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Re: st: Re: using coefficients from model


From   Chris Witte <eljefespeaks@yahoo.com>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: Re: using coefficients from model
Date   Mon, 30 Mar 2009 13:13:09 -0700 (PDT)

thanks a million Martin!... Now if I only had the last few hours back.



----- Original Message ----
From: Martin Weiss <martin.weiss1@gmx.de>
To: statalist@hsphsun2.harvard.edu
Sent: Monday, March 30, 2009 3:00:34 PM
Subject: st: Re: using coefficients from model

<>


*****
sysuse auto, clear
reg price weight rep tr
g y=_b[_cons]+_b[weight]*weight+_b[rep78]*rep+_b[trunk]*trunk
*compare to built-in solution
predict pred
compare pred y
*****


HTH
Martin
_______________________
----- Original Message ----- From: "Chris Witte" <eljefespeaks@yahoo.com>
To: <statalist@hsphsun2.harvard.edu>
Sent: Monday, March 30, 2009 9:55 PM
Subject: st: using coefficients from model


> 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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> *  http://www.ats.ucla.edu/stat/stata/
> 

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