Modeling intecept by T0 measure PreviousNext
Mplus Discussion > Growth Modeling of Longitudinal Data >
 Bill Dudley posted on Monday, December 15, 2008 - 11:04 am
I am modeling change over 6 time points with a quadratic model conditioned on type of treatment where treatment is indicated by two dummy codes. Fit is only ok, CFI = .91 and RMSEA = .07. Mod indicies indicate that fit would be inproved by freeing i by the observation at time 0.

When I do this CFI jumps to .97 and RMSEA drops to < .05. However I am not sure how to interpret the parameters in order to graph the three curves. The I BY Time 0 path is .73. and the intercept increases by about 50% (1.7 to 2.4).

So my understand is that the intercept is now indicated both within the growth model and directly by the time 0 observation.

Does this different specification alter how I would graph the curves (using the parameters for i s q shown under intercepts and the estimates of the effect on the two dummy codes on I S and Q?
 Bengt O. Muthen posted on Monday, December 15, 2008 - 11:20 am
Freeing that parameter makes you fall outside the growth modeling framework. It indicates that you don't have a quadratic growth model, but that the first time point needs some other treatment. Often there is a big initial drop, so that you need a 2-piece growth model.
 Bill Dudley posted on Thursday, December 18, 2008 - 9:17 am
Thank you for the clarification. I can live with the only ok model fit.

However, I have one question about interpretation of one of my loadings. In one dummy coded condition the effect of the dummy code on the intercept although not significant would take the intercept for that condition to a negative value which is meaningless.

As I graph the data I can simply fix this to zero but I wonder how I can constrain the intercept for one group to be zero in the analyses so that the the slope and quadratic terms are estimated correctly?

 Bengt O. Muthen posted on Thursday, December 18, 2008 - 2:00 pm
Are you sure that a negative intercept is meaningless? It is not the same as a negative mean.
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