I'm trying to determine n, the sample size.

We have $N=580$ and we want to estimate population proportion, $p$, at a 95%-CI and a margin of error of $0.10$, then I get

$$n=Npq / (N-1)D + pq$$

where $D= B^2 / 4$ to be about $n=86$.

But what if instead we keep that same margin of error at $0.10$ and now only want a 90%-CI, if we use the same formula don't we get the same $n$?

Am I using the formula wrong here?

  • 1
    $\begingroup$ Proportion of what? Is this binomial data? What's B? Why do you want a smaller N than 580? Are you sampling from the original 580 or is it a new random sample? What are you trying to find out from the sample? $\endgroup$ – John Oct 3 '13 at 6:51

The formulas concerning the calculation of the sample size to estimate a proportion $p$ in for finite populations are provided on this website. It also contains the derivations.

In short, the sample size necessary for estimating a population proportion $p$ of a finite population with $(1-\alpha)100\%$ confidence and a margin of error no larger than $\epsilon$ is:

$$ n = \frac{m}{1+\frac{m-1}{N}} $$ where $$ m=\frac{z_{1-\alpha/2}^{2}\hat{p}(1-\hat{p})}{\epsilon^{2}}. $$ $N$ denotes the population size, $z_{1-\alpha/2}$ the $(1-\alpha/2)$-quantile of the standard normal distribution and $\hat{p}$ the estimated proportion.

For $N=580, \alpha=0.05, \epsilon=0.1, \hat{p}=0.5$ (i.e. 95% confidence), we get $n\approx 83$. If we take $\alpha = 0.1$ which corresponds to 90% confidence, we get $n\approx 61$.

The more precise we want our estimate of the popultion proportion to be, the higher our sample size needs to be.

This means that the lower $\alpha$, the higher the necessary sample size will be. The following graph illustrates this (for $N=580, \epsilon=0.1, \hat{p}=0.5$): Sample size alpha

The necessary sample size will also increase with decreasing margin of error $\epsilon$ (note the reversed $x$-axis; graph for $N=580, \alpha=0.05, \hat{p}=0.5$):

Sample size margin of error

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  • $\begingroup$ Why, mathematically, do I need to know the estimated population proportion ex-ante to determine sample size? I have been looking for the reasoning, but also the site you linked does not explan that and just assumes a p-hat of 0.5. Do we always assume p-hat of 0.5 when we have not taken a pre-sample? Does our estimated required sample size change during the sampling process the closer we get to the population p? So if we have no clue about p-hat and after some sample n we have not yet found a success, we just stop and conclude p is between 0 and the error we initially set? $\endgroup$ – iraserd Mar 2 '17 at 10:14
  • $\begingroup$ And, should I post the comment above in a separate question? $\endgroup$ – iraserd Mar 2 '17 at 10:15
  • $\begingroup$ Also, is this true for small p? And what is a small p? 10%? 3%? 0.3%? $\endgroup$ – iraserd Mar 2 '17 at 10:19
  • $\begingroup$ @iraserd $\hat{p}=0.5$ is usually taken because it's the worst-case scenario. The standard error is largest when $p=0.5$. So by taking $\hat{p}=0.5$ we're on the safe side. If you have an a-priori estimate of the population proportion, you could take that instead. You certainly could design a sequential approach, where you update your estimate on the fly. I assume that Bayesian methods would be suited for that. This approach is also valid for small $p$, as far as I'm aware. What a small $p$ is is subjective. $\endgroup$ – COOLSerdash Mar 2 '17 at 11:03

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