| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Standard 2×3 contingency table |
| Difficulty | Moderate -0.3 This is a standard chi-squared test of independence with a 2×3 contingency table requiring routine calculation of expected frequencies, test statistic, and comparison with critical value. While it involves multiple computational steps, it follows a completely standard procedure with no conceptual challenges or novel insights required, making it slightly easier than average for A-level. |
| Spec | 5.06a Chi-squared: contingency tables |
| Reliability | ||||
| \cline { 3 - 5 } \multicolumn{2}{|c|}{} | Good | Fair | Poor | |
| \multirow{2}{*}{Supplier} | \(A\) | 65 | 63 | 33 |
| \cline { 2 - 5 } | \(B\) | 51 | 44 | 44 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0\): Reliability is independent | B1 | State null hypothesis |
| Expected values: 62.253, 57.423, 41.323, 53.747, 49.577, 35.677 | M1 A1 | Find expected values to 1 d.p. (lose A1 if rounded to integers) |
| \(\chi^2 = 0.1212 + 0.5416 + 1.6764 + 0.1404 + 0.6274 + 1.9417\) | Calculate value of \(\chi^2\) | |
| \(= 5.05\) | M1 A1 | (allow 5.03 if 1 d.p. expected values used) |
| \(\chi^2_{2,\ 0.95} = 5.99\) [1] | B1 | State or use correct tabular value |
| Accept \(H_0\) if \(\chi^2 <\) tabular value | M1 | Valid method for conclusion |
| Reliability is independent of supplier | A1 | Correct conclusion from correct values |
## Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0$: Reliability is independent | B1 | State null hypothesis |
| Expected values: 62.253, 57.423, 41.323, 53.747, 49.577, 35.677 | M1 A1 | Find expected values to 1 d.p. (lose A1 if rounded to integers) |
| $\chi^2 = 0.1212 + 0.5416 + 1.6764 + 0.1404 + 0.6274 + 1.9417$ | | Calculate value of $\chi^2$ |
| $= 5.05$ | M1 A1 | (allow 5.03 if 1 d.p. expected values used) |
| $\chi^2_{2,\ 0.95} = 5.99$ [1] | B1 | State or use correct tabular value |
| Accept $H_0$ if $\chi^2 <$ tabular value | M1 | Valid method for conclusion |
| Reliability is independent of supplier | A1 | Correct conclusion from correct values |
**Total: 8**
---6 The reliability of the broadband connection received from two suppliers, $A$ and $B$, is classified as good, fair or poor by a random sample of householders. The information collected is summarised in the following table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
\multicolumn{2}{|c|}{} & \multicolumn{3}{|c|}{Reliability} \\
\cline { 3 - 5 }
\multicolumn{2}{|c|}{} & Good & Fair & Poor \\
\hline
\multirow{2}{*}{Supplier} & $A$ & 65 & 63 & 33 \\
\cline { 2 - 5 }
& $B$ & 51 & 44 & 44 \\
\hline
\end{tabular}
\end{center}
Test, at the 5\% significance level, whether reliability is independent of supplier.
\hfill \mbox{\textit{CAIE FP2 2015 Q6 [8]}}