Answer
$y ̂=13.738x+150.798$, where $x$ is fat content and $y ̂$ is calories.
Work Step by Step
$x ̅ =\frac{20+39+27+29+26+47+35+38}{8}=32.625$
$s_x=\sqrt {\frac{(20-32.625)^2+(39-32.625)^2+(27-32.625)^2+(29-32.625)^2+(26-32.625)^2+(47-32.625)^2+(35-32.625)^2+(38-32.625)^2}{8-1}}=8.700$
$y ̅=\frac{430+750+480+540+510+760+690+632}{8}=599$
$s_y=\sqrt {\frac{(430-599)^2+(750-599)^2+(480-599)^2+(540-599)^2+(510-599)^2+(760-599)^2+(690-599)^2+(632-599)^2}{8-1}}=126.613$
$r=\frac{Σ(\frac{x_i-x ̅}{s_x})(\frac{y_i-y ̅}{s_y})}{n-1}=\frac{(\frac{20-32.625}{8.700})(\frac{430-599}{126.613})+(\frac{39-32.625}{8.700})(\frac{750-599}{126.613})+(\frac{27-32.625}{8.700})(\frac{480-599}{126.613})+(\frac{29-32.625}{8.700})(\frac{540-599}{126.613})+(\frac{26-32.625}{8.700})(\frac{510-599}{126.613})+(\frac{47-32.625}{8.700})(\frac{760-599}{126.613})+(\frac{35-32.625}{8.700})(\frac{690-599}{126.613})+(\frac{38-32.625}{8.700})(\frac{632-599}{126.613})}{8-1}=0.944$
The least-squares regression line:
$y ̂=b_1x+b_0$
$b_1=r\frac{s_y}{s_x}=0.944\times\frac{126.613}{8.700}=13.738$
$b_0=y ̅-b_1x ̅ =599-13.738\times32.625=150.798$
So:
$y ̂=13.738x+150.798$
