Chapter 10: Q18BSC (page 468)

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

Short Answer

Expert verified

(a)Explained Variation:3210.364

(b) Unexplained Variation:1087.191

Step by step solution

Given information

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

Regression equation

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:

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

Predicted values

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.

Predicted value at $\left( {{x_0}} \right)$

Substitute${x_0} = 800$in the regression equation to obtain the predicted value.

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

Formula of prediction interval

The prediction interval is obtained using the formula shown below:

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

Degrees of freedom and critical value

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.

Standard error of the estimate

The standard error of the estimate is computed below:

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

Value of $\bar x$

The value of $\bar x$ is computed as follows:

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

Prediction interval

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).