| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Test statistic given, complete the test |
| Difficulty | Moderate -0.8 This is a straightforward chi-squared test question requiring only standard recall and routine application. Part (i) asks for standard hypotheses (independence vs association), part (ii) tests textbook knowledge of Yates' correction (subtract 0.5 from |O-E|, decreases χ²), and part (iii) is a simple comparison with critical value at 1 d.f. No problem-solving, calculation, or conceptual insight required—pure recall and standard procedure. |
| Spec | 5.06a Chi-squared: contingency tables |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: A \text{ and } B \text{ are not associated}\) | \(B1\) | For both. Allow indpt., not indpt. |
| \(H_1: A \text{ and } B \text{ are associated}\) | \([1]\) |
| Answer | Marks | Guidance |
|---|---|---|
| Yates \(\chi^2 = \sum\frac{( | O - E | - 0.5)^2}{E}\) |
| which decreases the value | \(B1\) | |
| \([2]\) |
| Answer | Marks | Guidance |
|---|---|---|
| CV 5.024 seen | \(B1\) | Ft their CV |
| \(5.63 > \text{CV}\) and reject \(H_0\) | \(M1\) | Allow B1 if correct conclusion, but comparison not shown. CWO (ie from 5.024) |
| There is evidence at the \(2\frac{1}{2}\%\) SL of an association between \(A\) and \(B\) | \(A1\) | |
| \([3]\) |
### (i)
$H_0: A \text{ and } B \text{ are not associated}$ | $B1$ | For both. Allow indpt., not indpt.
$H_1: A \text{ and } B \text{ are associated}$ | $[1]$ |
### (ii)
Yates $\chi^2 = \sum\frac{(|O - E| - 0.5)^2}{E}$ | $B1$ | Dep '-0.5' seen.
which decreases the value | $B1$ |
| $[2]$ |
### (iii)
CV 5.024 seen | $B1$ | Ft their CV
$5.63 > \text{CV}$ and reject $H_0$ | $M1$ | Allow B1 if correct conclusion, but comparison not shown. CWO (ie from 5.024)
There is evidence at the $2\frac{1}{2}\%$ SL of an association between $A$ and $B$ | $A1$ |
| $[3]$ |In a test of association of two factors, $A$ and $B$, a $2 \times 2$ contingency table yielded $5.63$ for the value of $\chi^2$ with Yates' correction.
\begin{enumerate}[label=(\roman*)]
\item State the null hypothesis and alternative hypothesis for the test. [1]
\item State how Yates' correction is applied, and whether it increases or decreases the value of $\chi^2$. [2]
\item Carry out the test at the $2\frac{1}{2}\%$ significance level. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR S3 2012 Q1 [6]}}