certain if I could make the same assumption about sample size impacting pvalue or whether if it was advisable to use other fit statistics in making this determination. Thank you for any advice you may have on this matter. Sincerely, Nina 


1. Yes. 2. Yes. 

Nina Zuna posted on Sunday, October 01, 2006  5:26 pm



Thank you! 


I would like to test whether the fit of my model improves significantly when I add a quadratic term. I am using MLR. In order to do difference testing, my models must be nested and thus must include the same set of observed variables. My comparison (quadratic) model is: i s q  x1@0 x2@2 x3@4 x4@6 x5@8; Would my nested (linear) model be: i s q  x1@0 x2@2 x3@4 x4@6 x5@8; q ON x1@0 x2@0 x3@0 x4@0 x5@0; q@0; Thank you for your help 


A good way to see if a quadratic term is needed is to simply estimate the quadratic model i s q  x1@0 x2@2 x3@4 x4@6 x5@8; q@0; and see if the mean of q is significantly different from zero. The model you specified, Would my nested (linear) model be: i s q  x1@0 x2@2 x3@4 x4@6 x5@8; q ON x1@0 x2@0 x3@0 x4@0 x5@0; q@0; should not be used (it is not what you want). 


Thank you for your reply. So to clarify, I do not need to do chisquare difference testing to determine if the model fits better with a quadratic term? If I specify the model as you suggested above, and the mean of the quadratic term is indeed significant, what exactly does that mean? The model with a quadratic term fits significantly better than the linear model? Sorry for the elementary questions. 


When two models differ with respect to only one parameter, in this case the quadratic growth factor mean, the z test addresses the same question as the chisquare test you get from the loglikelihood difference. Squaring the z value you get the chi square value. Yes on your second question. 


Thank you very much for your quick reply. This discussion board has been a life saver! 


Hi I would like to conduct a chisquare difference test for a multigroup CFA model with correlated errors. All my variables are categorical. As far as I know, the MLR estimator cannot work when there are correlated errors in the model, and MLM does not work for noncontinuous variables. I tried to estimate the model using ML, but I get an error message that says "algorithm = integration is not available". How can I possibly estimate my models using ML? Do I have any other options besides MLR, MLM, or ML? Thank you 


You can include residual covariances in a model that is estimated using maximum likelihood. How to do this is shown in Example 7.16. Each residual covariance requires one dimension of integration. We recommend no more than four dimensions of integration. Alternatively, you can use weighted least squares where residual covariances are easily estimated. If you use WLSMV, you can use the DIFFTEST option for difference testing. 

Guy Cafri posted on Friday, November 07, 2008  6:32 pm



Hi, I am running a CFA with nested data (n=1089) with 100 clusters using the MLR estimator. I want to compare nested models but get negative cd values, and in turn a negative SatorraBentler chisquare difference test statistic. Here are my results: comparison: chisquare=403.22 df=83 scaledcorrection=1.18 nested: chisquare= 682.99 df=84 scaledcorrection=1.15 cd=1.34 SB test statistic=208.78 Am I doing something wrong with the calculations or is it as you noted in responses to others, the test fails with small sample sizes? If so, is there an alternative? You mentioned the wald test in passing. 


This test is asymptotic and can fail in a given sample. 2 ways out of this:  look up UCLA Statistics Series preprints where a new SatorrBentler paper describes how to use a modified version when negativity happens  since you have only a 1df difference, you can simply use the z test which with MLR is also robust to nonnormality (and z**2 is approx chisquare). 

Guy Cafri posted on Saturday, November 08, 2008  1:46 am



Great, thank you for the advice. 


I am running a multiple groups analysis with a continuous dv and a series of predictors, using MLR as an estimator. The data is complex with clustering and stratification. There are no latent variables. I am fitting nested models separately  e.g., a model with no constraints, a model constraining weights, a model constraining weights and intercepts, etc. Since I don't have the difftest option, I am trying to compute scaled difference tests, and I have two questions: 1. There are a total of 3 scaled correction factors (SCF) listed in each output: one under test of model fit, and two under the loglikelihood section (for Ho and H1). They all have different values. What is the difference between them? 2. Which exactly should I plug into the formula for computing the scaled difference test? For example, if I wanted to compare a patternconstrained model with a saturated model? Thanks in advance. This website is extremely helpful. I've already answered many of my questions my simply perusing the archives. 


