Question

For each exercise, determine the linear correlation coefficient by using

a. Definition 1.8 on page 183.

b. Formada 4.4 a pace 185.

Compare your answers in parts a) and (b).

Short answer

a) The linear coefficient by the definition =0.5

b) The linear coefficient by the formula =0.5

Step-by-step solution

Given Information (Part a)

To find the linear correlation coefficient by the definition.

Explanation(Part a)

The linear correlation coefficient using the definition is

$r=\frac{\frac{1}{n-1}\sum \left(x_i-\overline{x}\right)\left(y_i-\overline{y}\right)}{{\sigma }_x{\sigma }_y}$

The standard deviations are

${\sigma }_x=\sqrt{\frac{1}{n-1}\sum (x-\overline{x}{)}^2},{\sigma }_y=\sqrt{\frac{1}{n-1}\sum (y-\overline{y}{)}^2}$

Calculation:

Make a table of values

Find the standard deviations

${\sigma }_x=\sqrt{\frac{1}{n-1}\sum (x-\overline{x}{)}^2}=\sqrt{\frac{1}{3-1}\left(2\right)}=1$

${\sigma }_y=\sqrt{\frac{1}{n-1}\sum (y-\overline{y}{)}^2}=\sqrt{\frac{1}{3-1}\left(8\right)}=2$

Find the correlation coefficient

$r=\frac{\frac{1}{n-1}\sum \left(x_i-\overline{x}\right)\left(y_i-\overline{y}\right)}{{\sigma }_x{\sigma }_y}$

$=\frac{\frac{1}{3-1}\left(2\right)}{\left(1\right)\left(2\right)}$

$=0.5$

Hence,

The linear coefficient by the definition =0.5