Chapter 7.1, Problem 14E

A scatterplot contains four points: (–2, –2), (–1, –1), (0, 0), and (1, 1). A fifth point, (2, y), is to be added to the plot. Let r represent the correlation between x and y.

1. a. Find the value of y so that r = 1.
2. b. Find the value of y so that r = 0.
3. c. Find the value of y so that r = 0.5.
4. d. Find the value of y so that r = –0.5.
5. e. Give a geometric argument to show that there no value of y for which r = –1.

$\bar{x}=0$ and ${\displaystyle {\sum }_{i=1}^n{\left(x_i-\bar{x}\right)}^2}=10$.

$\bar{y}=\left[-2+\left(-1\right)+0+1+y\right]/5=\left(y-2\right)/5$, so $y=5\bar{y}+2$.

Express ${\displaystyle {\sum }_{i=1}^n\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)}$, ${\displaystyle {\sum }_{i=1}^n{\left(y_i-\bar{y}\right)}^2}$, and r in terms of $\bar{y}$ :

$\begin{array}{l}{\displaystyle {\sum }_{i=1}^n\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)}=-2\left(-2-\bar{y}\right)+\left(-1\right)\left(-1-\bar{y}\right)+0\left(0-\bar{y}\right)+1\left(1-\bar{y}\right)+2\left(5\bar{y}+2-\bar{y}\right)\\ =10\bar{y}+10\end{array}$.

${\displaystyle {\sum }_{i=1}^n{\left(y_i-\bar{y}\right)}^2}={\left(-2-\bar{y}\right)}^2+{\left(-1-\bar{y}\right)}^2+{\left(0-\bar{y}\right)}^2+{\left(1-\bar{y}\right)}^2+{\left(5\bar{y}+2-\bar{y}\right)}^2=20{\bar{y}}^2+20\bar{y}+10$.

Now $r=\frac{10\bar{y}+10}{\sqrt{10}\sqrt{20{\bar{y}}^2+20\bar{y}+10}}=\frac{\bar{y}+1}{\sqrt{{\bar{y}}^2+2\bar{y}+1}}$, so $\left(1-2r^2\right){\bar{y}}^2+\left(2-2r^2\right)\bar{y}+\left(1-r^2\right)=0$.

For any given value of r, substitute r into $\left(1-2r^2\right){\bar{y}}^2+\left(2-2r^2\right)\bar{y}+\left(1-r^2\right)=0$ and solve for $\bar{y}$, then compute $y=5\bar{y}+2$.

To determine

Find the value of y so that $r=1$.

Answer to Problem 14E

The value of y is 2.

Explanation of Solution

Calculation:

The given information is that the scatterplot contain the 4 points, $\left(-2,-2\right)$, $\left(-1,-1\right)$, $\left(0,0\right)$ and $\left(1,1\right)$.

The 5^th point that is added to the plot is $\left(2,y\right)$.

The mean for x is calculated as follows:

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x_i}}{n}\\ =\frac{\left(-2\right)+\left(-1\right)+0+1+2}{5}\\ =\frac{0}{5}\\ =0\end{array}$

The value of ${\displaystyle \sum _{i=1}^n\left(x_i-\overline{x}\right)}$ is calculated as follows:

$\begin{array}{c}{\displaystyle \sum _{i=1}^n\left(x_i-\overline{x}\right)}={\left(-2-0\right)}^2+{\left(-1-0\right)}^2+{\left(0-0\right)}^2+{\left(1-0\right)}^2+{\left(2-0\right)}^2\\ =4+1+0+1+4\\ =10\end{array}$

The mean for y is calculated as follows:

$\begin{array}{c}\overline{y}=\frac{{\displaystyle \sum y_i}}{n}\\ =\frac{\left(-2\right)+\left(-1\right)+0+1+y}{5}\\ =\frac{y-2}{5}\end{array}$

The value of y is,

$\begin{array}{c}\overline{y}=\frac{y-2}{5}\\ 5\overline{y}=y-2\\ y=5\overline{y}+2\end{array}$

Express ${\displaystyle \sum _{i=1}^n\left(x_i-\overline{x}\right)}\left(y_i-\overline{y}\right)$ in terms of $\overline{y}$.

