Introductory Econometrics
Home assignment # 1
(Suggested solutions)
1. Imagine you work in a bulb factory as a supervisory technician, who is asked to check
the lifespan of light bulbs. You know that the lifespan of light bulbs produced in
your factory is distributed N(200, 400), that is, normally distributed with a mean of
200 hours and a variance of 400 hours. What is the probability that you find a light
bulb with a life span longer than 245 hours? (Hint: convert the distribution to a
standard normal one and then refer to statistical tables.)
Solution:
We have X ∼ N(200, 400), and we are asked to compute P(X > 245).
We need to first convert X to a standard normal variable, which will be done by
substracting the mean (200) and dividing it by its standard deviation (
√
400):
Z =
X − 200
√
400
∼ N(0, 1) .
We have
P (X > 245) = 1 − P (X ≤ 245)
= 1 − P
X − 200
√
400
≤
245 − 200
√
400
= 1 − P Z ≤
45
20
= 1 − FZ (2.25) = 1 − Φ (2.25)
,
where Φ() is the cdf of standard normal distribution, whose values can be found in
statistical tables. We find in the statistical tables that
Φ (2.25) = 0.9878
and so we obtain
P(X > 245) = 1 − 0.9878 = 0.0122
Hence, the probability that you find a light bulb with a life span longer than 245
hours is slightly over 1%.
2. You receive a unique dataset that includes wages of all citizens of Ostrava as well as
their experience (number of years spent working). Obviously, you are very curious
about what is the effect of experience on wages. You run an OLS regression of
monthly wage in CZK on the number of years of experiences and obtain the following
results:
wagei = 14450 + 1135 · experi
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(a) Interpret the meaning of the coefficient of experi.
(b) Use the estimates to determine the average wage of a person with 1, 5, 20, and
40 years of experience. Do they seem as realistic?
(c) Use your answer to part (b) and explain what is the problem with your estimation.
Draw the theoretical diagram you’d expect for the relationship between
wagei and experi, then plot the estimated equation onto the same diagram.
Suggest a way to improve you estimation equation.
Solution:
(a) The coefficient is positive and the interpretation is that the increase of the
experience by 1 year increases the monthly wage by CZK 1135 per month.
(b) Average wages of individuals with 1, 5, 20, and 40 years of experience is the
following:
wagei(1 year) = 14450 + 1135 · 1 = 15585
wagei(5 years) = 14450 + 1135 · 5 = 20125
wagei(20 years) = 14450 + 1135 · 20 = 37150
wagei(40 years) = 14450 + 1135 · 40 = 59850
The predicted wages see to be realistic for people with low experience, but seems
to be way too high for individuals with 20 and 40 years of experience.
(c) The problem is, that we estimated this relationship as linear which obviously
is not true. The more probable relationship is non-linear, for example with
functional form wagei = β0 +β1 ·experi +β2 ·exper2
i . We would then expect the
β1 to be positive and β2 to be negative, which would ensure decreasing returns
to experience. The difference is plotted in Figure 1.
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Figure 1: Relation between wage and exper
3. Let us investigate the results of an experiment in broiler (poultry meat) production.
The average weight of an experimental lot of broilers and their corresponding level of
average feed consumption was tabulated over the time period in which they changed
from baby chickens to mature broilers ready for market. At time t, we denote the
average weight of the experimental group of broilers as the output yt, the total feed
consumed by the input xt. The observations are summarized in Table 1.
Let us suppose that the underlying model is
yt = β0 + β1xt + β2x2
t + εt .
Answer the following questions. Do not use statistical software (Stata etc.) for
the solution!!! You can use Excel or some mathematical software (e.g. Matlab) for
multiplication and inversion of the matrices in question1
, but otherwise you are asked
to do all estimation and computation “by hand” - meaning according to the formulae
we introduced on the lecture. Make sure you show in your solution all the matrices
you construct and compute.
(a) Using the Least Squares formula, find the coefficients β0, β1 and β2.
(b) Find and list the residuals of the model.
(c) What is the sign of β2? How do you interpret this coefficient?
(d) Based on your estimation result, find ∂yt
∂xt
, the marginal productivity of the feed
input. Interpret your your result.
(e) We denote the price of broilers by pb and the price of feed by pf . Explain why
∂yt
∂xt
=
pf
pb
. [Hint: Solve for the firm’s profit maximization problem.]
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For Excel, you may want to check out the commands =MMULT() and =MINVERSE()
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Average Weight Average Cumulative
End of of Broiler Feed Inputs
Time Period in Pounds in Pounds
t yt xt
1 0.57 1.00
2 1.01 2.00
3 1.20 3.00
4 1.27 4.00
5 1.91 5.00
6 2.52 6.00
7 2.55 7.00
8 2.92 8.00
9 3.38 9.00
10 3.43 10.00
11 3.53 11.00
12 4.11 12.00
13 4.26 13.00
14 4.33 14.00
15 4.41 15.00
Table 1: Data on poultry production
Solution:
(a) Let us rewrite the model and the data in matrix form:
y = Xβ + ε ,
where:
y =


























0.57
1.01
1.20
1.27
1.91
2.52
2.55
2.92
3.38
3.43
3.53
4.11
4.26
4.33
4.41


























X =


























1 1 1
1 2 4
1 3 9
1 4 16
1 5 25
1 6 36
1 7 49
1 8 64
1 9 81
1 10 100
1 11 121
1 12 144
1 13 169
1 14 196
1 15 225


























β =


β0
β1
β2

 .
4
To find the OLS estimate, we compute the following matrices:
X X =


15 120 1240
120 1240 14400
1240 14400 178312


(X X)
−1
=


0.7934 −0.2044 0.0110
−0.2044 0.0656 −0.0039
0.0110 −0.0039 0.0002


X y =


41.40
412.17
4682.65

 ,
which gives us
β =


β0
β1
β2

 =


0.059
0.426
−0.009

 ,
and so the estimated model is
y = 0.059 + 0.426x − 0.009x2
.
(b) The residuals (ei = yi − 0.059 − 0.426x + 0.009x2
) are:
e =


























0.094
0.133
−0.060
−0.356
−0.065
0.213
−0.072
0.000
0.180
−0.034
−0.180
0.170
0.108
−0.017
−0.115


























(c) β2 = −0.009 < 0, which means that the model exhibits decreasing returns to
scale. This means that broilers do not gain in weight at constant pace and
that the weight gain becomes slower over time. Hence, when a broiler achieves
certain weight, it does not pay to feed him any more (if our interest is purely
economic, i.e. if we just want to sell our broilers for meat).
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(d) Since
y = 0.059 + 0.426x − 0.009x2
,
we can say that
∂yt
∂xt
= 0.426 − 0.009 · 2xt = 0.426 − 0.018xt .
(e) Let us consider the firm’s maximization problem, where we subtract the costs
(price of feed) from the revenues (price of broilers sold), remembering that the
weight of broilers (y) is a function of the feed (x) and so the feed (x) is the
variable over which we maximize:
max
xt
π = max
xt
pbyt(xt) − pf xt .
First order condition gives us
pb
∂yt
∂xt
− pf = 0
∂yt
∂xt
=
pf
pb
,
which is what we had to show.
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