| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Standard 3×3 contingency table |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with a 3×3 contingency table. Students must calculate expected frequencies, compute the test statistic, find degrees of freedom (4), and compare to critical value. While it requires careful arithmetic and understanding of the test procedure, it follows a completely routine algorithm with no conceptual challenges or novel insights required—making it slightly easier than average for Further Maths. |
| Spec | 5.06a Chi-squared: contingency tables |
| \cline { 3 - 5 } \multicolumn{2}{c|}{} | Colour of car | |||
| \cline { 3 - 5 } \multicolumn{2}{c|}{} | Silver | Blue | Red | |
| \multirow{3}{*}{Type of car} | Hatchback | 53 | 36 | 41 |
| \cline { 2 - 5 } | Saloon | 29 | 40 | 31 |
| \cline { 2 - 5 } | Estate | 28 | 24 | 18 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): Colour is independent of type, or no association between colour and type | B1 (AEF) | State (at least) null hypothesis in full |
| \(E_i\): \(47.67\ \ 43.33\ \ 39\) / \(36.67\ \ 33.33\ \ 30\) / \(25.67\ \ 23.33\ \ 21\) (to 3 s.f.) | M1 A1 | Find expected values \(E_i\) (A0 if rounded to integers) |
| \(X^2 = 0.597 + 1.241 + 0.103 + 1.603 + 1.333 + 0.033 + 0.212 + 0.019 + 0.429 = 5.57\) (to 3 s.f.) | M1 A1 | Find value of \(\chi^2\) from \(\sum(E_i - O_i)^2/E_i\) [or \(\sum O_i^2/E_i - n\)] (allow 5.64 for this A1 if integer values of \(E_i\) used) |
| \(\chi^2_{4,\,0.9} = 7.779\) or \(7.78\) | B1 | State or use correct tabular \(\chi^2\) value |
| Accept \(H_0\) if \(X^2 <\) tabular value | M1 (AEF) | Valid method for reaching conclusion |
| \(5.57\ [\pm 0.01] < 7.78\), so independent or no association | A1 (AEF) | Correct (abbreviated) conclusion, from approx. correct values |
## Question 8:
| $H_0$: Colour is independent of type, or no association between colour and type | B1 (AEF) | State (at least) null hypothesis in full |
|---|---|---|
| $E_i$: $47.67\ \ 43.33\ \ 39$ / $36.67\ \ 33.33\ \ 30$ / $25.67\ \ 23.33\ \ 21$ (to 3 s.f.) | M1 A1 | Find expected values $E_i$ (A0 if rounded to integers) |
| $X^2 = 0.597 + 1.241 + 0.103 + 1.603 + 1.333 + 0.033 + 0.212 + 0.019 + 0.429 = 5.57$ (to 3 s.f.) | M1 A1 | Find value of $\chi^2$ from $\sum(E_i - O_i)^2/E_i$ [or $\sum O_i^2/E_i - n$] (allow 5.64 for this A1 if integer values of $E_i$ used) |
| $\chi^2_{4,\,0.9} = 7.779$ or $7.78$ | B1 | State or use correct tabular $\chi^2$ value |
| Accept $H_0$ if $X^2 <$ tabular value | M1 (AEF) | Valid method for reaching conclusion |
| $5.57\ [\pm 0.01] < 7.78$, so independent or no association | A1 (AEF) | Correct (abbreviated) conclusion, from approx. correct values |
---8 A manufacturer produces three types of car: hatchbacks, saloons and estates. Each type of car is available in one of three colours: silver, blue and red. The manufacturer wants to know whether the popularity of the colour of the car is related to the type of car. A random sample of 300 cars chosen by customers gives the information summarised in the following table.
\begin{center}
\begin{tabular}{ | l | l | c | c | c | }
\cline { 3 - 5 }
\multicolumn{2}{c|}{} & \multicolumn{3}{|c|}{Colour of car} \\
\cline { 3 - 5 }
\multicolumn{2}{c|}{} & Silver & Blue & Red \\
\hline
\multirow{3}{*}{Type of car} & Hatchback & 53 & 36 & 41 \\
\cline { 2 - 5 }
& Saloon & 29 & 40 & 31 \\
\cline { 2 - 5 }
& Estate & 28 & 24 & 18 \\
\hline
\end{tabular}
\end{center}
Test at the $10 \%$ significance level whether the colour of car chosen by customers is independent of the type of car.\\
\hfill \mbox{\textit{CAIE FP2 2018 Q8 [8]}}