CODEWITHSUNDEEP
rr-lavaanstructural-equation-model
Pascal
Pascal

Reputation: 1

Specifying a latent variable with a single indicator in lavaan

I would like to run a SEM in lavaan with two multiple indicator variables and one single indicator variable. The single indicator is a person ability parameter from an IRT model.

My model looks somewhat like this:

model1 <- '
   Factor.A =~ a1 + a2 + a3 + a4
   Factor.B =~ b1 + b2 + b3 + b4
   Factor.C =~ c1
   c1~~c1(1-0.89)*c1

C1 is the single indicator. The reliability of the availity parameter is 0.89.

Question 1: Is the specification of the single indicator latent variable (Factor.C) correct? And Question 2: How would one interprete the factor loading of the single indicator? In my model, I get 0.93 as the factor loading. Is it problematic that it is so high?

I looked in different forums, papers and other R support instruments but I couldnt find a satisfying answer to this issue. I wuould be very grateful if someone knows and likes to share how to deal with this issue.

Upvotes: 0

Views: 596

Answers (1)

Terrence
Terrence

Reputation: 1250

Is the specification of the single indicator latent variable (Factor.C) correct?

No, you specify the righthand-side variable both before and after the operator. Operators are only recognized when placed before the righthand-side variable (see ?model.syntax)

c1 ~~ .11*c1

And unless your c1 sample variance is exactly 1, then you can use labels to constrain the common- and unique-factor variances to the ratio of (un)reliability:

Factor.C =~ 1*c1
Factor.C ~~ commonVar*Factor.C
c1       ~~ uniqueVar*c1
## constraint consistent with commonVar/uniqueVar = .89 / .11
uniqueVar == commonVar * .11 / .89

How would one interpret the factor loading of the single indicator?

With a single-indicator construct, it is more straight-forward to think of it as a variance decomposition. That is reflected by fixing the loading to 1 in my syntax above, so the variance of c1 is simply the sum of common- and unique-factor variances.

When you fix Factor.C's variance to 1 and estimate the loading, it is statistically equivalent, but the loading is the square-root of the common-factor variance component (i.e., the SD).

Upvotes: 0

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