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.

Answer 1

Let X₁ represent the variable "Number of TV commercials" and X₂ represent the variable "car sales"

To calculate the Pearson correlation coefficient you have to apply the following formula:

`r=\frac{\Sigma x_1x_x-((\Sigma x_1)(\Sigma x_2))/(n)}{\sqrt[]{\lbrack\Sigma x^2_1-((\Sigma x_1)^2)/(n)\rbrack\lbrack\Sigma x^2_2-((\Sigma x_2)^2)/(n)\rbrack}}`

First, you have to calculate the sums:

`\begin{gathered} \Sigma x_1=3+7+12+16+18_{} \\ \Sigma x_1=56 \end{gathered}`
`\begin{gathered} \Sigma x^2_1=3^2+7^2+12^2+16^2+18^2_{} \\ \Sigma x^2_1=9+49+144+256+324 \\ \Sigma x^2_1=782 \end{gathered}`
`\begin{gathered} \Sigma x_2=2+3+9+8+9 \\ \Sigma x_2=31 \end{gathered}`
`\begin{gathered} \Sigma x^2_2=2^2+3^2+9^2+8^2+9^2 \\ \Sigma x^2_2=4+9+81+64+81 \\ \Sigma x^2_2=239 \end{gathered}`
`\begin{gathered} \Sigma x_1x_2=3\cdot2+7\cdot3+12\cdot9+16\cdot8+18\cdot9 \\ \Sigma x_1x_2=6+21+108+128+162 \\ \Sigma x_1x_2=425 \end{gathered}`

Now you can calculate the correlation coefficient:

`\begin{gathered} r=\frac{425-(56\cdot31)/(5)}{\sqrt[]{\lbrack782-(56^2)/(5)\rbrack\lbrack239-(31^2)/(5)\rbrack}} \\ r=\frac{425-347.20}{\sqrt[]{\lbrack782-627.20\rbrack\lbrack239-192.20\rbrack}} \\ r=\frac{77.8}{\sqrt[]{154.80\cdot46.80}} \\ r=\frac{77.8}{\sqrt[]{7244.64}} \\ r=0.91405\approx0.914 \end{gathered}`

The correlation coefficient, rounded to three decimal places, is r=0.914