# Margin of error, and comparing proportions in the same sample

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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 is

The standard 95% confidence interval, derived from a Gaussian approximation of the binomial distribution is

The largest value is obtained when p is 1/2, and then we have a worst case confidence interval (an upper bound) which is

So with a margin of error means that . 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 and denote empirical frequencies we have obtained from the sample, based on observations. Then since

and

we have

Thus, a

*natural*margin of error on the difference between the two proportion is here

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

- the “average one”

- the more accurate one we just obtained,

> 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 (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 and where is the average proportion, i.e. 44.75%. Theni.e. =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, 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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