Question

an automobile company is running a new television commercial in five cities with approximately the same population. the following table shows the number of times the commercial is run on tv in each city and the number of car sales (in hundreds). find the pearson correlation coefficient r for the data given in the table. round any intermediate calculations to no less than six decimal places, and round your final answer to three decimal places.

number of tv commercials, x	3	5	13	15	18
car sales, y (in hundreds)	2	3	9	8	6

answer
r =

Explanation:

Step1: Calculate the means

Let $x = [3, 5, 13, 15, 18]$ and $y=[2, 3, 9, 8, 6]$.
The mean of $x$, $\bar{x}=\frac{3 + 5+13+15+18}{5}=\frac{54}{5} = 10.8$.
The mean of $y$, $\bar{y}=\frac{2 + 3+9+8+6}{5}=\frac{28}{5}=5.6$.

Step2: Calculate 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 $i = 1$: $(3 - 10.8)(2-5.6)=(-7.8)\times(-3.6)=28.08$, $(3 - 10.8)^2=(-7.8)^2 = 60.84$, $(2 - 5.6)^2=(-3.6)^2 = 12.96$.
For $i = 2$: $(5 - 10.8)(3 - 5.6)=(-5.8)\times(-2.6)=15.08$, $(5 - 10.8)^2=(-5.8)^2=33.64$, $(3 - 5.6)^2=(-2.6)^2 = 6.76$.
For $i = 3$: $(13 - 10.8)(9 - 5.6)=2.2\times3.4 = 7.48$, $(13 - 10.8)^2=2.2^2=4.84$, $(9 - 5.6)^2=3.4^2 = 11.56$.
For $i = 4$: $(15 - 10.8)(8 - 5.6)=4.2\times2.4 = 10.08$, $(15 - 10.8)^2=4.2^2 = 17.64$, $(8 - 5.6)^2=2.4^2=5.76$.
For $i = 5$: $(18 - 10.8)(6 - 5.6)=7.2\times0.4 = 2.88$, $(18 - 10.8)^2=7.2^2 = 51.84$, $(6 - 5.6)^2=0.4^2 = 0.16$.

Sum of $(x_i-\bar{x})(y_i - \bar{y})$: $S_{xy}=28.08+15.08+7.48+10.08+2.88 = 63.6$.
Sum of $(x_i-\bar{x})^2$: $S_{xx}=60.84+33.64+4.84+17.64+51.84 = 168.8$.
Sum of $(y_i - \bar{y})^2$: $S_{yy}=12.96+6.76+11.56+5.76+0.16 = 37.2$.

Step3: Calculate the correlation coefficient

The Pearson correlation coefficient $r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{63.6}{\sqrt{168.8\times37.2}}$.
$168.8\times37.2 = 6289.76$, $\sqrt{6289.76}\approx79.308007$, $r=\frac{63.6}{79.308007}\approx0.802$.

Answer:

$0.802$