Question

find the correlation coefficient, r, of the data described below. grape growers in france think they will have a massive crop this year and want to know how this will influence demand for grape juice. they compared how many grapes were produced (in millions of metric tons), x, and the amount of french grape juice that was sold in previous years (in billions of liters), y. grape production grape juice sold 3.57 5.32 4.03 5.76 5.03 5.73 5.51 6.00 5.93 5.88 round your answer to the nearest thousandth.

Explanation:

Step1: Calculate the means of x and y

Let $x = [3.57,4.03,5.03,5.51,5.93]$ and $y=[5.32,5.76,5.73,6.00,5.88]$.
The mean of $x$, $\bar{x}=\frac{3.57 + 4.03+5.03+5.51+5.93}{5}=\frac{24.07}{5}=4.814$.
The mean of $y$, $\bar{y}=\frac{5.32 + 5.76+5.73+6.00+5.88}{5}=\frac{28.69}{5}=5.738$.

Step2: Calculate the numerator of the correlation - coefficient formula

The numerator $S_{xy}=\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})$.
$(3.57 - 4.814)(5.32 - 5.738)+(4.03 - 4.814)(5.76 - 5.738)+(5.03 - 4.814)(5.73 - 5.738)+(5.51 - 4.814)(6.00 - 5.738)+(5.93 - 4.814)(5.88 - 5.738)$
$=(- 1.244)\times(-0.418)+(-0.784)\times0.022+(0.216)\times(-0.008)+(0.696)\times0.262+(1.116)\times0.142$
$=0.519992-0.017248 - 0.001728+0.182352+0.158472$
$=0.83184$.

Step3: Calculate the denominator of the correlation - coefficient formula

The standard - deviation of $x$, $S_{x}=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$.
$\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=(3.57 - 4.814)^{2}+(4.03 - 4.814)^{2}+(5.03 - 4.814)^{2}+(5.51 - 4.814)^{2}+(5.93 - 4.814)^{2}$
$=(-1.244)^{2}+(-0.784)^{2}+(0.216)^{2}+(0.696)^{2}+(1.116)^{2}$
$=1.547536+0.614656+0.046656+0.484416+1.245456$
$=3.93872$.
$S_{x}=\sqrt{\frac{3.93872}{4}}\approx0.9923$.
The standard - deviation of $y$, $S_{y}=\sqrt{\frac{\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}{n - 1}}$.
$\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}=(5.32 - 5.738)^{2}+(5.76 - 5.738)^{2}+(5.73 - 5.738)^{2}+(6.00 - 5.738)^{2}+(5.88 - 5.738)^{2}$
$=(-0.418)^{2}+(0.022)^{2}+(-0.008)^{2}+(0.262)^{2}+(0.142)^{2}$
$=0.174724+0.000484+0.000064+0.068644+0.020164$
$=0.26408$.
$S_{y}=\sqrt{\frac{0.26408}{4}}\approx0.2569$.
The denominator $S_{x}S_{y}=0.9923\times0.2569\approx0.2549$.

Step4: Calculate the correlation coefficient

The correlation coefficient $r=\frac{S_{xy}}{S_{x}S_{y}}=\frac{0.83184}{0.2549}\approx0.326$.

Answer:

$0.326$