| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Paired t-test |
| Difficulty | Challenging +1.2 This S4 question requires understanding of paired t-test assumptions and critical values, but the calculation is straightforward: compute differences, find mean and standard deviation, then use t-tables with 7 degrees of freedom to find bounds. While it's a Further Maths topic (making it inherently harder), the execution is mechanical with no novel insight required—just standard hypothesis testing procedure. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Student | A | \(B\) | C | D | \(E\) | \(F\) | G | \(H\) |
| Reaction time at the start of the school day | 10.8 | 7.2 | 8.7 | 6.8 | 9.4 | 10.9 | 11.1 | 7.6 |
| Reaction time at the end of the school day | 10 | 6.1 | 8.8 | 5.7 | 8.7 | 8.1 | 9.8 | 6.8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Assumption that underlying distribution of the difference in reaction times is normally distributed | B1 (1) | Must mention "differences" and "normal" distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Differences (Start - end): 0.8, 1.1, -0.1, 1.1, 0.7, 2.8, 1.3, 0.8 | M1 | Attempting differences |
| \(\bar{d} = \frac{8.5}{8} = (\pm)1.0625\) | M1 | Attempt to find \(\bar{d} = \frac{\sum\text{"their }d\text{"} }{8}\) |
| \(s^2 = \frac{8}{7}\left(\frac{13.73}{8} - 1.0625^2\right) = 0.67125\) or \(s^2 = \frac{1}{7}\left(13.73 - \frac{8.5^2}{8}\right) = 0.67125\) | M1 | Attempting \(s\) or \(s^2\frac{1}{7}\left(\sum\text{"their }d^2\text{"} - \frac{(\sum\text{"their }d\text{"})^2}{8}\right)\); \(s = 0.8192...\) |
| \(t = \frac{\text{"1.0625"} - m}{\sqrt{\frac{\text{"0.67125"}}{8}}}\) | M1 | Correct test statistic \(\frac{\bar{d}-m}{\frac{s}{\sqrt{8}}}\), allow any letter |
| Critical values: \(t_7(2.5\%) = \pm2.365\), \(t_7(0.005\%) = \pm3.499\) | B1 | Both critical values correct (ignore sign) |
| \(\frac{1.0625-m}{\sqrt{\frac{0.67125}{8}}} = \pm2.365\) and \(\frac{1.0625-m}{\sqrt{\frac{0.67125}{8}}} = \pm3.499\) | M1d, A1ft | Dependent on previous M; pair of equations with same sign, one of each CV |
| \(0.049 < m < 0.377\) and \(1.748 < m < 2.076\) | A1 A1 (9) | awrt \(0.049 < m <\) awrt \(0.377\); awrt \(1.75 < m <\) awrt \(2.08\); allow \(\leq\) instead of \(<\) |
## Question 3:
### Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Assumption that underlying distribution of the **difference** in reaction times is **normally distributed** | B1 (1) | Must mention "differences" and "normal" distribution |
### Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Differences (Start - end): 0.8, 1.1, -0.1, 1.1, 0.7, 2.8, 1.3, 0.8 | M1 | Attempting differences |
| $\bar{d} = \frac{8.5}{8} = (\pm)1.0625$ | M1 | Attempt to find $\bar{d} = \frac{\sum\text{"their }d\text{"} }{8}$ |
| $s^2 = \frac{8}{7}\left(\frac{13.73}{8} - 1.0625^2\right) = 0.67125$ or $s^2 = \frac{1}{7}\left(13.73 - \frac{8.5^2}{8}\right) = 0.67125$ | M1 | Attempting $s$ or $s^2\frac{1}{7}\left(\sum\text{"their }d^2\text{"} - \frac{(\sum\text{"their }d\text{"})^2}{8}\right)$; $s = 0.8192...$ |
| $t = \frac{\text{"1.0625"} - m}{\sqrt{\frac{\text{"0.67125"}}{8}}}$ | M1 | Correct test statistic $\frac{\bar{d}-m}{\frac{s}{\sqrt{8}}}$, allow any letter |
| Critical values: $t_7(2.5\%) = \pm2.365$, $t_7(0.005\%) = \pm3.499$ | B1 | Both critical values correct (ignore sign) |
| $\frac{1.0625-m}{\sqrt{\frac{0.67125}{8}}} = \pm2.365$ and $\frac{1.0625-m}{\sqrt{\frac{0.67125}{8}}} = \pm3.499$ | M1d, A1ft | Dependent on previous M; pair of equations with same sign, one of each CV |
| $0.049 < m < 0.377$ and $1.748 < m < 2.076$ | A1 A1 (9) | awrt $0.049 < m <$ awrt $0.377$; awrt $1.75 < m <$ awrt $2.08$; allow $\leq$ instead of $<$ |
**Total: 10**
---\begin{enumerate}
\item A random sample of 8 students is selected from a school database.
\end{enumerate}
Each student's reaction time is measured at the start of the school day and again at the end of the school day. The reaction times, in milliseconds, are recorded below.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline
Student & A & $B$ & C & D & $E$ & $F$ & G & $H$ \\
\hline
Reaction time at the start of the school day & 10.8 & 7.2 & 8.7 & 6.8 & 9.4 & 10.9 & 11.1 & 7.6 \\
\hline
Reaction time at the end of the school day & 10 & 6.1 & 8.8 & 5.7 & 8.7 & 8.1 & 9.8 & 6.8 \\
\hline
\end{tabular}
\end{center}
(a) State one assumption that needs to be made in order to carry out a paired $t$-test.\\
(1)
The random variable $R$ is the reaction time at the start of the school day minus the reaction time at the end of the school day. The mean of $R$ is $\mu$.
John uses a paired $t$-test to test the hypotheses
$$\mathrm { H } _ { 0 } : \mu = m \quad \mathrm { H } _ { 1 } : \mu \neq m$$
Given that $\mathrm { H } _ { 0 }$ is rejected at the 5\% level of significance but accepted at the 1\% level of significance,\\
(b) find the ranges of possible values for $m$.
\hfill \mbox{\textit{Edexcel S4 2018 Q3 [10]}}