| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Wilcoxon rank-sum test (Mann-Whitney U test) |
| Difficulty | Standard +0.3 This is a straightforward application of the Wilcoxon rank-sum test with clear data, standard hypotheses, and a provided critical value table. Students must rank combined data, sum ranks for one group, and compare to the table—all routine procedures for Further Statistics. The question requires no conceptual insight beyond following the standard algorithm, making it slightly easier than average for A-level Further Maths. |
| Spec | 5.07d Paired vs two-sample: selection |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Ranking table: X values ranked 1–9, Y values ranked 1–10 (all 19 values ranked together): \(X\): 20.8→1, 21.3→3, 21.4→4, 21.5→5, 22.3→7, 22.7→8, 22.9→9, 24.6→15, 25.3→18; \(Y\): 20.9→2, 21.9→6, 23.4→10, 23.5→11, 23.8→12, 24.0→13, 24.1→14, 24.8→16, 25.1→17, 25.8→19 | M1 | Attempt at ranking from 1 to 19. |
| All ranks correct as shown | A1 | A1 for all correct. |
| Test statistic: 70 | A1 | If ranks reversed, then sum of ranks = 110, and then \(9(9+10+1)-110=70\). |
| \(H_0\): population median of \(X\) = population median of \(Y\); \(H_1\): population median of \(X\) < population median of \(Y\) | B1 | Accept \(H_0: m_x = m_y\), \(H_1: m_x < m_y\). Do not accept any other notation. |
| Critical value for \(m=9\), \(n=10\) is 61 | B1 | Using normal approximation implies last 3 marks are B0M0A0, max 4 out of 7. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(70 > 61\), accept \(H_0\) | M1 | '70' must come from ranks 1–19, '61' must be a critical value from the correct table. ft conclusion for their 70 and their 61. Condone Reject \(H_1\). 'Accept \(H_0\)' can be implied by a conclusion in context consistent with their 70 and their 61. |
| Insufficient evidence to suggest that population medians of \(X\) and \(Y\) are not equal or Insufficient evidence to suggest that population median of \(X\) is less than median of \(Y\) or Insufficient evidence to support the interviewer's belief | A1 | Correct work only, ignoring their hypotheses, conclusion in context with level of uncertainty in language. Not 'prove' or equivalent. Condone 'no sufficient' 'not enough'. Do not accept statements such as 'there is sufficient evidence to suggest….'. |
| Total: 7 |
## Question 1:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Ranking table: X values ranked 1–9, Y values ranked 1–10 (all 19 values ranked together): $X$: 20.8→1, 21.3→3, 21.4→4, 21.5→5, 22.3→7, 22.7→8, 22.9→9, 24.6→15, 25.3→18; $Y$: 20.9→2, 21.9→6, 23.4→10, 23.5→11, 23.8→12, 24.0→13, 24.1→14, 24.8→16, 25.1→17, 25.8→19 | M1 | Attempt at ranking from 1 to 19. |
| All ranks correct as shown | A1 | A1 for all correct. |
| Test statistic: 70 | A1 | If ranks reversed, then sum of ranks = 110, and then $9(9+10+1)-110=70$. |
| $H_0$: population median of $X$ = population median of $Y$; $H_1$: population median of $X$ < population median of $Y$ | B1 | Accept $H_0: m_x = m_y$, $H_1: m_x < m_y$. Do not accept any other notation. |
| Critical value for $m=9$, $n=10$ is 61 | B1 | Using normal approximation implies last 3 marks are B0M0A0, max 4 out of 7. |
# Question 1:
| Answer | Mark | Guidance |
|--------|------|----------|
| $70 > 61$, accept $H_0$ | M1 | '70' must come from ranks 1–19, '61' must be a critical value from the correct table. ft conclusion for their 70 and their 61. Condone Reject $H_1$. 'Accept $H_0$' can be implied by a conclusion in context consistent with their 70 and their 61. |
| Insufficient evidence to suggest that population medians of $X$ and $Y$ are not equal **or** Insufficient evidence to suggest that population median of $X$ is less than median of $Y$ **or** Insufficient evidence to support the interviewer's belief | A1 | Correct work only, ignoring their hypotheses, conclusion in context with level of uncertainty in language. Not 'prove' or equivalent. Condone 'no sufficient' 'not enough'. Do not accept statements such as 'there is sufficient evidence to suggest….'. |
| **Total: 7** | | |
---1 A college uses two assessments, $X$ and $Y$, when interviewing applicants for research posts at the college. These assessments have been used for a large number of applicants this year.
The scores for a random sample of 9 applicants who took assessment $X$ are as follows.
$$\begin{array} { l l l l l l l l l }
21.4 & 24.6 & 25.3 & 22.7 & 20.8 & 21.5 & 22.9 & 21.3 & 22.3
\end{array}$$
The scores for a random sample of 10 applicants who took assessment $Y$ are as follows.
$$\begin{array} { l l l l l l l l l l }
20.9 & 23.5 & 24.8 & 21.9 & 23.4 & 24.0 & 23.8 & 24.1 & 25.1 & 25.8
\end{array}$$
The interviewer believes that the population median score from assessment $X$ is lower than the population median score from assessment $Y$.
Carry out a Wilcoxon rank-sum test, at the $1 \%$ significance level, to test whether the interviewer's belief is supported by the data.\\
\includegraphics[max width=\textwidth, alt={}, center]{b5ff998a-fcb6-4a1b-ae86-ec66b0dccc3c-02_2714_37_143_2008}
\begin{center}
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\hfill \mbox{\textit{CAIE Further Paper 4 2024 Q1 [7]}}