Chapter 4.1, Problem 4.13E

(a)

To determine

To construct the scatterplot of the data given in the question and by examining the scatterplot tell about the correlation.

Explanation of Solution

The scatterplot of the data is as follows:

  

By looking at the scatterplot we can say that, since the scatterplot slopes upwards we can expect a positive relation and since the points are far apart we expect it to be to $-1$ .

(b)

To determine

To calculate the correlation r of the height of the men and women.

Answer to Problem 4.13E

The correlation is $r=0.5653$ .

Explanation of Solution

The mean of $x$ is:

  $\begin{array}{l}\overline{x}=\frac{66+64+66+65+70+65}{6}\\ \overline{x}=66\end{array}$

Now to calculate standard deviation of $x$ :

Consider the following table-

  

Then the sum of the numbers in the last column to get;

  ${\displaystyle \sum {(x_i-\overline{X})}^2=22}$

Calculate $\sigma$ using the above formula;

  $\begin{array}{l}\sigma =\sqrt{\frac{{\displaystyle \sum {(x_i-\overline{X})}^2}}{n-1}}\\ \sigma =\sqrt{\frac{22}{6-1}}\\ \sigma \approx 2.0976\end{array}$

The mean of $y$ :

  $\begin{array}{l}\overline{y}=\frac{72+68+70+68+71+65}{6}\\ \overline{y}=69\end{array}$

The standard deviation of $y$ :

Consider the table-

  

Now we calculate the sum of numbers in the last column to get;

  ${\displaystyle \sum {(y_i-\overline{Y})}^2=32}$

Calculate $\sigma$ using the above formula;

  $\begin{array}{l}\sigma =\sqrt{\frac{{\displaystyle \sum {(y_i-\overline{Y})}^2}}{n-1}}\\ \sigma =\sqrt{\frac{32}{6-1}}\\ \sigma \approx 2.5298\end{array}$

The correlation is determined with the formula:-

  $r=\frac{1}{n-1}{\displaystyle \sum [(\frac{x-\overline{x}}{s_x})(\frac{y-\overline{y}}{s_y})]}$

Now, we will find the value of $x.y,x^2,y^2$ as,

  $\begin{array}{l}{\displaystyle \sum x}=396\\ {\displaystyle \sum y}=414\\ x.y=66\times 72+64\times 68+66\times 70+65\times 68+70\times 71+65\times 65\\ {\displaystyle \sum xy}=27339\\ x^2={66}^2+{64}^2+{66}^2+{65}^2+{70}^2+{65}^2\\ {\displaystyle \sum x^2}=26158\\ y^2={72}^2+{68}^2+{70}^2+{68}^2+{71}^2+{65}^2\\ {\displaystyle \sum y^2}=28598\end{array}$

Now computing in the correlation formula we get,

  $\begin{array}{l}r=\frac{1}{n-1}{\displaystyle \sum [(\frac{x-\overline{x}}{s_x})(\frac{y-\overline{y}}{s_y})]}\\ r=\frac{n.{\displaystyle \sum xy-{\displaystyle \sum x}{\displaystyle \sum y}}}{\sqrt{[n{\displaystyle \sum x}-{({\displaystyle \sum x})}^2].[n{\displaystyle \sum y}-{({\displaystyle \sum y})}^2]}}\\ r=\frac{6\times 27339-396\times 414}{\sqrt{[6\times 26158-{396}^2].[6\times 28598-{414}^2]}}\\ r=0.5653\end{array}$