Hello. I've read with great interest your 2002 paper about sample size calculations using monte-carlo simulations. I've noticed that in the paper the population model, and the extracted models have the same specs. I am trying to check whether my sample is sufficient for a CFA with no restrictions (except for indicators loaded on two factors in a predeterimened way). This is the model I want to test:
f1 BY y1-y10; f2 BY y11-y15;
Ofcourse, for the population model I restrain all parameters, following previous findings, including exact loadings, correlation between two factors, and exact residuals.
My question is - whether I have to write the same in the MODEL section, even if in the actual CFA I don't specify all of these parameters, or do I have to write the simple model as I actually test it in the CFA (without specifying actual loadings, correlation and residuals)
Thanks for the quick response. The question is whether I should write in the MODEL section in the Monte-Carlo simulation, the same exact model as the one I actually intend to test (i.e. without specifying loadings and residuals)?
Another question is whether there is some method to calculate the robustness of the other fit indices (RMSE, CFI, TLI, SRMR) using monte-carlo simulation?
Itzik posted on Wednesday, March 11, 2015 - 1:39 am
Great. Just to verify that I used the right input and interpret the results correctly:
I used a WLSMV estimation for 15 ordinal indicators with 5 categories each, and set parametarization = THETA (I don't really understand the meaning of that but Mplus required me to do so).
In order to test the bias of the fit indices (Chi square, RMSEA, WRMR), I simply have to compare the expected distribution (percentiles and their proporions) to observed sampling distribution, right? So for example, if I use a cutoff of <.05 for the RMSEA, and results show that in the observed distribution the proportion of percentiles above .051 is .014, I can conclude there is less then 1.4% chance that the results we actually got in our study (RMSEA=0.120) are by chance, given the H0 (of fit as reported in previous studies).