CHAPTER 5: ANSWER KEY
Case Exercises
1. Valuing an MBA for Yourself
a. For a given pre-MBA salary, post-MBA salary will be higher on average by $1,732 if you go to
school B rather than school A.
b. For $15,000, at school A you can expect on average to make $(30.0 + 1.70*(15)) thousand
when you leave, i.e., $55,500, while coming out from school B you can expect to make
$(30.0 +1.70*(15) -7.31*1 + .232*(15)) thousand, i.e., $51,670. So all else being equal, it looks
like you should choose school A. The same calculations for a $65,000 income give figures of
$140,500 from school A and $148,270 from school B, suggesting you should choose school B.
c. The predicted income is $98,171. We expect 5% of these students to fall on each side of the
90% prediction interval, so about 3 will make less than $80,000 (more exactly, less than
$79,710).
d. It seems that for recent graduates of these two programs, 83.4% of the variation in salary can
be explained by considering what their salary was on entry and which of the two schools they
attended.
e. The difference across these two schools in post-MBA income depends on pre-MBA income.
In particular, school A does better by those with low pre-MBA incomes, while school B does
better by those with high pre-MBA incomes.
2. Valuing an MBA for Your Employer
a. Substituting in the regression equation we can see that the estimate of the average billing of a
MBA with 24 months of experience 44.1 + (9.07 - 1.43)*24 + 68.4 = 295.86, which is $295,860.
The estimate of the average billing of a non-MBA with 24 months of experience is 44.1 +
9.07*24 = 261.78, which is $261,780.
b. The difference between MBAs and non-MBAs regarding the value added to the company is
given by  2 + 3 * experience, where 2 is the true coefficient on MBA and 3 is the true
coefficient on the slope dummy “experience*MBA.” The only way that this difference could
change over the time the MBA is with the company is if 3 is not zero. The t-statistic and the pvalue associated with the test
H0: 3 = 0
H1: 3  0
is reported by Stata. The p-value is 0.022. So, we can reject the null with 97.8% confidence (or at
a 2.2% significance level) but not with 99% confidence (or not at a 1% significance level). So we
are not able to claim that the extra value to the company of an MBA as compared to a non-MBA
changes over the time the MBA is with the company.
c. With 120 months (ten years) experience, the expected billings for MBAs is 44.1 + (9.07 1.43)*120 + 68.4 = $1,029,300. The expected billings for non-MBAs with 120 months
experience is 44.1 + (9.07 )*120 = $1,132,500. Thus, the estimate of the expected difference in
billings is 68.4 - 1.43*120 = 68.4 - 171.6 = - 103.2. That is, $103,200 in favor of non-MBAs.
d. This prediction is one for values of experience which are far from those in our sample. We
would expect to have large prediction and confidence intervals as a result. We might be hesitant
to use the model to predict for values so far away from our sample. Furthermore, although it
seems reasonable that the advantage of having an MBA could diminish over time, you might
think it unlikely that the MBA ultimately becomes a disadvantage.
Problems
1a & b. The regression equation tells us the answers to both parts a & b.
. regress Sales Competitor
Source |
SS
df
MS
-------------+-----------------------------Model |
2525434.07
1
2525434.07
Number of obs =
F(
1,
Prob > F
50
48) =
26.22
=
0.0000
Residual |
4622481.48
48
96301.6976
R-squared
-------------+-----------------------------Total |
7147915.55
49
145875.828
=
0.3533
Adj R-squared =
0.3398
Root MSE
310.33
=
-----------------------------------------------------------------------------Sales |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------Competitor |
-455.349
88.91875
-5.12
0.000
-634.1321
-276.5659
_cons |
1105.434
67.7185
16.32
0.000
969.2773
1241.592
------------------------------------------------------------------------------
When there is no competition, the competitor variable equals 0 and the estimated sales equal 1105
– 455(0) = 1105.
When there is competition, the competitor variable equals 1 and the estimated sales will equal
1105 – 455(1) = 650.
c. The difference between these estimates is 455 dollars per day which you can tell by looking at
the coefficient on the dummy variable. The very low p-value of 0.000 tells us that this difference
is statistically significant, and it is a huge percentage of sales making it practically significant as
well.
d. The 95% confidence interval for the coefficient of „competition‟ is -455 ± 2.0106 ∙ 88.92 or
about [-634, -277]. Stata automatically reports this confidence interval, which is [-634.1321, 276.5659].
e. The R-squared of 0.3533 tells us that 35.33% of the variance in sales can be explained using
only the competitor variable.
2a & b. The regression equation tells us the answers to both parts a & b.
