back transform square-root linear regression

[Sonia Zam]

Hi all,


I have conducted a linear regression where outcome measures were square-root transformed (one of them is measured as a percentage, and the other is measured as centimetres squared) to improve the normality of the residuals. The predictor variables are categorical.

Now, I want to back transform the coefficients relative to a predetermined reference level of 25% for the percentage variable, and 140 cm² for centimetres squared variable so that the effect estimates could be directly compared between variables. I am not sure how to do this.

Any help with this will be highly appreciated.

[Rich Goldstein]

back transforms are not as straightforward as one might think; see Miller, DM (1984), "Reducing transformation bias in curve fitting," The American Statistician, 38(2): 124-126; note that an alternative is to use glm with a power link; see

help glm

[John Mullahy]

I agree with Rich that glm may be a preferred strategy. If you want to proceed with square-root transformation, however, this paper indicates how the retransformation may be performed for square-root transformations. https://www.jstor.org/stable/3768008

[FernandoRios]

this may be more complicated if trying to recover the standard errors, but wouldn't it be possible to "rescale" the coefficients using simple calculus?
for example:

model:
y^d=a+b*x 
Total Difference
d*y^(d-1)*dy=b*dx
dy/dx=b/(d*y^(d-1))

alternatively, more to the core of your question. There is no need to worry about the normality of the residuals. OLS will still be consistent and unbiased if the residuals are not normal.

Fernando

[Sonia Zam]

Thanks all for your prompt help,

I want to run a linear regression and back transform the mean differences to the original values, and report the absolute differences relative to the median dependent variable (the reference value of the dependent variable is 30%). I want to do this so the effect estimates can be directly compared between variables because otherwise, the absolute difference would depend on the average of the dependent variable in the reference group.

The dependent variable is measured as a percentage.

generate sqrtdepend=sqrt(dependent)
regress sqrtdepend i.indep1 i.indep2 i.indep3 i.indep4
margins indep1 indep2, expression(predict(xb)^2)

Not sure the correct syntax to report absolute differences relative to the median of the dependent variable (of 30%).

I will read more about glm but for now, I want to proceed with square-root transformation.

Thanks in advance.

[Maarten Buis]

FernandoRios unfortunately no, because squaring is a nonlinear transformation.

[FernandoRios]

Maarten Buis Thank you. I wasnt completely sure about that. It seemed to work with a toy example, but I know it is harder to do when the dependent variable goes through a transformation.
Best Regards