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How large a sample do we need to achieve a required level of precision?

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

What the margin of error is, how it depends on confidence level and sample size, and how to rearrange the formula to find the sample size needed for a required precision.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

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

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 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

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 practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20223 marksCalculator-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 20222 marksCalculator-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 20213 marksCalculator-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 $239$ .