Chapter 12, Problem 12.1P

Why does voter turnout vary from election to election? For municipal elections in five different cities, information has been gathered on the percentage of citizens who voted (the dependent variable) and three different independent variables: unemployment rate, average years of education for the city, and the percentage of all political ads that used "negative campaigning” (personal attacks, negative portrayals of the opponent’s record, etc.). For each relationship between turnout (Y) and the independent variables,

a. Compute the slope (b) and find the Y intercept (a). (HINT: Remember to compute b before computing a. A computing table such as table 12.3 is highly recommended.)

b. State the least-square regression line and predict the voter turnout for a city in which the unemployment rate was 12%, a city in which the average years of schooling was 11%, and an election in which 90% of the ads were negative.

c. Compute $r$ and $r^2$. (HINT: A computing table such as Table 12.4 is highly recommended. If you constructed one for computing b, you already have most of the quantities you will need to solve for r.)

d. Describe the strength and direction of each relationship in a sentence or two. Which of the independent variables had the strongest effect on turnout?

City	Turnout (Y)	Unemployment Rate ( $X_1$)	Average Years of School ( $X_2$)	Percentage of Negative Ads ( $X_3$)
A	55	5	11.9	60
B	60	8	12.1	63
C	65	9	12.7	55
D	68	9	12.8	53
E	70	10	13.0	48

To determine

(a)

To find:

The slope and $Y$ intercept.

Answer to Problem 12.1P

Solution:

The values of intercept are 39, $-94.77$ and 113.8

The values of slope are 3, 12.67 and $-0.9$.

Explanation of Solution

Given:

The three different independent variables: unemployment rate, average years of education for the unemployment rate, average years of school and the percentage of all political ads that used negative campaigning is given table below.

City	Turnout (Y)	Unemployment Rate ($X_1$)	Average Years of School ($X_2$)	Percentage of Negative Ads ($X_3$)
A	55	5	11.9	60
B	60	8	12.1	63
C	65	9	12.7	55
D	68	9	12.8	53
E	70	10	13.0	48

Approach:

The Y intercept (a) is the point of intersection of the regression line and the y axis.

The slope (b) of the regression line is the amount of change in the dependent variable (Y) by a unit change in the independent variable (X).

If the two variables are unrelated, the regression line would be parallel to the x axis.

Formula used:

The model for the least square regression line is defined as,

$Y=a+bX$

Where Y is the dependent variable,

$b=\frac{{\displaystyle \sum \left(X-\overline{X}\right)\left(Y-\overline{Y}\right)}}{{\displaystyle \sum {\left(X-\overline{X}\right)}^2}}$

Where Y is the dependent variable and X is the independent variable,

b is the slope of the regression line.

And $\overline{X}$ and $\overline{Y}$ are their respective means.

The $Y$ intercept (a) is given by,

$a=\overline{Y}-b\overline{X}$

Where, $\overline{X}$ is the means of $X$, $\overline{Y}$ is the means of $Y$ and $b$ is the slope.

Calculation:

From the given information,

The turnout is the dependent variable. Thus, it is represented as Y.

The Unemployment rate is the independent variables. Thus, it is as $X_1$.

Consider the following table of sum of squares of turnout and Unemployment rate.

City	Unemployment Rate ($X_1$)	$\left(X_1-\overline{X_1}\right)$	${\left(X_1-\overline{X_1}\right)}^2$	Turnout (Y)	$\left(Y-\overline{Y}\right)$	$\left(X_1-\overline{X_1}\right)\left(Y-\overline{Y}\right)$
A	5	$-3.2$	10.24	55	$-8.6$	27.52
B	8	$-0.2$	0.04	60	$-3.6$	0.72
C	9	0.8	0.64	65	1.4	1.12
D	9	0.8	0.64	68	4.4	3.52
E	10	1.8	3.24	70	6.4	11.52
Total	41		$\begin{array}{l}{\displaystyle \sum {\left(X_1-\overline{X}\right)}^2}\\ =14.8\end{array}$	318		$\begin{array}{l}{\displaystyle \sum \left(X_1-\overline{X}\right)\left(Y-\overline{Y}\right)}\\ =44.4\end{array}$

The value of $\overline{X_1}$ is 8.2 and $\overline{Y}$ is 63.6.

