| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Confidence intervals (paired data) |
| Difficulty | Standard +0.3 This is a straightforward paired t-test confidence interval calculation with clear data and standard procedure. Students must find differences, calculate mean and standard deviation, then apply the t-distribution formula. Part (b) requires interpreting whether zero lies in the interval. While it involves multiple steps, it's a routine textbook exercise requiring only standard technique with no novel insight or complex reasoning. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.07b Sign test: and Wilcoxon signed-rank |
| Engineer | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) |
| January | 17 | 19 | 22 | 26 | 15 | 28 | 18 | 21 |
| July | 19 | 18 | 25 | 24 | 17 | 25 | 16 | 19 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Differences \(d\) (Jan \(-\) Jul): \(-2, 1, -3, 2, -2, 3, 2, 2\) | M1 | Correct differences (either way) |
| \(\sum d = 3\), \(\bar{d} = \frac{3}{8} = 0.375\) | A1 | Correct mean of differences |
| \(\sum d^2 = 4+1+9+4+4+9+4+4 = 39\) | M1 | Correct sum of squares |
| \(s_d^2 = \frac{1}{7}\left(39 - \frac{9}{8}\right) = \frac{1}{7}(39 - 1.125) = \frac{37.875}{7} = 5.411\) | A1 | Correct variance |
| \(s_d = 2.327\) | A1 | Correct standard deviation |
| \(t_{7}\) at 95%: \(t = 2.365\) | B1 | Correct \(t\) value with 7 df |
| CI: \(0.375 \pm 2.365 \times \frac{2.327}{\sqrt{8}} = 0.375 \pm 1.945\) | M1 | Correct structure of CI |
| \((-1.570,\ 2.320)\) | A1 | Correct interval (allow rounding) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \mu_d = 0\), \(H_1: \mu_d \neq 0\) | B1 | Both hypotheses stated |
| Since \(0\) is contained within the confidence interval \((-1.570, 2.320)\) | M1 | Correct reasoning using CI |
| There is no evidence of a change in mean assembly time. | A1 | Correct conclusion in context |
# Question 2:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Differences $d$ (Jan $-$ Jul): $-2, 1, -3, 2, -2, 3, 2, 2$ | M1 | Correct differences (either way) |
| $\sum d = 3$, $\bar{d} = \frac{3}{8} = 0.375$ | A1 | Correct mean of differences |
| $\sum d^2 = 4+1+9+4+4+9+4+4 = 39$ | M1 | Correct sum of squares |
| $s_d^2 = \frac{1}{7}\left(39 - \frac{9}{8}\right) = \frac{1}{7}(39 - 1.125) = \frac{37.875}{7} = 5.411$ | A1 | Correct variance |
| $s_d = 2.327$ | A1 | Correct standard deviation |
| $t_{7}$ at 95%: $t = 2.365$ | B1 | Correct $t$ value with 7 df |
| CI: $0.375 \pm 2.365 \times \frac{2.327}{\sqrt{8}} = 0.375 \pm 1.945$ | M1 | Correct structure of CI |
| $(-1.570,\ 2.320)$ | A1 | Correct interval (allow rounding) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \mu_d = 0$, $H_1: \mu_d \neq 0$ | B1 | Both hypotheses stated |
| Since $0$ is contained within the confidence interval $(-1.570, 2.320)$ | M1 | Correct reasoning using CI |
| There is no evidence of a change in mean assembly time. | A1 | Correct conclusion in context |
---2. Every 6 months some engineers are tested to see if their times, in minutes, to assemble a particular component have changed. The times taken to assemble the component are normally distributed. A random sample of 8 engineers was chosen and their times to assemble the component were recorded in January and in July. The data are given in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | }
\hline
Engineer & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ \\
\hline
January & 17 & 19 & 22 & 26 & 15 & 28 & 18 & 21 \\
\hline
July & 19 & 18 & 25 & 24 & 17 & 25 & 16 & 19 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Calculate a $95 \%$ confidence interval for the mean difference in times.
\item Use your confidence interval to state, giving a reason, whether or not there is evidence of a change in the mean time to assemble a component. State your hypotheses clearly.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2013 Q2 [10]}}