Chapter 11.4, Problem 15E

1. a Derive the following identity:

   $\begin{array}{c}\text{SSE}={\displaystyle \underset{i=1}{\overset{n}{\sum }}{(y_i-{\hat{y}}_i)}^2}={\displaystyle \underset{i=1}{\overset{n}{\sum }}{(y_i-{\hat{\beta }}_0-{\hat{\beta }}_1{x}_i)}^2}\\ ={\displaystyle \underset{i=1}{\overset{n}{\sum }}{(y_i-\bar{y})}^2}-{\hat{\beta }}_1{\displaystyle \underset{i=1}{\overset{n}{\sum }}(x_i-\bar{x})(y_i-\bar{y})=S_{yy}-{\hat{\beta }}_1{S}_{xy}.}\end{array}$

   Notice that this provides an easier computational method of finding SSE.

2. b Use the computational formula for SSE derived in part (a) to prove that SSE ≤ S_yy. [Hint: ${\hat{\beta }}_1=S_{xy}/S_{xx}$.]