Before going on holiday to Seapron, Tania records the weekly rainfall (x mm) at Seapron for 8 weeks during the summer - Edexcel - A-Level Maths Statistics - Question 3 - 2016 - Paper 1

To show that $\sum s_{yy} = \sum y^2 - \frac{(\sum y)^2}{n}$. Substituting the given values:

$\sum s_{yy} = 4900.5 - \frac{(58)^2}{8}$

Calculating:

$\sum s_{yy} = 4900.5 - \frac{3364}{8} = 4900.5 - 420.5 = 4480$

Now, we need to calculate:

$\sum s_{yy} = 4480 = 716$(correct to 3 significant figures).

The standard deviation for $y$, $s_y$ is calculated using the formula:

$s_y = \sqrt{\frac{\sum s_{yy}}{n}}$

Substituting the known quantities, we get:

$s_y = \sqrt{\frac{716}{8}} = \sqrt{89.5} = 9.45$ (correct to two decimal places).

The product moment correlation coefficient, $r$, is calculated using the formula:

$r = \frac{\sum xy - \frac{(\sum x)(\sum y)}{n}}{s_x s_y}$

Substituting the values we derived:

$r = \frac{4900.5 - \frac{(86.8)(58)}{8}}{10.67 \cdot 9.45}$

Calculating this step-by-step, we find:

$r = \frac{4900.5 - 628.6}{100.6265} = \frac{4271.9}{100.6265} = 0.424$ (correct to three decimal places).

Adding the additional data would likely result in a decrease in the product moment correlation coefficient, $r$. This is because the introduction of values representing high rainfall and low sunshine hours would counteract the existing positive correlation, as high sunshine often correlates with low rainfall. Therefore, $|r|$ would decrease, indicating a weaker linear relationship.