Nested CFAs create linear dependency? PreviousNext
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 Nicholas Aramovich posted on Friday, September 01, 2006 - 8:43 pm
My objective is to compare the fit of a 6-factor CFA with a 3-Factor CFA using Chi-Square difference testing. In order to create a 3-Factor model that is nested under the 6-Factor, I believe I must keep all 6 factors, but constrain the correlations between some factors to 1.0.

Below is my input.

In the model, I would like to specify that factors F2, F3, F4, and F5 are 1 factor, therefore all correlations between these factors are constrained to 1.0. Additionally, to identify the model, all factors are standardized (e.g F1@1), which makes the 1.0 a correlation (as opposed to covariance).

All indicators are continuous

Estimation = ML

MODEL:
F1 by a1* a2 a3 a4 a5;
F2 by b1* b2 b3 b4 b5 b6 b7;
F3 by c1* c2 c3 c4 c5;
F4 by d1* d2 d3 d4;
F5 by e1* e2 e3 e4;
F6 by g1* g2 g3 g4;
F1@1;
F2@1;
F3@1;
F4@1;
F5@1;
F6@1;
F2 with F3-F5@1;
F3 with F4-F5@1;
F4 with F5@1;

Please see next post for questions...
 Nicholas Aramovich posted on Friday, September 01, 2006 - 8:44 pm
However, when I run this model I get an error message:

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, OR A LINEAR DEPENDENCY AMONG MORE THAN TWO LATENT VARIABLES.

CHECK THE TECH4 OUTPUT FOR MORE INFORMATION. PROBLEM INVOLVING VARIABLE F3.

I inspected the Tech 4 output and it is not the case that I have a negative variance or residual variance.

It is the case that I have a correlation = 1 between two latent variables, but this is of course by design.

So, the question becomes whether constraining factor correlations to 1.0 creates a linear dependency...

And further, if this is the case, how may one create nested CFA models in order to conduct Chi-square difference tests?

Much thanks,
Nick
 Richard E. Zinbarg posted on Saturday, September 02, 2006 - 2:57 am
have you tried simply combining factors 2, 3, 4 and 5 from your 6 factor model into one factor? This is conceptually identical to constraining the correlations among them to equal 1 and I suspect will eliminate your warning.
 Nicholas Aramovich posted on Saturday, September 02, 2006 - 7:48 pm
Richard, I appreciate the suggestion, but I don't believe the model you suggest is nested under a six factor model, and therefore a chi-square difference test may not be conducted.
 Bengt O. Muthen posted on Monday, September 04, 2006 - 10:10 pm
Answer to Aramovich: Restricting the F2-F5 correlations to 1 forces correlations to be on the border of their admissible parameter space. This violates a key assumption for the test of nested models to be chi-square distributed, so I don't think chi-square difference testing works for this situation.
 Nicholas Aramovich posted on Tuesday, September 05, 2006 - 1:33 pm
Dr. Muthen,

Thanks for your response...

If I understand correctly, however, this would mean that guidelines for creating nested CFA models (e.g. Anderson & Gerbing, 1988) should not be applied because fixing several correlations to 1.0 results in the situation you describe above...
 Richard E. Zinbarg posted on Tuesday, September 05, 2006 - 8:01 pm
Nicholas, I believe that model I suggested is a reparamertization of the model you tried running, I think this could be relatively easily seen if you reconceptualize your model as a higher-order model in which factors 2 through 5 all load 1.0 on a 2nd-order factor and then do a Schmid-Leiman transformation of that higher-order model and you will end up with the model I suggested (the residualized factors in the Schmid-Leiman transformation disappear because the residuals are zero). If the model you suggested and the model I suggested are equivalent ways of parameterizing the same model (which I believe they are as above) then if your model is nested under the 6 factor model (which I believe it is), so would mine. I believe Bentler makes a similar point in an old paper on model comparisons.
 Nicholas Aramovich posted on Wednesday, September 06, 2006 - 3:17 am
Richard,

I'm not totally familiar with S-L transformation, but I think I understand what you are saying. I did a little research and seems as though the following reference may be helpful to me:

On the Relationship between the Higher-Order Factor Model and the Hierarchical Factor Model. Yung, Yiu-Fai; Thissen, David; McLeod, Lori D.; Psychometrika, v64 n2 p113-28 Jun 1999.

Anyway, thanks for weighing-in on the matter.
 Richard E. Zinbarg posted on Thursday, September 07, 2006 - 2:19 am
yes the Yung, Thissen and McLeod may be helpful (albeit a bit dense) - Loehlin also has a nice introduction to the S-L transformation in his latent variables book
 Doveh Etti posted on Wednesday, September 27, 2006 - 10:44 am
Thanks for the interesting discussion above. Can anyone summarize, for the example that started the discussion above, what should be done practically when one wants to check if factors F2, F3, F4, and F5 are 1 factor.
i.e. how should we rephrase the following mplus program:

Estimation = ML

MODEL:
F1 by a1* a2 a3 a4 a5;
F2 by b1* b2 b3 b4 b5 b6 b7;
F3 by c1* c2 c3 c4 c5;
F4 by d1* d2 d3 d4;
F5 by e1* e2 e3 e4;
F6 by g1* g2 g3 g4;
F1@1;
F2@1;
F3@1;
F4@1;
F5@1;
F6@1;
F2 with F3-F5@1;
F3 with F4-F5@1;
F4 with F5@1;
 Richard E. Zinbarg posted on Saturday, February 17, 2007 - 3:53 am
sure, I would "rephrase" as below and then compare the fit to the six factor model (in which none of the factors are constrained to have correlations of 1.0)

Estimation = ML

MODEL:
F1 by a1* a2 a3 a4 a5;
F2 by b1* b2 b3 b4 b5 b6 b7 c1 c2 c3 c4 c5
d1 d2 d3 d4 e1 e2 e3 e4;
F3 by g1* g2 g3 g4;
F1@1;
F2@1;
F3@1;
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