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Re: st: Hierarchical CFA problem


From   John Antonakis <John.Antonakis@unil.ch>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: Hierarchical CFA problem
Date   Mon, 22 Apr 2013 13:21:39 +0200

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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