Suppose you fit a least squares line where n = 20, $\mathbf{\sum }\mathit{y}\mathbf{=}\mathbf{176}\mathbf{.}\mathbf{11}$, $\mathbf{\sum }{\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{1602}\mathbf{.}\mathbf{097}$,$\mathit{S}{\mathit{S}}_{\mathbf{x}\mathbf{y}}\mathbf{=}\mathbf{5}\mathbf{,}\mathbf{365}\mathbf{.}\mathbf{0735}$, and ${\hat{\mathbf{\beta }}}_{\mathbf{1}}\mathbf{=}\mathbf{.}\mathbf{0087}$.

a. Calculate the estimated standard error for the regression model.

b. Interpret the estimation value calculated in part a.

The estimated standard deviation of a statistical sample population is represented by the standard error (SE) of a statistic. “Standard deviation” is a statistical term that assesses the precision with which a sample distribution represents a population.

\(\begin{aligned}{l}S{S_{yy}} &= \sum {y^2} - \frac{{{{\left( {\sum y} \right)}^2}}}{n}\\ &= 1602.097 - \frac{{{{\left( {176.11} \right)}^2}}}{{20}}\\ &= 1602.097 - \frac{{31014.7321}}{{20}}\\ &= 1602.097 - 1550.74\\ &= 51.36\end{aligned}\)

\(\begin{aligned}{c}\widehat {{\beta _1}} &= \frac{{{\rm{S}}{{\rm{S}}_{{\rm{xy}}}}}}{{{\rm{S}}{{\rm{S}}_{{\rm{xx}}}}}}\\.0087 &= \frac{{{\rm{5365}}{\rm{.0735}}}}{{{\rm{S}}{{\rm{S}}_{{\rm{xx}}}}}}\\{\rm{S}}{{\rm{S}}_{{\rm{xx}}}} &= \frac{{5365.0735}}{{.0087}}\\S{S_{xx}} &= 616675.1149\end{aligned}\)

\(\begin{aligned}{c}{\rm{SSE }} &= S{S_{yy}} - \widehat {{\beta _1}}S{S_{xy}}\\ &= 51.36 - \left( {.00875 \times 365.0735} \right)\\ &= 51.36 - 46.68\\ &= 4.68\end{aligned}\)

The calculation of the standard error for the regression model is given below.

\(\begin{aligned}{c}s &= \sqrt {\frac{{SSE}}{{n - 2}}} \\ &= \sqrt {\frac{{4.68}}{{20 - 2}}} \\ &= \sqrt {\frac{{4.68}}{{18}}} \\ &= \sqrt {0.26} \\ &= 0.51\end{aligned}\)

Therefore, the standard error is 0.51.

The observed value,0.51, isfar from the regression line on average.