Message/Author
 Jim Prisciandaro posted on Wednesday, February 14, 2007 - 1:20 pm
Hello Drs. Muthen

I am attempting to fit a latent growth curve model with one continuous measure assessed at 9 time points. When the model is estimated with an intercept and linear slope, fit is marginally acceptable and there are no heywood cases. Adding a quadratic growth factor to the model improves model fit substantially.

However, when a quadratic growth factor is added to the model, there are several numbers in the stdyx columns of the linear slope and the quadratic slope that are greater than 1. There are no R-square values greater than 1, and no negative residual variances. Would you recommend sticking with the intercept + linear slope only model in this case? I would like to keep the quadratic factor if possible because of its benefit to model fit and its conceptual importance.
 Linda K. Muthen posted on Wednesday, February 14, 2007 - 3:59 pm
Are you saying Stdyx for the fixed time scores? Which parameters exactly are you referring to?
 Jim Prisciandaro posted on Thursday, February 15, 2007 - 8:05 am
I think so. Below is selected output so there will be no confusion. (I didn't paste intercept to save space)
Estimates S.E. Est./S.E. Std StdYX
SB |
B0 0.000 0.000 0.000 0.000 0.000
B1 1.000 0.000 0.000 2.449 0.261
B2 2.000 0.000 0.000 4.898 0.508
B3 3.000 0.000 0.000 7.347 0.682
B4 4.000 0.000 0.000 9.796 0.861
B5 5.000 0.000 0.000 12.245 1.040
B6 6.000 0.000 0.000 14.694 1.267
B7 7.000 0.000 0.000 17.143 1.432
B8 8.000 0.000 0.000 19.592 1.521
QB |
B0 0.000 0.000 0.000 0.000 0.000
B1 1.000 0.000 0.000 0.273 0.029
B2 4.000 0.000 0.000 1.091 0.113
B3 9.000 0.000 0.000 2.455 0.228
B4 16.000 0.000 0.000 4.364 0.383
B5 25.000 0.000 0.000 6.819 0.579
B6 36.000 0.000 0.000 9.819 0.846
B7 49.000 0.000 0.000 13.365 1.116
B8 64.000 0.000 0.000 17.456 1.355
 Linda K. Muthen posted on Thursday, February 15, 2007 - 9:03 am
The standardized values have no meaning for the fixed time scores.