| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| 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 straightforward chi-squared test of independence with a 2×2 contingency table requiring standard calculations (expected frequencies, test statistic, comparison with critical value), followed by a simple interpretation question about causation vs association. The setup is clear and the method is routine for Further Maths Statistics students. |
| Spec | 5.06a Chi-squared: contingency tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): There is no association between driving test result and driving test centre; \(H_1\): There is an association between driving test result and driving test centre | B1 | States both hypotheses using correct language; variables must be included in at least the null hypothesis |
| Observed table: \(A\): \(P=60\), \(F=42\), Total\(=102\); \(B\): \(P=80\), \(F=30\), Total\(=110\); Total: \(140\), \(72\), \(212\) | M1 | Constructs correct contingency table with frequencies shown (PI); condone one slip |
| Expected: \(A\): \(P=67.4\), \(F=34.6\); \(B\): \(P=72.6\), \(F=37.4\) | M1 | Finds expected frequencies (PI); AWRT 3 significant figures |
| \(ts = \dfrac{(\ | 60-67.4\ | -0.5)^2}{67.4} + \dfrac{(\ |
| \(4.0\) (AWRT) | A1 | Calculates \(\chi^2\) test statistic correctly |
| \(\chi^2\) cv for 1 df \(= 3.84\); \(4.0 > 3.84\); Reject \(H_0\) | R1 | Evaluates by comparing ts with cv (AWRT 3.84) or \(p = \) AWFW \(0.045\) to \(0.047\) with \(0.05\) |
| Infers \(H_0\) rejected | E1F | Follow through on their test statistic |
| Some evidence to suggest driving test result and driving test centre are not independent | E1F | Concludes in context based on their hypotheses (not definite); consistent with decision |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| No, may be another underlying cause for the higher pass rate at B | E1 | States disagreement with Rebecca's claim; gives valid reason or example of alternative possible cause |
# Question 6:
## Part 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: There is no association between driving test result and driving test centre; $H_1$: There is an association between driving test result and driving test centre | B1 | States both hypotheses using correct language; variables must be included in at least the null hypothesis |
| Observed table: $A$: $P=60$, $F=42$, Total$=102$; $B$: $P=80$, $F=30$, Total$=110$; Total: $140$, $72$, $212$ | M1 | Constructs correct contingency table with frequencies shown (PI); condone one slip |
| Expected: $A$: $P=67.4$, $F=34.6$; $B$: $P=72.6$, $F=37.4$ | M1 | Finds expected frequencies (PI); AWRT 3 significant figures |
| $ts = \dfrac{(\|60-67.4\|-0.5)^2}{67.4} + \dfrac{(\|42-34.6\|-0.5)^2}{34.6} + \dfrac{(\|80-72.6\|-0.5)^2}{72.6} + \dfrac{(\|30-37.4\|-0.5)^2}{37.4} = 4.0$ | M1 | Calculates $\chi^2$ test statistic with observed and expected frequencies; condone no Yates correction; condone missing modulus sign; accept use of $\dfrac{212(\|60\times30 - 42\times80\|-0.5\times212)^2}{140\times72\times102\times110}$ |
| $4.0$ (AWRT) | A1 | Calculates $\chi^2$ test statistic correctly |
| $\chi^2$ cv for 1 df $= 3.84$; $4.0 > 3.84$; Reject $H_0$ | R1 | Evaluates by comparing ts with cv (AWRT 3.84) or $p = $ AWFW $0.045$ to $0.047$ with $0.05$ |
| Infers $H_0$ rejected | E1F | Follow through on their test statistic |
| Some evidence to suggest driving test result and driving test centre are not independent | E1F | Concludes in context based on their hypotheses (not definite); consistent with decision |
## Part 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| No, may be another underlying cause for the higher pass rate at B | E1 | States disagreement with Rebecca's claim; gives valid reason or example of alternative possible cause |
---6 During August, 102 candidates took their driving test at centre $A$ and 60 passed. During the same month, 110 candidates took their driving test at centre $B$ and 80 passed.
6
\begin{enumerate}[label=(\alph*)]
\item Test whether the driving test result is independent of the driving test centre using the $5 \%$ level of significance.
6
\item Rebecca claims that if the result of the test in part (a) is to reject the null hypothesis then it is easier to pass a driving test at centre $B$ than centre $A$.
State, with a reason, whether or not you agree with Rebecca's claim.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2019 Q6 [9]}}