1. (Hypothetical) A random sample of 1000 daily newspapers found that on average, each
edition contained 3.01 local crime stories per day, with a standard deviation of 2.65.
Answer the following questions assuming that the distribution of crime stories in
EC507 PS 7 Solution
1.(10) Since
By Markov Inequality
, we can use Markov Inequality.
and let
,
We get
2.(20) a. By Chebyshevs inequality
P(53.1<X<108.3)=1P(X80.7 27.6) 19.22/27.62=0.889
b.
c. By Chebyshevs inequality
we need
d. i.
ii. we need
3.(5)
EC507 PS6 Solution
1. (15)
(a)
(b)
Since they are independent for ij it is satisfied
Thus one can find that
Also note that for i=1,2,3
Therefore
(c) It is the same as finding the mean of
2. (10)
(a)Let X, Y, and Z denote the number of times the three men
EC507 A1: Statistics for Economists
Spring 2016
Problem Set 9 Solutions
1. (20)
a)
S2=
58.75
b) Because the R.V are normal, a 98% CI is (
)
c) Recall that
A 98% CI for the variance is
(
) where
Therefore the CI is (
, 176.65
d) you need it for part b) and
Hypothesis Testing Solution PS10
1. H0:
H1:
n=15
a)
=P(
=P(
= P(
=1P(
=10.9015=0.0985
b)

=p(
=P(
= P(
=0.3897
c)

=p(
=P(
= P(
=0.7019
d) For
0.01=p(X
= p(
=p(
)
Hence
In a similar way, if
e) When
RR:
For p=0.7
, find k from

=p(
=P(
= P(
=0.80
When
EC507 PS4 Solution
1. (a) We have 5 men and 5 women who are ranked according to their scores on an exam, where no
two scores are alike and all 10! possible rankings are equally likely. Let X denote the highest
ranking achieved by a woman. Clearly, since 6
EC507 A1: Statistics for Economists
Fall 2014
Problem Set 8 Solution
1. [10']
a. All the estimators are unbiased.
b. is relatively more efficient since it uses more information than any other unbiased estimators.
2.[10']
a. Write likelihood function
. So
EC 507 PS3 Solution
1. The files are removed with replacement, thus, there is 3/5 probability to draw a red
file and 2/5 probability to draw a blue file. The first draw is independent from the
second draw.
(a) P(A) = P(red & red) + P(blue & blue) =
P(B) =
EC507 PS2 Solution
1 a)False.
A contradicting example:
Let A=cfw_1, B=cfw_1 and C=cfw_1, then
However,
and
and
.
1
.
b) False.
:
A contradicting example:
Let A=cfw_1,2, B=cfw_1,2,3 and C=cfw_0,1, then
However,
and
.
, thus
2. a) P(A B)=P(A)+P(B)P(A B).
S