Hi,

I'm afraid this problem will sound ridiculously elementary to mathematicians, but I'm dumb enough to be unable to get around it. I've searched the web for answers, but I get either too complicated stuff, or stuff I already know but which looks unrelated to my actual question (e.g. the normal distribution).

 

After an election, N (closed) voting papers are found when opening the total of the ballot boxes.

Then people start opening them, counting the votes and hence calculating the provisional results. So far, no problems.

However, if at a given moment we've opened S (randomly chosen) voting papers out of the total N, is there a formula, allowing us to calculate the 'confidence interval' of the results? I'm not sure whether that's the right name for it, what I mean is the analogue of the mean standard deviation for physical measures.

 

I guess the formula exists, because in fact in 'real' elections, before all the papers have been opened they DO say something like 'the candidate got 30% of the votes +/- 5%'.

But I can't figure out if this 5% is simply a function of N, S (and possibly an arbitrary parameter like the confidence level used in statistical tests), approaching 0% for S-->N, or if it also depends on the provisional results.

 

What puzzles me most is the fact that in physical measures your SD comes from errors in measuring a single, 'continuous' variable, whereas here the variables are necessarily more than one (the %s of the single candidates plus blank, void...), and they are discrete (rational with denominator N).

 

Can anyone help me with that? Thanks

It is described by the Bell Curve. There is a 95% chance that the results will be accurate to within 2 standard deviations. A standard deviation equals the square root of(npq). n is the nunmer of people being polled p is the probablity of candidate 1 being voted for, q is the probablty of candidate 2 being voted for. The standard deviation is similiar for all .2<p,q<.8 so it equals approximately sqrt(.5*.5*n).

Thanks for your reply. However, it doesn't justify either the fact that the uncertainty decreases as more and more voting papers are opened, or the fact that opened voting papers are never recounted. I.e. your sample s of the population n is such that s<<n.

 

I had some further reading about this problem, and in fact I found the answer in a statistics book.

 

msd^2 = p*(1-p)/s * (n-s)/(n-1)

 

Here, as s approaches n the uncertainty tends to 0, as it should be.

 

This raises another question, which I have posted in a new thread.