Chapter 9.1, Problem 4EA

To determine

Check whether the correlation changes or not, if double each value of x.

Answer to Problem 4EA

The correlation between y and x is the same as the correlation between x and y.

.

Explanation of Solution

From the Exercise 2, the values of x are 1, 2, 3, 4, 5

Double each value of x, and the doubled values of x are, 2, 4, 6, 8, and 10.

The average of x is obtained below:

$\begin{array}{c}\text{Average=}\frac{\text{Sum of data values}}{\text{Total number of data values}}\\ =\frac{2+4+6+8+10}{5}\\ =\frac{30}{5}\\ =6\end{array}$

The standard deviation of x is obtained as follows:

$\begin{array}{c}SD=\sqrt{\frac{\text{Sum of }\left({\left(\text{Entry}-\text{Average}\right)}^2\right)}{\text{Total number of data values }}}\\ =\sqrt{\frac{{\left(2-6\right)}^2+{\left(4-6\right)}^2+{\left(6-6\right)}^2+{\left(8-6\right)}^2+{\left(10-6\right)}^2}{5}}\\ =\sqrt{\frac{16+4+0+4+16}{5}}\\ =\sqrt{8}\\ =2.82\end{array}$

The average of y is obtained below:

$\begin{array}{c}\text{Average=}\frac{\text{Sum of data values}}{\text{Total number of data values}}\\ =\frac{2+3+1+5+6}{5}\\ =\frac{17}{5}\\ =3.4\end{array}$

The standard deviation of y is obtained as follows:

$\begin{array}{c}SD=\sqrt{\frac{\text{Sum of }\left({\left(\text{Entry}-\text{Average}\right)}^2\right)}{\text{Total number of data values}}}\\ =\sqrt{\frac{{\left(2-3.4\right)}^2+{\left(3-3.4\right)}^2+{\left(1-3.4\right)}^2+{\left(5-3.4\right)}^2+{\left(6-3.4\right)}^2}{5}}\\ =\sqrt{\frac{1.96+0.16+5.76+2.56+6.76}{5}}\\ =\sqrt{3.44}\\ =1.85\end{array}$

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)
2	2	$\frac{2-6}{2.82}=-1.42$	$\frac{2-3.4}{1.85}=-0.76$	1.08
4	3	$\frac{4-6}{2.82}=-0.71$	$\frac{3-3.4}{1.85}=-0.22$	0.15
6	1	$\frac{6-6}{2.82}=0$	$\frac{1-3.4}{1.85}=-1.30$	0
8	5	$\frac{8-6}{2.82}=0.71$	$\frac{5-3.4}{1.85}=0.86$	0.61
10	6	$\frac{10-6}{2.82}=1.42$	$\frac{6-3.4}{1.85}=1.41$	2.00

$\begin{array}{c}r=\text{Average of (x in standard units) }\times \text{ (y in standard units)}\\ \text{=}\frac{\left(1.08\right)+\left(0.15\right)+\left(0\right)+\left(0.61\right)+\left(2.00\right)}{5}\\ =\frac{3.84}{5}\\ =0.76\end{array}$

Thus, the value of r is 0.76

Thus, the correlation between y and x is the same as the correlation between x and y.