| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Standard 2×2 contingency table |
| Difficulty | Moderate -0.3 This is a straightforward chi-squared test of independence with a 2×2 contingency table. Students need to state hypotheses, calculate expected frequencies, compute the test statistic, and compare to critical value—all standard procedures with no conceptual challenges. The calculation is routine and the context is simple, making it slightly easier than average. |
| Spec | 5.06a Chi-squared: contingency tables |
| Pass | Fail | |
| Male | 23 | 27 |
| Female | 32 | 18 |
| Answer | Marks | Guidance |
|---|---|---|
| Observed Frequencies | ||
| Pass | Fail | Total |
| Male | 23 | 27 |
| Female | 32 | 18 |
| Total | 55 | 45 |
| Expected Frequencies | ||
| Pass | Fail | Total |
| Male | 27.5 | 22.5 |
| Female | 27.5 | 22.5 |
| Total | 55 | 45 |
| \(H_0\) : No association between gender and test result | B1 | |
| \(H_1\) : Association between gender and test result | B1 | |
| \(\sum \frac{(O-E)^2}{E} = \frac{(23-27.5)^2}{27.5} + \ldots + \frac{(18-22.5)^2}{22.5}\) | M1 A1 | Use of \(\sum \frac{(O-E)^2}{E}\) |
| \(= 3.27\) | A1 | |
| \(\alpha = 0.10 \Rightarrow \chi^2 > 2.705\) | B1 | \(\nu = 1\) |
| B1 | 2.705 | |
| Since 3.27 is in the critical region there is evidence of association between gender and test result. | A1 ft | |
| (11 marks) |
| | Observed Frequencies | | | |
|---|---|---|---|---|
| | Pass | Fail | Total | |
| Male | 23 | 27 | 50 | |
| Female | 32 | 18 | 50 | |
| Total | 55 | 45 | 100 | |
| | Expected Frequencies | | | |
|---|---|---|---|---|
| | Pass | Fail | Total | Use of $\frac{R_T \times C_T}{100}$ |
| Male | 27.5 | 22.5 | 50 | M1 |
| Female | 27.5 | 22.5 | 50 | A1 |
| Total | 55 | 45 | 100 | A1 |
$H_0$ : No association between gender and test result | B1 |
$H_1$ : Association between gender and test result | B1 |
$\sum \frac{(O-E)^2}{E} = \frac{(23-27.5)^2}{27.5} + \ldots + \frac{(18-22.5)^2}{22.5}$ | M1 A1 | Use of $\sum \frac{(O-E)^2}{E}$
$= 3.27$ | A1 |
$\alpha = 0.10 \Rightarrow \chi^2 > 2.705$ | B1 | $\nu = 1$
| | B1 | 2.705
Since 3.27 is in the critical region there is evidence of association between gender and test result. | A1 ft |
| (11 marks) |A survey in a college was commissioned to investigate whether or not there was any association between gender and passing a driving test. A group of 50 male and 50 female students were asked whether they passed or failed their driving test at the first attempt. All the students asked had taken the test. The results were as follows.
\begin{center}
\begin{tabular}{|l|c|c|}
\hline
& Pass & Fail \\
\hline
Male & 23 & 27 \\
\hline
Female & 32 & 18 \\
\hline
\end{tabular}
\end{center}
Stating your hypotheses clearly test, at the 10\% level, whether or not there is any evidence of an association between gender and passing a driving test at the first attempt. [11]
\hfill \mbox{\textit{Edexcel S3 Q7 [11]}}