Chapter 16, Problem 20CAP

(a)

To determine

Find the equation of the regression line.

Answer to Problem 20CAP

The equation of the regression line is, $\widehat{Y}=-1.41X+15.43$.

Explanation of Solution

Calculation:

The information given that is the psychologist records the time (in minutes) that parents spend with their children during a 20-minute session with the factors birth order (X) time attending to the child (Y).

The procedure of straight line by using the method of least square is shown below:

• Compute preliminary calculations.
• Calculate the slope $\left(b\right)$.
• Calculate the y-intercept $\left(a\right)$.

The formula for slope is,

$\text{slope}\left(b\right)=\frac{S{S}_{XY}}{S{S}_X}$

Where,

$S{S}_{XY}={\displaystyle \sum \left(X-M_X\right)\left(Y-M_Y\right)}$

$S{S}_X={\displaystyle \sum {\left(X-M_X\right)}^2}$

In the formulas, $M_X$ denotes the mean of factor X, $M_Y$ denotes the mean of factor Y.

The formula for y-intercept is,

$a=M_Y-b{M}_X$

The regression equation is, $\widehat{Y}=bX+a$.

The value of $S{S}_X$:

The value of mean for X is,

$\begin{array}{c}{M}_X=\frac{1+1+5+2+3+2+4+5}{8}\\ =\frac{23}{8}\\ =2.875\end{array}$

X	Y	$\left(X-M_X\right)$	${\left(X-M_X\right)}^2$
1	12	$\left(1-2.875\right)=-1.875$	3.516
1	14	$\left(1-2.875\right)=-1.875$	3.516
5	9	$\left(5-2.875\right)=2.125$	4.516
2	10	$\left(2-2.875\right)=\left\{0.875\right\}$	0.766
3	15	$\left(3-2.875\right)=0.125$	0.016
2	16	$\left(2-2.875\right)=-0.875$	0.766
4	8	$\left(4-2.875\right)=1.125$	1.266
5	7	$\left(5-2.875\right)=2.125$	4.516
			${\displaystyle \sum {\left(X-M_X\right)}^2}=18.875$

Table 1

The value of $S{S}_X$ is 18.875.

The value of $S{S}_{XY}$:

The value of mean for Y is,

$\begin{array}{c}{M}_Y=\frac{12+14+9+10+15+16+8+7}{8}\\ =\frac{91}{8}\\ =11.375\end{array}$

X	Y	$\left(X-M_X\right)$	$\left(X-M_Y\right)$
1	12	$\left(1-2.875\right)=-1.875$	$\left(12-11.375\right)=0.625$
1	14	$\left(1-2.875\right)=-1.875$	$\left(14-11.375\right)=2.625$
5	9	$\left(5-2.875\right)=2.125$	$\left(9-11.375\right)=-2.375$
2	10	$\left(2-2.875\right)=\left\{0.875\right\}$	$\left(10-11.375\right)=-1.375$
3	15	$\left(3-2.875\right)=0.125$	$\left(15-11.375\right)=3.625$
2	16	$\left(2-2.875\right)=-0.875$	$\left(16-11.375\right)=4.625$
4	8	$\left(4-2.875\right)=1.125$	$\left(8-11.375\right)=-3.375$
5	7	$\left(5-2.875\right)=2.125$	$\left(7-11.375\right)=-4.375$
$\left(X-M_X\right)\left(X-M_Y\right)$
–1.172
–4.922
–5.047
1.203
0.453
–4.047
–3.797
–9.297
${\displaystyle \sum \left(X-M_X\right)\left(X-M_Y\right)}=-26.625$

Table 2

The value of $S{S}_{XY}$ is –26.625.

Slope:

Substitute, $S{S}_X=18.875$ and $S{S}_{XY}=-26.625$ in slope formula

$\begin{array}{c}\text{slope}\left(b\right)=\frac{-26.625}{18.875}\\ =-1.41\end{array}$

Hence, the value of slope is –1.41.

Substitute, $M_Y=11.375,M_X=2.875,b=-1.41$ in y-intercept formula.

$\begin{array}{c}a=M_Y-b{M}_X\\ =11.375-\left(-1.41\right)2.875\\ =11.375+4.054\\ =15.43\end{array}$

Hence, the value of y-intercept is 15.43.

Substitute, $a=15.43,b=-1.41$ in regression equation formula.

$\begin{array}{c}\widehat{Y}=\left(-1.41\right)X+15.43\\ =-1.41X+15.43\end{array}$

The equation of the regression line is, $\widehat{Y}=-1.41X+15.43$.

(b)

To determine

Find the predicted amount of time a parent spends with his or her second-born child.

Answer to Problem 20CAP

The predicted amount of time a parent spends with his or her second-born child is 12.61 minutes.

Explanation of Solution

Calculation:

The given information is that the amount of time a parent spends with his or her second-born child.

The equation of the regression line is $\widehat{Y}=-1.41X+15.43$.

Substitute, $X=2$ in regression line.

$\begin{array}{c}\widehat{Y}=-1.41\left(2\right)+15.43\\ =-2.82+15.43\\ =12.61\end{array}$

Hence, the predicted amount of time a parent spends with his or her second-born child is 12.61 minutes.