Interpretation of quadratic factor PreviousNext
Mplus Discussion > Growth Modeling of Longitudinal Data >
 Anonymous posted on Saturday, March 03, 2001 - 2:23 pm
Dear Bengt and Linda

In linear LGM, the two factors indicate initial status and growth rate. In quadratic LGM, I'd like to know the interpretaion of quadratic factor. If its variance is significant, what does that mean? I know it means that there are ind differences in quadratic parameter values. Any better interpretations? If time-invariant covariates are positively related with linear and quadratic factors, how can I interpret that?
 Bengt O. Muthen posted on Sunday, March 04, 2001 - 6:10 pm
Simply put, the quadratic factor describes the upturn or downturn over time beyond what is predicted by the linear factor. However, the quadratic and linear factors are to some extent confounded. This is especially clear with time scores for the linear factor slopes of 0, 1, 2, ..., where the linear and quadratic factors describe the same development from time 1 to time 2 (time 2 scores are 1 also for the quadratic). I therefore find it hard to give a separate interpretation of effects of covariates for the linear and the quadratic factors. Interpretation may be easier with piece-wise growth modeling and with centering at different time points. On the other hand, an interpretation of covariate effects may not be needed; perhaps it is sufficient to include the effects so the model is correctly specified.
 Harry Garst posted on Monday, February 18, 2002 - 7:48 am
Dear Linda and Bengt,

My model consists of two separate growth curves based on the latent factors (a multivariate or cross-domain model using two separate second-order models - one for each growth curve - because the measurement models is included).
This is the first time I used quadratic factors (in this model: two quadratic factors, one for each growth curve). The results are rather disappointing: the correlations between the slope factors and quadratic factors are almost perfectly negative (r < -.92) and the asymptotic variances of the parameter estimates are huge for the quadratic factor variances. The asymptotic correlations of the parameter estimates of the slope/quadratic covariance is also huge, resulting in a high SE (even a moderate covariance is not significant). My conclusion is that either this model is misspecified (although the goodness of fit measures are superior to the linear growth model) or there is too less information to estimate the parameters of this quadratic model (collinearity). Do you agree? Are these problems typical for quadratic growth curve models? Is there an alternative way? I have heard of orthogonal polynominals. Can I use these? As you can understand the term orthogonal sounds very promising to me. I don't want to use nonlinear models based on freed factor loadings, because these models have their own problems.
 bmuthen posted on Monday, February 18, 2002 - 11:55 am
Quadratic growth modeling does not always give these types of results. However, it sounds as if it might be worthwhile for you to try to center at the mean time. That is, use

y_it = a_i + b_i * (x_t - xbar) + c_i * (x_t - xbar)^2 + e_it

where a, b, and c are the growth factors and xbar is the average time value. So your loading values are x_t - xbar and their squares. This defines the intercept growth factor a as the growth curve value at the mean time xbar. Typically, this makes parameter estimates less highly correlated.
 Jeff Cookston posted on Monday, October 23, 2006 - 11:12 pm
We are conducting a cohort sequential design in which we have two cohorts that are each measured at three waves. Given the nature of our cohort sequential design and our longitudinal data collection plan, we'll get data at 8 time points (e.g., fall 7th grade, spring 7th grade, fall 8th grade, etc.). Although each participant is only measured 3 times will we be able to test for quadratic and higher order trends given our extra time points or are we bound by the number of data points we have for each participant?
 Jeff Cookston posted on Monday, October 23, 2006 - 11:19 pm
Thinking more about what I posted above, if we put the cohorts together and use missing data then we should have the full 8 time points at our disposal, correct?
 Linda K. Muthen posted on Tuesday, October 24, 2006 - 7:31 am
Yes, your model has 8 time points.
 Annie Desrosiers posted on Thursday, October 26, 2006 - 10:38 am
Hi Linda,
I have a question about fitting quadratic model. I have a liner model and I want to know if the model could fit better in quadratic. So I change de s (slope) for a q (quadratic) in the input but the results still the same...
I plot the result but its not quadratic curve??
Thank you for telling me what is wrong with my input :

variable: names are id y1-y3 a1-a3;
usevariables are y1-y3;
auxiliary are a1-a3;
classes = c(6);
missing = . ;

analysis: type = mixture missing;
starts = 500 20;

model: %overall%
i q | y1@0 y2@1 y3@2;

plot: type = plot3;
series = y1-y3(q);

 Linda K. Muthen posted on Thursday, October 26, 2006 - 10:58 am
You can't identify a quadratic growth model with only three timepoints. If you had more timepoints, you would need three growth factors not two. You would need:

i s q |
 Sylvana Robbers posted on Tuesday, March 11, 2008 - 1:32 am
I have a question on how to specify the regressions on the covariate when using
i s q | in a growth mixture model.

