Reply #3: David, tell your father he is wrong. [View All]

[TruthIsAll]

Use 1.96 standard deviations for a 95% confidence level.

The margin of error formula for a sample size = n is
MoE = 1.96* Sqrt(p(1-p)/n)/2

where
p = winning vote percentage
1-p = losing vote percentage

As the spread widens, the MoE gets tighter.

For instance, assume p=.6, 1-p=.4:
.6 *.4 = .24
Sqrt (.24) = .4898

Compare to p=.5, 1-p=.5:
Sqrt (.5 *.5) = .5000

So the MoE is slightly lower for a 60-40% poll result then 50-50%.

Let's assume p= .5, 1-p=.5 (an even election)
n= 13047 = Pristine (Raw?) Nat. Exit poll respondents.

Sqrt (13047) = 114.2

MoE = 1.96 *.5 / Sqrt(13047) =.98 / 114.2 = 0.875%

The confidence level is 95%. That means that 19 times out of 20, the exit poll result will fall with +/- 0.875% of the true population mean.

So if Kerry has a 51-48% exit poll margin, the chances are greater than 95% that the exit poll would match the vote to within 0.875%.
That means there is a 95% probability that Kerry's true mean percentage was in the range {50.125-51.875}

The probability that Kerry would get 48% of the vote after winning the exit poll by 51-48% is:

p = 1 - NORMDIST(0.51,0.48,0.00875/1.96,TRUE)
p = 0.0000000000091

or 1 in 109,444,820,165

You:
One question I have about the national poll is whether the sample sizes in each state were adjusted in proportion to their populations or was some other algorithm used?

Me:
Don't know.

You:
And also, I'm not sure if the state polls, which had a much larger sample size than the 13,660, have been combined to arrive at a national popular vote total to compare to the 13,660.


Me:
The National Poll is a subsample (13000) of the state polls (70000+)

FL sample size was 2862 (MoE = 1.8%)
OH was approx 2200 (2.1% MoE)