Chapter 9.2, Problem 9P

(a)

To determine

Construct a scatter diagram for the data.

Answer to Problem 9P

The scatter diagram for data is,

Explanation of Solution

Calculation:

The variable x denotes the weight of the car (in hundreds of pounds) and y denotes the miles per gallon (mpg).

Step by step procedure to obtain scatter plot using MINITAB software is given below:

• Choose Graph > Scatterplot.
• Choose Simple. Click OK.
• In Y variables, enter the column of x.
• In X variables, enter the column of y.
• Click OK.

(b)

To determine

Verify the values of ${\displaystyle \sum x}$, ${\displaystyle \sum x^2}$, ${\displaystyle \sum y}$, ${\displaystyle \sum y^2}$, ${\displaystyle \sum xy}$ and r.

Explanation of Solution

Calculation:

The formula for correlation coefficient is,

$r=\frac{n{\displaystyle \sum xy}-\left({\displaystyle \sum x}\right)\left({\displaystyle \sum y}\right)}{\sqrt{n{\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\sqrt{n{\displaystyle \sum y^2}-{\left({\displaystyle \sum y}\right)}^2}}$

In the formula, n is the sample size.

The values are verified in the table below,

x	y	$x^2$	$y^2$	xy
27	30	729	900	810
44	19	1936	361	836
32	24	1024	576	768
47	13	2209	169	611
23	29	529	841	667
40	17	1600	289	680
34	21	1156	441	714
52	14	2704	196	728
${\displaystyle \sum x}=299$	${\displaystyle \sum y}=167$	${\displaystyle \sum x^2}=11,887$	${\displaystyle \sum y^2}=3,773$	${\displaystyle \sum xy}=5,814$

Hence, the values are verified.

The number of data pairs are $n=8$. The value of correlation r is,

$\begin{array}{c}r=\frac{n{\displaystyle \sum xy}-\left({\displaystyle \sum x}\right)\left({\displaystyle \sum y}\right)}{\sqrt{n{\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\sqrt{n{\displaystyle \sum y^2}-{\left({\displaystyle \sum y}\right)}^2}}\\ =\frac{8\left(5,814\right)-\left(299\times 167\right)}{\sqrt{8\left(11,887\right)-{\left(299\right)}^2}\sqrt{8\left(3,773\right)-{\left(167\right)}^2}}\\ =\frac{46,512-49,933}{\sqrt{5,695}\sqrt{2,295}}\\ =\frac{-3,421}{75.4652\times 47.9061}\\ =\frac{-3,421}{3,615.2434}\\ =-0.946\end{array}$

Hence, the value of r is verified as approximately –0.946.

(c)

To determine

Find the value of $\overline{x}$.

Find the value of $\overline{y}$.

Find the value of a.

Find the value of b.

Find the equation of the least-squares line.

Answer to Problem 9P

The value of $\overline{x}$ is 37.375.

The value of $\overline{y}$ is 20.875 mpg.

The value of a is 43.326.

The value of b is –0.6007.

The equation of the least-squares line is $\widehat{y}=43.326-0.6007x$.

Explanation of Solution

Calculation:

From part (b), the values are ${\displaystyle \sum x}=299$, ${\displaystyle \sum y}=167$, ${\displaystyle \sum x^2}=11,887$, ${\displaystyle \sum y^2}=3,773$, ${\displaystyle \sum xy}=5,814$ and $n=8$

The value of $\overline{x}$ is,

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{299}{8}\\ =37.375\end{array}$

Hence, the value of $\overline{x}$ is 37.375.

The value of $\overline{y}$ is,

$\begin{array}{c}\overline{y}=\frac{{\displaystyle \sum y}}{n}\\ =\frac{167}{8}\\ =20.875\end{array}$

Hence, the value of $\overline{y}$ is 20.875 mpg.

The value of b is,

$\begin{array}{c}b=\frac{n{\displaystyle \sum xy}-\left({\displaystyle \sum x}\right)\left({\displaystyle \sum y}\right)}{n{\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\\ =\frac{8\left(5,814\right)-\left(299\times 167\right)}{8\left(11,887\right)-{\left(299\right)}^2}\\ =\frac{46,512-49,933}{95,096-89,401}\\ =\frac{-3,421}{5,695}\\ =-0.6007\end{array}$

Hence, the value of b is –0.6007.

The value of a is,

$\begin{array}{c}a=\overline{y}-b\overline{x}\\ =20.875-\left(-0.6007\right)\left(37.375\right)\\ =20.875+22.451\\ =43.326\end{array}$

Hence, the value of a is 43.326.

The equation of the least-squares line is,

$\begin{array}{c}\widehat{y}=a+bx\\ =43.326-0.6007x\end{array}$

Hence, the equation of the least-squares line is $\widehat{y}=43.326-0.6007x$.

(d)

To determine

Construct a scatter diagram with least squares line.

Locate the point $\left(\overline{x},\overline{y}\right)$ on the line.

Answer to Problem 9P

The scatter diagram with least squares line with point $\left(\overline{x},\overline{y}\right)$ is,

Explanation of Solution

Calculation:

In the dataset of eight weights and miles per gallon also include the point $\left(\overline{x},\overline{y}\right)$. That is, $\left(37.375,20.875\right)$.

Step by step procedure to obtain scatter plot using MINITAB software is given below:

• Choose Graph > Scatterplot.
• Choose With regression. Click OK.
• In Y variables, enter the column of x.
• In X variables, enter the column of y.
• Click OK.

(e)

To determine

Find the value of the coefficient of determination $r^2$.

Mention percentage of the variation in y that can be explained by variation in x.

Mention percentage of the variation in y that cannot be explained by variation in x.

Answer to Problem 9P

The value of the coefficient of determination $r^2$ is 0.896.

The percentage of the variation in y that can be explained by variation in x is 89.4%.

The percentage of the variation in y that cannot be explained by variation in x is 10.6%.

Explanation of Solution

Calculation:

Coefficient of determination$\left(r^2\right)$:

The coefficient of determination $\left(r^2\right)$ determines the percent of the variation in response variable that is explained by the predictor variables. A larger value of $r^2$ indicates that the model is a good fit.

From part (b), the value of correlation coefficient r is –0.946. The value of $r^2$ is,

$\begin{array}{c}r=-0.946\\ r^2={\left(-0.946\right)}^2\\ =0.894\end{array}$

Hence, the value of the coefficient of determination $r^2$ is 0.894.

About 89.4% of the variation in y (miles per gallon (mpg)) is explained by x (weight of the car (in hundreds of pounds)). Since the value of $r^2$ is quite large the model is a good fit for the data.

Hence, the percentage of the variation in y that can be explained by variation in x is 89.4%.

About 10.6% $\left(=100\%-89.4\%\right)$ of the variation in y (miles per gallon (mpg)) cannot be explained by x (weight of the car (in hundreds of pounds)).

Hence, the percentage of the variation in y that cannot be explained by variation in x is 10.6%.

(f)

To determine

Find the number of miles per gallon for car weight $x=38$ hundred pounds.

Answer to Problem 9P

The number of miles per gallon for car weight $x=38$ hundred pounds is 20.4994 mpg.

Explanation of Solution

Calculation:

From part (c), the equation of the least-squares line is $\widehat{y}=43.326-0.6007x$.

Substitute $x=38$ in equation of the least-squares line.

$\begin{array}{c}\widehat{y}=43.326-0.6007\left(38\right)\\ =43.326-22.8266\\ =20.4994\end{array}$

Hence, the number of miles per gallon for car weight $x=38$ hundred pounds is 20.4994 mpg.