Chapter 4.1, Problem 4.16E

(a)

To determine

To construct a scatterplot of the data given in the question.

Explanation of Solution

The scatterplot of the data is as follows:

  

(b)

To determine

To show that the correlation is zero.

Explanation of Solution

The mean of $x$ is:

  $\begin{array}{l}\overline{x}=\frac{-5-3+0+3+5}{5}\\ \overline{x}=0\end{array}$

Now to calculate standard deviation of $x$ :

Consider the following table-

  

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

  ${\displaystyle \sum {(x_i-\overline{X})}^2=68}$

Calculate $\sigma$ using the above formula;

  $\begin{array}{l}\sigma =\sqrt{\frac{{\displaystyle \sum {(x_i-\overline{X})}^2}}{n-1}}\\ \sigma =\sqrt{\frac{68}{5-1}}\\ \sigma \approx 4.123\end{array}$

The mean of $y$ :

  $\begin{array}{l}\overline{y}=\frac{0+4+5+4+0}{5}\\ \overline{y}=2.6\end{array}$

The standard deviation of $y$ :

Consider the table-

  

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

  ${\displaystyle \sum {(y_i-\overline{Y})}^2=23.2}$

Calculate $\sigma$ using the above formula;

  $\begin{array}{l}\sigma =\sqrt{\frac{{\displaystyle \sum {(y_i-\overline{Y})}^2}}{n-1}}\\ \sigma =\sqrt{\frac{23.2}{5-1}}\\ \sigma \approx 2.4083\end{array}$

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,

  $\begin{array}{l}{\displaystyle \sum x}=0\\ {\displaystyle \sum y}=13\\ x.y=-5\times 0+(-3\times 4)+0\times 5+3\times 4+5\times 0\\ {\displaystyle \sum xy}=0\\ x^2={(}^{-}+{(}^{-}+0^2+3^2+5^2\\ {\displaystyle \sum x^2}=68\\ y^2=0^2+4^2+5^2+4^2+0^2\\ {\displaystyle \sum y^2}=57\end{array}$

Now computing in the correlation formula we get,

  $\begin{array}{l}r=\frac{1}{n-1}{\displaystyle \sum [(\frac{x-\overline{x}}{s_x})(\frac{y-\overline{y}}{s_y})]}\\ r=\frac{n.{\displaystyle \sum xy-{\displaystyle \sum x}{\displaystyle \sum y}}}{\sqrt{[n{\displaystyle \sum x}-{({\displaystyle \sum x})}^2].[n{\displaystyle \sum y}-{({\displaystyle \sum y})}^2]}}\\ r=\frac{5\times 0-0\times 13}{\sqrt{[5\times 68-0^2].[5\times 57-{13}^2]}}\\ r=0\end{array}$

Thus, the correlation is zero.

(c)

To determine

To explain that how correlation is zero as the scatterplot shows strong association between $x\text{ and }y$ .

Explanation of Solution

As the scatterplot shows strong association between $x\text{ and }y$ but in this case the correlation is zero because the strong association is quadratic and NOT linear.