Learn9MA0Paper 3 · Statistics and MechanicsStatistical hypothesis testingTests for a population mean

Statistical hypothesis testing · Subtopic 2 of 3

Tests for a Population Mean

Has the average changed? Assume it hasn't, work out how many standard errors the sample mean is from the claimed mean, and compare that $z$ with the critical value for your significance level.

Cambridge 9709 Paper 6Edexcel YMA01 WST03Edexcel 9MA0 Paper 3Edexcel 9FM0 OptionAQA 7357 Paper 3≈ 40 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

State hypotheses about a population mean $\mu$ .
Calculate the test statistic $z=\dfrac{\bar x-\mu_0}{\sigma/\sqrt n}$ .
Compare $z$ with critical values (one- and two-tailed), or find a critical value of $\bar x$ .
Use the CLT and an unbiased variance estimate for large samples.

Pause and think

Bags should hold $200$ g on average, with standard deviation $12$ g. A sample of $16$ bags averages $205$ g. Is $5$ g above the target a lot?

Section 2 of 14

Core idea

Start hereAre the bags underweight?

Crisp bags should average 50 g. A sample of 36 averages 49.2 g. Is the machine under-filling, or is this normal sampling wobble? The sample mean's distribution tells us how big a wobble to expect.

The big idea

Test statistic $z=\dfrac{\bar x-\mu_0}{\sigma/\sqrt n}$ . Compare with critical values ( $1.645$ one-tailed 5%, $\pm1.96$ two-tailed 5%), or compare the tail probability with the level.

1. Method

One-tailed 5% test: reject H₀ if z > 1.645

Our test statistic

2.1

lies in the shaded critical region, so we reject

H_0

Worked exampleOne-tailed

$H_0:\mu=50$ , $H_1:\mu>50$ , $\sigma=10$ , $n=36$ , $\bar x=53.5$ , 5%.

Goal: Decision.
Strategy: $z$ vs $1.645$ .

$z=\frac{3.5}{10/6}=2.1$
Standard error $\frac{10}6$ .
$2.1>1.645$
In the critical region.

AnswerReject

H_0

: evidence the mean exceeds 50.

✓ Check

P(Z>2.1)=0.018<0.05

✓.

Worked exampleTwo-tailed

$H_1:\mu\neq50$ , $\bar x=47$ , $n=25$ , $\sigma=8$ , 5%.

Goal: Decision.
Strategy: $|z|$ vs $1.96$ .

$z=\frac{-3}{1.6}=-1.875$
Compute.
$|{-1.875}|<1.96$
Not in either tail.

AnswerDo not reject

H_0

✓ CheckA one-tailed test (

\mu<50

) would reject (

-1.875<-1.645

): the tails matter ✓.

Edge caseUnknown σ, large n Use

s

in place of

\sigma

; the CLT justifies normality.

Pause and think

One-tailed 1% critical value?

Summary

$z=(\bar x-\mu_0)/(\sigma/\sqrt n)$ .
Critical values by tail and level.

Test statistic: Value computed from the sample to decide.

HistoryGuinness and the t-test. In 1908 William Gosset, a chemist at the Guinness brewery in Dublin, worked out how to test means from very small samples, when

\sigma

is unknown and the normal approximation is poor. Guinness did not allow staff to publish, so he wrote under the pen name "Student": his

t

-test is still called "Student's

t

". (Cambridge P6 uses only the large-sample and known-

\sigma

cases.)

Fun factStatistically significant ≠ important. With a huge sample, even a tiny difference becomes "significant": a sample of a million bags could detect a mean of

200.05

g. Significance tells you a difference is probably real, not that it matters. Good reports give the size of the effect as well.

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