Message/Author 

Anonymous posted on Tuesday, May 16, 2000  12:18 pm



Does Mplus provide RMSEA, CFI, NNFI etc? How about missing data situation? Thanks. 


RMSEA is currently provided when all outcomes are continuous. This includes the missing data situation. 

Anonymous posted on Thursday, May 16, 2002  8:31 am



When running separate analyses on two nested models using FIML to handle missing data, is it still legitimate to do a chisquare difference test? 

bmuthen posted on Thursday, May 16, 2002  9:39 am



Yes, but if you use MLR you need to use the chisquare difference testing procedure described in ChiSquare Difference Test for MLM under Special Analyses with Mplus. 

Yifu Chen posted on Monday, December 06, 2004  11:53 am



Hi, Dr. Muthen, I was running a growth model analysis with three timevarying variables. I also used TYPE=MISSING to handle the missing data. I found that MPLUS 3.11 only reported limited model fit indices (AIC, BIC...). I wonder if I can get chisquare and other fit indices in for the fitted model. Thanks 

bmuthen posted on Monday, December 06, 2004  3:04 pm



You get fit indices if you add H1, saying Type = Missing H1; 

Anonymous posted on Sunday, January 23, 2005  9:58 am



Do I understand from the comments above that it is reasonable to do a ChiSquare difference test with ChiSquare values that result from MLR estimation if the correction procedure for MLM estimation is used? Thanks. 

bmuthen posted on Sunday, January 23, 2005  3:32 pm



Yes. 


Hello Drs. Muthen, I am running a 1factor CFA model with the following features: TYPE IS MISSING H1 COMPLEX ESTIMATOR IS MLR My 3 dependent variables (indicators) are count variables (I have specified this, as well as my missing specification, weight strata and cluster variables in the "VARIABLE" section). My problem is that I am only getting limited fit indices (AIC, BIC...). Is there a way for me to get more fit indices (RMSEA, CFI...)? If not, which specification above is preventing me from getting these indices? Thanks, Jim 


With count variables, means, variances, and covariances are not sufficient statistics for model estimation. Therefore you will not obtain chisquare and other related test statistics. 


I am using Mplus to estimate a multiple group path model using FIML. Because it is a path model, the Chi square of the baseline (freely estimated) model is 0. So, when I constrain a parameter to be equal across groups how do I test to see the impact the constraint has on fit. Doing a chi square difference test between the baseline and a constrained model means comparing the constrained model to 0. Is this appropriate? Do you know of any references regarding how to test for group differences in path models? Thanks, Alicia 


The answer to your first question is yes. General SEM ref's apply. 


I was testing the path model below: USEVAR ARE sex zacap zaca zach zscho zstu; MISSING IS ALL (99.00); cluster = scho; define: int1=zach*zscho; int2=zach*zstu; ANALYSIS: type=complex; type=missing H1; estimator =MLR; MODEL: zaca ON sex zach zscho zstu; zacap ON sex zaca zach zscho zstu; model indirect: zacap IND zach; OUTPUT: STANDARDIZED; I already added the text "H1" to my syntax to get information about the overall fit of my model, but I only got this: ChiSquare Test of Model Fit Value 0.000* Degrees of freedom 0 Pvalue 0.0000 Scaling Correction Factor 1.000 for MLR Is this because I have a recursive/saturated model? And how do I report the overall model fit? Can I do this through chi square difference testing, i.e., comparing with a model with no predictors? Thank you for your time! 


You cannot test the fit of a model with zero degrees of freedom. The models would not be nested if you get rid of the covariates. You could try the following: MODEL: zaca ON sex@0 zach@0 zscho@0 zstu@0; zacap ON sex@0 zaca@0 zach@0 zscho@0 zstu@0; 


I had a similar problem as Sofie Wouters (April 30, 2009) with my freely estimated (full/saturated) path model: USEVARIABLES ARE alc2 cn0 gp1 cn2 gp2; CLUSTER = IDYRFAM; ANALYSIS: TYPE = COMPLEX; MODEL: gp1 gp2 cn2 alc2 ON cn0; gp2 cn2 alc2 ON gp1; alc2 ON gp2; alc2 ON cn2; gp2 WITH cn2; OUTPUT: SAMPSTAT STANDARDIZED; standardized mod(3.84); ChiSquare Test of Model Fit Value 0.000* Degrees of freedom 0 Pvalue 0.0000 Scaling Correction Factor 1.000 for MLR Using your suggestion to add @0, I was able to get a ChiSquare Test of Model Fit with actual values, but I am confused as to why I would want to do that. The UCLA Academic Technology Services explanation suggests that @0 sets the structural paths to 0. If I am using this as my full, comparison model with all paths being freely estimated and then trying to compare it to nested models with more constraints, how would I do that if the paths have been set to 0? Thanks! 


When you fix the parameters to zero, the chisquare you obtain is not a test of the saturated model. It is a chisquare difference test between the two models. There is no way to assess the fit of a saturated model. 


