Margin of error and sample size (SACE Stage 2 Mathematical Methods)

Syllabus dot point

“The margin of error sets the precision of an estimate; rearranging it determines the sample size needed for a target margin.”

8 min readUpdated 2 Jun 2026

Quick answer

The margin of error is $E=z^*\dfrac{\sigma}{\sqrt{n}}$ , the half-width of a confidence interval. To achieve a required margin $E$ , rearrange to $n=\left(\dfrac{z^*\sigma}{E}\right)^2$ and round up to the next whole number. Smaller $E$ or higher confidence both demand a larger sample.

Jump to a section

What this dot point is asking
What the margin of error depends on
Finding the required sample size
Margin of error for a proportion
Reading the precision requirement
Common errors
Why it matters
Exam-style questions

What this dot point is asking

The margin of error $E$ is how far the confidence interval extends each side of the sample mean - it measures the precision of the estimate. A small margin means a precise estimate; a large margin means a vague one.

What the margin of error depends on

Margin of error

E=z^*\frac{\sigma}{\sqrt{n}}.

Three levers control it:

Confidence level ( $z^*$ ): higher confidence ⇒ larger $z^*$ ⇒ larger $E$ .
Population spread ( $\sigma$ ): more variable data ⇒ larger $E$ .
Sample size ( $n$ ): bigger sample ⇒ smaller $E$ (via $\sqrt{n}$ ).

Only $n$ is usually under your control, so questions ask: how big must $n$ be to get a margin no larger than some target?

Finding the required sample size

Rearrange $E=z^*\dfrac{\sigma}{\sqrt{n}}$ to make $n$ the subject:

Sample size for a target margin

n=\left(\frac{z^*\sigma}{E}\right)^2.

Always round the result up to the next whole number, so the margin is at most

E

Always round up

A sample size must be a whole number, and rounding down would make the margin slightly larger than required. So even $n=64.1$ rounds up to $65$ - round up regardless of the decimal.

Sample size for a required precision

A researcher wants to estimate a population mean to within $E=5$ units with $95\%$ confidence, where $\sigma=30$ . This example finds the minimum sample size needed.

Step 1: identify the values and the sample-size formula

The sample-size formula comes from rearranging the margin-of-error equation. The inputs are: the required margin $E=5$ , the critical value $z^*=1.96$ for $95\%$ confidence, and the population standard deviation $\sigma=30$ :

n=\left(\frac{z^*\sigma}{E}\right)^2.

Step 2: substitute the values and calculate

Placing the numbers into the formula and evaluating step by step:

n=\left(\frac{1.96\times 30}{5}\right)^2=\left(\frac{58.8}{5}\right)^2=(11.76)^2=138.3.

Step 3: round up to the nearest whole number

A sample size must be a whole number, and rounding down would leave the margin slightly larger than $5$ units. Always round up to guarantee the required precision:

n=139.

Final answer: a sample of at least $139$ observations guarantees a margin of error within $5$ units.

Higher precision needs many more observations

Using the same population ( $\sigma=30$ , $95\%$ confidence), this example finds how many observations are needed when the margin is halved to $E=2.5$ .

Step 1: note the new target margin

The required precision is now $E=2.5$ , half the previous value of $E=5$ . All other inputs remain the same: $z^*=1.96$ and $\sigma=30$ .

Step 2: compute the required sample size

Substituting $E=2.5$ into the sample-size formula:

n=\left(\frac{1.96\times 30}{2.5}\right)^2=\left(\frac{58.8}{2.5}\right)^2=(23.52)^2=553.2.

Rounding up gives

n=554

Step 3: compare with the previous result and explain the quadrupling

Halving the margin from $5$ to $2.5$ increased the sample size from $139$ to $554$ , roughly four times as many observations. This happens because $n$ depends on $1/E^2$ : cutting $E$ by a factor of $2$ multiplies $1/E^2$ by $4$ , so the required sample size quadruples. Precision is expensive.

Final answer: $n=554$ observations are needed for a margin of error of $2.5$ units.

Margin of error for a proportion

The same logic applies when estimating a population proportion rather than a mean. The margin of error becomes $E=z^*\sqrt{\dfrac{p^*(1-p^*)}{n}}$ , where $p^*$ is the sample proportion. Rearranging for $n$ gives $n=\left(\dfrac{z^*}{E}\right)^2 p^*(1-p^*)$ . The structure is identical: the required sample size depends on the square of $\dfrac{z^*}{E}$ , so halving the margin again quadruples the sample. When no estimate of $p^*$ is available, using $p^*=0.5$ maximises $p^*(1-p^*)$ and therefore gives the safest (largest) sample size, a worst-case planning value examiners often expect.

