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Re: st: Adjust command


From   Ricardo Ovaldia <ovaldia@yahoo.com>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: Adjust command
Date   Fri, 18 Apr 2003 14:13:19 -0700 (PDT)

Thank you Ken. 

So let me see if I understand, when I type -adjust
mpg,by( foreign)- , mpg is set to the overall mean for
mpg, and if I type -adjust,by(foreign)-, (without
specifying mpg) then mpg is set to the mean of mpg
within each foreign level. Correct?

Regarding my first question, I understand that it
depends on the question that I want answered, what I
don't understand is what question is being answered by
each of the two approaches. Specifically if I type:

. regress  price mpg foreign
. adjust , by( foreign) se
or
. regress  price mpg
. adjust, by( foreign) se

What are these two adjusts telling me? I am sorry if
this is to trivial, but I am confused about the
interpretation of the second approach.

Best,
Ricardo.



the adjusted 
--- khigbee@stata.com wrote:
> Ricardo Ovaldia <ovaldia@yahoo.com> asks:
> 
> > I have two questions about -adjust- after
> -regress-.
> > First, under what circumstances should the -by()-
> > variable be included or not included in the
> regression
> > model. ( I get different adjust outputs),
> 
> The answer depends on what question you are trying
> to ask of your
> data and regression model.  The adjust output will
> vary depending
> on what you do, as it should.
> 
> The way I would approach it is to first think of the
> appropriate
> regression model and run it.  Then think about what
> kinds of
> summaries you wish to obtain from that model (i.e.,
> what question
> are you trying to answer).  This is what guides how
> -adjust-
> should be used.
> 
> 
> > Second, according to the help file, the -se-
> option is
> > "equivalent to the stdp option of predict".
> Therefore,
> > I should be able to get what -adjust- reports
> using
> > -predict-, but I can't. Using the auto data and
> these
> > commands:
> > 
> > qui regress  price mpg foreign
> > predict yhat
> > predict se, stdp
> > adjust , by( foreign) se
> > by  foreign, sort:sum  yhat se
> > 
> > I get that mean(yhat)= adjusted(mean) as expected,
> but
> > I can't get the adjusted(se). What am I missing?
> 
> 
> Try the following experiment to understand what
> Stata's -adjust-
> command is presenting in it's table.
> 
> Using the auto dataset as the example,
> 
>     . sysuse auto
>     . regress price mpg foreign
> 
> Now add two observations to the bottom of the
> dataset
> 
>     . set obs 76
> 
> And set foreign to 0 and 1 in the last two obs.
> 
>     . replace foreign = 0 in 75
>     . replace foreign = 1 in 76
> 
> And set mpg to the mean of mpg in the last two obs
> when foreign
> is 0 or 1 respectively.
> 
>     . summarize mpg if foreign==0 in 1/74
>     . replace mpg = r(mean) in 75
>     . summarize mpg if foreign==1 in 1/74
>     . replace mpg = r(mean) in 76
> 
> Now run your predicts
> 
>     . predict xb
>     . predict se, stdp
> 
> And look at what it produced in the last two obs.
> 
>     . list foreign mpg xb se in 75/76
> 
>         
> +------------------------------------------+
>          |  foreign       mpg       yhat         se
> |
>         
> |------------------------------------------|
>      75. | Domestic   19.8269   6072.423    350.979
> |
>      76. |  Foreign   24.7727   6384.682   539.5994
> |
>         
> +------------------------------------------+
> 
> And then compare that with what -adjust- produced.
> 
>     . adjust in 1/74 , by(foreign) se
> 
>    
>
----------------------------------------------------------
>          Dependent variable: price     Command:
> regress
>         Variable left as is: mpg
>    
>
----------------------------------------------------------
> 
>     ----------------------------------
>      Car type |         xb        stdp
>     ----------+-----------------------
>      Domestic |    6072.42   (350.979)
>       Foreign |    6384.68   (539.599)
>     ----------------------------------
>          Key:  xb    =  Linear Prediction
>                stdp  =  Standard Error
> 
> 
> The reason why what you did was not equivalent for
> the stdp, was
> that the mean of the stdp is not necessarily the
> same as the stdp
> evaluated at the mean.
> 
> 
> Ken Higbee    khigbee@stata.com
> StataCorp     1-800-STATAPC
> 
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