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Reply #131: You are wrong, once again. If exit poll data is not normally distributed.. [View All]

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TruthIsAll Donating Member (1000+ posts) Send PM | Profile | Ignore Mon Nov-29-04 07:10 PM
Response to Reply #126
131. You are wrong, once again. If exit poll data is not normally distributed..
Edited on Mon Nov-29-04 07:31 PM by TruthIsAll
than what does your MOE represent?

What is the distribution if not Normal?

The MOE is related to the standard deviation which is input to the normal distribution function to compute probabilities.

Your argument that the exit polls are not normally distributed HAS NO BASIS IN FACT. You are creating pure mathematical fiction.
What do you KNOW about exit polls? Even Mitofsky talks about the MOE for his exit polls.

You keep throwing out strawmen, and I will keep knocking them down.

You ask, where do I get the MOE from? As a statistician, you should know that the MOE is the interval which coincides with 95% confidence limits. The 95% confidence interval is exactly 1.96 standard deviations on either side of the sample mean.
So the MOE = Stdev * 1.96 and the Stdev = MOE / 1.96

Along with your MOE's, how about some probability calculations? Again, show ALL the data for ALL the states. Just like I did. In fact, why don't you reproduce my analysis with your MOE's. Be sure to include the probabilities of deviaitions (within AND beyond the MOE) for each state, as well.

Finally, assuming your contention that "only" 9 states fell beyond the MOE (which I don't accept) the odds that this occurrence is due to chance alone is:

0.00004766% OR 1 out of 2,098,096.

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