Yates' correctionAQA A-Level Further Maths: Revision notes

The chi-squared test statistic $\sum\frac{(O-E)^2}{E}$ is only an approximation: the observed frequencies are discrete whereas the chi-squared distribution is continuous. When the degrees of freedom are small this approximation tends to make the statistic too large, so the test rejects $H_0$ too often. Yates' correction (continuity correction) reduces this effect.

Apply Yates' correction when the table is a $2\times2$ contingency table, so that $\nu=(2-1)(2-1)=1$. For larger tables ($\nu>1$) it is not used. The convention that every expected frequency should be greater than 5 still applies.

$X^2=\sum\frac{(|O_i-E_i|-0.5)^2}{E_i}.$ Find each expected frequency as usual, take the modulus of $O-E$, subtract $0.5$, then square and divide by $E$. In a $2\times2$ table all four values of $|O-E|$ are equal, which gives a quick check. Compare the result with the critical value for $\nu=1$: $3.841$ at $5\%$ and $6.635$ at $1\%$.

Method X: 17 pass, 8 fail. Method Y: 10 pass, 15 fail. Totals: 27 pass, 23 fail, $N=50$.

• $E=\frac{25\times27}{50}=13.5$ (pass), $\frac{25\times23}{50}=11.5$ (fail), for each method.
• $|O-E|=3.5$ in every cell.
• Corrected: $3^2\left(\frac{2}{13.5}+\frac{2}{11.5}\right)=2.90$.
• $2.90<3.841$, so do not reject $H_0$. Without the correction the statistic would be $3.95$, which would wrongly reject $H_0$.

Yates' correction always gives a smaller statistic than the uncorrected formula, so it is more cautious about claiming an association. State the conclusion in context, for example 'there is insufficient evidence that the pass rate depends on the training method'. If the corrected statistic is close to the critical value, say that the result is borderline and that more data would help.