| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Wilcoxon matched-pairs signed-rank test |
| Difficulty | Standard +0.3 This is a straightforward application of the Wilcoxon matched-pairs signed-rank test with clear paired data. Students must calculate differences, rank absolute values, sum positive/negative ranks, and compare to critical values from tables. The procedure is standard with no conceptual complications, though it requires careful arithmetic. Part (ii) tests basic understanding of when to use matched-pairs vs independent samples tests—a routine conceptual check. Slightly easier than average due to being a textbook application of a prescribed procedure. |
| Spec | 5.07b Sign test: and Wilcoxon signed-rank |
| Species | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| Number of flowers at low altitude | 5 | 3 | 4 | 7 | 2 | 9 | 6 | 5 | 4 | 11 | 2 |
| Number of flowers at high altitude | 1 | 6 | 10 | 8 | 14 | 16 | 20 | 21 | 15 | 2 | 12 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: m_d = 0\), \(H_1: m_d > 0\) (where \(d\) = high \(-\) low) | B1 | Or \(H_0: m_H = m_L\) etc. Medians |
| \(D\): \(-4\ 3\ 6\ 1\ 12\ 7\ 14\ 16\ 11\ -9\ 10\) | M1 | |
| Rank: \(-3\ 2\ 4\ 1\ \ 9\ 5\ 10\ 11\ 8\ -6\ 7\) | A1 | Ranking top down, \(-9, -10, 8\), ..M1A0 |
| \(P = 57\), \(Q = 9\) | B1 | \(T = 15\) B0 |
| \(T = 9\) | [SR last 3 marks: \(z = -2\); 09 B1 | |
| \(CV = 13\) | B1 | \(< -1.96\) etc M1A1] |
| \(9 < CV\) so reject \(H_0\) | M1 | Or equivalent |
| There is sufficient evidence at the 5% significance level to support the botanist's belief | A1 ft 7 | ft T |
| [8] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| The rank sum test is for independent samples; the \(H\) and \(L\) values are correlated | B1 1 | Accept data paired |
## Question 2:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: m_d = 0$, $H_1: m_d > 0$ (where $d$ = high $-$ low) | B1 | Or $H_0: m_H = m_L$ etc. Medians |
| $D$: $-4\ 3\ 6\ 1\ 12\ 7\ 14\ 16\ 11\ -9\ 10$ | M1 | |
| Rank: $-3\ 2\ 4\ 1\ \ 9\ 5\ 10\ 11\ 8\ -6\ 7$ | A1 | Ranking top down, $-9, -10, 8$, ..M1A0 |
| $P = 57$, $Q = 9$ | B1 | $T = 15$ B0 |
| $T = 9$ | | [SR last 3 marks: $z = -2$; 09 B1 |
| $CV = 13$ | B1 | $< -1.96$ etc M1A1] |
| $9 < CV$ so reject $H_0$ | M1 | Or equivalent |
| There is sufficient evidence at the 5% significance level to support the botanist's belief | A1 ft **7** | ft T |
| | **[8]** | |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| The rank sum test is for independent samples; the $H$ and $L$ values are correlated | B1 **1** | Accept data paired |
---2 A botanist believes that some species of plants produce more flowers at high altitudes than at low altitudes. In order to investigate this belief the botanist randomly samples 11 species of plants each of which occurs at both altitudes. The numbers of flowers on the plants are shown in the table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | c | c | }
\hline
Species & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\
\hline
Number of flowers at low altitude & 5 & 3 & 4 & 7 & 2 & 9 & 6 & 5 & 4 & 11 & 2 \\
Number of flowers at high altitude & 1 & 6 & 10 & 8 & 14 & 16 & 20 & 21 & 15 & 2 & 12 \\
\hline
\end{tabular}
\end{center}
(i) Use the Wilcoxon signed rank test at the 5\% significance level to test the botanist's belief.\\
(ii) Explain why the Wilcoxon rank sum test should not be used for this test.
\hfill \mbox{\textit{OCR S4 2011 Q2 [8]}}