ESEM example 5.27 PreviousNext
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 Daniel E Bontempo posted on Wednesday, September 26, 2012 - 7:13 pm
In the user guide for example 5.27, it says

"fixing the factor variances at one in one group. For the other group, factor variances are free to be estimated."

But inspecting the tech1 output for the example shows no parameters in the PSI matrix for the 2nd group. And, the variances in the output are 1.0 for both groups.

Also, the 4th part of the example seeks to constrain variances across groups, but this seems to already be in place.

Am I misreading something? Are the 2nd group free variances expressed in some other parameter?
 Linda K. Muthen posted on Wednesday, September 26, 2012 - 10:08 pm
It is the part c output that estimates the model where factor variances are fixed to one in one group and are free in the other.

The equality in the fourth part of the example is on the factor covariance. The factor variances must be fixed at one in both groups here for model identification.
 ljc posted on Sunday, December 29, 2013 - 3:50 am
I am also confused by this example:

On page 99 it says

"fixing the factor variances at one
in one group. For the other group, factor variances are free to be
estimated. "

then on page 100 it says

"The variances of the factors are fixed at one in both groups."

Are both of these referring to the model on page 99?

Thank you.
 Linda K. Muthen posted on Sunday, December 29, 2013 - 4:07 pm
Page 100 is not correct. Page 99 is.
 ljc posted on Friday, January 31, 2014 - 8:18 pm
I am on step 4 of this example, but I have 4 factors instead of 2 and I am getting an error. Could you point out what I am doing wrong? Thank you.

MODEL:
f1-f4 BY egi1-egi28 (*1);
![f1-f4@0]; !factor means 0 in both groups;
f1 WITH f2 (1);
f1 WITH f3 (2);
f1 WITH f4 (3);
f2 WITH f3 (4);
f2 WITH f4 (5);
f3 WITH f4 (6);
f1-f4@1;
MODEL nofault:
!f1-f4 BY egi1-egi28 (*1); !free factor loadings in group 2;
F2 BY EGI28 ;

[EGI28];

![egi1-egi28]; !free intercepts in group 2;
OUTPUT: TECH1 stdyx STANDARDIZED MODINDICES(3.84);

EGI Fault ;

*** FATAL ERROR
IMPROPER PARAMETER CONSTRAINT FOR EFA MEASUREMENT SPECIFICATION.
(Error Code: 1020)
 Linda K. Muthen posted on Saturday, February 01, 2014 - 8:07 pm
You must either have all factor loadings equal across groups or all factor loadings not equal across groups.
 ljc posted on Sunday, February 02, 2014 - 6:31 pm
Is it because this is EFA instead of CFA that you can't have partial invariance?

I was also following the youtube video edu231e_lan and the example there has partial invariance at this step but it is CFA instead of EFA.

So, in order to achieve invariance in EFA, would I need to drop the item altogether?

Thank you.
 ljc posted on Sunday, February 02, 2014 - 6:52 pm
Never mind my comment above. I realize that edu231e_lan only had intercept non-invariance not factor loading non-invariance.

Is it possible with CFA or EFA to have partial invariance with respect to the loadings?
 Bengt O. Muthen posted on Sunday, February 02, 2014 - 8:48 pm
You can use CFA to specify partial invariance.
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