Is it sufficient to do:

i s ON x

or should I specify:

i s q ON x ?

I have fixed q@0 to avoid non-definite psi-matrices.

 Linda K. Muthen posted on Tuesday, March 11, 2008 - 6:29 am
I would say:

i s q ON x;

q@0 fixes the residual variance of q to 0 but it can still change as a function of x.
 Sylvana Robbers posted on Wednesday, March 12, 2008 - 2:43 am
Thank you very much for your clear answer.

I have a following question on this:

q is a significant growthfactor, but i s ON x gives a better model fit than i s q ON x. Should I eliminate q from this regression or is q really necessary here for the overall interpretation of the model?

 Linda K. Muthen posted on Wednesday, March 12, 2008 - 9:57 am
If you have a significant mean for your quadratic factor, I find it hard to understand how a linear model can fit better. Please send your input, data, output, and license to if you have further questions.
 Harald Gerber posted on Friday, April 11, 2008 - 10:32 am
hello all,

i have also some problems with this quadratic issue. here they are:

a. i have a unconditional quadratic growth model where my quadratic factor has no significant variance. Doese it make sense to retain that variance (not setting it to zero) when it comes to conditional models and predict q with the covariates? I've read this in a paper, but it is a bit counterintuitive for me.
b. In addition my linear factor of the quadratic model has a sig. variance but no signifcant mean. What does that mean for predicting that growth factor? Am I predicting nothing (no increase of that factor)?
c. What does it mean in general, when i have an effect of covariates on the linear factor but not on the quadratic factor in the quadratic model?
 Bengt O. Muthen posted on Friday, April 11, 2008 - 3:30 pm
a. Yes, retain the (residual) variance because adding covariates gives more power to detect slope variation.

b. You are predicting the variation. It doesn't matter that the variation is around a mean of zero. Some people are still above zero and some below and you want to predict that variation.

c. That's harder to disentangle. But with centering at the first time point, typically the linear influence has to do with early development and quadratic with later.
 Harald Gerber posted on Sunday, April 13, 2008 - 9:39 am
thank's a lot for your prompt answers!

a. I retained the variance but there were no effects of my covariates on this quadratic factor. In addition, some effects on the linear slope of the quadratic model faded away, as compared to the former "q=0 model". The residual variance of q was insignificant. In this case, would it be better to set the q-variance to zero?

b. In terms of a prevention effect on the development of aggression. How could one formulate this outcome in one sentence? E. g., "X buffered the early increase of aggression" ? (but for this statement, one needs a significant linear factor mean of the quadratic model, in my understanding). Btw, is there a chance to detect a significance of this linear factor mean in a conditional model (more power, as with the variances) and what is the indicator for that?

c. o. k., I adopted your interpretation as you can see above :-). Does this interpretation holds also for alternative parameterizations? I use one, in which the means of the linear and quadratic factors describe the increase between the first and the last measurement point (5 time points).

thank's so far!
 Bengt O. Muthen posted on Monday, April 14, 2008 - 8:56 am
a. No, keep it and report it as is. No need for "trimming" the model.

b. What you need for this is a significant slope in the regression of the growth slope factors (linear and/or quadratic) on x. The means of the growth slope factors do not need to be significant. When x is included you do have a conditional model.

This is a big topic - come and learn about it at our August growth modeling course at Hopkins (see home page).
 Harald Gerber posted on Monday, April 14, 2008 - 11:03 am
b. just to clarify: so one could in fact speak of an buffering effect of the treatment, despite of the insignificant mean? I only need a sig. regression coefficient (on the slope) to state such an interpretation? It's hard to imagine, because I'm always thinking of this zero mean....

c. yes, I'm still looking for some funding to get the opportunity to visit one of your courses! But, do you have a short comment available regarding this special parameterization and quadratic + linear factors? Would be very helpful!
 Bengt O. Muthen posted on Monday, April 14, 2008 - 11:27 am
b. That's right. Think of problem behavior development with x = 0 for Ctrls and 1 for Treatment. The mean slope of the development can be zero for Ctrls (say high, but flat problematic development) and the mean for Treatment negative. This happens if the slope in the regression of the growth factor on x is negative. Just like in regular linear regression with a dummy x variable shifting the mean by shifting the intercept.

