| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Standard CI with summary statistics |
| Difficulty | Standard +0.8 This is a two-sample t-test problem with unequal variances (Welch's t-test) requiring calculation of sample means and variances from summary statistics, construction of a 90% CI, and a one-tailed hypothesis test. While the individual steps are standard Further Maths Statistics techniques, the combination of summary statistics processing, unequal variance handling, and both CI and hypothesis test makes this moderately challenging with multiple opportunities for computational error. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s_x^2 = \frac{1}{49}\!\left(23480 - \frac{1080^2}{50}\right)(=3.102)\); \(s_y^2 = \frac{1}{39}\!\left(22220 - \frac{940^2}{40}\right)(=3.333)\) | M1 | \(\frac{152}{9}\), \(\frac{10}{3}\), Both. |
| \(s^2 = \frac{3.102}{50} + \frac{3.333}{40}\ (=0.1454)\) | M1 | |
| \(s^2 = \frac{2137}{14700} = 0.1454\) | A1 | |
| CI: \(\frac{1080}{50} - \frac{940}{40} \pm 1.645\sqrt{s^2}\) or \(\frac{940}{40} - \frac{1080}{50} \pm 1.645\sqrt{s^2}\) | M1 B1 | Use of correct formula with a \(z\) value. B1 for 1.645 (accept 1.64 or 1.65) used in CI formula. |
| \([1.27, 2.53]\) or \([-2.53, -1.27]\) | A1 | \((1.27, 2.53)\) A1. \(1.27 \leqslant \mu_Y - \mu_X \leqslant 2.53\) A1. Condone \([2.53, 1.27]\) or \([-1.27, -2.53]\) A1. Final answer \(1.9 \pm 0.627\) A0. Pooled estimate: max score M1 M0A0 M1B1A0. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z = \frac{1.9 - 1.1}{s}\) | M1 | \(z = \frac{-1.9-1.1}{s}\) or \(-7.87\) implies M1. |
| \(2.098\) | A1 | AWRT 2.10 |
| \(2.098 > 1.28(2)\) Reject \(H_0\); or \(\Phi(2.098) = 0.9821 > 0.9\) Reject \(H_0\); or \(1-\Phi(2.098)=0.0179 < 0.1\) Reject \(H_0\) | M1 | '2.098' from their attempt at \(z\); \(1.28(2)\) must be correct. Consistent signs in comparison. FT conclusion from their 2.098. Condone 'Accept \(H_1\)'. 'Reject \(H_0\)' can be implied by conclusion in context. |
| Sufficient evidence to suggest that the lengths of tails of wallabies in region \(Y\) are, on average, more than 1.1 cm longer than the tails of wallabies in region \(X\) | A1 | Correct work only; conclusion in context with level of uncertainty. Not 'prove' or equivalent. Accept 'enough evidence'. Condone 'some evidence'. |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s_x^2 = \frac{1}{49}\!\left(23480 - \frac{1080^2}{50}\right)(=3.102)$; $s_y^2 = \frac{1}{39}\!\left(22220 - \frac{940^2}{40}\right)(=3.333)$ | M1 | $\frac{152}{9}$, $\frac{10}{3}$, Both. |
| $s^2 = \frac{3.102}{50} + \frac{3.333}{40}\ (=0.1454)$ | M1 | |
| $s^2 = \frac{2137}{14700} = 0.1454$ | A1 | |
| CI: $\frac{1080}{50} - \frac{940}{40} \pm 1.645\sqrt{s^2}$ or $\frac{940}{40} - \frac{1080}{50} \pm 1.645\sqrt{s^2}$ | M1 B1 | Use of correct formula with a $z$ value. B1 for 1.645 (accept 1.64 or 1.65) used in CI formula. |
| $[1.27, 2.53]$ or $[-2.53, -1.27]$ | A1 | $(1.27, 2.53)$ A1. $1.27 \leqslant \mu_Y - \mu_X \leqslant 2.53$ A1. Condone $[2.53, 1.27]$ or $[-1.27, -2.53]$ A1. Final answer $1.9 \pm 0.627$ A0. Pooled estimate: max score M1 M0A0 M1B1A0. |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = \frac{1.9 - 1.1}{s}$ | M1 | $z = \frac{-1.9-1.1}{s}$ or $-7.87$ implies M1. |
| $2.098$ | A1 | AWRT 2.10 |
| $2.098 > 1.28(2)$ Reject $H_0$; or $\Phi(2.098) = 0.9821 > 0.9$ Reject $H_0$; or $1-\Phi(2.098)=0.0179 < 0.1$ Reject $H_0$ | M1 | '2.098' from their attempt at $z$; $1.28(2)$ must be correct. Consistent signs in comparison. FT conclusion from their 2.098. Condone 'Accept $H_1$'. 'Reject $H_0$' can be implied by conclusion in context. |
| Sufficient evidence to suggest that the **lengths** of tails of wallabies **in** region $Y$ are, **on average**, **more than 1.1 cm longer than** the tails of wallabies **in** region $X$ | A1 | Correct work only; conclusion in context with level of uncertainty. Not 'prove' or equivalent. Accept 'enough evidence'. Condone 'some evidence'. |6 Seva is investigating the lengths of the tails of adult wallabies in two regions of Australia, $X$ and $Y$. He chooses a random sample of 50 adult wallabies from region $X$ and records the lengths, $x \mathrm {~cm}$, of their tails. He also chooses a random sample of 40 adult wallabies from region $Y$ and records the lengths, $y \mathrm {~cm}$, of their tails. His results are summarised as follows.
$$\sum x = 1080 \quad \sum x ^ { 2 } = 23480 \quad \sum y = 940 \quad \sum y ^ { 2 } = 22220$$
It cannot be assumed that the population variances of the two distributions are the same.
\begin{enumerate}[label=(\alph*)]
\item Find a $90 \%$ confidence interval for the difference between the population mean lengths of the tails of adult wallabies in regions $X$ and $Y$.\\
\includegraphics[max width=\textwidth, alt={}, center]{b5ff998a-fcb6-4a1b-ae86-ec66b0dccc3c-10_2718_38_141_2010}
The population mean lengths of the tails of adult wallabies in regions $X$ and $Y$ are $\mu _ { X } \mathrm {~cm}$ and $\mu _ { Y } \mathrm {~cm}$ respectively.
\item Test, at the $10 \%$ significance level, the null hypothesis $\mu _ { Y } - \mu _ { X } = 1.1$ against the alternative hypothesis $\mu _ { Y } - \mu _ { X } > 1.1$. State your conclusion in the context of the question.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2024 Q6 [10]}}