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What controls the width of a confidence interval, and how large must a sample be to achieve a required precision?

Relate the margin of error to confidence level and sample size, and determine the sample size needed for a required margin of error

WACE Year 12 Mathematics Methods Unit 4 margin of error and sample size: the margin of error formula, how confidence level and sample size affect width, and solving for the sample size needed for a required precision, with worked examples.

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
The margin of error
The square-root relationship
Why $\hat{p}=0.5$ is the worst case
Determining the sample size

What this dot point is asking

SCSA Unit 4 develops the margin of error as the half-width of a confidence interval for a proportion. This dot point asks you to relate it to the confidence level and sample size, and to find the sample size required for a given precision. It is examined in both sections, with sample-size calculations common in the calculator-assumed section.

The margin of error

A confidence interval for a proportion has the form $\hat{p}\pm E$ , where $E$ is the margin of error.

Two levers control $E$ : the confidence level through $z$ , and the sample size through $n$ . A higher confidence level needs a larger $z$ and so a wider interval. A larger sample shrinks the standard error and so narrows the interval.

Margin of error and the effect of confidence level

In a sample of $n=500$ with $\hat{p}=0.5$ , find the margin of error at $95\%$ and at $99\%$ .

Step 1: Calculate the standard error (shared by both confidence levels)

The standard error depends only on $\hat{p}$ and $n$ , not on the confidence level, so it is the same for both intervals.

\mathrm{SE} = \sqrt{\frac{0.5\times 0.5}{500}}=\sqrt{0.0005}\approx 0.02236

Step 2: Apply each $z$ -multiplier to get the margin of error

A higher confidence level requires a larger $z$ -score, which stretches the margin of error. At $95\%$ the multiplier is $1.96$ ; at $99\%$ it is $2.576$ .

At $95\%$ : $E=1.96\times 0.02236\approx 0.0438$ .

At $99\%$ : $E=2.576\times 0.02236\approx 0.0576$ .

Final answer: raising the confidence from $95\%$ to $99\%$ widens the margin from about $0.044$ to $0.058$ .

The square-root relationship

Because $E$ depends on $\dfrac{1}{\sqrt{n}}$ , precision improves slowly with sample size. To halve the margin of error you must roughly quadruple $n$ , since $\sqrt{4}=2$ . This is why large gains in precision become expensive.

The effect of quadrupling the sample

How the margin responds to a larger sample

A $95\%$ confidence interval with $\hat{p}=0.5$ and $n=400$ has margin of error $E=1.96\sqrt{\dfrac{0.25}{400}}=1.96\times 0.025=0.049$ . Suppose the sample is increased to $n=1600$ .

The new standard error is $\sqrt{\dfrac{0.25}{1600}}=0.0125$ , exactly half the old $0.025$ , because $\sqrt{1600}=2\sqrt{400}$ . So the new margin of error is

E=1.96\times 0.0125=0.0245,

half the original. Quadrupling the sample halved the margin of error, illustrating the square-root law directly. Markers reward recognising that

n\to 4n

gives

E\to\tfrac{1}{2}E

Why $\hat{p}=0.5$ is the worst case

The product $\hat{p}(1-\hat{p})$ inside the formula is a downward parabola in $\hat{p}$ , largest at $\hat{p}=0.5$ where it equals $0.25$ . Because a larger $\hat{p}(1-\hat{p})$ means a larger margin of error for the same $n$ , and a larger required $n$ for the same target margin, using $\hat{p}=0.5$ when no estimate is available guarantees the sample is big enough whatever the true proportion turns out to be. If a reliable prior estimate of $\hat{p}$ exists, using it gives a smaller, cheaper sample, but the conservative $0.5$ is never too small.

Determining the sample size

Rearranging the margin-of-error formula for $n$ lets you find the sample size needed for a target margin $E$ .

Finding the required sample size

What sample size gives a $95\%$ margin of error of at most $0.03$ , using $\hat{p}=0.5$ ?

Step 1: Choose the conservative value of $\hat{p}$ and identify the parameters

When no prior estimate of the proportion is available, using $\hat{p}=0.5$ gives the largest possible margin of error for any $n$ , so the resulting sample size is guaranteed to be large enough regardless of the true proportion. Here $z=1.96$ for $95\%$ confidence and $E=0.03$ .

Step 2: Substitute into the sample-size formula

Rearranging $E = z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$ for $n$ gives $n = \dfrac{z^{2}\hat{p}(1-\hat{p})}{E^{2}}$ .

n = \frac{1.96^{2}\times 0.5\times 0.5}{0.03^{2}} = \frac{3.8416\times 0.25}{0.0009} = \frac{0.9604}{0.0009} \approx 1067.1

Step 3: Round up to ensure the margin is met

Because $n$ must be a whole number and rounding down would give a margin slightly larger than $0.03$ , always round up.

Final answer: $n = 1068$ .

Exam-style practice questions

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

WACE 20216 marksCalculator-assumed. A market researcher wants a

95\%

confidence interval for a proportion with a margin of error no greater than

0.04

. (a) Using

\hat{p}=0.5

, find the minimum sample size. (b) If a pilot study suggests

\hat{p}\approx 0.2

, find the minimum sample size and comment.

Show worked answer →

A sample-size determination with two scenarios.

(a) $n=\dfrac{z^{2}\hat{p}(1-\hat{p})}{E^{2}}=\dfrac{1.96^{2}(0.5)(0.5)}{0.04^{2}}=\dfrac{3.8416\times 0.25}{0.0016}=600.25$ , round up to $601$ .

(b) With $\hat{p}=0.2$ : $n=\dfrac{3.8416\times 0.16}{0.0016}=384.16$ , round up to $385$ . Because $\hat{p}(1-\hat{p})$ is smaller away from $0.5$ , fewer respondents are needed; using $0.5$ gives the safe worst-case size.

Markers reward the rearranged formula, rounding up, and the comment that $0.5$ is the conservative choice.

WACE 20234 marksCalculator-free. A

95\%

confidence interval for a proportion has margin of error

0.05

from a sample of size

400

with

\hat{p}=0.5

. (a) Explain how the margin of error would change if the sample size were increased to

1600

. (b) State the new margin of error.

Show worked answer →

A square-root-law reasoning question.

(a) The margin of error is proportional to $\dfrac{1}{\sqrt{n}}$ . Increasing $n$ from $400$ to $1600$ multiplies $n$ by $4$ , so $\sqrt{n}$ doubles and the margin of error is halved.

(b) Half of $0.05$ is $0.025$ .

Markers reward the $\dfrac{1}{\sqrt{n}}$ relationship and the halved value $0.025$ .