Chapter 16, Problem 21P

21. Problem 18 in Chapter 15 (p. 526) presented data showing the number of crime, the amount spend on crime prevention, and the population for a set n = 12 cities. At that time, we used a partial correlation to evaluate the relationship between the amount spent on crime prevention and the number of crimes crime while controlling population. It is possible to use multiple regression to accomplish essentially the same purpose. The data are as follows:

Number of

crimesAmount for PreventionPopulation Size 3 6 1 4 7 1 6 3 1 1 4 1 8 11 2 9 12 2 11 8 2 l2 9 2 l3 16 3 l4 17 3 l6 13 3 l7 14 3

1. Find the multiple regression equation for predict­ ing the number of crimes using the amount spent on prevention and population as the two predictor variables.

Find the value of R²for the regression equation.

18. A researcher records the annual number of serious crimes and the amount spent on crime prevention for several small towns, medium cities, and large cities across the country. The resulting data show a strong positive correlation between the number of serious crimes and the amount spent on crime prevention. However, the researcher suspects that the positive correlation is actually caused by population; as population increases, both the amount spent on crime prevention and the number of crimes will also increase. If population is controlled, there probably should be a negative correlation between the amount spent on crime prevention and the number of serious crimes. The following data show the pattern of results obtained. Note that the municipalities are coded in three categories. Use a partial correlation holding population constant to measure the true relationship between crime rate and the amount spent on prevention.

Number of

crimesAmount for PreventionPopulation Size 3 6 1 4 7 1 6 3 1 7 4 1 8 11 2 ,9 12 2 11 8 2 12 9 2 13 16 3 14 l7 3 16 l3 3 17 l4 3

To determine

1. Find the multiple regression equation for predicting the number of crimes using the amount spent on prevention and population size.
2. Find the value of $R^2$.

Answer to Problem 21P

Solution:

1. The regression equation is $Y=-0.8X_1+9X_2$.

   The value of $R^2$ is 0.953.

Explanation of Solution

Given:

Let Number of Crimes be $Y$, Amount for Prevention be $X_1$ and Population Size be $X_2$. Here, $n=12$ because there are 12 observations. The data is shown below:

Number of Crimes	Amount for Prevention	Population Size
$Y$	$X_1$	$X_2$
3	6	1
4	7	1
6	3	1
7	4	1
8	11	2
9	12	2
11	8	2
12	9	2
13	16	3
14	17	3
16	13	3
17	14	3

Formula used:

The general form of the multiple-regression equation is:

$\widehat{Y}=b_1{X}_1+b_2{X}_2+a$

The values for the multiple regression equation are:

$\begin{array}{l}{b}_1=\frac{\left(S{P}_{X_{1}Y}\right)\left(S{S}_{X_2}\right)-\left(S{P}_{X_1{X}_2}\right)\left(S{P}_{X_{2}Y}\right)}{\left(S{S}_{X_1}\right)\left(S{S}_{X_2}\right)-{\left(S{P}_{X_1{X}_2}\right)}^2}\\ b_2=\frac{\left(S{P}_{X_{2}Y}\right)\left(S{S}_{X_1}\right)-\left(S{P}_{X_1{X}_2}\right)\left(S{P}_{X_{1}Y}\right)}{\left(S{S}_{X_1}\right)\left(S{S}_{X_2}\right)-{\left(S{P}_{X_1{X}_2}\right)}^2}\\ a=M_Y-b_1{M}_{X_1}-b_2{M}_{X_2}\end{array}$

Where:

