He following data set contains data on Blood Alcohol Content (BAC) for a set of

Question

He following data set contains data on Blood Alcohol Content (BAC) for a set of men who drank a particular number of beers over a one hour period. We wish to predict BAC from the number of beers consumed using simple regression. The commands you need are in Minitab under Stat>Regression. To do a prediction, click on the "Options" box in the regression command. Type the value of the X variable into the box "Prediction Intervals for new observations."

Beers BAC

5 0.1

2 0.03

9 0.19

8 0.12

3 0.04

7 0.095

3 0.07

5 0.06

3 0.02

5 0.05

4 0.07

6 0.1

5 0.085

7 0.09

1 0.01

4 0.05

A.) On average, the predicted number of beers a person has drunk increases by_________________for every 1% increase in BAC. Round to two decimal places.

B.) The percentage of variation in Beers explained by BAC is:

Explanation / Answer

A)

Using technology, we get              
              
slope =    0.017963762          
intercept =    -0.012700604          
              
Thus, the regression line is              
              
BAC =    0.017963762Beers   -   0.012700604

Hence, for every 0.01 change in BAC, we expect an increase of

change in Beer = change in BAC/slope = 0.01/0.017963762 = 0.556676269

Hence, we expect an increase of 0.556676269 in Beers. [ANSWER: 0.556676269]

********************************

b)

Also, getting the correlation,  
  
r =    0.894338148
  
Thus, the coefficient of determination is  
  
r^2 =    0.799840723

Hence, around 79.98% of variation in Beers is explained by BAC.   

[ANSWER: 79.98%]

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