[EM] CONFIRMATION SAMPLE SIZE
gervase at group.force9.co.uk
Sat Nov 22 15:51:02 PST 2003
> From: "Joe Weinstein" <jweins123 at hotmail.com>
> Date: Thu, 20 Nov 2003 00:10:11 -0800
> Subject: [EM] CONFIRMATION SAMPLE SIZE
> CONFIRMATION SAMPLE SIZE (WAS Re: Re: Touch Screen Voting Machines)
> THE QUESTION. In EM message 12737, Wed. 19 Nov 03, Ken Johnson asked:
> "Suppose you have a two-candidate election with 10,000,000 voters, and
> the computer
> says that candidate A beats candidate B 51% to 49%. How many
> randomly-selected ballots would you need ...to confirm the election
> result with 99.99% confidence?"
Thanks for doing the calculation. I was thinking of replying to this
post. It's a good thing you did because I think I definitely would have
done a completely wrong statistical test.
You made a case for saying that for the above election, it would be better
to have approximately 150,000 randomly chosen people to vote in this
situation rather than get people to voluntarily vote. My thinking was if
the result is going to be a landslide (i.e. A = 1, B = 0), why bother
having 150,000 randomly chosen people? Why not add one person at a time
until the required accuracy is obtained?
I completed my own calculations but then noticed I made a mistake. I
forgot that take into account that the standard deviation S = SQRT
(X(1-X)/N) (i.e varies with X). As A = 1, therefore X = 1. This makes
the S = 0. And this is where I get stuck.
So, what should the expression be for N (the required number of voters)
given that the election is a landslide? I thought of using the Binomial
distribution, but I don't think that's right.
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