Answer :
To determine the Pearson correlation coefficient [tex]\( r \)[/tex] for the given data, we need to follow these steps:
1. Calculate the means of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ \text{Mean of } x = \bar{x} = \frac{6 + 8 + 12 + 14 + 18}{5} = \frac{58}{5} = 11.6 \][/tex]
[tex]\[ \text{Mean of } y = \bar{y} = \frac{5 + 2 + 9 + 8 + 9}{5} = \frac{33}{5} = 6.6 \][/tex]
2. Compute the deviations from the mean:
For [tex]\( x \)[/tex]:
[tex]\[ x - \bar{x} = \{ 6 - 11.6, 8 - 11.6, 12 - 11.6, 14 - 11.6, 18 - 11.6 \} = \{ -5.6, -3.6, 0.4, 2.4, 6.4 \} \][/tex]
For [tex]\( y \)[/tex]:
[tex]\[ y - \bar{y} = \{ 5 - 6.6, 2 - 6.6, 9 - 6.6, 8 - 6.6, 9 - 6.6 \} = \{ -1.6, -4.6, 2.4, 1.4, 2.4 \} \][/tex]
3. Find the product of the deviations for each pair:
[tex]\[ (x - \bar{x})(y - \bar{y}) = \{ (-5.6 \times -1.6), (-3.6 \times -4.6), (0.4 \times 2.4), (2.4 \times 1.4), (6.4 \times 2.4) \} \][/tex]
[tex]\[ = \{ 8.96, 16.56, 0.96, 3.36, 15.36 \} \][/tex]
4. Sum the products of the deviations:
[tex]\[ \sum{(x - \bar{x})(y - \bar{y})} = 8.96 + 16.56 + 0.96 + 3.36 + 15.36 = 45.2 \][/tex]
5. Calculate the squares of the deviations:
For [tex]\( x \)[/tex]:
[tex]\[ (x - \bar{x})^2 = \{ (-5.6)^2, (-3.6)^2, (0.4)^2, (2.4)^2, (6.4)^2 \} = \{ 31.36, 12.96, 0.16, 5.76, 40.96 \} \][/tex]
For [tex]\( y \)[/tex]:
[tex]\[ (y - \bar{y})^2 = \{ (-1.6)^2, (-4.6)^2, (2.4)^2, (1.4)^2, (2.4)^2 \} = \{ 2.56, 21.16, 5.76, 1.96, 5.76 \} \][/tex]
6. Sum the squares of the deviations:
For [tex]\( x \)[/tex]:
[tex]\[ \sum{(x - \bar{x})^2} = 31.36 + 12.96 + 0.16 + 5.76 + 40.96 = 91.2 \][/tex]
For [tex]\( y \)[/tex]:
[tex]\[ \sum{(y - \bar{y})^2} = 2.56 + 21.16 + 5.76 + 1.96 + 5.76 = 37.2 \][/tex]
7. Compute the Pearson correlation coefficient [tex]\( r \)[/tex]:
[tex]\[ r = \frac{\sum{(x - \bar{x})(y - \bar{y})}}{\sqrt{\sum{(x - \bar{x})^2} \sum{(y - \bar{y})^2}}} \][/tex]
With the given summations:
[tex]\[ r = \frac{45.2}{\sqrt{91.2 \times 37.2}} \][/tex]
8. Calculate the denominator:
[tex]\[ \sqrt{91.2 \times 37.2} = \sqrt{3394.64} \approx 58.246373 \][/tex]
9. Final calculation of [tex]\( r \)[/tex]:
[tex]\[ r = \frac{45.2}{58.246373} \approx 0.776 \][/tex]
So, the Pearson correlation coefficient [tex]\( r \)[/tex] for the given data is approximately [tex]\( 0.776 \)[/tex], rounded to three decimal places.
1. Calculate the means of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ \text{Mean of } x = \bar{x} = \frac{6 + 8 + 12 + 14 + 18}{5} = \frac{58}{5} = 11.6 \][/tex]
[tex]\[ \text{Mean of } y = \bar{y} = \frac{5 + 2 + 9 + 8 + 9}{5} = \frac{33}{5} = 6.6 \][/tex]
2. Compute the deviations from the mean:
For [tex]\( x \)[/tex]:
[tex]\[ x - \bar{x} = \{ 6 - 11.6, 8 - 11.6, 12 - 11.6, 14 - 11.6, 18 - 11.6 \} = \{ -5.6, -3.6, 0.4, 2.4, 6.4 \} \][/tex]
For [tex]\( y \)[/tex]:
[tex]\[ y - \bar{y} = \{ 5 - 6.6, 2 - 6.6, 9 - 6.6, 8 - 6.6, 9 - 6.6 \} = \{ -1.6, -4.6, 2.4, 1.4, 2.4 \} \][/tex]
3. Find the product of the deviations for each pair:
[tex]\[ (x - \bar{x})(y - \bar{y}) = \{ (-5.6 \times -1.6), (-3.6 \times -4.6), (0.4 \times 2.4), (2.4 \times 1.4), (6.4 \times 2.4) \} \][/tex]
[tex]\[ = \{ 8.96, 16.56, 0.96, 3.36, 15.36 \} \][/tex]
4. Sum the products of the deviations:
[tex]\[ \sum{(x - \bar{x})(y - \bar{y})} = 8.96 + 16.56 + 0.96 + 3.36 + 15.36 = 45.2 \][/tex]
5. Calculate the squares of the deviations:
For [tex]\( x \)[/tex]:
[tex]\[ (x - \bar{x})^2 = \{ (-5.6)^2, (-3.6)^2, (0.4)^2, (2.4)^2, (6.4)^2 \} = \{ 31.36, 12.96, 0.16, 5.76, 40.96 \} \][/tex]
For [tex]\( y \)[/tex]:
[tex]\[ (y - \bar{y})^2 = \{ (-1.6)^2, (-4.6)^2, (2.4)^2, (1.4)^2, (2.4)^2 \} = \{ 2.56, 21.16, 5.76, 1.96, 5.76 \} \][/tex]
6. Sum the squares of the deviations:
For [tex]\( x \)[/tex]:
[tex]\[ \sum{(x - \bar{x})^2} = 31.36 + 12.96 + 0.16 + 5.76 + 40.96 = 91.2 \][/tex]
For [tex]\( y \)[/tex]:
[tex]\[ \sum{(y - \bar{y})^2} = 2.56 + 21.16 + 5.76 + 1.96 + 5.76 = 37.2 \][/tex]
7. Compute the Pearson correlation coefficient [tex]\( r \)[/tex]:
[tex]\[ r = \frac{\sum{(x - \bar{x})(y - \bar{y})}}{\sqrt{\sum{(x - \bar{x})^2} \sum{(y - \bar{y})^2}}} \][/tex]
With the given summations:
[tex]\[ r = \frac{45.2}{\sqrt{91.2 \times 37.2}} \][/tex]
8. Calculate the denominator:
[tex]\[ \sqrt{91.2 \times 37.2} = \sqrt{3394.64} \approx 58.246373 \][/tex]
9. Final calculation of [tex]\( r \)[/tex]:
[tex]\[ r = \frac{45.2}{58.246373} \approx 0.776 \][/tex]
So, the Pearson correlation coefficient [tex]\( r \)[/tex] for the given data is approximately [tex]\( 0.776 \)[/tex], rounded to three decimal places.
