Help interpreting model with quadrati... PreviousNext
Mplus Discussion > Growth Modeling of Longitudinal Data >
 Nicole Watkins posted on Monday, June 03, 2019 - 9:54 am
I am running an unconditional growth model of depression over time with 6 time points. I've run a linear model, and then tested to see if a model with a quadratic term fit better. The chi square difference was significant. However, I am confused about my results from the model with the quadratic term:

1) MY TLI is 1.009

2) The I-S covar is significant, the I-Q covar is significant, and the S-Q covar is significant...

I -0.028 0.010 -2.649 0.008

I 0.004 0.002 2.166 0.030
S -0.005 0.002 -2.794 0.005

3) However the means for the Slope and Quadratic are NOT SIGNIFICANT.

I 0.592 0.011 53.939 0.000
S -0.006 0.009 -0.755 0.450
Q -0.002 0.002 -1.140 0.254

What would this mean? Is is still worth having the quadratic term in the model if the linear model overall fit is still very good?

Thank you,
 Bengt O. Muthen posted on Monday, June 03, 2019 - 2:49 pm
Note that if models differ with respect to a variance being free or fixed at zero, you cannot use a chi-square difference test to compare them but have to use BIC. I assume that your quadratic model has a free variance for the quadratic growth factor.

A growth factor mean not being significant doesn't mean that the growth factor should be removed if it still has variance.
 Nicole Watkins posted on Monday, June 03, 2019 - 6:20 pm
Dr. Muthen,
The variance is significant:
I 0.109 0.015 7.295 0.000
S 0.029 0.009 3.217 0.001
Q 0.001 0.000 2.552 0.011

However, the BIC is slightly larger than that of the linear trajectory.

I am sorry to ask but, I am confused as to why exactly I cannot use a chi-square difference test. Thank you for clarifying that for me.
 Bengt O. Muthen posted on Tuesday, June 04, 2019 - 5:30 pm
BIC becomes larger for the quadratic model because of the insignificant q mean - you add a parameter which doesn't improve the likelihood much. Still, you would ignore the insignificance of the q mean, ignore the higher BIC, and keep the quadratic because it has variance.

Likelihood-ratio chi-square difference testing is not ok when one of the models you are considering is a special case of the other by having parameter(s) on the border of their admissible space, such as a zero variance or a zero latent class probability. There is a large literature on that which you can google.
Back to top
Add Your Message Here
Username: Posting Information:
This is a private posting area. Only registered users and moderators may post messages here.
Options: Enable HTML code in message
Automatically activate URLs in message