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 $x_1 = 15,x_2=29,x_3 = 31,x_4=33,x_5 = 34$ and $y_1 = 12,y_2=10,y_3 = 11,y_4=6,y_5 = 7$.
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 denominator components

Calculate $(x_i-\bar{x})(y_i - \bar{y})$, $(x_i-\bar{x})^2$ and $(y_i-\bar{y})^2$ for each $i$.
For $i = 1$: $(x_1-\bar{x})(y_1-\bar{y})=(15 - 28.4)(12-9.2)=(- 13.4)\times2.8=-37.52$, $(x_1-\bar{x})^2=(-13.4)^2 = 179.56$, $(y_1-\bar{y})^2=(2.8)^2=7.84$.
For $i = 2$: $(x_2-\bar{x})(y_2-\bar{y})=(29 - 28.4)(10 - 9.2)=0.6\times0.8 = 0.48$, $(x_2-\bar{x})^2=(0.6)^2=0.36$, $(y_2-\bar{y})^2=(0.8)^2 = 0.64$.
For $i = 3$: $(x_3-\bar{x})(y_3-\bar{y})=(31 - 28.4)(11 - 9.2)=2.6\times1.8 = 4.68$, $(x_3-\bar{x})^2=(2.6)^2=6.76$, $(y_3-\bar{y})^2=(1.8)^2=3.24$.
For $i = 4$: $(x_4-\bar{x})(y_4-\bar{y})=(33 - 28.4)(6 - 9.2)=4.6\times(-3.2)=-14.72$, $(x_4-\bar{x})^2=(4.6)^2 = 21.16$, $(y_4-\bar{y})^2=(-3.2)^2 = 10.24$.
For $i = 5$: $(x_5-\bar{x})(y_5-\bar{y})=(34 - 28.4)(7 - 9.2)=5.6\times(-2.2)=-12.32$, $(x_5-\bar{x})^2=(5.6)^2=31.36$, $(y_5-\bar{y})^2=(-2.2)^2 = 4.84$.
Sum of $(x_i-\bar{x})(y_i - \bar{y})$: $S_{xy}=-37.52+0.48 + 4.68-14.72-12.32=-59.4$.
Sum of $(x_i-\bar{x})^2$: $S_{xx}=179.56+0.36+6.76+21.16+31.36=239.2$.
Sum of $(y_i-\bar{y})^2$: $S_{yy}=7.84+0.64+3.24+10.24+4.84=26.8$.

Step3: Calculate the correlation coefficient

The correlation coefficient $r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{-59.4}{\sqrt{239.2\times26.8}}=\frac{-59.4}{\sqrt{6410.56}}\approx\frac{-59.4}{80.066}\approx - 0.74$.

Answer:

$-0.74$