2021 Paper 1, Q19
A firm aims to have 95% confidence in estimating the proportion of office workers who respond to an email in less than an hour to within $\pm 0.05$.
A survey has never been undertaken before, so no past data is available. The firm believes that if the proportion is 0.5, then this will result in the largest variability in the sample proportion.
Based on this, determine the sample size needed using the approximate value of $z=2$ for the 95% confidence interval. Justify the choice of 0.5 for the proportion.
The worst case is right in the middle
The sample size you need depends on the proportion you are trying to measure. Sweep that proportion and the requirement peaks at 0.5, so planning for 0.5 is planning for the worst case.
Solve the margin formula for n, then defend the 0.5
$0.05=2\sqrt{\frac{0.5(1-0.5)}{n}}$ becomes $n=\frac{2^2\times 0.5(0.5)}{0.05^2}=400$. The unknown moved from the interval to the sample.
$n$ grows with $\hat p(1-\hat p)$, and that product is a downward parabola with its maximum at $\hat p=0.5$. Planning for the largest $n$ guarantees the margin for any true proportion.
$f(p)=p-p^2$, $f'(p)=1-2p=0$ at $p=\frac12$, and $f''=-2<0$, so it is a maximum. QCAA accepts either argument. But the nature of the point must be stated.
Halving the margin needs four times the sample
$n$ depends on the square of the margin, so the curve is brutal near the left-hand end. The question’s point, $\pm 0.05$ needing 400, sits on the gentle part.
Why does using 0.5 make the answer safe?
QCAA marking guide · 4 marks
$$\text{interval margin}=z\sqrt{\frac{\hat p(1-\hat p)}{n}}$$
Marker: correctly selects the interval margin formula. Alternative statements accepted, and this mark may be implied by subsequent working.
$$0.05=2\sqrt{\frac{0.5(1-0.5)}{n}}$$
Marker: substitutes values into the formula. Substitution before or after rearrangement is accepted; FT marks allowed.
$$n=\frac{2^2\times 0.5(0.5)}{(0.05)^2}=400$$
Survey 400 office workers
Marker: determines the sample size $n$. The approximation $z=2$ is what makes this a technology-free question. With $z=1.96$ it would be 384.16, i.e. 385 people.
The largest sample size results when $\hat p(1-\hat p)$ is maximised in the numerator, and that maximum occurs at $\hat p=0.5$. Planning for it guarantees the $\pm 0.05$ margin whatever the true proportion turns out to be.
Marker: verifies the firm’s decision to use $\hat p=0.5$ using mathematical reasoning. Equivalent methods accepted, e.g. solving $f'(x)=0$ including a statement about the nature of that point.
What a sample of 400 gives you
±0.05 costs 400 people. ±0.025 costs 1600. ±0.01 costs 10 000. Precision is bought in squares, which is why national polls sit around a 3% margin.
If the true proportion were 0.2, only 256 people would be needed. Planning for 0.5 costs 144 extra surveys and removes all risk of missing the margin.
A sample size must be a whole number, and rounding down would breach the margin the firm promised. Here 400 is exact, but with $z=1.96$ the 384.16 has to become 385.
Half of this question is a rearrangement, and the firm has even told you which proportion to use. The unfamiliar half is the last sentence, which asks you to justify the choice. I see a lot of students treat that sentence as optional, and they lose a quarter of the marks. It is really an optimisation problem in disguise. You need to show that $p(1-p)$ has a maximum at $p=\frac12$, either from the symmetry of the parabola or by differentiating, and then say why the largest variability is the one worth planning for. The QCAA were checking whether you could explain your reasoning, not just calculate.
Question wording and marking-guide steps are from the 2021 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.