Chapter 4.3, Problem 23E

(a)

>

To determine

The least squares regression line for the given data set.

Answer to Problem 23E

$y=-21.8708x+740.3765$

Explanation of Solution

Given information:

Below table represents the temperature, in degrees Fahrenheit, and barometric pressure, in inches of mercury, on August $15$ at $12$

noon in Macon, Georgia, over a nine-year period:

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:

$\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x}}{n}=\frac{30.16+30.09+29.99+29.83+29.83+30.00+29.98+29.85+30.01}{9}\\ \overline{x}=\frac{269.77}{9}\\ \overline{x}=\mathrm{29.97444...}\end{array}$

The mean of $y$ is given by:

$\begin{array}{l}\overline{y}=\frac{{\displaystyle \sum y}}{n}=\frac{80.1+85.5+87.0+94.5+84.0+83.2+80.1+84.9+84.0}{9}\\ \overline{y}=\frac{763.3}{9}\\ \overline{y}=\mathrm{84.81111...}\end{array}$

The data can be represented in tabular form as:

Hence, the standard deviation is given by:

$\begin{array}{l}{s}_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}}\\ s_x=\sqrt{\frac{\mathrm{0.09942...}}{9}}\\ s_x=\sqrt{\mathrm{0.01104...}}\end{array}$

And,

$\begin{array}{l}{s}_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n}}\\ s_y=\sqrt{\frac{\mathrm{147.44888...}}{9}}\\ s_y=\sqrt{\mathrm{16.38320...}}\end{array}$

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,

$\begin{array}{l}r=\frac{1}{n}\frac{{\displaystyle \sum \left(x-\overline{x}\right)\left(y-\overline{y}\right)}}{s_x{s}_y}\\ r=\frac{1}{9}\cdot \frac{-\mathrm{2.17444...}}{\left(\sqrt{\mathrm{0.01104...}}\right)\left(\sqrt{\mathrm{16.38320...}}\right)}\\ r=-\mathrm{0.56791...}\end{array}$

Putting the values to obtain $b_1$,

$\begin{array}{l}{b}_1=r\frac{s_y}{s_x}\\ b_1=\left(-\mathrm{0.56791...}\right)\frac{\left(\sqrt{\mathrm{16.38320...}}\right)}{\left(\sqrt{\mathrm{0.01104...}}\right)}\\ b_1=-\mathrm{21.87080...}\\ b_1\approx -21.8708\end{array}$

Putting the values to obtain $b_0$,

$\begin{array}{l}{b}_0=\overline{y}-b_1\overline{x}\\ b_0=\left(\mathrm{84.81111...}\right)-\left(-\mathrm{21.87080...}\right)\left(\mathrm{29.97444...}\right)\\ b_0=\mathrm{740.37646...}\\ b_0\approx 740.3765\end{array}$

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

$\begin{array}{l}y=b_0+b_{1}x\\ y=\left(740.3765\right)+\left(-21.8708\right)x\\ y=-21.8708x+740.3765\end{array}$

Therefore, the least squares regression line for the given data set is $y=-21.8708x+740.3765$

(b)

>

To determine

The coefficient of determination.

Answer to Problem 23E

$0.3225$.

Explanation of Solution

Given information:

Same as part $a$.

Calculation:

From part $a$ of this exercise, the correlation coefficient is $r=-\mathrm{0.56791...}$

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,

$r^2={\left(-\mathrm{0.56791...}\right)}^2=\mathrm{0.32253...}\approx 0.3225$

Therefore, the Coefficient of Determination is $0.3225$.

(c)

>

To determine

A scatter plot of the temperature (y) versus the barometric pressure (x).

Answer to Problem 23E

Explanation of Solution

Given information:

Same as part $a$.

Calculation:

Consider pressure as $x$ and temperature as $y$.

