Variance of a higher-order factor is ... PreviousNext
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 Kaigang Li posted on Tuesday, July 28, 2009 - 12:26 am
I am conducting a higher order CFA with 8 1st-order factors and 3 2rd-order factors.

I ran the following input and get “WARNING: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE DEFINITE. …. ”

Model:
fSE by Q6SEE2 ...;

fSSF1 by Q17SSF1 ...;
fSSFa2 by Q22SSFA1 ...;
fGP by ...;
Q33P2 with Q34P3;
fENA1 by ...;
fENN2 by...;
fEN by fENA1 fENN2;
!fEn@1.0;
fEXP1 by ...;
fEXN2 by ...;
fEX by fEXP1 fEXN2;

The variance of 2rd-order factor(FEN)is negative.

Variances
FEN -0.037 0.075 -0.489 0.625

When fix the variance of FEN to 0 I got the same warning.

fEN by fENA1 fENN2;
fEn@0;

When I fixed FEN to 1.0 and freed fENA1 as
fEN by fENA1* fENN2;
fEn@1.0;

I got the same warning and another error
"THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES MAY NOT BE
TRUSTWORTHY ...."

But when I did not free the first factor with only fixing the FEN to 1.0, the model fit well.

fEN by fENA1 fENN2;
fEn@1.0;

I am not sure if it makes sense. Would you give any suggestion? Thank you very much!

Kaigang
 Linda K. Muthen posted on Tuesday, July 28, 2009 - 7:58 am
If the second-order factor has no variance, the model is not appropriate for the data.

The following statements should not be used:

fEN by fENA1 fENN2;
fEn@1.0;

In them, you fix the metric of the factor twice by fixing a factor loading to one and the factor variance to one. This is not correct.
 Kaigang Li posted on Tuesday, July 28, 2009 - 9:11 am
Hi Professor Muthen,

Thanks for your quick response. Would you tell me what would be the reason that results in the zero variance since the EFA produced two factors for the FEN scale?

I have other second-order factors which have variances as follows:

Variances
FSE 1.180 0.148 7.984 0.000
FSS 0.542 0.154 3.528 0.000
FGP 2.136 0.166 12.889 0.000
FEN -0.037 0.075 -0.489 0.625
FEX 5.840 2.548 2.292 0.022

May I break down just FEN into two factors without using fEN by fENA1 fENN2; any more, but keep other seond-order factors?

Thanks a lot.

Kaigang
 Linda K. Muthen posted on Tuesday, July 28, 2009 - 12:54 pm
How did you do a second-order EFA?
 Kaigang Li posted on Tuesday, July 28, 2009 - 7:06 pm
I don't think I have done second-order EFA, instead I just did first-order EFA. Should I refer to the example 4.5 and 4.6 in the Mplus guide book for second-order EFA? Any suggestions for other reference would greatly appreciated. Thank you very much!

Kaigang
 Linda K. Muthen posted on Wednesday, July 29, 2009 - 1:42 pm
I asked because you implied that you had done a second-order EFA and therefore the second-order CFA should fit. Given the one second-order factor has no variance. It should not be included in the model.
 Kaigang Li posted on Wednesday, July 29, 2009 - 2:02 pm
Sorry about the confusion. I am considering to remove that second-order factor. Thanks!
 Kaigang Li posted on Wednesday, July 29, 2009 - 10:04 pm
Professor Muthen,

I got another question. I read a very good discussion regarding fit indices for categorical outcomes at http://www.statmodel.com/discussion/messages/23/26.html?1244640406 posted 10 ages ago. I am carrying out SEM with a binary outcome variable using MLR. It seems that so far Mplus still does not provide fit indices for categorical outcomes. I am wondering there is any conclusion on that or there is any alternative way to diagnose model fit with categorical outcomes. Any suggestion and useful references are appreciated.

Kaigang
 Linda K. Muthen posted on Thursday, July 30, 2009 - 5:47 am
Chi-square and other related fit statistics are not available when means, variances, and covariances are not sufficient statistics for model estimation. These fit statistics are available for categorical outcomes with weighted least squares estimation.
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