My main intention is to know if four variables form two constructs or just one. The problem is that two of these four variables are highly correlated (r. 97), however, these two variables form part of a same construct and the other two form part of another construct. I want to know if these highly correlated variables can be employed in CFA despite the problem of high correlation. If so, can you give me any reference? I have always heard that in CFA, multicollinearity might be a problem. I have tried Principal component analysis where multicollinearity is not a problem but I have a problem that one construct is dependent on the other. Thanks a lot for your help Marina
bmuthen posted on Monday, November 14, 2005 - 5:58 am
Multicollinearity is a problem if you use the two constructs in the prediction of another variable. If for example they are used as dependent variables, the problem would not arise.
I am running an multi-group model (cluster is 0 and 1) with TYPE=COMPLEX. I suspect that there is some degree of multicollinearity in group 1 that doesn't exist in group 0.
I have 2 questions that I'm hoping you can help me with.
1) Is there a way of incorpoating a "ridge regression" into the model to stablize the SE estimates, or is there a method of adjusting the SE's for multicollinearity? I am using the default MLR estimator, is there a better estimator when there are near multicollinearity cases?
2) Can I use the MODEL 1: function to add a second order factor behind the highly correlated factors in group 1 to try to control for this? If so, how do I specify a different factor structure in model 1 than in Model 0?
1. Ridging is only done for observed covariates and then it is automatic with no user input possible.
2. A second-order factor model can help by predicting from the second-order factor instead of the first-order factors. Each group can have a different model. You specify the full model in the general part and then you set second-order factor parameters to zero in the group-specific part for the group that doesn't have the second-order factor.