hierarchical model with two first-order factors and one second-order factor

[Mayke]

Hi! I am new to R and to CFA as well, but I succeeded in running the analysis so far... I got stuck at testing a hierarchical model:

# usability =~ usable + learning

 

SUShierarchicalmodel <- 'usable =~ S1 + S2 + S3 + S5 + S6 + S7 + S8 + S9

learning =~ S4 + S10

usability =~ usable + learning

'

#model fit

 

fit_hierarchical <- cfa(SUShierarchicalmodel, data=Final.df.items10, estimator = "WLSMV", std.lv=TRUE, missing="pairwise")

I received the following messsage:

Warning messages:

1: In lav_model_vcov(lavmodel = lavmodel, lavsamplestats = lavsamplestats,  :

  lavaan WARNING:

    Could not compute standard errors! The information matrix could

    not be inverted. This may be a symptom that the model is not

    identified.

2: In lav_test_satorra_bentler(lavobject = NULL, lavsamplestats = lavsamplestats,  :

  lavaan WARNING: could not invert information matrix

Is this R telling me that trying to test a hierarchical model with only two first-order factors (of which one only contains two items) and one single-order factor is stupid/incorrect/impossible? 

Many thanks!

Mayke 

[Edward Rigdon]

A basic rule for statistical identification of constrained factor models is that every factor must have at least 3 strong indicators, or at least 2 if the factor covaries with something else in the model. For your second order factor, the two first order factors are the indicators, but there is nothing else for the second order factor to covary with. Moreover, you must make sure that the variance of each common factor is determined, either by directly fixing it to one (the std.lv=T option works) or by constraints on indicator loadings (which lavaan does by default for first order factors, but maybe not for higehr order factors).

So you have two problems. First, regardless of numbers of indicators, make sure that a constraints is in place to define the variance of each factor. Second, with only two indicators for the second order factor, you must impose an additional constraint. You might, for example, constrain the paths from the second order factor to the first order factors to equality.

Both of these are limitations of factor models generally and not specific to software.

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[Terrence Jorgensen]

Is this R telling me that trying to test a hierarchical model with only two first-order factors (of which one only contains two items) and one single-order factor is stupid/incorrect/impossible? 

Yes.  2-indicator factors are not identified unless they are substantially correlated with other factors (unless residual correlations make it unidentified again).  You should fix both factor loadings to 1, in which case the estimated factor variance is equivalent to the covariance between indicators (i.e., the higher-order model does not give you any new information, and only could if it had at least 4 first-order factors as indicators).

Terrence D. Jorgensen

Assistant Professor, Methods and Statistics

Research Institute for Child Development and Education, the University of Amsterdam

UvA web page: http://www.uva.nl/profile/t.d.jorgensen

 

[Mayke]

Thank you very much for the quick and helpful replies! 

Op maandag 1 april 2019 15:37:51 UTC+2 schreef Mayke: