| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2024 |
| 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 straightforward paired t-test with clearly paired data, testing a one-sided hypothesis about mean difference. Students must calculate differences, find sample statistics, and compare to critical value from tables. The setup is standard with no conceptual tricks, though it requires careful handling of the direction (reduction > 1 second) and correct use of one-tailed test at 10% level. Slightly above average difficulty due to being Further Maths content and requiring precise execution of multiple steps, but remains a textbook application of the paired t-test procedure. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Student | A | B | C | D | \(E\) | \(F\) | G | \(H\) | I | J |
| Time before course | 54.2 | 47.4 | 52.1 | 59.0 | 55.3 | 51.0 | 48.9 | 52.2 | 58.4 | 51.4 |
| Time after course | 50.1 | 46.3 | 52.5 | 58.8 | 51.4 | 48.4 | 49.5 | 48.7 | 58.3 | 51.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Signed differences: \(4.1\ \ 1.1\ \ {-0.4}\ \ 0.2\ \ 3.9\ \ 2.6\ \ {-0.6}\ \ 3.5\ \ 0.1\ \ 0\) | M1 | Allow one error, must be signed differences. |
| \(\sum d = 14.5,\quad \sum d^2 = 52.81\quad s^2 = \frac{1}{9}\!\left(52.81 - \frac{14.5^2}{10}\right) = 3.532\) | M1 | |
| \(H_0: \mu_B - \mu_A = 1\); \(H_1: \mu_B - \mu_A > 1\) | B1 | Accept \(H_0: \mu_d = 1\), \(H_1: \mu_d > 1\). |
| \(t = \dfrac{1.45 - 1}{\sqrt{\dfrac{s^2}{10}}}\) | M1 | The \(-1\) must be present in the numerator. |
| \(0.757\) | A1 | |
| Critical value is \(1.383\): \(0.757 < 1.383\), Accept \(H_0\) | M1 | '0.757' must come from a paired sample \(t\)-calculation, 1.383 must be correct. Correct ft conclusion for their 0.757 and 1.383. Condone Reject \(H_1\). 'Accept \(H_0\)' can be implied by a conclusion consistent with their 0.757 and 1.383. |
| Insufficient evidence to support Jade's claim | A1 | Correct work only, ignoring their hypotheses, conclusion in context with level of uncertainty in language. Not 'prove'. Condone 'no sufficient' 'not enough'. Do not accept statements such as 'there is sufficient evidence to suggest….' |
| Total: 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Distribution of population differences is normal | B1 | Underlying distribution of differences normal. 'data' implies B0. |
| Total: 1 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Signed differences: $4.1\ \ 1.1\ \ {-0.4}\ \ 0.2\ \ 3.9\ \ 2.6\ \ {-0.6}\ \ 3.5\ \ 0.1\ \ 0$ | M1 | Allow one error, must be signed differences. |
| $\sum d = 14.5,\quad \sum d^2 = 52.81\quad s^2 = \frac{1}{9}\!\left(52.81 - \frac{14.5^2}{10}\right) = 3.532$ | M1 | |
| $H_0: \mu_B - \mu_A = 1$; $H_1: \mu_B - \mu_A > 1$ | B1 | Accept $H_0: \mu_d = 1$, $H_1: \mu_d > 1$. |
| $t = \dfrac{1.45 - 1}{\sqrt{\dfrac{s^2}{10}}}$ | M1 | The $-1$ must be present in the numerator. |
| $0.757$ | A1 | |
| Critical value is $1.383$: $0.757 < 1.383$, Accept $H_0$ | M1 | '0.757' must come from a paired sample $t$-calculation, 1.383 must be correct. Correct ft conclusion for their 0.757 and 1.383. Condone Reject $H_1$. 'Accept $H_0$' can be implied by a conclusion consistent with their 0.757 and 1.383. |
| Insufficient evidence to support Jade's claim | A1 | **Correct work only**, ignoring their hypotheses, conclusion in context with level of uncertainty in language. Not 'prove'. Condone 'no sufficient' 'not enough'. Do not accept statements such as 'there is sufficient evidence to suggest….' |
| **Total: 7** | | |
---
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Distribution of **population differences** is normal | B1 | Underlying distribution of differences normal. 'data' implies B0. |
| **Total: 1** | | |
---6 Jade is a swimming instructor at a sports college. She claims that, as a result of an intensive training course, the mean time taken by students to swim 50 metres has reduced by more than 1 second. She chooses a random sample of 10 students. The times taken, in seconds, before and after the training course are recorded in the table.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|}
\hline
Student & A & B & C & D & $E$ & $F$ & G & $H$ & I & J \\
\hline
Time before course & 54.2 & 47.4 & 52.1 & 59.0 & 55.3 & 51.0 & 48.9 & 52.2 & 58.4 & 51.4 \\
\hline
Time after course & 50.1 & 46.3 & 52.5 & 58.8 & 51.4 & 48.4 & 49.5 & 48.7 & 58.3 & 51.4 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Test, at the 10\% significance level, whether Jade's claim is justified.
\item State an assumption that is necessary for this test to be valid.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2024 Q6 [8]}}