Here is a plot of the data points
$(0, 8.5), \ \ (1,10), \ \ (2,10), \ \ (3,11):$
The observed values of $y$ are
$8.5,\ 10,\ 10,$ and $11$.
If we approximate the data by the linear equation $\hat{y} = %0x + %1$, then the predicted values are

**Residuals and sum of squares error**
Using the above data and linear equation, we get the following residuals:

Squaring and adding the residuals above gives

$\hat{y}_1 = %0(0) + %1 = %1$ | ||

$\hat{y}_2 = $BOX | ||

$\hat{y}_3 = $BOX | ||

$\hat{y}_4 = $BOX |

Residual at $x = x_1: \ \ y_1-\hat{y}_1 = 8.5 - %1 = %3$ | |||

Residual at $x = x_2:$ BOX | |||

Residual at $x = x_3:$ BOX | |||

Residual at $x = x_4:$ BOX |

SSE = BOX