| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Paired t-test |
| Difficulty | Challenging +1.2 This is a standard Wilcoxon signed-rank test application requiring students to calculate differences, subtract 0.1, rank absolute values, sum ranks, and compare to critical values. While it involves multiple computational steps and proper hypothesis formulation, it follows a completely routine procedure with no conceptual challenges beyond textbook application. The 10 marks reflect the length of working rather than conceptual difficulty. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Runner | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) |
| Indoor time | 21.5 | 21.8 | 20.9 | 21.2 | 21.4 | 21.4 | 21.2 | 21.0 |
| Outdoor time | 21.1 | 21.7 | 20.7 | 20.9 | 21.3 | 21.0 | 21.1 | 20.8 |
| Answer | Marks |
|---|---|
| 9 | State both hypotheses (A.E.F.): H0: µI – µO = 0⋅1, H1: µI – µO > 0⋅1 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Correct conclusion (AEF, dep *A1, *B1): Coach’s suspicion is not correct (B1 max 7) | 10 | [10] |
Question 9:
9 | State both hypotheses (A.E.F.): H0: µI – µO = 0⋅1, H1: µI – µO > 0⋅1 B1
State valid assumption for paired-sample test: Popln. of diffs. has Normal distn. B1
Consider differences eg: 0⋅4 0⋅1 0⋅2 0⋅3 0⋅1 0⋅4 0⋅1 0⋅2 M1
Calculate sample mean: d = 1⋅8 / 8 [= 0⋅225] M1
Estimate population variance: s 2 = (0⋅52 – 1⋅8 2 /8) / 7
(allow biased: 0⋅0144 or 0⋅120 2 ) [= 0⋅0164 or 0⋅128 2 ] M1
Calculate value of t (to 2 dp): t = (d – 0⋅1) / s√(1/8) = 2⋅76 M1*A1
Compare with correct tabular t value: t7, 0.975 = 2⋅36[5] *B1
χ2
Valid method for reaching conclusion: Reject H0 if > tabular value M1
Correct conclusion (AEF, dep *A1, *B1): Coach’s suspicion is correct A1
S.R.: State both hypotheses: H0: µI – µO = 0⋅1, H1: µI – µO > 0⋅1 (B1)
State valid assumption for 2-sample test: Both poplns. have Normal distns.
and a common variance (B1)
Calculate sample means: 170⋅4/8, 168⋅6/8 [= 21⋅3, 21⋅075]
and estimate population variance: s 2 = (3630⋅1 – 170⋅4 2 /8 + 3553⋅94
– 168⋅6 2 /8)/14 [= 0⋅09107] (M1)
Calculate value of t (to 2 dp): (0⋅225 – 0⋅1)/s√(1/8 + 1/8) = 0⋅828 (M1*A1)
Compare with correct tabular t value: t14, 0.975 = 2⋅14[5] (*B1)
Correct conclusion (AEF, dep *A1, *B1): Coach’s suspicion is not correct (B1 max 7) | 10 | [10]A national athletics coach suspects that, on average, 200-metre runners' indoor times exceed their outdoor times by more than 0.1 seconds. In order to test this, the coach randomly selects eight 200-metre runners and records their indoor and outdoor times. The results, in seconds, are shown in the table.
\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|c|c|}
\hline
Runner & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ \\
\hline
Indoor time & 21.5 & 21.8 & 20.9 & 21.2 & 21.4 & 21.4 & 21.2 & 21.0 \\
\hline
Outdoor time & 21.1 & 21.7 & 20.7 & 20.9 & 21.3 & 21.0 & 21.1 & 20.8 \\
\hline
\end{tabular}
\end{center}
Stating suitable hypotheses and any necessary assumption that you make, test the coach's suspicion at the 2.5% level of significance. [10]
\hfill \mbox{\textit{CAIE FP2 2010 Q9 [10]}}