Variation and Prediction Intervals. In Exercises 17–20, find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. In each case, there is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions.

Town Courts Listed below are amounts of court income and salaries paid to the town justices (based on data from the Poughkeepsie Journal). All amounts are in thousands of dollars, and all of the towns are in Dutchess County, New York. For the prediction interval, use a 99% confidence level with a court income of $800,000.

Court Income	65	404	1567	1131	272	252	111	154	32
Justice Salary	30	44	92	56	46	61	25	26	18

Data are given fortwo variables, “Court Income” and “Justice Salary”.

Let x denote the variable “Court Income.”

Let y denote the variable “Justice Salary.”

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).