Answer :
To find the linear correlation coefficient, also known as Pearson’s correlation coefficient [tex]r[/tex], we need to use the formula:
[tex]r = \frac{n(\sum xy) - (\sum x)(\sum y)}{\sqrt{[n \sum x^2 - (\sum x)^2][n \sum y^2 - (\sum y)^2]}}[/tex]
where:
- [tex]n[/tex] is the number of data points,
- [tex]\sum xy[/tex] is the sum of the product of paired scores,
- [tex]\sum x[/tex] is the sum of [tex]x[/tex]-scores,
- [tex]\sum y[/tex] is the sum of [tex]y[/tex]-scores,
- [tex]\sum x^2[/tex] is the sum of the squares of [tex]x[/tex]-scores,
- [tex]\sum y^2[/tex] is the sum of the squares of [tex]y[/tex]-scores.
First, let's list the data:
[tex]x: 22.6, 36.6, 15.6, 35.0, 17.5[/tex]
[tex]y: 7, 6, 6, 2, 6[/tex]
Next, calculate the necessary sums:
[tex]\sum x = 22.6 + 36.6 + 15.6 + 35.0 + 17.5 = 127.3[/tex]
[tex]\sum y = 7 + 6 + 6 + 2 + 6 = 27[/tex]
[tex]\sum xy = (22.6 \times 7) + (36.6 \times 6) + (15.6 \times 6) + (35.0 \times 2) + (17.5 \times 6) = 158.2 + 219.6 + 93.6 + 70 + 105 = 646.4[/tex]
[tex]\sum x^2 = (22.6^2) + (36.6^2) + (15.6^2) + (35.0^2) + (17.5^2) = 510.76 + 1339.56 + 243.36 + 1225 + 306.25 = 3624.93[/tex]
[tex]\sum y^2 = 7^2 + 6^2 + 6^2 + 2^2 + 6^2 = 49 + 36 + 36 + 4 + 36 = 161[/tex]
Substitute these values into the formula:
[tex]r = \frac{5(646.4) - (127.3)(27)}{\sqrt{[5(3624.93) - 127.3^2][5(161) - 27^2]}}[/tex]
[tex]r = \frac{3232 - 3437.1}{\sqrt{[18124.65 - 16210.29][805-729]}}[/tex]
[tex]r = \frac{-205.1}{\sqrt{1914.36 \times 76}}[/tex]
[tex]r = \frac{-205.1}{\sqrt{145892.16}}[/tex]
[tex]r = \frac{-205.1}{381.906} \approx -0.537[/tex]
The calculated value of the linear correlation coefficient [tex]r[/tex] is approximately [tex]-0.537[/tex], which corresponds to the negative correlation between the [tex]x[/tex] and [tex]y[/tex] values. The negative sign indicates that as [tex]x[/tex] increases, [tex]y[/tex] tends to decrease. Thus, the correct answer is -0.537.
