Question

Find the Correlation of the following two variables X: 2, 3, 5, 6 Y: 1, 2, 4, 5

Answer 1

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

$M_x = \frac{\sum x}{n}$

$M_x = \frac{2 + 3+5+6}{4}$

$M_x = \frac{16}{4}$

$M_x = 4$

For y

$M_y = \frac{\sum y}{n}$

$M_y = \frac{1+2+4+5}{4}$

$M_y = \frac{12}{4}$

$M_y = 3$

Next, we determine the standard deviation of both

$S = \sqrt{\frac{\sum (x - Mean)^2}{n - 1}}$

For x

$S_x = \sqrt{\frac{\sum (x_i - Mx)^2}{n -1}}$

$S_x = \sqrt{\frac{(2-4)^2 + (3-4)^2 + (5-4)^2 + (6-4)^2}{4 - 1}}$

$S_x = \sqrt{\frac{-2^2 + (-1^2) + 1^2 + 2^2}{3}}$

$S_x = \sqrt{\frac{4 + 1 + 1 + 4}{3}}$

$S_x = \sqrt{\frac{10}{3}}$

$S_x = \sqrt{3.33}$

$S_x = 1.82$

For y

$S_y = \sqrt{\frac{\sum (y_i - My)^2}{n - 1}}$

$S_y = \sqrt{\frac{(1-3)^2 + (2-3)^2 + (4-3)^2 + (5-3)^2}{4 - 1}}$

$S_y = \sqrt{\frac{-2^2 + (-1^2) + 1^2 + 2^2}{3}}$

$S_y = \sqrt{\frac{4 + 1 + 1 + 4}{3}}$

$S_y = \sqrt{\frac{10}{3}}$

$S_y = \sqrt{3.33}$

$S_y = 1.82$

Find the N pairs as $(x-M_x)*(y-M_y)$

$(2 - 4)(1 - 3) = (-2)(-2) = 4$

$(3 - 4)(2 - 3) = (-1)(-1) = 1$

$(5 - 4)(4 - 3) = (1)(1) = 1$

$(6-4)(5-3) = (2)(2) = 4$

Add up these results;

$N = 4 + 1 + 1 + 4$

$N = 10$

Next; Evaluate the following

$\frac{N}{S_x * S_y} * \frac{1}{n-1}$

$\frac{10}{1.82* 1.82} * \frac{1}{4-1}$

$\frac{10}{3.3124} * \frac{1}{3}$

$\frac{10}{9.9372}$

$1.006$

Hence, The correlation of X and Y is 1.006