A SUMMARY EXIT POLL MATHEMATICAL FORMULATION

[TruthIsAll]

I welcome a mathematically-based critique.

Given: N pre-selected precincts P(i), i= 1,N
Each precinct is selected based on the HISTORICAL TREND OF UNBIASED, PRISTINE EXIT POLL SAMPLES and is weighted accordingly:

R(i)+ D(i) = 1
where
R(i) = expected Republican vote percentage
D(i) = expected Democratic vote percentage

Let X(i) = number of respondents in precinct P(i)
Then ND(i) = number of expected Democratic votes in precinct P(i):
ND(i)= D(i)* X(i)

S = TOTAL number of NATIONAL respondents:
S = SUM(X(i)), i= 1, N

Then the total number of EXPECTED Democratic votes TD in the National sample is given by:
TD = SUM (ND(i)), where i = 1, N

and the Expected Democratic National vote percentage is:
DP = TD/S

The Margin of Error for precinct P(i) is given by:
MoE(i)= 1.96* SQRT((D(i)*(1-D(i))/(X(i)))

The Margin of Error (MoE) for the National Democratic total is given by:
MoE = 1.96 * SQRT( (DP*(1-DP))/S)

Thus, the Democratic National vote percentage can be expected to fall within the range {DP-MoE, DP +MoE} 95% of the time.

This means that there is a 95% PROBABILITY that the ACTUAL NATIONAL DEMOCRATIC VOTE POPULATION MEAN falls within the range {DP-MoE, DP+ MoE}.

[TruthIsAll]

1. Applying the formulas: The Preliminary National Exit Poll (13047 sample)

The Democratic National vote percentage can be expected to fall within the range {DP-MoE, DP +MoE} 95% of the time.

This means that there is a 97.5% PROBABILITY that the ACTUAL NATIONAL DEMOCRATIC VOTE EXCEEDED THE LOWER LIMIT = DP - MoE.

For example:
The National Exit poll MOE is 1.0%.
DP= 51%
MOE =1%
DP-MOE = 50%

Kerry won the poll 51%-48%.
There is a 97.5% probability that his popular vote was over 50%.

The probability that Bush would exceed his 48% exit poll by 1.0% for 49% of the popular vote is only 2.50%.

The probability that Bush would exceed 50% is virtually ZERO.

[mod mom]

2. You're a hero and a patriot TIA! nt

[kiwi_expat]

3. Why no N(i) for Nader?

R(i)+ D(i)+ N(i) = 1
where
R(i) = expected Republican vote percentage
D(i) = expected Democratic vote percentage
N(i) = expected Nader vote percentage

[TruthIsAll]

5. You miss the point of the post. It's a simulation to show...

that large precinct MOE is of no consequence in the big national picture.

KISS

[kiwi_expat]

9. I assume that I should disregard this response too (see 7)....

since your response referred to a "simulation". Although, I am sure that you would still reject including Nader - in accordance with your KISS philosophy.

An equation should include all of the relevant variables, if only in a highly aggregated form. Equal means equal. An equation is a pure concept. (Thanks to eomer for hammering this home to me.)

You could show R(i)+ D(i)+ O(i) = 1
where
R(i) = expected Republican vote percentage
D(i) = expected Democratic vote percentage
O(i) = expected Other (third party) vote percentage

or you could show R(i)+ D(i)= 1
where
R(i) = expected Repub vote percentage of the Repub+Dem total
D(i) = expected Dem vote percentage of the Repub+Dem total

[TruthIsAll]

10. I have always used the two-party vote for pre and post-election

analysis, except in the case of the NEP, which is broken down
into Kerry, Bush and Nader votes. 

Yes, I agree. To be mathematically pure, R+D+O=1. But it is
much cleaner to work with just two unknowns for calculation of
the MOE's and binomial  probabilities.

Kerry won 65% of former Nader 2000 voters and Bush 13%,
according to the NEP. But the third-party 1% vote (only 1
million+ votes) is  small potatoes when you consider that
there were  21 million new voters which Kerry won by a 3-2
margin.

I used a two-party analysis in an election model to forecast
the final vote based on state polls (using Monte Carlo
Electoral Vote simulation) and an average of 18 national polls
(for the popular vote) with an undecided vote allocation
factored in.

My final projection was Kerry 51.8%, Bush 48.2%. Since the
3rd-party vote was 1%, subtracting it equally from Bush and
Kerry the final result is: Kerry 51.3%, Bush 47.7%, Other 1%.

[kiwi_expat]

4. Where in your series of equations are the precincts weighted?

By design, the data from different precincts are given different weights in the final calculation. Some arch typical precincts could be weighted quite heavily.

Lets say precinct 3 has a weight of 10.
Where does that happen in your series of equations?

For example, if
the number of respondents is X(3) = 60
the expected Dem % is D(3) = 0.50
the number of expected Dem. respondents is ND(3) = 0.50 x 60 = 30

S and TD seem to be simple sums of X and ND, respectively. With no weighting that I can see.

Where does the precinct 3 data get multiplied by 10?

Are you saying that the weighting occurs at the very beginning? That the number of respondents would be shown as X(3) = 600?

[TruthIsAll]

6. Once again, KISS. You miss the point of the simulation.

Yes, I assume equal number of respondents per precinct.
What the hell is wrong with that for the purpose of this simulation?

To apply fictitious weights would have been overkill and just cloud the issue.

And the crux of the issue is...
THE LAW OF LARGE NUMBERS
It overwhelms and diminishes the effect of the high individual precinct MOEs

Keep your eye on the ball.
Have you ever designed a computer simulation?

You make the typical error of those who lose focus of the original problem by attempting an overly complex model. In any model design "less is more" - as far as the number of variables required to provide a solution are concerned.

Variable precinct size will have no effect other than to unnecessarily complicate the issue.

[TruthIsAll]

7. Disregard my previous response: I was referring to the simulation

Sorry for the confusion on my part.

The answer to your question is: Yes.
I assume that the precincts are pre-weighted by the number of respondents.

Precinct P(i) has been designed to include X(i) respondents, and
X(i)/S is the weighting for the i'th precinct based on the demographic makeup.

[kiwi_expat]

8. But is that actually the case?

According to the Univ. of Mich. "raw" exit poll data (see "Ohio Exit Poll Raw Data" thread), the maximum respondent size is about 60 respondents.

I have assumed that the respondent size was determined primarily by the size of the precinct - and perhaps also by the number of refusals.

The respondents are then shown as weighted AFTER the original number of respondents is shown. According to minvis, the precincts with low numbers of respondents (20-30) seem to be the ones that are most weighted. I assume that this is the weighting for missing respondents and refusals, that Mitofsky refers to in the NEP methodology.

Mitofsky talks about two different types of weighting. I suspect that the relative weight for each precinct in the final aggregate is applied subsequently.

I haven't seen the data, myself. It is in a very difficult format to decipher.