Question

2. The correlation coefficient for the following data
n = 10, 2x = 140, E y = 150, Ex? = 1980, y = 2465, 2xy = 2160 is​

Answer 1

Answer:

2. The correlation coefficient for the following data

n = 10, 2x = 140, E y = 150, Ex? = 1980, y = 2465, 2xy = 2160 is2. The correlation coefficient for the following data

n = 10, 2x = 140, E y = 150, Ex? = 1980, y = 2465, 2xy = 2160 is

Answer 2

Note: The correct question will be n=10, ∑x=140, ∑y=150, ∑x²=1980, ∑y²=2465, ∑xy=2160

The correlation coefficient (r) assuming the values given above will be 0.914

Step-by-step explanation:

Given:

n=10,

∑x=140,

∑y=150,

∑x²=1980,

∑y²=2465,

∑xy=2160

To find:

Correlation coefficient (r)

Formula:

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

Solution:

First, we put all the values in the given formula

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

$r=\frac{10 (2160)-(140)(150)}{\sqrt{[10(1980{})-(140)^2][10(2465)-(150)^2] } }$

$r=\frac{(21600)-(21000)}{\sqrt{[(19800{})-(19600)][(24650)-(22500)]} }$

$r=\frac{600}{\sqrt{(200)(2150) } }$

$r=\frac{600}{\sqrt{430000 } }$

$r=\frac{600}{655.74}$

$r=0.914$

The linear correlation coefficient is near 1, therefore it has a strong linear connection.

Hence, the linear correlation assuming the correct values will be 0.914