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From |
"Feiveson, Alan H. (JSC-SK311)" <Alan.H.Feiveson@nasa.gov> |

To |
<statalist@hsphsun2.harvard.edu> |

Subject |
RE: st: Interval variables as independent variables |

Date |
Mon, 10 Nov 2008 08:05:09 -0600 |

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: owner-statalist@hsphsun2.harvard.edu [mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of Austin Nichols Sent: Sunday, November 09, 2008 10:19 AM To: statalist@hsphsun2.harvard.edu Subject: Re: st: Interval variables as independent variables "Jessica A.Jakubowski" <j-jakubowski@northwestern.edu>: 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 <maartenbuis@yahoo.co.uk> 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" <j-jakubowski@northwestern.edu> 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. > * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/ * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

**Follow-Ups**:**RE: st: Interval variables as independent variables***From:*Maarten buis <maartenbuis@yahoo.co.uk>

**References**:**st: Interval variables as independent variables***From:*"Jessica A.Jakubowski" <j-jakubowski@northwestern.edu>

**Re: st: Interval variables as independent variables***From:*Maarten buis <maartenbuis@yahoo.co.uk>

**Re: st: Interval variables as independent variables***From:*"Austin Nichols" <austinnichols@gmail.com>

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