From above table, substitute $5$ for $X_1$ and $8.2$ for $\overline{X_1}$ in $\left(X_1-\overline{X_1}\right)$.

$\begin{array}{rll}\left(X_1-\overline{X_1}\right)& =& \left(5-8.2\right)\\ & =& -3.2\end{array}$

Square the both sides of the equation.

$\begin{array}{rll}{\left(X_1-\overline{X_1}\right)}^2& =& {\left(-3.2\right)}^2\\ & =& 10.24\end{array}$

Proceed in the same manner to calculate ${\left(X_1-\overline{X_1}\right)}^2$ for all the $1\le i\le N$ for the rest data and refer table for the rest of the ${\left(X_1-\overline{X_1}\right)}^2$ values calculated. Then the value of ${{\displaystyle \sum \left(X_1-\overline{X_1}\right)}}^2$ is calculated as,

$\begin{array}{rll}{{\displaystyle \sum \left(X_1-\overline{X_1}\right)}}^2& =& 10.24+0.04+0.64+0.64+3.24\\ & =& 14.8\end{array}$

Substitute $44.4$ for ${\displaystyle \sum \left(X_1-{\overline{X}}_1\right)\left(Y-\overline{Y}\right)}$ and 14.8 for ${\displaystyle \sum {\left(X_1-\overline{X_1}\right)}^2}$ in the above mentioned formula.

$\begin{array}{rll}{b}_1& =& \frac{44.4}{14.8}\\ & =& 3\end{array}$

Substitute $8.2$ for $\overline{X_1}$, 63.6 for $\overline{Y}$ and 3 for b in the above mentioned formula to solve for intercept.

$\begin{array}{rll}{a}_1& =& 63.6-3\left(8.2\right)\\ & =& 39\end{array}$

The value of intercept is 39 and the value of slope is 3.

From the given information,

The turnout is the dependent variable. Thus, it is represented as Y.

The average years of school is the independent variables. Thus, it is as $X_2$.

Consider the following table of sum of squares of turnout and Average years of school.

City	Average Years of School ($X_2$)	$\left(X_2-\overline{X_2}\right)$	${\left(X_2-\overline{X_2}\right)}^2$	Turnout (Y)	$\left(Y-\overline{Y}\right)$	$\left(X_2-\overline{X_2}\right)\left(Y-\overline{Y}\right)$
A	11.9	$-0.6$	0.36	55	$-8.6$	5.16
B	12.1	$-0.4$	0.16	60	$-3.6$	1.44
C	12.7	0.2	0.04	65	1.4	0.28
D	12.8	0.3	0.09	68	4.4	1.32
E	13.0	0.5	0.25	70	6.4	3.2
Total	62.5		$\begin{array}{l}{\displaystyle \sum {\left(X_2-\overline{X_2}\right)}^2}\\ =0.9\end{array}$	318		$\begin{array}{l}{\displaystyle \sum \left(X_2-{\overline{X}}_2\right)\left(Y-\overline{Y}\right)}\\ =11.4\end{array}$

The value of $\overline{X_2}$ is 12.5 and $\overline{Y}$ is 63.6.

From above table, substitute $11.9$ for $X_2$ and 12.5 for $\overline{X_2}$ in $\left(X_2-\overline{X_2}\right)$.

$\begin{array}{rll}\left(X_2-\overline{X_2}\right)& =& \left(11.9-12.5\right)\\ & =& -0.6\end{array}$

Square the both sides of the equation.

$\begin{array}{rll}\left(X_2-\overline{X_2}\right)& =& {\left(-0.6\right)}^2\\ & =& 0.36\end{array}$

Proceed in the same manner to calculate ${\left(X_2-\overline{X_2}\right)}^2$ for all the $1\le i\le N$ for the rest data and refer table for the rest of the ${\left(X_2-\overline{X_2}\right)}^2$ values calculated. Then the value of ${{\displaystyle \sum \left(X_2-\overline{X_2}\right)}}^2$ is calculated as,

$\begin{array}{rll}{{\displaystyle \sum \left(X_2-\overline{X_2}\right)}}^2& =& 0.36+0.16+0.04+0.09+0.25\\ & =& 0.9\end{array}$

Substitute $11.4$ for ${\displaystyle \sum \left(X_2-{\overline{X}}_2\right)\left(Y-\overline{Y}\right)}$ and 0.9 for ${{\displaystyle \sum \left(X_2-\overline{X_2}\right)}}^2$ in the above mentioned formula.

