| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2020 |
| Session | Specimen |
| 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 before/after data. Students need to calculate differences, find mean and standard deviation of differences, compute the t-statistic, and compare to critical value. While it's Further Maths content, it's a routine textbook exercise requiring standard procedure execution with no novel insight or complex reasoning. |
| Spec | 5.07b Sign test: and Wilcoxon signed-rank5.07d Paired vs two-sample: selection |
| Employee | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) | \(I\) | \(J\) |
| Before | 42 | 35 | 96 | 74 | 20 | 5 | 78 | 45 | 146 | 0 |
| After | 34 | 32 | 100 | 72 | 31 | 2 | 61 | 35 | 140 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Assume that population differences are normally distributed | 1 | B1 |
| \(H_0: \mu_B - \mu_A = 0\) and \(H_1: \mu_B - \mu_A > 0\) | 1 | B1 |
| Differences, \(d\): \(8\ 3\ -4\ 2\ -11\ 3\ 17\ 10\ 6\ 0\) | 1 | M1 |
| \(\bar{d} = 3.4\) and \(s^2 = \frac{1}{9}\left(648 - \frac{34^2}{10}\right) = 59.16\) | 1 | M1 |
| Calculate test statistic \(t = \dfrac{\bar{d}}{s/\sqrt{10}}\) | 1 | M1 |
| \(t = 1.398\) | 1 | A1 |
| Use tabular value \(t_{9,\,0.9} = 1.383\) in a comparison | 1 | B1 |
| Reject \(H_0\) and conclude that hours of absence have decreased | 1 | B1FT |
| Total | 8 |
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Assume that population differences are normally distributed | 1 | B1 | |
| $H_0: \mu_B - \mu_A = 0$ and $H_1: \mu_B - \mu_A > 0$ | 1 | B1 | Both hypotheses required, in terms of population means or alternatively in terms of population mean difference |
| Differences, $d$: $8\ 3\ -4\ 2\ -11\ 3\ 17\ 10\ 6\ 0$ | 1 | M1 | |
| $\bar{d} = 3.4$ and $s^2 = \frac{1}{9}\left(648 - \frac{34^2}{10}\right) = 59.16$ | 1 | M1 | |
| Calculate test statistic $t = \dfrac{\bar{d}}{s/\sqrt{10}}$ | 1 | M1 | |
| $t = 1.398$ | 1 | A1 | Correct value to at least 3 sf |
| Use tabular value $t_{9,\,0.9} = 1.383$ in a comparison | 1 | B1 | |
| Reject $H_0$ and conclude that hours of absence have decreased | 1 | B1FT | Follow through their calculated $t = 1.398$ |
| **Total** | **8** | | |
---3 Employees at a particular company have been working seven hours each day, from 9 am to 4 pm. To try to reduce absence, the company decides to introduce 'flexi-time' and allow employees to work their seven hours each day at any time between 7 am and 9 pm. For a random sample of 10 employees, the numbers of hours of absence in the year before and the year after the introduction of flexi-time are given in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | c | }
\hline
Employee & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ & $I$ & $J$ \\
\hline
Before & 42 & 35 & 96 & 74 & 20 & 5 & 78 & 45 & 146 & 0 \\
\hline
After & 34 & 32 & 100 & 72 & 31 & 2 & 61 & 35 & 140 & 0 \\
\hline
\end{tabular}
\end{center}
Test, at the $10\%$ significance level, whether the population mean number of hours of absence has decreased following the introduction of flexi-time, stating any assumption that you make.
\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q3 [8]}}