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Tim Seifert posted on Wednesday, February 21, 2007 - 7:09 am
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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? |
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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. |
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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 |
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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. |
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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 |
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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. |
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EFried posted on Monday, February 06, 2012 - 8:04 am
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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 |
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Yes, in theory, but a quadratic growth model is not identified with three time points. |
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EFried posted on Wednesday, February 08, 2012 - 6:20 am
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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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sed48 posted on Monday, June 12, 2017 - 4:56 am
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Dear Muthen, I have 6 timepoints in my dataset. I have tried a variety of models, including linear, quadratic, and piecewise models (with 1 breaking point). I have tried two types of piecewise models, either where both slopes are linear, or where one is quadratic and the other linear. My question concerns the piecewise model with both a linear segment and quadratic segment. i s1 q1 | y0@0 y1@1 y2@2 y3@3 y4@3 y5@3 i s2 | y0@0 y1@0 y2@0 y3@0 y4@4 y5@5 I am wondering whether this is really too complicated a model for the number of timepoints I have and it would result is overfitting. I would appreciate your advice on this. Many thanks! |
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I would think that piecewise is more motivated when there is a substantive reason for a change in the growth. Perhaps you can instead try a cubic. |
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Dear all, would it make sense to constrain the residual variance of the observed variable to zero to allow the estimation of the quadratic term with three timepoints? I am more interested in the actual trajectory of the data, and an initial observation suggests that the quadratic curve is best representing the data. Would alternatively make sense if I would estimate a quadratic term without a linear one? Thank you! |
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Q1: No. Q2: No. Equal residual variances over time and no quadratic variance might do it. |
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Dear Dr. Muthen, thank you for your prompt answer. My research questions look at predictors of the slopes (linear and quadratic). I think I may not be able to answer them if I set no variation in the quadratic slope. Do you have any hint on the best statistical strategy to adopt? Thank you in advance! |
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You can regress quadratic on covariates even if the quadratic residual variance is zero. But a quadratic for 3 time points is always going to be difficult to defend. You may want to discuss this on SEMNET. |
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ksk posted on Tuesday, June 30, 2020 - 1:04 pm
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Dear Dr. Muthen, I'm trying to fit a quadratic growth model using 6 time points. I want to set the intercept at the third time point. I used the command below: i s q| GPK@-2 GK@-1 G1@0 G2@1 G3@2 G5@4; However, the loadings on the quadratic slope didn't look correct. It was shown 4, 1, 0, 1, 4, and 16. So, I generated a new command below and it seems correct: i s| GPK@-2 GK@-1 G1@0 G2@1 G3@2 G5@4; i q| GPK@-4 GK@-1 G1@0 G2@1 G3@4 G5@16; Would you please let me know which command should I use? Thank you! |
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Your statemment i s q| GPK@-2 GK@-1 G1@0 G2@1 G3@2 G5@4; says that the "s" loadings have the values you give. The "q" values are the square of those. Keep this way of doing it. |
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