$\begin{array}{c}{\displaystyle \sum _{i=1}^n\left(x_i-\overline{x}\right)}\left(y_i-\overline{y}\right)={\displaystyle \sum _{i=1}^n{x}_i}\left(y_i-\overline{y}\right)-{\displaystyle \sum _{i=1}^n\overline{x}}\left(y_i-\overline{y}\right)\\ ={\displaystyle \sum _{i=1}^n{x}_i}\left(y_i-\overline{y}\right)-{\displaystyle \sum _{i=1}^{n}0}\left(y_i-\overline{y}\right)\\ ={\displaystyle \sum _{i=1}^n{x}_i}\left(y_i-\overline{y}\right)-0\\ =\left[\begin{array}{l}\left(-2\right)\left(-2-\overline{y}\right)+\left(-1\right)\left(-1-\overline{y}\right)+\left(0\right)\left(0-\overline{y}\right)+\\ \left(1\right)\left(1-\overline{y}\right)+\left(2\right)\left(5\overline{y}+2-\overline{y}\right)\end{array}\right]\end{array}$

$\begin{array}{c}=4+\cancel{2\overline{y}}+1+\cancel{\overline{y}}+0+1\cancel{-\overline{y}}+10\overline{y}+4\cancel{-2\overline{y}}\\ =10+10\overline{y}\end{array}$

Express ${\displaystyle \sum _{i=1}^n{\left(y_i-\overline{y}\right)}^2}$ in terms of $\overline{y}$.

$\begin{array}{c}{\displaystyle \sum _{i=1}^n{\left(y_i-\overline{y}\right)}^2}={\left(-2-\overline{y}\right)}^2+{\left(-1-\overline{y}\right)}^2+{\left(0-\overline{y}\right)}^2+{\left(1-\overline{y}\right)}^2+{\left(5\overline{y}+2-\overline{y}\right)}^2\\ =\left(4+4\overline{y}+{\overline{y}}^2\right)+\left(1+\cancel{2\overline{y}}+{\overline{y}}^2\right)+\left({\overline{y}}^2\right)+1\cancel{-2\overline{y}}+{\overline{y}}^2+4+16\overline{y}+16{\overline{y}}^2\\ =20{\overline{y}}^2+20\overline{y}+10\end{array}$

Express r in terms of $\overline{y}$.

$\begin{array}{c}r=\frac{{\displaystyle \sum _{i=1}^n\left(x_i-\overline{x}\right)\left(y_i-\overline{y}\right)}}{\sqrt{{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}\sqrt{{\displaystyle \sum _{i=1}^n{\left(y_i-\overline{y}\right)}^2}}}\\ =\frac{10+10\overline{y}}{\sqrt{10}\sqrt{20{\overline{y}}^2+20\overline{y}+10}}\\ =\frac{\overline{y}+1}{\sqrt{2{\overline{y}}^2+2\overline{y}+1}}\end{array}$

Squaring on both sides,

$\begin{array}{c}{r}^2=\frac{{\left(\overline{y}+1\right)}^2}{2{\overline{y}}^2+2\overline{y}+1}\\ r^2\left(2{\overline{y}}^2+2\overline{y}+1\right)={\left(\overline{y}+1\right)}^2\\ 2r^2{\overline{y}}^2+2r^2\overline{y}+r^2={\overline{y}}^2+2\overline{y}+1\\ 0={\overline{y}}^2+2\overline{y}+1-2r^2{\overline{y}}^2-2r^2\overline{y}-r^2\\ 0={\overline{y}}^2\left(1-2r^2\right)+\overline{y}\left(2-2r^2\right)+\left(1-r^2\right)\end{array}$

Substitute $r=1$,

$\begin{array}{c}{\overline{y}}^2\left(1-2r^2\right)+\overline{y}\left(2-2r^2\right)+\left(1-r^2\right)=0\\ {\overline{y}}^2\left(1-2{\left(1\right)}^2\right)+\overline{y}\left(2-2{\left(1\right)}^2\right)+\left(1-{\left(1\right)}^2\right)=0\\ {\overline{y}}^2\left(1-2\right)+\overline{y}\left(2-2\right)+\left(1-1\right)=0\\ {\overline{y}}^2\left(-1\right)+\overline{y}\left(0\right)+\left(0\right)=0\end{array}$

                                                   $\begin{array}{c}-{\overline{y}}^2=0\\ \overline{y}=0\end{array}$

The value of y is,

$\begin{array}{c}y=5\overline{y}+2\\ =5\left(0\right)+2\\ =2\end{array}$

Thus, the value of y is 2.

