a. Rewrite the data in Example 1 with new \(x\)-coordinates in mean deviation form. Let \(X\) be the associated design matrix. Why are the columns of \(X\) orthogonal?

b. Write the normal equations for the data in part (a), and solve them to find the least-squares line, \(y = {\beta _0} + {\beta _1}x*\), where \(x* = x - 5.5\).

(a) The data in example 1 is: \(\left( {2,1} \right),\left( {5,2} \right),\left( {7,3} \right)\) and \(\left( {8,3} \right)\).

Find the mean of the data.

\(\begin{aligned}\bar x &= \frac{{\sum x }}{n}\\ &= \frac{{2 + 5 + 7 + 8}}{4}\\ &= \frac{{22}}{4}\\ &= 5.5\end{aligned}\)

So, the mean of the data is \(5.5\).

It is given that, \(x* = x - 5.5\). Find \(x* = x - 5.5\) for different values of \(x\) in the given data set.

\(\begin{aligned}x* = 2 - 5.5\\ = - 3.5\end{aligned}\)

\(\begin{aligned}x* = 5 - 5.5\\ = - 0.5\end{aligned}\)

\(\begin{aligned}x* = 7 - 5.5\\ = 1.5\end{aligned}\)

\(\begin{aligned}x* = 8 - 5.5\\ = 2.5\end{aligned}\)

So, the data in mean deviation form can be written as,

\(X = \left( {\begin{aligned}1&{ - 3.5}\\1&{ - 0.5}\\1&{1.5}\\1&{2.5}\end{aligned}} \right)\)

On adding the elements of the second column of \(X\), the value is 0, so the columns of \(X\) are orthogonal.

The equation of the general linear model is given as,

\({\bf{y}} = X\beta + \in \)

Here, \({\bf{y}} = \left( {\begin{aligned}{{y_1}}\\{{y_2}}\\ \vdots \\{{y_n}}\end{aligned}} \right)\) is an observational vector, \(X = \left( {\begin{aligned}1&{{x_1}}& \cdots &{x_1^n}\\1&{{x_2}}& \cdots &{x_2^n}\\ \vdots & \vdots & \ddots & \vdots \\1&{{x_n}}& \cdots &{x_n^n}\end{aligned}} \right)\) is the design matrix, \(\beta = \left( {\begin{aligned}{{\beta _1}}\\{{\beta _2}}\\ \vdots \\{{\beta _n}}\end{aligned}} \right)\) is parameter vector, and \( \in = \left( {\begin{aligned}{{ \in _1}}\\{{ \in _2}}\\ \vdots \\{{ \in _n}}\end{aligned}} \right)\) is a residual vector.

Write the design matrix and observational vector for the given data points.

Design matrix: \(X = \left( {\begin{aligned}1&{ - 3.5}\\1&{ - 0.5}\\1&{1.5}\\1&{2.5}\end{aligned}} \right)\)

Observational matrix: \({\bf{y}} = \left( {\begin{aligned}1\\2\\3\\3\end{aligned}} \right)\)

And the parameter vectorfor the given equation is,

\({\bf{\beta }} = \left( {\begin{aligned}{{\beta _0}}\\{{\beta _1}}\end{aligned}} \right)\)

The normal equation is given by,

\({X^T}X\beta = {X^T}{\bf{y}}\)

(b)

The normal equation is\({X^T}X\beta = {X^T}{\bf{y}}\), find \(\beta \) for the given data by using the normal equation.

\(\begin{aligned}\left( {{{\left( {\begin{aligned}1&{ - 3.5}\\1&{ - 0.5}\\1&{1.5}\\1&{2.5}\end{aligned}} \right)}^T}\left( {\begin{aligned}1&{ - 3.5}\\1&{ - 0.5}\\1&{1.5}\\1&{2.5}\end{aligned}} \right)} \right)\beta & = {\left( {\begin{aligned}1&{ - 3.5}\\1&{ - 0.5}\\1&{1.5}\\1&{2.5}\end{aligned}} \right)^T}\left( {\begin{aligned}1\\2\\3\\3\end{aligned}} \right)\\\left( {\begin{aligned}4&0\\0&{21}\end{aligned}} \right)\beta & = \left( {\begin{aligned}9\\{7.5}\end{aligned}} \right)\\\left( {\begin{aligned}{{\beta _1}}\\{{\beta _2}}\end{aligned}} \right) &= {\left( {\begin{aligned}4&0\\0&{21}\end{aligned}} \right)^{ - 1}}\left( {\begin{aligned}9\\{7.5}\end{aligned}} \right)\\ &= \frac{1}{{84}}\left( {\begin{aligned}{21}&0\\0&4\end{aligned}} \right)\left( {\begin{aligned}9\\{7.5}\end{aligned}} \right)\end{aligned}\)

Solve further,

\(\begin{aligned}\left( {{{\left( {\begin{aligned}1&{ - 3.5}\\1&{ - 0.5}\\1&{1.5}\\1&{2.5}\end{aligned}} \right)}^T}\left( {\begin{aligned}1&{ - 3.5}\\1&{ - 0.5}\\1&{1.5}\\1&{2.5}\end{aligned}} \right)} \right)\beta & = \left( {\begin{aligned}{\frac{1}{4}}&0\\0&{\frac{1}{{21}}}\end{aligned}} \right)\left( {\begin{aligned}9\\{7.5}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{\frac{9}{4}}\\{\frac{{7.5}}{{21}}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{\frac{9}{4}}\\{\frac{5}{{14}}}\end{aligned}} \right)\end{aligned}\)

Substitute the obtained values into \(y = {\beta _0} + {\beta _1}x*\).

\(y = \frac{9}{4} + \frac{5}{{14}}\left( {x - 5.5} \right)\)

Hence, the required least-squares line is \(y = \frac{9}{4} + \frac{5}{{14}}\left( {x - 5.5} \right)\).