Using the information in Question 17, another guy - Dudley, has the regression equation written on a scrap of paper in his pocket. Dudley goes out drinking and has 4 beers. He calculates that he is under the legal limit (0.08) so he decides to drive to another bar. Unfortunately Dudley gets pulled over and confidently submits to a road-side blood alcohol test. He scores a blood alcohol of 0.085 and gets himself arrested. Obviously, Dudley skipped the lecture about residual/error variation. Dudley’s residual is:

The relationship between number of beers consumed (x) and blood alcohol content (y) was studied in 16 male college students by using least squares regression. The following regression equation was obtained from this study:

y= -0.0127 + 0.0180x

Another guy, his name Dudley, has the regression equation written on a scrap of paper in his pocket. Dudley goes out drinking and has 4 beers. He calculates that he is under the legal limit (0.08) so he decides to drive to another bar. Unfortunately Dudley gets pulled over and confidently submits to a road-side blood alcohol test. He scores a blood alcohol of 0.085 and gets himself arrested. Obviously, Dudley skipped the lecture about residual variation. Dudley's residual is:

a. +0.005

b. -0.005

c. +0.0257

d. -0.0257

Solution:

In regression, the difference amid the observed-value of the dependent-variable (y) and the predicted-value ([tex]\hat y[/tex]) is known as the residual (e).

[tex]e=y-\hat y[/tex]

The least square regression equation for the number of beers consumed (x) and blood alcohol content (y) is:

[tex]\hat y= -0.0127 + 0.0180x[/tex]

It is provided that Dudley goes out drinking and has x = 4 beers.

After taking the road-side blood alcohol test, he scores a blood alcohol content of 0.085.

Compute the estimated value as follows:

[tex]\hat y= -0.0127 + 0.0180x\\\\=-0.0127+(0.0180\times4)\\\\=0.0593[/tex]

Compute Dudley’s residual as follows:

[tex]e=y-\hat y[/tex]

[tex]=0.085-0.0593\\\\=0.0257[/tex]

Thus, Dudley’s residual is 0.0257.