| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Expected frequencies partially provided |
| Difficulty | Standard +0.3 This is a straightforward chi-squared test of independence requiring standard calculations: computing expected frequencies using row×column/total formula, calculating the test statistic, and comparing to critical values. Part (a) involves routine arithmetic, and part (b) follows a standard hypothesis testing procedure with no novel insights required. While it's a Further Maths topic, the execution is mechanical and below average difficulty even for FM students. |
| Spec | 5.06a Chi-squared: contingency tables |
| \multirow{2}{*}{} | Number of claims | |||||
| 0 | 1 | 2 | 3 or more | Total | ||
| \multirow[b]{3}{*}{Type of insurance policy} | Car | 9 | 10 | 11 | 5 | 35 |
| Motorbike | 19 | 13 | 8 | 5 | 45 | |
| Total | 28 | 23 | 19 | 10 | 80 | |
| \multirow{2}{*}{} | Number of claims | ||||
| 0 | 1 | 2 | 3 or more | ||
| \multirow{2}{*}{Type of insurance policy} | Car | 10.0625 | 4.375 | ||
| Motorbike | 10.6875 | ||||
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| One correct missing expected value, e.g. 0 and car \(= \frac{28 \times 35}{80}\) | M1 (AO1.1a) | |
| All expected values correct: \(C\): 12.25, 10.0625, 8.3125, 4.375; \(MB\): 15.75, 12.9375, 10.6875, 5.625 | A1 (AO1.1b) | Must not be rounded CAO. Condone 12.937 for 12.9375 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): There is no association between number of claims and type of insurance policy; \(H_1\): There is an association between number of claims and type of insurance policy | B1 (AO2.5) | States both hypotheses using correct language. Accept equivalent wording |
| Combines expected and observed values for 2 and 3 or more correctly: \(C\): Observed 16, Expected 12.6875; \(MB\): Observed 13, Expected 16.3125 | B1F (AO1.1b) | Follow through their answers to part (a) |
| \(\sum \frac{(O-E)^2}{E} = \frac{(9-12.25)^2}{12.25} + \frac{(19-15.75)^2}{15.75} + \frac{(10-10.0625)^2}{10.0625} + \frac{(13-12.9375)^2}{12.9375} + \frac{(16-12.6875)^2}{12.6875} + \frac{(13-16.3125)^2}{16.3125}\) | M1 (AO1.1a) | Can be awarded even if 2 and 3 or more are not combined |
| \(= 3.07\) | A1 (AO1.1b) | Condone given to 1 s.f. CAO |
| \(\chi^2\) cv for 2 df \(= 4.605\) \((p = 0.215)\); \(3.07 < 4.605\) \((0.215 > 0.1)\); Accept \(H_0\) | B1F (AO1.1b) | States critical value (or p-value, follow through their \(\chi^2\) value). Condone \(\chi^2\) cv for 3 df \(= 6.251\) if 2 and 3 or more are not combined |
| Evaluates \(\chi^2\)-test statistic by comparing cv with their ts (or their p value with 0.1) | R1 (AO3.5a) | No significant evidence to suggest that there is an association between number of claims and type of insurance policy |
| Infers \(H_0\) not rejected | E1 (AO2.2b) | |
| Concludes in context (not definite) | E1 (AO3.2a) | Accept equivalent wording |
| Total | 10 | |
| TOTAL | 40 |
# Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| One correct missing expected value, e.g. 0 and car $= \frac{28 \times 35}{80}$ | M1 (AO1.1a) | |
| All expected values correct: $C$: 12.25, 10.0625, 8.3125, 4.375; $MB$: 15.75, 12.9375, 10.6875, 5.625 | A1 (AO1.1b) | Must not be rounded CAO. Condone 12.937 for 12.9375 |
---
# Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: There is no association between number of claims and type of insurance policy; $H_1$: There is an association between number of claims and type of insurance policy | B1 (AO2.5) | States both hypotheses using correct language. Accept equivalent wording |
| Combines expected and observed values for 2 and 3 or more correctly: $C$: Observed 16, Expected 12.6875; $MB$: Observed 13, Expected 16.3125 | B1F (AO1.1b) | Follow through their answers to part (a) |
| $\sum \frac{(O-E)^2}{E} = \frac{(9-12.25)^2}{12.25} + \frac{(19-15.75)^2}{15.75} + \frac{(10-10.0625)^2}{10.0625} + \frac{(13-12.9375)^2}{12.9375} + \frac{(16-12.6875)^2}{12.6875} + \frac{(13-16.3125)^2}{16.3125}$ | M1 (AO1.1a) | Can be awarded even if 2 and 3 or more are not combined |
| $= 3.07$ | A1 (AO1.1b) | Condone given to 1 s.f. CAO |
| $\chi^2$ cv for 2 df $= 4.605$ $(p = 0.215)$; $3.07 < 4.605$ $(0.215 > 0.1)$; Accept $H_0$ | B1F (AO1.1b) | States critical value (or p-value, follow through their $\chi^2$ value). Condone $\chi^2$ cv for 3 df $= 6.251$ if 2 and 3 or more are not combined |
| Evaluates $\chi^2$-test statistic by comparing cv with their ts (or their p value with 0.1) | R1 (AO3.5a) | No significant evidence to suggest that there is an association between number of claims and type of insurance policy |
| Infers $H_0$ not rejected | E1 (AO2.2b) | |
| Concludes in context (not definite) | E1 (AO3.2a) | Accept equivalent wording |
| **Total** | **10** | |
| **TOTAL** | **40** | |8 An insurance company groups its vehicle insurance policies into two categories, car insurance and motorbike insurance.
The number of claims in a random sample of 80 policies was monitored and the results summarised in contingency Table 1.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 1}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{5}{|c|}{Number of claims} \\
\hline
& & 0 & 1 & 2 & 3 or more & Total \\
\hline
\multirow[b]{3}{*}{Type of insurance policy} & Car & 9 & 10 & 11 & 5 & 35 \\
\hline
& Motorbike & 19 & 13 & 8 & 5 & 45 \\
\hline
& Total & 28 & 23 & 19 & 10 & 80 \\
\hline
\end{tabular}
\end{center}
\end{table}
The insurance company decides to carry out a $\chi ^ { 2 }$-test for association between number of claims and type of insurance policy using the information given in Table 1.
8
\begin{enumerate}[label=(\alph*)]
\item The contingency table shown in Table 2 gives some of the exact expected frequencies for this test.
Complete Table 2 with the missing exact expected values.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 2}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{4}{|c|}{Number of claims} \\
\hline
& & 0 & 1 & 2 & 3 or more \\
\hline
\multirow{2}{*}{Type of insurance policy} & Car & & 10.0625 & & 4.375 \\
\hline
& Motorbike & & & 10.6875 & \\
\hline
\end{tabular}
\end{center}
\end{table}
8
\item Carry out the insurance company's test, using the $10 \%$ level of significance.\\
\includegraphics[max width=\textwidth, alt={}, center]{313cd5ce-07ff-4781-a134-565b8b221145-12_2488_1719_219_150}
Additional page, if required.\\
Write the question numbers in the left-hand margin.
Additional page, if required.\\
Write the question numbers in the left-hand margin.
Additional page, if required.\\
Write the question numbers in the left-hand margin.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2018 Q8 [10]}}