1) In this question, we can start setting our table:
2) Now let's find the regression line by finding the slope and the linear coefficient, according to these formulas below:
So
[tex]\begin{gathered} m=\frac{41.86-12.3\times24.2}{6\times32.79-(32.79)^2} \\ m=0.2911\approx0.29 \\ b=\frac{24.2-0.29\times12.3}{6} \\ b=3.4388\approx b=3.44\text{ } \end{gathered}[/tex]
Therefore we can write the regression line as:
[tex]y=0.29x\text{ +3.44 }[/tex]
3) Now let's find the correlation coefficient through another formula:
[tex]\begin{gathered} r=\frac{n\Sigma xy\text{ - (}\Sigma x)(\Sigma y)}{\sqrt[]{\lbrack n(\Sigma x)^2-(\Sigma x)^2\lbrack}n\Sigmay^2-\Sigmay^2)\text{ }} \\ r=\frac{6\times41.86-12.3\times24.2}{\sqrt[]{(6\times(12.3)^2)(6\times(24.2)^2-24.2)}} \\ r=-0.0261 \end{gathered}[/tex]
