| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Wilcoxon matched-pairs signed-rank test |
| Difficulty | Standard +0.3 This is a straightforward application of the Wilcoxon matched-pairs signed-rank test with clear paired data. Students must calculate differences, rank absolute values, sum ranks for positive/negative differences, and compare to critical values. While it requires careful execution of the procedure, it's a standard textbook application with no conceptual challenges beyond following the algorithm correctly. |
| Spec | 5.07b Sign test: and Wilcoxon signed-rank5.07d Paired vs two-sample: selection |
| Student | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) |
| Paper I (\%) | 70 | 70 | 84 | 80 | 64 | 65 | 65 | 90 |
| Paper II (\%) | 64 | 76 | 72 | 74 | 68 | 64 | 58 | 76 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(d\): 6, \(-6\), 12, 6, \(-4\), 1, 7, 14 | M1 | Realising paired \(t\)-test; finding differences \((\pm)\); at least 3 correct |
| \(\bar{d} = \pm 4.5\), \(s_d = \sqrt{50.285...} = 7.09...\) | M1 | Correct method for finding \(\bar{d}\) and \(s_d\) |
| \(H_0: \mu_d = 0\), \(H_1: \mu_d \neq 0\) | B1 | Correct model for difference; both hypotheses using \(\mu_d\) or \(\mu\); condone \(\mu_I = \mu_{II}\) and \(\mu_I \neq \mu_{II}\) |
| \(t = \pm\frac{\text{"4.5"}\sqrt{8}}{\text{"7.09..."}}\) | M1 | Using correct method; \(t = \pm\frac{\text{"their 4.5"}\sqrt{8}}{\text{"their 7.09..."}}\) |
| \(= \pm 1.7948...\) awrt \(\pm 1.79/1.8\) | A1 | |
| Critical value \(t_7 = \pm 3.499\) | B1 | Correct critical value with compatible sign |
| Insufficient evidence that papers are of different difficulty / Alexa's belief is correct | A1ft | Correct inference in context using their CV and value of \(t\); NB difference of means test gets M0M0B1M0A0B0A0 |
# Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $d$: 6, $-6$, 12, 6, $-4$, 1, 7, 14 | M1 | Realising paired $t$-test; finding differences $(\pm)$; at least 3 correct |
| $\bar{d} = \pm 4.5$, $s_d = \sqrt{50.285...} = 7.09...$ | M1 | Correct method for finding $\bar{d}$ and $s_d$ |
| $H_0: \mu_d = 0$, $H_1: \mu_d \neq 0$ | B1 | Correct model for difference; both hypotheses using $\mu_d$ or $\mu$; condone $\mu_I = \mu_{II}$ and $\mu_I \neq \mu_{II}$ |
| $t = \pm\frac{\text{"4.5"}\sqrt{8}}{\text{"7.09..."}}$ | M1 | Using correct method; $t = \pm\frac{\text{"their 4.5"}\sqrt{8}}{\text{"their 7.09..."}}$ |
| $= \pm 1.7948...$ awrt $\pm 1.79/1.8$ | A1 | |
| Critical value $t_7 = \pm 3.499$ | B1 | Correct critical value with compatible sign |
| Insufficient evidence that **papers** are of different difficulty / **Alexa's** belief is correct | A1ft | Correct inference in context using their CV and value of $t$; NB difference of means test gets M0M0B1M0A0B0A0 |
---5 Alexa believes that students are equally likely to achieve the same percentage score on each of two tests, paper I and paper II. She randomly selects 8 students and gives them each paper I and paper II. The percentage scores for each paper are recorded.
The following paired data are collected.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | }
\hline
Student & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ \\
\hline
Paper I (\%) & 70 & 70 & 84 & 80 & 64 & 65 & 65 & 90 \\
\hline
Paper II (\%) & 64 & 76 & 72 & 74 & 68 & 64 & 58 & 76 \\
\hline
\end{tabular}
\end{center}
Test, at the $1 \%$ significance level, whether or not there is evidence to support Alexa's belief. State your hypotheses clearly and show your working.
\hfill \mbox{\textit{Edexcel FS2 2019 Q5 [7]}}