Contingency tables and the chi-squared testAQA A-Level Further Maths: Revision notes

A contingency table shows observed frequencies for two categorical variables, with $n$ rows and $m$ columns. The chi-squared test checks whether the variables are associated.

• $H_0$: there is no association between the variables (they are independent).
• $H_1$: there is an association. Always state the hypotheses in the context of the question.

If $H_0$ is true, the expected frequency in each cell is $E=\frac{\text{row total}\times\text{column total}}{\text{grand total}}.$ Worked example: 150 employees, 40 walk, 70 live in town. Expected number who walk and live in town $=\frac{40\times70}{150}=18.67$. Check that the expected frequencies have the same row and column totals as the observed table.

The test statistic is $X^2=\sum\frac{(O_i-E_i)^2}{E_i},$ which is approximately chi-squared when $H_0$ is true. For an $n\times m$ table the degrees of freedom are $\nu=(n-1)(m-1).$ Large values of $X^2$ mean the observed frequencies are far from those expected under $H_0$. Compare with the upper-tail critical value: at the $5\%$ level these are $3.841$ ($\nu=1$), $5.991$ ($\nu=2$), $7.815$ ($\nu=3$) and $9.488$ ($\nu=4$). If $X^2$ exceeds the critical value, reject $H_0$.

The chi-squared approximation is only reliable when all expected frequencies are greater than 5. If any $E_i\le5$, combine adjacent categories (rows or columns) so that every expected frequency is greater than 5, recalculate, and use the reduced degrees of freedom for the new table. Example: expected frequencies of $4.2$, $3.6$ and $4.2$ in a 'not at all' column mean it should be combined with the neighbouring category, giving a table with fewer columns and $\nu$ reduced.

If $H_0$ is rejected, find which cells cause it by looking at each contribution $\frac{(O-E)^2}{E}$. The cells with the largest contributions are the main sources of association. Compare $O$ with $E$ for those cells and describe the pattern in context. Example: at clinic Z, 20 dissatisfied patients were observed against 11.7 expected (contribution $5.95$), so dissatisfaction is higher at Z than independence predicts, while a clinic with contributions close to zero follows the overall pattern.