Today is the anniversary of an archived DU thread from Nov. 26, 2004. It's very instructive and should be part of the permanent record.
http://www.democraticunderground.com/discuss/duboard.php?az=show_mesg&forum=203&topic_id=79760&mesg_id=79760
I present the thread now to show that virtually all the arguments vs. the exit polls (reluctant Bush voters, premature polling data, high MoE's, forgetful and untruthful Gore voters, etc.) were proposed immediately to explain the discrepancies. But smart, sane, analytical DUers were quick to refute.
Yet today, a full year later, naysayers are still citing the same tortured, debunked hypotheticals of the past. But unlike many of the the early doubters who have come to realize that the election was stolen, those who suffer from tunnel-vision or denial or plain ignorance, are relentless in their efforts to disparage the polls (pre and post election).
The fact is that over 130 pre-election AND exit polls confirm that Kerry won the election and the results agree to WITHIN A VERY SMALL MARGIN OF ERROR.
In addition, according to the Census 125.7 million voted, 3.4mm more than the 122.3mm recorded vote. Bit his is nothing new. It's also a FACT that in every election, Democratic presidential candidates find themselves in the hole by over a million votes (net) from a combination of spoiled ballots and voter disenfranchisement. Google Greg Palast for the details.
The naysayers claim that the poll all say that Kerry won is not circumstantial proof that the election was stolen. But yet they also accept the growing consensus that fraud did in fact occur and Kerry MAY indeed have won Ohio (which means he won the electoral vote). It's perverse logic and makes the argument that the exit polls were wrong, and the vote count was correct all the more ridiculous.
Ohio has historically lagged BEHIND the national Democratic vote. If Kerry won Ohio by 5%, as the state exit poll indicates, then HE MUST HAVE DONE EVEN BETTER OUTSIDE THE STATE... AND THEREFORE A SOLID WINNER OF THE NATIONAL POPULAR VOTE AS WELL.
The naysayers want to have it BOTH ways; they KNOW that eventually they will have to concede when the facts of the stolen election are exposed: Yes, they say, Kerry may have won the election but... the polls which showed he did are not proof.
Go figure. These are smart people. There must be a reason for this dichotomy of twisted logic. Follow the money.
If you believe Kerry won, you must believe the polls.
If you believe Bush won, you must disbelieve the polls.
Now, regarding the thread...
I came to realize later on that the probability calculation of 1 in 4.5 billion for the exit poll discrepancies exceeding the MoE in 16 states (all in favor of Bush) was incorrect.
.........The True Probability P is 1 in 19 TRILLION........
The parameters for the Excel Binomial distribution function should be:
P = 1 - BINOMDIST(15,50,0.025,TRUE)
P = 0.0000000000000524 and 1/P = 19 trillion.
Here are the probabilties for X states exceeding the MoE for Bush:
X Prob 1 in
2 35.64886895297020% 3
4 3.62043220383543% 28
6 0.15107732431885% 662
8 0.00320640032205% 31,188
10 0.00003920026633% 2,551,003
12 0.00000029870462% 334,778,890
14 0.00000000149595% 66,847,251,840
16 0.00000000000524% 19,083,049,268,519
17 0.00000000000039% 257,348,550,135,457
Regarding MoE:
The early argument that the Exit Poll Margin of Error used for the probability calculations were bogus due to an exit poll "cluster effect" has been thoroughly exposed as a non-starter. Assuming a given cluster effect, the probabilities are reduced in magnitude from one in trillions to one in billions to one in millions, depending on the assumed cluster.
The bottom line is that the naysayers now concede this major point:
The exit poll discrepancies could not have been due to chance.
In the thread which follows, the MoE calculation (at the 95% level of confidence) presented by c-macdonald is correct:
1) MOE = +/- 1.96* sqrt< (1-p)*p / n >
This is slightly more precise than the simple approximation I initially used:
2) MoE = 1/sqrt(N)
In fact, the formula MoE = .98/sqrt(N) is the same as (1) for p= 0.50
For sample size N=1000, (1) gives an MoE of 3.10% for p=.50 compared to an MoE of 3.16% for formula (2). Therefore, state exit polls are more likely to exceed the smaller MoE returned by (1). And this means the probabilities are even more remote.
c-macdonald's claim that only 9 of 16 states would exceed the MoE using his formula was incorrect - if anything, as stated above, more states would exceed the MoE, not less. As it turns out, however, the number exceeding the MoE didn't change. It's still 16 states.
For reference, this is an MoE table for various n-samples.
Formula (1) is shown for three vote splits p/(1-p):
.50/.50, .55/.45, .60/.40
1.96*sqrt(p*(1-p)/n)
p p p
n 1/sqrt(n) 0.50 0.55 0.60
1000 3.16% 3.10% 3.08% 3.04%
2000 2.24% 2.19% 2.18% 2.15%
3000 1.83% 1.79% 1.78% 1.75%
10000 1.00% 0.98% 0.98% 0.96%
13047 0.88% 0.86% 0.85% 0.84%
Once again, using the complex formula gives a lower MoE for a given n-sample. But it's a moot point; sixteen (16) states exceeded the MoE, regardless of the formula used.
I have consistently used the correct, longer formula, where applicable. It makes the case for fraud even stronger, because it lowers the MoE threshold, and the probability of any given discrepancy exceeding the MOE is more likely.
......................................................................
ONE OUT OF 4.5 BILLION!
TruthIsAll (1000+ posts) Fri Nov-26-04 01:16 AM
Those are the odds that Kerry's EXIT poll percentage would
EXCEED his ACTUAL reported vote percentage by MORE THAN THE
MARGIN OF ERROR in 16 out of 51 States by chance alone. That
is exactly what occurred on Nov. 2.
I know that is hard to fathom. But here is the data. And here
is the calculation, based on the number of individuals polled
in each state and the corresponding Margin of Error (MOE).
The chances of a given state falling outside the MOE = 1/20 =
.05. The calculation for the probability that 16 out of 51
states would fall outside the MOE is a simple one which uses
the BINOMIAL DISTRIBUTION FUNCTION:
The Probability (P) that at least 16 out of 51 states would
deviate beyond the exit poll MOE is:
P = 1-BINOMDIST(16,51,0.05,TRUE)
This returns P= 0.0000000218559% or 1 out of 4,575,415,347!
Here is how the odds decrease as the number of states (N)
exceeding the MOE increase:
N>MOE Odds
3: 1 out of 4
5: 1 out of 24
8: 1 out of 139
10: 1 out of 2707
12: 1 out of 1,037,879
14: 1 out of 57,503,169
15: 1 out of 490,978,624
16: 1 out of 4,575,415,347
The individual state probabilities are calculated using the
Normal distribution Function.
For example, consider Florida:
The probability that Kerry's 50.51% exit poll percentage
would decline to 47.47% in the actual vote (a 4.04%
deviation, far outside the 1.84% MOE) is equal to .06%.
This calculation is based on the FL exit poll sample size of
2846, which produces a MOE of 1.84%. The corresponding
standard deviation (StDev) is 0.94%. The StDev is plugged
into the normal distribution function, along with the exit
poll and reported vote percentages.
The probability that this deviation would occur due to chance
is: .06% = 100*NORMDIST(47.47%,50.51%,.94%,TRUE)
Here are the Exit Poll and Voting Results for all the states:
more.....
Click Here
Printer-friendly format
Email this thread to a friend
Bookmark this thread
(1000+ posts)
Send PM |
Profile |
Ignore
