ch02ans.pdf The Simple Linear Regression Model: Specification and Estimation

1
PRINCIPLES OF ECONOMETRICS
5TH
EDITION
ANSWERS TO ODD-NUMBERED
EXERCISES IN CHAPTER 2

Chapter 2, Exercise Answers, Principles of Econometrics, 5e 2
Copyright © 2018 Wiley
EXERCISE 2.1
(a)
i
x
 =5 i
y =
 10 ( )
i
x x
− =
 0 ( )
2
i
x x
− =
 10 ( )
y y
− =
 0 ( )( )
x x y y
− − =
 8
̅ = 1, = 2
(b)
( )( )
( )
2 2
8
0.8
10
x x y y
b
x x
− −
= = =
−


1 2 2 0.8 1 1.2
b y b x
= − = − × =
(c)
5
2
1
15
i
i
x
=
=

5
1
18
i i
i
x y
=
=

5
2 2
1
10
i
i
x Nx
=
− =

5
1
8
i i
i
x y Nxy
=
− =

(d)
i
x
 =5 i
y
 =10 î
y
 =10 î
e
 =0 2
î
e
 =3.6 ˆ
i i
x e
 =0
2
2.5
y
s =
2
2.5
x
s =
2
xy
s =
0.8
xy
r =
158.11388
x
CV =
median( ) 1
x =

(e)
(f) See figure above. The fitted line passes through the point of the means, ̅ = 1, = 2.
(g) = 2 , + ̅ = 2
(h) ˆ 2
y =
(i) 2
ˆ 1.2
σ =
(j) ( |x) = 0.12 and ( ) = 0.34641
EXERCISE 2.3
(a) We show the least squares fitted line.
(b) 2 2.285714
b = , 1 2
b =
0
1
2
3
4
-1 0 1 2 3
x
y Fitted values
Figure xr2.1 Observations and fitted line
5
10
15
20
1 2 3 4 5 6
x
y Fitted values

(c) 10
y = and 3.5
x = and 1 2
ˆ 10
y b b x
= + =
(d)
ˆi
e
1.71429
−2.57143
2.14286
−2.14286
−0.42857
1.28571
(e) ˆ 0
i
e =
 , and 2
ˆ 20.57143
i
e =

(f) ∑ ̂ = 0
EXERCISE 2.5
(a) SALES = 4000 + 4 × ADVERT
5
10
15
20
y
1 2 3 4 5 6
x
y yhat
Figure xr2.3 Observations and linear fitted line
5000
10000
15000
sales
0 1000 2000 3000
advert
Regression line
Figure xr2.5

EXERCISE 2.7
(a) 2
ˆ 697.82566
i
e =

(b) ∑( − ̅) = 1553.8833
(c) 2 1.02896
b = suggests that a 1% increase in the percentage of the population with a
bachelor’s degree or more will lead to an increase of $1028.96 in the mean income per
capita.
(d) 1 11.519745
b =
(e) 2
39722.17
i
x =

(f) ˆ 4.9231453
i
e = −
EXERCISE 2.9
(a) ( ) ( ) ( ) ( ) ( )
2, 2 1 2 1 2 1 2 1
ˆ | | 1 |
mean
E E y y x x x x E y y
β =  − −  =  −   − 
     
x x x
( ) [ ] [ ]
2 1 2 1
| | |
E y y E y E y
 −  = −
 
x x x
[ ] ( ) ( )
( )
6 6 6
2 1 2
4 4 4
6 6
1 2 1 2 1 2 2
4 4
1 1 1
| | |
3 3 3
1 1
3
3 3
i i i
i i i
i i
i i
E y E y E y x
x x x
= = =
= =
 
= = = β + β
 
 
 
= β + β = β + β = β + β
 
  
 
x x x
Similarly [ ]
1 1 2 1
|
E y x
= β + β
x . Then
( ) [ ] [ ] ( ) ( ) ( )
2 1 2 1 1 2 2 1 2 1 2 2 1
| | |
E y y E y E y x x x x
 −  = − = β + β − β + β = β −
 
x x x
Finally,
( ) ( ) ( ) ( ) ( )
( ) ( )
2, 2 1 2 1 2 1 2 1
2 1 2 2 1 2
ˆ | | 1 |
1
mean
E E y y x x x x E y y
x x x x
β =  − −  =  −   − 
     
