Question

find the correlation coefficient, r, of the data described below. finn hangs his laundry outside every saturday and has noticed that his clothes dry faster on some days than others. he is curious to know how the daily temperature influences the time it takes for his clothes to dry. over the last several saturdays, he recorded the temperature (in celsius), x, and the time it had taken for his clothes to dry (in hours), y. temperature drying time 15 12 29 10 31 11 33 6 34 7

Explanation:

Step1: Calculate the means

Let $x = [15,29,31,33,34]$ and $y=[12,10,11,6,7]$.
The mean of $x$, $\bar{x}=\frac{15 + 29+31+33+34}{5}=\frac{142}{5}=28.4$.
The mean of $y$, $\bar{y}=\frac{12 + 10+11+6+7}{5}=\frac{46}{5}=9.2$.

Step2: Calculate the numerator and denominators

The numerator $N=\sum_{i = 1}^{5}(x_i-\bar{x})(y_i - \bar{y})$.
$(15 - 28.4)(12-9.2)=(- 13.4)\times2.8=-37.52$.
$(29 - 28.4)(10 - 9.2)=0.6\times0.8 = 0.48$.
$(31 - 28.4)(11 - 9.2)=2.6\times1.8=4.68$.
$(33 - 28.4)(6 - 9.2)=4.6\times(-3.2)=-14.72$.
$(34 - 28.4)(7 - 9.2)=5.6\times(-2.2)=-12.32$.
$N=-37.52+0.48 + 4.68-14.72-12.32=-59.4$.

The denominator 1 $D_1=\sqrt{\sum_{i = 1}^{5}(x_i-\bar{x})^2}$.
$(15 - 28.4)^2=(-13.4)^2 = 179.56$.
$(29 - 28.4)^2=0.6^2=0.36$.
$(31 - 28.4)^2=2.6^2 = 6.76$.
$(33 - 28.4)^2=4.6^2=21.16$.
$(34 - 28.4)^2=5.6^2=31.36$.
$\sum_{i = 1}^{5}(x_i-\bar{x})^2=179.56+0.36 + 6.76+21.16+31.36=239.2$.
$D_1=\sqrt{239.2}\approx15.466$.

The denominator 2 $D_2=\sqrt{\sum_{i = 1}^{5}(y_i-\bar{y})^2}$.
$(12 - 9.2)^2=2.8^2 = 7.84$.
$(10 - 9.2)^2=0.8^2=0.64$.
$(11 - 9.2)^2=1.8^2 = 3.24$.
$(6 - 9.2)^2=(-3.2)^2=10.24$.
$(7 - 9.2)^2=(-2.2)^2=4.84$.
$\sum_{i = 1}^{5}(y_i-\bar{y})^2=7.84+0.64+3.24+10.24+4.84=26.8$.
$D_2=\sqrt{26.8}\approx5.177$.

The denominator $D = D_1\times D_2\approx15.466\times5.177\approx80$.

Step3: Calculate the correlation coefficient

The correlation coefficient $r=\frac{N}{D}=\frac{-59.4}{80}\approx - 0.74$.

Answer:

$-0.74$