compute r, the correlation coefficient, using the following data.\n| x | 2 | 8 | 1 | 5 | 6 | 4 |\n| y | 10 |…

Explanation:

Step1: Calculate the means

Let $x = [2,8,1,5,6,4]$ and $y=[10,2,9,7,6,6]$. The mean of $x$, $\bar{x}=\frac{2 + 8+1+5+6+4}{6}=\frac{26}{6}\approx4.33$. The mean of $y$, $\bar{y}=\frac{10 + 2+9+7+6+6}{6}=\frac{40}{6}\approx6.67$.

Step2: Calculate the numerator and denominator components

Calculate $(x_i-\bar{x})(y_i - \bar{y})$, $(x_i-\bar{x})^2$ and $(y_i-\bar{y})^2$ for each $i$. For $i = 1$: $(2 - 4.33)(10 - 6.67)=(-2.33)\times3.33=-7.76$. $(2 - 4.33)^2=(-2.33)^2 = 5.43$. $(10 - 6.67)^2=3.33^2 = 11.09$. For $i = 2$: $(8 - 4.33)(2 - 6.67)=3.67\times(-4.67)=-17.14$. $(8 - 4.33)^2=3.67^2 = 13.47$. $(2 - 6.67)^2=(-4.67)^2 = 21.81$. For $i = 3$: $(1 - 4.33)(9 - 6.67)=(-3.33)\times2.33=-7.76$. $(1 - 4.33)^2=(-3.33)^2 = 11.09$. $(9 - 6.67)^2=2.33^2 = 5.43$. For $i = 4$: $(5 - 4.33)(7 - 6.67)=0.67\times0.33 = 0.22$. $(5 - 4.33)^2=0.67^2 = 0.45$. $(7 - 6.67)^2=0.33^2 = 0.11$. For $i = 5$: $(6 - 4.33)(6 - 6.67)=1.67\times(-0.67)=-1.12$. $(6 - 4.33)^2=1.67^2 = 2.79$. $(6 - 6.67)^2=(-0.67)^2 = 0.45$. For $i = 6$: $(4 - 4.33)(6 - 6.67)=(-0.33)\times(-0.67)=0.22$. $(4 - 4.33)^2=(-0.33)^2 = 0.11$. $(6 - 6.67)^2=(-0.67)^2 = 0.45$. Sum of $(x_i-\bar{x})(y_i - \bar{y})$: $S_{xy}=-7.76-17.14 - 7.76+0.22-1.12 + 0.22=-32.34$. Sum of $(x_i-\bar{x})^2$: $S_{xx}=5.43+13.47+11.09+0.45+2.79+0.11=33.34$. Sum of $(y_i-\bar{y})^2$: $S_{yy}=11.09+21.81+5.43+0.11+0.45+0.45=39.34$.

Step3: Calculate the correlation coefficient

The formula for the correlation coefficient $r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}$. $r=\frac{-32.34}{\sqrt{33.34\times39.34}}=\frac{-32.34}{\sqrt{1312.59}}=\frac{-32.34}{36.23}\approx - 0.89$.

Answer:

$-0.89$