Answer:
The r-value of the linear regression that fits these data is -0.947110707.
Step-by-step explanation:
The linear regression equation is in the form of
[tex]y=bx+a[/tex]
Where,
[tex]b=\frac{n(\sum xy)-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2}[/tex]
[tex]a=\frac{(\sum y)(\sum x^2)-(\sum x)(\sum xy)}{n(\sum x^2)-(\sum x)^2}[/tex]
The formula of r is
[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]
The values are
[tex]\sum x=15,\sum y=2247.35,\sum x^2=55,\sum y^2=1010938.423,\sum xy=6656.18, n=5[/tex]
Using above formula, we get
[tex]r=-0.947110707[/tex]
Therefore the r-value of the linear regression that fits these data is -0.947110707.
