Margin of error, and comparing proportions in the same sample

October 15, 2010
By

(This article was first published on Freakonometrics - Tag - R-english, and kindly contributed to R-bloggers)

I recently tried to answer a simple question, asked by @adelaigue. Actually, I thought that the answer would be obvious... but it is a little bit more compexe than what I thought. In a recent pool about elections in Brazil, it was mentionned in a French newspapper that "Mme Rousseff, 62 ans, de 46,8% des intentions de vote et José Serra, 68 ans, de 42,7%" (i.e. proportions obtained from the survey). It is also mentioned that "la marge d'erreur du sondage est de 2,2% " i.e. the margin of error is 2.2%, which means (for the journalist) that there is a "grande probabilité que les 2 candidats soient à égalité" (there is a "large probability" to have equal proportions).
Usually, in sampling theory, we look at the margin of error of a single proportion. The idea is that the variance of widehat{p}, obtained from a sample of size http://perso.univ-rennes1.fr/arthur.charpentier/latex/multinomial15.png is

http://perso.univ-rennes1.fr/arthur.charpentier/latex/m201.png
thus, the standard error is
http://perso.univ-rennes1.fr/arthur.charpentier/latex/m202.png
The standard 95% confidence interval, derived from a Gaussian approximation of the binomial distribution is
http://perso.univ-rennes1.fr/arthur.charpentier/latex/m203.png
The largest value is obtained when p is 1/2, and then we have a worst case confidence interval (an upper bound) which is
http://perso.univ-rennes1.fr/arthur.charpentier/latex/m204.png
So with a margin of error http://perso.univ-rennes1.fr/arthur.charpentier/latex/m205.png means that http://perso.univ-rennes1.fr/arthur.charpentier/latex/m206.png. Hence, with a 5% margin of error, it means that n=400. While 2.2% means that n=2000:
> 1/.022^2
[1] 2066.116   

Classically, we compare proportions between two samples: surveys at two different dates, surveys in different regions, surveys paid by two different newpapers, etc. But here, we wish to compare proportions within the same sample. This has been consider in an "old" paper published in 1993 in the American Statistician,

It contains nice figures to illustrate the difference between the standard approach,

and the one we would like to study here.

This point is mentioned in the book by Kish, survey sampling (thanks Benoit for the reference),


Let http://perso.univ-rennes1.fr/arthur.charpentier/latex/multinomial05.png and http://perso.univ-rennes1.fr/arthur.charpentier/latex/multinomial06.png denote empirical frequencies we have obtained from the sample, based on http://perso.univ-rennes1.fr/arthur.charpentier/latex/multinomial15.png observations. Then since
http://perso.univ-rennes1.fr/arthur.charpentier/latex/multinomial07.png
http://perso.univ-rennes1.fr/arthur.charpentier/latex/multinomial08.png
and
http://perso.univ-rennes1.fr/arthur.charpentier/latex/multinomial09.png
we have
http://perso.univ-rennes1.fr/arthur.charpentier/latex/multinomial11.png
Thus, a natural margin of error on the difference between the two proportion is here
http://perso.univ-rennes1.fr/arthur.charpentier/latex/m207.png
which is here 4 points
> n=2000
> p1=46.8/100
> p2=42.7/100
> 1.96*sqrt((p1+p2)-(p1-p2)^2)/sqrt(n)
[1] 0.04142327
Which is exactly the difference we have here ! Hence, the probability of reaching such a value is quite small (2%)
> s=sqrt(p1*(1-p1)/n+p2*(1-p2)/n+2*p1*p2/n)
> (p1-p2)/s
[1] 1.939972
> 1-pnorm(p1-p2,mean=0,sd=sqrt((p1+p2)-(p1-p2)^2)/sqrt(n))
[1] 0.02619152

Actually, we can compare the three margin of errors we have so far,
  • the upper bound
http://perso.univ-rennes1.fr/arthur.charpentier/latex/m208.png
  • the "average one"
http://perso.univ-rennes1.fr/arthur.charpentier/latex/m209.png
where
http://perso.univ-rennes1.fr/arthur.charpentier/latex/m212.png
  • the more accurate one we just obtained,
http://perso.univ-rennes1.fr/arthur.charpentier/latex/m213.png
where http://perso.univ-rennes1.fr/arthur.charpentier/latex/m214.png.
> p=seq(0,.5,by=.01)
> ic1=rep(1.96/sqrt(4*n),length(p))
> ic2=1.96*sqrt(p*(1-p))/sqrt(n)
> delta=.01
> ic31=1.96*sqrt(2*p-delta^2)/sqrt(n)
> delta=.2
> ic32=1.96*sqrt(2*p-delta^2)/sqrt(n)
> plot(p,ic32,type="l",col="blue")
> lines(p,ic31,col="red")
> lines(p,ic2)
> lines(p,ic1,lty=2)
So on the graph below, the dotted line is the standard upper bound, the plain line in black being a more accurate one when the probability is http://perso.univ-rennes1.fr/arthur.charpentier/latex/multinomial19.png (the x-axis). The red line is the true margin of error with a large difference between candidates (20 points) and the blue line with a small difference (1 point).


Remark: an alternative is to consider a chi-square test, comparering two multinomial distributions, with probabilities http://perso.univ-rennes1.fr/arthur.charpentier/latex/m215.png and http://perso.univ-rennes1.fr/arthur.charpentier/latex/m216.png where http://perso.univ-rennes1.fr/arthur.charpentier/latex/multinomial19.png is the average proportion, i.e. 44.75%. Then
http://perso.univ-rennes1.fr/arthur.charpentier/latex/multinomial21.png
i.e.  http://perso.univ-rennes1.fr/arthur.charpentier/latex/multinomial22.png=3.71  
> p=(p1+p2)/2
> (x2=n*((p1-p)^2/p+(p2-p)^2/p))
[1] 3.756425
> 1-pchisq(x2,df=1)
[1] 0.05260495
Under the null hypothesis, http://perso.univ-rennes1.fr/arthur.charpentier/latex/multinomial22.png should have a chi-square distribution, with one degree of freedom (since the average is fixed here). Here the probability to reach that level is around 5% (which can be compared with the 2% we add before).

So finally, I would think that here, stating that there is a "large probability" is not correct....

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