Greetings! The 2factor model I am trying to fit is: VARIABLE: NAMES = finemp decihealth deciexpl deciexps visitfam prenatal tt instdel immun; MISSING = all(999); CATEGORICAL ARE finemp decihealth deciexpl deciexps visitfam prenatal tt instdel immun; ANALYSIS: ESTIMATOR =MLR; INTEGRATION=MONTECARLO; MODEL: mch BY prenatal tt instdel immun ; autonomy BY finemp deciexpl deciexps decihealth visitfam; mch ON autonomy; OUTPUT: STDYX ; I am getting only loglikelihood, AIC, BIC and ChiSquare Test of Model Fit for the Binary and Ordered Categorical (Ordinal) Outcomes. My N=46304, and degrees of freedom reported under the chisquare tests is 489. Is there any way for me to get CFI, TLI and RMSEA? Thank you! 


I had a followup: 1. If I run WLS or WLSMV as the estimator in the same model I noted above, then the model drops all cases where there is any missing data on xvariables. Is this normal? Is there a way to avoid this? 2. The standardized parameters across WLS and MLR results are extremely similar. With WLS, I even get CFI, TLI and RMSEA and they show very good model fit. Now if there were some way for WLS to run with all of my cases, and not do listwise deletion on my cases with missing data, then would I be better off just using WLS rather than MLR? Thank you so much. 


The chisquare you get with maximum likelihood and categorical outcomes is not the chisquare for the H0 model. It is the chisquare that compares observed and expected frequencies of the categorical outcomes. Chisquare and related fit statistics are not available in this situation. Yes, cases with missing on covariates are dropped with all estimators as the model is estimated conditioned on x. I would suggest imputing data sets using multiple imputation and then using WLSMV. 


I have got questions to the following problem: I conduct a path analysis with weighted least squares means and variance adjusted (WLSMV) estimation in Mplus, version 5.1. The WLSMV estimator was chosen automatically, because the variable “processing depth” was indicated as categorical. The information of 20 data sets (reached over imputation) was included to examine the model. For each descriptive quality criterium the average fit indices and standard deviations over the 20 data sets were computed. In the output of Mplus, the model fit indexes for descriptive quality criteria are indicated as means and standard deviations. I think, I must report these means of the fit indices to report the model fit? (The fit indices of my path model are (standard deviation are in parentheses): CFI = 1.00 (.00), TLI = 1.39 (.02), RMSEA = .00 (.00), and WRMR= .26 (.02)). I do not know, whether I must report the ChiSquare, because there is not the mean of chi square indicated. Furthermore I read that TLI and CFI have rather low power to reject a model with binary outcomes, while WRMR works well. And I read that “Recent studies indicate that a value less than .90 indicates good fit for WRMR.” (Linda K. Muthen posted on Thursday, March 08, 2001  3:14 pm). Can you please give me the references to these studies? Thank you very much for your help. 


It sounds like you are using TYPE=IMPUTATION for your analysis. We provide means for the fit indices over the imputed data sets. It is a research topic of how fit statistics should be handled with multiple imputation. So it is not known how to interpret these. 

Eric Teman posted on Sunday, June 24, 2012  11:45 am



When using WLSMV with multiple imputation in Mplus, is the model fit chisquare valid? 


No, we give the average value. 

wei w posted on Monday, January 27, 2014  12:37 pm



I am wondering how the chi square test statistic is computed with FIML. Is it computed based on the log likelihood difference between tested and saturated model or one of the test statistics proposed in Yuan and Bentler(2000)? If it is computed based on Yuan and Bentler(2000), which formula is used? The formula in equation 18 or 20. Yuan, K., & Bentler, P. (2000). Three likelihoodbased methods for mean and covariance structure analysis with nonnormal missing data. Sociological methodology, 30(1), 165200. Thanks! 


If you use estimator=ML then it is based on the loglikelihood difference between tested and saturated model. If you use estimator=MLR it is T2* from Yuan, K., & Bentler, P. (2000). 


I have the following problem: I want to do a CFA with one latent variable and three observed variables: MODEL: control BY m1hs1 m1hs2 m1hs3; OUTPUT: STDYX; But the output tells me that chisquare is zero. My observed variables are highly correlated though. This must be some kind of mistake: ChiSquare Test of Model Fit Value 0.000 Degrees of Freedom 0 PValue 0.0000 RMSEA (Root Mean Square Error Of Approximation) Estimate 0.000 90 Percent C.I. 0.000 0.000 Probability RMSEA <= .05 0.000 CFI/TLI CFI 1.000 TLI 1.000 ChiSquare Test of Model Fit for the Baseline Model Value 46.998 Degrees of Freedom 3 PValue 0.0000 


You have zero degrees of freedom. Your model is justidentified. Model fit cannot be assessed. 

Eric Teman posted on Wednesday, June 10, 2015  6:49 am



As of now, when using WLSMV with multiple imputation in Mplus, is the model fit chisquare valid? Or is the average still given? 


It is just the average. 

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