1. They are different because they apply to different statistics, one for chisquare and two for loglikelihoods. 2. You should use the one associated with the statistic you are using in the calculation of the test. You can do difference testing using either chisquare or loglikelihood. Both tests are described at http://www.statmodel.com/chidiff.shtml and in Chapter 13 of the Mplus User's Guide. 


I ran two regression models with 96 obs using MLR. In run #1 the dependent continuous variable was regressed on 7 continuous predictors. In run #2 I added the square of one of the predictors. Both models were estimated using FIML. Results: Rsquare quadratic = .250 Rsquare linear = .220. (Suggests quadratic model fit better) Linear model H1 LL = 598.242 Quadratic model H1 LL = 637.278 (Suggests linear model fit better 1) Could you please explain why examining fit with the Rsquare and loglikelihood statistics gives contradictory results about which model fit better? Which should I believe? Following www.statmodel.com/chidiff.shtml for the SatorraBentler diference test using loglikelihoods, my value of TRd was negative! This was because L0  L1 = (LL linear model  LL quadratic) = 598.242  637.278 = a positive value, which when multiplied by 2 yields a negative value of TRd. (2) Have I done something wrong or do I just use the absolute value of the TRd statistic to test with the Chisquare distribution? If I use the absolute value, I will be testing Trd = 105.5 with 1 df. (Details: H1 scaling factor for linear model = 1.16, for quadratic = 1.118; linear model parameters estimated = 9, quadratic model = 10 paramters). Thanks for your help! 


1) I assume you mean H0 and not H1 in your reporting of those 2 LLs. The LL's are not in a comparable metric due to having different covariate sets in the 2 runs. In this analysis you are in the SEM framework where the covariates contribute to the LL. You can avoid this and make the metrics comparable by saying Type = Random, which conditions on covariates so the metric is determined only by the DVs. 2) This question is resolved by the answer to 1). Note also that Rsquare is not a measure of model fit in the SEM sense. The regression model is justidentified (fits trivially). Overidentified models that fit well (in the SEM sense) can have poor Rsquares and vice versa. But of course Rsquare is informative about how well the regression equation represents the data. 


Thanks for your response. 1. I know that in the SEM sense, "fit" refers to how well the estimated variances and covariances match the sample variances and covariances. But to evaluate fit in terms of the accuracy of the predicted values from a multiple regression, should I use the Rsquare value or the reported loglikelihood (H1 or H0)? 2. With MLR reg models, are you saying that I can only compare the loglikelihoods of models that a) contain the same number of covariates or, b) contain exactly the same covariates (the predictor variables are exactly the same in both models)? For b) in terms of residuals, I would think the fits would be identical  they're the same model. 3. From your first comment, it seems I've used the wrong loglikelihoods in my attempt to use the SatorraBentler test. Could you please explain what the difference is between the H0 and H1 loglikelihoods and which loglikelihood (H0 or H1)I should use for the more restrictive model? For the less restrictive model? Your assistance is gratefully appreciated! 


1. You can use both in different ways. You use Rsquare in the usual way. You use the LL (taken from H0  which is your model) by computing 2 times the H0 LL difference between your two models: the model with the quadratic effects fixed at zero and the model with the quadratic effects freely estimated. This is a chisquare variate. 2. b) But if you say Type = Random you can compare even if you change the covariate set. 3. For a given run, H0 is your model. H1 is an "unrestrictive" model in the SEM sense  a free mean vector and covariance matrix model. For a simple linear regression model H0 and H1 are one and the same. 


Thanks for the clarification. It is very much appreciated. 


When I run a model using the ML estimator (on summary data), I do not get the the scaled chisquare values and am therefore wondering how to conduct the SatorraBentler difference test. Thank you in advance. 


