0
$\begingroup$

I recently performed a multiple linear regression using a standardized set of data, and I was wondering if it possible to convert the standardized coefficients from the regression into usable unstandardized coefficients. I found and attempted to implement the method found here: https://www3.nd.edu/~rwilliam/stats1/x92.pdf

but I got significantly different (and very wrong) results compared to what I got using the standardized coefficients and standardized data. All I did was convert each coefficient by multiplying by the ratio of the standard deviations of y and the corresponding x. To calculate the results, I then multiplied each new unstandardized coefficient by each corresponding unstandardized entry in the data set, summed them, and added the intercept (which I believe does not change through the unstandardization process). Is there something I'm doing wrong. Any help would be greatly appreciated!

$\endgroup$
  • $\begingroup$ Did you handle interaction variables properly? $\endgroup$ – Aksakal Dec 18 '14 at 18:50
  • $\begingroup$ I think so. Here is my data and calculations, sorry if it's a bit ugly: docs.google.com/spreadsheets/d/… $\endgroup$ – dwm8 Dec 18 '14 at 19:11
  • $\begingroup$ Where did your "Intercept" term come from? The spreadsheet suggests it is the constant term in a regression involving the standardized variables--but in that case it should have been zero rather than $-2.45$. To make progress on this question we really need to see the calculations you made to standardize the data and to perform the regression. $\endgroup$ – whuber Jun 16 '15 at 21:23
1
$\begingroup$

Let's say a model is: $y=1+x+xz+z^3$, and $\sigma_y,\sigma_x,\sigma_z$ - standard deviations of variables. You would transform the equation like follows: $Y=\sigma_y+\frac{\sigma_y}{\sigma_x}X+\frac{\sigma_y}{\sigma_x\sigma_z}XZ+\frac{\sigma_y}{\sigma_z^3}Z^3$

$\endgroup$
  • $\begingroup$ I believe that's what I did. My calculations are in the comment link above. $\endgroup$ – dwm8 Dec 18 '14 at 20:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.