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Question:For the conditions of Exercise \({\bf{12}}\), and the data in Table \({\bf{11}}{\bf{.2}}\), determine the values of \({\bf{Var}}\left( {{{{\bf{\hat \beta }}}_{\bf{0}}}} \right){\bf{,Var}}\left( {{{{\bf{\hat \beta }}}_{\bf{1}}}} \right){\bf{,Var}}\left( {{{{\bf{\hat \beta }}}_{\bf{2}}}} \right){\bf{,cov}}\left( {{{{\bf{\hat \beta }}}_{\bf{0}}}{\bf{,}}{{{\bf{\hat \beta }}}_{\bf{1}}}} \right){\bf{,cov}}\left( {{{{\bf{\hat \beta }}}_{\bf{0}}}{\bf{,}}{{{\bf{\hat \beta }}}_{\bf{2}}}} \right){\bf{,\;\;cov}}\left( {{{{\bf{\hat \beta }}}_{\bf{1}}}{\bf{,}}{{{\bf{\hat \beta }}}_{\bf{2}}}} \right)\).

Short Answer

Expert verified

The values are:

\(\begin{array}{*{20}{c}}{Var\left( {{{\hat \beta }_0}} \right) = 222.725{\sigma ^2},Var\left( {{{\hat \beta }_1}} \right) = 0.135{\sigma ^2},Var\left( {{{\hat \beta }_2}} \right) = 0.058{\sigma ^2}}\\{{\mathop{\rm cov}} \left( {{{\hat \beta }_0},{{\hat \beta }_1}} \right) = 4.832{\sigma ^2},{\mathop{\rm cov}} \left( {{{\hat \beta }_0},{{\hat \beta }_2}} \right) = - 3.598{\sigma ^2},{\mathop{\rm cov}} \left( {{{\hat \beta }_1},{{\hat \beta }_2}} \right) = - 0.079{\sigma ^2}}\end{array}\)

Step by step solution

01

Define linear statistical models

A linear model describes the correlation between the dependent and independent variables as a straight line.

\(y = {a_0} + {a_1}{x_1} + {a_2}{x_2} + \tilde A,\hat A1/4 + {a_n}{x_n}\)

Models using only one predictor are simple linear regression models. Multiple predictors are used in multiple linear regression models. For many response variables, multiple regression analysis models are used.

02

Design the matrix

All the necessary quantities can be found in the covariance matrix which can be found as,

The design matrix is

\(Z = \left( {\begin{array}{*{20}{c}}1&{{x_{1,1}}}&{{x_{1,2}}}\\1&{{x_{2,1}}}&{{x_{2,2}}}\\ \vdots & \vdots & \vdots \\1&{{x_{10,1}}}&{{x_{10,2}}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{1.9}&{66}\\1&{0.8}&{62}\\1&{1.1}&{64}\\1&{0.1}&{61}\\1&{ - 0.1}&{63}\\1&{4.4}&{70}\\1&{4.6}&{68}\\1&{1.6}&{62}\\1&{5.5}&{68}\\1&{3.4}&{66}\end{array}} \right]\)

This is because

\({\mathop{\rm cov}} (\hat \beta ) = \left[ {\begin{array}{*{20}{c}}{Var\left( {{{\hat \beta }_0}} \right)}&{{\mathop{\rm cov}} \left( {{{\hat \beta }_0},{{\hat \beta }_1}} \right)}&{{\mathop{\rm cov}} \left( {{{\hat \beta }_0},{{\hat \beta }_2}} \right)}\\{{\mathop{\rm cov}} \left( {{{\hat \beta }_1},{{\hat \beta }_0}} \right)}&{Var\left( {{{\hat \beta }_1}} \right)}&{{\mathop{\rm cov}} \left( {{{\hat \beta }_1},{{\hat \beta }_2}} \right)}\\{{\mathop{\rm cov}} \left( {{{\hat \beta }_2},{{\hat \beta }_0}} \right)}&{{\mathop{\rm cov}} \left( {{{\hat \beta }_2},{{\hat \beta }_1}} \right)}&{Var\left( {{{\hat \beta }_2}} \right)}\end{array}} \right]\)

Hence the desired matrix is \(\left( {\begin{array}{*{20}{c}}1&{1.9}&{66}\\1&{0.8}&{62}\\1&{1.1}&{64}\\1&{0.1}&{61}\\1&{ - 0.1}&{63}\\1&{4.4}&{70}\\1&{4.6}&{68}\\1&{1.6}&{62}\\1&{5.5}&{68}\\1&{3.4}&{66}\end{array}} \right)\)

03

Find the values

The desired values can be found in the upper triangle of the matrix found above.

We can find that,

Finally we have

Hence the values are \(\begin{array}{*{20}{c}}{Var\left( {{{\hat \beta }_0}} \right) = 222.725{\sigma ^2},Var\left( {{{\hat \beta }_1}} \right) = 0.135{\sigma ^2},Var\left( {{{\hat \beta }_2}} \right) = 0.058{\sigma ^2}}\\{{\mathop{\rm cov}} \left( {{{\hat \beta }_0},{{\hat \beta }_1}} \right) = 4.832{\sigma ^2},{\mathop{\rm cov}} \left( {{{\hat \beta }_0},{{\hat \beta }_2}} \right) = - 3.598{\sigma ^2},{\mathop{\rm cov}} \left( {{{\hat \beta }_1},{{\hat \beta }_2}} \right) = - 0.079{\sigma ^2}}\end{array}\)

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