Reading the precision requirement

Exam questions express the precision target in different words: "to within $5$ units" means $E=5$ ; "an interval of width $10$ " means $2E=10$ , so $E=5$ ; and "accurate to one decimal place" usually means $E=0.05$ . Translating the wording into a value of $E$ before substituting is the step most often fumbled. Always write down $E$ explicitly, halving the width when the question gives a full interval width rather than a one-sided margin.

Common errors

Rounding down, or forgetting to square

Two slips dominate. First, rounding the sample size to the nearest whole number or down - you must always round up, or the margin exceeds the target. Second, forgetting the square: $n=\left(\tfrac{z^*\sigma}{E}\right)^2$ , not $\tfrac{z^*\sigma}{E}$ . The square is why precision is expensive.

Sanity-check the direction

Before finalising, check the relationship makes sense: a smaller target margin or higher confidence should give a larger $n$ . If you tightened the requirement but got a smaller sample size, you have likely mis-rearranged the formula or dropped the square.

Why it matters

Sample-size planning is the practical pay-off of Topic 6 - it answers the real-world question "how much data do I need?" and is a reliable exam item. It also reinforces the inverse-square relationship between precision and sample size, a key insight for designing the data collection in your Mathematical Investigation.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style3 marks

Calculator-free. Given an estimate

p^*

, show that the sample size

n

required for a

95\%

confidence interval of width

w

for a population proportion is

n = \left(\dfrac{2 \times 1.96}{w}\right)^2 p^*(1 - p^*)

Show worked answer →

A $95\%$ interval for a proportion is $p^* \pm E$ where $E = 1.96\sqrt{\dfrac{p^*(1 - p^*)}{n}}$ .

The full width is twice the margin: $w = 2E = 2 \times 1.96\sqrt{\dfrac{p^*(1 - p^*)}{n}}$ .

Divide by $2 \times 1.96$ : $\dfrac{w}{2 \times 1.96} = \sqrt{\dfrac{p^*(1 - p^*)}{n}}$ .

Square: $\left(\dfrac{w}{2 \times 1.96}\right)^2 = \dfrac{p^*(1 - p^*)}{n}$ .

Rearrange: $n = \left(\dfrac{2 \times 1.96}{w}\right)^2 p^*(1 - p^*)$ , as required.

Marks: one for width $= 2E$ , one for substituting $E$ , one for the algebra isolating $n$ .

SACE 2022-style2 marks

Calculator-assumed. A farmer planted 100 seeds and observed 53 viable, giving

p^* = 0.53

. Using

n = \left(\dfrac{2 \times 1.96}{w}\right)^2 p^*(1 - p^*)

, calculate the smallest number of seeds needed for a

95\%

confidence interval of width

0.1

Show worked answer →

Substitute $p^* = 0.53$ and $w = 0.1$ .

Bracket: $\dfrac{2 \times 1.96}{0.1} = 39.2$ , and $39.2^2 = 1536.64$ .

Then $p^*(1 - p^*) = 0.53 \times 0.47 = 0.2491$ .

n = 1536.64 \times 0.2491 \approx 382.8.

Round up to guarantee the interval is no wider than required: $n = 383$ seeds.

Marks: one for $n \approx 382.8$ , one for rounding up to $383$ . Rounding down would give an interval wider than $0.1$ .

SACE 2021-style3 marks

Calculator-assumed. A population has standard deviation

\sigma = 24

. Find the smallest sample size needed to estimate the mean to within a margin of error

E = 4

with

99\%

confidence.

Show worked answer →

Use $n = \left(\dfrac{z^*\sigma}{E}\right)^2$ with $z^* = 2.576$ , $\sigma = 24$ , $E = 4$ :

n = \left(\frac{2.576 \times 24}{4}\right)^2 = \left(\frac{61.824}{4}\right)^2 = (15.456)^2 \approx 238.9.

Round up: $n = 239$ .

Marks: one for the formula with $z^* = 2.576$ , one for the value $238.9$ , one for rounding up to $239$ .

Practise this

Sources & how we know this

Stage 2 Mathematical Methods subject outline (2025 onwards) — SACE Board
Stage 2 Mathematical Methods: external assessment — SACE Board