With the given data, we need to find the quadratic regression equation best fits. We can apply the least-square method.
The equation we have to find is in the form:
[tex]y=ax^2+bx+c[/tex]
We need to find a, b and c.
So, let's start with the calculations:
Now, let's replace these sums into the following equations:
[tex]\begin{gathered} a\sum_^x_i^4+b\sum_^x_i^3+c\sum_^x_i^2=\sum_^x_i^2y_i \\ a*708+b*0+c*60=2673\text{ Equation 1} \end{gathered}[/tex][tex]\begin{gathered} a\sum_^x_i^3+b\sum_^x_i^2+c\sum_^x_i=\sum_^x_iy_i \\ a*0+b*60+c*0=73\text{ equation 2} \end{gathered}[/tex][tex]\begin{gathered} a\sum_^x_i^2+b\sum_^x_i+c*n_i=\sum_^y_i \\ a*60+b*0+c*9=240\text{ Equation 3} \end{gathered}[/tex]
Now, let's solve for b in equation 2:
[tex]\begin{gathered} 0+b*60+0=73 \\ b=\frac{73}{60} \\ b=1.22 \end{gathered}[/tex]
The next step is to isolate a from equation 3, and replace it into equation 1 to solve for c:
[tex]\begin{gathered} 60a+9c=240 \\ 60a=240-9c \\ a=\frac{240-9c}{60} \\ a=4-\frac{9}{60}c \\ \\ \text{ Equation 1:} \\ 708a+60c=2673 \\ 708(4-\frac{9}{60}c)+60c=2673 \\ \\ 2832-\frac{531}{5}c+60c=2673 \\ \\ -\frac{231}{5}c=-159 \\ \\ c=\frac{-159*5}{-231} \\ \\ c=3.44 \end{gathered}[/tex]
Finally, replace c into equation 3 and find a:
[tex]\begin{gathered} 60a+9*3.44=240 \\ 60a=240-30.97 \\ 60a=209.03 \\ a=\frac{209.03}{60} \\ a=3.48 \end{gathered}[/tex]
Finally, if we replace a, b and c into the quadratic regression equation we obtain:
[tex]y=3.48x^2+1.22x+3.44[/tex]
The answer is B.
