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RE: st: Accounting for measurement error in regression

From   "Jason Becker" <>
To   <>
Subject   RE: st: Accounting for measurement error in regression
Date   Thu, 21 Oct 2010 09:14:20 -0400

Thanks Stas for this response.  This helps clarify things a bit for me-- I've been told this was measurement error but that didn't match with what I recalled and I knew I had seen the formula before.

This clears up a bunch of confusion I had based on conversations of other folks around the Department here.
Jason Becker
Research Specialist
Office of Data Analysis and Research
Rhode Island Department of Education
255 Westminster Street Providence, RI 02903

-----Original Message-----
From: [] On Behalf Of Stas Kolenikov
Sent: Wednesday, October 20, 2010 5:06 PM
Subject: Re: st: Accounting for measurement error in regression

On Wed, Oct 20, 2010 at 3:47 PM, Jason Becker <> wrote:
> Hello,
> My data has measurement error which is generally modeled as following a
> Bernoulli distribution.  The data are percentages of students at a
> school who score above a cutoff point on an exam, and the error is
> modeled as sqrt((p)*(q)/N) where p = percentage of students above the
> cutoff, q = percentage of students below the cutoff, and N is the number
> of students).

p*(1-p)/N is the sampling variability: you believe there is a true
probability of something equal to p, and out of N sampled objects, you
will observe variance p*(1-p)*N in the number of positive responses.
You probably want to address the inaccuracy of the instrument, and
that is a far more complex thing to do.

Suppose statalist subscription were only open to people with IQ above
120, and there was an instrument that gives you 3 point margin of
accuracy (standard deviation of the scores in repeated testing, or
between people of identical ability). Then for a person with a given
IQ the probability of getting the highly sought statalist subscription
is p(IQ)=Prob[ N(IQ,3) > 120 ]. To quantify the overall measurement
error for a given university, you would want to sum all
p(IQ)*(1-p(IQ)) over the Stata users in this university; this should
give you something like a variance of the number of people eligible
for subscription. Note that this is conditional on the test score, so
I imagine that for tests with good psychometric properties
(reliability), this variance will be much smaller than p*(1-p)*N where
p = Prob[ N(100,15)>120 ], which is the measure that you think of

Stas Kolenikov, also found at
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