$\begin{array}{rll}{b}_2& =& \frac{11.4}{0.9}\\ & =& 12.67\end{array}$

Substitute 12.5 for $\overline{X}$, 63.6 for $\overline{Y}$ and 12.67 for b in the above mentioned formula to solve for intercept.

$\begin{array}{rll}{a}_2& =& 63.6-12.67\left(12.5\right)\\ & =& -94.77\end{array}$

The value of intercept is $-94.77$ and the value of slope is 12.67.

From the given information,

The turnout is the dependent variable. Thus, it is represented as Y.

The percentage of negative ads is the independent variables. Thus, it is as $X_3$.

Consider the following table of sum of squares of turnout and Average years of school.

City	Percentage of Negative Ads ($X_3$)	$\left(X_3-\overline{X_3}\right)$	${\left(X_3-\overline{X_3}\right)}^2$	Turnout (Y)	$\left(Y-\overline{Y}\right)$	$\left(X_3-\overline{X_3}\right)\left(Y-\overline{Y}\right)$
A	60	4.2	17.64	55	$-8.6$	$-36.12$
B	63	7.2	51.84	60	$-3.6$	$-25.92$
C	55	$-0.8$	0.64	65	1.4	$-1.12$
D	53	$-2.8$	7.84	68	4.4	$-12.32$
E	48	$-7.8$	60.84	70	6.4	$-49.92$
Total	279		$\begin{array}{l}{\displaystyle \sum {\left(X_3-\overline{X3}\right)}^2}\\ =138.8\end{array}$	318		$\begin{array}{l}{\displaystyle \sum \left(X_3-{\overline{X}}_3\right)\left(Y-\overline{Y}\right)}\\ =-125.4\end{array}$

The value of $\overline{X_3}$ is 55.8 and $\overline{Y}$ is 63.6.

From above table, substitute $60$ for $X_3$ and 55.8 for $\overline{X_3}$ in $\left(X_3-\overline{X_3}\right)$.

$\begin{array}{rll}\left(X_3-\overline{X_3}\right)& =& \left(60-55.8\right)\\ & =& 4.2\end{array}$

Square the both sides of the equation.

$\begin{array}{rll}{\left(X_3-\overline{X_3}\right)}^2& =& {\left(4.2\right)}^2\\ & =& 17.64\end{array}$

Proceed in the same manner to calculate ${\left(X_3-\overline{X_3}\right)}^2$ for all the $1\le i\le N$ for the rest data and refer table for the rest of the ${\left(X_3-\overline{X_3}\right)}^2$ values calculated. Then the value of ${{\displaystyle \sum \left(X_3-\overline{X_3}\right)}}^2$ is calculated as,

$\begin{array}{rll}{{\displaystyle \sum \left(X_3-\overline{X_3}\right)}}^2& =& -36.12+\left(-25.92\right)+\left(-1.12\right)+\left(-12.32\right)+\left(-49.92\right)\\ & =& -125.4\end{array}$

Substitute $-125.4$ for ${\displaystyle \sum \left(X_3-{\overline{X}}_3\right)\left(Y-\overline{Y}\right)}$ and 138.8 for ${{\displaystyle \sum \left(X_3-\overline{X_3}\right)}}^2$ in the above mentioned formula.

$\begin{array}{rll}{b}_2& =& \frac{-125.4}{138.8}\\ & =& -0.9\end{array}$

Substitute 55.8 for $\overline{X}$, 63.6 for $\overline{Y}$ and $-0.9$ for b in the above mentioned formula to solve for intercept.

$\begin{array}{rll}{a}_2& =& 63.6-\left(-0.9\right)\left(55.8\right)\\ & =& 113.82\end{array}$

The value of intercept is 113.82 and the value of slope is $-0.9$.

