| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| 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 clearly given test statistic (W=867) and sample size (n=50). Students need to state the symmetry assumption, compare to critical values from tables, and recall the sign test as an alternative. The calculation is minimal and the procedure is standard textbook material, making it slightly easier than average for an S4 question. |
| Spec | 5.07a Non-parametric tests: when to use5.07b Sign test: and Wilcoxon signed-rank |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Assumes salaries symmetrically distributed | B1 | In context |
| \(H_0\): \(m(\text{median}) = 19.5\), \(H_1\): \(m(\text{median}) \neq 19.5\) | B1 | For both; not \(\mu\); accept words |
| \(P = 867\) (or 408) | ||
| Using normal approximation | M1 | |
| \(\mu = \frac{1}{4} \times 50 \times 51 = 637.5\) | A1 | |
| \(\sigma^2 = 50 \times 51 \times 101/24 = 10731.25\) | A1 | |
| \(z = (a - 637.5)/\sqrt{10731.25}\) | M1 | \(a = 866.5, 867, 867.5\) (or \(408.5, 408, 407.5\)) |
| Use \(a = 866.5\) | A1 | |
| \(= 2.211\), or \(2.215\) or \(2.220\) (from 408) | A1 | |
| Compare their \(z\) with \(1.96\) and reject \(H_0\) | M1 | Or \(p\)-value rounding to \(0.026\) or \(0.027\) |
| There is sufficient evidence at the 5% SL that the median salary differs from £19 500 | A1ft [10] | Compare with 0.05 or equivalent; ft \(z\). Or find critical region |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use sign test when salary distribution is skewed | B1 [1] |
## Question 5:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Assumes salaries symmetrically distributed | B1 | In context |
| $H_0$: $m(\text{median}) = 19.5$, $H_1$: $m(\text{median}) \neq 19.5$ | B1 | For both; not $\mu$; accept words |
| $P = 867$ (or 408) | | |
| Using normal approximation | M1 | |
| $\mu = \frac{1}{4} \times 50 \times 51 = 637.5$ | A1 | |
| $\sigma^2 = 50 \times 51 \times 101/24 = 10731.25$ | A1 | |
| $z = (a - 637.5)/\sqrt{10731.25}$ | M1 | $a = 866.5, 867, 867.5$ (or $408.5, 408, 407.5$) |
| Use $a = 866.5$ | A1 | |
| $= 2.211$, or $2.215$ or $2.220$ (from 408) | A1 | |
| Compare their $z$ with $1.96$ and reject $H_0$ | M1 | Or $p$-value rounding to $0.026$ or $0.027$ |
| There is sufficient evidence at the 5% SL that the median salary differs from £19 500 | A1ft **[10]** | Compare with 0.05 or equivalent; ft $z$. Or find critical region |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use sign test when salary distribution is skewed | B1 **[1]** | |
---5 In order to test whether the median salary of employees in a certain industry who had worked for three years was $\pounds 19500$, the salaries $x$, in thousands of pounds, of 50 randomly chosen employees were obtained.\\
(i) The values $| x - 19.5 |$ were calculated and ranked. No two values of $x$ were identical and none was equal to 19.5 . The sum of the ranks corresponding to positive values of $( x - 19.5 )$ was 867. Stating a required assumption, carry out a suitable test at the $5 \%$ significance level.\\
(ii) If the assumption you stated in part (i) does not hold, what test could have been used?
\hfill \mbox{\textit{OCR S4 2010 Q5 [11]}}