| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Draw back-to-back stem-and-leaf diagram |
| Difficulty | Easy -1.8 This is a routine data representation task requiring students to construct a back-to-back stem-and-leaf diagram from given data sets. It involves basic data organization skills with no problem-solving or conceptual depth—purely mechanical execution of a standard statistical display technique. |
| Spec | 2.01a Population and sample: terminology2.02a Interpret single variable data: tables and diagrams |
| 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/Working | Marks | Guidance |
| Differences: \(8\ 3\ {-4}\ 2\ {-11}\ 3\ 17\ 10\ 6\ 0\) | M1 | Consider differences before – after |
| \(\bar{d} = 34/10 = 3.4\) | Calculate sample mean | |
| \(s^2 = (648 - 34^2/10)/9 = 2662/45\) or \(59.16\) or \(7.691^2\) | M1 | Estimate population variance (biased: \(53.24\) or \(7.297^2\) allowed) |
| \(H_0: \mu_B - \mu_A = 0\), \(H_1: \mu_B - \mu_A > 0\) | B1 | State hypotheses, A.E.F. |
| \(t = \bar{d}/(s/\sqrt{10}) = 1.398\) or \(1.4\) | M1 A1 | Calculate value of \(t\) |
| \(t_{9,\,0.9} = 1.38[3]\) (or compare \(\bar{d}\) with \(3.36[4]\)) | B1 | State or use correct tabular \(t\)-value |
| Reject \(H_0\); Hours of absence have decreased | B1\(\checkmark\) | Consistent conclusion, A.E.F. |
## Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Differences: $8\ 3\ {-4}\ 2\ {-11}\ 3\ 17\ 10\ 6\ 0$ | M1 | Consider differences before – after |
| $\bar{d} = 34/10 = 3.4$ | | Calculate sample mean |
| $s^2 = (648 - 34^2/10)/9 = 2662/45$ or $59.16$ or $7.691^2$ | M1 | Estimate population variance (biased: $53.24$ or $7.297^2$ allowed) |
| $H_0: \mu_B - \mu_A = 0$, $H_1: \mu_B - \mu_A > 0$ | B1 | State hypotheses, A.E.F. |
| $t = \bar{d}/(s/\sqrt{10}) = 1.398$ or $1.4$ | M1 A1 | Calculate value of $t$ |
| $t_{9,\,0.9} = 1.38[3]$ (or compare $\bar{d}$ with $3.36[4]$) | B1 | State or use correct tabular $t$-value |
| Reject $H_0$; Hours of absence have decreased | B1$\checkmark$ | Consistent conclusion, A.E.F. |
*Note: Wrong test can earn only B1 for hypotheses and B1 for conclusion*
---6 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 & $A$ & $B$ & \multicolumn{1}{|c|}{$C$} & $D$ & $E$ & $F$ & $G$ & $H$ & $I$ & $J$ \\
\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}
Use a paired sample $t$-test to test, at the $10 \%$ significance level, whether the population mean number of hours of absence has decreased, following the introduction of flexi-time.
\hfill \mbox{\textit{CAIE FP2 2014 Q6}}