Conclusion:

The values of intercept are 39, $-94.77$ and 113.8

The values of slope are 3, 12.67 and $-0.9$.

To determine

(b)

To find:

The least regression line and prestige score.

Answer to Problem 12.1P

Solution:

The least square regression line for unemployment rate and turnout is,

$Y_1=39+3X_1$

The least square regression line for average years of school and turnout model is,

$Y_2=-94.77+12.67X_2$

The least square regression line for percentage of negative ads and turnout model is,

$Y_3=113.8-0.9X_3$

The voter turnout is 75, 44.6 and 32.82.

Explanation of Solution

Given:

The three different independent variables: unemployment rate, average years of education for the unemployment rate, average years of school and the percentage of all political ads that used negative campaigning is given table below.

City	Turnout (Y)	Unemployment Rate ($X_1$)	Average Years of School ($X_2$)	Percentage of Negative Ads ($X_3$)
A	55	5	11.9	60
B	60	8	12.1	63
C	65	9	12.7	55
D	68	9	12.8	53
E	70	10	13.0	48

Approach:

The Y intercept (a) is the point of intersection of the regression line and the y axis.

The slope (b) of the regression line is the amount of change in the dependent variable (Y) by a unit change in the independent variable (X).

If the two variables are unrelated, the regression line would be parallel to the x axis.

Formula used:

The model for the least square regression line is defined as,

$Y=a+bX$

Where Y is the dependent variable, $a$ is the intercept of $Y$ and $b$ is the slope.

Calculation:

From the sub-part (a),

The values of intercept are 39, $-94.77$ and 113.8

The values of slope are 3, 12.67 and $-0.9$.

Substitute 39 for $a$ and 3 for $b$ in the above mentioned linear regression formula.

The least square regression line for unemployment rate and turnout is,

$Y_1=39+3X_1\mathrm{....}(1)$

The least square regression line for average years of school and turnout model is,

$Y_2=-94.77+12.67X_2\mathrm{......}(2)$

The least square regression line for percentage of negative ads and turnout model is,

$Y_3=113.8-0.9X_3\mathrm{....}(3)$

Substitute 12 for $X_1$ in equation $\left(1\right)$ and solve for the voter turnout score.

$\begin{array}{rll}{Y}_1& =& 39+3\left(12\right)\\ & =& 75\end{array}$

Substitute 11 for $X_2$ in equation $\left(2\right)$ and solve for the voter turnout score.

$\begin{array}{rll}{Y}_2& =& -94.77+12.67\left(11\right)\\ & =& 44.6\end{array}$

Substitute 90 for $X_3$ in equation $\left(3\right)$ and solve for the voter turnout score.

$\begin{array}{rll}{Y}_3& =& 113.8-0.9\left(90\right)\\ & =& 32.82\end{array}$

Conclusion:

The least square regression line for unemployment rate and turnout is,

$Y_1=39+3X_1$

The least square regression line for average years of school and turnout model is,

$Y_2=-94.77+12.67X_2$

The least square regression line for percentage of negative ads and turnout model is,

$Y_3=113.8-0.9X_3$

The voter turnout is 75, 44.6 and 32.82.

To determine

(c)

To find:

The coefficients $r$ and $r^2$.

Answer to Problem 12.1P

Solution:

The values of intercept are 29.71 and 35.6

The values of slope are 0.58 and 0.52.

Explanation of Solution

Given:

The three different independent variables: unemployment rate, average years of education for the unemployment rate, average years of school and the percentage of all political ads that used negative campaigning is given table below,

City	Turnout (Y)	Unemployment Rate ($X_1$)	Average Years of School ($X_2$)	Percentage of Negative Ads ($X_3$)
A	55	5	11.9	60
B	60	8	12.1	63
C	65	9	12.7	55
D	68	9	12.8	53
E	70	10	13.0	48

Approach:

The Y intercept (a) is the point of intersection of the regression line and the y axis.

The slope (b) of the regression line is the amount of change in the dependent variable (Y) by a unit change in the independent variable (X).

If the two variables are unrelated, the regression line would be parallel to the x axis.

