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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. |
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Are you saying Stdyx for the fixed time scores? Which parameters exactly are you referring to? |
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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 |
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The standardized values have no meaning for the fixed time scores. |
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