Re: st: Hierarchical CFA problem

[John Antonakis <John.Antonakis@unil.ch>]

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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