Answer:
The correlation of X and Y is 1.006
Step-by-step explanation:
Given
X: 2, 3, 5, 6
Y: 1, 2, 4, 5
n = 4
Required
Determine the correlation of x and y
Start by calculating the mean of x and y
For x
[tex]M_x = \frac{\sum x}{n}[/tex]
[tex]M_x = \frac{2 + 3+5+6}{4}[/tex]
[tex]M_x = \frac{16}{4}[/tex]
[tex]M_x = 4[/tex]
For y
[tex]M_y = \frac{\sum y}{n}[/tex]
[tex]M_y = \frac{1+2+4+5}{4}[/tex]
[tex]M_y = \frac{12}{4}[/tex]
[tex]M_y = 3[/tex]
Next, we determine the standard deviation of both
[tex]S = \sqrt{\frac{\sum (x - Mean)^2}{n - 1}}[/tex]
For x
[tex]S_x = \sqrt{\frac{\sum (x_i - Mx)^2}{n -1}}[/tex]
[tex]S_x = \sqrt{\frac{(2-4)^2 + (3-4)^2 + (5-4)^2 + (6-4)^2}{4 - 1}}[/tex]
[tex]S_x = \sqrt{\frac{-2^2 + (-1^2) + 1^2 + 2^2}{3}}[/tex]
[tex]S_x = \sqrt{\frac{4 + 1 + 1 + 4}{3}}[/tex]
[tex]S_x = \sqrt{\frac{10}{3}}[/tex]
[tex]S_x = \sqrt{3.33}[/tex]
[tex]S_x = 1.82[/tex]
For y
[tex]S_y = \sqrt{\frac{\sum (y_i - My)^2}{n - 1}}[/tex]
[tex]S_y = \sqrt{\frac{(1-3)^2 + (2-3)^2 + (4-3)^2 + (5-3)^2}{4 - 1}}[/tex]
[tex]S_y = \sqrt{\frac{-2^2 + (-1^2) + 1^2 + 2^2}{3}}[/tex]
[tex]S_y = \sqrt{\frac{4 + 1 + 1 + 4}{3}}[/tex]
[tex]S_y = \sqrt{\frac{10}{3}}[/tex]
[tex]S_y = \sqrt{3.33}[/tex]
[tex]S_y = 1.82[/tex]
Find the N pairs as [tex](x-M_x)*(y-M_y)[/tex]
[tex](2 - 4)(1 - 3) = (-2)(-2) = 4[/tex]
[tex](3 - 4)(2 - 3) = (-1)(-1) = 1[/tex]
[tex](5 - 4)(4 - 3) = (1)(1) = 1[/tex]
[tex](6-4)(5-3) = (2)(2) = 4[/tex]
Add up these results;
[tex]N = 4 + 1 + 1 + 4[/tex]
[tex]N = 10[/tex]
Next; Evaluate the following
[tex]\frac{N}{S_x * S_y} * \frac{1}{n-1}[/tex]
[tex]\frac{10}{1.82* 1.82} * \frac{1}{4-1}[/tex]
[tex]\frac{10}{3.3124} * \frac{1}{3}[/tex]
[tex]\frac{10}{9.9372}[/tex]
[tex]1.006[/tex]
Hence, The correlation of X and Y is 1.006