. regress Sales Income
Source |
SS
df
MS
-------------+-----------------------------Model |
4736343.66
1
4736343.66
Residual |
2411571.89
48
50241.081
-------------+------------------------------
Number of obs =
F(
1,
50
48) =
94.27
Prob > F
=
0.0000
R-squared
=
0.6626
Adj R-squared =
0.6556
Total |
7147915.55
49
145875.828
Root MSE
=
224.15
-----------------------------------------------------------------------------Sales |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------Income |
2.697244
.2777973
9.71
0.000
2.138695
3.255793
_cons |
48.72079
87.57192
0.56
0.581
-127.3544
224.7959
------------------------------------------------------------------------------
When income equals 200, the estimated sales equal 48.7 +2.697 (200) = 588.2
When income equals 300, the estimated sales equal 48.7 +2.697 (300) = 857.9
Or we could use Stata‟s prediction (confint) command to give us the same answers:
Income
Sales
Competitor
Predicted
200
--
--
588.1696
300
--
--
857.8939
c. As income increases by $1, the coefficient on income tells us that the sales increase by $2.697.
So, a $100 increase will increase sales by $269.7. You can see this difference in parts a & b by
comparing the two estimates of sales [857.9 – 588.2 = 269.7]
d. The 95% confidence interval for the estimate in part c. is 269.7 ± 2.0106 ∙ [0.2778*100] or
about [213.9, 325.6]. You can also multiply the reported 95% confidence interval for the
estimated coefficient on Income to get [2.138695*100, 3.255793*100] or about [213.9, 325.6].
e. The R-squared of 0.6626 tells us that 66.26% of the variance in sales can be explained using
only the income variable.
3.
Scatterplot
2000
Sales
1500
1000
500
0
100
200
300
Income
Sales
400
500
Fitted values
There seems to be two different lines here. Maybe one is for sales where a competitor exists and
the other is for sales without any competition? Let‟s check it out in problem 4.
4a. The variable competitor is not significant but that doesn‟t mean that competition doesn‟t
matter. The presence of a competitor affects the way that income affects sales. The high p-value
represents the fact that if we extended both lines all the way to the axis (where income equals
zero) then the two intercepts are pretty close together.
b & c. The prediction (confint) command in Stata makes this really easy:
Income
Sales
Competitor
CompInc
predicted
300
--
0
0
1070.75
300
--
1
300
687.0919
d. Without any competition, the estimate is 1070.8 while competition drops the estimate down to
687.1 giving a difference of 383.7. Obviously competition does matter. The only time it is
insignificant is when income equals zero which is obviously an extrapolation beyond the range of
our data set.
5. Stata‟s regression below helps answer this question:
. regress WageGrowth Unemployment Belgium BEUnemployment
Source |
SS
df
MS
Number of obs =
-------------+------------------------------
F(
3,
84
80) =
13.53
Model |
477.042849
3
159.014283
Prob > F
=
0.0000
Residual |
940.261318
80
11.7532665
R-squared
=
0.3366
-------------+-----------------------------Total |
1417.30417
83
17.0759538
Adj R-squared =
0.3117
Root MSE
3.4283
=
-----------------------------------------------------------------------------WageGrowth |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------Unemployment |
-.8269871
.1929112
-4.29
0.000
-1.210893
-.4430815
Belgium |
-.1532025
1.519459
-0.10
0.920
-3.177022
2.870618
BEUnemployment | .0940872
.2512231
0.37
0.709
-.4058627
.594037
1.023651
11.53
0.000
9.764992
13.83925
_cons |
11.80212
------------------------------------------------------------------------------
a. Wage Growth = 11.80 – 0.827 Un Rate – 0.153 Belgium + 0.094 BE * Un Rate
b. Wage Growth = 11.80 – 0.827 Un Rate – 0.153 (1) + 0.094 (1) * Un Rate
= 11.65 – 0.733 Un Rate
c. Wage Growth = 11.80 – 0.827 Un Rate – 0.153 (0) + 0.094 (0) * Un Rate
= 11.80 – 0.827 Un Rate
d. They are the same as they should be.
e. Wages go down by 0.73 on average.
f. Wages go down by 0.83 on average.
g. The slope dummy variable gives us the answer: 0.094
h. We can use the standard approach for finding a confidence interval for a coefficient: 0.094 ±
1.9901∙0.251 or [-0.41, 0.59]. We can also find this confidence interval directly from the
regression output, which is [-.4058627, 0.594037].
i & j. Stata‟s prediction (confint) command gives us both answers quickly:
Unemployment
Belgium
BEUnemployment
predicted
se_est_mean
se_ind_pred
CIlow
CIhigh
PIlow
3
1
3
9.45022
.7332073
3.505832
8.230072 10.67037 3.616079 15.28436
3
0
0
9.321161 .606981
3.481622
8.311069 10.33125 3.527308 15.11501
The two predictions are: 9.45 for BE and 9.32 for DK. The two 90% confidence intervals are:
[8.23, 10.67] and [8.31, 10.33].
