Austin -
But suppose one is trying to make a prediction model from actual income,
not an income range? Then wouldn't some adjustment have to be made for
the predictor variable being measured with error? If so, how?
Al Feiveson
-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Austin
Nichols
Sent: Sunday, November 09, 2008 10:19 AM
To: [email protected]
Subject: Re: st: Interval variables as independent variables
"Jessica A.Jakubowski" <[email protected]>:
Usually it is better to make indicators, e.g.
tab income, gen(d)
reg y d*
On Sat, Nov 8, 2008 at 4:57 PM, Maarten buis <[email protected]>
wrote:
> I can see two alternative strategies (there may well be more):
>
> 1) You can consider this a missing data problem and try multiple
> imputation, like in:
>
> Patrick Royston (2007) Multiple imputation of missing values: further
> update of ice, with an emphasis on interval censoring. The Stata
> Journal, 7(4): 445-464.
> http://stata-journal.com/article.html?article=st0067_3
>
> 2) Alternatively you can scale the income variable such that it
> optimally predicts the outcome variable like in:
>
> Maarten L. Buis (2008) Scaling levels of eduction.
> http://home.fsw.vu.nl/m.buis/
>
> and
>
> Usefulness and estimation of proportionality constraints: The
> propcnsreg package. presented at the 13th UK Stata Users Group Meeting
> on September 10 2007 in London. http://home.fsw.vu.nl/m.buis/
>
> Hope this helps,
> Maarten
>
> --- "Jessica A.Jakubowski" <[email protected]> wrote:
>> I have data that includes an income variable measured at the interval
>> level, e.g.:
>> 1= less than $10,000
>> 2= $10,000-$15,000
>> 3= $15,000-$20,000
>> 4= $20,000-$25,000
>> 5= $25,000-$35,000
>> 6= $35,000-$50,000
>> 7= $50,000-$75,000
>> 8= over $75,000
>>
>> Stata has a way to deal with interval data when it is a dependent
>> variable in a regression model with the command -intreg-
>>
>> I would like to know if Stata has a way to deal with interval
>> variables on the right-hand side of the regression equation, that is,
>> as an independent variable. I am using this interval income variable
>> in models with continuous (-regress-) and binary (-logit-) outcomes.
>> Right now, I am using the midpoint of the intervals (and the open top
>> interval*1.5), but I'm wondering if there's a better
> way.
>
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