Minimum n for a proportion

n ≥ z² p*(1 − p*)/m², rounded up; with no prior estimate use p* = 0.5, the worst case, because p(1 − p) peaks at 0.25.

Solve the margin for n

The margin of error for a proportion is $m = z_{α /2} p (1 - p) / n$ . Fix the margin you can live with, and solve for $n$ :

$n \geq \frac{z _{α /2}^{2} p ^{*} ( 1 - p ^{*} )}{m ^{2}}$

Round up to the next whole unit — a sample of 1,067.1 invoices is 1,068 invoices, because 1,067 would leave the margin a hair too wide. The awkward part is $p^{*}$ : the formula needs the proportion you are about to estimate. Two conventions:

A prior estimate is given (last year’s audit, a pilot sample): use it as $p^{*}$ .
No prior estimate: use $p^{*} = 0.5$ . The product $p (1 - p)$ is $0.25$ at $p = 0.5$ and smaller everywhere else — $0.12 \times 0.88 = 0.1056$ , $0.9 \times 0.1 = 0.09$ — so 0.5 gives the largest possible $n$ , guaranteeing the margin whatever $p$ turns out to be.

Worked example — how many invoices to audit?

Maple Ledger wants next year’s audit to estimate the invoice error rate to within 3 percentage points at 95 % confidence. How many invoices must be sampled?

With no prior estimate. $z = 1.96$ , $m = 0.03$ , $p^{*} = 0.5$ :

$n \geq \frac{1.9 6 ^{2} \times 0.5 \times 0.5}{0.0 3 ^{2}} = \frac{3.8416 \times 0.25}{0.0009} = 1067.1; \Rightarrow; n = 1, 068$

Using this year’s $\overset{p}{^} = 0.12$ as the prior.

$n \geq \frac{3.8416 \times 0.12 \times 0.88}{0.0009} = \frac{3.8416 \times 0.1056}{0.0009} = 450.8; \Rightarrow; n = 451$

The prior estimate cuts the sample by more than half. That is the value of last year’s audit: it is not just a number, it is a cheaper next audit. The risk is that the rate has drifted towards 0.5 since — if it has, 451 invoices give a margin wider than 3 points. When the prior is old or doubtful, a question will say so and expect $p^{*} = 0.5$ .

The margin is in proportion units

"Within 3 percentage points" is $m = 0.03$ , not $m = 3$ and not 3 % of $\overset{p}{^}$ . A margin of "3 % of the estimated rate" would be $0.03 \times 0.12 = 0.0036$ and need a sample seventy times larger; the paper does not ask for that, but it does expect you to write $m$ as a decimal.

Same shape as the mean

Module 9’s rule for a mean was $n \geq (z σ / m)^{2}$ . This is the same rule with $σ^{2} = p (1 - p)$ — the Bernoulli variance — and the same instructions: $z$ from the confidence level, $m$ from the question, round up. For a mean the difficulty was a $σ$ you had to guess; for a proportion the guess is bounded, which is why $p^{*} = 0.5$ is always available.