Edexcel A-level Further Maths coverage

Chi squared tests

Section FS1-6
1 spec leaf

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

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FS1-6.1

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

  • For goodness of fit, H0H_0 states that the proposed distribution fits; for a contingency table, H0H_0 states that the two classifications are independent.
  • Use χ2=∑(O−E)2E\chi^2=\sum\dfrac{(O-E)^2}{E}. For kk goodness-of-fit groups, df=k−1−mdf=k-1-m when mm parameters are estimated from the data; for an r×cr\times c table, df=(r−1)(c−1)df=(r-1)(c-1).
  • In a contingency table, each expected frequency is E=(row total)(column total)/(grand total)E=(\text{row total})(\text{column total})/(\text{grand total}).
  • A common error is to leave an expected frequency below 55; adjacent or sensible categories should be combined before calculating the statistic and degrees of freedom.

Tier 1 · Easy

3 marks
ORIGINAL

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.

Tier 2 · Standard

8 marks
ORIGINAL

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.

Tier 3 · Hard

10 marks
ORIGINAL

The observed counts in a 3×43\times4 contingency table are ABCDR24182018S16222022T10203020\begin{array}{c|rrrr}&A&B&C&D\\ \hline R&24&18&20&18\\ S&16&22&20&22\\ T&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.