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].
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].
