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Mplus Discussion > Structural Equation Modeling >
 Isaac Petersen posted on Wednesday, February 16, 2011 - 11:20 am
I am predicting a latent factor from four predictors. How can I estimate the individual contributions (effect sizes) of *two* of the four predictors on the outcome?

I'm not sure how to estimate effect sizes in Mplus, but it might be related to r-squared values or the percent of variance in the outcome accounted for by the predictors.

 Bengt O. Muthen posted on Wednesday, February 16, 2011 - 6:18 pm
I don't know how one can isolate percent variance contributed by some of a set of predictors when the predictors are correlated. Typically standardized values are used to gauge relative effects.
 Isaac Petersen posted on Wednesday, February 23, 2011 - 7:58 am
Unfortunately, standardized regression coefficients don't determine the relative importance of individual predictors (Willett, Singer, & Martin, 1998, p. 412). Usually, r-squared change (or f-squared) is used to assess effect sizes within the multiple regression framework.

Is there a way in Mplus to determine r-squared for an outcome? If so, then presumably I can determine the r-squared without a given predictor, and then calculate the r-squared with the predictor to determine the change in r-square with the addition of the predictor. Can this be done?

Thanks for your help!
 Linda K. Muthen posted on Wednesday, February 23, 2011 - 8:56 am
The STANDARDIZED option provides an R-square for each dependent variable based on all of its predictors.
 Dan Abner posted on Thursday, December 01, 2011 - 9:32 am
When I specify the STANDARDIZED option, I obtain both Std and StdYX. I, like Isaac, want a partial R-square type of measure of effect size for individual effects. For this purpose, which one (Std or StdYX) should I use?
 Linda K. Muthen posted on Thursday, December 01, 2011 - 3:04 pm
I don't think either of those would get you a partial R-square. I'm not sure how you would get this.
 Luke Brooks-Shesler posted on Sunday, April 01, 2012 - 1:28 pm
Hi Linda,
You mentioned earlier that the STANDARDIZED option provides an R-square for each dependent variable based on all of the predictors.
My SEM includes interactions that are created with the xwith command. Therefore, I believe that only UNSTANDARDIZED output is generated.
Is there any way to compute R-square using the UNSTANDARDIZED output, or is this not possible?
Thank you,
 Linda K. Muthen posted on Monday, April 02, 2012 - 9:12 am
The following FAQ on the website shows how to compute R-square when the model contains an interaction:

Latent variable interactions
 Luke Brooks-Shesler posted on Monday, April 02, 2012 - 11:00 am
Thank you, Linda.
 Anthony posted on Thursday, October 25, 2012 - 1:44 pm

I'm conducting a relatively simple path model and I know Mplus provides the total R-square value for each outcome variable in the model. However, my reviewers are asking for the variance accounted for by each path. Since each outcome has multiple paths, the R-square value doesn't work for this - would it simply be the sum of both the beta (i.e., STDXY estimate) value for the direct effect (none) added to the beta value for the sum of the indirect effects. For instance:

Beta1: X1->M1->M2->DV
Beta2: X1->M2->DV
Beta3: X2->M1->DV
Beta4: X2->M2->DV


Total variance in the DV accounted for by X1 is Beta1+Beta2?
Total variance in the DV accounted for by X2 is Beta3+Beta4?

Any insights you may be able to provide would be helpful.

Thank you!
 Linda K. Muthen posted on Friday, October 26, 2012 - 1:44 pm
Unless the correlations among the x's are zero, these values cannot be obtained.
 Benjamin Miller posted on Wednesday, July 31, 2013 - 5:54 pm
I would like to report the total variance explained (r-squared) for my latent factor outcome variable.

However, the residual variance of the latent outcome factor was a small, negative, and non-significant value. Thus, I fixed the negative residual variance of the latent outcome factor to 0 as recommended from previous posts.

Although the model ran, now I do not obtain an r-squared value for my latent outcome factor.

Is there any way to circumvent this problem?

 Linda K. Muthen posted on Thursday, August 01, 2013 - 10:38 am
When you fix the residual variance to zero, you are in effect saying that R-square is one.
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