Standard +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.
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 \(B\).
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]}}