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	8	12	16	19
car sales, y (in hundreds)	2	3	2	8	8

Explanation:

Step1: Calculate the means of \(x\) and \(y\)

Let \(x_1 = 3,x_2=8,x_3 = 12,x_4=16,x_5 = 19\) and \(y_1 = 2,y_2=3,y_3 = 2,y_4=8,y_5 = 8\).
The mean of \(x\), \(\bar{x}=\frac{3 + 8+12+16+19}{5}=\frac{58}{5}=11.6\).
The mean of \(y\), \(\bar{y}=\frac{2 + 3+2+8+8}{5}=\frac{23}{5}=4.6\).

Step2: Calculate the numerator of the correlation - coefficient formula

\[

$$\begin{align*} \sum_{i = 1}^{5}(x_i-\bar{x})(y_i - \bar{y})&=(3 - 11.6)(2-4.6)+(8 - 11.6)(3 - 4.6)+(12-11.6)(2 - 4.6)+(16-11.6)(8 - 4.6)+(19-11.6)(8 - 4.6)\\ &=(-8.6)\times(-2.6)+(-3.6)\times(-1.6)+(0.4)\times(-2.6)+(4.4)\times(3.4)+(7.4)\times(3.4)\\ &=22.36+5.76-1.04 + 14.96+25.16\\ &=67.2 \end{align*}$$

\]

Step3: Calculate the denominator of the correlation - coefficient formula

\[

$$\begin{align*} \sum_{i = 1}^{5}(x_i-\bar{x})^2&=(3 - 11.6)^2+(8 - 11.6)^2+(12-11.6)^2+(16-11.6)^2+(19-11.6)^2\\ &=(-8.6)^2+(-3.6)^2+(0.4)^2+(4.4)^2+(7.4)^2\\ &=73.96 + 12.96+0.16+19.36+54.76\\ &=161.2 \end{align*}$$

\]
\[

$$\begin{align*} \sum_{i = 1}^{5}(y_i-\bar{y})^2&=(2 - 4.6)^2+(3 - 4.6)^2+(2-4.6)^2+(8 - 4.6)^2+(8 - 4.6)^2\\ &=(-2.6)^2+(-1.6)^2+(-2.6)^2+(3.4)^2+(3.4)^2\\ &=6.76+2.56 + 6.76+11.56+11.56\\ &=39.2 \end{align*}$$

\]
The denominator is \(\sqrt{\sum_{i = 1}^{5}(x_i-\bar{x})^2\sum_{i = 1}^{5}(y_i-\bar{y})^2}=\sqrt{161.2\times39.2}=\sqrt{6319.04}\approx79.4924\).

Step4: Calculate the correlation coefficient \(r\)

\(r=\frac{\sum_{i = 1}^{5}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i = 1}^{5}(x_i-\bar{x})^2\sum_{i = 1}^{5}(y_i-\bar{y})^2}}=\frac{67.2}{79.4924}\approx0.845\)

Answer:

\(0.845\)