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HALUS posted on Thursday, June 11, 2009  2:20 pm



Hi, As the user guide said, in multiple group analysis, intercepts, thresholds, and factor loadings are held equal across groups as default. But does this also apply to secondorder factor model? I have a secondorder factor model: three secondorder factors and each secondorder factor has two firstorder factors (each firstorder factor has multiple observed variables). I have no problem constraining the firstorder factor loadings across groups but have difficulty holding the secondorder factor loadings equal across groups. 


These should be held equal as the default. You should not have to do anything to hold them equal. Please send the output and your license number to support@statmodel.com. 

Thomas Eagle posted on Thursday, December 04, 2014  9:14 am



I am having trouble fitting a multiple group model with 2 2nd order factors. I can fit the two groups separately just fine with the model structure I have, but when I attempt to fit them using the default multiple group mode setup I get nonconvergence and the error is related to the mean of one of the 2nd order factors (183 = alpha of 2nd order factor SpRelInt) THE STANDARD ERRORS ... THE MODEL MAY NOT BE IDENTIFIED... PROBLEM INVOLVING THE FOLLOWING: Parameter 183, Group NOT_ENGAGED: [SPRELINT] THE CONDITION NUMBER IS 0.348D14. THE ROBUST CHISQUARE... I thought the means of 2nd order factors were held constant across groups. I know I have to free up, or fix some values somewhere. When I tried to fix the mean of SpRelInt to zero the error message remains the same. Tom 


Perhaps you don't fix the intercepts of the 1st order factors to zero in your analysis. 

Thomas Eagle posted on Thursday, December 04, 2014  5:33 pm



Bengt, I figured out what I was doing wrong. When I set the 2nd order factor means/intercepts to zero for BOTH 2nd order factors the model converged nicely. I thought that was done automatically. This way I could leave the 1st order factor intercepts free. My bad. Thank you! You are always so patient and helpful! Tom 


But for scalar invariance, I think you would want the means of the secondorder factors estimated in all but one group so you can compare them. That then requires that the firstorder factor intercepts are zero in all groups and the factor indicator intercepts are held equal across all groups. 


I'm running a 3 group invariance model on a higher order factor model with 1 higher order factor and 4 first order factors. I first established metric invariance on the four factors, and now am trying to establish invariance on the higher order part of the model. I fix the variances of the higher order factor (SEHS) to be equal and am trying to test for mean differences, but I keep getting an error related to the higher order means. I tried fixing the first group mean to zero and freeing the other two, and holding two equal compared to the first, but I still get the following error: 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 THE FOLLOWING PARAMETER: Parameter 75, Group PAPER: [ SEHS ] THE CONDITION NUMBER IS 0.182D11. 


(continued) Below is my syntax. Any suggestions? Model: BiS by SE_mean SA_mean PER_mean; BiO by SS_mean FC_mean PS_mean; Emo by ER_mean EMP_mean SC_mean; Engaged by OPT_mean GRA_mean Z_mean; SEHS by BiS BiO(1) Emo(2) Engaged(3); [BiS](4); [BiO](5); [Emo](6); [Engaged](7); Model rand: SEHS by BiO(1) Emo(2) Engaged(3); [BiS](4); [BiO](5); [Emo](6); [Engaged](7); [SEHS](8); Model paper: SEHS by BiO(1) Emo(2) Engaged(3); [BiS](4); [BiO](5); [Emo](6); [Engaged](7); 


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