Hi Sanja, This page might solve your problem: http://www.statmodel.com/chidiff.shtml Sincerely, Amir 


Thanks Amir. I had in fact been using that page, but the formulas require regular chisquare values (t0 and t1, these are the ML scisquare values), and the scaled chisquare values (tr0 and tr1, which I cannot find in the output). 


Hi Sanja, The scaling factors are obtained when you use one of the robust estimators: MLM (Satorra Bentler) or MLR (asymptotically equivalent to Yuan Bentler) and not when you use the ML estimator... 


Hi Alexandre, Hm, I thought so. Do you know how I can conduct chisquare difference testing when using the ML estimator then? Thanks for replying, Sanja 


Hi Sanja, With regular ML, you do it the standard way... The difference (substraction) between 2 chi squares is like a chi square with degrees of freedom (df) corresponding to the differences between the df of both models. 


Thanks! 


But dont be "afraid" to use MLM or MLR if you suspect that you need a robut estimator (non normal data, clustered data, etc.). 


The thing is, I only have summary data and not individual data. For that I can only use ML, ULS, or GLS. The data are a summary of ordinal data, for which I would normally use WLSMV. However, since I ran the polychoric correlation matrix of this data through Mx to get a genetic factor model decomposition (a decomposition of covariance matrix into 3 matrices  one due to additive genetics (A), one due to common environment (C), and one due to unique environment (E)), what I end up with is a summary statistic, i.e. the A, C, and E covariance matrices. So I only have covariance matrices of ordinal data derived by decomposing a polychoric correlation into 3 components. Now I need to do a CFA on these summary data, and I can't use WLSMV anymore (nor do I know whether I should). I can only use ML, ULS, or GLS. The ML and ULS seem to provide identical results. But my chisquares are generally quite large, and other fit statistics indicate quite a bad fit as well (regardless of which model I specify in the CFA). I am wondering whether this is due to using ML on a covariance matrix of ordinal data. I am not simulating data to try to find this out, but if you have any insights I would greatly appreciate them, though this is a bit of a different topic from chisquare difference testing. Thanks a lot! 


Btw the original (ordinal) data are not normally distributed. 


It sounds like you are analyzing an estimated covariance/correlation matrix from an ACE model. You are right in suspecting this  your chisquare and SEs are not correct here. All you can trust are the parameter estimates. 

jks posted on Saturday, October 17, 2009  4:52 am



Hi Muthen, you said, "Stay with the MLR estimator. The MLR chisquare multiplied by the scaling correction factor is the ML chisquare." Instead of this, if I estimate the regular chisquares (T0 & T1) using ML estimator and then do the difference tests, am I ok? As you know that CFI is also used in measurement invariance tests (i.e., for metric, scalar, complete invariance). Is there any correction required to use the difference in CFI values for constrained and unconstrained models? 


You can do difference tests with ML the regular way. Note that ML is not robust to nonnormality. I am not aware of using CFI for difference testing. 

jks posted on Sunday, October 18, 2009  7:12 am



Thanks for your reply. The comparative fit index (CFI) has also been used to assess measurement equivalence (Yuan, 2005). It is recommended that changes in CFI of 0.01 or less indicate that null hypothesis of measurement equivalence should not be rejected (Bentler, 1990). So, my concern is can I apply this in my measurement equivalence study where I am using MLR estimator for complex survey data. 


If this is appropriate for ML, it would be appropriate for MLR. 


Hi, I want to perform Chisquare difference tests (with MLM estimator) between several models that I tested on one given set of data. I compare two 6 factor models, a 2factor model, and a 4factor model. My problem is that the two 6factor models have the same degrees of freedom, therefore I can't compute the cd index because it requires to divide by the difference of df (zero). Which alternative do I have to determine if the difference of fit between these two models is significant? Thanks in advance for your answer. 


The two six factor models are not nested so difference testing would not be appropriate. 