=  − β − = β
 
x x x
(b) ( ) ( ) ( )
2, 2, 2 2
ˆ ˆ |
mean mean
E E E E
 
β = β = β = β
 
x x
x

(c) ( ) ( ) ( ) ( ) [ ] [ ]
{ }
2 2
2, 2 1 2 1 2 1 2 1
ˆ
var | 1 var | 1 var | var |
mean x x y y x x y y
β =  −   −  =  −  +
     
x x x x
[ ] ( ) ( )
6 6 2 2
2 4 4
1 1 1
var | var | var | 3 3
3 9 9
i i
i i
y y y
= =
   
= = = σ = σ
   
 
 
x x x
Similarly [ ] 2
1
var | 3
y = σ
x . So that
( ) ( ) [ ] [ ]
{ } ( )
( )
2 2 2
2 2
2, 2 1 2 1 2 1 2
2 1
2
ˆ
var | 1 var | var | 1
3 3 3
mean x x y y x x
x x
 
σ σ σ
β =  −  + =  −  + =
 
   
−
 
x x x
We know that ( )
2,
ˆ
var |
mean
β x is larger than the variance of the least squares estimator because
2,
ˆ
mean
β is a linear estimator. To show this note that
( ) ( )
( ) ( ) ( )
6 3 6 3
4 1 4 1
2, 2 1 2 1
2 1 2 1 2 1
6
1
1
ˆ
3 3 3 3
i i i i
i i i i
mean
i i
i
y y y y
y y x x
x x x x x x
a y
= = = =
=
 
     
 
     
β = − − = − = −
   
− − −
   
     
 
=
   

Where
( )
1 2 3
2 1
1
3
a a a
x x
−
= = =
−
and
( )
4 5 6
2 1
1
3
a a a
x x
= = =
−
Furthermore 2,
ˆ
mean
β is an unbiased estimator. From the Gauss-Markov theorem we know that the
least squares estimator is the “best” linear unbiased estimator, the one with the smallest variance.
Therefore, we know that ( )
2,
ˆ
var |
mean
β x is larger than the variance of the least squares estimator.
EXERCISE 2.11
(a) We estimate that each additional $100 per month income is associated with an additional 52
cents per person expenditure, on average, on food away from home. If monthly income is
zero, we estimate that household will spend an average of $13.77 per person on food away
from home.
(b) ˆ 24.17
y = .
(c) ˆ 0.43
ε = .
(d) In this log-linear relationship, the elasticity is ( )
ˆ 0.007 20 0.14
ε = = .
(e) For x = 20, ˆ / 0.1860
dy dx = . For x = 30, ˆ / 0.1995
dy dx = . It is increasing at an increasing
rate. Also, the second derivative, the rate of change of the first derivative is
( )( )
2
2 2
ˆ / exp 3.14 0.007 0.007 0
d y dx x
= + > . A positive second derivative means that the
function is increasing at an increasing rate for all values of x.

(f) The number of zeros is 2334 – 2005 = 329. The reason for the reduction in the number of
observations is that the logarithm of zero is undefined and creates a missing data value. The
software throws out the row of data when it encounters a missing value when doing its
calculations.
EXERCISE 2.13
(a) We estimate that each additional 1000 FTE students increase real total academic cost per
student by $266, holding all else constant. The intercept suggests if there were no students
the real total academic cost per student would be $14,656.
(b) _ = 22.0907.
(c) ˆ 0.6877
e = − .
(d) 20.732975
ACA = .
EXERCISE 2.15
(a) 2 1
2 1
2 1 2 1 2 1
1 1
EZ i i
y y
b y y k y
x x x x x x
   
−
= = − =
   
− − −
   

where 1
2 1
1
k
x x
−
=
−
, 2
2 1
1
k
x x
=
−
, and k3 = k4 = ... = kN = 0
Thus, EZ
b is a linear estimator.
(b) ( ) ( ) ( ) ( )
2 1
1 2 2 1 2 1 2
2 1 2 1 2 1
1 1
| |
EZ EZ
y y
E b E x x E b
x x x x x x
 
−
= = β + β − β + β = β =
 
− − −
 
x x
(c) ( ) ( )
( )
2
2 2 2
2
2 1
2
var | var( | ) var |
EZ i i i i i
b k y k e k
x x
σ
= = = σ =
−
  
x x x
(d) If ( )
2
~ 0,
i
e N σ , then
( )
2
2 2
2 1
2
| ~ ,
EZ
b N
x x
 
σ
β
 
−
 
 
x
(e) To convince E.Z. Stuff that var(b2|x) < var(bEZ|x), we need to show that
( ) ( )
2 2
2 2
2 1
2
i
x x x x
σ σ
>
− −

or that ( )
( )
2
2 2 1
2
i
x x
x x
−
− >

Consider
( ) ( ) ( ) ( ) ( ) ( )( )
2
2 2 2
2 1
2 1 2 1 2 1
2
2 2 2
x x x x
x x x x x x x x x x
 − − − 
− − + − − − −
 
