Chapter 4.3, Problem 24E

(a)

>

To determine

The least squares regression line for the given data set.

Answer to Problem 24E

$y=0.8395x+12.6556$

Explanation of Solution

Given information:

The following table represents the ages of the last $12$

U.S. presidents and their wives on the first day of their presidencies:

Concepts Used:

The equation for least-square regression line:

$y=b_0+b_{1}x$

Where $b_1=r\frac{s_y}{s_x}$ is the slope and $b_0=\overline{y}-b_1\overline{x}$ is the $y$ -intercept.

The correlation coefficient of a data is given by:

$r=\frac{1}{n}\frac{{\displaystyle \sum \left(x-\overline{x}\right)\left(y-\overline{y}\right)}}{s_x{s}_y}$

Where,

$\overline{x},\overline{y}$ represent the mean of $x$ and $y$ respectively. $s_x,s_y$ represent the standard deviations of $x$ and $y$, $n$ represents the number of terms.

The standard deviations are given by:

$s_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}},s_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n}}$

Calculation:

The mean of $x$ is given by:

$\overline{x}=\frac{{\displaystyle \sum x}}{n}=\frac{45+54+45+55+59+49+56+56+50+31+56+60}{12}=\frac{616}{12}=51.33333$

The mean of $y$ is given by:

$\overline{y}=\frac{{\displaystyle \sum y}}{n}=\frac{47+54+46+64+69+52+61+56+55+43+62+60}{12}=\frac{669}{12}=55.75$

The data can be represented in tabular form as:

Hence, the standard deviation is given by:

$\begin{array}{l}{s}_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}}\\ s_x=\sqrt{\frac{720.66667}{12}}\\ s_x=\sqrt{60.05556}\end{array}$

And,

$\begin{array}{l}{s}_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n}}\\ s_y=\sqrt{\frac{680.25}{12}}\\ s_y=\sqrt{56.6875}\end{array}$

Consider, $r=\frac{1}{n}\frac{{\displaystyle \sum \left(x-\overline{x}\right)\left(y-\overline{y}\right)}}{s_x{s}_y}$

Putting the values in the formula,

$\begin{array}{l}r=\frac{1}{n}\frac{{\displaystyle \sum \left(x-\overline{x}\right)\left(y-\overline{y}\right)}}{s_x{s}_y}\\ r=\frac{1}{12}\cdot \frac{605}{\left(\sqrt{60.05556}\right)\left(\sqrt{56.6875}\right)}\\ r=0.86408\end{array}$

Putting the values to obtain $b_1$,

$\begin{array}{l}{b}_1=r\frac{s_y}{s_x}\\ b_1=\frac{1}{12}\cdot \frac{605}{\left(\sqrt{60.05556}\right)\left(\sqrt{56.6875}\right)}\frac{\left(\sqrt{56.6875}\right)}{\left(\sqrt{60.05556}\right)}\\ b_1=0.8395\end{array}$

Putting the values to obtain $b_0$,

$\begin{array}{l}{b}_0=\overline{y}-b_1\overline{x}\\ b_0=\left(55.75\right)-\left(0.8395\right)\left(51.33333\right)\\ b_0=12.6556\end{array}$

Hence, the least-square regression line is given by:

$\begin{array}{l}y=b_0+b_{1}x\\ y=\left(12.6556\right)+\left(0.8395\right)x\\ y=0.8395x+12.6556\end{array}$

Therefore, the least squares regression line for the given data set is $y=0.8395x+12.6556$

(b)

>

To determine

The coefficient of determination.

Answer to Problem 24E

$0.74663$.

Explanation of Solution

Given information:

Same as part $a$.

Calculation:

From part $a$ of this exercise, the correlation coefficient is $r=0.86408$

The coefficient of determination is given by:

Where $r$ is the correlation coefficient of the data.

Putting the values to obtain Coefficient of Determination,

$r^2={\left(0.86408\right)}^2=\mathrm{0.746634...}\approx 0.74663$

Therefore, the Coefficient of Determination is $0.74663$.

(c)

>

To determine

A scatter plot of the given data.

Answer to Problem 24E

Explanation of Solution

Given information:

Same as part $a$.

Calculation:

Consider age of the president as $x$ and age of the first lady as $y$.

The points representing the data would be given by:

$\begin{array}{l}\left(45,47\right)\hfill \\ \left(54,54\right)\hfill \\ \left(45,46\right)\hfill \\ \left(55,64\right)\hfill \\ \left(59,69\right)\hfill \\ \left(49,52\right)\hfill \\ \left(56,61\right)\hfill \\ \left(56,56\right)\hfill \\ \left(50,55\right)\hfill \\ \left(31,43\right)\hfill \\ \left(56,62\right)\hfill \\ \left(60,60\right)\hfill \end{array}$

Plotting the points to make a scatter plot:

(d)

>

To determine

The outlier points.

