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Student Name:
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Midterm 2: Form A
6E:071 Spring 2007
Please write legibly. There are two forms for this exam. If an answer from the other form
appears on this exam, we will consider that evidence that you copied from another exam and
you will receive a grade of zero on the entire exam. You may use any calculator that the College
Board allows and one cheat sheet, 8.5
x 11. Please be sure to sign the honor pledge
after
you
complete the exam.
1. Consider the data set from Assignment 5, which contained the return on Phillip Morris (MO)
stock and the return on the S&P 500 (SP). If we regress MO on SP, we obtain the following
results:
Predictor
Coef
SE Coef
T
P
Constant
0.3537
0.7612
0.46
0.643
S&P
1.1695
0.2106
5.55
0.000
If we regress SP on MO, we obtain
Predictor
Coef
SE Coef
T
P
Constant
0.8608
0.3286
2.62
0.011
MO
0.23575
0.04246
5.55
0.000
a. (6 points) If the linear regression line minimizes the squared difference between the
data and the regression line, why is the slope coefficient in the first regression different
from the slope coefficient in the second regression?
b. (2 points) What is the predicted value of MO if the return on the S&P is 4% (use
c. (2 points) What is the predicted value of S&P if the return on MO is 4% (use MO=4)?
The regression of MO on S&P minimizes the distance in a different direction than does the
regression of S&P on MO. In each case, we’ve assumed that one variable depends on the other
variable, and our assumption of which variable is dependent changes the regression coefficient.
MO = .3537 + 1.1695(4) = 5.0317
S&P = .8608 + .23575(4) = 1.8038
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View Full Document2. A person’s blood alcohol content (BAC) is determined by many factors, including how much
alcohol has been consumed, the person’s weight, and recent food consumption. We wish to
predict BAC from number of beers consumed. We have data on 16 people who drank between
1 and 9 beers. The Minitab output from a regression of BAC on Beer is:
The regression equation is
BAC =  0.0127 + 0.0180 Beers
Predictor
Coef
SE Coef
T
P
Constant
0.01270
0.01264
1.00
0.332
Beers
0.017964
0.002402
7.48
0.000
S = XXXXXXXX
RSq = XXXX%
RSq(adj) = 78.6%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
1
0.023375
0.023375
XXXXX
0.000
Residual Error
14
0.005850
0.000418
Total
15
0.029225
a. (4 points) Is the coefficient on Beers significantly different from zero? Explain how you
reached your answer.
b. (2 points each) Calculate the following values:
Fstatistic = MSR/MSE =55.921
Regression standard error, s = sqrt(MSE) = 0.0204
R
2
=SSR/SST = 79.98
Correlation(BAC, Beers)=
(Sx/Sy)*b = (2.198/.0441)*.017964 = 0.8953
c. (4 points) Provide an English interpretation of the coefficient on Beers.
d. (6 points) Does it make sense to use the regression to predict BAC when Beers = 0?
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 Spring '08
 BethIngram

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