| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Interpret association after test |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with clearly presented data in a 2×3 contingency table. Part (a) requires routine calculation of expected frequencies, test statistic, and comparison with critical value. Part (b) asks for contextual interpretation by comparing observed vs expected frequencies. While it requires careful arithmetic and understanding of the test, it follows a well-practiced procedure with no novel insight needed, making it slightly easier than average for Further Maths statistics. |
| Spec | 5.06a Chi-squared: contingency tables |
| Morning show | Afternoon show | Evening show | Total | |
| Enjoyed | 62 | 91 | 172 | 325 |
| Not enjoyed | 25 | 35 | 115 | 175 |
| Total | 87 | 126 | 287 | 500 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): There is no association between the show a customer attends and whether they enjoyed the show | B1 | States both hypotheses using correct language; variables stated in at least \(H_0\) |
| \(H_1\): There is an association between the show a customer attends and whether they enjoyed the show | — | — |
| Expected frequencies table: \(En\): MS = 56.55, AS = 81.90, ES = 186.55; \(NEn\): MS = 30.45, AS = 44.10, ES = 100.45 | M1 | Translates situation into expected contingency table for \(\chi^2\) model |
| \(\sum \frac{(O-E)^2}{E}\): \(En\): MS = 0.5252, AS = 1.0111, ES = 1.1348; \(NEn\): MS = 0.9755, AS = 1.8778, ES = 2.1075 | M1 | Uses \(\chi^2\) model to calculate test statistic |
| \(\sum \frac{(O-E)^2}{E} = 7.6\) | A1 | AWRT 7.6 |
| \(\chi^2\) cv for 2 df \(= 7.378\); \(7.6 > 7.378\), Reject \(H_0\) | B1 | AWRT 7.4 or corresponding probability AWRT 0.022 |
| Sufficient evidence to suggest there is an association between the show a customer attends and whether they enjoyed the show | R1 | Evaluates \(\chi^2\) test statistic by correctly comparing critical value with test statistic or probability with 0.025 |
| — | E1F | Infers \(H_0\) rejected; FT their comparison using \(\chi^2\) model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Largest sources of association: evening show/did not enjoy. \(\frac{(O-E)^2}{E} = 2.1\ldots\) | E1 | AO 2.4. Explains reasoning by considering \((O-E)\) or \(\frac{(O-E)^2}{E}\) to identify largest source of association. The value of \((O-E)\) or \(\frac{(O-E)^2}{E}\) does not need to be seen provided it is identified as the largest |
| People who attend the evening show did not enjoy the show more often than expected | E1 | AO 3.2a. Interprets the main source of association in context. OE: People who attend the evening show enjoyed the show less often than expected. Needs to be a comparison, not just listing observed and expected frequencies |
## Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: There is no association between the show a customer attends and whether they enjoyed the show | B1 | States both hypotheses using correct language; variables stated in at least $H_0$ |
| $H_1$: There is an association between the show a customer attends and whether they enjoyed the show | — | — |
| Expected frequencies table: $En$: MS = 56.55, AS = 81.90, ES = 186.55; $NEn$: MS = 30.45, AS = 44.10, ES = 100.45 | M1 | Translates situation into expected contingency table for $\chi^2$ model |
| $\sum \frac{(O-E)^2}{E}$: $En$: MS = 0.5252, AS = 1.0111, ES = 1.1348; $NEn$: MS = 0.9755, AS = 1.8778, ES = 2.1075 | M1 | Uses $\chi^2$ model to calculate test statistic |
| $\sum \frac{(O-E)^2}{E} = 7.6$ | A1 | AWRT 7.6 |
| $\chi^2$ cv for 2 df $= 7.378$; $7.6 > 7.378$, Reject $H_0$ | B1 | AWRT 7.4 or corresponding probability AWRT 0.022 |
| Sufficient evidence to suggest there is an association between the show a customer attends and whether they enjoyed the show | R1 | Evaluates $\chi^2$ test statistic by correctly comparing critical value with test statistic or probability with 0.025 |
| — | E1F | Infers $H_0$ rejected; FT their comparison using $\chi^2$ model |
## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Largest sources of association: evening show/did not enjoy. $\frac{(O-E)^2}{E} = 2.1\ldots$ | E1 | AO 2.4. Explains reasoning by considering $(O-E)$ or $\frac{(O-E)^2}{E}$ to identify largest source of association. The value of $(O-E)$ or $\frac{(O-E)^2}{E}$ does not need to be seen provided it is identified as the largest |
| People who attend the evening show did not enjoy the show more often than expected | E1 | AO 3.2a. Interprets the main source of association in context. OE: People who attend the evening show enjoyed the show less often than expected. Needs to be a comparison, not just listing observed and expected frequencies |
**Subtotal: 2 marks**
---7 A theatre has morning, afternoon and evening shows.
On one particular day, the theatre asks all of its customers to state whether they enjoyed or did not enjoy the show.
The results are summarised in the table.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
& Morning show & Afternoon show & Evening show & Total \\
\hline
Enjoyed & 62 & 91 & 172 & 325 \\
\hline
Not enjoyed & 25 & 35 & 115 & 175 \\
\hline
Total & 87 & 126 & 287 & 500 \\
\hline
\end{tabular}
\end{center}
The theatre claims that there is no association between the show that a customer attends and whether they enjoyed the show.
7
\begin{enumerate}[label=(\alph*)]
\item Investigate the theatre's claim, using a $2.5 \%$ level of significance.\\
7
\item By considering observed and expected frequencies, interpret in context the association between the show that a customer attends and whether they enjoyed the show.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2023 Q7 [10]}}