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 4.77 min answerUpdated 2026-05-29

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
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:

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.

2017 SACE Stage 23 marksGiven an estimate p*, show that the sample size n required to obtain a 95% confidence interval of width w for a population proportion is n = (2 x 1.96 / w)^2 x p*(1 - p*).

Show worked answer →

A 95% confidence interval for a proportion is p* plus or minus the margin of error E, where E = 1.96 times sqrt( p*(1 - p*) / n ).

The full width of the interval is twice the margin of error, so w = 2E = 2 times 1.96 times sqrt( p*(1 - p*) / n ).

Now solve for n. Divide both sides by 2 times 1.96:
w / (2 times 1.96) = sqrt( p*(1 - p*) / n ).

Square both sides:
( w / (2 times 1.96) )^2 = p*(1 - p*) / n.

Rearrange for n:
n = p*(1 - p*) / ( w / (2 times 1.96) )^2 = (2 times 1.96 / w)^2 times p*(1 - p*), as required.

Marks: one for writing width = 2 times margin of error, one for substituting the margin-of-error formula, and one for the algebra that isolates n.

2017 SACE Stage 22 marksA farmer planted 100 seeds and observed 53 viable, giving p* = 0.53. Using n = (2 x 1.96 / w)^2 x p*(1 - p*), calculate the smallest number of seeds she would need to plant to obtain a 95% confidence interval of width 0.1 for the proportion of viable seeds.

Show worked answer →

Substitute p* = 0.53 and w = 0.1 into the formula.

First the bracket: 2 times 1.96 / 0.1 = 3.92 / 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 = 382.78.

Sample size must be a whole number, and we always round up to guarantee the interval is no wider than required, so n = 383 seeds.

Marks: one for substituting correctly to get n = 382.78, one for rounding up to 383. Rounding down to 382 would give an interval slightly wider than 0.1, so rounding up is essential.