| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | June |
| Marks | 18 |
| 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 steps: calculate differences from the median, rank absolute differences, sum ranks, and compare to critical values. Part (ii) is routine confidence interval calculation with large sample justification, and part (iii) requires only basic conceptual understanding. Slightly above average difficulty due to the non-parametric test context, but all steps are mechanical with no novel problem-solving required. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution5.07b Sign test: and Wilcoxon signed-rank |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: m = 7.4\), \(H_1: m < 7.4\) where \(m\) is the population median time | B1 | Both. Accept hypotheses in words, but must include "population". Do NOT allow symbols other than \(m\) unless clearly stated to be a population median |
| Adequate definition of \(m\) to include "population" | B1 | |
| Subtract 7.4 to get differences; ranks of \ | diff\ | all correct (1–12) |
| \(W_+ = 2+4+5+7 = 18\) | B1 | \(W_- = 1+3+6+8+9+10+11+12 = 60\) |
| Refer to Wilcoxon single sample tables for \(n=12\) | M1 | No ft from here if wrong |
| Lower 5% point is 17 (or upper is 61 if 60 used) | A1 | 1-tail test. No ft from here if wrong |
| Result is not significant | A1 | ft only candidate's test statistic |
| Insufficient evidence to suggest that the median time has been reduced | A1 | ft only candidate's test statistic. Conclusion in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\bar{x} = 6.94\), \(s = 0.37\) | B1 | Accept \(s^2 = 0.1369\). Beware use of msd (0.13518875) or rmsd (0.3676(8)). Do not allow here or below |
| CI is given by \(6.94 \pm\) | M1 | ft candidate's \(\bar{x} \pm\) |
| \(1.96\) | B1 | 1.96 seen |
| \(\times \dfrac{0.37}{\sqrt{80}}\) | M1 | ft candidate's \(s\) but not rmsd |
| \(= 6.94 \pm 0.0811 = (6.859, 7.021)\) | A1 | c.a.o. Must be expressed as an interval. [rmsd gives \(6.94 \pm 0.0805(7) = (6.8594(2), 7.0205(7))\)] |
| Normal distribution can be used because the sample size is large enough for the Central Limit Theorem to apply | E1 | CLT essential |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Advantage: A 99% confidence interval is more likely to contain the true mean | E1 | o.e. |
| Disadvantage: A 99% confidence interval is less precise/wider | E1 | o.e. |
# Question 1:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: m = 7.4$, $H_1: m < 7.4$ where $m$ is the population median time | B1 | Both. Accept hypotheses in words, but must include "population". Do NOT allow symbols other than $m$ unless clearly stated to be a population median |
| Adequate definition of $m$ to include "population" | B1 | |
| Subtract 7.4 to get differences; ranks of \|diff\| all correct (1–12) | M1, M1, A1 | M1 for subtracting 7.4; M1 for ranking; A1 all correct, ft if ranks wrong |
| $W_+ = 2+4+5+7 = 18$ | B1 | $W_- = 1+3+6+8+9+10+11+12 = 60$ |
| Refer to Wilcoxon single sample tables for $n=12$ | M1 | No ft from here if wrong |
| Lower 5% point is 17 (or upper is 61 if 60 used) | A1 | 1-tail test. No ft from here if wrong |
| Result is not significant | A1 | ft only candidate's test statistic |
| Insufficient evidence to suggest that the median time has been reduced | A1 | ft only candidate's test statistic. Conclusion in context |
**[10]**
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{x} = 6.94$, $s = 0.37$ | B1 | Accept $s^2 = 0.1369$. Beware use of msd (0.13518875) or rmsd (0.3676(8)). Do not allow here or below |
| CI is given by $6.94 \pm$ | M1 | ft candidate's $\bar{x} \pm$ |
| $1.96$ | B1 | 1.96 seen |
| $\times \dfrac{0.37}{\sqrt{80}}$ | M1 | ft candidate's $s$ but not rmsd |
| $= 6.94 \pm 0.0811 = (6.859, 7.021)$ | A1 | c.a.o. Must be expressed as an interval. [rmsd gives $6.94 \pm 0.0805(7) = (6.8594(2), 7.0205(7))$] |
| Normal distribution can be used because the sample size is large enough for the Central Limit Theorem to apply | E1 | CLT essential |
**[6]**
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Advantage: A 99% confidence interval is more likely to contain the true mean | E1 | o.e. |
| Disadvantage: A 99% confidence interval is less precise/wider | E1 | o.e. |
**[2]**
---1 In the past, the times for workers in a factory to complete a particular task had a known median of 7.4 minutes. Following a review, managers at the factory wish to know if the median time to complete the task has been reduced.\\
(i) A random sample of 12 times, in minutes, gives the following results.
$$\begin{array} { l l l l l l l l l l l l }
6.90 & 7.23 & 6.54 & 7.62 & 7.04 & 7.33 & 6.74 & 6.45 & 7.81 & 7.71 & 7.50 & 6.32
\end{array}$$
Carry out an appropriate test using a $5 \%$ level of significance.\\
(ii) Some time later, a much larger random sample of times gives the following results.
$$n = 80 \quad \sum x = 555.20 \quad \sum x ^ { 2 } = 3863.9031$$
Find a $95 \%$ confidence interval for the true mean time for the task. Justify your choice of which distribution to use.\\
(iii) Describe briefly one advantage and one disadvantage of having a $99 \%$ confidence interval instead of a $95 \%$ confidence interval.
\hfill \mbox{\textit{OCR MEI S3 2013 Q1 [18]}}