1
Midterm Review Fall 2010
Exercises from previous years’ midterms – solutions (except for 2008 and 2009)
2003 #3. (5 points) In a poll of 1000 persons, you find that 44% of the respondents state that they are in
favor of a land tax.
Build a 95% confidence interval for the percentage of the population that is in favor of
the land tax.
Let
p
be the true percentage in the population.
We draw a random sample of 1000 persons.
Let
x
be a dummy variable equal to 1 if the person is
in favor of the tax, and to 0 if it is not.
We estimate
p
with
ˆ
p
x
=
.
( )
( )
1
var
p
p
x
n
−
=
which is estimated by
( )( )
.44 .56
.2464
.0002464
1000
1000
=
=
. Thus
( )
.0002464
.0157
se x
=
=
The 95% confidence interval is
( )
.44 1.96 .0157
±
or
( ) ( )
.44 .031;.44 .031
.41;.47
−
+
=
2003 #5 (15 points) In a study of turnover in the US labor market, annual data for the period 1950 to 1982
were used to estimate the following model:
( ) ( )
0
1
2
log
log
quit
unemp
year
u
β
=
+
+
+
where
quit
is the quit rate in manufacturing, defined as the number of people leaving their job voluntarily
per 100 employees,
unemp
is the
unemployment rate, and
year
is the year of the observation.
The estimated model is:
( )
s
( )
log
.08 0.34log
0.012
(.03) (0.12)
(0.007)
quit
unemp
year
=
−
−
2
0.54
33 observations
R
n
=
=
(iii) Test the hypothesis
2
0
=
against
2
0
<
at the 5% significance level. Interpret the result.
0
2
1
2
:
0
:
0
H
H
=
<
30
.012
1.71
.007
t
−
=
= −
The critical value for a 5% significance level, onetailed test at nk1=3321=30 degrees of freedom is
1.697.
We thus reject that
2
0
=
with a 5% chance of error.
The parameter
2
measures the
percentage change in quit rate for any additional year.
This is also called the time trend.
There is
evidence that the quit rate has been declining over time.
2004: #1. (5 points)
Denote by X the number of miles per gallon achieved by cars of a particular model.
You are told that
( )
2
20,
4
X
N
μ
σ
=
=
∼
.
What is the probability that for a random sample of 25 cars, the
average miles per gallon
X
will be between 19 and 21?
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1
n
i
i
X
X
n
=
=
∑
follows a Normal distribution of mean
20
μ
=
and variance
2
4
.16
25
n
σ
=
=
Therefore
( )
20
0,1
.4
X
z
N
−
=
∼
And the probability that average miles per gallon will be between 19 and 21 miles/gallon is:
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 Spring '11
 elizabeth
 Statistics, Normal Distribution, Standard Deviation

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