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RE: st: ttest


From   "Nick Cox" <n.j.cox@durham.ac.uk>
To   <statalist@hsphsun2.harvard.edu>
Subject   RE: st: ttest
Date   Wed, 21 Feb 2007 19:43:16 -0000

I have been on the look-out for textbook expositions 
for some years since someone pointed it out and I 
realised it was true and wondered why I had not
been aware of it previously. But it's still helpful
to know textbook sources to refer people to. 

The main idea follows from focusing on regression 
as averaging. I use -ave()- here as an operator rather than 
-E()- to underline that this applies to data, not models. 

The regression line a + bx is ave(y | x). If x is 0 or 1
then 

ave(y | 0) = a 

and 

ave(y | 1) = a + b 

and so 

ave(y | 1) - ave(y | 0) = b 

Thus the difference between the means is equal to
the absolute value of the numerator of the t 
statistic, and the only detail that remains
is the denominator. 

This isn't a proof in anybody's sense, but I've
found it useful in motivating the link to others. 

Nick 
n.j.cox@durham.ac.uk 

Ronán Conroy
 
> On 16 Feabh 2007, at 16:26, Daniel Hoechle wrote:
> 
> > The Stata-manual entry on [R] "ttest - Mean comparison 
> tests" (p.486)
> > states that Student's t-test for the equality of means can be
> > replicated by aid of a simple linear regression if the two groups'
> > variances are assumed to be unknown but identical.
> >
> > I would highly appreciate if anyone could give me a reference with a
> > formal proof of this relationship. Thanks a lot in advance.
> 
> Gosh - hard to think of a published paper that shows this. You will  
> find it demonstrated in many introductory textbooks. My favourite is
> 
> Judd, C. M., & McClelland, G. H. (1989). Data analysis: A model  
> comparison approach. New York: Harcourt Brace Jovanovich.
> 
> I'd give you a page reference but some scunner has made off with my  
> copy (my PhD student I think)

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