Formula used:

The formula for calculating the correlation coefficient r is given as,

$r=\frac{{\displaystyle \sum \left(X-\overline{X}\right)\left(Y-\overline{Y}\right)}}{\sqrt{\left[{\displaystyle \sum {\left(X-\overline{X}\right)}^2}\right]\left[{\displaystyle \sum {\left(Y-\overline{Y}\right)}^2}\right]}}$

Where X and Y are the two variables

And $\overline{X}$ and $\overline{Y}$ are their respective means.

Calculation:

From the given information,

The turnout is the dependent variable. Thus, it is represented as Y.

The Unemployment rate is the independent variables. Thus, it is as $X_1$.

Consider the following table of sum of squares of turnout and Unemployment rate.

City	Unemployment Rate ($X_1$)	$\left(X_1-\overline{X_1}\right)$	${\left(X_1-\overline{X_1}\right)}^2$	Turnout (Y)	$\left(Y-\overline{Y}\right)$	${\left(Y-\overline{Y}\right)}^2$	$\left(X_1-\overline{X_1}\right)\left(Y-\overline{Y}\right)$
A	5	$-3.2$	10.24	55	$-8.6$	73.96	27.52
B	8	$-0.2$	0.04	60	$-3.6$	12.96	0.72
C	9	0.8	0.64	65	1.4	1.96	1.12
D	9	0.8	0.64	68	4.4	19.36	3.52
E	10	1.8	3.24	70	6.4	40.96	11.52
Total	41		$\begin{array}{l}{\displaystyle \sum {\left(X_1-\overline{X}\right)}^2}\\ =14.8\end{array}$	318		$\begin{array}{l}{\displaystyle \sum {\left(Y-\overline{Y}\right)}^2}\\ =149.2\end{array}$	$\begin{array}{l}{\displaystyle \sum \left(X_1-\overline{X}\right)\left(Y-\overline{Y}\right)}\\ =44.4\end{array}$

The value of $\overline{X_1}$ is 8.2 and $\overline{Y}$ is 63.6.

From above table, substitute $5$ for $X_1$ and $8.2$ for $\overline{X_1}$ in $\left(X_1-\overline{X_1}\right)$.

$\begin{array}{rll}\left(X_1-\overline{X_1}\right)& =& \left(5-8.2\right)\\ & =& -3.2\end{array}$

Square the both sides of the equation.

$\begin{array}{rll}{\left(X_1-\overline{X_1}\right)}^2& =& {\left(-3.2\right)}^2\\ & =& 10.24\end{array}$

Proceed in the same manner to calculate ${\left(X_1-\overline{X_1}\right)}^2$ for all the $1\le i\le N$ for the rest data and refer table for the rest of the ${\left(X_1-\overline{X_1}\right)}^2$ values calculated. Then the value of ${{\displaystyle \sum \left(X_1-\overline{X_1}\right)}}^2$ is calculated as,

$\begin{array}{rll}{{\displaystyle \sum \left(X_1-\overline{X_1}\right)}}^2& =& 10.24+0.04+0.64+0.64+3.24\\ & =& 14.8\end{array}$

Substitute $44.4$ for ${\displaystyle \sum \left(X_1-{\overline{X}}_1\right)\left(Y-\overline{Y}\right)}$, 14.8 for ${\displaystyle \sum {\left(X_1-\overline{X}\right)}^2}$ and 149.2 for in ${\displaystyle \sum {\left(Y_1-\overline{Y}\right)}^2}$ the above mentioned formula.

$\begin{array}{rll}{r}_1& =& \frac{44.4}{\sqrt{\left(14.8\right)\left(149.2\right)}}\\ & =& 0.94\end{array}$

Square the above calculated value of $r$ and solve for $r^2$.

$\begin{array}{l}{r}_1{}^2={\left(0.94\right)}^2\\ r_1{}^2=0.89\end{array}$

The coefficient of $r$ and $r^2$ is 0.94 and 0.89 respectively.

From the given information,

The turnout is the dependent variable. Thus, it is represented as Y.

The average year of school is the independent variables. Thus, it is as $X_2$.

Consider the following table of sum of squares of turnout and Average years of school.

