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yunci, ye posted on Tuesday, June 28, 2011  11:25 am



Dear Dr. Muthen: I perform pairwise chisquared difference test to exam difference between the unconstrained model(coach and gt is free) and constrained model(coach and gt is fixed to 1). My constrained model(coach and gt is fixed to 1) for Mplus command is as follows. TITLE: CFA with Discriminant Validity DATA: FILE IS c:/mplus/gsw.dat; VARIABLE: NAMES ARE g1 g2 g3 g4 g5 c1 c2 c3 c4 c5 t1 t2 t3 t4 t5 h1 h2 h3; ANALYSIS: estimator = ml; iterations = 5000; MODEL: gt by g1* g2 g3 g4 g5; coach by c1* c2 c3 c4 c5; ath by t1* t2 t3 t4 t5; we by h1* h2 h3; gt@1 coach@1 ath@1 we@1 coach with gt@1; Output: sampstat; TECH4; But I run the command, and show error message is : ˇ§WARNING: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE DEFINITE. THIS COULD INDICATE A NEGATIVE VARIANCE/RESIDUAL VARIANCE FOR A LATENT VARIABLE, A CORRELATION GREATER OR EQUAL TO ONE BETWEEN TWO LATENT VARIABLES, ....PROBLEM INVOLVING VARIABLE COACHˇ¨. I donˇ¦t know what happen is? Is the command wrong, or others? I appreciate your response beforehand. Thanks. 


You should use MODEL TEST to test that the correlation is one. 


Dr. Muthen Could you tell me if the following syntax is correct to test the discriminant validity with the procedure mentioned above with the command MODEL TEST f1 by y1* y2 y3; f2 by y4* y5 y6; f3 by y7* y8 y9; f1 with f2 (a1); !covariance=1 in model test f1 with f3; f2 with f3; f1@1; f2@1; f3@1; Model test: a1 =1; Thank you. 


Testing that the correlation is not equal to one is one test of discriminant validity. 


Thank you, so a significative Wald test indicates that there is discriminant validity among these constructs. 


Yes. 

Dodam Park posted on Wednesday, April 05, 2017  3:38 am



Dear Dr. Muthen When testing discriminant validity of a scale I am interested in using another scale which doesn't satisfy the normality assumption, is there any serious problem? If it is, what kind of solution I can use in that situation? 


Ask SEMNET. 

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