Margin of error and sample-size calculations | Year 12

01 / LEARN

Understand the idea.

Rearrange the margin inequality

Choose a positive target margin $E$ and critical value before solving for sample size. For a mean with known spread, $n \geq {(\frac{z^{*} σ}{E})}^{2}$ . For a proportion with planning value $p_{0}$ , use $n \geq \frac{{(z^{*})}^{2} p_{0} (1 - p_{0})}{E^{2}}$ . Enter proportion margins on the 0–1 scale.

When no defensible prior proportion is available, choose $p_{0} = 0.5$ : the identity $p (1 - p) = \frac{1}{4} - {(p - \frac{1}{2})}^{2}$ shows that $\frac{1}{4}$ is its largest possible value. Another planning value may reduce the required size, but does not establish the true population proportion.

\begin{array}{c} n \geq {(\frac{z^{*} σ}{E})}^{2} \\ or \\ n \geq \frac{{(z^{*})}^{2} p_{0} (1 - p_{0})}{E^{2}} \end{array}

Round in the direction that meets the target

A sample-size calculation gives a lower bound for completed valid observations. Always round upwards after retaining full calculation precision. Rounding to the nearest integer can produce a margin larger than the target, even when the numerical difference looks small.

If only a fraction of contacts are expected to respond, divide the required completions by that expected response rate and round up to budget contacts. This is an expectation-based plan, not a guarantee of either the response count or representativeness; inspect the achieved responses.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Mean precision

Known SD is 12. Plan a 95% mean interval with margin at most 2.

Worked solution

1
Set the mean-margin requirement.
$1.96 \frac{12}{\sqrt{n}} \leq 2$
The 95% critical value, known population spread and desired maximum margin determine the inequality.
2
Rearrange before rounding.
$\begin{aligned} \sqrt{n} & \geq \frac{1.96 (12)}{2} \\ n & \geq {(\frac{1.96 (12)}{2})}^{2} = 138.2976 \end{aligned}$
All quantities are positive, so isolating the square root and squaring preserve the inequality.
3
Choose the least allowed integer.
$n_{min} = 139$
138 is below the required lower bound, so at least 139 valid independent observations are needed under the interval model.

Answer

139 valid independent observations.

EXAMPLE 02

Proportion without prior estimate

Plan a 95% proportion interval with margin at most 0.05.

Worked solution

1
Select the conservative planning proportion.
$p_{0} = 0.5, p_{0} (1 - p_{0}) = 0.25$
Without a prior estimate, this choice maximises the variance factor and gives the conservative normal-margin requirement.
2
Insert confidence and target margin.
$n \geq \frac{{1.96}^{2} (0.25)}{{0.05}^{2}} = 384.16$
The desired margin is 0.05 on the proportion scale, or five percentage points.
3
Round upwards.
$n_{min} = 385$
The plan needs at least 385 valid observations; 384 falls below the mathematical lower bound.
4
State what the plan assumes.
The result plans the normal-approximation margin for a valid sampling design; it does not guarantee completed responses or remove selection and non-response bias.

Answer

385 valid observations.

03 / YOUR TURN

Try it yourself.

Check your understanding

Use p₀ = 0.2, 95% confidence and margin 0.04. Find n.

Show answer & reasoning

385.

Using the supplied planning value, $n \geq \frac{{1.96}^{2} (0.2) (0.8)}{{0.04}^{2}} = 384.16$ . Since this is a lower bound for an integer count, round up to $385$ valid observations.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Known SD is 15. Plan a 95% mean interval with margin at most 3.
2 marks
Without prior $p$ , plan a 95% proportion interval with margin at most 0.04.
2 marks
For planning $p_{0} = 0.3$ , 95% confidence and margin 0.05, find $n$ .
2 marks
A study needs 385 completions and expects a 70% response rate. Give a contact-budget estimate and limitation.
2 marks
Why is $p_{0} = 0.5$ conservative?
2 marks
Why does rounding a bound of 120.02 down to 120 fail?
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Known SD is 15. Plan a 95% mean interval with margin at most 3.
2 marks
${(\frac{1.96 \times 15}{3})}^{2} = 96.04$ ; round up.
$97$ .
Without prior $p$ , plan a 95% proportion interval with margin at most 0.04.
2 marks
$\frac{{1.96}^{2} (0.25)}{{0.04}^{2}} = 600.25$ ; round up.
$601$ .
For planning $p_{0} = 0.3$ , 95% confidence and margin 0.05, find $n$ .
2 marks
$\frac{{1.96}^{2} (0.3) (0.7)}{{0.05}^{2}} = 322.6944$ ; round up.
$323$ .
A study needs 385 completions and expects a 70% response rate. Give a contact-budget estimate and limitation.
2 marks
$\frac{385}{0.70} = 550$ .
Monitor actual response and non-response bias.
550 contacts; no guarantee of count or representativeness.
Why is $p_{0} = 0.5$ conservative?
2 marks
The squared term is non-negative, so the maximum occurs at 0.5.
$p (1 - p) = 0.25 - {(p - 0.5)}^{2} \leq 0.25$ .
Why does rounding a bound of 120.02 down to 120 fail?
2 marks
The target margin inequality is not met below that bound.
120 is below the required lower bound; use 121.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: sampling distributions and confidence intervals for means and proportions. Mean tasks distinguish known population spread from sufficiently large-sample estimated-spread normal intervals. Repeated-interval computational simulation, approximation conditions, design and interpretation are included.

Official curriculum source ↗

In this lesson

Understand the idea.

Rearrange the margin inequality

Round in the direction that meets the target

See the method in action.

Mean precision

Set the mean-margin requirement.

Rearrange before rounding.

Choose the least allowed integer.

Proportion without prior estimate

Select the conservative planning proportion.

Insert confidence and target margin.

Round upwards.

State what the plan assumes.

Try it yourself.

Check your understanding

Keep practising