- #1

Sleeker

- 9

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As a concrete example, I'm going to use the latest Gallup survey:

Obama: 49%.

Romney: 44%.

Margin of error (95% confidence) = 2%.

As I've found out, the margin of error applies to each person individually, so Obama's share will be between 47% and 51% in 95 out of 100 cases, and Romney's share will be between 42% and 46% in 95 out of 100 cases.

I'm trying to find a formula that will give me the probability that one candidate (say Obama) is ahead of the other candidate. Each candidate's share of the vote is the peak of a normal distribution.

Now, I believe the following represents the probability that each candidate, individually, has at least x% of the vote:

[itex]P = \frac{1}{2} (1-erf (\frac{x-\mu}{\frac{MOE}{2} \sqrt{2}}))[/itex]

With

[itex]\mu = [/itex] The percentage in the poll for the candidate.

MOE = Stated margin of error.

I don't know if that helps, but that's what I have so far.

Basically, I'm trying to find a function that describes the probability that one candidate is ahead of the other

*by any amount*as a function of (the difference between the two candidates) and (the margin of error).