
\subsection{Polling}

Another good example of binomial distributions comes up in polling.
The Gallup polling service reported that on November 10, only 37\% of
American adults thought Louise Woodward was guilty.  Furthermore,
Gallup asserted that since they asked the opinions of 623 adults,
``one can say with 95 percent confidence that the error attributable
to sampling and other random effects is within plus or minus 4
percentage points''.  Can we confirm this claim?

We want to determine $p$, the fraction of Americans that think Louise
is guilty.  Our plan is to sample the opinion of $n$ people chosen
uniformly and at random, {\em with replacement}.  That is, we might
poll the same person twice!  This may seem inefficient, but it
simplifies the analysis.  If $G$ is the number of people in our sample
that say Louise is guilty, then we will claim that $p$ is about
$\frac{G}{n}$.

The problem is that our sample might not be representative.  For
example, maybe everyone in the country outside of North Dakota thinks
Louise is guilty, but by bad luck the sample contained mostly North
Dakotans.  Then our poll would give the wrong answer.

Let $\epsilon$ be the margin of error we can tolerate, and let
$\delta$ be the probability that our result lies outside this margin.
How many people must we poll so that our result is within $\epsilon$
percent of national opinion with probability at least $1-\delta$?  For
example, Gallup claims that for $\epsilon = 0.04$ and $\delta = 0.05$,
polling $623$ people is sufficient.

We can define $\delta$, the probability that our poll is off by more
than the margin of error $\epsilon$, as follows:

\begin{eqnarray*}
\delta
	& = &
	\underbrace{\Pr\left(\frac{G}{n} < p - \epsilon\right)}_{
\begin{array}{c}
\mbox{too many in sample} \\
\mbox{say ``not guilty''}
\end{array}}
	+ \underbrace{\Pr\left(\frac{G}{n} > p + \epsilon\right)}_{
\begin{array}{c}
\mbox{too many in sample} \\
\mbox{say ``guilty''}
\end{array}} \\
	& = & \Pr(G < (p - \epsilon) n) + \Pr(G > (p + \epsilon) n)
\end{eqnarray*}

Each term in the definition of $\delta$ can be evaluated using
Theorem~\ref{th:tail}.  In the second term, we must use the same trick
as in the Noisy Channel problem to ensure that $\alpha < p$.  We
observe that $\Pr(\frac{G}{n} > p + \epsilon) = \Pr(\frac{n - G}{n} <
1 - p - \epsilon)$, where $\frac{n - G}{n}$ is the fraction of people
polled who say that Louise is not guilty, and $1-p$ is the fraction of
all Americans who say that she is not guilty.  This gives:

\begin{eqnarray*}
\delta & \leq & F_{n,p}((p - \epsilon) n) + F_{n,1-p}((1-p-\epsilon) n)
\end{eqnarray*}

This is an expression for the probability that our poll is off by more
than the margin of error.  The problem is that the expression contains
$p$, the fraction of Americans that think Louise is guilty.  This is
the number we are trying to determine by polling!

Fortunately, we can upper bound $\delta$ by using the following fact:

\begin{fact}
For all $\epsilon$, the maximum value of $\delta$ occurs when $p =
\frac{1}{2}$.
\end{fact}

The fact implies that to get an upper bound on $\delta$, we can
pretend that half of the people think Louise is guilty.  This gives:

\begin{eqnarray*}
\delta & \leq & F_{n,\frac{1}{2}}((\frac{1}{2} - \epsilon) n) +
		F_{n,1-\frac{1}{2}}((1-\frac{1}{2}-\epsilon) n) \\
	& = &	2 F_{n}((\frac{1}{2} - \epsilon) n)
\end{eqnarray*}

Now suppose that we want a margin of error of $4\%$ as Gallup claimed.
Plugging in $\epsilon = 0.04$ gives:

\begin{eqnarray*}
\delta
	& \leq &  2 F_{n}(0.46 n) \\
	& \leq & 2 \cdot 6.75 \cdot \frac{2^{-0.004622\ldots \cdot n}}{1.2492\ldots \sqrt{n}}
\end{eqnarray*}

We want to poll enough people so that $\delta$ is less than $0.05$.
The easiest way is to plug in values for $n$, the number of people
polled:

\[
\begin{array}{c|l}
\mbox{$n$ = people polled} & \begin{array}{cc}
\mbox{upper bound on} \\
\mbox{probability poll is wrong}
\end{array} \\
\hline
500 &  9.7\% \\
600 &  6.4\% \\
623 &  5.9\% \quad \leftarrow \quad \mbox{Gallup's poll size} \\
650 &  5.3\% \\
662 &  5.0\% \quad \leftarrow \quad \mbox{our poll size} \\
700 &  4.3\% \\
\end{array}
\]

Gallup's poll size is just about right.  By our calculation, polling
662 people is sufficient to determine public opinion to within 4\%
with confidence of 95\%.  We can be certain that Gallup's poll of 623
people gives 95\% confidence with at most a 4.13\% margin of error.
But the real margin of error may be less, since we made several
approximations.  Still, our approximations must be quite good, since
4.13\% is quite close to Gallup's claim of 4\%.

The remarkable point is that the population of the country has no
effect on the poll size!  Whether there are a thousand people or a
billion in the country, polling only a few hundred is sufficient!