| Exam Board | OCR MEI |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2010 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Two-sample t-test |
| Difficulty | Standard +0.3 This is a straightforward application of standard hypothesis testing procedures. Part (i) requires routine calculation of an acceptance region for a two-sample z-test with known variances. Part (ii) applies the test and constructs a confidence interval using standard formulas. Part (iii) requires applying the Wilcoxon rank-sum test, which is a bookwork procedure at S4 level. All parts involve direct application of learned techniques with no novel problem-solving or insight required, making this slightly easier than average. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution5.07c Single-sample tests |
| First production line | 25.9 | 25.8 | 25.3 | 24.7 | 24.4 | 25.4 | ||
| Second production line | 23.8 | 25.6 | 24.0 | 23.5 | 24.1 | 24.5 | 24.3 | 25.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\) accepted if \(-1.96 <\) test statistic \(< 1.96\) | M1, B1 | Double inequality; 1.96 |
| i.e. \(-1.96 < \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{1.2^2}{8} + \frac{1.4^2}{10}}} < 1.96\) | M1, M1 | Numerator; denominator |
| i.e. \(-1.96 \times 0.6132 < \bar{x}_1 - \bar{x}_2 < 1.96 \times 0.6132\) | A1 | |
| i.e. \(-1.20(18) < \bar{x}_1 - \bar{x}_2 < 1.20(18)\) | A1 | Special case: allow 1 out of 2 A1s if 1.645 used provided all 3 M marks earned |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x}_1 - \bar{x}_2 = 1.4\) | B1 | FT if wrong |
| Which is outside the acceptance region, so \(H_0\) is rejected | M1, E1 | FT can's acceptance region if reasonable |
| CI for \(\mu_1 - \mu_2\): \(1.4 \pm (2.576 \times 0.6132)\) | M1, B1, M1 | |
| i.e. \(1.4 \pm 1.5796\), i.e. \((-0.18, 2.97)\) | A1 cao |
| Answer | Marks | Guidance |
|---|---|---|
| Wilcoxon rank sum test (or Mann-Whitney form of test) | M1 | |
| Ranks: First: \(14\ 13\ 10\ 8\ 6\ 11\); Second: \(2\ 12\ 3\ 1\ 4\ 7\ 5\ 9\) | M1, A1 | Combined ranking; allow up to 2 errors |
| \(W = 14+13+10+8+6+11 = 62\) [or \(41\) if M-W used] | B1 | |
| Refer to \(W_{6,8}\) [or \(MW_{6,8}\)] tables | M1 | No FT if wrong |
| Lower \(2\frac{1}{2}\)% critical point is \(29\) [or \(8\) if M-W used] | A1 | |
| Consideration of upper \(2\frac{1}{2}\)% point also needed | M1 | |
| Upper critical point is \(61\) [M-W: \(40\)] | A1 | |
| Result is significant; seems (population) medians may not be assumed equal | E1, E1 |
# Question 3:
## Part (i)
| $H_0$ accepted if $-1.96 <$ test statistic $< 1.96$ | M1, B1 | Double inequality; 1.96 |
|---|---|---|
| i.e. $-1.96 < \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{1.2^2}{8} + \frac{1.4^2}{10}}} < 1.96$ | M1, M1 | Numerator; denominator |
| i.e. $-1.96 \times 0.6132 < \bar{x}_1 - \bar{x}_2 < 1.96 \times 0.6132$ | A1 | |
| i.e. $-1.20(18) < \bar{x}_1 - \bar{x}_2 < 1.20(18)$ | A1 | Special case: allow 1 out of 2 A1s if 1.645 used provided all 3 M marks earned |
## Part (ii)
| $\bar{x}_1 - \bar{x}_2 = 1.4$ | B1 | FT if wrong |
|---|---|---|
| Which is outside the acceptance region, so $H_0$ is rejected | M1, E1 | FT can's acceptance region if reasonable |
| CI for $\mu_1 - \mu_2$: $1.4 \pm (2.576 \times 0.6132)$ | M1, B1, M1 | |
| i.e. $1.4 \pm 1.5796$, i.e. $(-0.18, 2.97)$ | A1 cao | |
## Part (iii)
| Wilcoxon rank sum test (or Mann-Whitney form of test) | M1 | |
|---|---|---|
| Ranks: First: $14\ 13\ 10\ 8\ 6\ 11$; Second: $2\ 12\ 3\ 1\ 4\ 7\ 5\ 9$ | M1, A1 | Combined ranking; allow up to 2 errors |
| $W = 14+13+10+8+6+11 = 62$ [or $41$ if M-W used] | B1 | |
| Refer to $W_{6,8}$ [or $MW_{6,8}$] tables | M1 | No FT if wrong |
| Lower $2\frac{1}{2}$% critical point is $29$ [or $8$ if M-W used] | A1 | |
| Consideration of upper $2\frac{1}{2}$% point also needed | M1 | |
| Upper critical point is $61$ [M-W: $40$] | A1 | |
| Result is significant; seems (population) medians may not be assumed equal | E1, E1 | |
---3 At a factory, two production lines are in use for making steel rods. A critical dimension is the diameter of a rod. For the first production line, it is assumed from experience that the diameters are Normally distributed with standard deviation 1.2 mm . For the second production line, it is assumed from experience that the diameters are Normally distributed with standard deviation 1.4 mm . It is desired to test whether the mean diameters for the two production lines, $\mu _ { 1 }$ and $\mu _ { 2 }$, are equal. A random sample of 8 rods is taken from the first production line and, independently, a random sample of 10 rods is taken from the second production line.\\
(i) Find the acceptance region for the customary test based on the Normal distribution for the null hypothesis $\mu _ { 1 } = \mu _ { 2 }$, against the alternative hypothesis $\mu _ { 1 } \neq \mu _ { 2 }$, at the $5 \%$ level of significance.\\
(ii) The sample means are found to be 25.8 mm and 24.4 mm respectively. What is the result of the test? Provide a two-sided $99 \%$ confidence interval for $\mu _ { 1 } - \mu _ { 2 }$.
The production lines are modified so that the diameters may be assumed to be of equal (but unknown) variance. However, they may no longer be Normally distributed. A two-sided test of the equality of the population medians is required, at the $5 \%$ significance level.\\
(iii) The diameters in independent random samples of sizes 6 and 8 are as follows, in mm .
\begin{center}
\begin{tabular}{ l l l l l l l l l }
First production line & 25.9 & 25.8 & 25.3 & 24.7 & 24.4 & 25.4 & & \\
Second production line & 23.8 & 25.6 & 24.0 & 23.5 & 24.1 & 24.5 & 24.3 & 25.1 \\
\end{tabular}
\end{center}
Use an appropriate procedure to carry out the test.
\hfill \mbox{\textit{OCR MEI S4 2010 Q3 [24]}}