
Message/Author 

cconnell posted on Thursday, January 12, 2006  10:16 am



I am working with data from a randomized treatment study and 3 measurement points (baseline, 6month, and 12month followup). Analyses include latent growth model of Tx effects on outcome of interest, as well as a number of Tx mediation models that examine Tx effects on growth trajectories for mediators and then intercept/slope effects of mediators on outcome. We are using the 2003 Structural Equation Modeling article by Cheong, MacKinnon, & Khoo as a guide for conducting parallel process LGM. One question  the data has previously been examined within a repeated measures ANOVA framework, giving a pretty clear indication that (for some of the variables) a linear trajectory is not appropriate (e.g., much of the change takes place from baseline to 6months and is then maintained). With only 3 time points, I don't have the parameters to add model covariates (e.g., Tx) and model both linear and quadratic slope components; and have had similar difficulty with a free trajectory approach. Is there any appropriate way to set factor loadings for a growth model in such instances? 


This is tough because with three time points, if you estimate growth factor means, variances, and the covariance, that is 5 parameters. These plus 3 residual variances come to 8 total parameters leaving you only one degree of freedom to play with. If you want to free time scores, you would need to hold residual variances equal or make some other model modifications. This is a difficult situation and why we recommend no fewer than 4 time points. 

mihyun park posted on Wednesday, September 28, 2011  8:59 am



I tried to do LGM. In output data, Means are follwing.. y11= 2.144, y12=2.118, y13=2.281, y14=2.393. But model fit of quadratic growth model is not good than it of molinear growth model. Is this result possible? As you see, the mean of y12 is decreased and means again increased. Why did the result happened? 


You could look at modification indices. I would do that for the linear model. It may be using a free time score would help. 

mihyun park posted on Wednesday, September 28, 2011  11:05 am



Dear Drs Muthen, Thank you for your answer. I used a free time score like this.. i s  y11@0 y12* y13* y14@1 ; The model fit could be more better. And then, in the model result section, S estimate of y12 was negative. Can linear model have negative S estimate? Is the interpretation of linear model similiar with free time score model? 


A negative variance makes the model inadmissible. If it is small and not significant, you can fix it at zero. I would suggest that you listen to the Topic 4 video which covers growth modeling. 

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