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I have a survey with a binary answer (yes/no) on a population of about 10,000 people. I'd like to calculate the required sample size for a x% margin of error (e.g., 5) at 95% confidence level. Usually, most people in this population answer yes (80%+).

I found this calculator but I don't understand how we can calculate the CI and SE without inputting the population size: http://sample-size.net/confidence-interval-proportion/

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  • $\begingroup$ The answer is different if (a) the 'success' probability is around $0.5$ and you want a CI inside $(0.45, 0.55)$ than if (b) the 'success' probability is around $0.1$ and and you want a CI inside $(0.05,0.15).$ Not sure what you mean by 'x% margin of error`. Can you provide some context and particular examples? $\endgroup$
    – BruceET
    Feb 24 '21 at 15:32
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margin of error will be approximately $1.96\sqrt{\frac{0.8 \times 0.2}{n}}$. If you want the margin of error to be $m$, then take $n=1.96^2\frac{0.8 \times 0.2}{m^2}$.

Example: 5% margin of error, $n=246$.

The Wikipedia article also has a section for finite population size correction, but your population is pretty large and you would not need to worry about that unless you want a very tiny margin of error.

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    $\begingroup$ I think you need to change the "$m = 1537$ to ${\pmb n} = 1537$. $\endgroup$ Feb 24 '21 at 16:14
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    $\begingroup$ @gung-ReinstateMonica Thanks. I corrected the m to n and I also changed to match the example given in the original problem, m=5%. With that example, it is more clear that finite population correction is not needed. $\endgroup$
    – John L
    Feb 24 '21 at 17:18

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