2
$\begingroup$

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?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
+50
$\begingroup$

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

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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
    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
    Oct 16, 2019 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.