Answer
$y ̂=2.2x+1.2$
Work Step by Step
$x ̅ =\frac{-2+(-1)+0+1+2}{5}=0$
$s_x=\sqrt {\frac{(-2-0)^2+(-1-0)^2+(0-0)^2+(1-0)^2+(2-0)^2}{5-1}}=1.581$
$y ̅=\frac{-4+0+1+4+5}{5}=1.2$
$s_y=\sqrt {\frac{(-4-1.2)^2+(0-1.2)^2+(1-1.2)^2+(4-1.2)^2+(5-1.2)^2}{5-1}}=3.564$
$r=\frac{Σ(\frac{x_i-x ̅}{s_x})(\frac{y_i-y ̅}{s_y})}{n-1}=\frac{(\frac{-2-0}{1.581})(\frac{-4-1.2}{3.564})+(\frac{-1-0}{1.581})(\frac{0-1.2}{3.564})+(\frac{0-0}{1.581})(\frac{1-1.2}{3.564})+(\frac{1-0}{1.581})(\frac{4-1.2}{3.564})+(\frac{2-0}{1.581})(\frac{5-1.2}{3.564})}{5-1}=0.976$
The least-squares regression line:
$y ̂=b_1x+b_0$
$b_1=r\frac{s_y}{s_x}=0.976\times\frac{3.564}{1.581}=2.200$
$b_0=y ̅-b_1x ̅ =1.2-2.200\times0=1.2$
So:
$y ̂=2.2x+1.2$
