Question

find the correlation coefficient, r, of the data described below.
a television sports commentator recently declared that basketball teams that rely on the three - point shot cannot win championships. to see if the commentator had a valid point, a sports analyst reviewed game statistics from several teams over the past season.
for each of the teams, the analyst calculated what percentage of attempted field goals had been three - pointers, x, and the number of wins in the season, y.
percentage of shots taken from three - point range|number of wins in a season
20|37
23|21
26|60
28|35
29|30

Explanation:

Step1: Calculate the means

Let $x = [20,23,26,28,29]$ and $y=[37,21,60,35,30]$.
The mean of $x$, $\bar{x}=\frac{20 + 23+26+28+29}{5}=\frac{126}{5}=25.2$.
The mean of $y$, $\bar{y}=\frac{37+21+60+35+30}{5}=\frac{183}{5}=36.6$.

Step2: Calculate the numerator and denominators of the correlation - coefficient formula

The formula for the correlation coefficient $r=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i - \bar{y})^2}}$.
Calculate $(x_i-\bar{x})(y_i - \bar{y})$ for each $i$:
For $i = 1$: $(20 - 25.2)(37-36.6)=-5.2\times0.4=-2.08$.
For $i = 2$: $(23 - 25.2)(21 - 36.6)=-2.2\times(-15.6) = 34.32$.
For $i = 3$: $(26 - 25.2)(60 - 36.6)=0.8\times23.4 = 18.72$.
For $i = 4$: $(28 - 25.2)(35 - 36.6)=2.8\times(-1.6)=-4.48$.
For $i = 5$: $(29 - 25.2)(30 - 36.6)=3.8\times(-6.6)=-25.08$.
$\sum_{i = 1}^{5}(x_i-\bar{x})(y_i - \bar{y})=-2.08 + 34.32+18.72-4.48-25.08=1.4$.
Calculate $(x_i-\bar{x})^2$ for each $i$:
$(20 - 25.2)^2=(-5.2)^2 = 27.04$.
$(23 - 25.2)^2=(-2.2)^2 = 4.84$.
$(26 - 25.2)^2=(0.8)^2 = 0.64$.
$(28 - 25.2)^2=(2.8)^2 = 7.84$.
$(29 - 25.2)^2=(3.8)^2 = 14.44$.
$\sum_{i = 1}^{5}(x_i-\bar{x})^2=27.04 + 4.84+0.64+7.84+14.44=54.8$.
Calculate $(y_i - \bar{y})^2$ for each $i$:
$(37 - 36.6)^2=(0.4)^2 = 0.16$.
$(21 - 36.6)^2=(-15.6)^2 = 243.36$.
$(60 - 36.6)^2=(23.4)^2 = 547.56$.
$(35 - 36.6)^2=(-1.6)^2 = 2.56$.
$(30 - 36.6)^2=(-6.6)^2 = 43.56$.
$\sum_{i = 1}^{5}(y_i - \bar{y})^2=0.16+243.36+547.56+2.56+43.56=837.2$.
The denominator is $\sqrt{54.8\times837.2}=\sqrt{45878.56}\approx214.19$.

Step3: Calculate the correlation coefficient

$r=\frac{1.4}{214.19}\approx0.0065$.

Answer:

$r\approx0.0065$