| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2020 |
| Session | November |
| 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 application with clear structure: calculate differences, find mean and standard deviation, apply the t-test formula, and compare to critical value. The setup is straightforward with given data, though it requires careful handling of the one-tailed hypothesis (testing if reduction ≥ 0.3s) and stating the normality assumption. Slightly above average difficulty due to the directional claim and need for precise hypothesis formulation, but remains a textbook Further Maths statistics question. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Athlete | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) | \(I\) |
| Time before | 48.8 | 48.2 | 50.3 | 49.6 | 49.4 | 48.9 | 47.6 | 50.3 | 48.4 |
| Time after | 47.9 | 47.8 | 49.6 | 49.1 | 49.6 | 48.9 | 47.7 | 49.1 | 48.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Assume (population) differences are normally distributed | B1 | |
| \(H_0: \mu_X - \mu_Y = 0.3\), \(H_1: \mu_X - \mu_Y > 0.3\) | B1 | |
| Diff: \(0.9, 0.4, 0.7, 0.5, -0.2, 0, -0.1, 1.2, 0.3\) | M1 | Signed differences |
| \(\sum d = 3.7\), \(\sum d^2 = 3.29\); \(\bar{d} = 0.411\), \(s^2 = \frac{1}{8}\left(3.29 - \frac{3.7^2}{9}\right) = 0.2211\) | M1 | \(s = 0.4702\) |
| \(t = \frac{0.411 - 0.3}{\sqrt{\frac{0.2211}{9}}} = 0.708\) or \(0.709\) | M1 A1 | |
| \(0.708 < 1.397\) | M1 | Compare their 0.708 with 1.397 |
| Accept \(H_0\); insufficient evidence to support claim | A1 FT | In context, except possibly hypotheses. Level of uncertainty in language used. No contradictions |
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Assume (population) differences are normally distributed | B1 | |
| $H_0: \mu_X - \mu_Y = 0.3$, $H_1: \mu_X - \mu_Y > 0.3$ | B1 | |
| Diff: $0.9, 0.4, 0.7, 0.5, -0.2, 0, -0.1, 1.2, 0.3$ | M1 | Signed differences |
| $\sum d = 3.7$, $\sum d^2 = 3.29$; $\bar{d} = 0.411$, $s^2 = \frac{1}{8}\left(3.29 - \frac{3.7^2}{9}\right) = 0.2211$ | M1 | $s = 0.4702$ |
| $t = \frac{0.411 - 0.3}{\sqrt{\frac{0.2211}{9}}} = 0.708$ or $0.709$ | M1 A1 | |
| $0.708 < 1.397$ | M1 | Compare their 0.708 with 1.397 |
| Accept $H_0$; insufficient evidence to support claim | A1 FT | In context, except possibly hypotheses. Level of uncertainty in language used. No contradictions |4 Members of the Sprints athletics club have been taking part in an intense training scheme, aimed at reducing their times taken to run 400 m . For a random sample of 9 athletes from the club, the times taken, in seconds, before and after the training scheme are given in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | }
\hline
Athlete & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ & $I$ \\
\hline
Time before & 48.8 & 48.2 & 50.3 & 49.6 & 49.4 & 48.9 & 47.6 & 50.3 & 48.4 \\
\hline
Time after & 47.9 & 47.8 & 49.6 & 49.1 & 49.6 & 48.9 & 47.7 & 49.1 & 48.1 \\
\hline
\end{tabular}
\end{center}
The organiser of the training scheme claims that on average an athlete's time will be reduced by at least 0.3 seconds.
Test at the 10\% significance level whether the organiser's claim is justified, stating any assumption that you make.\\
\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q4 [8]}}