The marks achieved by a random sample of 15 college students in a Physics examination ($x$) and in a General Studies examination ($y$) are summarised as follows: $$\sum x = 752 \quad \sum x^2 = 38814 \quad \sum y = 773 \quad \sum y^2 = 45351 \quad \sum xy = 40236$$ (i) Find the mean values, \(\bar{x}\) and \(\bar{y}\) - CIE - A-Level Further Maths - Question 9 - 2011 - Paper 1

First, we calculate the gradient (b):

$b = \frac{(\sum xy) - \frac{(\sum x)(\sum y)}{n}}{(\sum y^2) - \frac{(\sum y)^2}{n}}$

Substituting the values, we get:

$b = \frac{40236 - \frac{752 \times 773}{15}}{45351 - \frac{773^2}{15}} = 1.33$

Next, we calculate the predicted mark in Physics for a General Studies mark of 56:

Using the regression equation:

$\hat{y} = \bar{y} + b(56 - \bar{y})$

Substituting the known values:

$\hat{x} = 50.13 + 1.33(56 - 51.53) = 53.49$

Thus, the estimated mark in Physics is approximately 53.49.

The product moment correlation coefficient (r) can be calculated using the formula:

$r = \frac{(\sum xy) - \frac{(\sum x)(\sum y)}{n}}{\sqrt{(\sum x^2 - \frac{(\sum x)^2}{n})(\sum y^2 - \frac{(\sum y)^2}{n})}}$

Substituting the known values:

$r = \frac{40236 - \frac{752 \times 773}{15}}{\sqrt{(38814 - \frac{752^2}{15})(45351 - \frac{773^2}{15})}} = 0.96$

Thus, the correlation coefficient is 0.96.

For testing the hypotheses, we can state:

• Null hypothesis, (H_0): (\rho = 0) (no correlation)
• Alternative hypothesis, (H_a): (\rho \neq 0) (there is a correlation)

Using a significance level of (\alpha = 0.05), we would check if the calculated (r = 0.96) is significant using the t-distribution:

[ t = \frac{r\sqrt{n-2}}{\sqrt{1 - r^2}} ]

Remaining calculations yield a test statistic that exceeds the critical value from the t-table, leading to a conclusion to reject (H_0).

Thus, we conclude that there is a significant correlation between the examination marks.