Factorial invariance PreviousNext
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 Enis Dogan posted on Friday, August 17, 2007 - 9:43 pm
Trying to confirm factorial invariance (1 factor only) across 2 groups; observed are 6 item bundles (assumed continous).

I ran the model with the following constraints (1)equal factor loadings; (2) equal (factor loadings + variance/residuals).
Do you i need to run an unconstraint model to test if (1) and (2) fit the data no worse in order to be able to test the hytpothesis of factorial invariance?
If yes; how do i run an unconstraint model? Do I
a) Combine the two groups and, get rid of grouping variable, run the 1 factor CFA. b) Run the 1 factor CFA for each group separately and add chi-square and df from each

I hope i got the syntax for (1) and (2) right:
(1):
DATA: FILE IS data.dat;
VARIABLE: NAMES ARE group bundle1-bundle6;
GROUPING = group (0= g1 1=g2);
ANALYSIS: TYPE = mgroup ;
MODEL: f1 BY bundle1-bundle6;
OUTPUT: standardized ;
(2):
DATA: FILE IS data.dat;
VARIABLE: NAMES ARE group bundle1-bundle6;
GROUPING = group (0= g1 1=g2);
ANALYSIS: TYPE = mgroup ;
MODEL: f1 BY bundle1-bundle6;
f1 (1) bundle1 (2) bundle2 (3) bundle3 (4)
bundle4 (5) bundle5 (6) bundle6 (7) ;
OUTPUT: standardized ;
 Bengt O. Muthen posted on Saturday, August 18, 2007 - 12:03 am
A convenient way to do invariance testing is to do a 2-group analysis following the suggestions in the UG chapter in the section called "Testing for Measurement Invariance..."

Note that the first factor loading is fixed at 1 as the default so this parameter should not be referred to.
 Linda K. Muthen posted on Saturday, August 18, 2007 - 11:51 am
This is in Chapter 13 of the user's guide after the discussion of multiple group modeling.
 Enis Dogan posted on Monday, August 20, 2007 - 6:21 pm
That was helpful.
Thank you.
I am still having rouble running the baseline model (configural invariance only).
syntax looks like this:

DATA: FILE IS data.dat;
VARIABLE: NAMES ARE booklet bundle1-bundle4 bundle6 bundle7;
GROUPING = booklet (0= book1 1=book2);
ANALYSIS: TYPE = mgroup ;
MODEL: f1 BY bundle1-bundle4 bundle6 bundle7;
MODEL book2:
f1 BY bundle1-bundle4 bundle6 bundle7;
OUTPUT: standardized ;


The error message i get reads:

THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE
COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL.
PROBLEM INVOLVING PARAMETER 25.

Does the syntax look OK?

i also tried to run the same, this time by fixing the mean of the latent variable at zero. i am getting the same error message again:


DATA: FILE IS data.dat;
VARIABLE: NAMES ARE booklet bundle1-bundle4 bundle6 bundle7;
GROUPING = booklet (0= book1 1=book2);
ANALYSIS: TYPE = mgroup ;
MODEL: f1 BY bundle1-bundle4 bundle6 bundle7;
[f1@0];
MODEL book2:
f1 BY bundle1-bundle4 bundle6 bundle7;
[f1@0];
OUTPUT: standardized ;
 Linda K. Muthen posted on Thursday, August 23, 2007 - 5:16 pm
The problem is that you are freeing the first factor loading by mentioning it in the group-specific MODEL command. Instead of

MODEL book2:
f1 BY bundle1-bundle4 bundle6 bundle7;

try

MODEL book2:
f1 BY bundle2-bundle4 bundle6 bundle7;
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