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 Morayo Ayodele posted on Saturday, June 30, 2012 - 2:25 pm
Hello Dr. Muthen,

I am trying to run a third-order factor CFA with TWO second-order factors both of which have TWO first-order factors each. I understand that this model is not directly identified but I have substantive reasons to suggest such a model. Moreover, I have come across articles where similar models were run. Could you please advice on the appropriate syntax to use that would return an output without errors?

For the second-order model, this is the syntax I feel is most reasonable:

Model:
f1 by sglse41 sglse42 sglse67 sglse76 sglse100 sglse30 sglse40 sglse55 sglse83 sglse92 sglse97 sglse102;
f2 by sglse3 sglse17 sglse25 sglse68 sglse35 sglse58 sglse78 sglse84 sglse94 sglse96 sglse98;

f3 by f1* f2;
f3@1;
Output: Sampstat standardized mod;

For the third-order model, this is the syntax I feel is most reasonable:

Model:
f1 by sglse3 sglse17 sglse25 sglse68;
f2 by sglse41 sglse42 sglse67 sglse76 sglse100;
f3 by sglse35 sglse58 sglse78 sglse84 sglse94 sglse96 sglse98;
f4 by sglse30 sglse40 sglse55 sglse83 sglse92 sglse97 sglse102;

f5 by f1* f2;
f6 by f3* f4;
f7 by f5* f6;

f5@1;
f6@1;
f7@1;
Output: Sampstat standardized mod;
 Linda K. Muthen posted on Sunday, July 01, 2012 - 10:45 am
A second-order factor must have a minimum of three first-order factors to be identified unless you place a restriction on the model such as having both factor loadings equal to one. The model with three second-order factors may be identified by borrowing from other parts of the model.
 Morayo Ayodele posted on Tuesday, July 03, 2012 - 10:23 am
Thank you very much. Your suggestion solved the problem.
 Morayo Ayodele posted on Friday, July 06, 2012 - 8:57 am
Hello Dr. Muthen,

After constraining the two first-order factors to 1, I observed that the fit indices for both the two-factor model and the second-order model are exactly the same.

Does this matter or does it indicate an error somewhere?

Model:
f1 by sglse41 sglse42 sglse67 sglse76 sglse100 sglse30 sglse40 sglse55 sglse83 sglse92 sglse97 sglse102;
f2 by sglse3 sglse17 sglse25 sglse68 sglse35 sglse58 sglse78 sglse84 sglse94 sglse96 sglse98;

f3 by f1@1 f2@1;

Output: Sampstat standardized mod;
 Linda K. Muthen posted on Friday, July 06, 2012 - 11:06 am
There are no degrees of freedom for the second-order factor. You cannot asses the fit of it above the fit of the first-order factors. You would need at least four first-order indicators.
 Morayo Ayodele posted on Sunday, July 08, 2012 - 10:48 am
Thank you
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