Find the value of the linear correlation coefficient r. The paired data below consist of the temperatures on randomly chosen days and the amount a certain kind of plant grew (in millimeters). Temperature 62 76 50 51 71 46 51 44 79 Growth 36 39 50 13 33 33 17 6 16 Question options

Respuesta :

Answer:

n=9 [tex] \sum x = 530, \sum y = 243, \sum xy = 14615, \sum x^2 =32656, \sum y^2 =8245[/tex]  

And replacing the info we got:

[tex]r=\frac{9(14615)-(530)(243)}{\sqrt{[9(32656) -(530)^2][9(8245) -(243)^2]}}=0.1955[/tex]  

So then the correlation coefficient would be r =0.1955

Step-by-step explanation:

Data given

Temperature (x) 62 76 50 51 71 46 51 44 79

Growth (y) 36 39 50 13 33 33 17 6 16

Solution

The correlation formula is given by:

[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 the following sums:  

n=9 [tex] \sum x = 530, \sum y = 243, \sum xy = 14615, \sum x^2 =32656, \sum y^2 =8245[/tex]  

And replacing the info we got:

[tex]r=\frac{9(14615)-(530)(243)}{\sqrt{[9(32656) -(530)^2][9(8245) -(243)^2]}}=0.1955[/tex]  

So then the correlation coefficient would be r =0.1955