| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2011 |
| Session | June |
| 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. While S4 is Further Maths content, this question requires only routine execution: calculate differences, find mean and standard deviation, compute test statistic, and compare to critical value. No conceptual challenges or novel problem-solving required, making it slightly easier than average despite being Further Maths. |
| Spec | 5.07b Sign test: and Wilcoxon signed-rank5.07d Paired vs two-sample: selection |
| Orange | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Machine \(A\) | 60 | 58 | 55 | 53 | 52 | 51 | 54 | 56 |
| Machine \(B\) | 61 | 60 | 58 | 52 | 55 | 50 | 52 | 58 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(d = B - A\): 1, 2, 3, -1, 3, -1, -2, 2 | M1 | For attempting the \(d\)s |
| \(\bar{d} = 0.875\) | M1 | For attempting \(\bar{d}\) |
| \(s_d^2 = \frac{33 - 8 \times 0.875^2}{7} = (3.8392\ldots)\) | M1 | For attempting \(s_d\) or \(s_d^2\) |
| \(H_0: \mu_d = 0 \quad H_1: \mu_d > 0\) | B1 | |
| \(t_7 = \frac{0.875}{\frac{s_p}{\sqrt{8}}} = 1.263\ldots\) awrt 1.26 | M1A1 | For attempting correct test statistic |
| \(t_7(10\%)\) one tail critical value is 1.415 | B1 | |
| Not significant. There is insufficient evidence to support the claim of manufacturer \(B\) or machine \(B\) does not produce more juice (than machine \(A\)) | A1 | Contextual statement only required. Allow: the juice provided by machine \(A\) is the same as by machine \(B\) |
| Total: 8 marks | ||
| NB 2 sample test can score 3/8: M0 M0; M1 \(\frac{7\times9.27+7\times16.79}{14}\); B1 for \(H_0: \mu_A=\mu_B\), \(H_1:\mu_A<\mu_B\); M0 A0; B1 1.345; A0 |
# Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $d = B - A$: 1, 2, 3, -1, 3, -1, -2, 2 | M1 | For attempting the $d$s |
| $\bar{d} = 0.875$ | M1 | For attempting $\bar{d}$ |
| $s_d^2 = \frac{33 - 8 \times 0.875^2}{7} = (3.8392\ldots)$ | M1 | For attempting $s_d$ or $s_d^2$ |
| $H_0: \mu_d = 0 \quad H_1: \mu_d > 0$ | B1 | |
| $t_7 = \frac{0.875}{\frac{s_p}{\sqrt{8}}} = 1.263\ldots$ **awrt 1.26** | M1A1 | For attempting correct test statistic |
| $t_7(10\%)$ one tail critical value is **1.415** | B1 | |
| Not significant. There is insufficient evidence to support the claim of manufacturer $B$ or machine $B$ does not produce more juice (than machine $A$) | A1 | Contextual statement only required. Allow: the juice provided by machine $A$ is the same as by machine $B$ |
| **Total: 8 marks** | | |
| **NB** 2 sample test can score 3/8: M0 M0; M1 $\frac{7\times9.27+7\times16.79}{14}$; B1 for $H_0: \mu_A=\mu_B$, $H_1:\mu_A<\mu_B$; M0 A0; B1 1.345; A0 | | |
---3. Manuel is planning to buy a new machine to squeeze oranges in his cafe and he has two models, at the same price, on trial. The manufacturers of machine $B$ claim that their machine produces more juice from an orange than machine $A$. To test this claim Manuel takes a random sample of 8 oranges, cuts them in half and puts one half in machine $A$ and the other half in machine $B$. The amount of juice, in ml , produced by each machine is given in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | }
\hline
Orange & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
Machine $A$ & 60 & 58 & 55 & 53 & 52 & 51 & 54 & 56 \\
\hline
Machine $B$ & 61 & 60 & 58 & 52 & 55 & 50 & 52 & 58 \\
\hline
\end{tabular}
\end{center}
Stating your hypotheses clearly, test, at the $10 \%$ level of significance, whether or not the mean amount of juice produced by machine $B$ is more than the mean amount produced by machine $A$.\\
\hfill \mbox{\textit{Edexcel S4 2011 Q3 [8]}}