Modeling a quadratic term with three ... PreviousNext
Mplus Discussion > Growth Modeling of Longitudinal Data >
Message/Author
 Tim Seifert posted on Wednesday, February 21, 2007 - 7:09 am
Hello,
The data I would like to model is children's recognition of printed words. At pretest (t=0), posttest (2 months), and follow-up (8 months) the number of words children could identify was recorded. The pretest data was normally distributed, a little less so at the other time points. Two questions.

1) Should the data be modelled as Poisson or continuous distributions? If modelled as a continuous (normal) distribution, the intercept = the meannumber of words at pretest (5.5). When modelled as Poisson, I am not sure what the intercept represents since its value is about 1.3.

2) Examination of the data suggests a quadratic term would be useful. I noticed in previous posts that four time points are preferred to three, and in adding a quadratic term the number of free parameters is zero. But I tried constraining the intercept and linear means, variances and covariances, but still have no free parameters. Can a quadratic term be modelled with only three time points?
 Bengt O. Muthen posted on Saturday, February 24, 2007 - 4:17 pm
1) With high counts, Poisson is approximately normal so I would go with normal since it is simpler to work with. The Poisson means are on a log rate scale where rate is the "lambda" parameter of a Poisson distribution (see stat texts),

2)You can have a quadratic even with 3 time points, but you can't have all of its potential var-cov parameters free. You can for example have a random intercept and random linear slope but a fixed quadratic. That is just-identified. There is no use in restraining the model just to get df's.
 Johannes Meier posted on Sunday, July 31, 2011 - 11:13 am
Dear all,

I have estimated a "Fixed Quadratic, Random Linear Model" in Mplus using the following multilevel syntax:

! time-level model;

%WITHIN%
linear | y ON t_linear;
quadratic | y ON t_quadratic;

! person-level model;

%BETWEEN%
y*
linear*;
y WITH linear;
quadratic@0;

However, I didn't figure out how to specify this model using the Mplus growth curve syntax (i.e. i s q | y1@0 y2@1 y@2) . I would greatly appreciate your help.

Thanks,
Johannes Meier
 Linda K. Muthen posted on Monday, August 01, 2011 - 10:50 am
i s q | y1@0 y2@1 y@2

is the way to specify a quadratic growth model. The problem is that it is not identified with three time points. You need at least four.
 Johannes Meier posted on Monday, August 01, 2011 - 2:39 pm
Hi Linda,

Thanks for your answer.

I have over 20 time points and I just wondered how to specify a "Fixed Quadratic, Random Linear Model" in the regular MPlus growth curve syntax. In my example above I did it the multilevel way (which I am used to as a former Stata user).

If I would use just
i s q | y1@0 y2@1 y3@2 ... y20@19
I would estimate a "Random Quadratic, Random Linear Model".

Do you have any hint for me?

Best regards,
Johannes
 Linda K. Muthen posted on Tuesday, August 02, 2011 - 8:06 am
i s q | y1@0 y2@1 y3@2 ... y20@19
q@0;

I would divide the time scores by ten given that the quadratic times scores are the squared value of the linear time scores.
 EFried posted on Monday, February 06, 2012 - 8:04 am
Dear Dr Muthén,

when adding a quadratic term

i s q | y1@0 y2@1 y3@2

Does the term "q" also have to be added to the

i s (Q) ON x1 x2 x3;

and the

c%1
y1 y2 y3 i s (Q?);

statements?

Thank you
 Linda K. Muthen posted on Monday, February 06, 2012 - 9:11 am
Yes, in theory, but a quadratic growth model is not identified with three time points.
 EFried posted on Wednesday, February 08, 2012 - 6:20 am
I have 5 time points and just copied the example from above - apologies for the confusion.
Thank you for the answer, I'll include the quadratic term in my ".. ON .." statements!
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