| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Paired sample t-test |
| Difficulty | Standard +0.3 This is a standard paired t-test question requiring calculation of differences, their mean and standard deviation, then applying the t-test formula and comparing to critical values. While it involves multiple computational steps and understanding of hypothesis testing framework, it follows a completely routine procedure taught in Further Maths Statistics with no novel insight required. The paired design is clearly signposted by the data structure, making it slightly easier than average A-level difficulty overall. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Swimmer | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) |
| Outdoor time | 66.2 | 62.4 | 60.8 | 65.4 | 68.8 | 64.3 | 65.2 | 67.2 |
| Indoor time | 66.1 | 60.3 | 60.9 | 65.2 | 66.4 | 63.8 | 62.4 | 69.8 |
| Answer | Marks |
|---|---|
| 7 | Consider differences e.g. outdoor – indoor |
| Answer | Marks | Guidance |
|---|---|---|
| Correct conclusion (AEF, dep *A1, *B1): No difference between mean times B1 | 8 | [8] |
Question 7:
7 | Consider differences e.g. outdoor – indoor
times: 0⋅1 2⋅1 -0⋅1 0⋅2 2⋅4 0⋅5 2⋅8 –2⋅6 M1
Calculate sample mean: d = 5⋅4 / 8 = 0⋅675 M1
Estimate population variance: s 2 = (25⋅08 – 5⋅4 2 /8) / 7
(allow biased here: 2⋅679 or 1⋅637 2 ) = 3⋅062 or 1⋅750 2 M1
State hypotheses (A.E.F.): H0: µo – µi = 0, H1: µo – µi ≠ 0 B1
Calculate value of t (to 3 sf): t = d/(s/√8) = 1⋅09 M1 *A1
State or use correct tabular t value: t7, 0.975 = 2⋅36[5] *B1
(or can compared with 1⋅46[3])
Correct conclusion (AEF, dep *A1, *B1): No difference between mean times B1 | 8 | [8]A random sample of 8 swimmers from a swimming club were timed over a distance of 100 metres, once in an outdoor pool and once in an indoor pool. Their times, in seconds, are given in the following table.
\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|c|c|}
\hline
Swimmer & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ \\
\hline
Outdoor time & 66.2 & 62.4 & 60.8 & 65.4 & 68.8 & 64.3 & 65.2 & 67.2 \\
\hline
Indoor time & 66.1 & 60.3 & 60.9 & 65.2 & 66.4 & 63.8 & 62.4 & 69.8 \\
\hline
\end{tabular}
\end{center}
Assuming a normal distribution, test, at the 5% significance level, whether there is a non-zero difference between mean time in the outdoor pool and mean time in the indoor pool. [8]
\hfill \mbox{\textit{CAIE FP2 2012 Q7 [8]}}