Chapter 15.4, Problem 5WE

a.

To determine

To calculate: The value of the correlation coefficient.

Answer to Problem 5WE

The value of the correlation coefficientis $0.96$ .

Explanation of Solution

Given information:

The values $M_{xy}\approx 3226.33,M_x\approx 65.17,M_y\approx 49.33,{\sigma }_x\approx 2.85\text{ and }{\sigma }_y\approx 4.19$ .

Formula used:

The correlation coefficient r is denoted as $r=\frac{M_{xy}-M_x\cdot M_y}{{\sigma }_x\cdot {\sigma }_y}$ , where $M_{xy}$ is the mean of the product of values of x and y , $M_x$ is the mean of x -values, $M_y$ is the mean of y -values, ${\sigma }_x$ is standard deviation of x -values, ${\sigma }_y$ is the standard deviation of y -values.

Calculation:

Consider the provided values $M_{xy}\approx 3226.33,M_x\approx 65.17,M_y\approx 49.33,{\sigma }_x\approx 2.85\text{ and }{\sigma }_y\approx 4.19$ .

Recall that the correlation coefficient r is denoted as $r=\frac{M_{xy}-M_x\cdot M_y}{{\sigma }_x\cdot {\sigma }_y}$ , where $M_{xy}$ is the mean of the product of values of x and y , $M_x$ is the mean of x -values, $M_y$ is the mean of y -values, ${\sigma }_x$ is standard deviation of x -values, ${\sigma }_y$ is the standard deviation of y -values.

Substitute the values and apply it,

  $\begin{array}{c}r=\frac{3226.33-\left(65.17\right)\cdot \left(49.33\right)}{\left(2.85\right)\cdot \left(4.19\right)}\\ r=0.96\end{array}$

Thus, the value of the correlation coefficient is $0.96$ .

b.

To determine

To calculate: Theequation of regression line.

Answer to Problem 5WE

The equation of regression lineis $y=1.41x-42.55$ .

Explanation of Solution

Given information:

The values $M_{xy}\approx 3226.33,M_x\approx 65.17,M_y\approx 49.33,{\sigma }_x\approx 2.85\text{ and }{\sigma }_y\approx 4.19$ .

Formula used:

The correlation coefficient r is denoted as $r=\frac{M_{xy}-M_x\cdot M_y}{{\sigma }_x\cdot {\sigma }_y}$ , where $M_{xy}$ is the mean of the product of values of x and y , $M_x$ is the mean of x -values, $M_y$ is the mean of y -values, ${\sigma }_x$ is standard deviation of x -values, ${\sigma }_y$ is the standard deviation of y -values.

The regression line that relates x and y , contains the point $\left(M_x,M_y\right)$ has slope $r\left(\frac{{\sigma }_y}{{\sigma }_x}\right)$ and the equation is $y-M_y=r\left(\frac{{\sigma }_y}{{\sigma }_x}\right)\left(x-M_x\right)$ .

Calculation:

Consider the provided values $M_{xy}\approx 3226.33,M_x\approx 65.17,M_y\approx 49.33,{\sigma }_x\approx 2.85\text{ and }{\sigma }_y\approx 4.19$ .

Recall that the correlation coefficient r is denoted as $r=\frac{M_{xy}-M_x\cdot M_y}{{\sigma }_x\cdot {\sigma }_y}$ , where $M_{xy}$ is the mean of the product of values of x and y , $M_x$ is the mean of x -values, $M_y$ is the mean of y -values, ${\sigma }_x$ is standard deviation of x -values, ${\sigma }_y$ is the standard deviation of y -values.

Substitute the values and apply it,

  $\begin{array}{c}r=\frac{3226.33-\left(65.17\right)\cdot \left(49.33\right)}{\left(2.85\right)\cdot \left(4.19\right)}\\ r=0.96\end{array}$

Next, recall that the regression line that relates x and y , contains the point $\left(M_x,M_y\right)$ has slope $r\left(\frac{{\sigma }_y}{{\sigma }_x}\right)$ and the equation is $y-M_y=r\left(\frac{{\sigma }_y}{{\sigma }_x}\right)\left(x-M_x\right)$ .

Substitute the provided values in the expression provided above,

  $\begin{array}{c}y-49.33=0.96\left(\frac{4.19}{2.85}\right)\left(x-65.17\right)\\ y-49.33=\left(1.41\right)\left(x-65.17\right)\\ y=1.41x-91.88+49.33\\ y=1.41x-42.55\end{array}$

Thus, the equation of regression line is $y=1.41x-42.55$ .