An automobile company is running a new television commercial in five cities with approximately the same population. The table below 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 [tex]\( r \)[/tex] for the data given in the table. Round intermediate calculations to six decimal places and your final answer to three decimal places.

[tex]\[
\begin{array}{|c|c|c|c|c|c|}
\hline
\text{Number of TV commercials, } x & 3 & 5 & 11 & 16 & 18 \\
\hline
\text{Car sales, } y \text{ (in hundreds)} & 3 & 2 & 9 & 7 & 8 \\
\hline
\end{array}
\][/tex]



Answer :

To determine the Pearson correlation coefficient [tex]\( r \)[/tex] for the given dataset, we can follow these steps:

1. List the data:
- Number of TV commercials [tex]\( x \)[/tex]: 3, 5, 11, 16, 18
- Car sales [tex]\( y \)[/tex] (in hundreds): 3, 2, 9, 7, 8

2. Calculate the means of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ \bar{x} = \frac{3 + 5 + 11 + 16 + 18}{5} = \frac{53}{5} = 10.6 \][/tex]
[tex]\[ \bar{y} = \frac{3 + 2 + 9 + 7 + 8}{5} = \frac{29}{5} = 5.8 \][/tex]

3. Calculate the deviations from the mean:
- For [tex]\( x \)[/tex]:
[tex]\[ x_1 - \bar{x} = 3 - 10.6 = -7.6, \quad x_2 - \bar{x} = 5 - 10.6 = -5.6 \][/tex]
[tex]\[ x_3 - \bar{x} = 11 - 10.6 = 0.4, \quad x_4 - \bar{x} = 16 - 10.6 = 5.4 \][/tex]
[tex]\[ x_5 - \bar{x} = 18 - 10.6 = 7.4 \][/tex]

- For [tex]\( y \)[/tex]:
[tex]\[ y_1 - \bar{y} = 3 - 5.8 = -2.8, \quad y_2 - \bar{y} = 2 - 5.8 = -3.8 \][/tex]
[tex]\[ y_3 - \bar{y} = 9 - 5.8 = 3.2, \quad y_4 - \bar{y} = 7 - 5.8 = 1.2 \][/tex]
[tex]\[ y_5 - \bar{y} = 8 - 5.8 = 2.2 \][/tex]

4. Calculate the products of the deviations:
[tex]\[ (x_1 - \bar{x})(y_1 - \bar{y}) = (-7.6)(-2.8) = 21.28 \][/tex]
[tex]\[ (x_2 - \bar{x})(y_2 - \bar{y}) = (-5.6)(-3.8) = 21.28 \][/tex]
[tex]\[ (x_3 - \bar{x})(y_3 - \bar{y}) = (0.4)(3.2) = 1.28 \][/tex]
[tex]\[ (x_4 - \bar{x})(y_4 - \bar{y}) = (5.4)(1.2) = 6.48 \][/tex]
[tex]\[ (x_5 - \bar{x})(y_5 - \bar{y}) = (7.4)(2.2) = 16.28 \][/tex]

Sum of these products:
[tex]\[ \sum (x_i - \bar{x})(y_i - \bar{y}) = 21.28 + 21.28 + 1.28 + 6.48 + 16.28 = 66.60 \][/tex]

5. Calculate the squared deviations:
- For [tex]\( x \)[/tex]:
[tex]\[ (x_1 - \bar{x})^2 = (-7.6)^2 = 57.76, \quad (x_2 - \bar{x})^2 = (-5.6)^2 = 31.36 \][/tex]
[tex]\[ (x_3 - \bar{x})^2 = (0.4)^2 = 0.16, \quad (x_4 - \bar{x})^2 = (5.4)^2 = 29.16 \][/tex]
[tex]\[ (x_5 - \bar{x})^2 = (7.4)^2 = 54.76 \][/tex]

Sum:
[tex]\[ \sum (x_i - \bar{x})^2 = 57.76 + 31.36 + 0.16 + 29.16 + 54.76 = 173.20 \][/tex]

- For [tex]\( y \)[/tex]:
[tex]\[ (y_1 - \bar{y})^2 = (-2.8)^2 = 7.84, \quad (y_2 - \bar{y})^2 = (-3.8)^2 = 14.44 \][/tex]
[tex]\[ (y_3 - \bar{y})^2 = (3.2)^2 = 10.24, \quad (y_4 - \bar{y})^2 = (1.2)^2 = 1.44 \][/tex]
[tex]\[ (y_5 - \bar{y})^2 = (2.2)^2 = 4.84 \][/tex]

Sum:
[tex]\[ \sum (y_i - \bar{y})^2 = 7.84 + 14.44 + 10.24 + 1.44 + 4.84 = 38.80 \][/tex]

6. Calculate the Pearson correlation coefficient:
[tex]\[ r = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum (x_i - \bar{x})^2 \sum (y_i - \bar{y})^2}} = \frac{66.60}{\sqrt{173.20 \cdot 38.80}} \][/tex]
[tex]\[ r = \frac{66.60}{\sqrt{6720.96}} = \frac{66.60}{81.96} \approx 0.812427 \][/tex]

Rounded to three decimal places:
[tex]\[ r \approx 0.812 \][/tex]

Hence, the Pearson correlation coefficient [tex]\( r \)[/tex] for the given data is [tex]\( \boxed{0.812} \)[/tex].