Question

The correlation coefficient between \(x\) and \(y\) using the following information is (rounded off to two decimal places). \[ \sum_{i=1}^{100} x_i = 280, \quad \sum_{i=1}^{100} y_i = 60, \quad \sum_{i=1}^{100} x_i^2 = 2384, \quad \sum_{i=1}^{100} y_i^2 = 117, \quad \sum_{i=1}^{100} x_i y_i = 438 \]

Hint:

To calculate the correlation coefficient, use the formula that takes into account the sums of \(x\), \(y\), and their squares and products.

Answer 1

The formula for the correlation coefficient \(r\) is: \[ r = \frac{n \sum x_i y_i - \sum x_i \sum y_i}{\sqrt{[n \sum x_i^2 - (\sum x_i)^2][n \sum y_i^2 - (\sum y_i)^2]}} \] Where:
- \(n = 100\) (the number of data points),
- \(\sum x_i = 280\),
- \(\sum y_i = 60\),
- \(\sum x_i^2 = 2384\),
- \(\sum y_i^2 = 117\),
- \(\sum x_i y_i = 438\).
Substituting these values into the formula: \[ r = \frac{100 \times 438 - 280 \times 60}{\sqrt{[100 \times 2384 - 280^2][100 \times 117 - 60^2]}} \] Simplifying the numerator: \[ 100 \times 438 = 43800, \quad 280 \times 60 = 16800 \] \[ \text{Numerator} = 43800 - 16800 = 27000 \] Simplifying the denominator: \[ 100 \times 2384 = 238400, \quad 280^2 = 78400 \] \[ 100 \times 117 = 11700, \quad 60^2 = 3600 \] \[ \text{Denominator} = \sqrt{(238400 - 78400)(11700 - 3600)} = \sqrt{160000 \times 8100} = \sqrt{1296000000} \] \[ \text{Denominator} = 35960 \] Thus, the correlation coefficient \(r\) is: \[ r = \frac{27000}{35960} \approx 0.75 \] Final Answer: \[ \boxed{0.75} \]