CAIE FP2 2012 June — Question 7 8 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2012
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypePaired sample t-test
DifficultyStandard +0.3 This is a standard paired t-test application with clear structure: calculate differences, find mean and standard deviation of differences, apply the t-test formula, and compare to critical value. While it requires careful arithmetic and knowledge of the paired t-test procedure, it follows a routine template with no conceptual surprises or novel problem-solving required. The 8-mark allocation reflects computational work rather than difficulty. Slightly above average (0.0) because it's Further Maths content and requires accurate multi-step calculation, but still a textbook exercise.
Spec5.05c Hypothesis test: normal distribution for population mean

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.
Swimmer\(A\)\(B\)\(C\)\(D\)\(E\)\(F\)\(G\)\(H\)
Outdoor time66.262.460.865.468.864.365.267.2
Indoor time66.160.360.965.266.463.862.469.8
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]

Question 7:
AnswerMarks
7Consider 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 compared with 1⋅46[3])
AnswerMarks Guidance
Correct conclusion (AEF, dep *A1, *B1): No difference between mean times B18 [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 compared 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{tabular}{|c|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}

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]}}