| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Wilcoxon signed-rank test (single sample) |
| Difficulty | Standard +0.3 This is a straightforward application of the Wilcoxon signed-rank test with clear instructions. Students must calculate differences from the median (18.0), rank absolute differences, sum ranks, and compare to critical values. While it requires careful arithmetic and knowledge of the procedure, it's a standard textbook exercise with no conceptual challenges or novel problem-solving required. |
| Spec | 5.07b Sign test: and Wilcoxon signed-rank |
| 18.2 | 17.5 | 16.4 | 15.1 | 20.5 | 26.5 | 19.2 | 23.2 |
| 17.9 | 18.8 | 25.8 | 19.9 | 17.7 | 16.2 | 17.3 | 16.6 |
| 17.1 | 20.1 | 20.3 | 12.6 | 16.0 | 21.4 | 22.7 | 18.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Differences/rank table with values: \(0.2, -0.5, -1.6, -2.9, 2.5, 8.5, 1.2, 5.2\) and ranks \(2, -5, -11, -18, 17, 24, 9, 21\) and \(-0.1, 0.8, 7.8, 1.9, -0.3, -1.8, -0.7, -1.4\) and ranks \(-1, 7, 23, 13, -3, -12, -6, -10\) and \(-0.9, 2.1, 2.3, -5.4, -2.0, 3.4, 4.7, 0.4\) and ranks \(-8, 15, 16, -22, -14, 19, 20, 4\) | M1 A1 | Attempt at differences, allow 4 errors |
| \(Q = 110\), \(P = 190\) | A1 | Either sum correct |
| \(H_0\): population median \(= 18.0\) and \(H_1\): population median \(\neq 18.0\) | B1 | Must be 'population', allow \(m\) |
| Mean \(= \frac{1}{4}n(n+1) = \frac{1}{4} \times 24 \times 25 = 150\) | B1 | Normal approximation: mean |
| Variance \(= \frac{1}{24}n(n+1)(2n+1) = \frac{1}{24} \times 24 \times 25 \times 49 = 1225\) | B1 | |
| \(\dfrac{110.5 - 150}{\sqrt{1225}}\) | M1 | Attempt at test statistic, allow incorrect or missing cc, FT *their* 110 |
| \(-1.129\) | A1 | |
| \(10\%\) 2-tail, critical value is \(1.645\); \(-1.129 > -1.645\) or \(0.1296 > 0.05\), accept \(H_0\) | M1 | Compare with \(-1.645\), correct FT conclusion |
| Insufficient evidence to suggest that median is not \(18.0\) / Insufficient evidence to reject teacher's claim | A1 | Correct conclusion, in context. Level of uncertainty in language is used |
| Total | 10 |
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differences/rank table with values: $0.2, -0.5, -1.6, -2.9, 2.5, 8.5, 1.2, 5.2$ and ranks $2, -5, -11, -18, 17, 24, 9, 21$ and $-0.1, 0.8, 7.8, 1.9, -0.3, -1.8, -0.7, -1.4$ and ranks $-1, 7, 23, 13, -3, -12, -6, -10$ and $-0.9, 2.1, 2.3, -5.4, -2.0, 3.4, 4.7, 0.4$ and ranks $-8, 15, 16, -22, -14, 19, 20, 4$ | M1 A1 | Attempt at differences, allow 4 errors |
| $Q = 110$, $P = 190$ | A1 | Either sum correct |
| $H_0$: population median $= 18.0$ and $H_1$: population median $\neq 18.0$ | B1 | Must be 'population', allow $m$ |
| Mean $= \frac{1}{4}n(n+1) = \frac{1}{4} \times 24 \times 25 = 150$ | B1 | Normal approximation: mean |
| Variance $= \frac{1}{24}n(n+1)(2n+1) = \frac{1}{24} \times 24 \times 25 \times 49 = 1225$ | B1 | |
| $\dfrac{110.5 - 150}{\sqrt{1225}}$ | M1 | Attempt at test statistic, allow incorrect or missing cc, FT *their* 110 |
| $-1.129$ | A1 | |
| $10\%$ 2-tail, critical value is $1.645$; $-1.129 > -1.645$ or $0.1296 > 0.05$, accept $H_0$ | M1 | Compare with $-1.645$, correct FT conclusion |
| Insufficient evidence to suggest that median is not $18.0$ / Insufficient evidence to reject teacher's claim | A1 | Correct conclusion, in context. Level of uncertainty in language is used |
| **Total** | **10** | |6 A teacher at a large college gave a mathematical puzzle to all the students. The median time taken by a random sample of 24 students to complete the puzzle was 18.0 minutes. The students were then given practice in solving puzzles. Two weeks later, the students were given another mathematical puzzle of the same type as the first. The times, in minutes, taken by the random sample of 24 students to complete this puzzle are as follows.
\begin{center}
\begin{tabular}{ l l l l l l l l }
18.2 & 17.5 & 16.4 & 15.1 & 20.5 & 26.5 & 19.2 & 23.2 \\
17.9 & 18.8 & 25.8 & 19.9 & 17.7 & 16.2 & 17.3 & 16.6 \\
17.1 & 20.1 & 20.3 & 12.6 & 16.0 & 21.4 & 22.7 & 18.4 \\
\end{tabular}
\end{center}
The teacher claims that the practice has not made any difference to the average time taken to complete a puzzle of this type.
Carry out a Wilcoxon signed-rank test, at the 10\% significance level, to test whether there is sufficient evidence to reject the teacher's claim.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.\\
\hfill \mbox{\textit{CAIE Further Paper 4 2022 Q6 [10]}}