Chapter 14, Problem 24P

Although you might suspect that dissatisfied people would be the most likely individuals to participate in political activities in an attempt to change things, research tends to show just the opposite. Flavin and Keane (2011) found a positive relationship between life satisfaction and political participation, which included activities such as attending rallies, contributing to candidates, and displaying a yard sign. Following are data similar to those obtained in the study.

Life Satisfaction	Political Participation
5	4
8	7
3	2
6	9
3	5
1	3
4	6
2	4

• a. Find the regression equation for predicting political participation from life satisfaction.
• b. Using α = .05, test the significance of the regression equation.

To determine

The regression equation for predicting political participation from life satisfaction.

Answer to Problem 24P

The regression equation of the data is $Y=0.723X+2.108$.

Explanation of Solution

Given info:

The given data are shown below,

Life satisfaction	Political participation
5	4
8	7
3	2
6	9
3	5
1	3
4	6
2	4

Calculation:

Generally the linear regression equation is defined as

$Y=bX+a$

Formula to calculate $b$ is,

$b=\frac{SP}{S{S}_x}$

Formula to calculate the $a$ is,

$a=M_Y-b{M}_X$

Mean of the variable $X$ is,

$\begin{array}{c}{M}_X=\frac{5+8+3+6+3+1+4+2}{8}\\ =\frac{32}{8}\\ =4\end{array}$

Thus the mean of the variable $X$  is 4.

Mean of the variable $Y$ is,

$\begin{array}{c}{M}_Y=\frac{4+7+2+9+5+3+6+4}{8}\\ =\frac{40}{8}\\ =5\end{array}$

Thus the mean of the variable y is 5.

For calculating $S{S}_x$ and $SP$ the table is shown below,

Sr.no	$X$	$Y$	$X-M_X$	$Y-M_Y$	${\left(X-M_X\right)}^2$	${\left(Y-M_Y\right)}^2$	$\left(X-M_X\right)\left(Y-M_Y\right)$
1	5	4	1	-1	1	1	-1
2	8	7	4	2	16	4	8
3	3	2	-1	-3	1	9	3
4	6	9	2	4	4	16	8
5	3	5	-1	0	1	0	0
6	1	3	-3	-2	9	4	6
7	4	6	0	1	0	1	0
8	2	4	-2	-1	4	1	2
Total			0	0	36	36	26

Where, $M_Y$ is the mean of the variable $Y$ and $M_X$ is the mean of variable $X$. Substitute 26 for $SP$ and 36 for $S{S}_X$ in the equation for calculating b to get the slop coefficient.

$\begin{array}{c}b=\frac{26}{36}\\ =0.723\end{array}$

Substitute 0.723 for $b$, 4 for $M_X$ and 5 for $M_Y$ In the equation (3) to get the coefficient.

$\begin{array}{c}a=5-0.723\times 4\\ =5-2.892\\ =2.108\end{array}$

Substitute 0.723 for $b$ and 2.108 for $a$ in equation for calculating a to get the regression equation.

$Y=0.723X+2.108$

Thus, the regression equation is $\underset{\_}{Y=0.723X+2.108}$.

To determine

The significance of the regression equation.

Answer to Problem 24P

The regression equation is significance at 5% level of significance.

Explanation of Solution

Calculation:

As per part (a) the regression equation is $Y=0.723X+2.108$.

Formula to calculate the Pearson correlation is,

$r=\frac{SP}{\sqrt{S{S}_{x}S{S}_Y}}$

Where variability of the variable x and y is $SP$, $S{S}_X$ is the variability of the x variable and $S{S}_Y$ is the variability of the y variable.

Mean of the variable $X$ is,

$\begin{array}{c}{M}_X=\frac{5+8+3+6+3+1+4+2}{8}\\ =\frac{32}{8}\\ =4\end{array}$

Thus the mean of the variable $X$  is 4.

