Question

Refer to Exercise 11.14 (p. 653). Calculate SSE and s for the least-squares line. Use the value of s to determine where most of the errors of prediction lie.

Short answer

SSE is 41.9533, ${\mathit{s}}^{\mathbf{2}}$ 8.39066, and s is 2.89667.

Step-by-step solution

Introduction

The most often reported standard error type is the standard error of the mean (SE or SEM). Other statistics, such as medians or proportions, can also be found with the standard error. The difference between a population parameter and a sample statistic is measured by the standard error.

Find SSE s2 and s

\(\begin{aligned}{c}S{S_{xy}} &= \sum {{x_i}} {y_i} - \frac{{\sum {{x_i}} \sum {{y_i}} }}{n}\\ &= 80 - \frac{{24x31}}{7}\\ &= 80 - \frac{{744}}{7}\\ &= \frac{{560 - 744}}{7}\end{aligned}\)

\(\begin{aligned}{l} &= - \frac{{184}}{7}\\ &= - 26.29\end{aligned}\)

\(\begin{aligned}{c}S{S_{xx}} &= {\sum {{x_i}} ^2} - \frac{{{{\left( {\sum {{x_i}} } \right)}^2}}}{n}\\ &= 116 - \frac{{{{\left( {24} \right)}^2}}}{7}\\ &= 116 - \frac{{576}}{7}\\ &= \frac{{812 - 576}}{7}\end{aligned}\)

\(\begin{aligned}{l} &= \frac{{236}}{7}\\ &= 33.71\end{aligned}\)

\(\begin{aligned}{c}\widehat {{\beta _1}} &= \frac{{S{S_{xy}}}}{{S{S_{xx}}}}\\ &= - \frac{{26.29}}{{33.71}}\\ &= - 0.77\end{aligned}\)

\(\begin{aligned}{c}S{S_{yy}} &= \sum {y^2} - \frac{{{{\left( {\sum y} \right)}^2}}}{n}\\ &= 159 - \frac{{{{\left( {31} \right)}^2}}}{7}\\ &= 159 - \frac{{961}}{7}\\ &= 159 - 137.28\end{aligned}\)

\( = 21.71\)

\(\begin{aligned}{c}SSE &= S{S_{yy}} - \widehat {{\beta _1}}S{S_{xy}}\\ &= 21.71 - ( - 0.77)(26.29)\\ &= 21.71 + 20.2433\\ &= 41.9533\end{aligned}\)

\(\begin{aligned}{c} {s^2} &= \frac{{SSE}}{{n - 2}}\\ &= \frac{{41.9533}}{{7 - 2}}\\ &= \frac{{41.9533}}{5}\\ &= 8.39066\end{aligned}\)

\(\begin{aligned}{c}s &= \sqrt {{s^2}} \\ &= \sqrt {8.39066} \\ &= 2.89667\end{aligned}\)

Therefore,SSE is 41.9533, ${\mathit{s}}^{\mathbf{2}}$8.39066, and s is 2.89667.

It shows the variation in predicted values. It is preferable if it is small.