c. I don't understand your specific parameterization under c.
 Harald Gerber posted on Tuesday, April 15, 2008 - 12:53 am
thank's, that was very helpful!

sorry, using 4 time points:

I switched from the common parameterization for a quadratic development "0 1 9 25" (using my particualar time points) to
"0 0.04 0.36 1" ("0 0.20 0.60 1", would be a linear trend), where the mean of both growth factors (lin and quad) represent the average increase between the first and the last measurement point. My Question is, if your interpretation of the quadratic factor (later developement) still holds for this kind of parameterization.
 Linda K. Muthen posted on Tuesday, April 15, 2008 - 8:54 am
What you have done is rescale your time scores. Rescaling the time scores does not change the model and will not change the interpretation of the results.
 Harald Gerber posted on Wednesday, May 07, 2008 - 10:30 am
Ok. But is this rescaling of time scores may be an issue in mixture modeling? The funny thing is, I get "different" solutions when using one or the other time scores parameterization in mixture modeling. The solutions differ with regard to the run one special group is extracted. 3 group and 5 group solutions have the same groups and size in both timescore scaling methods. But in a 4 group run mplus extracts a group (under one kind of time score scaling) which is extracted under the other kind of time score scaling in the run with 6 groups and vice versa. So groups and their timepoint of extraction are mixed up a little bit, depending on time score scaling.
Anyway, the solution with my rescaled time scores fits my theory better. So the question is clear: is it in principal possible to use this sort of time scores scaling (mean of slope is average change between first and last measurement point) in mixture modeling?
 Linda K. Muthen posted on Wednesday, May 07, 2008 - 5:55 pm
It is hard to say much without seeing more information. Rescaling should not result in different results. Please send the two outputs where the two scalings are shown and your license number to
 Vincent May posted on Saturday, August 09, 2008 - 7:07 am
Hello! I have fairly the same problem like in the post april 14. 11.27 am.
I'm a little bit confused and may be it is not a big problem and I'm thinking too much.

My treatment is coded 0 = intervention and 1 = control. I have a treatment effect (coefficient sign is positive) on the linear slope of substance abuse, but the slope mean is negative and insignificant (I have tested significance of the slope mean in an unconditional model).

Does that suggest that the intervention has an iatrogenic effect on the slope of substance abuse or is it a positive and desirable effect? How would you characterize this effect?
Many thanks!!!!
 Bengt O. Muthen posted on Saturday, August 09, 2008 - 9:17 am
Positive effect means "higher" slope. Think of the growth line as a hand on a clock where the hand can turn counter-clock wise or clock wise. A positive effect on the slope implies a counter-clock wise turn (no matter where the hand started from). So, yes, an iatrogenic effect is suggested.
 Vincent May posted on Sunday, August 10, 2008 - 6:56 am
Thank you, very good example. But my treatment variable is coded: treatment = 0 and control = 1, so belonging to control group means that you have a "higher" slope?!
Is this correct? I was irritated by the negative, however, insignificant slope. But this plays no role, as I understand your posting.
 Bengt O. Muthen posted on Sunday, August 10, 2008 - 9:58 am
I see - yes, if you have control=1 and tx=0, then your tx effect is beneficial, turning the slope down.
 Renee Thompson posted on Monday, September 01, 2008 - 1:01 pm
On an earlier post, it was noted that > 3 times points are required to include a quadratic growth factor. Why is this? For example, I have three points of data and the data look quadratic (e.g., time 1 = 3, time 2 = 7, time 4 = 2). Perhaps I am using the term incorrectly? I would like to run an analysis like example 6.13 growth model for two parallel processes for continuous outcomes, but did not think I could because of the way the data are looking.

