how do political polls calculate confidence intervals?
When statistical techniques are used to calculate confidence intervals for political polls?
Say that p is the true proportion, expressed as a fraction, of those who support political party A. This is what you would like to know but can’t know with certainty until there is a real election.
You ask a number of people whether they support A and suppose m say yes out of n definite replies. (Ignore don’t knows and people who don’t want to tell you.)
If you did the sampling process many times you would get different values of m for the same n value. This variation would be expected to relate to the Binomial distribution which can be approximated by the Normal distribution with large numbers such as >1000 in political sampling.
The mean of the Binomial is np and the standard deviation
sqrt[np(1 - p)]. Of course we don’t know p but m/n is a good estimate of it if the sample was large. This gives a mean of m and standard deviation of sqrt[m(1 - m/n)].
Confidence intervals are expressed as so many standard deviations usually 2. If n = 2400 and m = 960 say this gives you a standard deviation of 24. So 2s.d = 48 or about 2% either way.
Note that you need to 4 times the sample size to halve the confidence interval.
February 15th, 2010 at 2:27 am
Say that p is the true proportion, expressed as a fraction, of those who support political party A. This is what you would like to know but can’t know with certainty until there is a real election.
You ask a number of people whether they support A and suppose m say yes out of n definite replies. (Ignore don’t knows and people who don’t want to tell you.)
If you did the sampling process many times you would get different values of m for the same n value. This variation would be expected to relate to the Binomial distribution which can be approximated by the Normal distribution with large numbers such as >1000 in political sampling.
The mean of the Binomial is np and the standard deviation
sqrt[np(1 - p)]. Of course we don’t know p but m/n is a good estimate of it if the sample was large. This gives a mean of m and standard deviation of sqrt[m(1 - m/n)].
Confidence intervals are expressed as so many standard deviations usually 2. If n = 2400 and m = 960 say this gives you a standard deviation of 24. So 2s.d = 48 or about 2% either way.
Note that you need to 4 times the sample size to halve the confidence interval.
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