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Mplus Discussion > Growth Modeling of Longitudinal Data >
 Matt Keough posted on Thursday, March 26, 2015 - 2:32 pm

I am attempting to look at the unconditional model for growth of alcohol-related cognition in a longitudinal data set with 4 time points. I observed that the linear slope provided poor fit to the data and thus I added the quadratic slope term and received the following error:


I looked and there are no negative residual variances and the tech 4 output has a 999 for the correlations among the latent quadratic, linear, and intercept variables. Can you suggest a way to fix this issue? Thanks kindly,

 Bengt O. Muthen posted on Thursday, March 26, 2015 - 3:30 pm
999 is typically printed for correlations when you have a negative variance and therefore cannot get the correlation, but you claim that this is not the case, so please send output to support so we can see the whole story.
 Lisa Gudenkauf posted on Thursday, March 26, 2015 - 6:40 pm

I am having a similar problem, and I also have additional questions.

I am trying to run a basic LGM of a psychological outcome variable measured at four different timepoints (0 months, 2 months, 6 months, & 12 months) in a sample of 181 participants. I am hoping to determine whether a linear model or a quadratic model better fits the data before proceeding with my other aims of (a) including group condition as a predictor of change over time, and (b) testing for moderator effects. I would really appreciate if you could help me with questions that have arisen.

First, would you be able to tell me which information I should be using to decide between the linear and the quadratic models. It has been suggested to me that linear and quadratic models based on the same 4 timepoints are not nested, so I cannot use a chi-square difference test. I was told I must, instead, compare AIC and BIC for the linear and quadratic models. Is this correct?

Secondly, should I be interpreting the “model results” or the “STDYX standardization results” for this LGM, particularly when I add predictors?

Thirdly, I have been struggling with model fit for the linear and the quadratic models, and I am hoping you might be able to help me. I can specify these in a separate post due to space.
 Lisa Gudenkauf posted on Thursday, March 26, 2015 - 7:05 pm
I ran the following linear model: MODEL: i s | T1@0 T2@2 T3@6 T4@12;

The MODEL FIT INFORMATION for this linear model is as follows:
Free parameters = 9
Ho = -613.702; H1 = -593.453
AIC = 1245.403; BIC = 1274.190; Sample-size Adjusted BIC = 1245.686
Chi-Square = 40.497, df = 5, p = 0.0000
RMSEA = 0.198
CFI = 0.918; TLI = 0.902
Chi-Square for Baseline Model = 438.510, df = 6, p = 0.0000
SRMR = 0.179

My understanding is that this indicates poor model fit. I am having difficulty determining how to make the model fit better. I have tried freely estimating timepoints, but the estimates provided are not close to the actual timepoint (e.g., freeing T2 results in an estimate of T2 = 21 months). I have also tried constraining different timepoint variances to be equal, but this has not improved fit. I’m not exactly sure what the suggested “ON” (e.g., T1 on T1) and “BY” statements suggested by Modification Indices would actually mean in the context of this LGM model or if they would be helpful.
 Lisa Gudenkauf posted on Thursday, March 26, 2015 - 7:06 pm
I also ran the following quadratic model: MODEL: i s q | T1@0 T2@2 T3@6 T4@12;

The output states that the standard errors couldn't be computed, which appears to be due to a negative residual variance for T4 of -1.776. I tried adding “T4@0;” to the model, but I got a Non-Positive Definite error message. It seems the slope has a negative variance of -0.004. Next, I tried adding “s@0.”

Now the model runs, but I’m no longer sure if I can be confident in this quadratic model given the model specification changes.

The quadratic MODEL FIT INFORMATION is as follows:
Free parameters = 9; Ho = -610.618; H1 = -593.453
AIC = 1239.236 / BIC = 1268.023
Sample-size Adjusted BIC = 1239.519
Chi-Square = 34.330, df = 5, p = 0.0000
RMSEA = 0.180; CFI = 0.932; TLI = 0.919
Chi-Square for Baseline Model= 438.510, df = 6, p = 0.0000; SRMR = 0.196

Based on AIC and BIC, it doesn't seem that the quadratic model is significantly better than the linear model. My sense is that I should go with the linear model and try to improve model fit for the linear model before adding predictors, but I am feeling stumped. Any input or feedback you may have would be greatly appreciated!
 Bengt O. Muthen posted on Friday, March 27, 2015 - 11:25 am
Please note that we request doing only 1-window posts.

Q1. I would use BIC to settle on the model.

Q2. Stand'd or not is your choice and depending on your purpose - ask on SEMNET if you like.

Q3. See our handout and video from Topic 3, slides 49-51.

Often, a quadratic model with fixed-zero variance, but free non-zero mean is a good alternative.
 Lisa Gudenkauf posted on Monday, March 30, 2015 - 3:07 pm
Thank you very much! This is my first time posting, so I apologize for my earlier long post. I will definitely only do 1-window posts in the future. I appreciate your help with answering my questions.

I'm glad to hear that fixing the slope variance to zero is acceptable for a quadratic model. Since I have now tried the two recommended model modifications from Pg. 51 of the handout (i.e., freeing time scores for slope growth factors and fixing residual covariances) and the model still does not fit, I am not sure how to proceed. Are there any other options for improving model fit? Would any of the "BY" or "ON" statements from the Mod Indices make sense in the context of this LGM model?

One other question I have relates to one of the covariates I am trying to include in my model - a binary ethnicity variable. The linear model runs with this covariate, but the quadratic model does not run properly. (Neither model fits well.) The output shows problems with the relationship between the intercept and this binary covariate. Do you happen to have ideas as to why this might be?

Would it be helpful for me to send an output via email to help clarify my question?
 Bengt O. Muthen posted on Wednesday, April 01, 2015 - 11:37 am
Fist, use the plot menu to look at the individual observed trajectories to see if a linear or a quadratic model looks suitable. Perhaps your development has quite a different shape.

If that doesn't help, send your outputs to, both the model without a covariate that you have trouble fitting and the quadratic model with a binary covariate. Be sure to include sufficient Output options inluding Sampstat.
 Lisa Gudenkauf posted on Sunday, April 12, 2015 - 10:02 am
Thank you for your response and your help!

I believe you're right that the development has a different shape. It appears more cubic, with two bends. However, with only 4 timepoints, I don't think I have enough timepoints to model this shape. Do you happen to have recommendations as to how to examine differential change over time between my three groups, given these limitations?
 Bengt O. Muthen posted on Sunday, April 12, 2015 - 5:23 pm
You can try a quadratic with one time score free to be estimated. Or you can try to fit only the first 3 timepoints. You also want to look at whether the variances of the outcomes get larger or smaller over time. Data where the means increase and the variances decrease over time are often hard to fit with regular growth models.

You may also ask on SEMNET and/or Multilevelnet.
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