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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 Likertscale [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 


In crosssectional 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 crosssectional 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' yxstandardized factor loading and then calculate the mean and the std. dev. for each factor? Best Regards, Christian 


This is not how Mplus handles factors in crosssectional studies, it is the conventional way to do this. A factor mean in a crosssectional 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. 


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. 


Dear Linda, If I just want to test the configural model, is it still recommend to do that? Thank you. 


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. 


Dear Linda, From your first reply in this post, I understand that estimated means for latent variables in a crosssectional 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: 1213)? The numerator would always be zero... Thanks in advance for your help! 


What are "informative indices"? 


I'm sorry that my question was unclear. According to Asparouhov (2004: 12), an informative index is essentially a tstatistic 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. 


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 pvalue associated with the mean at time2 in the output as in the example below  for example, the mean score associated with variable 1 at followup (Follow1) is .118 which is associated with a pvalue 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 


With two time points, the ztest 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 followup (vs. baseline time0). Does the estimate (i.e., 0.118) indicate that at followup the scores are .12 points higher? should I use standardised estimates? Thanks! 


Yes, but it is not a t test. It is a ztest 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? 


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!!! 


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 ztest 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! 


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. 

JW posted on Tuesday, July 29, 2014  4:23 am



Hi Linda, I have been looking at this a bit more and was discussing it with someone at work. This conversation has brought to my attention the fact that I am not sure how MPLUS constructs the factors and how the free mean at time 2 is calculated, especially since the latent factor consists of questionnaires with different scales. Grateful for your response and please feel free to point me to any relevant reference. Thanks 


See the Topic 4 course handout on the website under multiple indicator growth. 

JW posted on Wednesday, July 30, 2014  2:10 am



Thanks Linda! 

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