RichardKennaway comments on Politics Discussion Thread December 2012 - All
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It isn't, at all. Back of envelope calculation:
Standard deviation of binomial distribution not too far from 50-50 is sqrt(N)/2 = sqrt(18)/2 or about 2 out of 18 = 11%. The Moldbuggians are 8% different from all of LW answering the poll, clearly an insignificant difference.