Q. 4.134 - Elementary Statistics

Question

In exercises, 4.134 - 4.143, we repeat data from exercises in section 4.2. For each section, determine the linear correlation coefficient by using,

(a) Definition 4.8 on page 183.

(b) Definition 4.5 on page 185

Compare your results in part (a) and (b).

Step-by-Step Solution

Answer

(a) The correlation coefficient is 0.5.

(b) The correlation coefficient is 0.5. 

Part (a) Step 1. Given Information.

A table of values.

Part (a) Step 2. Construct the table.

Construct the table using the given values:

Part (a) Step 3. Find the mean and standard deviation.

The mean value of x is,

$$\begin{gathered} \overline{x}=\frac{{\displaystyle \sum _{}}x}{n} \\ =\frac{2+4+3}{3} \\ =3 \\ \overline{y}=\frac{{\displaystyle \sum _{}y}}{n} \\ =\frac{3+5+7}{3} \\ =5 \end{gathered}$$

The standard deviation is given by,

$$\begin{gathered} s_x=\sqrt{\frac{1}{n-1}\sum _{}{(x_i-\overline{x})}^2} \\ =\sqrt{\frac{1}{3-1}\sum _{}{(x_i-\overline{x})}^2} \\ =\sqrt{1} \\ =1 \\ {s}_x=\sqrt{\frac{1}{n-1}\sum _{}{(y_i-\overline{y})}^2} \\ =\sqrt{\frac{1}{3-1}\sum _{}{(y_i-\overline{y})}^2} \\ =\sqrt{4} \\ =2 \end{gathered}$$

Part (a) Step 4. Find the correlation coefficient.

$$\begin{gathered} r=\frac{{\displaystyle \frac{1}{n-1}}\sum _{}(x_i-\overline{x})(y_i-\overline{y})}{s_x{s}_y} \\ =\frac{1}{2} \\ =0.5 \end{gathered}$$

Part (b) Step 1. Construct the table.

The table is given by:

Part (b) Step 2. Find the correlation coefficient.

From the table,

$$\begin{gathered} r=\frac{{\displaystyle \sum _{}}{x}_i{y}_i-{\displaystyle \frac{{\displaystyle \sum _{}}{x}_i{\displaystyle \underset{}{\sum y_i}}}{n}}}{\sqrt{[{x_i}^2-{\displaystyle \frac{{\displaystyle \underset{}{(\sum }}{x}_i{)}^2}{n}}][{y_i}^2-\frac{{\displaystyle \underset{}{(\sum }{y}_i{)}^2}}{n}]}} \\ =0.5 \end{gathered}$$