Re: Election-polling math
Lucky for you we just took a quiz on polling data in statistics about 5 hours ago. Basically, they take the half-interval margin of error (kind of a copout from what I can see) defined as:
z(alpha/2)*sqrt(pq/n)
where z is the t-distribution at an infinite degree of freedom, p is the probability of something happening, q is 1-p, and n is the sample size. For simplicity, they round z to 2, 95% confidence interval has z=1.96. Also, before the data you can assume that p=q=.5 since that will give you the highest possible margin of error. This simplifies a lot, and you end up getting:
MoE = 1/sqrt(n)
So, for a margin of error of 4%, you need >= 625 people. Of course, this all really depends on how they pick the participants. I hope that's right, because I put 625 down on the quiz.
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