Chapter 9.6, Problem 6RE

(i)

To determine

Check whether the program is working correctly or not for data set (i).

Answer to Problem 6RE

Yes, the program is working as correctly.

Explanation of Solution

The average of x is obtained below:

$\begin{array}{c}\text{Average=}\frac{\text{Sum of data values}}{\text{Total number of data values}}\\ =\frac{1+2+3+4+5+6+7}{7}\\ =\frac{28}{7}\\ =4\end{array}$

The standard deviation of x is obtained as follows:

$\begin{array}{c}SD=\sqrt{\frac{\text{Sum of }\left({\left(\text{Entry}-\text{Average}\right)}^2\right)}{\text{Total number of data values}}}\\ =\sqrt{\frac{{\left(1-4\right)}^2+{\left(2-4\right)}^2+{\left(3-4\right)}^2+{\left(4-4\right)}^2+{\left(5-4\right)}^2+{\left(6-4\right)}^2+{\left(7-4\right)}^2}{7}}\\ =\sqrt{\frac{9+4+1+0+1+4+9}{7}}\\ =\sqrt{4}\\ =2\end{array}$

The average of y is obtained below:

$\begin{array}{c}\text{Average=}\frac{\text{Sum of data values}}{\text{Total number of data values}}\\ =\frac{2+1+4+3+7+5+6}{7}\\ =\frac{28}{7}\\ =4\end{array}$

The standard deviation of y is obtained as follows:

$\begin{array}{c}SD=\sqrt{\frac{\text{Sum of }\left({\left(\text{Entry}-\text{Average}\right)}^2\right)}{\text{Total number of data values}}}\\ =\sqrt{\frac{{\left(2-4\right)}^2+{\left(1-4\right)}^2+{\left(4-4\right)}^2+{\left(3-4\right)}^2+{\left(7-4\right)}^2+{\left(5-4\right)}^2+{\left(6-4\right)}^2}{7}}\\ =\sqrt{\frac{4+9+0+1+9+1+4}{7}}\\ =\sqrt{4}\\ =2\end{array}$

The value of r is obtained below:

x	y	x in standard units	y in standard units	(x in standard units) × (y in standard units)
1	2	$\frac{1-4}{2}=-1.5$	$\frac{2-4}{2}=-1.0$	1.5
2	1	$\frac{2-4}{2}=-1.0$	$\frac{1-4}{2}=-1.5$	1.5
3	4	$\frac{3-4}{2}=-0.5$	$\frac{4-4}{2}=0$	0
4	3	$\frac{4-4}{2}=0.0$	$\frac{3-4}{2}=-0.5$	0
5	7	$\frac{5-4}{2}=0.5$	$\frac{7-4}{2}=1.5$	0.75
6	5	$\frac{6-4}{2}=1.0$	$\frac{5-4}{2}=0.5$	0.5
7	6	$\frac{7-4}{2}=1.5$	$\frac{6-4}{2}=1.0$	1.5

$\begin{array}{c}r=\text{Average of (x in standard units) }\times \text{ (y in standard units)}\\ \text{=}\frac{\left(1.5\right)+\left(1.5\right)+\left(0\right)+\left(0\right)+\left(0.75\right)+\left(0.50\right)+\left(1.5\right)}{7}\\ =\frac{5.75}{7}\\ =0.8214\end{array}$

Thus, the value of r is 0.8214

The given correlation value (0.8214) and the estimated correlation value (0.8214) both are same. Therefore, yes the program is working as correctly.

(ii)

To determine

Explain whether the program is working as correctly or not for data set (ii).

Answer to Problem 6RE

No, the program is not working correctly.

Explanation of Solution

The average of x is obtained below:

$\begin{array}{c}\text{Average=}\frac{\text{Sum of data values}}{\text{Total number of data values}}\\ =\frac{1+2+3+4+5+6+7}{7}\\ =\frac{28}{7}\\ =4\end{array}$

The standard deviation of x is obtained as follows:

$\begin{array}{c}SD=\sqrt{\frac{\text{Sum of }\left({\left(\text{Entry}-\text{Average}\right)}^2\right)}{\text{Total number of data values}}}\\ =\sqrt{\frac{{\left(1-4\right)}^2+{\left(2-4\right)}^2+{\left(3-4\right)}^2+{\left(4-4\right)}^2+{\left(5-4\right)}^2+{\left(6-4\right)}^2+{\left(7-4\right)}^2}{7}}\\ =\sqrt{\frac{9+4+1+0+1+4+9}{7}}\\ =\sqrt{4}\\ =2\end{array}$

The average of y is obtained below:

$\begin{array}{c}\text{Average=}\frac{\text{Sum of data values}}{\text{Total number of data values}}\\ =\frac{5+4+7+6+10+8+9}{7}\\ =\frac{49}{7}\\ =7\end{array}$

The standard deviation of y is obtained as follows:

$\begin{array}{c}SD=\sqrt{\frac{\text{Sum of }\left({\left(\text{Entry}-\text{Average}\right)}^2\right)}{\text{Total number of data values}}}\\ =\sqrt{\frac{{\left(5-7\right)}^2+{\left(4-7\right)}^2+{\left(7-7\right)}^2+{\left(6-7\right)}^2+{\left(10-7\right)}^2+{\left(8-7\right)}^2+{\left(9-7\right)}^2}{7}}\\ =\sqrt{\frac{4+9+0+1+9+1+4}{7}}\\ =\sqrt{4}\\ =2\end{array}$

The value of r is obtained below:

x	y	x in standard units	y in standard units	(x in standard units) × (y in standard units)
1	5	$\frac{1-4}{2}=-1.5$	$\frac{5-7}{2}=-1.0$	1.5
2	4	$\frac{2-4}{2}=-1.0$	$\frac{4-7}{2}=-1.5$	1.5
3	7	$\frac{3-4}{2}=-0.5$	$\frac{7-7}{2}=0$	0
4	6	$\frac{4-4}{2}=0.0$	$\frac{6-7}{2}=-0.5$	0
5	10	$\frac{5-4}{2}=0.5$	$\frac{10-7}{2}=1.5$	0.75
6	8	$\frac{6-4}{2}=1.0$	$\frac{8-7}{2}=0.5$	0.50
7	9	$\frac{7-4}{2}=1.5$	$\frac{9-7}{2}=1.0$	1.5

$\begin{array}{c}r=\text{Average of (x in standard units) }\times \text{ (y in standard units)}\\ \text{=}\frac{\left(1.5\right)+\left(1.5\right)+\left(0\right)+\left(0\right)+\left(0.75\right)+\left(0.50\right)+\left(1.50\right)}{7}\\ =\frac{5.75}{7}\\ =0.8214\end{array}$

Thus, the value of r is 0.8214.

The given correlation value (0.7619) and the estimated correlation value (0.8614) both are different. Therefore, no the program is not working correctly.