CAIE Further Paper 4 2021 November — Question 1 7 marks

Exam BoardCAIE
ModuleFurther Paper 4 (Further Paper 4)
Year2021
SessionNovember
Marks7
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TopicLinear combinations of normal random variables
TypeStandard CI with summary statistics
DifficultyStandard +0.3 This is a straightforward two-sample confidence interval problem with summary statistics provided. Students need to calculate sample means and variances from the summaries, then apply the standard formula for a CI for difference in means. While it requires careful arithmetic and knowledge of the correct formula, it involves no conceptual challenges or novel problem-solving—just direct application of a standard procedure from the syllabus.
Spec5.05d Confidence intervals: using normal distribution
1 The number, \(x\), of pine trees was counted in each of 40 randomly chosen regions of equal size in country \(A\). The number, \(y\), of pine trees was counted in each of 60 randomly chosen regions of the same equal size in country \(B\). The results are summarised as follows. $$\sum x = 752 \quad \sum x ^ { 2 } = 14320 \quad \sum y = 1548 \quad \sum y ^ { 2 } = 40200$$ Find a 95\% confidence interval for the difference between the mean number of pine trees in regions of this size in countries \(A\) and \(B\).
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(s_x^2 = \frac{1}{39}\left(14320 - \frac{752^2}{40}\right) = 4.677 = \frac{304}{65}\)M1 One correct unsimplified.
\(s_y^2 = \frac{1}{59}\left(40200 - \frac{1548^2}{60}\right) = 4.434 = \frac{1308}{295}\)A1 Both correct to at least 3sf.
\(s^2 = \frac{4.6769}{40} + \frac{4.4339}{60}\)M1
\(0.1908 \left(= \frac{3659}{19175}\right)\)A1
\(\text{CI} = \left(\frac{752}{40} - \frac{1548}{60}\right) \pm 1.96s\)M1A1 Correct form with a \(z\)-value with \(1.96\ s\) can be unsimplified.
\(-7.0 \pm 0.856 = [-7.86,\ -6.14]\)A1 Accept in either form, ISW. Accept \([6.14, 7.86]\). Accept inequality form.
7 
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $s_x^2 = \frac{1}{39}\left(14320 - \frac{752^2}{40}\right) = 4.677 = \frac{304}{65}$ | M1 | One correct unsimplified. |
| $s_y^2 = \frac{1}{59}\left(40200 - \frac{1548^2}{60}\right) = 4.434 = \frac{1308}{295}$ | A1 | Both correct to at least 3sf. |
| $s^2 = \frac{4.6769}{40} + \frac{4.4339}{60}$ | M1 | |
| $0.1908 \left(= \frac{3659}{19175}\right)$ | A1 | |
| $\text{CI} = \left(\frac{752}{40} - \frac{1548}{60}\right) \pm 1.96s$ | M1A1 | Correct form with a $z$-value with $1.96\ s$ can be unsimplified. |
| $-7.0 \pm 0.856 = [-7.86,\ -6.14]$ | A1 | Accept in either form, ISW. Accept $[6.14, 7.86]$. Accept inequality form. |
| | **7** | |
1 The number, $x$, of pine trees was counted in each of 40 randomly chosen regions of equal size in country $A$. The number, $y$, of pine trees was counted in each of 60 randomly chosen regions of the same equal size in country $B$. The results are summarised as follows.

$$\sum x = 752 \quad \sum x ^ { 2 } = 14320 \quad \sum y = 1548 \quad \sum y ^ { 2 } = 40200$$

Find a 95\% confidence interval for the difference between the mean number of pine trees in regions of this size in countries $A$ and $B$.\\

\hfill \mbox{\textit{CAIE Further Paper 4 2021 Q1 [7]}}
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