In applied literature about the required sample size such as "Sampling for effective internal auditing" we find formulas for the required sample size to detect a proportion of a feature or an error rate as (I cite and use the same variable names) $$ n = \frac{C^2 \rho q}{P^2}, $$ where $n$ is the required sample size, $C$ is the confidence level, $P$ is some measure of precision, $\rho = 1-q$ is the expected error rate.

Where do these formulas from auditing come from? I understand that their origin must be some Binomial distribution consideration with normal approximation.

Is there a publication where these sample size requirements are derived in a rigorous mathematical way? I find quite a few publications citing formulae as the above but what is a good source for at the same time a rigorous mathematical treatment and applications in audit?


1 Answer 1


I am not familiar with the literature for interal auditing, but this looks quite similar to Normal approximation/Wald confidence interval (see wikipedia) and was published in
Laplace, Pierre Simon (1812). Théorie analytique des probabilités (in French). p. 283. For a newer article about some of its problems: Brown, Lawrence D.; Cai, T. Tony; DasGupta, Anirban. Interval Estimation for a Binomial Proportion. Statist. Sci. 16 (2001), no. 2, 101--133. doi:10.1214/ss/1009213286. https://projecteuclid.org/euclid.ss/1009213286

For a specific citation for Audits somebody else might be able to help.
The formula of the Normal approximation interval is as follows : $$p=\hat{p}\pm z*\sqrt{\dfrac{p(1-p)}n} $$

if we subtract $\hat{p}$ and square afterwards we get: $$(p-\hat{p})²=\dfrac{z^2 *p(1-p)}n $$

Now we solve for n: $$n= \dfrac{z^2 *p(1-p)}{(p-\hat{p})²} $$

This is exactly your formula. $z^2$ is $C^2$ and $(p-\hat{p})^2$ is $P^2$ in your equation

  • $\begingroup$ Thanks for this reference! Thus if I expect and error rate $p$ (e.g. $5\%$) and I want to detect differences on a sample with confidence related to the normal quantile $z$ (e.g. $95\%$ and $z = z_{0.95}$) and I tolerate some error $P = \hat{p}-p$ (e.g. $6\%-5\%$) then I need a sample size of $n$ as calculated above ... does this make sense? $\endgroup$
    – Richi W
    Commented Oct 16, 2019 at 19:17
  • 1
    $\begingroup$ @Ric yes that makes sense! However, i confused the name of the interval. It is the normal approximation interval. I changed it above and also edited the correct citations. You should note, that the approximation interval tends to be less reliable for small n or extreme high or low p. (see the see the Brown et al. article above). Therefore, your estimated n might also be unreliable in these cases. $\endgroup$
    – Mr Pi
    Commented Oct 16, 2019 at 19:37

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