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I am using lavaan to run path analysis to predict PC from WA and LC. These are multi-group models. Described here is the first step (identifying best-fitting paths); it is followed by constraining parameters to be equal across the three groups (group.equal = c("intercepts", "regressions")), which is not described here.

My question has to do with the equivalence of models. A reviewer of the manuscript indicated that these three models should be equivalent, but I am not sure. I appreciate any advice the forum can offer.

Model 1 models WA and LC predictors separately but allows (residual) covariance.

Mod_1  <- '
PC ~ c(c1, c2, c3) * WA  + c(d1, d2, d3) * LC  
WA ~~ LC
'

Model 2 models an indirect effect via WA.

Mod_2  <- '
PC ~ c(c1, c2, c3) * WA  + c(d1, d2, d3) * LC  
WA ~ c(a1, a2, a3) * LC 
'

Model 3 models an indirect effect via LC.

Mod_3  <- '
PC ~ c(c1, c2, c3) * WA  + c(d1, d2, d3) * LC  
LC ~ c(a1, a2, a3) * WA 
'

All three models were run with this syntax (models would not converge without intercepts being constrained):

fit<- sem(mod, data = new_df,  group = "grade", missing="ML",
               group.equal = c("intercepts")) 

Here are the model fit statistics:

Text

The models appear to be different. I have read that allowing exogenous variables to covary can be equivalent to modeling a path between them (see for example this article), but these results appear to indicate otherwise. Plus, in the mediated model, there is only 1 exogenous variable, so I am not sure how that applies to these models.

With regard to models 2 and 3, the only difference is directionality of the indirect effect. The only estimated parameter that differs between them is whether WA is predicted by LC (Model 2), or vice versa (Model 3). Across groups, the estimates of the slopes for WA~LC are greater for model 2 (.25-.32) than LC~WA in model 3 (.20-.23). Given that the other slope estimates are equal across models, I believe the model with greater variance explained will have better fit statistics. Yet, the reviewer indicates that these models are equivalent.

I would appreciate any clarification about the equivalence of models. I also posted my question at https://groups.google.com/g/lavaan/c/VIdHuxqRKM4.

Many thanks, Jeanne Sinclair

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1 Answer 1

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If you do not constrain intercepts to equality, then these models should be statistically equivalent. I'm not sure why you would be interested in constraining intercepts at all, but it especially does not make sense to do so before the slopes are constrained to equality.

The reason the models are not equivalent when the intercepts are held equal is because the intercepts don't mean the same thing across models. In all 3 models, the PC intercept has the same meaning (the expected value of PC when WA and LC = 0). But when there is no mediator, the WA and LC intercepts are their grand means (note that in Mod_1, you overwrite the default setting fixed.x=TRUE by explicitly specifying the WA ~~ LC covariance should be estimated, so all 3 variables' intercepts are constrained, giving you df = 6). Conversely, a mediator's intercept is its expected value when the exogenous predictor = 0, but the predictor's intercept would be its grand mean. However, under the default setting fixed.x=TRUE, both mediation models do not estimate the exogenous predictor's mean/variance (both are merely fixed at the observed summary statistics), so you only constrain endogenous intercepts across groups, giving you df = 4. Either setting fixed.x=FALSE or removing the exogenous WA ~~ LC covariance from Mod_1 would give all models the same df, but they are still not equivalent. They might with complete data, if they were mean-centered (so the intercept is the grand mean), but mean-centering doesn't work like you expect with missing data.

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