| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2024 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample t-test equal variance |
| Difficulty | Standard +0.8 This is a two-sample t-test with equal variances (pooled variance) from Further Statistics. While it requires multiple computational steps (calculating sample statistics, pooled variance, test statistic, and critical value comparison), it follows a standard procedure taught in the syllabus. The question is straightforward in setup with clearly stated hypotheses, though the calculations are somewhat involved. It's moderately harder than average due to being Further Maths content and requiring careful arithmetic, but doesn't require novel insight or unusual problem-solving approaches. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s_x^2 = \frac{1}{7}\left(519.46 - \frac{64.2^2}{8}\right) [= 0.607857]\) | M1 | \(\frac{851}{1400}\) |
| \(s_y^2 = \frac{1}{8}\left(567.13 - \frac{71.1^2}{9}\right) [= 0.680]\) | \(\frac{17}{25}\), both | |
| \(s^2 = \frac{8 \times 0.680 + 7 \times 0.607857}{9 + 8 - 2} [= 0.64633]\) | M1 A1 | Correct formula for pooled variance used, \(\frac{1939}{3000}\) |
| \(H_0: \mu_X - \mu_Y = 0\); \(H_1: \mu_X - \mu_Y > 0\) | B1 | |
| \(t = \frac{8.025 - 7.9}{s\sqrt{\frac{1}{8} + \frac{1}{9}}} = \frac{0.125}{0.3906} = 0.320\) | M1 A1 | Dependent on pooled variance being used |
| \(0.320 < 1.341\) accept \(H_0\) | M1 | Compare with 1.341 and conclusion |
| Insufficient evidence to support Ansal's suspicion / Insufficient evidence to suggest that wingspan from region \(X\) is greater than wingspan from region \(Y\) | A1 | Correct work only, except possibly B1, in context, level of uncertainty in language |
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s_x^2 = \frac{1}{7}\left(519.46 - \frac{64.2^2}{8}\right) [= 0.607857]$ | M1 | $\frac{851}{1400}$ |
| $s_y^2 = \frac{1}{8}\left(567.13 - \frac{71.1^2}{9}\right) [= 0.680]$ | | $\frac{17}{25}$, both |
| $s^2 = \frac{8 \times 0.680 + 7 \times 0.607857}{9 + 8 - 2} [= 0.64633]$ | M1 A1 | Correct formula for pooled variance used, $\frac{1939}{3000}$ |
| $H_0: \mu_X - \mu_Y = 0$; $H_1: \mu_X - \mu_Y > 0$ | B1 | |
| $t = \frac{8.025 - 7.9}{s\sqrt{\frac{1}{8} + \frac{1}{9}}} = \frac{0.125}{0.3906} = 0.320$ | M1 A1 | Dependent on pooled variance being used |
| $0.320 < 1.341$ accept $H_0$ | M1 | Compare with 1.341 and conclusion |
| Insufficient evidence to support Ansal's suspicion / Insufficient evidence to suggest that wingspan from region $X$ is greater than wingspan from region $Y$ | A1 | Correct work only, except possibly B1, in context, level of uncertainty in language |6 Ansal is investigating the wingspans of Monarch butterflies in two different regions, $X$ and $Y$. He takes a random sample of 8 Monarch butterflies from region $X$ and records their wingspans, $x \mathrm {~cm}$. His results are as follows.
$$\begin{array} { l l l l l l l l }
8.2 & 7.0 & 7.3 & 8.8 & 7.8 & 8.5 & 9.2 & 7.4
\end{array}$$
Ansal also takes a random sample of 9 Monarch butterflies from region $Y$ and records their wingspans, $y \mathrm {~cm}$. His results are summarised as follows.
$$\sum y = 71.10 \quad \sum y ^ { 2 } = 567.13$$
Ansal suspects that the mean wingspan of Monarch butterflies from region $X$ is greater than the mean wingspan of Monarch butterflies from region $Y$. It is known that the wingspans of Monarch butterflies in regions $X$ and $Y$ are normally distributed with equal population variances.
Test, at the 10\% significance level, whether Ansal's suspicion is supported by the data.\\
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If you use the following page to complete the answer to any question, the question number must be clearly shown.\\
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\hfill \mbox{\textit{CAIE Further Paper 4 2024 Q6 [8]}}