| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2010 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Wilcoxon matched-pairs signed-rank test |
| Difficulty | Standard +0.3 This is a standard S3 question testing two routine non-parametric methods (Wilcoxon signed rank test and chi-squared goodness of fit). Part (a) requires straightforward application of the Wilcoxon test with clear data, and part (b) is an incomplete chi-squared test. Both are textbook applications with no novel insight required, making this slightly easier than average for A-level statistics. |
| Spec | 5.06a Chi-squared: contingency tables5.07b Sign test: and Wilcoxon signed-rank |
| Authority | A | B | C | D | E | F | G | H | I |
| Before | 76 | 98 | 88 | 81 | 86 | 84 | 83 | 93 | 80 |
| After | 82 | 97 | 93 | 77 | 83 | 95 | 91 | 95 | 89 |
| Digit | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| Probability | 0.301 | 0.176 | 0.125 | 0.097 | 0.079 | 0.067 | 0.058 | 0.051 | 0.046 |
| Digit | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| Frequency | 55 | 34 | 27 | 16 | 15 | 17 | 12 | 15 | 9 |
| Answer | Marks |
|---|---|
| B1 | [1] |
| Answer | Marks | Guidance |
|---|---|---|
| B1 | Both. Accept hypotheses in words. | |
| B1 | Adequate definition of \(m\) to include "population". | |
| Diff (After − Before) | 6 | −1 |
| Rank of \(\ | \text{diff}\ | \) |
| Answer | Marks |
|---|---|
| M1 | For differences. ZERO in this section if differences not used. |
| M1 | For ranks. |
| A1 | FT from here if ranks wrong |
| B1 | |
| M1 | No ft from here if wrong. |
| A1 | i.e. a 1-tail test. No ft from here if wrong. |
| A1 | If only c's test statistic. |
| A1 | If only c's test statistic. |
| Answer | Marks | Guidance |
|---|---|---|
| Prob | 0.301 | 0.176 |
| Exp f | 60.2 | 35.2 |
| Obs f | 55 | 34 |
| M1 | Probs \(\times\) 200 for expected frequencies. All correct. | |
| M1 | Calculation of \(\chi^2\). |
| Answer | Marks |
|---|---|
| A1 | c.a.o. |
| M1 | Allow correct df (= cells − 1) from wrongly grouped table and fit. Otherwise, no ft if wrong. |
| Answer | Marks |
|---|---|
| A1 | \(P(\chi^2 > 4.53059) = 0.80636.\) No ft from here if wrong. |
| A1 | If only c's test statistic. |
| A1 | If only c's test statistic. |
## Part (a)(i)
Use paired data in order to eliminate differences between authorities.
| B1 | [1]
## Part (a)(ii)
$H_0: m = 0$
$H_1: m > 0$
where $m$ is the population median difference.
| B1 | Both. Accept hypotheses in words. |
| B1 | Adequate definition of $m$ to include "population". |
| Diff (After − Before) | 6 | −1 | 5 | −4 | −3 | 11 | 8 | 2 | 9 |
|---|---|---|---|---|---|---|---|---|---|
| Rank of $\|\text{diff}\|$ | 6 | 1 | 5 | 4 | 3 | 9 | 7 | 2 | 8 |
$W = 1 + 3 + 4 + 8$ (or $2 + 5 + 6 + 7 + 8 + 9 = 37$)
Refer to tables of Wilcoxon paired (single sample) statistic for $n = 9$.
Lower 5% point is 8 (or upper is 37 if $W_-$ used).
Result is significant.
Evidence suggests the percentage has been raised (on the whole).
| M1 | For differences. ZERO in this section if differences not used. |
| M1 | For ranks. |
| A1 | FT from here if ranks wrong |
| B1 | |
| M1 | No ft from here if wrong. |
| A1 | i.e. a 1-tail test. No ft from here if wrong. |
| A1 | If only c's test statistic. |
| A1 | If only c's test statistic. |
[10]
## Part (b)
$H_0:$ Stock market prices can be modelled by Benford's Law.
$H_1:$ Stock market prices can not be modelled by Benford's Law.
| Prob | 0.301 | 0.176 | 0.125 | 0.097 | 0.079 | 0.067 | 0.058 | 0.051 | 0.046 |
|---|---|---|---|---|---|---|---|---|---|
| Exp f | 60.2 | 35.2 | 25.0 | 19.4 | 15.8 | 13.4 | 11.6 | 10.2 | 9.2 |
| Obs f | 55 | 34 | 27 | 16 | 15 | 17 | 12 | 15 | 9 |
| M1 | Probs $\times$ 200 for expected frequencies. All correct. |
| M1 | Calculation of $\chi^2$. |
$\chi^2 = 0.44917 + 0.04091 + 0.16 + 0.59588 + 0.04051 + 0.96716 + 0.01379 + 2.25882 + 0.00435 = 4.5305(9)$
Refer to $\chi_8^2$.
| A1 | c.a.o. |
| M1 | Allow correct df (= cells − 1) from wrongly grouped table and fit. Otherwise, no ft if wrong. |
Upper 5% point is 13.36.
Not significant.
Suggests Benford's Law provides a reasonable model in the context of share prices.
| A1 | $P(\chi^2 > 4.53059) = 0.80636.$ No ft from here if wrong. |
| A1 | If only c's test statistic. |
| A1 | If only c's test statistic. |
[7]
---\begin{enumerate}[label=(\alph*)]
\item In order to prevent and/or control the spread of infectious diseases, the Government has various vaccination programmes. One such programme requires people to receive a booster injection at the age of 18. It is felt that the proportion of people receiving this booster could be increased and a publicity campaign is undertaken for this purpose. In order to assess the effectiveness of this campaign, health authorities across the country are asked to report the percentage of 18-year-olds receiving the booster before and after the campaign. The results for a randomly chosen sample of 9 authorities are as follows.
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|}
\hline
Authority & A & B & C & D & E & F & G & H & I \\
\hline
Before & 76 & 98 & 88 & 81 & 86 & 84 & 83 & 93 & 80 \\
\hline
After & 82 & 97 & 93 & 77 & 83 & 95 & 91 & 95 & 89 \\
\hline
\end{tabular}
This sample is to be tested to see whether the campaign appears to have been successful in raising the percentage receiving the booster.
\begin{enumerate}[label=(\roman*)]
\item Explain why the use of paired data is appropriate in this context. [1]
\item Carry out an appropriate Wilcoxon signed rank test using these data, at the 5\% significance level. [10]
\end{enumerate}
\item Benford's Law predicts the following probability distribution for the first significant digit in some large data sets.
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|}
\hline
Digit & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
Probability & 0.301 & 0.176 & 0.125 & 0.097 & 0.079 & 0.067 & 0.058 & 0.051 & 0.046 \\
\hline
\end{tabular}
On one particular day, the first significant digits of the stock market prices of the shares of a random sample of 200 companies gave the following results.
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|}
\hline
Digit & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
Frequency & 55 & 34 & 27 & 16 & 15 & 17 & 12 & 15 & 9 \\
\hline
\end{tabular}
Test at the 10\% level of significance whether Benford's Law provides a reasonable model in the context of share prices. [7]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S3 2010 Q3 [18]}}