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Edexcel A-level Further Maths revision notes

Chi squared tests

Section 6
Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.
1 specification point

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section 6

Checked against Edexcel 9FM0 section 6. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

In the exam: Formulae booklet provided · calculator allowed in every paper

Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

Goodness of fit tests and contingency tables. The null and alternative hypotheses. Degrees of freedom.

Notes
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Explanation

  • For goodness of fit, H0H_0 states in context that the proposed distribution fits; for a contingency table, both hypotheses must name the two classifications, for example H0H_0: travel method is independent of year group and H1H_1: travel method is associated with year group. Use χ2=∑(O−E)2E\chi^2=\sum\dfrac{(O-E)^2}{E}.
  • Goodness-of-fit degrees of freedom are k−1−mk-1-m after combining, where mm parameters were estimated from the data. For an r×cr\times c contingency table, df=(r−1)(c−1)df=(r-1)(c-1) and each expected frequency is row total times column total divided by grand total.
  • Cells should be combined sensibly when E<5E<5.
  • In a critical-value test, state the degrees of freedom and the critical value explicitly before comparing it with χ2\chi^2; the critical value can carry its own mark.
  • Finish with a contextual conclusion.
Worked example

A 2×32\times3 contingency table has row totals 60,4060,40, column totals 20,30,5020,30,50 and grand total 100100. Find the expected frequencies and degrees of freedom for a test of independence.

  1. 1.First row expectations: 60(20,30,50)/100=(12,18,30)60(20,30,50)/100=(12,18,30).
  2. 2.Second row expectations: 40(20,30,50)/100=(8,12,20)40(20,30,50)/100=(8,12,20).
  3. 3.df=(2−1)(3−1)=2df=(2-1)(3-1)=2.

Answer: Expected table (12183081220)\begin{pmatrix}12&18&30\\8&12&20\end{pmatrix}; 22 degrees of freedom.

Common mistakes

  • Don't use observed rather than expected frequency in the denominator of each contribution.
  • Don't calculate contingency-table degrees of freedom as rc−1rc-1.
  • Don't combine a cell with E<5E<5 but forget to recalculate the number of groups and degrees of freedom.

Exam tip

Show one expected-frequency calculation and state the degrees-of-freedom formula before using the calculator.

Tier 1 · Easy

ORIGINAL

1.

A goodness-of-fit test has four categories with model probabilities 0.1,0.2,0.3,0.40.1,0.2,0.3,0.4 and a sample size of 120120. No parameter is estimated. Find the expected frequencies and the degrees of freedom.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

After fitting a one-parameter model from the same data, five ordered categories have observed frequencies 41,33,17,7,241,33,17,7,2 and expected frequencies 40,32,18,7,340,32,18,7,3. Carry out a goodness-of-fit test at the 5%5\% level of significance. The critical value for 22 degrees of freedom is 5.9915.991.

(8)

(Total for Question 1 is 8 marks)

Tier 3 · Hard

ORIGINAL

1.

The observed counts for travel method by year group are walkcyclebuscarYear 1024182018Year 1116222022Year 1210203020\begin{array}{c|rrrr}&\text{walk}&\text{cycle}&\text{bus}&\text{car}\\ \hline \text{Year 10}&24&18&20&18\\ \text{Year 11}&16&22&20&22\\ \text{Year 12}&10&20&30&20\end{array}. Test independence at the 5%5\% level of significance. The critical value for 66 degrees of freedom is 12.59212.592.

(10)

(Total for Question 1 is 10 marks)

Your progress and exam materials

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