Thanks in advance!
 Linda K. Muthen posted on Monday, September 01, 2008 - 3:58 pm
With three time points, the unrestricted H1 model has 9 free parameters: three means and 6 variances/covariances. The H0 quadratic growth model has 12 parameters: three means for the intercept, slope, and quadratic growth factors; three variances for the intercept, slope, and quadratic growth factors; three covariances among the growth factors; and three residual variances of the outcome. For three time points, at least three restrictions would have to be imposed to identify the model.
 Kirsten Bank posted on Thursday, September 25, 2008 - 12:29 pm
I have a conditional quadratic growth model with a) a positive slope and a negative quadratic factor for self-concept (5 measurement points) and b) a positive effect of a covariate (achievement) on the slope and a negative effect of the same covariate on the negative quadratic trend. How do I interpret that? Is it right that a) means that there is first an increase in self-concept and later a decrease? Or does the negative quadratic trend mean that the increase later is not as fast as at the beginning but no decrease? My interpretation of b) is that a higher achievement has a positive effect on the increase of self-concept. Can I say so? And what about the negative effect on the negative quadratic factor? Does it mean that the braking / decrease of the self-concept trajectory is less strong if you have a higher achievement? Or that it is even stronger (maybe because of a ceiling effect)?
 Bengt O. Muthen posted on Thursday, September 25, 2008 - 10:05 pm
Regarding "that a) means that there is first an increase in self-concept and later a decrease? Or does the negative quadratic trend mean that the increase later is not as fast as at the beginning but no decrease?"
- this choice cannot be determined unless you plot the estimated mean growth curve.

Regarding "My interpretation of b) is that a higher achievement has a positive effect on the increase of self-concept. Can I say so?"
- the answer is yes.

Regarding "And what about the negative effect on the negative quadratic factor?" - that means that the quadratic factor value is even more negative as achievement increases.
 hanneke creemers posted on Monday, November 03, 2008 - 2:09 am

I have a question on your reply-post of September 1, about fitting a quadratic model on a growth path with three time points.

Do you mean with imposing at least three restrictions that it is possible to fit a quadratic model by, for example, restricting q@0? And do you have a reference about this issue?

Thanks in advance.
 Linda K. Muthen posted on Monday, November 03, 2008 - 8:00 am
You would need to impose three restrictions on a quadratic model for three time points for it to be identified. I do not think this is a good idea because you have no way to test if the restrictions are valid. I would not fit a quadratic model with less than four time points. I know of no reference for this issue.
 Thierno Diallo posted on Tuesday, May 05, 2009 - 7:36 am
Dear Dr. Muthèn:
I’m running growth curve analyses with continuous variables (5 time-points; from grade 7 to 11). One of my curves is best represented by a quadratic slope. When I fix the time-points to 0, 1, 2, 3, and 4, the correlation between the linear and quadratic factors is quite large (-.87).
I read on the Mplus discussion site that one way to reduce the collinearity between the linear and quadratic slopes is to center at the mean time by using this formula: y_it = a_i + b_i * (x_t - xbar) + c_i * (x_t - xbar)^2 + e_it .
Here, this would result in using -2, -1, 0, 1, and 2 as time-points.
The problem is that when I try to predict outcomes with this growth curve, there are significant associations with the slope when I fixed the growth parameters from 0 to 4. However, when I fixed them from -2 to 2, the results are not significant anymore.
Interestingly, when I move the zero point to the left, my results remain significant (-1, 0, 1, 2, and 3) but when I move it to the right (-2, -1, 0, 1, 2; -3, -2, -1, 0, 1), they are not.
Is it normal that when changing the growth parameters, the results change quite dramatically (from significant to nonsignificant)? Can this highlight a major problem with collinearity in the data?
Thank you very much for your time and help
 Bengt O. Muthen posted on Tuesday, May 05, 2009 - 7:55 pm
It could be that the early change is what predicts. If you predict from only the linear slope the high correlation between the linear and the quadratic doesn't matter.

You can also consider predicting from the intercept growth factor alone, centering at different time points - perhaps it is only the height of the curve that is predictive.
 Michael Spaeth posted on Friday, July 03, 2009 - 5:03 am
I have an associated LGM and let the growth factors correlate between the growth curves. However I have two problems.

1)In one of the growth curves there is a significant q-mean, however no variance. To symplify the model I set the q-variance to zero and I get the impression that this small variance is (partly) soaken up by the linear factor "s" (variance became bigger). Am I right? How do I interpret covariance of this "altered" s-factor variance (after q variance is set to zero) with growth factor variance from associated LGMs?

2)The mean of the s-factor is not significant (however, the variance is sig.). How do I interpret significant positive covariances with other growth factors? Does your statement "b" from "april, 11, 2008, 13.30h" still apply, also to correlations between growth factors?

 Bengt O. Muthen posted on Friday, July 03, 2009 - 11:13 am
1) If there is a small but not negative q variance I would keep it in the model. I don't think you want to enter into interpretations about altered s variance.