Answer to Problem 24E

$\left(31,43\right)$

Explanation of Solution

Given information:

Same as part $a$.

Calculation:

From the given table, it can be seen that among all the $y$ values $43$ stands away from all other values.

Therefore, the outlier point would be $\left(31,43\right)$

(e)

>

To determine

The least squares regression line for the given data set by excluding the outlier.

Answer to Problem 24E

$y=1.1949x-6.6365$

Explanation of Solution

Given information:

Same as part $a$.

Concepts used:

The equation for least-square regression line:

$y=b_0+b_{1}x$

Where $b_1=r\frac{s_y}{s_x}$ is the slope and $b_0=\overline{y}-b_1\overline{x}$ is the $y$ -intercept.

The correlation coefficient of a data is given by:

$r=\frac{1}{n}\frac{{\displaystyle \sum \left(x-\overline{x}\right)\left(y-\overline{y}\right)}}{s_x{s}_y}$

Where,

$\overline{x},\overline{y}$ represent the mean of $x$ and $y$ respectively. $s_x,s_y$ represent the standard deviations of $x$ and $y$, $n$ represents the number of terms.

The standard deviations are given by:

$s_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}},s_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n}}$

Calculation:

From part $d$, the outlier point is $\left(31,43\right)$.

Excluding the outlier,

The mean of $x$ is given by:

$\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x}}{n}=\frac{45+54+45+55+59+49+56+56+50+56+60}{11}\\ \overline{x}=\frac{585}{11}\\ \overline{x}=53.18182\end{array}$

The mean of $y$ is given by:

$\begin{array}{l}\overline{y}=\frac{{\displaystyle \sum y}}{n}=\frac{47+54+46+64+69+52+61+56+55+62+60}{11}\\ \overline{y}=\frac{626}{11}\\ \overline{y}=56.90909\end{array}$

The data can be represented in tabular form as:

Hence, the standard deviation is given by:

$\begin{array}{l}{s}_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}}\\ s_x=\sqrt{\frac{269.63636}{11}}\\ s_x=\sqrt{24.51240}\end{array}$

And,

$\begin{array}{l}{s}_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n}}\\ s_y=\sqrt{\frac{502.90909}{11}}\\ s_y=\sqrt{45.71901}\end{array}$

Consider,

Putting the values in the formula,

$\begin{array}{l}r=\frac{1}{n}\frac{{\displaystyle \sum \left(x-\overline{x}\right)\left(y-\overline{y}\right)}}{s_x{s}_y}\\ r=\frac{1}{11}\cdot \frac{322.18182}{\left(\sqrt{24.51240}\right)\left(\sqrt{45.71901}\right)}\\ r=0.87492\end{array}$

Putting the values to obtain $b_1$,

$\begin{array}{l}{b}_1=r\frac{s_y}{s_x}\\ b_1=\frac{1}{11}\cdot \frac{322.18182}{\left(\sqrt{24.51240}\right)\left(\sqrt{45.71901}\right)}\frac{\left(\sqrt{45.71901}\right)}{\left(\sqrt{24.51240}\right)}\\ b_1=1.1949\end{array}$

Putting the values to obtain $b_0$,

$\begin{array}{l}{b}_0=\overline{y}-b_1\overline{x}\\ b_0=\left(56.90909\right)-\left(1.1949\right)\left(53.18182\right)\\ b_0=-6.6365\end{array}$

Hence, the least-square regression line is given by:

$\begin{array}{l}y=b_0+b_{1}x\\ y=\left(-6.6365\right)+\left(1.1949\right)x\\ y=1.1949x-6.6365\end{array}$

Therefore, the least squares regression line for the given data set by removing the outlier is $y=1.1949x-6.6365$

(f)

>

To determine

Whether the outlier is influential.

Answer to Problem 24E

The outlier is influential.

Explanation of Solution

Given information:

Same as part $a$.

Calculation:

From part $a$ of this exercise, the least squares regression line for the given data set is $y=0.8395x+12.6556$

From part $a$ of this exercise, the least squares regression line for the given data set excluding the outlier is $y=1.1949x-6.6365$

From the equations, it can be seen that removing the outlier creates a great difference in the equation of the least square regression line.

Therefore, the outlier is influential.

(g)

>

To determine

The coefficient of determination for the data set with the outlier removed.

Answer to Problem 24E

Coefficient of Determination is $0.7655$.

The proportion of variation is more without the outlier.

Explanation of Solution

Given information:

Same as part $a$.

Calculation:

From part $e$ of this exercise, the correlation coefficient with the outlier removed is $r=0.87492$

The coefficient of determination is given by:

$r^2$

Where $r$ is the correlation coefficient of the data.

Plugging the values to obtain Coefficient of Determination,

$r^2={\left(0.87492\right)}^2=0.765480\approx 0.7655$

Therefore, the Coefficient of Determination is $0.7655$.

Here the coefficient of determination increased without the outlier.

Hence, the proportion of variance explained is more without the outlier.