In order to fill the observed time, let's use the data from the table, and to fill the predicted time, we use the equation of the best fit.
So, for a time t = 11.9, the observed time is 13.00, and the predicted time is:
[tex]\begin{gathered} y=-0.4x+21.22\\ \\ y=-0.4\cdot11.9+21.22\\ \\ y=-4.76+21.22\\ \\ y=16.46 \end{gathered}[/tex]
The residual is the difference between the observed and the predicted values:
[tex]residual=13-16.46=-3.46[/tex]
Now, for t = 18.1, the observed time is 17.50, and the predicted time is:
[tex]\begin{gathered} y=-0.4(18.1)+21.22\\ \\ y=-7.24+21.22\\ \\ y=13.98 \end{gathered}[/tex]
The residual is:
[tex]residual=17.5-13.98=3.52[/tex]
