2- Using the Pearson coefficient correlation formula on top of page 85 of the book which you can also find in "Files", please calculate the coefficient correlation using the following variables: PLEASE SHOW CALCULATION; no credit for answer without calculation. Exercise minutes GPA 25 3.6 30 4.0 20 3.8 60 3.0 45 3.7 90 3.9

Given Data:
- Exercise minutes (x): 25, 30, 20, 60, 45, 90
- GPA (y): 3.6, 4.0, 3.8, 3.0, 3.7, 3.9

We need to calculate the Pearson correlation coefficient using the provided formula:

r = \frac{n(\sum xy) - (\sum x)(\sum y)}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}}

First, let's calculate the necessary sums and products:
1. Number of pairs ( n ): 6
2. Sum of x (Exercise minutes) = 25 + 30 + 20 + 60 + 45 + 90 = 270
3. Sum of y (GPA) = 3.6 + 4.0 + 3.8 + 3.0 + 3.7 + 3.9 = 22.0
4. Sum of xy (Product of pairs) = (25 * 3.6) + (30 * 4.0) + (20 * 3.8) + (60 * 3.0) + (45 * 3.7) + (90 * 3.9) = 983.5
5. Sum of x^2 (Squared exercise minutes) = 25^2 + 30^2 + 20^2 + 60^2 + 45^2 + 90^2 = 625 + 900 + 400 + 3600 + 2025 + 8100 = 15650
6. Sum of y^2 (Squared GPA) = 3.6^2 + 4.0^2 + 3.8^2 + 3.0^2 + 3.7^2 + 3.9^2 = 12.96 + 16 + 14.44 + 9 + 13.69 + 15.21 = 81.3

Now, plug these values into the Pearson correlation coefficient formula:

r = \frac{6(983.5) - (270)(22.0)}{\sqrt{[6(15650) - (270)^2][6(81.3) - (22.0)^2]}}

r = \frac{5901 - 5940}{\sqrt{[93900 - 72900][487.8 - 484.0]}}

r = \frac{-39}{\sqrt{21000 * 3.8}}

r = \frac{-39}{\sqrt{79800}}

r \approx -0.1381

Therefore, the Pearson correlation coefficient for the given data is approximately -0.1381.

\boxed{r \approx -0.1381}