CAIE FP2 2019 June — Question 8 8 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2019
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 with clear structure: calculate differences, find mean and standard deviation, apply the t-test formula, and compare to critical value. While it requires multiple computational steps and understanding of hypothesis testing, it follows a routine procedure taught in Further Statistics with no conceptual surprises or novel problem-solving required.
Spec5.05c Hypothesis test: normal distribution for population mean

8 A large number of runners are attending a summer training camp. A random sample of 6 runners is chosen and their times to run 1500 m at the beginning of the camp and at the end of the camp are recorded. Their times, in minutes, are shown in the following table.
Runner\(A\)\(B\)\(C\)\(D\)\(E\)\(F\)
Time at beginning of camp3.823.623.553.713.753.92
Time at end of camp3.723.553.523.683.543.73
The organiser of the training camp claims that a runner's time will improve by more than 0.05 minutes between the beginning and end of the camp. Assuming that differences in time over the two runs are normally distributed, test at the \(10 \%\) significance level whether the organiser's claim is justified. [8]

Question 8:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0: \mu_b - \mu_e = 0.05\), \(H_1: \mu_b - \mu_e > 0.05\) (AEF)B1 State both hypotheses (B0 for \(\bar{x}\ldots\))
\(d_i\): \(0.10\ \ 0.07\ \ 0.03\ \ 0.03\ \ 0.21\ \ 0.19\) (or in sec)M1 Consider differences \(d_i\) from e.g. \(x_b - y_e\)
\(\bar{d} = 0.63/6 = 0.105\) (or 6.3 sec)B1 Find sample mean
\(s^2 = (0.0969 - 0.63^2/6)/5 = 123/20000\) or \(0.00615\) or \(0.0784^2\) (or 22.14)M1 Estimate population variance (allow biased: \([41/8000\) or \(0.005125\) or \(0.0716^2]\))
\(t_{5,\,0.9} = 1.476\) (to 3 sf)B1 State or use correct tabular \(t\)-value
\(t = (\bar{d} - 0.05)/(s/\sqrt{6}) = 1.72\)M1 A1 Find value of \(t\) (or compare \(\bar{d}-0.05 = 0.055\) with \((t_{5,\,0.9})s/\sqrt{6} = 0.0473\))
[Reject \(H_0\):] Evidence for organiser's belief or times improve by more than 0.05 min (AEF)B1 FT on both \(t\)-values. Consistent conclusion. SC Wrong type of hypothesis test: B1 for hypotheses, B1FT for conclusion (max 2/8)
8
## Question 8:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu_b - \mu_e = 0.05$, $H_1: \mu_b - \mu_e > 0.05$ (AEF) | B1 | State both hypotheses (B0 for $\bar{x}\ldots$) |
| $d_i$: $0.10\ \ 0.07\ \ 0.03\ \ 0.03\ \ 0.21\ \ 0.19$ (or in sec) | M1 | Consider differences $d_i$ from e.g. $x_b - y_e$ |
| $\bar{d} = 0.63/6 = 0.105$ (or 6.3 sec) | B1 | Find sample mean |
| $s^2 = (0.0969 - 0.63^2/6)/5 = 123/20000$ or $0.00615$ or $0.0784^2$ (or 22.14) | M1 | Estimate population variance (allow biased: $[41/8000$ or $0.005125$ or $0.0716^2]$) |
| $t_{5,\,0.9} = 1.476$ (to 3 sf) | B1 | State or use correct tabular $t$-value |
| $t = (\bar{d} - 0.05)/(s/\sqrt{6}) = 1.72$ | M1 A1 | Find value of $t$ (or compare $\bar{d}-0.05 = 0.055$ with $(t_{5,\,0.9})s/\sqrt{6} = 0.0473$) |
| [Reject $H_0$:] Evidence for organiser's belief or times improve by more than 0.05 min (AEF) | B1 | **FT** on both $t$-values. Consistent conclusion. **SC** Wrong type of hypothesis test: B1 for hypotheses, B1FT for conclusion (max 2/8) |
| | **8** | |
8 A large number of runners are attending a summer training camp. A random sample of 6 runners is chosen and their times to run 1500 m at the beginning of the camp and at the end of the camp are recorded. Their times, in minutes, are shown in the following table.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Runner & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ \\
\hline
Time at beginning of camp & 3.82 & 3.62 & 3.55 & 3.71 & 3.75 & 3.92 \\
\hline
Time at end of camp & 3.72 & 3.55 & 3.52 & 3.68 & 3.54 & 3.73 \\
\hline
\end{tabular}
\end{center}

The organiser of the training camp claims that a runner's time will improve by more than 0.05 minutes between the beginning and end of the camp. Assuming that differences in time over the two runs are normally distributed, test at the $10 \%$ significance level whether the organiser's claim is justified. [8]\\

\hfill \mbox{\textit{CAIE FP2 2019 Q8 [8]}}