$\begin{array}{l}{M}_Y=\frac{{\displaystyle \sum Y}}{n}\\ M_{X_1}=\frac{{\displaystyle \sum X_1}}{n}\\ M_{X_2}=\frac{{\displaystyle \sum X_2}}{n}\\ S{P}_{X_{1}Y}={\displaystyle \sum X_{1}Y}-\frac{\left({\displaystyle \sum X_1}\right)\left({\displaystyle \sum Y}\right)}{n}\\ S{P}_{X_{2}Y}={\displaystyle \sum X_{2}Y}-\frac{\left({\displaystyle \sum X_2}\right)\left({\displaystyle \sum Y}\right)}{n}\\ S{P}_{X_1{X}_2}={\displaystyle \sum X_1{X}_2}-\frac{\left({\displaystyle \sum X_1}\right)\left({\displaystyle \sum X_2}\right)}{n}\\ S{S}_{X_1}={{\displaystyle \sum \left(X_1-M_{X_1}\right)}}^2\\ S{S}_{X_2}={{\displaystyle \sum \left(X_2-M_{X_2}\right)}}^2\\ S{S}_Y={{\displaystyle \sum \left(Y-M_Y\right)}}^2\end{array}$

The formula of $R^2$ is:

$R^2=\frac{b_{1}S{P}_{X_{1}Y}+b_{2}S{P}_{X_{2}Y}}{S{S}_Y}$

Calculation:

1. Consider the following table:

   	$Y$	$X_1$	$X_2$	$X_{1}Y$	$X_{2}Y$	$X_1{X}_2$
   	3	6	1	18	3	6
   	4	7	1	28	4	7
   	6	3	1	18	6	3
   	7	4	1	28	7	4
   	8	11	2	88	16	22
   	9	12	2	108	18	24
   	11	8	2	88	22	16
   	12	9	2	108	24	18
   	13	16	3	208	39	48
   	14	17	3	238	42	51
   	16	13	3	208	48	39
   	17	14	3	238	51	42
   SUM	120	120	24	1376	280	280

The values for the multiple regression equation are:

$\begin{array}{c}{b}_1=\frac{\left(S{P}_{X_{1}Y}\right)\left(S{S}_{X_2}\right)-\left(S{P}_{X_1{X}_2}\right)\left(S{P}_{X_{2}Y}\right)}{\left(S{S}_{X_1}\right)\left(S{S}_{X_2}\right)-{\left(S{P}_{X_1{X}_2}\right)}^2}\\ =\frac{\left(176\right)\left(8\right)-\left(40\right)\left(40\right)}{\left(230\right)\left(8\right)-{\left(40\right)}^2}\\ =\frac{-192}{240}\\ =-0.8\end{array}$

$\begin{array}{c}{b}_2=\frac{\left(S{P}_{X_{2}Y}\right)\left(S{S}_{X_1}\right)-\left(S{P}_{X_1{X}_2}\right)\left(S{P}_{X_{1}Y}\right)}{\left(S{S}_{X_1}\right)\left(S{S}_{X_2}\right)-{\left(S{P}_{X_1{X}_2}\right)}^2}\\ =\frac{\left(40\right)\left(230\right)-\left(40\right)\left(176\right)}{\left(230\right)\left(8\right)-{\left(40\right)}^2}\\ =\frac{2160}{240}\\ =9\end{array}$

$\begin{array}{c}a=M_Y-b_1{M}_{X_1}-b_2{M}_{X_2}\\ =10-\left(-0.8\right)\left(10\right)-9\left(2\right)\\ =10+8-18\\ =0\end{array}$

The multiple-regression equation is:

$\widehat{Y}=-0.8X_1+9X_2$

2. The value of $R^2$ is:

   $\begin{array}{c}{R}^2=\frac{b_{1}S{P}_{X_{1}Y}+b_{2}S{P}_{X_{2}Y}}{S{S}_Y}\\ =\frac{-0.8\left(176\right)+9\left(40\right)}{230}\\ =\frac{219.2}{230}\\ =0.953\end{array}$

Conclusion:

1. The multiple-regression equation is: $\widehat{Y}=-0.8X_1+9X_2$
2. The value of $R^2$ is 0.953.

Justification:

Since there are more than one explanatory variable, use multiple regression.