| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| 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 with clear data presentation requiring routine calculation of differences, their mean and standard deviation, followed by application of the t-test formula and comparison with critical values. While it involves multiple computational steps, it follows a well-established procedure with no conceptual challenges or novel insights required, making it slightly easier than average for Further Maths statistics. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Cyclist | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) |
| Course 1 | 18.5 | 17.8 | 19.2 | 22.3 | 16.5 | 20.0 |
| Course 2 | 20.2 | 20.4 | 18.1 | 20.6 | 18.5 | 20.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1.7\ 2.6\ -1.1\ -1.7\ 2.0\ 0.5\) | M1 | Consider differences e.g. course 2 – course 1 |
| \(\bar{d} = 4/6 = 0.6667\) | M1 | Calculate sample mean |
| \(s^2 = (18 - 4^2/6)/5 = 46/15\) or \(3.067\) or \(1.751^2\) | M1 | Estimate population variance (allow biased: \(23/9\) or \(2.556\) or \(1.599^2\)) |
| \(H_0: \mu_1 - \mu_2 = 0,\ H_1: \mu_1 - \mu_2 \neq 0\) | B1 | State hypotheses (A.E.F.) |
| \(t = \bar{d}/(s/\sqrt{6}) = 0.932[5]\) | M1 A1 | Calculate value of \(t\) (to 3 s.f.) |
| \(t_{5,\ 0.95} = 2.01[5]\) | *B1 | State or use correct tabular \(t\) value (or can compare \(\bar{d}\) with 1.44) |
| No difference between mean times | \(B1\sqrt{}\) | Correct conclusion (AEF, \(\sqrt{}\) on \(t\), dep *B1) |
## Question 7:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1.7\ 2.6\ -1.1\ -1.7\ 2.0\ 0.5$ | M1 | Consider differences e.g. course 2 – course 1 |
| $\bar{d} = 4/6 = 0.6667$ | M1 | Calculate sample mean |
| $s^2 = (18 - 4^2/6)/5 = 46/15$ or $3.067$ or $1.751^2$ | M1 | Estimate population variance (allow biased: $23/9$ or $2.556$ or $1.599^2$) |
| $H_0: \mu_1 - \mu_2 = 0,\ H_1: \mu_1 - \mu_2 \neq 0$ | B1 | State hypotheses (A.E.F.) |
| $t = \bar{d}/(s/\sqrt{6}) = 0.932[5]$ | M1 A1 | Calculate value of $t$ (to 3 s.f.) |
| $t_{5,\ 0.95} = 2.01[5]$ | *B1 | State or use correct tabular $t$ value (or can compare $\bar{d}$ with 1.44) |
| No difference between mean times | $B1\sqrt{}$ | Correct conclusion (AEF, $\sqrt{}$ on $t$, dep *B1) |
**Total: 8 marks**
---7 Each of a random sample of 6 cyclists from a cycling club is timed over two different 10 km courses. Their times, in minutes, are recorded in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Cyclist & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ \\
\hline
Course 1 & 18.5 & 17.8 & 19.2 & 22.3 & 16.5 & 20.0 \\
\hline
Course 2 & 20.2 & 20.4 & 18.1 & 20.6 & 18.5 & 20.5 \\
\hline
\end{tabular}
\end{center}
Assuming that differences in time over the two courses are normally distributed, test at the $10 \%$ significance level whether the mean times over the two courses are different.
\hfill \mbox{\textit{CAIE FP2 2013 Q7 [8]}}