Respuesta :
Answer:
[tex]r=\frac{10(2892)-(34.5)(780)}{\sqrt{[10(151.25) -(34.5)^2][10(62100) -(780)^2]}}=0.9975[/tex]
The final answer would be the correlation coefficient r =0.9975
Step-by-step explanation:
For this case we have the following data given:
x: 0.5 1 2 2.5 3.5 4 4.5 5 5.5 6
y: 60 63 69 72 78 81 84 87 90 96
And we want to calculate the correlation coefficient, and we have the following formula to do this:
[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=10 [tex] \sum x = 34.5, \sum y = 780, \sum xy = 2892, \sum x^2 =151.25, \sum y^2 =62100[/tex]
And replacing we got:
[tex]r=\frac{10(2892)-(34.5)(780)}{\sqrt{[10(151.25) -(34.5)^2][10(62100) -(780)^2]}}=0.9975[/tex]
The final answer would be the correlation coefficient r =0.9975
