Standard deviation of the residuals (s)

The value $s$ answers a practical question: when you predict with this line, how far off are you usually? It is $s = \sqrt{\dfrac{\sum (y - \hat{y})^2}{n - 2}}$, so you square the residuals, average them using $n - 2$ because the line costs you two estimated coefficients, then take the square root, which returns you to the units of the response variable. For example, five residuals of 2, -1, 0, 1, and -2 give $\sum (y - \hat{y})^2 = 4 + 1 + 0 + 1 + 4 = 10$ and $s = \sqrt{10/3} \approx 1.83$, so predictions typically miss by about 1.83 units of $y$. Software prints it as s, and a smaller $s$ for the same response values means a tighter fit, the same tightness a residual plot shows visually.