91Ó°ÊÓ

The equation for the least-squares line is \(y = {\beta _0} + {\beta _1}x\).

The formula for the intercept and slope coefficients \({\beta _0}\) and \({\beta _1}\) are given as:

\(\begin{array}{l}{{\hat \beta }_1} = \frac{{\sum\limits_{i = 1}^n {\left( {{x_i} - \bar x} \right)} \left( {{y_i} - \bar y} \right)}}{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} }}\\{{\hat \beta }_0} = \bar y - {{\hat \beta }_1}\bar x\end{array}\)

Thus, the least-squares line can be re-written as,

\({\bf{y = }}\left( {{\bf{\bar y - }}{{{\bf{\hat \beta }}}_{\bf{1}}}{\bf{\bar x}}} \right){\bf{ + }}{{\bf{\hat \beta }}_{\bf{1}}}{\bf{x}}\)

To show that the point \((\bar x,\bar y)\)satisfies the above equation.

In other words, least-squares line passes through the point. \((\bar x,\bar y)\).

Substitute the point \(x = \bar x\)into the above equation.

\(\begin{array}{c}y = \left( {\bar y - {{\hat \beta }_1}\bar x} \right) + {{\hat \beta }_1}\bar x\\ = \bar y - {{\hat \beta }_1}\bar x + {{\hat \beta }_1}\bar x\\ = \bar y\end{array}\)

Thus, the point \((\bar x,\bar y)\)actually satisfies the above equation.

Therefore, the point \((\bar x,\bar y)\)lies on the least square line.