We want to find the best prediction of the average number of hours a person spends at work every week if that person spends an average of 10 hours on recreational activities every week.
We will construct a line that adapts to the system by simple linear regression, and then we will find the x-value that makes the line take y=10.
First, we have the data:
We remember that in a simple regression model, we want to write an equation of the form:
[tex]y=\hat{\alpha}+\hat{\beta}x[/tex]
where:
[tex]\begin{gathered} \hat{\alpha}=\bar{y}-\hat{\beta}\bar{x} \\ \hat{\beta}=\frac{nS_{xy}-S_xS_y}{nS_{xx}-S^2_x}_{} \end{gathered}[/tex]
And the Sx, Sy and Sxx are the sums over all the x-values, the y-values and the multiplication of the x-values and y-values (respectively).
We will find those values:
[tex]\begin{gathered} S_x=\sum ^{13}_{i=1}x_i=370 \\ S_y=\sum ^{13}_{i=1}y_i=336.5 \end{gathered}[/tex]
Also, we have:
[tex]\begin{gathered} S_{xx}=\sum ^{13}_{i=1}x^2_i=12600 \\ S_{xy}=\sum ^{13}_{i=1}x_iy_i=8680_{}_{} \end{gathered}[/tex]
And applying the formula, having in mind that n=13, we get:
[tex]\begin{gathered} \hat{\beta}=\frac{nS_{xy}-S_xS_y}{nS_{xx}-S^2_x}_{} \\ =\frac{13(8680)-(370)(336.5)}{13(12600)-(370^2)} \\ =\frac{-11665}{26900} \\ \approx-0.4336 \end{gathered}[/tex]
And, for alpha:
[tex]\begin{gathered} \hat{\alpha}=\frac{1}{n}S_y-\hat{\beta}\frac{1}{n}S_x \\ =\frac{1}{13}(336.5)-(-0.4336)\frac{1}{13}(370) \\ \approx38.2255 \end{gathered}[/tex]
This means that the linear regression equation will be:
[tex]y=38.2255-0.4336x[/tex]
For finding the x-value that will have 10 hours of recreational activities, we replace the 10 value on y, and clear out the variable x:
[tex]10=38.2255-0.4336x[/tex]
And thus,
[tex]\begin{gathered} 10-38.2255=-0.4336x \\ \frac{-28.2255}{-0.4336}=x \\ 65.09=x \end{gathered}[/tex]
This means that when a person works 65 hours approximately, he will have 10 hours of recreational activities every week.
