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Re: st: Hierarchical CFA problem


From   W Robert Long <W.R.Long@leeds.ac.uk>
To   "statalist@hsphsun2.harvard.edu" <statalist@hsphsun2.harvard.edu>
Subject   Re: st: Hierarchical CFA problem
Date   Mon, 22 Apr 2013 13:45:45 +0100

Dear all

After discussions with my senior colleague, we found that using the "difficult" option in Stata gets to the same result as with R - whereas without the option Stata is getting stuck at a lower log likelihood.

Maybe this has to do with different optimisation algorithms ?

Thanks
Robert Long


On 22/04/2013 12:54, W Robert Long wrote:
Hi John and thanks for your reply

Yes, I forgot to include the SDs in my post - I am working with raw data and it was an oversight - apologies for that, here they are:

ssd set sd 1.167432 1.177451 1.130979 1.164116 1.290472 1.095603 1.089534 1.012615 1.009152

FWIW, here are the means:

ssd set means 4.93577 6.08804 2.250415 3.060908 4.340532 2.185572 4.185902 5.527076 5.374123

Regarding your second point, I was aware of the equivalence of models. Nevertheless I still wanted to fit the hierarchical model, as outlined in Kline. Actually the issue came up because one of my students tried to fit the model in class: it wouldn't converge and she wanted to know why. So far I haven't been able to find a reason - only the work-around of changing the unit loading variable (or to fit it in R)

Thanks again
Robert Long

On 22/04/2013 12:21, John Antonakis wrote:
Hi Robert:

There is one issues to deal with before trying to reproduce the results
and another issue about the estimation per se.

First issue is that you have not set the means and standard deviations;
was that an oversight? There is not much point in estimating the model
with a correlation matrix:

Bentler, P. M., & Savalei, V. (2010). Analysis of correlation
structures: Current status and open problems. In S. Kolenikov, L. Thombs
& D. Steinley (Eds.), Recent Methodological Developments in Social
Science Statistics (pp. 1-36). Hoboken, NJ Wiley.
Browne, M. W. (1984). Asymptotically distribution-free methods for the
analysis of covariance structures. British Journal of Mathematical and
Statistical Psychology, 37, 62-83.
Cudeck, R. (1989). Analysis of correlation matrices using covariance
structure models. Psychological Bulletin, 105(2), 317-327.
Steiger, J. H. (2001). Driving fast in reverse - The relationship
between software development, theory, and education in structural
equation modeling. Journal of the American Statistical Association,
96(453), 331-338.

Second issue about the esestimation. A model with three first order
factors is equivalent to the model with a higher order factor predicting
the three factors (if you work out the DF by hand you'll see that they
are the same). So, you are not testing anything with the hierarchical
CFA beyond a first-order three factor theory.

Rindskopf, D. & Rose, T. 1988. Some Theory and Applications of
Confirmatory Second-Order Factor Analysis. Multivariate Behavioral
Research, 23: 51-67.

Best,
J.

__________________________________________

John Antonakis
Professor of Organizational Behavior
Director, Ph.D. Program in Management

Faculty of Business and Economics
University of Lausanne
Internef #618
CH-1015 Lausanne-Dorigny
Switzerland
Tel ++41 (0)21 692-3438
Fax ++41 (0)21 692-3305
http://www.hec.unil.ch/people/jantonakis

Associate Editor
The Leadership Quarterly
__________________________________________

On 22.04.2013 11:31, W Robert Long wrote:
Hi all

I'm working with the hierarchical CFA model of cognitive ability
described on p199 of Kline "Principles and Practice of Structural
Equation Modeling", 2nd edition - or p249 in the 3rd edition. I have
reproduced summary statistics so that people who don't have access to
the data will be able to follow:

clear all
ssd init x1 x2 x3 x4 x5 x6 x7 x8 x9

ssd set correlations ///
1.0000 \ ///
0.2973   1.0000 \ ///
0.4407   0.3398   1.0000 \ ///
0.3727   0.1529   0.1586   1.0000 \ ///
0.2934   0.1394   0.0772   0.7332   1.0000 \ ///
0.3568   0.1925   0.1977   0.7045   0.7200   1.0000 \ ///
0.0669  -0.0757   0.0719   0.1738   0.1020   0.1211   1.0000 \ ///
0.2239 0.0923 0.1860 0.1069 0.1387 0.1496 0.4868 1.0000 \ ///
0.3903   0.2060   0.3287   0.2078   0.2275   0.2142   0.3406 0.4490
1.0000

ssd set observations 301

The model is very straight forward:

sem (L1 -> x1 x2 x3) ///
     (L2 -> x4 x5 x6) ///
     (L3 -> x7 x8 x9) ///
     (G -> L1@1 L2 L3)

However, it fails to converge. It does however, converge if the G ->
L2 path is constrained to 1 instead of the G -> L1 path

I am trying to figure out what the problem is. L1 is the largest
loading on G , but using the iterate() option I don't see what the
problem is.

I have successfully fitted the model using R, with the lavaan
package,  and the model with the G -> L2 path constrained has
identical output in Stata and R, so I believe this must be a issue
with Stata, but  I do not know how to track the problem down. FWIW,
here is the output from R for the model which doesn't converge in Stata:

                    Estimate  Std.err  Z-value  P(>|z|)
Latent variables:
   L1 =~
     x1                1.000
     x2                0.554    0.100    5.554    0.000
     x3                0.729    0.109    6.685    0.000
   L2 =~
     x4                1.000
     x5                1.113    0.065   17.014    0.000
     x6                0.926    0.055   16.703    0.000
   L3 =~
     x7                1.000
     x8                1.180    0.165    7.152    0.000
     x9                1.082    0.151    7.155    0.000
   G =~
     L1                1.000
     L2                0.662    0.173    3.826    0.000
     L3                0.425    0.118    3.602    0.000

Variances:
     x1                0.549    0.114
     x2                1.134    0.102
     x3                0.844    0.091
     x4                0.371    0.048
     x5                0.446    0.058
     x6                0.356    0.043
     x7                0.799    0.081
     x8                0.488    0.074
     x9                0.566    0.071
     L1                0.192    0.170
     L2                0.709    0.107
     L3                0.272    0.069
     G                 0.617    0.183


I would be grateful for any help or advice.

Thanks
Robert Long





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