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Edited on Fri Dec-10-04 11:20 AM by TruthIsAll
THESE PROBABILITIES HAVE NOT BEEN REFUTED; IN FACT, THEY HAVE BEEN CONFIRMED BY OTHER MATHEMATICIANS HERE AT DU.
These are the probabilities that Bush's tallies could have deviated as they did from the exit polls based on the MOE and calculated using the binomial distribution:
1) If you assume the calculated MOE, based on sample size (as for standard, non-exit polls), then the odds of Bush's vote tallies exceeding the exit poll MOE in at least 16 states is 1 out of 13.5 TRILLION.
2) If you assume a 2% MOE (more likely for exit polls), then the odds of Bush's vote tallies exceeding the 2% Exit Poll MOE in at least 23 states is ZERO.
If N = number of states exceeding the MOE, then the probability is: Prob = 1-BINOMDIST(N-1,51, MOE/2, TRUE)
For (1) and (2) above: P (1) = 1-BINOMDIST(15,51,.015, TRUE) ; the odds are 1/ P(1) P (2) = 1-BINOMDIST(22,51, .01, TRUE) ; the odds are 1/ P(2)
It's as simple as that.
Try it yourself in Excel. Then show the results to the Mystery Man.
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