Anonymous posted on Tuesday, May 04, 2004 - 4:14 pm
I'm trying to specify a growth model of my contiuous DV measured yearly from age 13 to age 22. I want to specify a piecewise model that captures linear growth from age 13-15, 16-19, and 20-22. Can you help me to determine how I should specify this model in mplus?
I am creating a similar model and have a couple of questions.
I have seven timepoints. I am interested in understanding the effect of an intervention which would have occured between timepoints 6 and 7. The outcome is the percentage of students in a school that enter college.
Overall college enrollment rates have been slowly increasing in my state each year. I want to seperate this effect from the possible effect of the treatment. Should I specify my time scores as:
0 1 2 3 4 5 6 0 0 0 0 0 0 1
0 1 2 3 4 5 5 0 0 0 0 0 0 1 ?
Finally, will the model be identified given that the second slope will apply to only a single timepoint?
I have been recently trying to specify a piecewise paramterization. The observations are in months/10 and are at initiation, 3, 6, 12 ,18, 24, 30, and 36 months. I have been trying to specify two parts, with four observations each, one over the first year and the second over the remaining two years. Is the following correct?
Note that you need to mention the intercept growth factor in both parts of the growth model. Just use the same name. The way you have your second model specified, Mplus would interpret it as a linear model because it has two growth factors. In Chapter 16, there is a table that shows a piecewise growth model. I would specify the timescores for the piecewise growth model as follows.
I have some intervention data collected at pre, post 6, 12, 24, 36 month follow-ups. I want to compare two randomized intervention conditions on various externalizing outcomes. The problem is there is a large drop in externalizing problems from pre to post, then only modest linear change in the outcome across the follow-up assessments. A quadratic does not fit the data well. It is more like a piecewise model with the first slope consisting of change from pre-post, and the second consisting of linear change occuring after the intervention has been completed.
Because there are only two timepoints for the first piece the model it is not identified. I have considered leaving out the baseline assessment since random assignment makes the two treatment groups of interest equal prior to treatment. However, I was wondering if you had any suggestions for dealing with this problem.
One approach that I have used is to work with piecewise growth as you say, where the first piece has only a random intercept (centered at the baseline) but a fixed slope. This is then identified from 2 time points. The second piece can have its own intercept and slope which are both random. So part of the intervention effect is on the fixed slope of the first piece and part is on the random intercept and slope of the second piece.
I would not leave out the baseline assessment from the growth modeling.
So would the interpretation for the first slope (which is fixed) be the the average pre-post intervention effect for all participants, and the second intercept would represent individual deviations from this average effect.
I am using a piecewise growth model to describe late elementary school and the transition to middle school. The linear model for each piece estimates fine, but when I add a quadratic function, the model will not converge.
To ask for the quadratic function, I am adding a "q" as I would in a traditional growth model. Do I need to do something different? Here is an example of the syntax...
A second question: if both slopes of a piecewise model appear similar, is there a way to test (through some kind of constraint) whether a single slope, rather than two slopes, fits the model just as well as two processes? Would you consider these models nested?
If you have no data for certain timepoints, you reflect this through the time scores, for example, if you measured in years 1, 2, and 4, the time scores would be 0, 1, and 3. You don't do this through missing data.
Usually a piecewise model is used to break a growth curve that is not linear into linear pieces so I don't understand why you are adding a quadratic growth factor.
You can test if the slope of the two pieces are equal by chi-square difference testing of the two nested models or by using MODEL TEST.
Bill Dudley posted on Thursday, July 03, 2008 - 9:26 am
I have longitudinal data on recovery from cancer treatment. There are six time points (0, 6, 12, 18, 24 and 36). The time = 0 is prior to treatment and then starting with 6 months we have the recovery phase. The means show a jump between T1 and t2 so it makes sense to fit a two piece model. The first linear piece has only two time points so I have looked at the notes below re fixing s1 variance and the s1, i covariance. I assume that in the second piece that the time 2 serves as the intercept for that curve. I also want to include covariates - age and treatment type (a binary variable). Question1 - So I wonder if I should model an i2 as well as i. Question2 - If s1 and i are fixed then it makes no sense to look for predictors of s1…. Correct? Question 3 - I also can fit a nice quadratic curve to the data without the piecewise approach. I assume that although this fits, it is not valid given the nature of the data collection and the mean and that I should stick with the piecewise… Correct?
Q2 You can. Think of a gender dummy variable where s1 is different for males and females - that can be identified. But the residual variance for s1 then needs to still be zero.
Q3 If the second piece does not need a quadratic and if having only one piece that is quadratic fits the data well I would do one piece quadratic. My experience is that 2 pieces are needed - and they might have different covariate influence.
