If I want to achieve a margin-of-error of <= 5 % for a representative population sample, how large a sample do I need when: The interviewees are picked from X regions and from Y age groups? That is, how many samples from each age group & region?

I know how many people live in each region and how they are distributed in the Y age groups in each region.

  • $\begingroup$ just to clarify, margin-of-error of what? Of calculating the average? $\endgroup$
    – mpiktas
    Mar 9, 2011 at 11:39
  • $\begingroup$ I need a sample to have, with 95 % probability, the same distribution as the population has (the distribution between regions and their age groups). Sorry if my terminology or explanations are subpar. $\endgroup$
    – Figaro
    Mar 9, 2011 at 11:50
  • $\begingroup$ So when the question is how many people needs to be interviewed in each region and age group, this is what I want? Underlying reasoning: To assure that the poll results will not be skewed due to poor sampling, I want to know the minimum required sample size for this 5 % margin. $\endgroup$
    – Figaro
    Mar 9, 2011 at 12:06

2 Answers 2


Your problem can be solved using the info in the Wikipedia article for 'Margin of error'. As it says, the margin of error is largest when the proportion is 0.5. The maximum margin of error at 95% confidence is $m = 0.98 / \sqrt{n}.$ Rearranging gives $n = (0.98/m)^2.$ So if you want a margin of error of 5% = 0.05, $n = (0.98/0.05)^2 = 384$. So you need around 400 subjects for each group in which you want to estimate the proportion of 'yes' responses.

  • $\begingroup$ Thanks, realized this yesterday evening myself. The problem boils down to a quite simple one in the end. $\endgroup$
    – Figaro
    Mar 10, 2011 at 7:21

If you weight your measurements (proportion of subpopulation/proportion of subpopulation in sample), your estimates will be unbiased. I assume this is what you meant by "poll results being skewed".

If I interpret your question correctly, your goal is the simultaneous estimation of multiple population proportions, where your proportions are

P_1 = proportion of population voting yes on poll question 1

P_2 = proportion of population voting yes on poll question 2

etc. (Let's work with one region at a time for now.) These can be represented in a proportion vector $P= (P_1, P_2, ...)$. We will denote a point estimate of $P$ by $\hat{P}$.

By what you want is probably not a point estimate, but a 95% confidence interval. This is an interval $(P_1 \pm t, P_2 \pm t, ...)$ where $t$ is your tolerance. (What 95% confidence means is a tricky issue which is hard to explain and easy to misunderstand, so I'll skip it for now.)

The thing is, it is always possible to construct a 95% confidence set no matter how small your sample size is. For your problem to be properly defined you need to specify the $t$, which is how accurate you require your estimates to be. The more accuracy you require the more samples you will need. In the problem as I have set it up it is possible to find the minimum number of samples given $t$, but there you can get better results if you can estimate the variability of your respective subpopulations ahead of time (which does not seem to be the case in your problem.)

Please give further clarification to your problem, though.

  • $\begingroup$ Thanks! So let's say that each person in the poll is asked a yes or no question. How do I go about and select a sample size that produces a result which is, with 95 % certainty, within a confidence interval of 5 %-units? $\endgroup$
    – Figaro
    Mar 9, 2011 at 12:35
  • $\begingroup$ I suppose that doing this for each age group in each region will provide me the sample-size I need (?). $\endgroup$
    – Figaro
    Mar 9, 2011 at 12:44

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