= =
Thus, we need to show that
( ) ( ) ( ) ( )( )
2 2 2
2 1 2 1
1
2 2
N
i
i
x x x x x x x x x x
=
− > − + − − − −

or that

( ) ( ) ( )( ) ( )
2 2 2
1 2 2 1
3
2 2 0
N
i
i
x x x x x x x x x x
=
− + − + − − + − >

or that
( ) ( ) ( )
2 2
1 2
3
2 0.
N
i
i
x x x x x x
=
 − + −  + − >
  
This last inequality clearly holds. Thus, EZ
b is not as good as the least squares estimator.
Rather than prove the result directly, as we have done above, we could also refer Professor
E.Z. Stuff to the Gauss Markov theorem.
EXERCISE 2.17
(a)
Figure xr2.11(a) Price (in $1,000s) against square feet for houses (in 100s)
(b) The fitted linear relationship is
= −115.4236 + 13.40294
( ) (13.0882) (0.4492)
We estimate that an additional 100 square feet of living area will increase the expected home
price by $13,402.94 holding all else constant. The estimated intercept −115.4236 would
imply that a house with zero square feet has an expected price of $−115,423.60.
0
500
1000
1500
Price,
$1000
0 20 40 60 80 100
Sqft, 100s
Figure xr2.17a Collegetown: Price and Square Foot

Figure xr2.17(b) Observations and fitted line
(c) The fitted quadratic model is
= 93.5659 + 0.1845
( ) (6.0722) (0.00525)
We estimate that an additional 100 square feet of living area for a 2000 square foot home
will increase the expected home price by $7,380.80 holding all else constant.
(d)
Figure xr2.17(d) Observations and quadratic fitted line
(e) ̂ = 0.882
(f) The residual plots are
0
500
1000
1500
Price,
$1000
0 20 40 60 80 100
Sqft, 100s
selling price of property ($1000) Fitted values
Figure xr2.17b Observations and fitted line
0
500
1000
1500
2000
Price,
$1000
0 20 40 60 80 100
Sqft, 100s
selling price of property ($1000) Fitted values
tangent
Figure xr2.17d Observations and quadratic fitted line

Figures xr2.17(f) Residuals from linear and quadratic relations
In both models, the residual patterns do not appear random. The variation in the residuals
increases as SQFT increases, suggesting that the homoskedasticity assumption may be
violated.
(g) The sum of square residuals linear relationship is 5,262,846.9. The sum of square residuals
for the quadratic relationship is 4,222,356.3. In this case the quadratic model has the lower
SSE. The lower SSE means that the data values are closer to the fitted line for the quadratic
model than for the linear model.
EXERCISE 2.19
(a)
Figure xr2.19(a) Scatter plot of selling price and living area
(b) The estimated linear relationship is
= −35.9664 + 9.8934
( ) (3.3085) (0.1912)
We estimate that an additional 100 square feet of living area will increase the expected home
price by $9,893.40 holding all else constant. The estimated intercept −35.9664 would imply
that a house with zero square feet has an expected price of $−35,966.40. This estimate is not
-400
-200
0
200
400
Residuals,
linear
fit
0 20 40 60 80 100
Sqft, 100s
Figure xr2.17 Residuals from linear relation
-400
-200
0
200
400
Residuals,
quadratic
fit
0 20 40 60 80 100
Sqft, 100s
Figure xr2.17 Residuals from quadratic relation
0
200
400
600
Price,
$1000
10 20 30 40 50
Sqft, 100s
Figure xr2.19a Selling price vs. square feet

meaningful in this example. The reason is that there are no data values with a house size
near zero.
Figure xr2.19(b) Fitted linear relation
(c) The estimated quadratic equation is
= 56.4572 + 0.2278
( ) (1.6955) (0.0043)
We estimate that an additional 100 square feet of living area for a 1500 square foot home
will increase the expected home price by $6,834 holding all else constant.
(d)
Figure xr2.19(d) Fitted linear and quadratic relations
The sum of squared residuals for the linear relation is SSE = 1,879,826.9948. For the
quadratic model the sum of squared residuals is SSE = 1,795,092.2112. In this instance, the
sum of squared residuals is smaller for the quadratic model, one indicator of a better fit.
(e) If the quadratic model is in fact “true,” then the results and interpretations we obtain for the
linear relationship are incorrect, and may be misleading.
0
200
400
600
Price,
$1000
10 20 30 40 50
Sqft, 100s
selling price of home, $1000 dollars Fitted values
Figure xr2.19b Fitted linear relation
0
200
400
600
Price,
$1000
10 20 30 40 50
Sqft, 100s
Fitted values
Figure xr2.19d Fitted linear and quadratic