Use the 1-piece quadratic if it fits well. If it doesn't, use 2 pieces.
Tim Stump posted on Friday, October 30, 2009 - 8:12 am
I'm specifying a linear growth model with 5 unequally-spaced time points--baseline and then followup at 2, 4, 8, and 12 wks from baseline.
I'd like to specify a piecewise model since most of the change in individuals occurs between baseline and 2 wks and then they don't exhibit as much change after that. Would the timescores in the following statements be correct:
Tim Stump posted on Monday, November 02, 2009 - 6:53 am
Thanks for your reply to my post on 30oct2009. I have a follow-up question. By specifying s1@0, I'm setting the variance of the slope factor s1 to be zero. Can I still make comparisons on the mean of the slope factor s1? Say I have a binary covariate named group coded 0 or 1. If I add: i s1 s2 on group; to the model, does it still make sense to test for differences in the mean s1 across group even though the variance is contstrained to zero?
I would like to continue with this thread by asking some additional questions:
I would like to run a piecegrowth model. The reason I'd like to do this is because I have four time points, but a linear model does not fit the data. Specifically, there is a big change between time 1 and time 2. The change between time 2-4 is less dramatic. Substantively, time 1-2 represents the transition into elementary school and var is achievement. The data points are measured in two-year increments.
I have found I have to free the time 4 factor loading to get the model to run. The estimated time score for this model is 2.067.
Here are my questions:
First, should this var4 time score alarm me? If I run a gc from times 2-4 and free the time 4 loading, the model runs and has good model fit. The estimated time 4 factor loading for this model is 2.65. If I estimate a quadratic using times 1-4, I get an error message.
Secondly, I’d like to allow the intercepts to be different (specifying i1 and i2), but this model does not run. Could you suggest why this might be and how I can fix it?
Also, you mention that with only two measures, I cannot identify the variances of the slope growth factors. This is why I need to set s1@0, which I understand, but as I think about it, is beginning to seem problematic because I'd like to look at the covariance of s1 with s2. Moreover, I am assuming that everyone has the same s1, which is not the correct.
Are there other ways of dealing with this issue that you might suggest? Perhaps I could calculate a deviation score between t1 and t2, and use that to predict i1 and s2? As I mentioned in my previous post, I was considering a piecewise approach because there is considerable growth between t1 and t2, and more modest growth between t2-t4.
I would like to do a GMM identifying latent classes of individuals who have differential responses to treatment (e.g., responders, non-responders). I have intervention data collected at pre, post 6, 12, 24, 36 month follow-ups. When I initially modeled the data using a traditional growth curve approach, a quadratic growth did not fit the data well. Instead, I found that a two process model fit well with the first process consisting of change from pre-post, and the second process of linear change occuring after the intervention has been completed (see below).
Now that I am moving toward a GMM, I am not sure how to include both processes to define different latent classes across all assessment points. Is it possible to use both processes to define a latent trajectory classes.
Jon Elhai posted on Thursday, May 19, 2011 - 6:04 pm
I'm not sure if piecewise growth modeling would be appropriate... But I have 4 timepoints in a growth model, each of which are equidistant (measured every 2 weeks). But for the scores over the four weeks, the slope looks unlike the slopes I usually see in the Mplus forum. In this example, the observed scores would take the shape of something like: 100 (time1), 50 (time2), 80 (time3), 65 (time4). Would a piecewise model be appropriate, or some transformation of the time scores?
Hi, I am trying to model data on students' well-being across the school year, and have 6 time points. What I am finding is that well-being decreases from September-December, then jumps back up in January, and decreases again throughout the second semester (January-April). My means are as follows:
I am running into a similar problem as a post above. I have three time points (time 1-time 2 with traditional instruction and time 2-time 3 with an intervention). I wanted to compare the direction and rate of change for significance, but I don't think there are enough time points to do this. Is there any work around for this? I saw above a suggestion for setting up an intervention dummy variable, but I don't know how to set this up. Thank you.
I am running multivariate piecewise growth models. I have been using a single intercept with 2 slopes, but someone has recommended I use 2 intercepts instead. Are there any advantages/disadvantages to using 2 intercepts rather than 1 intercept in a piecewise model?
you obtain the first model as a special case where the mean and variance of s1 are the same as for s2, but the covariance for s1 and s2 would need to lead to correlation 1 which is on the border of the admissible parameter space. So I don't think nested chi-square testing works. Use BIC instead to compare and absolute fit (for each model) to decide which one is good enough.
Anne Chan posted on Saturday, August 09, 2014 - 4:38 am
Hello! Thanks for your reply. May I ask what is "absolute fit" for each model? Are there any indices on absolute fit showing on the output files?
