--- Dan Weitzenfeld <[email protected]> wrote:
> I've reviewed the assumptions of OLS, etc., and I can't find a
> glaring objection to using a z-score as a dependent variable in a
> linear regression. However, it feels strange, especially because the
> units of a z-score are not uniform, in that the difference in
> probability between a z-scores 0 and .5 is not the same as the
> difference in probability beween z-scores .5 and 1. In other words,
> I feel that by using the z-score, I am not giving adequate weight to
> observations with very high and very low z-scores.
This sounds to me like a very common misunderstanding. Turning a
variable into a z-scores does not make that variable normally
distributed. It is a linear transformations of your variable, nothing
more. So each value retains the same relative weight as in your
original variable. You are only changing the way you interpret the
result but your are not changing any of the underlying information. The
way to realize that is to see that you can get those standardized
estimates using a normal regression on your unstandardized variables
without estimating a new model, as you can see in the example below:
*-------------------- begin example --------------------
sysuse auto, clear
reg pri mpg
sum pri
local standardized = _b[mpg]/r(sd)
sum mpg
local standardized = `standardized'*r(sd)
di `standardized'
reg pri mpg, beta
sum pri
gen double z_pri = (pri-r(mean))/r(sd)
sum mpg
gen double z_mpg = (mpg-r(mean))/r(sd)
reg z_pri z_mpg
*---------------------- end example -----------------------
(For more on how to use examples I sent to the Statalist, see
http://home.fsw.vu.nl/m.buis/stata/exampleFAQ.html )
-- Maarten
-----------------------------------------
Maarten L. Buis
Department of Social Research Methodology
Vrije Universiteit Amsterdam
Boelelaan 1081
1081 HV Amsterdam
The Netherlands
visiting address:
Buitenveldertselaan 3 (Metropolitan), room Z434
+31 20 5986715
http://home.fsw.vu.nl/m.buis/
-----------------------------------------
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