CIs: unstandardized vs. standardized PreviousNext
Mplus Discussion > Structural Equation Modeling >
 Jeff Wardell posted on Wednesday, May 27, 2009 - 2:36 pm
Dear Dr. Muthen.

I am running a path analysis and I have requested estimates of specific indirect effects. For one of my indirect pathways, the unstandardized estimate is significant (the 95% confidence interval does not contain zero), but the standardized estimate is NOT significant (the 95% confidence interval DOES contain zero). I obtained the standardized confidence interval from the section of output labelled "Confidence intervals of standardized total, total indirect, specific indirect, and direct effects." I was under the impression that standardized estimates are merely in a different metric than unstandardized estimates and that standardization should not change the level of significance of the estimate. Could you please advise me as to why my results differ when I look at the standardized CIs? Thank you very much for your time.
 Linda K. Muthen posted on Wednesday, May 27, 2009 - 2:56 pm
The ratio or the parameter estimate to its standard error and therefore confidence interval can differ slightly for raw and standardized coefficients. The standardized standard errors are not rescaled raw standard errors. See

Standardized Coefficients and Their Standard Errors

on the website under Technical Appendices.
 Jeff Wardell posted on Thursday, May 28, 2009 - 11:55 am
Thank you Dr. Muthen for your prompt response, which was very helpful. I am now curious whether you would recommend reporting the unstandardized or the standardized solution, given that they provide a different pattern of results.

Thank you.
 Bengt O. Muthen posted on Thursday, May 28, 2009 - 12:14 pm
I would go with the unstandardized SEs and CIs (but you can still report the standardized parameter point estimates). But I would assume that the CIs are not very different so that significance in the unstandardized case means that the CI barely excludes zero - in which case I wouldn't make a big deal out of that significance.
 Shiny posted on Saturday, September 20, 2014 - 9:41 am
I Regressed a binary y on a latent continuous x.The unstandardized estimate is significant yet standardized is not. Which estimate shall I Report? I wonder why the standardized is not significant. Is it due to the fact that I have categorical data that is skewed, and SE is larger?

Here are values:

Unstandardized Estimate S.E. Est./S.E. p-Value

y on x -0.488 0.241 -2.020 0.043

STDYX Estimate S.E. Est./S.E. p-Value

y on x -0.239 0.328 -0.726 0.468

 Bengt O. Muthen posted on Saturday, September 20, 2014 - 1:57 pm
Unstandardized and standardized versions of estimated coefficients have different sampling distributions. This means that the normality assumption of the estimate assumed by the usual z-score test can be better approximated in one version or the other. Significance testing using z-scores can therefore have different outcomes for the two versions.

One way to check the distribution of the estimated coefficient is to do Bayesian estimation and looking at the posterior distribution. The version that best approximates normality presumably has the most trustworthy z-score. With Bayes estimation, the normality assumption of the estimates is not needed and the 95% credibility interval is trustworthy for both versions.
 Shiny posted on Saturday, September 20, 2014 - 2:49 pm
Thank you very much for the Explanation! I did Bayesian estimation in the one Group Analysis and it worked. I am also interested in a multiple Group Analysis and check the standardized coefficient. The System suggested me to specify KNOWNCLASS and Type = mixture. I tried but got an error.

This analysis is only available with the Mixture or Combination Add-On.

Would you kindly give me a hint how this can be done in multiple Group Analysis? I do not have much experience with mixture models or bayes. So I just checked the mplus material on bayes (the 2011 short course) and did the test. Thank you!

Syntax is attached:

x med mod y
KNOWNCLASS = mod (1=js 2=ijs);
 Linda K. Muthen posted on Saturday, September 20, 2014 - 2:56 pm
It sounds like you don't have the mixture or combination add-on. You would need that to do this.
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