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How do you test goodness of fit and independence using the chi-squared distribution?

Goodness of fit tests, contingency tables and tests for independence using the chi-squared statistic, expected frequencies, degrees of freedom, and Yates' correction.

A focused answer to the Edexcel A-Level Further Mathematics Further Statistics content on chi-squared tests, covering goodness of fit tests, contingency tables and tests for independence, calculating expected frequencies, choosing degrees of freedom, and applying Yates' correction.

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  1. What this dot point is asking
  2. The chi-squared statistic
  3. Goodness of fit and degrees of freedom
  4. Contingency tables and independence
  5. Examples in context
  6. Try this

What this dot point is asking

Edexcel Further Statistics wants you to carry out chi-squared goodness of fit tests against a proposed distribution, test for independence using a contingency table, compute expected frequencies, choose the correct degrees of freedom, apply Yates' correction where required, and reach a conclusion in context. A full hypothesis-test structure (hypotheses, statistic, comparison, contextual conclusion) is expected for the marks.

The chi-squared statistic

Both the goodness of fit test and the test of independence compare observed counts with the counts expected under a null hypothesis, using the same statistic. A large value means observed and expected diverge sharply, giving evidence against the null hypothesis; a small value is consistent with it. The statistic is then compared with a tabulated critical value at the chosen significance level and degrees of freedom.

Goodness of fit and degrees of freedom

A goodness of fit test checks whether data are consistent with a proposed distribution (uniform, binomial, Poisson and so on). The degrees of freedom start as the number of classes minus one (because the totals must agree), then lose one further degree for each parameter you estimated from the data, such as a Poisson mean estimated from the sample.

Contingency tables and independence

A contingency table cross-classifies a sample by two factors, and the chi-squared test of independence checks whether the two factors are associated. The expected frequency in each cell, assuming independence, is the product of its row and column totals divided by the grand total.

Examples in context

Chi-squared tests are the inferential capstone of Further Statistics, drawing on the distributions studied earlier. Goodness of fit tests are most often applied to the Poisson and binomial models of the poisson-and-binomial dot point (with the mean estimated from the data, costing a degree of freedom) and to the discrete distributions whose expectations you compute elsewhere. The expected-frequency calculation for contingency tables uses the multiplication of probabilities for independent events. The combining of small classes connects to the practical requirement that the chi-squared approximation to the discrete sampling distribution be reliable.

Try this

Q1. State the degrees of freedom for a 3×43 \times 4 contingency table. [1 mark]

  • Cue. (3−1)(4−1)=6(3 - 1)(4 - 1) = 6.

Q2. Write down the chi-squared test statistic formula. [1 mark]

  • Cue. χ2=∑(O−E)2E\chi^2 = \sum \frac{(O - E)^2}{E}.

Q3. A Poisson goodness of fit test has 77 classes and the mean was estimated from the data. State the degrees of freedom. [2 marks]

  • Cue. 7−1−1=57 - 1 - 1 = 5 (one lost for the total, one for the estimated mean).

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20198 marksA die is rolled 120120 times. The observed frequencies of the scores 11 to 66 are 14,22,18,20,26,2014, 22, 18, 20, 26, 20. Test, at the 5%5\% significance level, whether the die is fair. The critical value of χ2\chi^2 with 55 degrees of freedom at 5%5\% is 11.0711.07.
Show worked answer →

State hypotheses, compute expected frequencies, the test statistic, and compare with the critical value.

H0H_0: the die is fair (each score equally likely); H1H_1: the die is not fair (B1). Under H0H_0, each expected frequency is E=1206=20E = \frac{120}{6} = 20 (M1 A1).

χ2=∑(O−E)2E=(−6)2+22+(−2)2+02+62+0220=36+4+4+0+36+020=8020=4\chi^2 = \sum\frac{(O - E)^2}{E} = \frac{(-6)^2 + 2^2 + (-2)^2 + 0^2 + 6^2 + 0^2}{20} = \frac{36 + 4 + 4 + 0 + 36 + 0}{20} = \frac{80}{20} = 4 (M1 A1 A1).

Degrees of freedom =6−1=5= 6 - 1 = 5. Since 4<11.074 < 11.07, we do not reject H0H_0 (M1). There is insufficient evidence at the 5%5\% level to conclude the die is unfair (A1).

Edexcel 20227 marksA survey of 200200 people records gender against preferred drink in a 2×32 \times 3 contingency table. Explain how to find the expected frequencies and state the degrees of freedom. Given the test statistic is χ2=9.2\chi^2 = 9.2 and the critical value at 5%5\% is 5.9915.991, state the conclusion.
Show worked answer →

Describe the expected-frequency rule, give the degrees of freedom, then compare and conclude.

H0H_0: drink preference is independent of gender; H1H_1: they are not independent (B1).

Each expected frequency is E=row total×column totalgrand totalE = \frac{\text{row total} \times \text{column total}}{\text{grand total}} (M1 A1).

Degrees of freedom =(r−1)(c−1)=(2−1)(3−1)=2= (r - 1)(c - 1) = (2 - 1)(3 - 1) = 2 (M1 A1).

Since χ2=9.2>5.991\chi^2 = 9.2 > 5.991, reject H0H_0 (M1). There is evidence at the 5%5\% level that drink preference is associated with gender (A1).

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