| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2006 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Critical region or test statistic properties |
| Difficulty | Standard +0.3 This is a standard S3 question testing routine application of chi-squared goodness of fit test and Wilcoxon signed-rank test. Part (i) requires mechanical calculation of chi-squared statistic with given expected frequencies (no parameter estimation complications). Part (ii)(A) is straightforward discussion of normality/sample size for t-tests. Part (ii)(B) involves standard Wilcoxon procedure with small sample. All techniques are textbook applications with no novel insight required, though the 18 marks indicate it's somewhat lengthy. Slightly easier than average due to the procedural nature. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Length \(x\) (mm) | Observed frequency |
| \(x \leq 298\) | 1 |
| \(298 < x \leq 300\) | 30 |
| \(300 < x \leq 301\) | 62 |
| \(301 < x \leq 302\) | 70 |
| \(302 < x \leq 304\) | 34 |
| \(x > 304\) | 3 |
| Length \(x\) (mm) | Expected frequency |
| \(x \leq 298\) | 1.49 |
| \(298 < x \leq 300\) | 37.85 |
| \(300 < x \leq 301\) | 55.62 |
| \(301 < x \leq 302\) | 58.32 |
| \(302 < x \leq 304\) | 44.62 |
| \(x > 304\) | 2.10 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\chi^2 = 1.7681 + 0.7318 + 2.3392 + 2.0222 = 6.86\) | M, A1 | for grouping. Allow the M1 for correct method from wrongly grouped or ungrouped table. |
| Refer to \(\chi_1^2\) | M | Allow correct df (= cells − 3) from wrongly grouped or ungrouped table, and FT. Otherwise, no FT if wrong. |
| Upper 5% point is 3.84. Significant. Suggests Normal model does not fit | A1, E1, E1 | No ft from here if wrong. ft only c's test statistic. ft only c's test statistic. |
| Answer | Marks | Guidance |
|---|---|---|
| \(t\) test unwise ... | E1 | |
| ... because underlying population appears non-Normal | E1 | FT from result of candidate's work in (i) |
| Answer | Marks | Guidance |
|---|---|---|
| Data | Median 301 | Difference |
| 301.3 | 0.3 | |
| 301.4 | 0.4 | |
| 299.6 | −1.4 | |
| 302.2 | 1.2 | |
| 300.3 | −0.7 | |
| 303.2 | 2.2 | |
| 302.6 | 1.6 | |
| 301.8 | 0.8 | |
| 300.9 | −0.1 | |
| 300.8 | −0.2 | |
| M | for differences. ZERO in this section if differences not used. | 9 marks |
| \(T = 1 + 2 + 5 + 8 = 16\) (or \(3+4+6+7+9+10 = 39\)) | B1 | |
| Refer to tables of Wilcoxon single sample (/paired) statistic | M | |
| Lower (or upper if 39 used) 5% tail is needed | M | |
| Value for \(n = 10\) is 10 (or 45 if 39 used) | A1 | |
| Result is not significant. No evidence against median being 301 | E1, E1 |
## (i)
$\chi^2 = 1.7681 + 0.7318 + 2.3392 + 2.0222 = 6.86$ | M, A1 | for grouping. Allow the M1 for correct method from wrongly grouped or ungrouped table. | 7 marks
Refer to $\chi_1^2$ | M | Allow correct df (= cells − 3) from wrongly grouped or ungrouped table, and FT. Otherwise, no FT if wrong. |
Upper 5% point is 3.84. Significant. Suggests Normal model does not fit | A1, E1, E1 | No ft from here if wrong. ft only c's test statistic. ft only c's test statistic. |
## (ii)(A)
$t$ test unwise ... | E1 | | 2 marks
... because underlying population appears non-Normal | E1 | FT from result of candidate's work in (i) |
## (ii)(B)
| Data | Median 301 | Difference | Rank of [diff] |
|------|-----------|-----------|-----------------|
| 301.3 | | 0.3 | 3 |
| 301.4 | | 0.4 | 4 |
| 299.6 | | −1.4 | 8 |
| 302.2 | | 1.2 | 7 |
| 300.3 | | −0.7 | 5 |
| 303.2 | | 2.2 | 10 |
| 302.6 | | 1.6 | 9 |
| 301.8 | | 0.8 | 6 |
| 300.9 | | −0.1 | 1 |
| 300.8 | | −0.2 | 2 |
| M | for differences. ZERO in this section if differences not used. | 9 marks
$T = 1 + 2 + 5 + 8 = 16$ (or $3+4+6+7+9+10 = 39$) | B1 | |
Refer to tables of Wilcoxon single sample (/paired) statistic | M | |
Lower (or upper if 39 used) 5% tail is needed | M | |
Value for $n = 10$ is 10 (or 45 if 39 used) | A1 | |
Result is not significant. No evidence against median being 301 | E1, E1 | |Quality control inspectors in a factory are investigating the lengths of glass tubes that will be used to make laboratory equipment.
\begin{enumerate}[label=(\roman*)]
\item Data on the observed lengths of a random sample of 200 glass tubes from one batch are available in the form of a frequency distribution as follows.
\begin{tabular}{|c|c|}
\hline
Length $x$ (mm) & Observed frequency \\
\hline
$x \leq 298$ & 1 \\
$298 < x \leq 300$ & 30 \\
$300 < x \leq 301$ & 62 \\
$301 < x \leq 302$ & 70 \\
$302 < x \leq 304$ & 34 \\
$x > 304$ & 3 \\
\hline
\end{tabular}
The sample mean and standard deviation are 301.08 and 1.2655 respectively.
The corresponding expected frequencies for the Normal distribution with parameters estimated by the sample statistics are
\begin{tabular}{|c|c|}
\hline
Length $x$ (mm) & Expected frequency \\
\hline
$x \leq 298$ & 1.49 \\
$298 < x \leq 300$ & 37.85 \\
$300 < x \leq 301$ & 55.62 \\
$301 < x \leq 302$ & 58.32 \\
$302 < x \leq 304$ & 44.62 \\
$x > 304$ & 2.10 \\
\hline
\end{tabular}
Examine the goodness of fit of a Normal distribution, using a 5\% significance level. [7]
\item It is thought that the lengths of tubes in another batch have an underlying distribution similar to that for the batch in part (i) but possibly with different location and dispersion parameters. A random sample of 10 tubes from this batch gives the following lengths (in mm).
301.3 \quad 301.4 \quad 299.6 \quad 302.2 \quad 300.3 \quad 303.2 \quad 302.6 \quad 301.8 \quad 300.9 \quad 300.8
\begin{enumerate}[label=(\Alph*)]
\item Discuss briefly whether it would be appropriate to use a $t$ test to examine a hypothesis about the population mean length for this batch. [2]
\item Use a Wilcoxon test to examine at the 10\% significance level whether the population median length for this batch is 301 mm. [9]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S3 2006 Q4 [18]}}