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Re: st: z-score as a dependent variable in a linear regression


From   Ana Gabriela Guerrero Serdan <[email protected]>
To   [email protected]
Subject   Re: st: z-score as a dependent variable in a linear regression
Date   Fri, 16 Nov 2007 10:08:36 -0800 (PST)

Hm....

Sorry to jump into this but I dont see whay Dan cannot
use z-scores as a dependent variable, even if z-scores
are standarized. Wouldnt it depend on which are the
independent variables used?

Gaby  


--- Austin Nichols <[email protected]> wrote:

> Maarten et al.--
> The issue is that Dan is standardizing using
> individual variation, I
> think, so these models are not equivalent to
> standardized
> coefficients, and the interpretation is a bit
> odd--something like,
> what is the effect of X on y relative to past
> variability in y at the
> individual level, rather than what is the effect of
> a one-unit change
> in X on y measured in SD units of y?  This example
> measuring the
> "return to experience" may clarify:
> 
> webuse psidextract, clear
> g w=lw if t<4
> egen sdw=sd(w), by(id)
> egen mw=mean(w), by(id)
> g iswage=(lw-mw)/sdw if t>3
> la var isw "Indiv stdized ln wage"
> keep if if t>3
> * Transformed data
> reg iswage exp
> xtreg isw exp, fe
> * Transformed coefs give dy/dx in SD units
> su lw if e(sample)
> loc sd=r(sd)
> reg lw exp
> di _b[exp]/`sd'
> xtreg lw exp, fe
> di _b[exp]/`sd'
> * Alternative: downweight high Var cases
> reg lw exp [aw=sdw]
> xtreg lw exp [aw=sdw], fe
> 
> All of the above have different interpretations. 
> Does another year of
> experience on the job increase wages by a constant
> percent, in
> absolute terms or relative to individuals' past
> wages, or does it
> increase wages by some constant shift in the
> individual's own
> distribution of past wages?  Or do we just get less
> information about
> the avg percent increase in wages due to another
> year of experience on
> the job from the experiences of individuals who have
> higher
> variability in their past earnings histories?
> 
> On 11/15/07, Maarten buis <[email protected]>
> wrote:
> > --- 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)
> *
> *   For searches and help try:
> *  
> http://www.stata.com/support/faqs/res/findit.html
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> *   http://www.ats.ucla.edu/stat/stata/
> 


Gaby Guerrero Serdan 

Deparment of Economics
Royal Holloway, University of London
TW20 OEX
Egham, Surrey
England, UK
http://www.rhul.ac.uk/economics/About-Us/postgrads.html
http://www.flickr.com/photos/49939890@N00/

Tel: +44 7912657259


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