Nick Lee posted on Friday, April 09, 2010  3:12 pm



Dear Bengt/Linda I am testing a multigroup complex sample model for latent mean invariance across genders, using MLR. As have many others, I have encountered the negative TRd issue using the loglikelihood chisq difference test for MLR. I have read the Satorra/Bentler paper using the modification. However, before putting this into practice I'd like to ask a question. Specifically, if I can gain the ML chisq values by multiplying the MLR ones by the correction factor given in the MLR output, am I able to use the (rather simpler) method for computing the scaled chisq test, rather than the loglikelihood? Of course, I can't gain the ML chisq by running an ML model, because the COMPLEX model defaults to MLR. Judging by some of the earlier comments on this thread, it seems implied that I can indeed create ML chisq from the MLR one, which would enable this. Or have I got something backwards here? Thanks, Nick 


It is true that the MLR chisquare can be converted to ML using the scaling corrections factor. However, using ML to do the difference test and using the MLR standard errors would not be correct. You should just estimate the model with ML if you want to do the ML difference test. 

Nick Lee posted on Tuesday, April 13, 2010  4:08 pm



Hi Linda Thanks for the advice. The problem is that using the complex samples option does not allow me to use ML at all. Thus I can't do this. So am I correct to draw the conclusion that using complex samples (therefore prohibiting the use of ML) means the only way to do the difference test is to use the loglikelihood method (which in my case gives a negative Trd), even though I can get the ML chisquares using the correction factor? Thanks, Nick 


You can try using the Wald test in MODEL TEST. See the user's guide. 


I have a nested models problem. I have 3 models: One factor model x x x x x x x x x Two factor model x 0 x 0 x 0 0 x 0 x 0 x 0 x 0 x 0 x Three factor model x 0 0 x 0 0 x 0 0 0 x 0 0 x 0 0 x 0 0 x 0 0 0 x 0 0 x I thought these models are not nested, but a reviewer suggests they are. Can you tell me what am I missing to make these into nested models? Thanks! 

JackBox posted on Thursday, April 15, 2010  11:05 am



I’m also having the negative cd problem with chisquare difference test in group invariance testing with MLR. I’ve read the newest Santorra Bentler (2009) article “Ensuring Positiveness of the Scaled Difference ChiSquare Test Static” and I wonder is it possible to do the new scaled difference test with the present version of Mplus? The article says that “This can be obtained by creating a model setup M10 that contains the parameterization of M1 with start values taken from the output of model M0. Model M10 is run with zero iterations, so that the parameter values do not change before output including test statistics is produced”. It seems that the issue of the starting values is simple, but what could be the solution in Mplus for the zero iterations? Is it possible to use zero iterations in Mplus and if so which iteration procedure should be constrained to zero in this case? 


We will post a note about how to do this after Version 6 comes out. You can't have zero iterations but instead should have extremely lax convergence criteria so they are fulfilled already by the starting values. 

Hana Shin posted on Sunday, May 23, 2010  7:32 pm



Hello Drs. Muthen & Muthen, I'm new to Mplus and have run a model using ML with BOOTSTRAP, although I have a positively skewed continuous dependent variable. I had hoped that bootstrapping would assist with the observed nonnormal distribution, but would MLR be a more appropriate estimator (since MLR is robust to nonnormality but does not allow for bootstrapping)? Many thanks for your support. 


I think it is more straightforward to use MLR. This will give you proper standard errors and chisquare test of model fit. 

Jen posted on Tuesday, June 22, 2010  4:38 pm



Hello, I wondered if there is any solution when the MLR difference test yields a very large (seemingly unreasonable) chisquare value due to big differences in the SCFs. I am working with a relatively small sample, and this difference test results when I constrain a path that is clearly positive and significant to 0 (I am constraining it purely for purposes of model comparison). The chisquares of the two models are Model 1: 14.011, 4 df, SCF=1.088 Model 2: 2.042, 3 df, SCF=1.368 Which results in cd=.248 and a chisquare difference of 50.20(!). I feel that reporting this 50.20 might not go over so well (and am also suspicious of such a large SCF in the case of Model 2). Thanks for any advice! 


