

Measurement invariance bifactor grow... 

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Xu, Man posted on Friday, January 18, 2013  6:26 am



The Latent State Trait (LST) model often is modelled with a bifactor or secondorder structure without means. I want to fit a LST model that incorporates trait growth parameters. Based on example ex6.14, it's: sf1 BY y11 y21y31 (12); sf2 BY y12 y22y32 (12); sf3 BY y13 y23y33 (12); [y11 y12 y13] (3); [y21 y22 y23] (4); [y31 y32 y33] (5); i by y11@1 y21@1 y31@1 y12@1 y22@1 y32@1 y13@1 y23@1 y33@1; s by y11@0 y21@0 y31@0 y12@1 y22@1 y32@1 y13@2 y23@2 y33@2; [i@0]; [s]; sf1 with sf2@0 sf3@0 i@0 s@0; sf2 with sf3@0 i@0 s@0; sf3 with i@0 s@0; 

Xu, Man posted on Friday, January 18, 2013  6:34 am



Continued from the previous post (sorry!). For this model, the growth parameters (i, s) seem close enough to those from ex6.14, df was same although fit was not exactly the same  similar though. Now, my question is about measurement invariance (MI) under this specification. I gather that if this LST growth model is what I want. Then it is probably no longer suitable to start the MI tests based on the CFA with free means (as taught in Topic 4), because the LST growth model specification is like a bifactor (maybe depends on the schmidleiman transformation too?). Now, I was wondering if it is better if in this model, I free up the loadings of the slope from one of the three waves? Like s by y11@0 y21@0 y31@0 y12@1 y22@1 y32@1 y13* y23* y33* Would this then allow a better model for testing MI, especially for the item intercepts (a free latent mean structure, without linear restriction on the slope)? Thanks! Kate 

Xu, Man posted on Friday, January 18, 2013  6:56 am



Oh, I should have constrained the last three loading menas equal, like: s by y11@0 y21@0 y31@0 y12@1 y22@1 y32@1 y13 y23 y33 (a); 


I'm afraid that I am not familiar with the meaning of this LST model, so you may be better off asking this on SEMNET. It looks peculiar to me. Where has it been proposed? 

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