Chapter 4.3, Problem 23E

Hot enough for you? The following table presents 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.

1. Compute the least-squares regression line for predicting temperature from barometric pressure.
2. Compute file coefficient of determination.
3. Construct a scatterplot of due temperature (y) versus the barometric pressure (x).
4. Which point is an outlier?
5. Remove outlier and compute least-squares regression line for predicting temperature from barometric pressure.
6. Is the outlier Explain.
7. Compute the coefficient of determination for the data set with the outlier removed. Is the proportion of variation explained by the least-squares regression fine greater, less, or about the same without die outlier? Explain.

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:

Belowtable 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:

  

Formula 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 $y=-21.8708x+740.3765$

To determine

The coefficient of determination.

Answer to Problem 23E

  $0.3225$

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

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.

Putting 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 $0.3225$ .

To determine

Construct the scatterplot of the temperature $\left(y\right)$ versus the barometric pressure $\left(x\right)$ .

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:

  

Calculation:

The scatter plot can be drawn with the help of the given data. The Pressure will be taken on horizontal axis and the temperature will betaken on vertical axis.

  

From the scatterplot, it is observed that there is a moderate association between the “variable temperature” and “barometric pressure”.

To determine

The points which are outliers.

Answer to Problem 23E

  $\left(29.83,94.5\right)$

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Calculation:

  

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

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

To determine

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

Answer to Problem 23E

  $y=-7.8999x+320.5388$

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Formula 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 $d$ of this exercise outlier is $\left(29.83,94.5\right)$

Excluding the outlier,

The mean of $x$ is given by:

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

The mean of $y$ is given by:

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

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 $y=-7.8999x+320.5388$

To determine

Whether the outlier is influential.

Answer to Problem 23E

The outlier is influential.

Explanation of Solution

Given information:

Same as part $a$ .

Calculation:

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

From part $e$ , the least squares regression line for the given data set by removing 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.

To determine

The coefficient of determination for the data set excluding the outliers.

Answer to Problem 23E

  $0.1133$

The proportion of variation is less without the outlier.

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Calculation:

The coefficient of determination is given by:

  $r^2$

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

From part $d$ of this exercise, the correlation coefficient with the outlier removed is $r=-\mathrm{0.33658...}$

Putting 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 reduced without the outlier.

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