\problemdata      
        {?} % latex-friendly label for the prob.
        {Probability-Variance} % The topic(s) of the problem content.
        {?} % Source (if known)
        {F01, PS12; S98, PS10} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).

\begin{problem}
{\bf Harry Potter mania}
We want to determine the percentage $r$ of people who intend to watch
the {\em Harry Potter} movie.   Since we are better at probability 
than magic, we conduct a poll.  Say that we poll a total of 800 people. 
What is the confidence level
that our estimated value is within 4 percentage points
of the actual value $r$?  Compute two lower bounds on the confidence 
level, one using Chebyshev's Inequality 
and the other using the binomial distribution
approximation.


\solution{
For this problem, we have the number of people polled $n=800$, the
error $\epsilon=0.04$, and the actual percentage of people who intend
to watch the movie $p=r$.  Let $\delta$ be the probability that the
error is greater than $\epsilon$, and correspondingly, let $c =
1-\delta$ be the confidence level. 

From Lecture Notes 12, equation (15), we know that
 
$$
Pr(more than \epsilon error)   \leq    F_{n,p} ((p-\epsilon)n)  
                                        + F_{n,1-p} ((1-p-\epsilon)n)
$$

We don't know $p$, but we do know from Fact 3.2 that the RHS is
maximized when $p=1/2$.  Therefore

\begin{eqnarray*}
\delta  & \leq  &  F_{n,1/2} ((1/2-\epsilon)n) + F_{n,1/2} ((1 - 1/2 \epsilon)n) \\
          & =     &  2 F_{n,1/2} ((1/2 - \epsilon) n)  \\ 
          & \leq  &  k_1         
\end{eqnarray*}

Thus we have $c = 1-\delta \geq k_1$.


Similarly, using Lecture Notes 13, we know that

$$\delta  \leq  \frac{\sigma^2}{n e^2}$$

We know that the variance, $\sigma^2 = p(1-p)$, but again, we do not know $p$.
However, as the Lecture Notes explain, $\sigma$ is maximized at $p=1/2$, and therefore we can say that

$$\delta  \leq  \frac{1/4}{n e^2}  = k_2$$.
 
Thus we obtain our second estimate $c = 1 - \delta \geq  k_2$.
}

\end{problem}
