what is the correlation coefficient, r, for this data?\n|x|y|\n|2|68|\n|3|78|\n|5|74|\n|7|43|\n|10|47|\nround…

Explanation:

Step1: Calculate the means

Let $n = 5$. $\bar{x}=\frac{2 + 3+5+7+10}{5}=\frac{27}{5}=5.4$ $\bar{y}=\frac{68 + 78+74+43+47}{5}=\frac{310}{5}=62$

Step2: Calculate the numerator and denominator components

For the numerator: [ \begin{align*} &\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})\ =&(2 - 5.4)(68 - 62)+(3 - 5.4)(78 - 62)+(5 - 5.4)(74 - 62)+(7 - 5.4)(43 - 62)+(10 - 5.4)(47 - 62)\ =&(- 3.4)\times6+(-2.4)\times16+(-0.4)\times12+(1.6)\times(-19)+(4.6)\times(-15)\ =&-20.4-38.4 - 4.8-30.4 - 69\ =&-163 \end{align*} ]

For the denominator: [ \begin{align*} &\sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i - \bar{y})^2}\ \sum_{i = 1}^{n}(x_i-\bar{x})^2=&(2 - 5.4)^2+(3 - 5.4)^2+(5 - 5.4)^2+(7 - 5.4)^2+(10 - 5.4)^2\ =&(-3.4)^2+(-2.4)^2+(-0.4)^2+(1.6)^2+(4.6)^2\ =&11.56 + 5.76+0.16+2.56+21.16\ =&41.2\ \sum_{i = 1}^{n}(y_i - \bar{y})^2=&(68 - 62)^2+(78 - 62)^2+(74 - 62)^2+(43 - 62)^2+(47 - 62)^2\ =&6^2+16^2+12^2+(-19)^2+(-15)^2\ =&36+256+144 + 361+225\ =&1022\ &\sqrt{41.2\times1022}=\sqrt{42106.4}\approx205.2 \end{align*} ]

Step3: Calculate the correlation coefficient

$r=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i - \bar{y})^2}}=\frac{-163}{205.2}\approx - 0.8$

Answer:

$-0.8$