Question

advanced mathematics decision making sem 2
3.4.3 quiz: assessing data models
what is the value of $r^{2}$ for the following data to three decimal places?

Explanation:

Step1: Calculate the means of x and y

Let $x = [1,2,5,7,10]$ and $y=[2,4,7,10,20]$.
The mean of $x$, $\bar{x}=\frac{1 + 2+5+7+10}{5}=\frac{25}{5} = 5$
The mean of $y$, $\bar{y}=\frac{2 + 4+7+10+20}{5}=\frac{43}{5}=8.6$

Step2: Calculate the numerator of the correlation - coefficient formula

$S_{xy}=\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})$
$(1 - 5)(2 - 8.6)+(2 - 5)(4 - 8.6)+(5 - 5)(7 - 8.6)+(7 - 5)(10 - 8.6)+(10 - 5)(20 - 8.6)$
$=(-4)\times(-6.6)+(-3)\times(-4.6)+0\times(-1.6)+2\times1.4 + 5\times11.4$
$=26.4+13.8+0 + 2.8+57$
$=100$

Step3: Calculate the denominator of the correlation - coefficient formula

$S_{xx}=\sum_{i = 1}^{n}(x_{i}-\bar{x})^2=(1 - 5)^2+(2 - 5)^2+(5 - 5)^2+(7 - 5)^2+(10 - 5)^2$
$=(-4)^2+(-3)^2+0^2+2^2+5^2$
$=16 + 9+0+4+25$
$=54$
$S_{yy}=\sum_{i = 1}^{n}(y_{i}-\bar{y})^2=(2 - 8.6)^2+(4 - 8.6)^2+(7 - 8.6)^2+(10 - 8.6)^2+(20 - 8.6)^2$
$=(-6.6)^2+(-4.6)^2+(-1.6)^2+1.4^2+11.4^2$
$=43.56+21.16+2.56+1.96+129.96$
$=199.2$
$S_{xx}S_{yy}=54\times199.2 = 10756.8$
$\sqrt{S_{xx}S_{yy}}=\sqrt{10756.8}\approx103.715$

Step4: Calculate the correlation coefficient $r$

$r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{100}{103.715}\approx0.964$

Step5: Calculate $r^{2}$

$r^{2}=(0.964)^2\approx0.929$

Answer:

$0.929$