Answer :
In order to calculate the best airfare for 1300 miles, we need to fit a line to represent these points using linear regression.
The line can be calculated using the following equations:
[tex]\begin{gathered} f=\hat{\alpha}+\hat{\beta}x \\ \hat{\beta}=\frac{\sum ^n_{i=1}(x_i-\bar{x})(y_i-\bar{y})}{\sum ^n_{i=1}(x_i-\bar{x})^2} \\ \hat{\alpha}=\bar{y}-\hat{\beta}\bar{x} \end{gathered}[/tex]Calculating the average value of x and y, we have:
[tex]\begin{gathered} \bar{x}=\frac{170+370+410+590+610+720+940+1210+1500}{9}\approx724 \\ \bar{y}=\frac{108+125+160+162+144+180+180+260+215}{9}\approx170 \end{gathered}[/tex]Now, calculating the value of Beta, we have:
[tex]\begin{gathered} \hat{\beta}=\frac{(-554)(62)+(-354)(45)+(-314)(10)+(-134)(8)+(-114)(26)+(-4)(-10)+(216)(-10)+(486)(90)+(776)(45)}{(-554)^2+(-354)^2+(-314)^2+(-134)^2+(-114)^2+(-4)^2+(216)^2+(486)^2+(776)^2} \\ \hat{\beta}=\frac{19086}{1446824}=0.01319 \end{gathered}[/tex]Calculating Alpha, we have:
[tex]\hat{\alpha}=170-0.01319\cdot724=160.45[/tex]So our function is:
[tex]f=160.45+0.01319x[/tex]Using the value of x = 1300, we have:
[tex]\begin{gathered} f=160.45+0.01319\cdot1300 \\ f=177.6 \end{gathered}[/tex]So the value that best represent the cost is $180, therefore the answer is the first option.