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Re: st: using coefficients from model


From   Chris Witte <eljefespeaks@yahoo.com>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: using coefficients from model
Date   Mon, 30 Mar 2009 13:33:29 -0700 (PDT)

Maarten,

I want to report the regression equation in my manuscript so the reader can plug in their independent variable values and get the correct estimate.  I thought that calculating the difference between the _fracpred_ estimated values and the values estimated using the coefficients from the _fracpoly_ output would be the fastest way to find the "real" constant for reporting purposes.  Is there a better (i.e. easier) way?  Am I misunderstanding the process?

Chris


----- Original Message ----
From: Maarten buis <maartenbuis@yahoo.co.uk>
To: statalist@hsphsun2.harvard.edu
Sent: Monday, March 30, 2009 3:09:57 PM
Subject: Re: st: using coefficients from model


You can access the parameter estimates using _b[varname], 
but I don't understand why you would want to do this. 
Isn't it much easier to just read the manual entry for 
-fracpoly-? This extensively documents how the variables
are transformed.

-- Maarten

-----------------------------------------
Maarten L. Buis
Institut fuer Soziologie
Universitaet Tuebingen
Wilhelmstrasse 36
72074 Tuebingen
Germany

http://home.fsw.vu.nl/m.buis/
-----------------------------------------


--- On Mon, 30/3/09, Chris Witte <eljefespeaks@yahoo.com> wrote:

> From: Chris Witte <eljefespeaks@yahoo.com>
> Subject: st: using coefficients from model
> To: statalist@hsphsun2.harvard.edu
> Date: Monday, 30 March, 2009, 8:55 PM
> I am using stata 9.2.  I would like to know how to call
> the coefficients from the estimation results matrix e(b) and
> use them in an equation.  For example, if my matrix e(b)
> has values for  "_cons" and coefficient
> "test", I would like to calculate the answer to an
> equation y = [_cons] + x*[test].  How can I do this?
> 
> The reason why I want to do this is because I am trying to
> find the value _fracpoly_ adjusts its estimates by.  If I
> use this output from my _fracpoly_ estimation:
> 
> study1twsg       Coef.   Std. Err.     
> t    P>t     [95% Conf. Interval]
> Itemp__1    15093.12   4454.121     3.39  
> 0.012     4560.802    25625.45
> Itemp__2   -10666.18    3136.84    -3.40  
> 0.011    -18083.63   -3248.734
> Itemp__3    1945.114   569.9302     3.41  
> 0.011     597.4436    3292.785
> _cons    .0053322   .0008244     6.47  
> 0.000     .0033828    .0072815
> Deviance:  -113.24. Best powers of temp among 164 models
> fit: -2 -2 -2.
> 
> and I plug in these coefficients with my 'x'
> having a value of 25, my linear regression equation for
> predicting y is:
> 
> y=.0053322+(15093.12*25^-2)+(-10666.18*25^-2*log(25))+1945.114*25^-2*(log(25))^2
> 
> However, this equation does not yield the same value as
> _predict_ (or _fracpred_), and this is because fracpoly
> secretly adjusts its constant.  If I can calculate the
> answer to this equation and put it into a variable, I can
> then find the difference between this variable and the value
> that _fracpred_ gives.
> 
> Thank you for your help.  Hopefully this isn't too
> confusing.
> 
> Chris
> 
> 
> 
>      
> 
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