2) The average trend can be zero while still having individual variation around that trend. So a positive correlation means that the higher a person's slope value, the higher the person's ....
 nina chien posted on Wednesday, November 18, 2009 - 5:49 pm
I have four timepoints, and tried to fit a quadratic growth model.


So I fixed one of my parameters (s@0) and then the quadratic model ran fine.

Question 1) To fit a quadratic growth curve model with only 4 timepoints, do I always have to impose one (or more) restrictions? I had thought not because the example in the mplus manual (6.9) has only 4 timepoints for a quadratic growth curve model.

Question 2) If I do have to impose restrictions, is there a recommendation for a good place to impose such a restriction? Or should I always inspect the model output before deciding which parameter to fix?

Thanks so much.
 Linda K. Muthen posted on Wednesday, November 18, 2009 - 5:54 pm
A quadratic model is identified for four time points. Please send your full output and license number to
 Kara Thompson posted on Wednesday, November 10, 2010 - 10:09 am

We are interested in the co-development of anxiety, depression and ODD in adolescent females. We have 4 waves of data (ages 12-24)and are using time scores. The univariate models indicate that all three processes are quadratic (significant quadratic means). Therefore, we ran a quadratic parallel process model. We could use some help interpreting this quadratic parallel process model. We are interested in how these constructs are related over time, so we want to know the intercept-intercept covariance, slope-slope covariance and the residual-residual covariance. We are aware of how to interpret this in a linear parallel model (it was linear for males in our sample). However, we are unsure how to interpret the model when we have the addition of a quadratic slope because we know that the linear slope and quadratic slopes are confounded with one another.Do you have any suggestions on how we should be interpreting this model?
 Bengt O. Muthen posted on Wednesday, November 10, 2010 - 4:12 pm
That's hard to separate out as you say. How about instead doing a series of runs where you center the time scores at each of the 4 waves? So a variation on the Muthen-Muthen (2000) theme. The intercept then captures the systematic part of the growth at that time point. Then you only have to look at the intercept covariation.
 Kara Thompson posted on Thursday, November 18, 2010 - 3:15 pm
Thank you for your quick response. Could you please clarify further how interpeting the intercept-intercept covariance at each time point helps us get around the confounded slope-slope covariances?

Does a signficant intercept-intercept covariance at each time point suggest that these two constructs "travel" together, or only that they are related at each time point? Does doing this allow us to say anything about how these constructs change together over time?

Thank you
 Bengt O. Muthen posted on Thursday, November 18, 2010 - 5:37 pm
Take a look at Muthen-Muthen (2000) on our website.

The approach I suggest says how much they are related at different time points. It is hard to show that the processes travel together - certainly with a quadratic model where you have this confounding of effects. Orthogonal polynomials can be attempted, but interpretations are not easy.

Some researchers try using an underlying slope factor - a second-order factor - that lies behind the slopes of all the processes.
 Bradley Jorgensen posted on Sunday, July 08, 2012 - 11:46 pm
Hello Dr Muthen,

I have 4 quarters of household water consumption data. The data are fairly consistent with seasonal periods (i.e. Quarter 1 - Winter; Quarter 2 - Spring; Quarter 3 - Summer; Quarter 4 - Fall). given temperature and rainfall variations across seasons, the consumption means rise from Winter to Summer and then decrease to a level in Quarter 4 that lies between consumption in Quarters 1 and 2. I ran a LGC model in which the slope loading for Quarter 4 was free. The loadings for Quarters 1 to 3 were 0, 1, 2, respectively. The model fit was good. I then ran another LGC model which added a quadratic factor and I fixed the Quarter 4 loading on the slope factor to equal 3.

MODEL: i s q | QRT1@0 QRT2@1 QRT3@2 QRT4@3;

The model fit was poor. The theta and psi matrices were both not positive definite. I'm interpreting this to mean that the quadratic model doesn't fit well and that I should go with the original optimal curve model.

I also ran both models with covariates and the outcome was the same. The two-factor model fitted well and the three-factor (quadratic) model output reported non-positive definite matrices.

Regards, Brad.
 Linda K. Muthen posted on Monday, July 09, 2012 - 12:02 pm
You could also consider:

i s | QRT1@0 QRT2* QRT3* QRT4@1;

MODEL: i s q | QRT1@0 QRT2@1 QRT3@2 QRT4*;
 sojung park  posted on Friday, November 23, 2012 - 2:29 pm
Dear Linda and Bengt,

I was trying to model multi-group growth curve with quadratic term-

out of the 5 groups, I got warnings for the two groups as follows-

when I get this warnings only for some of the groups, what should I do?