City	Average Years of School ($X_2$)	$\left(X_2-\overline{X_2}\right)$	${\left(X_2-\overline{X_2}\right)}^2$	Turnout (Y)	$\left(Y-\overline{Y}\right)$	${\left(Y-\overline{Y}\right)}^2$	$\left(X_2-\overline{X_2}\right)\left(Y-\overline{Y}\right)$
A	11.9	$-0.6$	0.36	55	$-8.6$	73.96	5.16
B	12.1	$-0.4$	0.16	60	$-3.6$	12.96	1.44
C	12.7	0.2	0.04	65	1.4	1.96	0.28
D	12.8	0.3	0.09	68	4.4	19.36	1.32
E	13.0	0.5	0.25	70	6.4	40.96	3.2
Total	62.5		$\begin{array}{l}{\displaystyle \sum {\left(X_2-\overline{X_2}\right)}^2}\\ =0.9\end{array}$	318		$\begin{array}{l}{\displaystyle \sum {\left(Y-\overline{Y}\right)}^2}\\ =149.2\end{array}$	$\begin{array}{l}{\displaystyle \sum \left(X_2-{\overline{X}}_2\right)\left(Y-\overline{Y}\right)}\\ =11.4\end{array}$

The value of $\overline{X_2}$ is 12.5 and $\overline{Y}$ is 63.6.

From above table, substitute $11.9$ for $X_2$ and 12.5 for $\overline{X_2}$ in $\left(X_2-\overline{X_2}\right)$.

$\begin{array}{rll}\left(X_2-\overline{X_2}\right)& =& \left(11.9-12.5\right)\\ & =& -0.6\end{array}$

Square the both sides of the equation.

$\begin{array}{rll}\left(X_2-\overline{X_2}\right)& =& {\left(-0.6\right)}^2\\ & =& 0.36\end{array}$

Proceed in the same manner to calculate ${\left(X_2-\overline{X_2}\right)}^2$ for all the $1\le i\le N$ for the rest data and refer table for the rest of the ${\left(X_2-\overline{X_2}\right)}^2$ values calculated. Then the value of ${{\displaystyle \sum \left(X_2-\overline{X_2}\right)}}^2$ is calculated as,

$\begin{array}{rll}{{\displaystyle \sum \left(X_2-\overline{X_2}\right)}}^2& =& 0.36+0.16+0.04+0.09+0.25\\ & =& 0.9\end{array}$

Substitute 11.4 for ${\displaystyle \sum \left(X_2-{\overline{X}}_2\right)\left(Y-\overline{Y}\right)}$, 0.9 for ${\displaystyle \sum {\left(X_2-\overline{X_2}\right)}^2}$ and 561.5 for in ${\displaystyle \sum {\left(Y-\overline{Y}\right)}^2}$ the above mentioned formula.

$\begin{array}{rll}{r}_2& =& \frac{11.4}{\sqrt{\left(0.9\right)\left(149.5\right)}}\\ & =& 0.98\end{array}$

Square the above calculated value of $r$ and solve for $r^2$.

$\begin{array}{rll}{r}_2{}^2& =& {\left(0.98\right)}^2\\ & =& 0.96\end{array}$

The coefficient of $r$ and $r^2$ is 0.98 and 0.96 respectively.

From the given information,

The turnout is the dependent variable. Thus, it is represented as Y.

The percentage of negative ads is the independent variables. Thus, it is as $X_3$.

Consider the following table of sum of squares of turnout and Average years of school.

City	Percentage of Negative Ads ($X_3$)	$\left(X_3-\overline{X_3}\right)$	${\left(X_3-\overline{X_3}\right)}^2$	Turnout (Y)	$\left(Y-\overline{Y}\right)$	${\left(Y-\overline{Y}\right)}^2$	$\left(X_3-\overline{X_3}\right)\left(Y-\overline{Y}\right)$
A	60	4.2	17.64	55	$-8.6$	73.96	$-36.12$
B	63	7.2	51.84	60	$-3.6$	12.96	$-25.92$
C	55	$-0.8$	0.64	65	1.4	1.96	$-1.12$
D	53	$-2.8$	7.84	68	4.4	19.36	$-12.32$
E	48	$-7.8$	60.84	70	6.4	40.96	$-49.92$
Total	279		$\begin{array}{l}{\displaystyle \sum {\left(X_3-\overline{X3}\right)}^2}\\ =138.8\end{array}$	318		$\begin{array}{l}{\displaystyle \sum {\left(Y-\overline{Y}\right)}^2}\\ =149.2\end{array}$	$\begin{array}{l}{\displaystyle \sum \left(X_3-{\overline{X}}_3\right)\left(Y-\overline{Y}\right)}\\ =-125.4\end{array}$

