LearnYMA01WST03 · Statistics 3Estimation, confidence intervals and testsConfidence intervals

Estimation, confidence intervals and tests · Subtopic 2 of 2

Confidence Intervals

A single estimate is almost certainly a little wrong. A confidence interval gives a range, $\bar x\pm z\dfrac{\sigma}{\sqrt n}$ , built so that in, say, $95\%$ of samples the range would capture the true mean.

Cambridge 9709 Paper 6Edexcel YMA01 WST03Edexcel 9FM0 OptionAQA 7367 Paper 3≈ 40 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Calculate a confidence interval for $\mu$ : $\bar x\pm z\dfrac\sigma{\sqrt n}$ .
Use the right $z$ for $90\%$ , $95\%$ and $99\%$ intervals.
Calculate an approximate confidence interval for a proportion.
Interpret an interval correctly and find the sample size for a required width.

Pause and think

A poll says a party has $40\%$ support "plus or minus $3\%$ ". What do you think the $\pm3\%$ is telling you?

Section 2 of 14

Core idea

Start hereA net, not a spear

Estimating the population mean by $\bar x$ alone is like throwing a spear at an invisible fish. A confidence interval throws a net: a range that, by its construction, catches the true mean 95% of the time.

The big idea

95% confidence interval for $\mu$ : $\bar x\pm1.96\dfrac{\sigma}{\sqrt n}$ . For a proportion: $\hat p\pm z\sqrt{\dfrac{\hat p(1-\hat p)}n}$ .

1. Building the interval

Turning the statement round

$P\left(-1.96<\frac{\bar X-\mu}{\sigma/\sqrt n}<1.96\right)=0.95$
Standard normal.
$P\left(\bar X-1.96\tfrac\sigma{\sqrt n}<\mu<\bar X+1.96\tfrac\sigma{\sqrt n}\right)=0.95$
Rearrange each inequality for $\mu$ .

The random interval captures

\mu

with probability

0.95

12 samples, 12 intervals: one misses the true mean

Each line is

\bar x\pm1.96

. The 7th sample happened to give an unusually high mean, so its interval misses

50

Worked example $n=25$ , $\bar x=52.3$ , $\sigma=10$

95% CI for $\mu$ .

Goal: Interval.
Strategy: Formula.

$\tfrac{10}{5}=2$
Standard error.
$52.3\pm3.92$
$1.96\times2$ .

Answer

(48.4,\ 56.2)

✓ CheckCentred on

\bar x

✓.

Worked exampleSample size for a given width

Width at most 4 at 95%, $\sigma=10$ . Least $n$ ?

Goal: $n$ .
Strategy: Width $=2\times1.96\frac\sigma{\sqrt n}$ .

$\frac{39.2}{\sqrt n}\le4\Rightarrow\sqrt n\ge9.8$
Rearrange.
$n\ge96.04$
Square.

Answer

n=97

✓ CheckRound up to guarantee the width ✓.

Worked exampleProportion

72 of 200 prefer brand A. 95% CI for $p$ .

Goal: Interval.
Strategy: $\hat p=0.36$ .

$\sqrt{\tfrac{0.36\times0.64}{200}}=0.0339$
Standard error.
$0.36\pm0.0665$
$1.96\times0.0339$ .

Answer

(0.293,\ 0.427)

✓ CheckWithin

[0,1]

✓.

Common misconceptionInterpretation "95% of such intervals contain

\mu

", not "

\mu

has a 95% chance of being in this particular interval" (

\mu

is fixed).

Pause and think

Wider interval: more or less confidence?

Summary

$\bar x\pm z\sigma/\sqrt n$ .
Proportion version.

Confidence interval: Range likely to contain the parameter.

HistoryNeyman's intervals. The Polish statistician Jerzy Neyman introduced confidence intervals in 1937, carefully defining them by what happens over repeated sampling. He did much of this work in London at University College, alongside Egon Pearson, the son of Karl Pearson.

Fun fact19 out of 20. A

95\%

confidence level means that, on average,

1

in every

20

properly calculated intervals misses the true value. If a newspaper reports

20

polls with their margins of error, you should expect about one of them to be "wrong" through pure chance.

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