 Linda K. Muthen posted on Saturday, November 24, 2012 - 10:34 am
You should fit the growth model separately in each group to see that the same model fits in all groups before doing a multiple group analysis.

If the problem is the variance of s is small an negative, you can fix it to zero.
 Dustin Pardini posted on Thursday, April 11, 2013 - 2:41 pm

I'm running a growth curve model that fits best with a quadratic included, to predict an outcome. The collinearity between the slope and quadratic is leading to largely inflated estimates in this prediction. I've considered 1) setting the quadratic variance to 0 (however the variance estimate is significant) or 2) recentering to the middle time point, which fixes the collinearity issue but leads to problems in interpreting change overtime as a predictor of the outcome (i.e.-slope is no longer a significant predictor). Any thoughts/alternative suggestions would be greatly appreciated.

 Linda K. Muthen posted on Friday, April 12, 2013 - 10:34 am
It is very hard to predict from slopes. You might want to take a look at the following paper which is available on the website:

Muthén, B. & Muthén, L. (2000). The development of heavy drinking and alcohol-related problems from ages 18 to 37 in a U.S. national sample. Journal of Studies on Alcohol, 61, 290-300.

In this paper, we look at the models centered at each time point and focus on the intercept growth factor.
 Dustin Pardini posted on Friday, April 12, 2013 - 1:42 pm
Thanks Linda. Having looked at our model and recentered the intercept at each time point as suggested in Muthén, B. & Muthén, L. (2000), it looks as if the early time time points are driving prediction.

If I'm following along correctly from the 2000 paper as well as from Bengt's post from May 05, 2009 above, would I be correct to state that having a quadratic growth model but only using the intercept and/or slope as predictors in the regression portion of the model would be okay (i.e.- not including the quadratic term in the regression part of the model)? Thanks again.
 Bengt O. Muthen posted on Friday, April 12, 2013 - 4:51 pm
Yes, although you simplify the question answered. The intercept represents the systematic part of the growth (so minus the time-specific residual) at the time point where the time score is zero. So you are predicting from the level where the person is at that point, not how he got there (so not a function of the slope, steep or not steep).

If you want to predict from the shape of how a person got to where he is, it is probably better to use growth mixture modeling and predict from the latent class variable - that's what I have chosen to do.
 Samuli Helle posted on Wednesday, June 12, 2013 - 2:22 am
We are constructing a parallel growth model with quardatic time scores in two growth processes (with continuous response variables). The quadratic model seems to fit the data better than the linear model. We have some problems interpreting the results.

1) There is a negative residual variance for the first observed variable that is not statistically significant. Is it ok to fix it to zero?

2) For one of the processes, the linear and quadratic slope factor variances are negative and non-significant. Should these as well be fixed to zero?

3) There are some statistically non-significant covariances between intercept and rate factors. Should these be set to zero? And does it matter whether these are between or within the two growth processes?

4) Should we try orthogonal polynomials (we have six unequally measured time scores) to potentially fix the above mentioned concerns?

5) Our current model has 39 free parameters, but we only have 212 subjects. Are we overfitting here?

Thank you very much for your help!
 Linda K. Muthen posted on Wednesday, June 12, 2013 - 1:14 pm
1. Yes.
2. Fix the quadratic first.
3. No.
4. This is not needed.
5. Probably not for repeated measures.
 Samuli Helle posted on Monday, June 17, 2013 - 12:38 pm
Thank you! I'd like to ask a follow-up question relating to my previous question no. 3. So, you would not recommend fixing non-significant covariances to zero? I'm currently reading Byrne's book "SEM with Mplus" and there, for the sake of parsimony, non-significant covariances between intercept and slope factors of parallel LGC-model were fixed to zero. I'm thus a bit confused what would be the best strategy to proceed? The current model doesn't fit the data well, so I suppose some modification are in order (adding residual covariances?)...
 Bengt O. Muthen posted on Monday, June 17, 2013 - 1:15 pm
No, I would not recommend that. Just report them. My general preference is to not do that sort of model trimming.
 Samuli Helle posted on Tuesday, June 18, 2013 - 8:05 am
Thank you for your helpful answers. I still have a few more follow-up questions.

