Q.Using the same advertising expenditure and sales data as Worked Example 1, find Pearson's correlation coefficient between the two variables.

Pearson's $r$ is the covariance (already found in Worked Example 1) divided by the product of the two standard deviations.

$\sum(x-\bar{x})^2=10$, $\sum(y-\bar{y})^2=6$

$r = \dfrac{6}{\sqrt{10\times6}} = \dfrac{6}{\sqrt{60}} \approx 0.775$

We use $r = \dfrac{\sum(x-\bar{x})(y-\bar{y})}{\sqrt{\sum(x-\bar{x})^2}\sqrt{\sum(y-\bar{y})^2}}$, reusing the deviations already tabulated in Worked Example 1 ($\bar{x}=3,\bar{y}=4$).

Step 1 — Recall the cross-product sum. From Worked Example 1, $\sum(x-\bar{x})(y-\bar{y}) = 6$.

Step 2 — Find $\sum(x-\bar{x})^2$ and $\sum(y-\bar{y})^2$.

Step 3 — Apply the formula.

Independent check (via covariance and standard deviations). $\text{Var}(x) = 10/5 = 2$, $\text{Var}(y) = 6/5 = 1.2$, so $\sigma_x = \sqrt{2} \approx 1.414$, $\sigma_y = \sqrt{1.2} \approx 1.095$. Using $\text{Cov}(x,y) = 1.2$ from Worked Example 1, $r = 1.2/(1.414\times1.095) = 1.2/1.549 \approx 0.775$ — matches exactly.

Dividing covariance by the variances instead of the standard deviations (i.e. forgetting to take the square root) is a common error — always check that the denominator, $\sigma_x\sigma_y$, has the same units as the numerator so the final $r$ is a pure, unit-free number between $-1$ and $1$.