Answer:
[tex]r=\frac{6(702000)-(29000)(152)}{\sqrt{[6(150000000) -(29000)^2][6(4000) -(152)^2]}}=-0.852[/tex]
So then the correlation coefficient would be r =-0.852
D) r = −0.85
Step-by-step explanation:
We have the follwoing dataset:
X: 3000, 5500, 6500, 6000,4500, 3500
Y: 34,22,18,26,24,28
n=6
The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.
And in order to calculate the correlation coefficient we can use this formula:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
For our case we have this:
n=6 [tex] \sum x = 29000, \sum y = 152, \sum xy = 702000, \sum x^2 =150000000, \sum y^2 =4000[/tex]
And if we replace we got:
[tex]r=\frac{6(702000)-(29000)(152)}{\sqrt{[6(150000000) -(29000)^2][6(4000) -(152)^2]}}=-0.852[/tex]
So then the correlation coefficient would be r =-0.852
So then the correct option would be:
D) r = −0.85