1. What is the interpretation of the intercept in the case that the first one or two time points for a subject are missing values? Does FIML estimate missing values and the value at time 0 is the intercept?

2. I want to modify the growth model to make it fit better to the data. In which order should I proceed?

a. Set residual variances of the time points to be equal?

b. Estimate residual covariances between different time points?

3. Based on AIC values a quadratic growth model is clearly better than a linear one. However, a plot of sample and estimated means shows a very much linear relation, and the fit indices of the quadratic model are not great. Is it ok to use a linear model?

4. I am trying to model an interaction between factor intercept and linear growth factor, which seems to mean that I have to use numerical integration. However, model fit indices are not available. Is there a way of assessing absolute model fit of the model to the data?
 Bengt O. Muthen posted on Tuesday, June 18, 2013 - 3:03 pm
1. Yes, implicitly.

2. Big topic; see pour Topic 3 handout and video on our website.

3. I would be guided by chi-square testing rather than AIC/BIC when possible.

4. No, only by comparing to other models using likelihood-ratio chi-square.
 Samuli Helle posted on Monday, July 01, 2013 - 6:03 am
Thankyou once again for your prompt answers. I have yet again a few questions regarding the model I am constructing.

1. Since I am using numerical integration, and cannot assess the absolute fit of the model, should I first attempt to modify the latent growth model portion of the model, which does not require numerical integration (the interaction is not used here) or assess the relative fit of the whole model to other candidate models?

2. In order to improve the fit of our model to the data I modified the model on the basis of modification indices by allowing error covariances between some adjacent time points. However, the modification indices suggested some error covariances that were far apart in terms of time (for example between the first and last time points, I had a total of six time points). Is it ok to estimate these covariances even though their interpretation does not make sense to me?

3. Related to the above question, by estimating the suggested covariances at some point the factor slope variance changes from being significantly positive to significantly negative. I know from a random regression model that there is significant variation between subjects. What’s going on?
 Linda K. Muthen posted on Monday, July 01, 2013 - 7:48 am
1. Both approaches sound reasonable.
2-3. You should never add parameters to the model based on modification indices unless there is a theoretical reason for adding them.
 Samuli Helle posted on Thursday, July 04, 2013 - 4:41 am
There seems to be a general linear trend in growth in my data, except that there is a slight upward hump between two time points in the middle of our data. I have freed one time score as shown in the code below:

INT6 SLO6 | PL6_99@0 PL6_01@2 PL6_02@3 PL6_03* PL6_05@6 PL6_06@7;

1. Is it ok to estimate the hump as a free time score? This seems to improve the fit of the model to the data considerably.

2. Does this change the interpretation of my linear slope factor because my growth model is not linear after freeing a time point in the middle?

3. If it does change the interpretation of my linear slope factor then how should I interpret it?

Thanks once again for your help!
 Linda K. Muthen posted on Thursday, July 04, 2013 - 2:08 pm
1. Yes.
2. Yes.
3. The mean of the slope growth factor applies to only the linear part of the model not the estimated time score.
 Samuli Helle posted on Monday, July 08, 2013 - 12:13 pm
Thanks a lot again! Some additional questions...

1. Does the interpretation of the slope growth factor change if I free more time scores? For example, using the code below:

INT6 SLO6 | PL6_99@0 PL6_01* PL6_02* PL6_03* PL6_05* PL6_06@1;

2. I'm also interested in how to interpret a parallel growth curve model where both growth processes include free time scores. Given e.g. the code above (both processes modelled in the same way), does the covariance/regression coefficient between the two slope growth factors indicate a linear association between the processes (although the individual processes have also non-linear parts)?
 Linda K. Muthen posted on Tuesday, July 09, 2013 - 8:50 am
The means, variances, and covariances of the growth factors apply only to the first and last time points. Using so many free time scores makes for a difficult to interpret growth model. I would free time scores sparingly.
 Markus Martini posted on Wednesday, August 14, 2013 - 12:39 am
I have to questions:
(1) When estimating a quadratic model a linear latent slope faktor is also estimated. Is it allowed to estimate the model without the linear slope faktor?
If no, in which other cases do I need a linear slope faktor too (cubic?, inverse?, exponential?).

(2) When I estimate a kurvilinar model is it possible to estimate the individual breakpoint of the curve? To be more precise, I would like to predict the different breakpoints by several covariates.

