Minimum sample size for a mean

Solving the margin for n

The margin of error is $m=z_{\alpha /2}\sigma /\sqrt{n}$. Planning a study turns this around: the analyst fixes the margin they can live with and the confidence they want, and asks how many observations that costs. Solve for $n$:

$$\sqrt{n}=\frac{z_{\alpha /2}\sigma }{m}⟹n\ge {\left(\frac{z_{\alpha /2}\sigma }{m}\right)}^2$$

Two rules ride with the formula. Always $z$, never $t$: a $t$ critical value needs df $=n-1$, and $n$ is the unknown — there is no row to read. Use a planning value for $\sigma$ from history, a pilot sample or a range/4 guess, and $z$ from the memorised trio. Always round up: $n$ is a count of invoices, and the inequality says at least — 109.27 invoices means 110, never 109, because 109 leaves the margin fractionally above target.

Worked example — planning Kitchener Ridge's audit sample

Kitchener Ridge wants to estimate mean days-to-pay to within $m=1.5$ days at 95 % confidence, with $\sigma =8$ from its history:

$$n\ge {\left(\frac{1.96\times 8}{1.5}\right)}^2=(10.453{)}^2=109.27⟹n=110$$

What if 90 % is enough? $z_{0.05}=1.645$:

$$n\ge {\left(\frac{1.645\times 8}{1.5}\right)}^2=76.97⟹n=77$$

Dropping from 95 % to 90 % confidence saves a third of the sample. What if the margin must be 0.75 days — half as wide — at 95 %?

$$n\ge {\left(\frac{1.96\times 8}{0.75}\right)}^2=437.09⟹n=438$$

Halving the margin quadruples the sample ($110\to 438$, allowing for rounding), because $m$ sits squared in the denominator. Precision is expensive at an accelerating rate — the square-root law of L8.2 seen from the buyer's side.

Reading the question

The paper phrases the target several ways: "within 1.5 days", "to within ±1.5 days", "a margin of error of 1.5", "an interval no wider than 3 days". The first three all give $m=1.5$; the last gives a width of 3, so $m=1.5$ too. Read for the half-width. And use the $\sigma$ the question supplies — if it gives a variance, take the square root before it goes into the formula; ${\sigma }^2=64$ is $\sigma =8$, and squaring 64 by mistake produces an $n$ in the thousands.