The value of $\overline{X_3}$ is 55.8 and $\overline{Y}$ is 63.6.

From above table, substitute $60$ for $X_3$ and 55.8 for $\overline{X_3}$ in $\left(X_3-\overline{X_3}\right)$.

$\begin{array}{rll}\left(X_3-\overline{X_3}\right)& =& \left(60-55.8\right)\\ & =& 4.2\end{array}$

Square the both sides of the equation.

$\begin{array}{rll}{\left(X_3-\overline{X_3}\right)}^2& =& {\left(4.2\right)}^2\\ & =& 17.64\end{array}$

Proceed in the same manner to calculate ${\left(X_3-\overline{X_3}\right)}^2$ for all the $1\le i\le N$ for the rest data and refer table for the rest of the ${\left(X_3-\overline{X_3}\right)}^2$ values calculated. Then the value of ${{\displaystyle \sum \left(X_3-\overline{X_3}\right)}}^2$ is calculated as,

$\begin{array}{rll}{{\displaystyle \sum \left(X_3-\overline{X_3}\right)}}^2& =& -36.12+\left(-25.92\right)+\left(-1.12\right)+\left(-12.32\right)+\left(-49.92\right)\\ & =& -125.4\end{array}$

Substitute $-125.4$ for ${\displaystyle \sum \left(X_3-{\overline{X}}_3\right)\left(Y-\overline{Y}\right)}$, 138.8 for ${\displaystyle \sum {\left(X_3-\overline{X_3}\right)}^2}$ and 561.5 for in ${\displaystyle \sum {\left(Y-\overline{Y}\right)}^2}$ the above mentioned formula.

$\begin{array}{rll}{r}_3& =& \frac{-125.4}{\sqrt{\left(138.8\right)\left(149.5\right)}}\\ & =& -0.87\end{array}$

Square the above calculated value of $r$ and solve for $r^2$.

$\begin{array}{rll}{r}_3{}^2& =& {\left(-0.87\right)}^2\\ & =& 0.76\end{array}$

The coefficient of $r$ and $r^2$ is $-0.87$ and 0.76 respectively.

Conclusion:

The coefficients of $r$ are 0.94, 0.98 and $-0.87$.

The coefficients of $r^2$ are 0.89, 0.96 and 0.76

To determine

(d)

To explain:

The strength, direction and impact of given independent variable on voter’s turnout.

Answer to Problem 12.1P

Solution:

The required explanation is stated.

Explanation of Solution

Given:

The three different independent variables: unemployment rate, average years of education for the unemployment rate, average years of school and the percentage of all political ads that used negative campaigning is given table below.

City	Turnout (Y)	Unemployment Rate ($X_1$)	Average Years of School ($X_2$)	Percentage of Negative Ads ($X_3$)
A	55	5	11.9	60
B	60	8	12.1	63
C	65	9	12.7	55
D	68	9	12.8	53
E	70	10	13.0	48

Calculation:

From sub-part (c), the positive value of correlation between unemployment rate and voter’s turnout shows that there is positive relation between unemployment rate and voter turnout.

The coefficient of determination shows that 89% of variation in the voter’s turnout will be explained by unemployment rate.

The positive value of correlation between average years of school and voter’s turnout shows that there is positive relation between average years of school and voter’s turnout.

The coefficient of determination shows that 96% of variation in the voter’s turnout will be explained by average years of school.

The negative value of correlation between negative campaigning and voter’s turnout shows that there is negative relation between negative campaigning and voter’s turnout

The coefficient of determination shows that 76% of variation in the voter’s turnout will be explained by negative campaigning.

Conclusion:

The required explanation is stated.