To determine

Find the value of y so that $r=0$.

Answer to Problem 14E

The value of y is –3.

Explanation of Solution

Calculation:

The value of y is calculated as follows:

Substitute $r=1$ into ${\overline{y}}^2\left(1-2r^2\right)+\overline{y}\left(2-2r^2\right)+\left(1-r^2\right)=0$

$\begin{array}{c}{\overline{y}}^2\left(1-2r^2\right)+\overline{y}\left(2-2r^2\right)+\left(1-r^2\right)=0\\ {\overline{y}}^2\left(1-2{\left(0\right)}^2\right)+\overline{y}\left(2-2{\left(0\right)}^2\right)+\left(1-{\left(0\right)}^2\right)=0\\ {\overline{y}}^2+2\overline{y}+1=0\\ \overline{y}=-1\end{array}$

The value of y is,

$\begin{array}{c}y=5\overline{y}+2\\ =5\left(-1\right)+2\\ =-3\end{array}$

Thus, the value of y is –3.

To determine

Find the value of y so that $r=0.5$.

Answer to Problem 14E

If $r=0.5$, the value of y is –1.169875.

Explanation of Solution

Calculation:

The value of y is calculated as follows:

Substitute $r=0.5$ into ${\overline{y}}^2\left(1-2r^2\right)+\overline{y}\left(2-2r^2\right)+\left(1-r^2\right)=0$

$\begin{array}{c}{\overline{y}}^2\left(1-2r^2\right)+\overline{y}\left(2-2r^2\right)+\left(1-r^2\right)=0\\ {\overline{y}}^2\left(1-2{\left(\frac{1}{2}\right)}^2\right)+\overline{y}\left(2-2{\left(\frac{1}{2}\right)}^2\right)+\left(1-{\left(\frac{1}{2}\right)}^2\right)=0\\ {\overline{y}}^2\left(1-\frac{1}{2}\right)+\overline{y}\left(2-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)=0\\ {\overline{y}}^2\left(\frac{1}{2}\right)+\overline{y}\left(\frac{3}{2}\right)+\left(\frac{3}{4}\right)=0\end{array}$

On simplification,

$\begin{array}{c}\overline{y}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(\frac{3}{2}\right)\pm \sqrt{{\left(\frac{3}{2}\right)}^2-4\left(\frac{1}{2}\right)\left(\frac{3}{4}\right)}}{2\left(\frac{1}{2}\right)}\\ =-\left(\frac{3}{2}\right)\pm \sqrt{\frac{9}{4}-\frac{12}{8}}\\ =-1.5\pm 0.866025\end{array}$

$=\left(-\text{2}\text{.366025},-\text{0}\text{.633975}\right)$

If $r=0.5$, the value of y is

$\begin{array}{c}y=5\overline{y}+2\\ =5\left(-0.633975\right)+2\\ =-\text{1}\text{.169875}\end{array}$

If $r=-0.5$, the value of y is

$\begin{array}{c}y=5\overline{y}+2\\ =5\left(-2.366025\right)+2\\ =-\text{9}\text{.830125}\end{array}$

Thus, the value of y is –3.

To determine

Find the value of y so that $r=-0.5$.

Answer to Problem 14E

If $r=-0.5$, the value of y is –9.830125.

Explanation of Solution

Calculation:

From part c., if $r=-0.5$, the value of y is

$\begin{array}{c}y=5\overline{y}+2\\ =5\left(-2.366025\right)+2\\ =-\text{9}\text{.830125}\end{array}$

Thus, the value of y is –9.830125.

To determine

Show that there is no value y for which $r=-1$

Explanation of Solution

Conclusion:

If the correlation coefficient is equal to –1, then the point lies on a straight line with negative slope. In this case, there is no value for y.