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 Shige Song posted on Sunday, July 30, 2006 - 1:42 am
I have a variable that have been repeatedly measured for four times. Now I want to generate a time-invariant variable our of these four measures and put it into a a larger models. I first used CFA to extract one factor. The result did not look very good; so I did a EFA to see how well the factor summarize the four repeated measurements. Here is what I got:

---------------------------------------
EXPLORATORY ANALYSIS WITH 1 FACTOR(S) :


CHI-SQUARE VALUE 2.462
DEGREES OF FREEDOM 2
PROBABILITY VALUE 0.2920

RMSEA (ROOT MEAN SQUARE ERROR OF APPROXIMATION) :
ESTIMATE (90 PERCENT C.I.) IS 0.032 ( 0.000 0.138)
PROBABILITY RMSEA LE 0.05 IS 0.477


ROOT MEAN SQUARE RESIDUAL IS 0.0547


ESTIMATED FACTOR LOADINGS
1
________
WR1991 -0.048
WR1993 -0.153
WR1997 0.239
WR2000 0.424


ESTIMATED RESIDUAL VARIANCES
WR1991 WR1993 WR1997 WR2000
________ ________ ________ ________
1 0.998 0.977 0.943 0.821
-----------------------------------------

Apparently there is something wrong because the factor loadings have opposite signs. This means that one factor does not adequately capture all the information embedded in the four repeated measurements, correct? In this case, what else can I do to improve model fit (with four factor indicators, is it possible to specify two different latent factors?)

Thanks!

Shige
 Linda K. Muthen posted on Sunday, July 30, 2006 - 6:11 pm
A couple of things come to mind when I read your post. Why do you want to make a factor out of the repeated measure of the same variable? Why not have any intercept only growth model or a growth model with both an intercept and a slope and include those in the larger model.

The second thing is that one one-factor CFA and a one factor EFA should be the same. The sign changes you see are not important. This can happen.
 Shige Song posted on Monday, July 31, 2006 - 5:29 am
Hi Linda,

Thanks for the post. I have been experimenting for quite a while now. As for factor model vs. growth model as measurement model, I find that factor model generally fits data better than growth model, maybe because the factor model has less constraints. My question is: does this (model fit based on CFI/TLI or RMSEA) matter, especially when trying to extract time-invariant information out of repeated measurements?

Thanks!

Shige
 Shige Song posted on Monday, July 31, 2006 - 5:51 am
Also, when saving factor scores as external files, is it possbile to put something that is not in the model in the external data file? For example, I have an ID variable which I need to use to merge the factor scores back into the main data set. I cannot put the ID variable in the model because it messes up the results; but I cannot put it in the output file without having it in the model, is there a way to handle this? Thanks!

Shige
 Linda K. Muthen posted on Monday, July 31, 2006 - 7:47 am
See the IDVARIABLE and AUXILIARY options of the VARIABLE command.
 Shige Song posted on Monday, July 31, 2006 - 9:07 am
Thought I'd read the manual pretty well... Thanks!

Shige
 Shige Song posted on Monday, July 31, 2006 - 10:57 am
Hi Linda,

About CFA vs. growth model, I have one more question. I understand the advantage of using growth model over CFA when continuous variables are repeatedly measured because the mean structure is retained. What if the variable is being measured is binary? Does growth model provide more information that CFA in this case?

Shige
 Linda K. Muthen posted on Monday, July 31, 2006 - 11:31 am
I can't see why it would.
 Sebastian Therman posted on Monday, February 11, 2013 - 4:36 am
Dear Drs. Muthén,

I'd like to do an EFA of a set of linear and categorical variables measured at two timepoints from the same individuals, assuming intercept and slope invariance over time. There are a lot of missing values, but they're MAR. How do I specify this model and obtain factor scores for both timepoints?

Kind regards,

Sebastian
 Linda K. Muthen posted on Monday, February 11, 2013 - 6:39 am
See Example 5.26.
 Sebastian Therman posted on Tuesday, February 12, 2013 - 1:30 am
Thank you for your swift reply!

This was almost exactly what I needed; my mistake was searching in the v5 manual...

Example 5.26:
f1-f2 BY y1-y6 (*t1 1);
f3-f4 BY y7-y12 (*t2 1);
y1-y6 PWITH y7-y12;

In the example (and by default), factor means are fixed at zero at both timepoints.

Since change in the overall latent trait level is likely in this particular case, the intercepts would need to be held equal over time and the factor means not.

to relax the factor means at t2, Do I simply add
[f3-f4];

...and can I constrain intercepts to be equal with the following?
[y1-y6] (2-7);
[y7-y12] (2-7);
 Linda K. Muthen posted on Tuesday, February 12, 2013 - 1:03 pm
One factor mean must be fixed to zero and the other free. So

[f4];

Yes, that constrains the intercepts.
 Jennifer DeCuir posted on Thursday, September 12, 2013 - 12:08 pm
Hi,
I am working on developing a stigma measure from 7 items that were measured in the same individuals at 2 time points. I conducted EFAs on the items from each time point separately to get an idea of the underlying factor structure at each time point, and then combined the data from the two time points, and ran a multiple group CFA (using time point as the group identifier) to ensure that the factor structure remained stable over time. Is this an appropriate strategy for assessing factor structure and model fit?

Thanks so much for your help!
 Linda K. Muthen posted on Thursday, September 12, 2013 - 1:24 pm
You can't use groups in your case because the groups do not include different individuals. You should test measurement invariance across time instead in a single group analysis. The steps to do this are shown in either the Topic 3 or Topic 4 course handout and video on the website under multiple indicator growth. The first part shows how to test for measurement invariance across time.
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