Mean of the variable $Y$ is,

$\begin{array}{c}{M}_Y=\frac{4+7+2+9+5+3+6+4}{8}\\ =\frac{40}{8}\\ =5\end{array}$

Thus the mean of the variable y is 5.

For calculating $S{S}_x$ and $SP$ the table is shown below,

Sr.no	$X$	$Y$	$X-M_X$	$Y-M_Y$	${\left(X-M_X\right)}^2$	${\left(Y-M_Y\right)}^2$	$\left(X-M_X\right)\left(Y-M_Y\right)$
1	5	4	1	-1	1	1	-1
2	8	7	4	2	16	4	8
3	3	2	-1	-3	1	9	3
4	6	9	2	4	4	16	8
5	3	5	-1	0	1	0	0
6	1	3	-3	-2	9	4	6
7	4	6	0	1	0	1	0
8	2	4	-2	-1	4	1	2
Total			0	0	36	36	26

Where, $M_Y$ is the mean of the variable $Y$ and $M_X$ is the mean of variable $X$. Substitute 26 for $SP$, 36 for $S{S}_X$ and 36 for $S{S}_Y$ in the general linear equation to get the Pearson correlation of the given data.

$\begin{array}{c}r=\frac{26}{\sqrt{36\times 36}}\\ =\frac{26}{36}\\ =0.722\end{array}$

Thus, the Pearson correlation is 0.722.

Set the null hypothesis

$H_0:b=0$

The alternative hypothesis

$H_1:b\ne 0$

Formula to calculate the $F$ ratio is,

$F=\frac{M{S}_{regression}}{M{S}_{residual}}\text{with}\text{df}=1,n-2$

Where, $M{S}_{regression}=\frac{S{S}_{regression}}{d{f}_{regression}}\text{with}df=1$ and $M{S}_{residual}=\frac{S{S}_{residual}}{d{f}_{residual}}\text{with}df=n-2$.

$S{S}_{regression}=r^2\times S{S}_Y$

$S{S}_{residual}=\left(1-r^2\right)S{S}_Y$

Substitute 0.722 for $r$ and 36 for $S{S}_Y$ in the equation (3) to get the $S{S}_{regression}$,

$\begin{array}{c}S{S}_{regression}={(}^{0.722}\times 36\\ =0.5212\times 36\\ =18.766\end{array}$

Substitute 0.722 for $r$ and 36 for $S{S}_Y$ in the equation (4) to get the $S{S}_{residual}$,

$\begin{array}{c}S{S}_{residual}=\left(1-{\left(0.722\right)}^2\right)\times 36\\ =\left(1-0.5212\right)\times 36\\ =0.4788\times 36\\ =17.2368\end{array}$

Thus the $S{S}_{residual}$ is 17.2368

$\begin{array}{c}M{S}_{regression}=\frac{S{S}_{regression}}{df}\\ =\frac{18.766}{1}\\ =18.766\end{array}$

Thus the $M{S}_{regression}$ is 18.2376

$\begin{array}{c}M{S}_{residual}=\frac{S{S}_{residual}}{d{f}_{residual}}\\ =\frac{17.2368}{6}\\ =2.8728\end{array}$

Thus the $M{S}_{residual}$ is 2.8728

Substitute 18.766 for $M{S}_{regression}$ and 2.8728 for $M{S}_{residual}$ in the equation of F-ratio,

$\begin{array}{c}F=\frac{18.766}{2.8728}\text{with}\text{df}=1,6-2\\ =6.532\end{array}$

Thus the $F$ ratio is 6.532.

From the F­-table, the value of $F_{tab}$ at 5% level of significance with $df$ 1 and $n-2=6$ is 5.99.

Decision rule:

If the calculated F-value is less than table F-value, then there is enough evidence to accept to reject the null hypothesis.

As $F_{cal}$ is greater than the $F_{tab}$ , thus there is enough evidence to reject the null hypothesis that is $H_0:b=0$ and conclude that the regression equation does account for a significance portion of the variance for the $Y$ score.

Thus, the regression equation is significance at 5% level of significance.