| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2024 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Percentages given, table construction required |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with straightforward setup. Part (a) requires basic percentage calculations to complete a contingency table using given constraints, which is algebraically simple. Part (b) is a routine application of the chi-squared test procedure. While it's a Further Maths topic, the question involves no conceptual challenges or novel problem-solving—just methodical calculation and standard hypothesis testing steps. |
| Spec | 5.06a Chi-squared: contingency tables |
| \multirow{2}{*}{} | Rating | |||
| Good | Satisfactory | Poor | ||
| \multirow{3}{*}{Shop} | A | |||
| B | ||||
| C | ||||
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Calculates one correct row or column of observed frequencies | M1 | AO 3.1b |
| Calculates two correct rows or two correct columns or one correct row and one correct column | A1 | AO 1.1b |
| All correct observed frequencies: \(A\): 65, 90, 95; \(B\): 80, 100, 70; \(C\): 65, 71, 114 | A1 | AO 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0\): There is no association between shop and rating; \(H_1\): There is an association between shop and rating | B1 | States both hypotheses using correct language; variables needed in at least null hypothesis; AO 2.5 |
| Expected table: \(A\): 70, 87, 93; \(B\): 70, 87, 93; \(C\): 70, 87, 93 | B1 | Correct expected contingency table for \(\chi^2\) model; PI; AO 3.3 |
| Attempts to calculate \(\sum \dfrac{(O-E)^2}{E}\); condone slips if intent clear; PI | M1 | AO 3.4 |
| \(\sum \dfrac{(O-E)^2}{E} = \dfrac{(65-70)^2+(80-70)^2+(65-70)^2}{70} + \dfrac{(90-87)^2+(100-87)^2+(71-87)^2}{87} + \dfrac{(95-93)^2+(70-93)^2+(114-93)^2}{93} = 17.60\) | A1 | AWRT 17.6; AO 1.1b |
| \(\chi^2\) critical value for 4 df \(= 13.277\); or \(p =\) AWRT 0.001 | B1 | AWRT 13.3; AO 1.1b |
| \(17.60 > 13.277\); evaluates \(\chi^2\) test statistic by comparing with critical value or \(p\) with 0.01 | M1 | AO 3.5a |
| Reject \(H_0\); FT their comparison using \(\chi^2\) model; condone Accept \(H_1\) | A1F | AO 2.2b |
| Sufficient evidence to suggest that there is an association between shop and rating | R1 | Concludes in context referring to association between shop and rating; conclusion must not be definite; AO 3.2a |
## Question 9(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Calculates one correct row or column of observed frequencies | M1 | AO 3.1b |
| Calculates two correct rows or two correct columns or one correct row and one correct column | A1 | AO 1.1b |
| All correct observed frequencies: $A$: 65, 90, 95; $B$: 80, 100, 70; $C$: 65, 71, 114 | A1 | AO 1.1b |
**Subtotal: 3 marks**
---
## Question 9(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: There is no association between shop and rating; $H_1$: There is an association between shop and rating | B1 | States both hypotheses using correct language; variables needed in at least null hypothesis; AO 2.5 |
| Expected table: $A$: 70, 87, 93; $B$: 70, 87, 93; $C$: 70, 87, 93 | B1 | Correct expected contingency table for $\chi^2$ model; PI; AO 3.3 |
| Attempts to calculate $\sum \dfrac{(O-E)^2}{E}$; condone slips if intent clear; PI | M1 | AO 3.4 |
| $\sum \dfrac{(O-E)^2}{E} = \dfrac{(65-70)^2+(80-70)^2+(65-70)^2}{70} + \dfrac{(90-87)^2+(100-87)^2+(71-87)^2}{87} + \dfrac{(95-93)^2+(70-93)^2+(114-93)^2}{93} = 17.60$ | A1 | AWRT 17.6; AO 1.1b |
| $\chi^2$ critical value for 4 df $= 13.277$; or $p =$ AWRT 0.001 | B1 | AWRT 13.3; AO 1.1b |
| $17.60 > 13.277$; evaluates $\chi^2$ test statistic by comparing with critical value or $p$ with 0.01 | M1 | AO 3.5a |
| Reject $H_0$; FT their comparison using $\chi^2$ model; condone Accept $H_1$ | A1F | AO 2.2b |
| Sufficient evidence to suggest that there is an association between shop and rating | R1 | Concludes in context referring to association between shop and rating; conclusion must not be definite; AO 3.2a |
**Subtotal: 8 marks**
---9 A company owns three shops, A, B and C, which are based in different towns.
Each shop gives a questionnaire to 250 of their customers, and every customer completes the questionnaire.
One of the questions asks whether the customer rates the shop as good, satisfactory or poor.
For shop A, 26\% of customers rate the shop as good and 38\% of customers rate the shop as poor.
For shop B, 32\% of customers rate the shop as good and 40\% of customers rate the shop as satisfactory.
Altogether, there are 210 good ratings and 261 satisfactory ratings.
9
\begin{enumerate}[label=(\alph*)]
\item Complete the following table with the observed frequencies.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{3}{|c|}{Rating} \\
\hline
& & Good & Satisfactory & Poor \\
\hline
\multirow{3}{*}{Shop} & A & & & \\
\hline
& B & & & \\
\hline
& C & & & \\
\hline
\end{tabular}
\end{center}
9
\item Carry out a test for association between shop and rating, using the 1\% level of significance.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2024 Q9 [11]}}