CAIE Further Paper 4 2020 Specimen — Question 4 7 marks

Exam BoardCAIE
ModuleFurther Paper 4 (Further Paper 4)
Year2020
SessionSpecimen
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBivariate data
TypeHypothesis test for correlation
DifficultyStandard +0.8 This is a hypothesis test for correlation in bivariate data from Further Maths Statistics. While the OCR corruption makes the exact values unclear, this requires calculating product moment correlation coefficient and conducting a formal hypothesis test. It's a standard Further Maths procedure but more sophisticated than typical A-level statistics, involving multiple steps and understanding of correlation theory.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration

4 The number, \(x\), of a certain type of sea shell was counted at 60 randomly chosen sites, each one metre square, along the coastline in country \(A\). The number, \(y\), of the same type of sea shell was counted at 50 randomly chosen sites, each one metre square, along the coastline in country \(B\). The results are summarised as follows, where \(\bar{x}\) and \(\bar{y}\) denote the sample means of \(x\) and \(y\) respectively. $$\bar{x} = 29.2 \quad \Sigma(x - \bar{x})^{2} = 4341.6 \quad \bar{y} = 24.4 \quad \Sigma(y - \bar{y})^{2} = 3732.0$$ Find a \(95\%\) confidence interval for the difference between the mean number of sea shells, per square metre, on the coastlines in country \(A\) and in country \(B\).

Question 4:
AnswerMarks Guidance
AnswerMarks Guidance
Sample variances are \(\frac{4341.6}{59}\) and \(\frac{3732.0}{49}\)1 M1
\(s_x^2 = 73.59\) and \(s_y^2 = 76.16\)1 A1
Variance for difference of means: \(s^2 = \dfrac{s_x^2}{60} + \dfrac{s_y^2}{50}\)1 M1
\(s^2 = 2.750\) or \(s = 1.658\)1 A1
Evaluate \((\bar{x} - \bar{y}) \pm zs\)1 M1
Use of correct \(z = 1.96\)1 A1
End-points of confidence interval are 1.55 and 8.051 A1
Total7
## Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Sample variances are $\frac{4341.6}{59}$ and $\frac{3732.0}{49}$ | 1 | M1 | Either attempted |
| $s_x^2 = 73.59$ and $s_y^2 = 76.16$ | 1 | A1 | Both correct to at least 2 dp |
| Variance for difference of means: $s^2 = \dfrac{s_x^2}{60} + \dfrac{s_y^2}{50}$ | 1 | M1 | |
| $s^2 = 2.750$ or $s = 1.658$ | 1 | A1 | Correct value to at least 4 sf |
| Evaluate $(\bar{x} - \bar{y}) \pm zs$ | 1 | M1 | Any $z$-value, and using their value of $s$ |
| Use of correct $z = 1.96$ | 1 | A1 | |
| End-points of confidence interval are 1.55 and 8.05 | 1 | A1 | Accept e.g. $(1.55, 8.05)$ or $1.55 < \mu_A - \mu_B < 8.05$ |
| **Total** | **7** | | |

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4 The number, $x$, of a certain type of sea shell was counted at 60 randomly chosen sites, each one metre square, along the coastline in country $A$. The number, $y$, of the same type of sea shell was counted at 50 randomly chosen sites, each one metre square, along the coastline in country $B$. The results are summarised as follows, where $\bar{x}$ and $\bar{y}$ denote the sample means of $x$ and $y$ respectively.

$$\bar{x} = 29.2 \quad \Sigma(x - \bar{x})^{2} = 4341.6 \quad \bar{y} = 24.4 \quad \Sigma(y - \bar{y})^{2} = 3732.0$$

Find a $95\%$ confidence interval for the difference between the mean number of sea shells, per square metre, on the coastlines in country $A$ and in country $B$.

\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q4 [7]}}