Given the following four points, determine whether the correlation coefficient will be high or low for the least squares line with the best goodness of fit, and plot the best fit point for x = 5.

(2, 2)

(3, 5)

(4, 4)

(5, 1)

The correlation coefficient is a measure of the strength between two different variables and can exist between -1 and 1. -1 indicates a strong negative relationship, 0 indicates no association, and 1 indicates a strong positive relationship. 

All of our needed equations to calculate both the line of least squares as well as the correlation coefficient exist across pages 69 and 70 of our FE Reference Handbook:

Based on these equations, we first want to start by finding the sum of squares of x, the sum of squares of y, and the sum of x-y products:

$$\begin{gathered} S_{xx}=\sum _{i=1}^n{x_i}^2-\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n$}\right.\right){\left(\sum _{i=1}^n{x}_i\right)}^2 \\ {S}_{xx}=(2^2+3^2+4^2+5^2)-\left(\frac{1}{4}{(2+3+4+5)}^2\right) S_{xx}=(4+9+16+25)-\left(\frac{1}{4}\right(196\left)\right) \\ {S}_{xx}=\left(54\right)-\left(49\right)=5 \\ {S}_{yy}=\sum _{i=1}^n{y_i}^2-\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n$}\right.\right){\left(\sum _{i=1}^n{y}_i\right)}^2 \\ {S}_{yy}=(2^2+5^2+4^2+1^2)-\left(\frac{1}{4}{(2+5+4+1)}^2\right) \\ {S}_{yy}=\left(46\right)-\left(\frac{1}{4}\right(144\left)\right) \\ {S}_{yy}=\left(46\right)-\left(44\right)=2 \\ {S}_{xy}=\sum _{i=1}^n{x}_i{y}_i-\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n$}\right.\right)\left(\sum _{i=1}^n{x}_i\right)\left(\sum _{i=1}^n{y}_i\right) \\ {S}_{xy}=\left(\right(2\left)\right(2)+(3\left)\right(5)+(4\left)\right(4)+(5\left)\right(1\left)\right)-\left(\frac{1}{4}\right(2+3+4+5\left)\right(2+5+4+1\left)\right) \\ {S}_{xy}=(4+15+16+5)-\left(\frac{1}{4}\right(14\left)\right(12\left)\right) \\ {S}_{xy}=(4+15+16+5)-\left(\frac{1}{4}\right(14\left)\right(12\left)\right) = 40 - 42 = -2 \end{gathered}$$

Now that we have these three numbers, let's find our correlation coefficient:

$R=\frac{S_{xy}}{\sqrt{S_{xx}{S}_{yy}}}=\frac{-2}{\sqrt{\left(5\right)\left(2\right)}}=\frac{-2}{\sqrt{10}}\approx -.632$

Since -.632 > -.5, this coefficient represents a strong (high) negative correlation.

Let's now find our least squares line:

$$\begin{gathered} \overbrace{y}=\overbrace{a}+\overbrace{b}x \\ \overbrace{b}=\raisebox{1ex}{$S_{xy}$}\!\left/ \!\raisebox{-1ex}{$S_{xx}={\displaystyle \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$}\right. \\ \overbrace{a}=\overline{y}-\overbrace{b}\overline{x} \\ \overbrace{a}=(\frac{2+5+4+1}{4}-(\frac{-2}{5}\left(\frac{2+3+4+5}{4}\right)\left)\right) \\ \overbrace{a}=(3-(\frac{-28}{20}\left)\right) \\ \overbrace{a}=\left(\frac{88}{20}\right) =\frac{22}{5} \\ \overbrace{y}=\frac{22}{5}-\frac{2x}{5} \end{gathered}$$

Solving for $\overbrace{y}$ at x=5:

$$\begin{gathered} \overbrace{y}=\frac{22}{5}-\frac{2x}{5} \\ \overbrace{y}=\frac{22}{5}-\frac{2\left(5\right)}{5} \\ \overbrace{y}=\frac{22}{5}-\frac{10}{5}=\frac{12}{5} \end{gathered}$$

Therefore, our point is at (5, 12/5). Everything is plotted visually below: