Calculate SSE andfor each of the following cases:

a. n = 20,$\mathit{S}{\mathit{S}}_{\mathbf{y}\mathbf{y}}\mathbf{=}\mathbf{95}$,$\mathit{S}{\mathit{S}}_{\mathbf{x}\mathbf{y}}\mathbf{=}\mathbf{50}$,$\widehat{{\mathbf{\beta }}_{\mathbf{1}}}\mathbf{=}\mathbf{.}\mathbf{75}$

b. n = 40,$\mathbf{\sum }{\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{860}$ , $\mathbf{\sum }\mathit{y}\mathbf{=}\mathbf{50}$, $\mathit{S}{\mathit{S}}_{\mathbf{x}\mathbf{y}}\mathbf{=}\mathbf{2700}$, $\widehat{{\mathbf{\beta }}_{\mathbf{1}}}\mathbf{=}\mathbf{.}\mathbf{2}$

c. n = 10, $\mathbf{\sum }{\left(y_i-\overline{y}\right)}^{\mathbf{2}}\mathbf{=}\mathbf{58}$,$\mathit{S}{\mathit{S}}_{\mathbf{x}\mathbf{y}}\mathbf{=}\mathbf{91}$,$\mathit{S}{\mathit{S}}_{\mathbf{x}\mathbf{x}}\mathbf{=}\mathbf{170}$

The overall deviation of the response values from the response values' fit is calculated using this statistic. The summed square of residuals is abbreviated as SSE.

\(\begin{aligned}{c}SSE &= S{S_{yy}} - \widehat {{\beta _1}}S{S_{xy}}\\ &= 95 - (.75)(50)\\ &= 95 - 37.5\\ &= 57.5\end{aligned}\)

\(\begin{aligned}{c} {s^2} &= \frac{{SSE}}{{n - 2}}\\ &= \frac{{57.5}}{{20 - 2}}\\ &= \frac{{57.5}}{{18}}\\ &= 3.195\end{aligned}\)

Therefore,SSE is 57.5 and \( {s^2}\) 3.195.

\(\begin{aligned}{c}S{S_{yy}} &= \sum {y^2} - \frac{{{{\left( {\sum y} \right)}^2}}}{n}\\ &= 860 - \frac{{{{\left( {50} \right)}^2}}}{{40}}\\ &= 860 - \frac{{2500}}{{40}}\\ &= 860 - 62.5\end{aligned}\)

\( = 797.5\)

\(\begin{aligned}{c}SSE &= S{S_{yy}} - \widehat {{\beta _1}}S{S_{xy}}\\ &= 797.5 - (.2)(2700)\\ &= 797.5 - 540\\ &= 257.5\end{aligned}\)

\(\begin{aligned}{c} {s^2} &= \frac{{SSE}}{{n - 2}}\\ &= \frac{{257.5}}{{40 - 2}}\\ &= \frac{{257.5}}{{38}}\\ &= 6.78\end{aligned}\)

Therefore, SSE is 257.5 and ${\mathit{s}}^{\mathbf{2}}$6.78.

\(S{S_{yy}} = \sum {\left( {{y_i} - \overline y } \right)^2} = 58\)

\(\begin{aligned}{c}\widehat {{\beta _1}} &= \frac{{S{S_{xy}}}}{{S{S_{xx}}}}\\ &= \frac{{91}}{{170}}\\ &= 0.5\end{aligned}\)

\(\begin{aligned}{c}SSE &= S{S_{yy}} - \widehat {{\beta _1}}S{S_{xy}}\\ &= 58 - (.5)(91)\\ &= 58 - 45.5\\ &= 12.5\end{aligned}\)

\(\begin{aligned}{c} {s^2} &= \frac{{SSE}}{{n - 2}}\\ &= \frac{{12.5}}{{10 - 2}}\\ &= \frac{{12.5}}{8}\\ &= 1.56\end{aligned}\)

Therefore, SSE is 12.5 and ${\mathit{s}}^{\mathbf{2}}$1.56.