CAIE Further Paper 4 2024 November — Question 1 6 marks

Exam BoardCAIE
ModuleFurther Paper 4 (Further Paper 4)
Year2024
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeStandard CI with summary statistics
DifficultyStandard +0.3 This is a straightforward two-sample confidence interval question requiring calculation of sample statistics from summaries, then applying the standard formula for difference of means. While it involves multiple steps (computing mean and variance for sample B, then constructing the CI), each step is routine application of standard formulas with no conceptual challenges or novel insights required.
Spec5.04b Linear combinations: of normal distributions5.05c Hypothesis test: normal distribution for population mean

1 Ellie is investigating the heights of two types of beech tree, \(A\) and \(B\), in a certain region. She has chosen a random sample of 60 beech trees of type \(A\) in the region, recorded their heights, \(x \mathrm {~m}\), and calculated unbiased estimates for the population mean and population variance as 35.6 m and \(4.95 \mathrm {~m} ^ { 2 }\) respectively. Ellie also chooses a random sample of 50 beech trees of type \(B\) in the region and records their heights, \(y \mathrm {~m}\). Her results are summarised as follows. $$\sum y = 1654 \quad \sum y ^ { 2 } = 54850$$ Find a \(95 \%\) confidence interval for the difference between the population mean heights of type \(A\) and type \(B\) beech trees in the region.

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(s_x^2 = \frac{1}{49}\left(54850 - \frac{1654^2}{50}\right) [= 2.769]\)B1 \(\frac{3392}{1225}\)
\(s^2 = \frac{4.95}{60} + \frac{2.769}{50} [= 0.1379]\)M1 A1
CI: \(35.6 - \frac{1654}{50} \pm 1.96s\)M1 Correct formula with a \(z\) value.
B11.96 seen.
\([1.79, \ 3.25]\)A1 Accept inequality.
Total: 6
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $s_x^2 = \frac{1}{49}\left(54850 - \frac{1654^2}{50}\right) [= 2.769]$ | **B1** | $\frac{3392}{1225}$ |
| $s^2 = \frac{4.95}{60} + \frac{2.769}{50} [= 0.1379]$ | **M1 A1** | |
| CI: $35.6 - \frac{1654}{50} \pm 1.96s$ | **M1** | Correct formula with a $z$ value. |
| | **B1** | 1.96 seen. |
| $[1.79, \ 3.25]$ | **A1** | Accept inequality. |
| **Total: 6** | | |

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1 Ellie is investigating the heights of two types of beech tree, $A$ and $B$, in a certain region. She has chosen a random sample of 60 beech trees of type $A$ in the region, recorded their heights, $x \mathrm {~m}$, and calculated unbiased estimates for the population mean and population variance as 35.6 m and $4.95 \mathrm {~m} ^ { 2 }$ respectively.

Ellie also chooses a random sample of 50 beech trees of type $B$ in the region and records their heights, $y \mathrm {~m}$. Her results are summarised as follows.

$$\sum y = 1654 \quad \sum y ^ { 2 } = 54850$$

Find a $95 \%$ confidence interval for the difference between the population mean heights of type $A$ and type $B$ beech trees in the region.\\

\hfill \mbox{\textit{CAIE Further Paper 4 2024 Q1 [6]}}