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?
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.
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@0y2@1y3@2 ... y20@19 I would estimate a "Random Quadratic, Random Linear Model".