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Reply #56: did you understand my post, or TIA's? [View All]

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foo_bar Donating Member (1000+ posts) Send PM | Profile | Ignore Wed Nov-29-06 10:52 AM
Response to Reply #42
56. did you understand my post, or TIA's?
Assuming a 1.5% MoE, the probability that the Democratic vote share would decline 5.1% from the Generic Poll (56.4%) to the actual vote (51.3%) is 1 in 76 billion. The probability is calculated using the Excel Normal Distribution function:

Prob = 1.310E-11 = NORMDIST(0.513,0.564,0.015/1.96,TRUE)
or 1 in 76,326,375,571

http://www.democraticunderground.com/discuss/duboard.php?az=view_all&address=364x2775205

The bottom line is that the Democrats won by a much bigger margin than indicated the reported vote count indicates. If the Wikipedia reorted vote is correct, then the Democrats received 57.7% + X%, where X =3-5%.

Using all of TIA's faulty assumptions (1.5% MoE, discount of all non-sampling error and "house effects", all 116 polls having a 95% confidence interval, the different wording in the different polls, and a generic "Republican or Democratic" poll not being a predictor of named R vs. D contests), we would plug =1-NORMDIST(0.607; 0.564; (0.015/1.96); 1) into our freeware spreadsheet program (this is TIA's most conservative case in the aforelinked para, 60.7%). The odds of TIA being right, with strict deference to TIA's assumptions about his generic poll-of-polls, are .00000000962205948251693000 or 103,927,854 to 1. If we use 4% to represent TIA's "3 to 5%" fudge factor (=1-NORMDIST(0.617; 0.564; (0.015/1.96); 1)), we get .00000000000217470486063576 or 459.8 billion to one odds of TIA's assumptions today matching his assumptions last week.

It gets worse: TIA was arguing for a <1.0% MoE, "MoE = 0.97% =1.96*standard error = 1.96*SQRT((1-p)*p)/N) So I'm being conservative when I use a 1.5% MoE". Thus the mistaken spreadsheet function would be (=1-NORMDIST(0.607; 0.564; (0.0097/1.96); 1)), which has so many zeroes that OpenOffice can't find a significant digit (not quite infinity to one, but smaller than a 64-bit float can measure, so in the neighborhood of <= 1/18,446,744,073,709,551,616 depending on the mantissa and sign bit in question). So TIA created a rock even TIA can't push; either all of his assumptions from the last 2 years are bogus, and/or today's opus du jour doesn't employ the TIA Method.

Unlike others here who make absolutist claims, TIA rarely does, notice all the "if's and then's".

Understatement of the year.

Either you trust Diebold, or you don't.

Either you're with us or you're with the terrorists. In fact, it's logically consistent for both TIA and Diebold to be full of it.


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