EXERCISE 2.21
(a)
= 152.6144 − 0.9812
( ) (3.3473) (0.0949)
We estimate that a house that is new, AGE = 0, will have expected price $152,614.40.
We estimate that each additional year of age will reduce expected price by $981.20,
other things held constant. The expected selling price for a 30-year-old house is
= $123,177.70 .
(b)
Figure xr2.21(b) Observations and linear fitted line
The data show an inverse relationship between house prices and age. The data on newer
houses is not as close to the fitted regression line as the data for older homes.
(c) ln( ) = 4.9283 − 0.0075
( ) (0.0205) (0.0006)
We estimate that each additional year of age reduces expected price by about 0.75%,
holding all else constant.
0
200
400
600
Selling
Price
0 20 40 60 80 100
Age
Figure xr2.21b Observations and linear fitted line

(d)
Figure xr2.21(c) Observations and log-linear fitted line
The fitted log-linear model is not too much different than the fitted linear relationship.
(e) The expected selling price of a house that is 30 years old is = $110,370.32.
(f) For the estimated linear relationship ∑ − = 5,580,871. For the
log-linear model ∑ − = 5,727.332. The sum of squared
differences between the data and fitted values is smaller for the estimated linear relationship,
by a small margin. In this case, based on fit alone, we might choose the linear relationship
rather than the log-linear relationship.
EXERCISE 2.23
(a)
Figure xr2.23(a) Vote against Growth
There appears to be a positive association between VOTE and GROWTH.
0
200
400
600
Selling
Price
0 20 40 60 80 100
Age
selling price of home, $1000 dollars spricehat2
Figure xr2.21d Observations and log-linear fitted line
35
40
45
50
55
60
democratic
share
of
presidential
vote
-10 -5 0 5 10 15
Growth
Figure xr2.23a Vote vs Growth

(b) The estimated equation for 1916 to 2012 is
= 48.6160 + 0.9639
( ) (0.9043) (0.1658)
The coefficient 0.9639 suggests that for a 1 percentage point increase in a favorable growth
rate of GDP in the 3 quarters before the election there is an estimated increase in the share
of votes of the democratic party of 0.9639 percentage points.
We estimate, based on the fitted regression intercept, that that the Democratic party’s
expected vote is 48.62% when the growth rate in GDP is zero. This suggests that when
there is no real GDP growth, the Democratic party is expected to lose the popular vote. A
graph of the fitted line and data is shown in the following figure.
Figure xr2.23(a) Vote vs Growth fitted
(c) In 2016 the actual growth rate in GDP was 0.97% and the predicted expected vote in favor
of the Democratic party was = 49.55, or 49.55%. The actual popular vote in favor
of the Democratic party was 50.82%.
(d) The figure below shows a plot of VOTE against INFLATION. It is difficult to see if there
is positive or inverse relationship.
Figure xr2.23(d) Vote against Inflat
30
40
50
60
-10 -5 0 5 10 15
Growth
democratic share of presidential vote Fitted values
Figure xr2.23b Vote vs Growth fitted
35
40
45
50
55
60
dem
ocratic
share
of
presidential
vote
-10 -5 0 5 10
Inflation
Figure xr2.23d Vote vs Inflat

(e) The estimated equation (plotted in the figure below) is
= 49.6229 + 0.2616
( ) (1.4188) (0.3907)
We estimate that a 1 percentage point increase in inflation during the party’s first 15
quarters increases the share of Democratic party’s vote by 0.2616 percentage points.
The estimated intercept suggests that when inflation is at 0% for that party’s first 15
quarters, the expected share of votes won by the Democratic party is 49.6%.
Figure xr2.23(e) Vote vs Inflat fitted
(f) The actual inflation value in the 2016 election was 1.42%. The predicted vote in favor of
the Democratic candidate (Clinton) was = 49.99, or 49.99%.
EXERCISE 2.25
(a)
Figure xr2.25(a) Histogram of foodaway
The mean of the 1200 observations is 49.27, the 25th
, 50th
and 75th
percentiles are 12.04,
32.56 and 67.60.
35
40
45
50
55
60
-10 -5 0 5 10
Inflation
democratic share of presidential vote Fitted values
Figure xr2.23e Vote vs Inflat fitted
0
20
40
60
Percent
0 500 1000 1500
food away from home expenditure per month per person past quarter, $
Figure xr2-25a Histogram of FOODAWAY