By absolute fit I mean the fit of your H0 model compared to the unrestricted H1 model, that is, the usual chi-square/RMSEA/CFI testing - as opposed to relative fit measured by BIC where you compare two different H0 models.
Anne Chan posted on Monday, August 18, 2014 - 7:29 am
Hello! May I ask if the following two models are nested or not:
Jon Heron posted on Thursday, December 11, 2014 - 4:57 am
So you want consistent linear change with a vertical jump midway through the process? that's a little unusual. If you're happy for everyone to have the same vertical jump and you don't actually want to be able to predict it with covariates you could capture this through a non-zero but constrained equal intercept for the last two measures.
Heather - I think your version looks ok. But do you really get identification problems unless you say intcpt2@0?
Our Topic 3 handout, slide 122 refers to your situation. That slide shows two slopes (one for each piece), but you could hold their means and variances equal. Your situation is a little harder because the second piece has just 2 time points.
Hello, I am fitting a piecewise model with two intercepts to a 6 timepoints (baseline, 4 months, 8 months, 12 months, 18 months, and 24 months) assessment wheere there is a steep decrease from the first to second assesment followed by an increase from T2 to T6. Here are the sample means:
T1= 60.662 T2=56.489 T3=68.144 T4=70.304 T5= 71.445 and T6= 71.472
The model fit is excellent, but the estimated intercepts have means at s1 = 60.649 and i2 67.204.
I cannot make sense of why the second intercept is not corresponding to the second assement which would be lower than the first asessment. The plot of sample and estimated means show excellent correspoendence.
Andrea Ted posted on Wednesday, August 19, 2015 - 10:58 am
I have 6 time points, the first 4 sequential, separated by the latter 2 by 48 hours. The treatment is introduced at time 2 and is effective until time 4. I'd estimate a piecewise model with each piece having its own intercept.
Could you kindly detail on why the first model is not set up correctly?
For the second piece, would it be appropriate to fix both the slope and the intercept variance to 0?
For the first piece I specified linear growth. Would it be adequate to freely estimate timepoints 2 and 3 if this renders a better fit? Then, I also want to add a quadratic factor for the treatment group.
But to answer your question, the top approach computes the mean of i2 over all 6 outcomes, while the bottom approach computes the mean over those 3 outcomes. So the definitions of the intercept are different.
fred posted on Monday, October 05, 2015 - 12:53 am
Dear Bengt, when you say "I think you are asking for 3 pieces. So try something like this where the s1 and s2 pieces share the same intercept:
fred posted on Tuesday, October 06, 2015 - 1:32 am
Thank you Bengt! Implementing the 3 slopes model gives a fairly good model fit (CFI=994), although with a Chi2 still being signficant (at p=0.019). The slope S3 from cQl2 to dQl2 and onward is close to 0 and flat (mean for S3 is 0.097 and non-signficant). By allowing an error covariance between cQl2 and dQl2 the modell indices turn to excellent. Would you consider this modification reasonable when the slope is closie to zero?
Hello! I have a question regarding sequential parallel process modeling. I have run the three step process for an LCA with baseline data (4 classes). We would like to look at developmental differences (growth) within each of these classes. Our data are in accelerated longitudinal design - 14 years old to 24 years old
we would also like to look at developmental shifts in parallel growth processes within each class. We can, conceptually, look at growth during adolescence and then growth during young adulthood (starting at 19 years old).
Our question: 1) would we set our growth models up with two separate intercepts and 2) if so, what would the intercept and slope regressions look like within each class?
There are a lot of possibilities here. First, I think you mean regress icYA on sbADO and not the other way around (so sbADO predicts icYA). I think you worry about the time overlap at 18 and 19 - so why have that overlap? Also, you can also let the intercept defined at the end of the earlier process be the predictor of the later process.
in your last sentence do you mean in the parallel process model we should also let the adolescent intercept (ibADO) predict the adolescent slope (sbADO) ? if that is the case , you are right we are facing numerous possibilities - especially across 4 latent classes
it may be a moot point. When estimating simple growth within each class I get the following: THE ESTIMATED COVARIANCE MATRIX FOR THE Y VARIABLES IN CLASS 3 COULD NOT BE INVERTED. PROBLEM INVOLVING VARIABLE BINGE1. COMPUTATION COULD NOT BE COMPLETED IN ITERATION 11. CHANGE YOUR MODEL AND/OR STARTING VALUES. THIS MAY BE DUE TO A ZERO ESTIMATED VARIANCE, THAT IS, NO WITHIN-CLASS VARIATION FOR THE VARIABLE.