16 subjects in an experiment were assigned to drink a certain number of beers, a

Question

16 subjects in an experiment were assigned to drink a certain number of beers, and after a period of time their blood alcohol content was measured. The data and the plot are given below. (Sorry, the graph is not showing up in the question for some reason.)

1) Give the means and standard deviations for both the number of beers (X) and the BAC (Y).

2) Find the value of the correlation r and give the value of R2.

3) Give the equation of the regression equation to predict BAC from the number of beers.

4) Give the value of the standard error (use the Excel command =STEYX).   

5) How much is the predicted BAC for subject #1?

6) How much is the prediction error for this subject?

7) Do you think the equation does a good job of predicting someone's BAC if you know how many beers they have had? Justify your answer by using all of the above results that you think are relevant.

Subject Beers BAC 1 5 0.1 2 2 0.03 3 9 0.19 4 8 0.12 5 3 0.04 6 7 0.095 7 3 0.07 8 5 0.06 9 3 0.02 10 5 0.05 11 4 0.07 12 6 0.1 13 5 0.085 14 7 0.09 15 1 0.01 16 4 0.05

Explanation / Answer

(1) Beer: x-bar = 4.8125, s = 2.1975

BAC: x-bar = 0.0738, s = 0.0441

(2) r = 0.894, r^2 = 0.8

(3) BAC = -0.0127 + 0.018 * Beers

(4) 0.020

(5) 0.0771

(6) 0.1 - 0.0771 = 0.0229

(7) Since the values of r and r^2 are high, the model is a good fit to the data and can be used for making predictions.

Regression Analysis r² 0.800 n   16 r   0.894 k   1 Std. Error   0.020 Dep. Var. BAC ANOVA table Source SS   df   MS F p-value Regression 0.0234 1   0.0234 55.94 2.97E-06 Residual 0.0058 14   0.0004 Total 0.0292 15   Regression output confidence interval variables coefficients std. error    t (df=14) p-value 95% lower 95% upper std. coeff. Intercept -0.0127 0.000 Beers 0.0180 0.0024 7.480 2.97E-06 0.0128 0.0231 0.894 Predicted values for: BAC 95% Confidence Interval 95% Prediction Interval Beers Predicted lower upper lower upper Leverage 5 0.077118 0.066115 0.088121 0.031917 0.122319 0.063

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