(b)
N Mean Median
ADVANCED = 1 257 73.15 48.15
COLLEGE = 1 369 48.60 36.11
NONE 574 39.01 26.02
(c)
Figure xr2.25(c) Histogram of ln(foodaway)
There are 178 fewer values of ln(FOODAWAY) because 178 households reported spending
$0 on food away from home per person, and ln(0) is undefined. It creates a “missing value”
which software cannot use in the regression.
(d) The estimated model is
ln( ) = 3.1293 + 0.0069
( ) (0.0566) (0.0007)
We estimate that each additional $100 household income increases food away expenditures
per person of about 0.69%, other factors held constant.
0
5
10
15
Percent
0 2 4 6 8
lfoodaway
Figure xr2-25c Histogram of ln(FOODAWAY)

(e)
Figure xr2.25(e) Observations and log-linear fitted line
The plot shows a positive association between ln(FOODAWAY) and INCOMEs.
(f)
Figure xr2.25(f) Residuals vs. income
The OLS residuals do appear randomly distributed with no obvious patterns. There are fewer
observations at higher incomes, so there is more “white space.”
0
2
4
6
8
ln(foodaway)
0 50 100 150 200
Income
lfoodaway Fitted values
Figure xr2.25e Observations and log-linear fitted line
-4
-2
0
2
4
OLS
residuals
0 50 100 150 200
Income
Figure xr2.25f Residuals vs. Income

EXERCISE 2.27
(a)
Figure xr2.27(a) Motel_pct vs. 100relprice
There seems to be an inverse association between relative price and occupancy rate.
(b)
_ = 166.6560 − 1.2212
( ) (43.5709) (0.5835)
Based economic reasoning we anticipate a negative coefficient for RELPRICE. The slope
estimate is interpreted as saying, the expected model occupancy rate falls by 1.22% given a
1% increase in relative price, other factors held constant.
(c)
Figure xr2.27(c) OLS residuals
The residuals are scattered about zero for the first 16 observations but for observations 17-
23 all but one of the residuals is negative. This suggests that the occupancy rate was lower
than predicted by the regression model for these dates.
(d) _ = 79.3500 − 13.2357
( ) (3.1541) (5.9606)
40
60
80
100
Motel
Occupancy
Rate
65 70 75 80 85
100*Relative price
Figure xr2.27a motel_pct vs. relprice
-40
-20
0
20
OLS
residuals
0 5 10 15 20 25
Time
Figure xr2.27c OLS residuals

We estimate that during the non-repair period the expected occupancy rate is 79.35%.
During the repair period, the expected occupancy rate is estimated to fall by 13.24%, other
things held constant, to 66.11%.
EXERCISE 2.29
(a)
variable N mean median min max skewness kurtosis
ln(WAGE) 1200 2.9994 2.9601 1.3712 5.3986 0.2306 2.6846
Figure xr2.29(a) Histogram and statistics for ln(WAGE)
The histogram shows the distribution of ln(WAGE) to be almost symmetrical. Note that the
mean and median are similar, which is not the case for skewed distributions. The skewness
coefficient is not quite zero. Similarly, the kurtosis is not quite three, as it should be for a
normal distribution.
(b) The OLS estimates are
ln( ) = 1.5968 + 0.0987
( ) (0.0702) (0.0048)
We estimate that each additional year of education predicts a 9.87% higher wage, all else
held constant.
(c) For someone with 12 years of education the predicted value is = 16.1493 and for
someone with 16 years of education it is = 23.9721.
(d) For individuals with 12 and 16 years of education, respectively, these values are $1.1850
and $1.5801.
(e)
0
.2
.4
.6
.8
Density
1 2 3 4 5
ln(wage)
Figure xr2.29a Histogram of ln(wage)

Figure xr2.29(e) Observations with linear and loglinear fitted lines
The log-linear model fits the data better at low levels of education.
(f) For the log-linear model this value is 228,573.5 and for the linear model 220,062.3. Based
on this measure the linear model fits the data better than the linear model.
0
50
100
150
200 0 5 10 15 20
years of education
earnings per hour, $ Fitted values
wagehat
Figure xr2.29e Observations with linear and loglinear fitted line

ch02ans.pdf The Simple Linear Regression Model: Specification and Estimation

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ch02ans.pdf The Simple Linear Regression Model: Specification and Estimation