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 Albert E. Mannes posted on Friday, February 20, 2009 - 9:19 pm
Hello,

A straightforward CFA question for anyone who can help.

I have 50 continuous indicators and a 6 factor model which is estimated without issue in mplus 5.1. I want to test whether 2 of the factors are distinct, so I added the following line to the model command: f1 with f2@1;

The model now does not converge due to a non-invertible covariance matrix. If instead I simply drop f2 and have its indicators also load on f1, the model converges.

I'm not sure why the 1st approach doesn't work. Do I have to constrain the other covariances as well? That is, do I also have to specify that the covariance between f1 and f3 is now equal to the covariance of f2 and f3, etc.?

Sorry if confusing, and thanks for the help.

Al
 Linda K. Muthen posted on Saturday, February 21, 2009 - 2:49 pm
If you want to test if the correlation is one be sure you have all factor loadings free and the factor variance fixed to one. If not, you are testing that the covariance is zero.

Instead of fixing the correlation to one which may be a misspecification that makes convergence problems, use MODEL TEST, for example,

MODEL:
f1 WITH f2 (p1);

MODEL TEST:
0 = 1 - p1;
 Albert E. Mannes posted on Saturday, February 21, 2009 - 3:41 pm
That did it. Thank you.
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