| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Standard 2×3 contingency table |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with a 2×3 contingency table. Students must calculate expected frequencies, compute the test statistic, find critical value from tables, and state conclusion. While it requires multiple computational steps, it follows a completely routine procedure with no conceptual challenges or novel elements—slightly easier than average due to its algorithmic nature. |
| Spec | 5.06a Chi-squared: contingency tables |
| \cline { 3 - 5 } \multicolumn{2}{c|}{} | Method of voting | |||
| \cline { 3 - 5 } \multicolumn{2}{c|}{} | Post | Text | ||
| \multirow{2}{*}{Gender} | Male | 10 | 12 | 38 |
| \cline { 2 - 5 } | Female | 5 | 21 | 14 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): No association or method is independent of gender (AEF) | B1 | State (at least) null hypothesis |
| \(E_i\): \(9.0\ 19.8\ 31.2\ 6.0\ 13.2\ 20.8\) | M1 A1 | Find expected values \(E_i\) (A0 if rounded to integers) |
| \(\chi^2 = 0.111 + 3.073 + 1.482 + 0.167 + 4.610 + 2.223 = 11.7\) (or \(12.3\)) (to 2 d.p.) | 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 \(12.3\) if integer values of \(E_i\) used) |
| \(\chi^2_{2,\,0.99} = 9.21\) | B1 | State or use correct tabular \(\chi^2\) value |
| Reject \(H_0\) if \(\chi^2 >\) tabular value (AEF) | M1 | Valid method for reaching conclusion |
| \(11.7\ [\pm 0.1] > 9.21\) so there is an association | A1 | Correct conclusion, from correct values |
## Question 8:
$H_0$: No association or method is independent of gender (AEF) | B1 | State (at least) null hypothesis
$E_i$: $9.0\ 19.8\ 31.2\ 6.0\ 13.2\ 20.8$ | M1 A1 | Find expected values $E_i$ (**A0** if rounded to integers)
$\chi^2 = 0.111 + 3.073 + 1.482 + 0.167 + 4.610 + 2.223 = 11.7$ (or $12.3$) (to 2 d.p.) | 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 $12.3$ if integer values of $E_i$ used)
$\chi^2_{2,\,0.99} = 9.21$ | B1 | State or use correct tabular $\chi^2$ value
Reject $H_0$ if $\chi^2 >$ tabular value (AEF) | M1 | Valid method for reaching conclusion
$11.7\ [\pm 0.1] > 9.21$ so there is an association | A1 | Correct conclusion, from correct values
**Total: 8**
---8 Members of a Statistics club are voting to elect a new president of the club. Members must choose to vote either by post or by text or by email. The method of voting chosen by a random sample of 60 male members and 40 female members is given in the following table.
\begin{center}
\begin{tabular}{ | c | l | c | c | c | }
\cline { 3 - 5 }
\multicolumn{2}{c|}{} & \multicolumn{3}{|c|}{Method of voting} \\
\cline { 3 - 5 }
\multicolumn{2}{c|}{} & Post & Text & Email \\
\hline
\multirow{2}{*}{Gender} & Male & 10 & 12 & 38 \\
\cline { 2 - 5 }
& Female & 5 & 21 & 14 \\
\hline
\end{tabular}
\end{center}
Test, at the $1 \%$ significance level, whether there is an association between method of voting and gender.\\
\hfill \mbox{\textit{CAIE FP2 2017 Q8 [8]}}