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In local news [1] comes with information that 58% of people is for something. This information is based on option pool that ask 500 people from 10.5 million about that question. What is confidence interval for people that are for that something?

To solve this I try following. The $n$ people from $10^7$ is for something. I randomly select 500 of them and I will try to find $n_1,n_2$ for which:

$$ n_1: \frac{\binom{n_1}{290}\binom{10^7-n_1}{210}}{\binom{10^7}{500}}<0.05 $$

$$ n_2: \frac{\binom{n_2}{290}\binom{10^7-n_2}{210}}{\binom{10^7}{500}}>0.95 $$

But this is not easy to solve. So I am looking for some solvable algorithm.

[1] link

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  • $\begingroup$ Physicist's back-of-the-envelope calculation: $500\ll10.5M\approx\infty$, so it is a Poisson distribution, then count error is like $\sqrt{500}\approx22$ people, so 4.4 in per cent points. Thus CI is like 54%-62%. $\endgroup$
    – user88
    Commented Jul 12, 2013 at 11:31
  • $\begingroup$ @mbq I urge you to re-think that comment, because it appears to address a different problem. The value of 500 is not random but is selected by the researcher. It is not the realization of a Poisson variate and your answer has no meaning in this context. $\endgroup$
    – whuber ♦
    Commented Jul 12, 2013 at 14:27
  • $\begingroup$ This question is extensively discussed and answered here, in many guises. This search will reveal a few hundred relevant posts. Enjoy! $\endgroup$
    – whuber ♦
    Commented Jul 12, 2013 at 14:31
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    $\begingroup$ @whuber I agree, I posted that as a math trick how to use binomial->Poisson asymptotic to quickly approximate solution to such a problem. That's why its a comment and starts with a disclaimer. $\endgroup$
    – user88
    Commented Jul 12, 2013 at 17:50
  • $\begingroup$ @mbq I appreciate that: but your approach, as you have outlined it, is conceptually wrong. The question asks for a confidence limit for $p$ in a sample of $n=500$ where $\hat{p}=0.58$. It is an accident that $\sqrt{n}$ is even the right order of magnitude in this case. There is no Poisson distribution in sight, either: it's a Binomial distribution problem. For $\hat{p} \approx 1/2$ and a $95$% CI it *is* the case that two SEs is approximately equal to $\sqrt{n}$, but that is not because $n$ (a constant) has a Poisson distribution! $\endgroup$
    – whuber ♦
    Commented Jul 12, 2013 at 18:18

1 Answer 1

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Computing confidence intervals for binomial proportions is an interesting problem that has attracted quite a bit of research. None of the proposed confidence intervals take the population size into account. You can start here: http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval

Some of the proposed confidence intervals for your problem are (computed using Stata):

. cii 500 290, wald

                                                         -- Binomial Wald ---
    Variable |        Obs        Mean    Std. Err.       [95% Conf. Interval]
-------------+---------------------------------------------------------------
             |        500         .58    .0220726        .5367385    .6232615

. cii 500 290, exact

                                                         -- Binomial Exact --
    Variable |        Obs        Mean    Std. Err.       [95% Conf. Interval]
-------------+---------------------------------------------------------------
             |        500         .58    .0220726        .5353716    .6236769

. cii 500 290, wilson

                                                         ------ Wilson ------
    Variable |        Obs        Mean    Std. Err.       [95% Conf. Interval]
-------------+---------------------------------------------------------------
             |        500         .58    .0220726        .5362895    .6224906

. cii 500 290, agresti

                                                         -- Agresti-Coull ---
    Variable |        Obs        Mean    Std. Err.       [95% Conf. Interval]
-------------+---------------------------------------------------------------
             |        500         .58    .0220726        .5362852    .6224949

. cii 500 290, jeffreys

                                                         ----- Jeffreys -----
    Variable |        Obs        Mean    Std. Err.       [95% Conf. Interval]
-------------+---------------------------------------------------------------
             |        500         .58    .0220726        .5363799    .6226969
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  • $\begingroup$ It would be nice to see all prerequisites that must be fulfilled to make this approach to be correct for actual opinion poll analysis. $\endgroup$
    – boucekv
    Commented Jul 15, 2013 at 9:59
  • $\begingroup$ @boucekv : a discussion of all prerequisites would require a book. The short answer is: it depends on what you exactly want to do, as that defines what correct means. $\endgroup$
    – Maarten Buis
    Commented Aug 5, 2013 at 7:06

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