Two Variables

Data: $(1, 2), (2, 3), (3, 5), (4, 4), (5, 6)$. Means $\bar x = 3$, $\bar y = 4$. Deviations: $x - \bar x = -2, -1, 0, 1, 2$ and $y - \bar y = -2, -1, 1, 0, 2$. Products sum to $4 + 1 + 0 + 0 + 4 = 9$; squares sum to $10$ for $x$ and $10$ for $y$. So $r = \dfrac{9}{\sqrt{10\cdot10}} = 0.9$: a strong positive linear pattern. (Dividing each deviation by its SD and averaging the products with $n - 1$ gives the same number.)

Points tightly along a line rising to the right: $r \approx 0.9$. A loose cloud falling to the right: $r \approx -0.4$. A circle of points: $r \approx 0$. The sign is the direction; the size is how closely the points hug a line — not the steepness of the line.

Data: $x = -2, -1, 0, 1, 2$ with $y = x^2 = 4, 1, 0, 1, 4$. Here $\bar x = 0$ and $\bar y = 2$, and $\sum (x - \bar x)(y - \bar y) = (-2)(2) + (-1)(-1) + 0 + (1)(-1) + (2)(2) = 0$. So $r = 0$ although $y$ is completely determined by $x$. Correlation measures LINEAR association only.

$\bar x = 10$, $s_x = 2$, $\bar y = 50$, $s_y = 8$, $r = 0.75$. Slope $b = 0.75\cdot8/2 = 3$. The line passes through the point of means, so $a = 50 - 3\cdot10 = 20$. Line: $\hat y = 20 + 3x$. Predicted $y$ at $x = 12$: $20 + 36 = 56$.

For $(1, 2), (2, 3), (3, 5), (4, 4), (5, 6)$: $b = \dfrac{\sum(x - \bar x)(y - \bar y)}{\sum(x - \bar x)^2} = \dfrac{9}{10} = 0.9$ (the same as $r\,s_y/s_x$, since $s_x = s_y$ here). Then $a = 4 - 0.9\cdot3 = 1.3$: $\hat y = 1.3 + 0.9x$. The predictions $2.2, 3.1, 4.0, 4.9, 5.8$ leave residuals $-0.2, -0.1, 1.0, -0.9, 0.2$, which sum to $0$ — the line balances the points above it against those below.

With $\hat y = 1.3 + 0.9x$, the prediction at $x = 6$ is $1.3 + 5.4 = 6.7$ — a mild extrapolation just past the data. At $x = 50$ the line says $46.3$, a number with no support at all: nothing in five points between $x = 1$ and $x = 5$ says the pattern continues that far.

Residuals $-0.2, -0.1, 1.0, -0.9, 0.2$ have squares summing to $0.04 + 0.01 + 1 + 0.81 + 0.04 = 1.9$. The total variation of $y$ is $\sum(y - 4)^2 = 10$. So $r^2 = 1 - 1.9/10 = 0.81$, which is indeed $0.9^2$. The line explains $81\%$ of the variation in $y$.

Fit a line to $(1, 1), (2, 4), (3, 9), (4, 16), (5, 25)$: $b = 60/10 = 6$, $a = 11 - 18 = -7$, $\hat y = -7 + 6x$. The residuals are $2, -1, -2, -1, 2$: positive, negative, positive — a U. And $r^2 = 1 - 14/374 \approx 0.96$. The fit looks excellent by $r^2$, yet the residual plot shows the points follow a curve; predictions from the line will be systematically wrong at both ends.