Anonymous posted on Friday, May 27, 2005 - 6:33 am
How can I compare a sem mixture model (with k classes) and the same model in which one of the independant variables x, is a covariate of the categorical latent variable c ? %overall% c#1 ON x; What is the recommended test ? Thank you for your help
These models are not directly comparable using the likelihood since they have different observed variables. The recommended test is the z ratio of Est/SE that is provided in the output. If you have several x's, you can do 2 runs, one with the c on x slopes fixed at zero and one with them free and then do a 2*loglikelihood difference chi-square test.
Anonymous posted on Friday, May 27, 2005 - 12:51 pm
To follow-up on your response regarding model comparison, (1)is it not appropriate to compare a multilevel mixture model with a covariate to a multilevel mixture model without a covariate using any of the information criteria's? (2)Is it appropriate to compare single and multilevel models using any of the information criteria's? For example, could a 4-item 1-factor single level CFA be compared to the same 4-item 1-factor model at both the within and between level using AIC, BIC, aBIC?
I have two variables that I assume are Poisson distributed within class. I have three classes of girls and three classes of boys. I want to compare two models: In the first model, the distribution of the two variables for girls and boys are independent of eachother in each class. In the second model, the relations between the means among girls and boys respectively are the same across classes. For example; the ratio between the means of Variable 1 for girls and for boys in Class 1 is the same as the ratio between the mean of Variable 1 for girls and for boys in Class 2. My question is: How can I compare these models?
You can analyze boys and girls jointly using Knownclass. Then you give labels to the means and in one run you impose the parameter constraints on the means using Model Constraint and in the other run you don't. Then compute 2*the loglikelihood difference as an LRT chi-square test of the constraints. And/or, do the testing via Model Test in a single run where you don't impose the parameter constraints, giving a Wald chi-square test.