Thank your very much indeed!
 Linda K. Muthen posted on Wednesday, August 14, 2013 - 9:19 am
1. It is customary to have a linear growth factor in quadratic and cubic models.

2. No.

You should listen to the Topic 3 and Topic 4 course videos on the website.
 Markus Martini posted on Friday, August 16, 2013 - 2:56 am
Thank you very much for your reply!

If I have three timepoints and I see that my graph looks like a roof because the values at T2 are always higher than at T1 and T3 (e.g. means T1:420 ms, T2:470ms, T3:400ms) then I could specify my parameters as follows:

t1@0 t2@1 t3@-1. Is that right?

Concerning T2, I could also set my centering point at T2 and it could be intepreted as the breaking point that can be predicted by several covariates?
 Linda K. Muthen posted on Friday, August 16, 2013 - 6:46 am
No. Please listen to the Topic 3 course video where time scores are explained.
 Matt Easterbrook posted on Friday, January 17, 2014 - 6:36 am
Hi, I am running a quadratic growth model with 18 time points. I had rescaled my time scores to help with convergence, so now they read xt1@0 xt2@0.01 et3@0.03...etc. rather than xt@0 xt2@1 xt3@3...etc.

The model fits well, but I noticed that the output for the quadratic growth goes like this:

That doesn't seem right to me as it is not progressing quadratically - does this seem incorrect and could it be due to the rescaling?

I have tried the model with normal time scaling (i.e. xt2@1...) and the fit statistics are identical but the estimations for the S and Q parameters and covariances are different. Can you not rescale like this for quadratic models?

Many thanks
 Bengt O. Muthen posted on Friday, January 17, 2014 - 8:57 am
Isn't this just a matter of rounding? Why don't you make your time scores 10 times larger: 0, 0.1, 0.3, etc You are allowed to rescale this way.
 Yadira Taboada posted on Saturday, April 11, 2015 - 10:34 pm
In need of very basic help, but help nonetheless.

I am interested in looking at Oral reading fluency trajectories for 2 groups and have 4 time points.

A "multiple group" linear growth model shows poor fit at all global indices while a "multiple group" quadratic model has a good fit as per the CFI and SRMR indices. (I ran individual models for each group to ensure the quadratic was the best fit for each individually.)

However, when looking at the Means for intercept, slope, and quadratic (Model results) for one of the groups i get the following:

Means Est/p-value
INT 5.138/0.000
SLOPE 7.752/0.000
QUAD 0.125/0.741

What does a non-significant quadratic estimate mean? When calculating expected scores at given time points is the quadratic estimate used even if it is non-significant?

With gratitude.
 Bengt O. Muthen posted on Sunday, April 12, 2015 - 5:10 pm
I don't know which parameters you hold equal across the groups to motivate a multiple-group analysis.

If the quad mean is insignificant in a group it means you probably don't need a quadratic in that group (although the variance can be needed).
 Daniel Lee posted on Thursday, September 29, 2016 - 10:29 am
Hi Dr. Muthen,

Why might the correlation between my slp and quadratic be so highly correlated? the sample means look quadratic, so I'm a little confused.

Also, is there a remedy for the issue that allows me to keep the quadratic term in the model?

Thank you!
 Bengt O. Muthen posted on Thursday, September 29, 2016 - 5:44 pm
Answered elsewhere.
 hal9000 posted on Thursday, December 22, 2016 - 8:47 am
I have a growth model with slope and quadratic which I've centered at the end point.
-3 -2 -1 0
I'd like the quadratic to be
-9 -4 -1 0
but when it's squared, it yields (obviously)
9 4 1 0

So this yields slightly different results than the same model centered at the baseline, is there a way around this?
 Bengt O. Muthen posted on Thursday, December 22, 2016 - 5:52 pm
This is a good question for SEMNET.
 hal9000 posted on Friday, December 23, 2016 - 11:15 am
Ok thanks! To be clear, this is only a problem that I have when I use Mplus.

I write:
i s q | x1@-3 x2@-2 x3@-1 x4@0;

And the output is:
s |
x1 -3
x2 -2
x3 -1
x4 0

q |
x1 9
x2 4
x3 1
x4 0

In other programs where I can manually enter the linear and quadratic time points it's not an issue.

I will search SEMNET.
 Linda K. Muthen posted on Friday, December 23, 2016 - 11:49 am
You can use BY statements for the growth model instead of the | statement. See Chapter 17 of the user's guide where the relationship between the BY and | specifications are described.
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