I am running measurement models for my data. One of my models runs, but I get a chi-square model value of 0.000. What does this mean? Is this a poor model? Below are my output and input; the data is secure so I cannot send it
Chi-Square Test of Model Fit Value 0.000* Degrees of Freedom 0 P-Value 0.0000 Scaling Correction Factor 1.000 for MLR
Chi-Square Test of Model Fit for the Baseline Model Value 45.180 Degrees of Freedom 3 P-Value 0.0000
CFI/TLI CFI 1.000 TLI 1.000
Loglikelihood H0 Value -21243.164 H1 Value -21243.164
Number of Free Parameters 9 Akaike (AIC) 42504.327 Bayesian (BIC) 42548.379 Sample-Size Adjusted BIC 42519.795 RMSEA 0.000 SRMR 0.000
DATA: file is "R:\Users\Sarah Strand\public\Data\MPlus Data\subequalgroups2.txt"; format is free; type is individual;
VARIABLE: NAMES ARE AID FEMALE ... etc [omitted to preserve space];
IDVAR = aid; USEVARIABLES ARE par_app perc_map perc_dap; MISSING ARE .; WEIGHT IS GSWGT3; CLUSTER IS PSUNUM;
Your model is just identified. It has zero degrees of freedom. In this case, model fit cannot be assessed.
Yellowdog posted on Wednesday, October 31, 2012 - 2:37 am
Dear Linda, we want to test a path model (N=189) with the following observed trait variables: - four predictors (IV1 to IV4) - two mediators (M1 and M2) that are, as expected a priori, strongly negatively correlated with each other (r = -.56) - with quality of life (QoL) as DV We specified the following model:
qol on iv1 iv2 iv3 iv3 m1 m2; m1 on iv1 iv2 iv3 iv4; m2 on iv1 iv2 iv3 iv4;
Our question refers to how to model the relationship between M1 and M2. There is equilibrium, but we cannot make assumptions on a direction of causality from one to the other. When we specify a nonrecursive feedback loop (m1 on m2 (p31); m2 on m1 (p32);), model estimation fails (see output below). If we only specify that M1 and M2 are correlated (m1 with m2 (p31);), model estimation fails too (see output below).
Chi-Square Test of Model Fit Value 0.000 Degrees of Freedom 0 P-Value 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
How can we fix the problem? Many thanks for your help, Mario
The models are not failures. They are just-identified with zero degrees of freedom. Model fit cannot be assessed in this case. I would covary m1 with m2.
Yellowdog posted on Thursday, November 01, 2012 - 3:51 am
thank you for your reply.
We understand that with df=0, fit indices are not available. Refocusing our question, we are wondering whether the model outlined above and its results (with df=0) are valid, although we do not get information on how the model fits the data.
Can we go ahead and report path coefficients from this analysis as a final result? Or should we try to change the model specification until df>0 (e.g., using MODINDICES) ?
Hi, I am wondering why df=0 is not uncommon in path analysis. My model should not be just-identified because I have a sample size of 390 and am only trying to estimate 4 free parameters but my fit indices come up with a df=0. I'm wondering why this is and if there is any way to have positive df using path analysis. Even in the output examples you have on the website for linear regression they are 0 but this limits the usefulness of the model.
In situations where path analysis does not yield model fit indices how can we compare different models. Can we use an average of the R-squares? Is it too crude? Please can you suggests some ways around this.
Why does the model have no fit indices? Does it have zero degrees of freedom? R-square is not a fit measure. It is a measure of variance explained. Fit compares the observed covariance matrix to the model estimated covariance matrix.
Hello, I have a model with all observed variables and N=209, but I'm concerned about my fit statistics. I have 2 df. Based on reading the posts, am not sure what the problem is (I'm assuming low df?), or how to fix it. Is this problematic, or could I report these results for publication?
Model: GMREL ON Partner CommScore WGDabs;
GMSEX ON Partner Variety WGDabs CommScore;
GMREL WITH GMSEX;
Variety ON CommScore WGDabs;
OUTPUT: Chi-Square Test of Model Fit Value 0.008 Degrees of Freedom 2 P-Value 0.9960
RMSEA (Root Mean Square Error Of Approximation Estimate 0.000 90 Percent C.I. 0.000 0.000 Probability RMSEA <= .05 0.998
CFI/TLI CFI 1.000 TLI 1.045
Chi-Square Test of Model Fit for the Baseline Model Value 278.161 Degrees of Freedom 12 P-Value 0.0000
SRMR (Standardized Root Mean Square Residual) Value 0.001
Chi-Square test of model fit is for the model you specify in the MODEL command.
Hasina posted on Saturday, August 26, 2017 - 12:16 am
Dear Dr. Muthen,
I would like to know if it is normal to see a chi-square fit to be better (not reject the true model) than alternative fits (Ex, CFI, AGFI etc.) in a simulation study either in small or high sample size?
I see where you are coming from. It all depends on the hypothesis. Regular statistics and SEM have the opposite starting point in this regard. Usually, in statistics you hope to reject your null hypothesis that there is no effect because you hope there is an effect (of the treatment that you've come up with for example). In contrast, in a SEM setting, you hope to not reject your null hypothesis because the null refers to the model you have specified. So with SEM, you want large p-values because a p-value less than say 0.05 would say that you reject your null.
I am a bit confused Sir. Some articles report that, "Chi-Square Test of Model Fit" is determined only if the P-value is less than "0.05" But here, "Chi-Square Test of Model Fit" of my results have the P-value = 0.3623. So, should I understand the "Chi-square Test of Model fit" as statistically significant or not?
Simply put, in SEM testing of overall test of model fit, p less than 0.05 is taken as poor fit (model can be rejected) and p greater than 0.05 is taken as good fit. If you have seen the opposite, I'd like to hear what the reference is. It doesn't make sense to take a low p-value as an indication of model fit because the hypothesis is that the H0 model is correct so we don't want it to have a low p-value.