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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

Learn7367Paper 3 · Statistics optionChi-squared testsGoodness of fit and contingency tables

Chi-squared tests · Subtopic 1 of 1

Chi-squared Tests

Compare observed frequencies $O$ with those expected under $H_0$ , $E$ , using $X^2=\sum\frac{(O-E)^2}E$ . Large values mean poor fit. Compare with $\chi^2_\nu$ where $\nu=$ number of classes $-1-$ parameters estimated, or $(r-1)(c-1)$ for a contingency table. Expected frequencies should be at least $5$ .

Cambridge 9231 Paper 4Edexcel YMA01 WST03Edexcel 9FM0 OptionAQA 7367 Paper 3≈ 50 minRevision 1 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Calculate expected frequencies and the $\chi^2$ statistic.
Find degrees of freedom and use $\chi^2$ tables.
Carry out goodness-of-fit and contingency-table tests.

Pause and think

A die gives sixes 14 times in 60 rolls instead of 10. Biased?

Section 2 of 14

Core idea

Start hereIs the die loaded?

Roll a die 60 times and you won't get exactly ten of each face. Some wobble is just chance. A chi-squared test adds up all the mismatches between what you saw and what a fair die should give, and asks whether the total is too big to blame on luck.

The big idea

$X^2=\sum\frac{(O-E)^2}E$ measures total mismatch. Under $H_0$ it approximately follows $\chi^2_\nu$ . Large $X^2$ means poor fit, so the test is upper-tailed.

1. Goodness of fit

Worked exampleFair die

60 rolls: $8,12,9,11,6,14$ . Test at 5% whether the die is fair.

Goal: Decision.
Strategy: $H_0$ : fair, $E=10$ each.

$O-E:\ -2,2,-1,1,-4,4$
Differences.
$X^2=\frac{4+4+1+1+16+16}{10}=4.2$
Sum of $\frac{(O-E)^2}E$ .
$\nu=6-1=5$
Six classes, total fixed.
$\chi^2_5(5\%)=11.07$
Critical value.

Answer

4.2<11.07

: insufficient evidence that the die is biased.

✓ CheckMean of

\chi^2_5

5

;

4.2

is typical ✓.

Counting degrees of freedom

$k\text{ classes}$
Start with the number of cells after combining.
$-1$
Expected totals are forced to match the observed total.
$-m$
Subtract one for each parameter estimated from the data (e.g. $\lambda$ , or $\mu$ and $\sigma$ ).

\nu=k-1-m

2. Contingency tables

E_{ij}=\frac{R_iC_j}{N},\quad\nu=(r-1)(c-1)

$R_i,C_j$: row and column totals.

Worked exampleExpected frequency

$N=100$ , row totals $40,60$ , column totals $30,50,20$ . Find $E$ for row 1, column 2, and $\nu$ .

Goal: $E_{12}$ , $\nu$ .
Strategy: Independence: $P(\text{row 1 and col 2})=\frac{40}{100}\cdot\frac{50}{100}$ .

$E_{12}=100\times0.4\times0.5$
Expected under independence.
$=20$
Same as $\frac{40\times50}{100}$ .
$\nu=(2-1)(3-1)=2$
Free cells once totals are fixed.

Answer

E_{12}=20

\nu=2

✓ CheckRow 1 expected:

12+20+8=40

✓.

χ² test procedure

Drawing diagram…

The same steps every time.

Common misconceptionHypotheses for tables

H_0

: the two factors are independent.

H_1

: they are not independent. Avoid "related" without saying how.

Edge case2×2 tables With

\nu=1

, AQA uses Yates' correction:

\sum\frac{(|O-E|-0.5)^2}E

Pause and think

Why is a very small $X^2$ suspicious?

Summary

$X^2=\sum\frac{(O-E)^2}E$ .
$\nu=k-1-m$ or $(r-1)(c-1)$ .
Combine so $E\ge5$ .
Reject for large $X^2$ .

Goodness of fit: How well data match a model.
Contingency table: Two-way frequency table.
Degrees of freedom: Number of independent cells.

HistoryKarl Pearson. Pearson introduced the

\chi^2

test in 1900, one of the first formal significance tests. Fisher later corrected his degrees of freedom.

