Question

2) Many people believe that gender, weight, drinking habits, and many other factors are much more important in predicting blood alcohol content (BAC) than simply considering the number of drinks a person consumed. Here we examine data from sixteen student volunteers at Ohio State University who each drank a randomly assigned number of cans of beer. These students were evenly divided between men and women, and they differed in weight and drinking habits. Thirty minutes later, a police officer measured their blood alcohol content (BAC) in grams of alcohol per deciliter of blood. The regression equation is y=-0.0127+0.0180X (a) One of the volunteers in the Ohio State University study drank 5 cans of beer and had a blood alcohol content of 10 grams of alcohol per deciliter. Calculate the residual and explain the meaning of the residual value. (b) One of the volunteers in the Ohio State University study drank 5 beers and had a blood alcohol content of 0.05 grams of alcohol per deciliter. Calculate the residual and explain the meaning of the residual value. Solution to (a): 0.15 BAC (grams per deciliter) 0.10 0.05 2 Cans of beer Solution to (b):

Answer 1

The estimated regression equation is $\hat{y}=-0.0127+0.0153*x$

Solution to (a). The residual=0.0362 g/dL. This is the amount of BAC that the equation could not accounted for.

Solution to (b). The residual=-0.0138. This is the amount of the BAC value that the linear regression overestimated.

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(a). The estimated BAC for the volunteer who drank 5 cans of beer, $\hat{y}=-0.0127+0.0153*5=0.0638.$ His actual BAC y=0.10 g/dL. Therefore the residual is $y-\hat{y}=0.1-0.0638=0.0362$ g/dL.

This is the amount of BAC that the equation could not accounted for.

(b). The volunteer drank 5 cans and hence the estimate as before is 0.0638. His actual BAC is y=0.05 g/dL. The residual is therefore $y-\hat{y}=0.05-0.0638=-0.0138$ .