Solution Guide Autumn 2018


Question A1

(a)(5pts)
- The data step constructs a data set called “library”
- The input statement constructs 3 variables
- 1 variable is numeric, 2 variables are characters
- The cards statement and subsequent lines contain the data in raw form
- The print procedure prints the data set in the output window and gives it a title.

(b)(5pts)
- The upper two graphs are histograms. The bars represent the number of
observations in each pre-specified bin.
- Each histogram contains an overlay of a normal density curve.
- The bottom graph is a kernel density estimate. The curve represents a
“smoothed” histogram.
- The top two histograms look different because observations are grouped into
different numbers of bins.
- The bi-modal property is well captured by the top-left and bottom graphs
because they do not “over-smooth” the distribution.


Question A2

(a)(5pts)
- (3pts) The code reads the data set “library”; computes the summation of
“number” in each textbook category; prints the category name and summation
in ascending order of the category name.
- (2pts) The output is
o ARITHMETIC 4
o ART 3
o ENGLISH 5
o SCIENCE 4

(b)(5pts)
- (3pts) The code reads the data sets “library” and “people”; perform a left join
of “library” on “people” using name as the merging variable; prints the name
and age in the ascending order of name.
- (2pts) The output is
o BARBARA .
o CAROL 30
o CAROL 30
o DONALD .
o JAMES .
o MARY 35
o WILLIAM 29


Question A3

(a)(5pts)
- (2pts) �3 51 1� �12� = �2010�
- (3pts)
proc iml;
A = {3 5, 1 1};
d = {20, 10};
iA = inv(A);
x = iA * d;
print x;
quit;


(b)(5pts)
- The solution from the IML code will be the identical to that in part (a). This is
because there is not enough machine precision to store
1.0000000000000000001. The computer will treat this number as if this is 1.


Question A4

(a)(5pts)
- The number is positive. The exponent is (11) -> 3. The fractional part of the
mantissa is (1) -> 0.5.
- Therefore, the number is 23*1.5 = 12.

(b)(5pts)
- (3pts) Linear congruential generator; Starting from seed i=8, it uses a
remainder formula to generate pseudo-random numbers; the pseudo-random
numbers presumably follow a uniform distribution between 0 and 1.
- (2pts) The numbers generated may exhibit deterministic patterns (repeated
sequences) or autocorrelation.




Question A5

(a)(5pts)
- (2pts) The data step generates 5000 different samples of a random variable that
follows a uniform distribution (between 0 and 1). Each sample has 100
observations.
- (2pts) The summary procedure computes the median of each sample and exports
the median to a data set called “meddist”
- (1pt) The univariate procedure uses “meddist” to plot a histogram of the median in
each sample.


(b)(5pts)
- Draw 100 observations from “unif_data” with replacement to form a bootstrap
sample.
- Compute the median in this bootstrap sample.
- Repeat #1 and #2 500 times.
- Analyze the distribution of the median across bootstrapped samples
- The 90% confidence interval is [5th percentile, 95th percentile] in this distribution.






Question B1 (15pts)

(a)(5pts)
- (2pts) Missing number in first row: 70+10+75 = 155; Missing number in
second row: 200 – 100 – 5 = 95
- (3pts)
a11 = 10/155 = 0.0645
a21 = 100/155 = 0.645
a12 = 70/200 = 0.35
a22 = 5/200 = 0.025

= �0.0645 0.350.645 0.025�

(b)(5pts)
A*A = �0.23 0.030.06 0.23� ( − )−1 ≈ + + ∗ = �1 00 1� + �0.0645 0.350.645 0.025� + �0.23 0.030.06 0.23� = �1.29 0.380.71 1.26�


(c)(5pts)
The level change in final demand for good 2 is 95*10%=9.5.

The level change in total output is ( − )−1 � 09.5� = �1.29 0.380.71 1.26� � 09.5� = � 3.6111.97�.

The percent change of output in sector 1 is 3.61/155 = +2.3%
The percent change of output in sector 2 is 11.97/200 = +6.0%


Question B2 (15pts)

(a)(5pts)
�


�
+
= �

� �


�



(b)(5pts)
Number of individuals entering = ax1t
Number of new individuals leaving = bx2t


(c)(5pts)
� 01 � ∗ � 01 � = � 2 0 + 2�;
� 01 � ∗ � 01 � ∗ � 01 � = � 2 0 + 2� ∗ � 01 � = � 3 02 + + 2 3�;

Repeated iteration gives
� 01 � = � 0�−1−+1
=1

�

Therefore,

�
1
2
�
+
= � 0
�−1−+1

=1

� �
1
2
�

= � 1
� −1−+1

=1
� 1 + 2�




Question C1 (20pts)

(a)(5pts)
- DMU1 and DMU3 are efficient.
- DMU1 has the highest Y1 and DMU3 has the highest Y2. There are no virtual
DMUs that can produce such output.

(b)(5pts)
- DMU1 and DMU3 constitute the relevant efficiency frontier for DMU2.
- A virtual DMU is constructed from a linear combination of DMU1 and DMU3
only (hence the weights).
- The inequality constraints imply that the virtual DMU must produce at least as much as
DMU2.
- We assume 1 + 3 = 1 under variable returns to scale.
- The LP solution gives a virtual DMU that uses the least input relative to DMU2 (minimizes
the value of the objective function).


(c)(5pts)

3points for correctly labeled constraints, 2 points for correct feasible area.


(d)(5pts)
- The objective function S=0.6w1+0.3w3 can be rewritten as w3 = S-2w1. This line
is steeper than the slope of the feasible line in the graph.
- Therefore, the optimal solution is on the left end of the feasible line.
- Solve for the system of linear equations: w1+w3=1; 10w1+4w3=5
- The solution is w1*=1/6, w3*=5/6.
- The efficiency score is (0.6)w1 + (0.3)w3 = (0.6)(1/6) + (0.3)(5/6) = 35%.