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SCI199Y: October 10, 1995
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\noindent{\bf Required for two weeks from now}
\begin{itemize}
\item Project 1
\end{itemize}

\noindent{\bf Some technicalities on polls}

\noindent{\it Allocating the undecideds}

In the poll reported on September 30 in 
the Globe and Mail (front page),
the results were given as 53.2\% for no, and 46.8\% for yes.  Later
in the article, the results were given as:

\begin{tabular}{rrrr}
no&yes&undec&won't say\\
45.1&43.8&6.4&4.7
\end{tabular}

According to the reporter, the undecideds and won't says are grouped,
and their votes are allocated in the ratio 70\% to no, 30\% to yes.
(In fact my calculation shows that 76\% was assigned to no.)

Later in the same article, a study from three weeks earlier is quoted
as giving the results 49.8\% for no and 50.2\% for yes.  The detailed
breakdown was:

\begin{tabular}{rrrr}
no&yes&undec&won't say\\
42.9&43.8&7.5&5.8
\end{tabular}
In this poll about 50\% of the undecideds and won't says were 
were allocated to the no group.  So a large part of the shift is due
to a change in the method of allocation.

\bigskip
\noindent{\it The margin of error}

For the poll quoted above (the first one), it is also stated in the
article that 1,006 people were polled, and that the poll has a margin
of error of 3.1\%, 19 times out of 20.  The way to compute the margin
of error is simple:
$$
margin~of~error = 2\times \sqrt {\frac {(percentage~Yes)\times (percentage~No)}
{number~polled}} .
$$
In this poll, that is $2\times \sqrt {53.2\times 46.8 \div 1006} = 3.1\%$.
The theory behind this is sketched on the next page.

\bigskip
\noindent{\it The polling trend}

In an article on September 26 in the Globe and Mail, headlined "Polls show
trend", the following results were given:

\begin{tabular}{rrrrrr}
Date&number& & &undecided/&margin\\
of poll&polled&No&Yes&won't say&of error \\ \hline 
Sept 7-8&959&42.9&43.8&13&3.1\\
Sept. 8-12&1003&45.0&37.0&18&3.2\\
Sept. 11-14&500&40.0&36.0&24&4.5\\
Sept. 15-19&1004&46.2&38.8&15&3.1\\
\\
Sept. 21-25&1003&45.1&43.8&11.1&3.1\\
\end{tabular}

(I've added the poll reported on Sept. 30 myself.)
Does it look like a trend to you?

\bigskip

\noindent{\bf In the news this week...}
\begin{itemize}
\item "Prostate cancer epidemic looms": G \&M, Sept. 27 (A10)
Summarizes an article in the latest issue of the Canadian Journal of Public 
Health.
\item "Canadian banks the most robbed": G \& M, Sept. 28 (A2)
But the take per robbery is the lowest... [with accompanying table]
\item "High school graduates caught in job vacuum": G \& M, Oct.3 
[with accompanying graphic]
\item "Eating disorders on the increase": G \& M, Oct. 5 (A14)
New studies suggest that both anorexia and bulimia occur up to twice
as frequently as reported in earlier studies.
\end{itemize}

\noindent{\bf Monty Hall again...}
This rather elegant solution was posted on the internet.  I think it's okay...
\begin{quotation}
There are two strategies:
\begin{enumerate}
\item You stay with your initial pick all the time.
\item You switch doors every time.
\end{enumerate}
In case 1, the probability of winning is 1/3.
In case 2, the probability of losing  is 1/3, because you will
lose only if you had picked the winning door first.
\end{quotation}

\noindent{\bf Theory behind the margin of error}
\begin{itemize}
\item We assume there is a proportion of the total voting population
that will vote no (often called the true proportion).
\item The proportion of the sample that will vote no is assumed to
{\it estimate} the true proportion.   This estimate has some sampling error.
\item If the error in the poll is only due to sampling (this is a big if),
then this error can itself be estimated.  Denote the true proportion
by $p$, and the sample size by $n$. 
\item The theory behind the {\it binomial distribution} shows that the
error in the estimate is given
by $\sqrt{p(1-p)/n}$.  We don't know $p$, but use the sample estimate again.
\item The interval given by
$$
sample~proportion \pm 2 \times sampling~error
$$
catches the true proportion 95\% of the time (19 times out of 20).
The factor 2, the 95\%
and the $\pm$ comes from the 'bell curve', or normal
distribution, used here as an approximation to the binomial distribution.
\end{itemize}

 
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