Standard deviation for adjusted effect size estimates I am trying to calculate the effect size for a trait adjusted for my set of nuisance variables. My models looks like this:
Trait ~ Dx + Age + Sex

Where "Trait" is a continuous variable and "Dx" is a binary group membership variable. 
I'm using this basic formula to get an effect size estimate:
(mean(Trait[which(Dx==0)])-mean(Trait[which(Dx==1)]))/SDpooled

but I am mostly interested in the effect size after correcting for Age and Sex.
I calculated the marginal means (i.e. corrected for Age and Sex) for the Trait split by Dx group as shown here:  on page 15, section 7.5 (PDF) 
But I want to know which standard deviation I should use. Is it required to also adjust the pooled standard deviation by the covariates? Is there some better way to get effect size estimates adjusted by a set of nuisance variables?
 A: A simple way would be to use linear regression with the formula you gave, which provides a standard error (which is equivalent to the standard deviation of the adjusted mean difference).
lm(trait ~ Dx + Age + Sex)
If you insist on doing it "by hand", you can take the residuals from a regression of trait on Age and Sex, and calculate the mean difference according to trait and its standard deviation like this (supposing Dx has levels 0 and 1)
resids = resid(lm(trait ~ Age + Sex))
Dx0 = Dx == 0; Dx1 = Dx == 1
mean.diff = mean(resids[Dx1]) - mean(resids[Dx0])
pooled.sd = sqrt(sd(resids[Dx1])^2/length(resids[Dx1])+sd(resids[Dx0])^2/length(resids[Dx0]))
T = mean.diff/pooled.sd

using this formula:

But this involves an unsolved problem in statistics (see this). You might observe that the two approaches give very similar results if the explanatory variables are uncorrelated.
A: I think you can estimate the SD from SE (SD = SE × square root of sample size). Then you can use the SD and the estimated marginal means to estimate the effect size.