6. Stata‟s regression below helps answer this question:
. regress WageGrowth Unemployment Germany DEUnemployment
Source |
SS
df
MS
Number of obs =
-------------+------------------------------
F(
80) =
40.53
Model |
2150.3685
3
716.789501
Prob > F
=
0.0000
Residual |
1414.889
80
17.6861125
R-squared
=
0.6031
Adj R-squared =
0.5883
Root MSE
4.2055
-------------+-----------------------------Total |
3565.2575
83
42.9549096
3,
84
=
-----------------------------------------------------------------------------WageGrowth |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------Unemployment |
-1.069147
.2261388
-4.73
0.000
-1.519178
-.6191168
Germany |
-10.47289
1.875114
-5.59
0.000
-14.20448
-6.741293
DEUnemployment | .1329423
.3091755
0.43
0.668
-.4823365
.7482211
1.498199
13.58
0.000
17.36949
23.33251
_cons |
20.351
------------------------------------------------------------------------------
a. Wage Growth = 20.35 – 1.069 Un Rate – 10.47 Germany + 0.133 DE * Un Rate
b. Wage Growth = 20.35 – 1.069 Un Rate – 10.47 (1) + 0.133 (1) * Un Rate
= 9.88 – 0.936 Un Rate
c. Wage Growth = 20.35 – 1.069 Un Rate – 10.47 (0) + 0.133 (0) * Un Rate
PIhigh
= 20.35 – 1.069 Un Rate
d. They are the same as they should be.
e. Wages go down by 0.94 on average.
f. Wages go down by 1.07 on average.
g. The slope dummy variable gives us the answer: 0.133
h. We can use the standard approach for finding a confidence interval for a coefficient: 0.133 ±
1.9901∙0.309 or [-0.48, 0.75]. We can also find this confidence interval directly from the
regression output, which is [-.4823365, 0.7482211].
i & j. Stata‟s prediction (confint) command gives us both answers quickly:
Unemployment
Germany
DEUn~ment
predicted
se_est_mean
se_ind_pred
CIlow
CIhigh
PIlow
PIhigh
3
1
3
7.0695
.710628
4.265103
5.886927
8.252074
-.0281629
14.16716
3
0
0
17.14356
.9341422
4.307985
15.58903
18.69809
9.974539
24.31259
The two predictions are: 7.07 for DE and 17.14 for Greece. The two 90% confidence intervals
are: [5.89, 8.25] and [15.59, 18.70].
7. Stata‟s regression below helps answer this question:
. regress WageGrowth Unemployment Spain ESUnemployment
Source |
SS
df
MS
Number of obs =
-------------+------------------------------
F(
3,
84
80) =
30.50
Model |
1681.09807
3
560.366023
Prob > F
=
0.0000
Residual |
1469.57181
80
18.3696476
R-squared
=
0.5336
Adj R-squared =
0.5161
-------------+-----------------------------Total |
3150.66988
83
37.9598781
Root MSE
=
4.286
-----------------------------------------------------------------------------WageGrowth |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------Unemployment |
-.8915849
.1795224
-4.97
0.000
-1.248846
-.5343239
Spain |
3.914415
1.711688
2.29
0.025
.5080459
7.320783
ESUnemployment | .2433135
_cons |
13.91131
.20256
1.20
0.233
-.1597938
.6464209
1.349091
10.31
0.000
11.22653
16.59609
------------------------------------------------------------------------------
a. Wage Growth = 13.91 – 0.892 Un Rate + 3.91 Spain + 0.243 ES * Un Rate
b. Wage Growth = 13.91 – 0.892 Un Rate + 3.91 (1) + 0.243 (1) * Un Rate
= 17.82 – 0.648 Un Rate
c. Wage Growth = 13.91 – 0.892 Un Rate + 3.91 (0) + 0.243 (0) * Un Rate
= 13.91 – 0.892 Un Rate
d. They are the same as they should be.
e. Wages go down by 0.65 on average.
f. Wages go down by 0.89 on average.
g. The slope dummy variable gives us the answer: 0.243
h. We can use the standard approach for finding a confidence interval for a coefficient: 0.243 ±
1.9901∙0.203 or [-0.16, 0.65]. We can also find this confidence interval directly from the
regression output, which is [-.1597938, 0.6464209].
i & j. Stata‟s prediction (confint) command gives us both answers quickly:
Unemployment
Spain
ESUn~ment
predicted
se_est_mean
se_ind_pred
CIlow
CIhigh
PIlow
PIhigh
3
1
3
15.88091
.8528976
4.370021
14.46158
17.30024
8.608651
23.15317
3
0
0
11.23656
.9184387
4.383284
9.708158
12.76495
3.942224
18.53089
The two predictions are: 15.88 for ES and 11.24 for FR. The two 90% confidence intervals are:
[14.46, 17.30] and [9.71, 12.76].