It may be that the largesample approximation needed for this difference test is not good in this smallsample setting. Because the models differ by only 1 parameter, you can compare to what you get by the approximate z test printed as Est/SE. The SE here derived by MLR also takes into account the nonnormality. 

Syd posted on Monday, July 26, 2010  10:12 am



Hi, I am trying to test the comparative fit of two models. As I have interaction terms in the model, I am using the MLR estimator with numerical integration. Thus, I need to use loglikelihood difference testing. It is stated on the website that I need to know p0 (number of parameters in the nested model) and p1 (number of parameters in the comparison model). I wasn't sure whether p0 and p1 were referring to the "Number of Free Parameters" noted under the Information Criteria heading, as this is the only parameter noted in the output. I would greatly appreciate it if you could confirm this. Thank you, 


Yes, it is. 

Syd posted on Sunday, August 01, 2010  5:34 am



Hi Linda, You had noted that in loglikelihood difference testing, the p0 and p1 values referred to the number of free parameters noted in the output. I was just looking at two nested models estimated using MLR, for which both chisquare and loglikelihood values were provided by Mplus. If I do the chisquare difference testing, TRd=339.90 with d0d1 = 192. If I do the loglikelihood difference testing, TRd=2659.98 with p0p1 = 21. I thought both tests were used to test change in fit. Yet, using the formulas provided at http://www.statmodel.com/chidiff.shtml, I'm calculating different results for the tests. Am I doing something wrong here? The nested model is: f6 ON f5 f4 f2 f1; The comparison model is: f3 ON f2 f1; f6 ON f5 f4 f3 f2 f1; The values for the two models are as follows. Nested Model: MLR Chisq=527.965 Chisq scaling corr. factor=1.082 df=336 Loglikelihood H0 value=18915.632 H0 scaling corr. factor=1.522 Number of free parameters=101 Comparison Model: MLR Chisq=863.315 Chisq scaling corr. factor=1.013 df=528 Loglikelihood H0 value=26294.418 H0 scaling corr. factor=2.215 Number of free parameters=122 


It is not clear from the information you give that the models are nested. Please send the two full outputs and your license number to support@statmodel.com. 

Joan W. posted on Thursday, March 10, 2011  9:48 pm



Dr. Muthen, Is the formula for chisquare difference test from http://www.statmodel.com/chidiff.shtml still valid when "algorithm=integration" is used with MLR? Thank you. 


Yes. 

Joe King posted on Friday, December 02, 2011  2:13 am



Drs. Muthen, I am running a multiple group comparison, one with a constraint on one of the paths, another model without the constraint. I want to see how the model fit differs (if significant). i know how to do the ratio test, but in the model fit section in the likelihood section theres a value for H0 and H1 for each of the models, but the H1 likelihood doesnt move at all between the models when the parameter is released but H0 does, why is this, and should i use H0 for model comparison? 


You should compare the H0 models. The H1 models are the same because they are the unrestricted model. 


Dear Drs. Muthén, I would like to compare two models  one nested within the other  in path analysis and I am using MLR as an estimator. I followed the procedure you describe in the web notes no.12 to compute the strictly positive SatorraBentler chisquare test in Mplus syntax but I think I am doing something wrong because it says in your text that the loglikelihood value of the M10 model should be the same as that for the M0 model. However, when I look at my outputs, it is rather the loglikelihood of the M1 model that is equal to the M10 value. Would you be able to tell me what I am doing wrong? Many thanks in advance for your help! Geneviève 


You would have to send your 3 outputs to Support. 


Thank you! I think I have figured out the problem. 


Dear Prof. Muthen, I want to compare a model without an interaction to a model with an interaction (between a factor and a manifest variable using XWITH option, MLR and algorithm is integration). The only difference in the models is the interaction term. In this regard I wondered whether these models are nested? Can I use a Loglikelihood ratiotest? Or should I compare the models based on the AIC or BIC? Thanks in advance, Kind regards, 


The significance of the interaction term is the same as doing a difference test between the model with the interaction and the model without the interaction. 

Back to top 