Fun factMendel’s peas. Fisher used

\chi^2

to show Mendel's pea data fitted his theory almost too well, suggesting the results had been tidied.

Section 3 of 14

Key formulae

Statistic

X^2=\sum\frac{(O-E)^2}{E}=\sum\frac{O^2}E-N

Goodness of fit

\nu=k-1-(\text{parameters estimated})

Contingency

E=\frac{\text{row total}\times\text{column total}}{N},\quad\nu=(r-1)(c-1)

TrickCombine small classes. If any

E<5

, merge it with a neighbour before calculating, and reduce

k

accordingly.

Section 4 of 14

Worked examples

Example 1 · Fair die0 / 3 steps

60 rolls:

8,12,9,11,6,14

. Test fairness at 5%.

Example 2 · Contingency0 / 2 steps

2\times3

table with row totals 40, 60 and column totals 30, 50, 20. Find

E

for row 1, column 2 and

\nu

Section 5 of 14

Explore the graph

χ² distributions

Loading the interactive graph…

Try this

The curve is $\chi^2_5$ ; its mean is $5$ , its degrees of freedom.
The line marks a critical value: the area to its right is the significance level.

Reject $H_0$ when $X^2$ lands in the right tail.

In the real worldMedical trials. Contingency tables test whether recovery depends on treatment: rows are treatments, columns outcomes.

Section 6 of 14

Common mistakes

IncorrectUsing percentages

CorrectUse frequencies

\chi^2

depends on the counts.

IncorrectForgetting estimated parameters

CorrectSubtract one

\nu

per parameter

Estimating

\lambda

from the data makes the fit easier.

IncorrectTwo-tailed test

CorrectUpper tail only

Only poor fit (large

X^2

) counts against

H_0

Observed ≈ expected

χ²: small. H₀: survives another day.

Section 7 of 14

Exam tip

Exam tipContext conclusions. End with words: "there is insufficient evidence that the die is biased", not just "accept

H_0

Section 8 of 14

Guided practice

SA-ST-CS-G012 marksL2 StandardCalculatorStatistic

A die is rolled 60 times giving

8,12,9,11,6,14

. Find

X^2

for a test of fairness.

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-ST-CS-0011 markL1 FoundationNon-calculatorStatistic

How many degrees of freedom does the die test have?

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SA-ST-CS-0022 marksL2 StandardNon-calculatorContingency

In a contingency table with

N=100

, a cell's row total is 40 and column total is 50. Find its expected frequency.

Salam Academy Original Practice · Proposed mark scheme

SA-ST-CS-0031 markL1 FoundationNon-calculatorContingency

Find the degrees of freedom for a

3\times4

contingency table.

Salam Academy Original Practice · Proposed mark scheme

SA-ST-CS-0042 marksL2 StandardNon-calculatorGoodness of fit

Data in 7 classes (after combining) are tested for fit to a Poisson distribution whose mean is estimated from the data. Find

\nu

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-ST-CS-0053 marksL3 ExamCalculatorGoodness of fit

200 people are classified by blood group: O 82, A 78, B 30, AB 10. Theory gives proportions

0.45, 0.40, 0.10, 0.05

. Find

X^2

(3 s.f.).

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

$X^2=\sum\frac{(O-E)^2}E$ .
$\nu$ : classes $-1-$ parameters, or $(r-1)(c-1)$ .
$E\ge5$ : combine.
Upper-tail test.

Section 12 of 14

Quick quiz

Q1 · SA-ST-CS-Q011 markL1 FoundationNon-calculatorStatistic

If an expected frequency is below 5 you should

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Section 13 of 14

Same topic, different board

Specification: 7367 A-level Further Mathematics
Where: Statistics · Chi-squared tests
Expected: Contingency tables and goodness of fit; expected frequencies; degrees of freedom; Yates’ correction for 2×2 tables.
Calculator: Permitted on all papers.
Typical style: AQA mixes short multiple-choice questions with longer structured ones.
Open in: 7367 · Paper 3

Section 14 of 14

Past paper questions

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Salam Academy original content · Content status: reviewed and published. Syllabus mapping: AQA 7367, First teaching 2017 (official specification).