| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Paired t-test |
| Difficulty | Standard +0.3 This is a straightforward paired t-test application with clear data and standard hypotheses. Students must calculate differences, find mean and standard deviation, compute the t-statistic, and compare to critical value. While it requires multiple computational steps and stating the normality assumption, it follows a standard textbook procedure with no conceptual challenges or novel insights required. Slightly easier than average due to small sample size (simple calculations) and explicit guidance on test type and significance level. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Athlete | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| Time on old track \(( s )\) | 12.2 | 10.3 | 11.5 | 13.0 | 11.8 | 11.7 | 11.9 |
| Time on new track \(( s )\) | 11.1 | 10.5 | 11.0 | 12.6 | 11.0 | 10.9 | 12.0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of paired \(t\) test | M1 | |
| \(d =\) old\(-\)new: \(1.1,\ -0.2,\ 0.5,\ 0.4,\ 0.8,\ 0.8,\ -0.1\) | ||
| \(\bar{d} = 0.471(4),\ s^2 = 0.2324\) | B1B1 | Allow 3.3/7, 33/70; 122/525 |
| \((H_0: \mu_d=0,\ H_1: \mu_d > 0)\) | ||
| Test statistic \(= 0.4714/(0.2324/7)^{0.5} = 2.587\) | M1, A1 | Must be \(/7\); or use \(\bar{d}/\sqrt{0.2347/7} > 2.447\) B1 |
| Critical value \(= 2.447\) | B1 | \(\bar{d} > 0.448\) |
| \(TS > CV\) so reject \(H_0\) | M1 | ft wrong TS, CV; \(0.4714 > 0.448\) so reject \(H_0\) M1 |
| There is sufficient evidence that the new track is faster | A1 | Allow from rounding errors, otherwise cwo. Or similar in context, not over-assertive. eg \(s^2=0.233\) (from using \(\bar{d}=0.471\)) TS=2.58(2) M1B1B0M1A0B1M1A1 |
| [8] |
# Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of paired $t$ test | M1 | |
| $d =$ old$-$new: $1.1,\ -0.2,\ 0.5,\ 0.4,\ 0.8,\ 0.8,\ -0.1$ | | |
| $\bar{d} = 0.471(4),\ s^2 = 0.2324$ | B1B1 | Allow 3.3/7, 33/70; 122/525 |
| $(H_0: \mu_d=0,\ H_1: \mu_d > 0)$ | | |
| Test statistic $= 0.4714/(0.2324/7)^{0.5} = 2.587$ | M1, A1 | Must be $/7$; or use $\bar{d}/\sqrt{0.2347/7} > 2.447$ B1 |
| Critical value $= 2.447$ | B1 | $\bar{d} > 0.448$ |
| $TS > CV$ so reject $H_0$ | M1 | ft wrong TS, CV; $0.4714 > 0.448$ so reject $H_0$ M1 |
| There is sufficient evidence that the new track is faster | A1 | Allow from rounding errors, otherwise cwo. Or similar in context, not over-assertive. eg $s^2=0.233$ (from using $\bar{d}=0.471$) TS=2.58(2) M1B1B0M1A0B1M1A1 |
| | **[8]** | |
---2 A new running track has been developed and part of the testing procedure involves 7 randomly chosen athletes. They each run 100 m on both the old and new tracks.\\
The results are as follows.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
Athlete & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
Time on old track $( s )$ & 12.2 & 10.3 & 11.5 & 13.0 & 11.8 & 11.7 & 11.9 \\
\hline
Time on new track $( s )$ & 11.1 & 10.5 & 11.0 & 12.6 & 11.0 & 10.9 & 12.0 \\
\hline
\end{tabular}
\end{center}
The population mean times on the old and new tracks are denoted by $\mu _ { \mathrm { O } }$ seconds and $\mu _ { \mathrm { N } }$ seconds respectively. Stating any necessary assumption, carry out a suitable $t$-test of the null hypothesis $\mu _ { \mathrm { O } } - \mu _ { \mathrm { N } } = 0$ against the alternative hypothesis $\mu _ { \mathrm { O } } - \mu _ { \mathrm { N } } > 0$. Use a $2 \frac { 1 } { 2 } \%$ significance level .
\hfill \mbox{\textit{OCR S3 2013 Q2 [8]}}