| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| 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 and standard procedure. Students must rank combined data, calculate test statistic, and compare to critical value. The justification part requires understanding when non-parametric tests are appropriate. While it involves careful ranking of 22 values, it's a routine S4 question with no conceptual surprises—slightly easier than average A-level maths due to its procedural nature. |
| Spec | 5.07b Sign test: and Wilcoxon signed-rank |
| Group \(A\) | 8.6 | 9.4 | 9.7 | 9.8 | 10.1 | 10.5 | 11.0 | 11.2 | 11.8 | 12.7 |
| Group \(B\) | 7.4 | 8.4 | 8.5 | 8.8 | 9.2 | 9.3 | 9.5 | 9.9 | 10.0 | 11.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Distribution of heights may not be normal/is unknown | B1 | Allow "No assumption required", but nothing else. Not "groups independent" unless something else as well |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: m_A = m_B\), \(H_1: m_A \neq m_B\) | B1 | Medians. Allow words in context. Not \(\mu\) unless "median" stated |
| A: 4, 8, 10, 11, 14, 15, 16, 18, 20, 21, 22; B: 1, 2, 3, 5, 6, 7, 9, 12, 13, 17, 19 | B1 | |
| \(m = n = 11\), \(R_m = 159\) or 94 | B1 | |
| Use normal approximation with mean \(126.5\ [=253/2]\) | M1 | allow \(\frac{1}{2}\times 11\times(11+11+1)\) |
| Variance \(231.92\ [=2783/12]\) | B1 | allow \(\frac{1}{12}\times 11\times 11\times(11+11+1)\) |
| \((\alpha)\ P(\leq 94) = \Phi((94.5-126.5)/\sqrt{231.92})\) or \(P(\geq 159) = 0.0178\); \(< 0.025\) and reject \(H_0\) | M1, A1, M1 | Standardising. Allow no/incorrect cc. Value. ft TS |
| \((\beta)\ z = (94.5-126.5)/\sqrt{231.92} = -2.101\); \(< -1.96\) so reject \(H_0\) | M1A1, M1 | Standardising; value. ft TS |
| There is evidence that salinity affects growth | A1 | Or equivalent in context. ft TS |
| [9] |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distribution of heights may not be normal/is unknown | B1 | Allow "No assumption required", but nothing else. Not "groups independent" unless something else as well |
| **[1]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: m_A = m_B$, $H_1: m_A \neq m_B$ | B1 | Medians. Allow words in context. Not $\mu$ unless "median" stated |
| A: 4, 8, 10, 11, 14, 15, 16, 18, 20, 21, 22; B: 1, 2, 3, 5, 6, 7, 9, 12, 13, 17, 19 | B1 | |
| $m = n = 11$, $R_m = 159$ or 94 | B1 | |
| Use normal approximation with mean $126.5\ [=253/2]$ | M1 | allow $\frac{1}{2}\times 11\times(11+11+1)$ |
| Variance $231.92\ [=2783/12]$ | B1 | allow $\frac{1}{12}\times 11\times 11\times(11+11+1)$ |
| $(\alpha)\ P(\leq 94) = \Phi((94.5-126.5)/\sqrt{231.92})$ or $P(\geq 159) = 0.0178$; $< 0.025$ and reject $H_0$ | M1, A1, M1 | Standardising. Allow no/incorrect cc. Value. ft TS |
| $(\beta)\ z = (94.5-126.5)/\sqrt{231.92} = -2.101$; $< -1.96$ so reject $H_0$ | M1A1, M1 | Standardising; value. ft TS |
| There is evidence that salinity affects growth | A1 | Or equivalent in context. ft TS |
| **[9]** | | |
---4 The effect of water salinity on the growth of a type of grass was studied by a biologist. A random sample of 22 seedlings was divided into two groups $A$ and $B$, each of size 11 .\\
Group $A$ was treated with water of $0 \%$ salinity and group $B$ was treated with water of $0.5 \%$ salinity. After three weeks the height (in cm) of each seedling was measured with the following results, which are ordered for convenience.
\begin{center}
\begin{tabular}{ r r r r r r r r r r r }
Group $A$ & 8.6 & 9.4 & 9.7 & 9.8 & 10.1 & 10.5 & 11.0 & 11.2 & 11.8 & 12.7 \\
Group $B$ & 7.4 & 8.4 & 8.5 & 8.8 & 9.2 & 9.3 & 9.5 & 9.9 & 10.0 & 11.1 \\
\end{tabular}
\end{center}
Jeffery was asked to test whether the two treatments resulted, on average, in a difference in growth. He chose the Wilcoxon rank sum test.\\
(i) Justify Jeffery's choice of test.\\
(ii) Carry out the test at the $5 \%$ significance level.
\hfill \mbox{\textit{OCR S4 2013 Q4 [10]}}