The points representing the data would be given by:

$\begin{array}{l}\left(30.16,80.1\right)\\ \left(30.09,85.5\right)\\ \left(29.99,87.0\right)\\ \left(29.83,94.5\right)\\ \left(29.86,84.0\right)\\ \left(30.00,83.2\right)\\ \left(29.98,80.1\right)\\ \left(29.85,84.9\right)\\ \left(30.01,84.0\right)\end{array}$

Plotting the points to make a scatter plot:

(d)

>

To determine

The outliers point.

Answer to Problem 23E

$\left(29.83,94.5\right)$

Explanation of Solution

Given information:

Same as part $a$.

Calculation:

Consider pressure as $x$ and temperature as $y$.

From above table, it can be observed that among all the $y$ values $94.5$ stands away from all other values.

Therefore, the outlier point is $\left(29.83,94.5\right)$

(e)

>

To determine

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

Answer to Problem 23E

$y=-7.8999x+320.5388$

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 $\left(29.83,94.5\right)$.

Excluding the outlier,

The mean of $x$ is given by:

$\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x}}{n}=\frac{30.16+30.09+29.99+29.83+30.00+29.98+29.85+30.01}{8}\\ \overline{x}=\frac{239.94}{8}\\ \overline{x}=29.9925\end{array}$

The mean of $y$ is given by:

$\begin{array}{l}\overline{y}=\frac{{\displaystyle \sum y}}{n}=\frac{80.1+85.5+87.0+84.0+83.2+80.1+84.9+84.0}{8}\\ \overline{y}=\frac{668.8}{8}\\ \overline{y}=83.6\end{array}$

The data can be represented in tabular form as:

Hence, the standard deviation is given by:

$\begin{array}{l}{s}_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}}\\ s_x=\sqrt{\frac{0.07595}{8}}\\ s_x=\sqrt{\mathrm{0.00949...}}\end{array}$

And,

$\begin{array}{l}{s}_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n}}\\ s_y=\sqrt{\frac{41.84}{8}}\\ s_y=\sqrt{5.23}\end{array}$

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,

$\begin{array}{l}r=\frac{1}{n}\frac{{\displaystyle \sum \left(x-\overline{x}\right)\left(y-\overline{y}\right)}}{s_x{s}_y}\\ r=\frac{1}{8}\cdot \frac{\left(-0.6\right)}{\left(\sqrt{\mathrm{0.00949...}}\right)\left(\sqrt{5.23}\right)}\\ r=-\mathrm{0.33658...}\end{array}$

Putting the values to obtain $b_1$,

$\begin{array}{l}{b}_1=r\frac{s_y}{s_x}\\ b_1=\left(-\mathrm{0.33658...}\right)\frac{\left(\sqrt{5.23}\right)}{\left(\sqrt{\mathrm{0.00949...}}\right)}\\ b_1=-\mathrm{7.89993...}\\ b_1\approx -7.8999\end{array}$

Plugging the values to obtain $b_0$,

$\begin{array}{l}{b}_0=\overline{y}-b_1\overline{x}\\ b_0=\left(83.6\right)-\left(-\mathrm{7.89993...}\right)\left(29.9925\right)\\ b_0=\mathrm{320.53877...}\\ b_0\approx 320.5388\end{array}$

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

$\begin{array}{l}y=b_0+b_{1}x\\ y=\left(320.5388\right)+\left(-7.8999\right)x\\ y=-7.8999x+320.5388\end{array}$

Therefore, the least squares regression line for the given data set by excluding the outlier is $y=-7.8999x+320.5388$

(f)

>

To determine

Whether outlier is influential.

Answer to Problem 23E

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 $y=-21.8708x+740.3765$

From part $a$ of this exercise, the least squares regression line for the given data set excluding the outlier is $y=-7.8999x+320.5388$

From above equations, it can be observed 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 23E

$0.1133$

The proportion of variation is less 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 $r=-\mathrm{0.33658...}$

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,

$r^2={\left(-\mathrm{0.33658...}\right)}^2=\mathrm{0.11328...}\approx 0.1133$

Therefore, the Coefficient of Determination is $0.1133$.

Here the coefficient of determination has reduced without the outlier.

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