Johnny Wu posted on Thursday, April 23, 2009 - 12:54 pm
Thank you very much.
Another question: How can exogenous predictors interact with time to predict growth factors, within the LGM framework?
You see, because on Preacher's website, time is treated as an predictor. Hence, a two-way interaction regressing slope on x1 with x2, would be a three-way interaction regression slope on x1 with x2 and time.
How do I obtain the regression coefficient for time?
It sounds like Preacher's approach to growth modeling is via two-level analysis with a random slope for a time variable - see the Mplus UG example 9.16. You get the same effect by letting the x1*x2 interaction variable influence the slope growth factor in the regular Mplus single-level wide approach to growth: the slope growth factor multiplies time to create the product of the 3 variables.
TECH3 gives covariances among the parameter estimates. The growth factors and covariates are variables. It would be difficult to get that without changing your model. Why not regress the growth factors on the covariates.
I have a related question. I am using Preacher's online tool to graph a multilevel interaction effect (http://www.quantpsy.org/interact/hlm2.htm) and I find myself unable to find the required asymptotic covariances using the tech3 output.
Elsewhere, it was stated that the order of the tech1 values can be used to interpret tech3. However, while the order of my tech1 output is Y, X1, X2, X1*X2, there is a 6*6 matrix in the tech3 output.
Is there a reference that explains how to interpret the respective tech3 values? Also, does tech3 contain the covariances between the predictors and the intercept y00?
I am interested in the answers to the last two questions that last poster raised. Specifically, does tech3 contain the covariances between the coefficients for the predictors and the intercept? If so, where? If not, is there another way to obtain these?
Also, I am unclear as to how to interpret the values in the tech3 output matrix. For instance:
What does the "D-06" in the above value mean?
Thank you in advance for your time and help! Lauren