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Data are given fortwo variables, 鈥淐ourt Income鈥� and 鈥淛ustice Salary鈥�.

Let x denote the variable 鈥淐ourt Income.鈥�

Let y denote the variable 鈥淛ustice Salary.鈥�

The regression equation of y on x has the following notation:

\(\hat y = {b_0} + {b_1}x\), where

\({b_0}\)is the intercept term and\({b_1}\)is the slope coefficient.

The following calculations are done to compute the intercept and the slope coefficient:

The y-intercept is computed below:

\(\begin{array}{c}{b_0} = \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{\left( {398} \right)\left( {4076640} \right) - \left( {3988} \right)\left( {262465} \right)}}{{9\left( {4076640} \right) - {{\left( {3988} \right)}^2}}}\\ = 27.701478\\ \approx 27.70\end{array}\)

The slope coefficient is computed below:

\(\begin{array}{c}{b_1} = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{\left( 9 \right)\left( {262465} \right) - \left( {3988} \right)\left( {398} \right)}}{{9\left( {4076640} \right) - {{\left( {3988} \right)}^2}}}\\ = 0.0372835\\ \approx 0.04\end{array}\)

Thus, the regression equation becomes as shown:

\(\begin{array}{l}\hat y = 27.701478 - 0.0372835x\\\hat y \approx 27.70 - 0.04x\end{array}\)

The mean value of observed y is computed below:

\(\begin{array}{c}\bar y = \frac{{\sum y }}{n}\\ = \frac{{398}}{9}\\ = 44.222\end{array}\)

The following table shows the predicted values (obtained by substituting the values of x in the regression equation) and other important calculations:

The value of the explained variation is shown below:

\(\sum {{{\left( {\hat y - \bar y} \right)}^2}} = 3210.364\)

Thus, the explained variation is 3210.364.

The value of the unexplained variation is shown below:

\(\sum {{{\left( {y - \hat y} \right)}^2}} = 1087.191\)

Thus, the unexplained variation is 1087.191.

Substitute\({x_0} = 800\)in the regression equation to obtain the predicted value.

\(\begin{array}{c}\hat y = 27.70 + 0.04x\\ = 27.70 + 0.04\left( {800} \right)\\ = 57.5283\\ \approx 58\end{array}\)

The prediction interval is obtained using the formula shown below:

\(\begin{array}{c}PI = \hat y \pm E\\ = \hat y \pm {t_{\frac{\alpha }{2}}}{s_e}\sqrt {1 + \frac{1}{n} + \frac{{n{{\left( {{x_0} - \bar x} \right)}^2}}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}} \end{array}\)

The following formula is used to compute the level of significance

\(\begin{array}{c}Confidence\;Level = 99\% \\100\left( {1 - \alpha } \right) = 99\\1 - \alpha = 0.99\\ = 0.01\end{array}\)

The degrees of freedom for computing the t-multiplier are shown below:

\(\begin{array}{c}df = n - 2\\ = 9 - 2\\ = 7\end{array}\)

The two-tailed value of the t-multiplier for 0.01 level of significance and 7 degrees of freedom is 3.4995.

The standard error of the estimate is computed below:

\(\begin{array}{c}{s_e} = \sqrt {\frac{{\sum {{{\left( {y - \hat y} \right)}^2}} }}{{n - 2}}} \\ = \sqrt {\frac{{1087.191}}{{9 - 2}}} \\ = 12.46247\end{array}\)

The value of \(\bar x\) is computed as follows:

\(\begin{array}{c}\bar x = \frac{{\sum x }}{n}\\ = \frac{{3988}}{9}\\ = 443.111\end{array}\)

Substitute the values obtained above to calculate the margin of error (E).

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}}{s_e}\sqrt {1 + \frac{1}{n} + \frac{{n{{\left( {{x_0} - \bar x} \right)}^2}}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}} \\ = \left( {3.4995} \right)\left( {12.46247} \right)\sqrt {1 + \frac{1}{9} + \frac{{9{{\left( {800 - 443.111} \right)}^2}}}{{9\left( {4076640} \right) - {{\left( {3988} \right)}^2}}}} \\ = 47.0986\end{array}\)

Thus, the prediction interval becomes as shown:

\(\begin{array}{c}PI = \left( {\hat y - E,\hat y + E} \right)\\ = \left( {57.5283 - 47.0986,57.5283 + 47.0986} \right)\\ = \left( {10.4297,04.6269} \right)\\ \approx \left( {10.4,104.6} \right)\end{array}\)

Therefore, the 99% prediction interval for the justice salary for thecourt income of $800,000is (10.4,104.6).