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 Christian S posted on Wednesday, August 25, 2010 - 11:51 am
Dear Drs. Muthen,

I have a CFA with latent factors (each measured with multiple Likert-scale [-3;3] indicators). For the descriptive statistics, I would like to list mean and std. deviation for each latent factor. The TECH 4 output gives me 0 as mean for every factor. What should I do to get the "real" means and the standard deviations? Does the fact that I get 0 as mean for each factor mean that something with my model is wrong?

I really appreciate your reply.

Best regards,

Christian
 Linda K. Muthen posted on Wednesday, August 25, 2010 - 2:44 pm
In cross-sectional studies, the means of latent variables are zero. In multiple group analysis or with repeated measures, the means of latent variables are zero in one group or at one time point and are estimated in the other groups or time points.
 Christian S posted on Wednesday, August 25, 2010 - 3:21 pm
Dear Dr. Muthen,

thank you very much for your reply. As far as I understand, you wrote how Mplus handles factors in cross-sectional studies.

However, in many publications, I find means and std. dev.s of latent factors in the descriptive statistics part.

When indicators are e.g. all skewed to the right (avg>0), then the mean of the latent variable should (e.g. in my example with Likert scales from -3 to 3) not be zero. Is there a way with Mplus to get this mean? Should I use an average of the indicators of each factor weighted by the indicators' yx-standardized factor loading and then calculate the mean and the std. dev. for each factor?

Best Regards,

Christian
 Linda K. Muthen posted on Wednesday, August 25, 2010 - 3:38 pm
This is not how Mplus handles factors in cross-sectional studies, it is the conventional way to do this. A factor mean in a cross-sectional study has no meaning. You can't compare it to other factor means because there is no basis for comparison. It makes sense to compare factor means only across groups or across time after measurement invariance has been established.

I would imagine in the studies you mention, factor score means are being reported. Factor scores are generally not good approximations of true factor values.

The mean of a factor indicator is equal to the intercept of the factor indicator plus the factor loading times the mean of the factor. When the factor mean is zero, the mean of the factor indicator is equal to its intercept. This is why the factor mean can be zero even when the observed variable indicator mean is not zero.
 Brewery Lin posted on Friday, April 06, 2012 - 12:12 am
In Byrne's book (2012), p.211
Appearing below these specifications, however, you will see the following:
[Fl@O F2@0 F3@0]......in structuring the input file for a configural model, it is necessary to void this default by fixing all factor means to zero.

Is it more appropriate to do that constriain? Thank you.
 Linda K. Muthen posted on Friday, April 06, 2012 - 8:27 am
To test if factor means are different across groups use a model where factor means are zero in all groups versus a model where factor means are zero in one group and free in the other groups.
 Brewery Lin posted on Sunday, April 08, 2012 - 7:42 pm
Dear Linda,
If I just want to test the configural model, is it still recommend to do that?

Thank you.
 Linda K. Muthen posted on Monday, April 09, 2012 - 8:21 am
The configural model is factor means free across groups, intercepts free across groups, and factor means zero in all groups. You may find the multiple group section of the Topic 1 course handout on the website useful. It shows all of the inputs needed to test for measurement invariance.
 Jorge Walter posted on Saturday, June 07, 2014 - 4:03 am
Dear Linda,

From your first reply in this post, I understand that estimated means for latent variables in a cross-sectional model are always zero.

Given this, how can I then calculate the informative indices as part of models with sampling weights discussed in Asparounov (2004: 12-13)? The numerator would always be zero...

Thanks in advance for your help!
 Bengt O. Muthen posted on Saturday, June 07, 2014 - 5:53 pm
What are "informative indices"?
 Jorge Walter posted on Sunday, June 08, 2014 - 4:27 am
I'm sorry that my question was unclear. According to Asparouhov (2004: 12), an informative index is essentially a t-statistic comparing weighted with unweighted estimates:

I = (weighted estimated mean of Y - unweighted estimated mean of Y) / sqrt (estimated variance of the weighted mean - estimated variance of the unweighted mean).

Source:

Asparouhov, T. 2004. Weighting for unequal probability of selection in multilevel modeling. Mplus Web Notes: No. 8.
 Linda K. Muthen posted on Sunday, June 08, 2014 - 11:11 am
I believe this formula is for observed not latent variables.
 JW posted on Tuesday, July 08, 2014 - 7:09 am
Hi Linda,

in the post from 6th April 2012 you say:

"To test if factor means are different across groups use a model where factor means are zero in all groups versus a model where factor means are zero in one group and free in the other groups."

I am unsure on how to obtain the test. I have 3 latent variables measured at 2 time points so I would like to compare the means at time2 vs. mean at time1 (which would be 0).

Is this provided by the p-value associated with the mean at time2 in the output as in the example below -

for example, the mean score associated with variable 1 at follow-up (Follow1) is .118 which is associated with a p-value of .019 - would I interpret this an increase over time:

Means
Base_var1 0.000 0.000 999.000 999.000
Base_var2 0.000 0.000 999.000 999.000
Base_var3 0.000 0.000 999.000 999.000

Follow1 0.118 0.050 2.349 0.019
Follow2 0.071 0.065 1.082 0.279
Follow3 0.049 0.062 0.794 0.427


Or how else can I assess this?

Grateful for your help!

J
 Linda K. Muthen posted on Tuesday, July 08, 2014 - 8:19 am
With two time points, the z-test is column three of the follow variables is a test of the difference in means across the two time points.
 JW posted on Tuesday, July 08, 2014 - 8:44 am
Hi Linda -

Thank you for your reply!

So in the example above, looking at the line corresponding with Follow1, I could report the results for measure 1 as an increase over time, t(1) = 2.3, p = .02 -

is that correct?

Thanks again,
J
 JW posted on Tuesday, July 08, 2014 - 9:00 am
oh my other question is - can I quantify the difference at follow-up (vs. baseline time0).

Does the estimate (i.e., 0.118) indicate that at follow-up the scores are .12 points higher? should I use standardised estimates?

Thanks!
 Linda K. Muthen posted on Wednesday, July 09, 2014 - 9:22 am
Yes, but it is not a t test. It is a z-test in large samples.

Yes, it is the difference between the two groups. It should be compared to its standard deviation to understand how large it is.
 JW posted on Thursday, July 10, 2014 - 2:05 am
Thanks again Linda,

should I specify

OUTPUT: STDX;

to be able to estimate how large it is?
 Linda K. Muthen posted on Thursday, July 10, 2014 - 6:23 am
You should take the square root of its variance to get the standard error.
 JW posted on Thursday, July 10, 2014 - 6:34 am
Thank you very much!!!
 JW posted on Monday, July 14, 2014 - 8:46 am
As I am writing my results up, it suddenly occurred to me: is diving the estimate by standard deviation equivalent to Cohen's d effect size?

Thanks again!!!
 Linda K. Muthen posted on Monday, July 14, 2014 - 4:12 pm
This is only true if the estimate is the difference between two means.
 JW posted on Tuesday, July 15, 2014 - 1:41 am
I am unsure whether this is true in my case where I am interested in comparing means of 2 latent variables. One assessed at time 0 (and set equal to 0) and the other assessed at time 1 (which is estimated). Does dividing the z-test value corresponding to the mean at time 1 by the variable's SD correspond to a cohen's d?

I am unsure as we are forcing the first mean to be 0...

Grateful for your help!
 Linda K. Muthen posted on Tuesday, July 15, 2014 - 11:40 am
This is a mean difference even if one mean is zero. The difference in means is the mean parameter that is not zero. You divide this value by the standard deviation of that latent variable which you can find in the results or TECH4.
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