The equation for the line of best fit is given by:
y = mx + b
In which m is the slope
They are given by:
[tex]m=\frac{n\sum^{}_{}xy-\sum^{}_{}x\sum^{}_{}y}{n\sum^{}_{}x^2-(\sum^{}_{}x)^2}[/tex][tex]b=\frac{\sum^{}_{}y-m\sum^{}_{}x}{n}[/tex]
Sum of x:
Sum of all values of x.
[tex]\sum ^{}_{}x=3\ast0+5+10+12+15+16+2\ast24+28+30+21+36_{}[/tex][tex]\sum ^{}_{}x=221[/tex]
Sum of y:
[tex]\sum ^{}_{}y=0+2\ast2+3+2\ast5+2\ast8+10+2\ast12+2\ast15+20[/tex][tex]\sum ^{}_{}y=117[/tex]
Sum of squares of x:
[tex]\sum ^{}_{}x^2=3\ast0^2+5^2+10^2+12^2+15^2+16^2+2\ast24^2+28^2+30^2+21^2+36^2_{}[/tex][tex]\sum ^{}_{}x^2=5323[/tex]
Sum of xy:
[tex]\sum ^{\infty}_{n\mathop=0}xy=0\ast(0+2+5)+5\ast3+10\ast5+12\ast8+15\ast10+16\ast2[/tex][tex]+24\ast(8+12)+28\ast15+30\ast15+21\ast20+36\ast12_{}[/tex][tex]\sum ^{}_{}xy=2545[/tex]
Slope:
14 students, so n = 14.
Then
[tex]m=\frac{n\sum^{}_{}xy-\sum^{}_{}x\sum^{}_{}y}{n\sum^{}_{}x^2-(\sum^{}_{}x)^2}=\frac{14\ast2545-(221\ast117)}{14\ast5323-221^2}=0.38[/tex][tex]b=\frac{\sum^{}_{}y-m\sum^{}_{}x}{n}=\frac{117-0.38\ast221}{14}=2.36[/tex]
The line of best fit is y = 0.38x + 2.36. This means that for a parents that smokes x cigarettes a day, the child is expect to miss 0.38x + 2.36 days of school during the quarter.
Graphic
