econ 325: introduction to empirical...
TRANSCRIPT
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Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Econ 325: Introduction to
Empirical Economics
Lecture 6
Sampling and
Sampling Distributions
Ch. 6-1
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Populations and Samples
� A Population is the set of all items or individuals of interest
� Examples: All likely voters in the next election
All parts produced today
All sales receipts for November
� A Sample is a subset of the population
� Examples: 1000 voters selected at random for interview
A few parts selected for destructive testing
Random receipts selected for audit
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Population vs. Sample
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a b c d
ef gh i jk l m n
o p q rs t u v w
x y z
Population Sample
b c
g i n
o r u
y
Ch. 6-3
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Why Sample?
� Less time consuming than a census
� Less costly to administer than a census
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Random Sampling
� Every object in the population has an
equal chance of being selected
� Objects are selected independently
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Inferential Statistics
� Making statements about a population by
examining sample results
Sample statistics Population parameters
(known) Inference (unknown, but can
be estimated from
sample evidence)
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SamplePopulation
Ch. 6-6
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Inferential Statistics
� Estimation
� e.g., Estimate the population mean
weight using the sample mean
weight
� Hypothesis Testing
� e.g., Use sample evidence to test
the claim that the population mean
weight is 120 pounds
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Drawing conclusions and/or making decisions concerning a population based on sample results.
Ch. 6-7
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Sampling Distribution
� Assume there is a population …
� Four types of people
� Random variable, X,
is age of individuals
� Possible Values of X:
18, 20, 22, 24 (years)
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A B C D
Ch. 6-8
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Sampling Distribution
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.25
018 20 22 24
A B C D
Uniform Distribution
P(x)
x
(continued)
Summary Measures for the Population Distribution:
µ =18+ 20 + 22 + 24
4= 21
σ 2=
(Xi∑ − µ)2
4= 2.236
Ch. 6-9
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Now consider all possible samples of size n = 2
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1st
2nd
Observation Obs 18 20 22 24
18 18,18 18,20 18,22 18,24
20 20,18 20,20 20,22 20,24
22 22,18 22,20 22,22 22,24
24 24,18 24,20 24,22 24,24
16 possible samples (sampling with replacement)
1st 2nd Observation
Obs 18 20 22 24
18 18 19 20 21
20 19 20 21 22
22 20 21 22 23
24 21 22 23 24
(continued)
Sampling Distribution
16 Sample Means
Ch. 6-10
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Sampling Distribution of All Sample Means
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1st 2nd Observation
Obs 18 20 22 24
18 18 19 20 21
20 19 20 21 22
22 20 21 22 23
24 21 22 23 24
18 19 20 21 22 23 240
.1
.2
.3
P(X)
X
Sample Means
Distribution16 Sample Means
_
Sampling Distribution(continued)
(no longer uniform)
_
Ch. 6-11
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Summary Measures of this Sampling Distribution:
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Developing aSampling Distribution
(continued)
Ch. 6-12
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Comparing the Population with its Sampling Distribution
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18 19 20 21 22 23 240
.1
.2
.3 P(X)
X18 20 22 24
A B C D
0
.1
.2
.3
Population
P(X)
X_
µX
= 21 σX
=1.582.236σ 21μ ==
Sample Means Distributionn = 2
_
Ch. 6-13
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Expected Value of Sample Mean
� Let X1, X2, . . . Xn represent a random sample from a
population
� The sample mean value of these observations is defined as
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∑=
=n
1i
iXn
1X
Ch. 6-14
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Standard Error of the Mean
� A measure of the variability in the mean from
sample to sample is given by the Standard
Error of the Mean:
� If n=2, then
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σX
=σ
n
Ch. 6-15
σ /σX
= 2 =1.4142
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Clicker Question 6-1
� Suppose a random sample of size n = 36 is
drawn from the population distribution with
mean μ = 8 and standard deviation σ = 3. What
is the standard deviation of the sample mean
�� =�
��∑ ��
���� ?
A). 3
B). 1/13
C). 1/2
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Clicker Question 6-2
� What will happen to the variance of the sample
mean �� =�
∑ ��
�� as the sample size n goes
to infinity?
A). Var(��) goes to 1 as → ∞B). Var(��) goes to 0 as → ∞C). Var(��) goes to infinity as → ∞
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Law of Large Numbers
Let �� =�
∑ ��
�� , where {��, �� , … , �} is a
random sample with finite mean and finite
variance.
Then, for any � > 0,
lim→�
� �� − � < � = 1
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Law of Large Numbers
We say that !"# converges in probability to $,which is denoted as
��%→ �
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What is the distribution of ?
� As → ∞, �� converges in probability to a
constant value � and, therefore, �� is not
random in the limit.
� However, when is finite, �� is a random
variable.
� What is the distribution of �� when is finite?
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If the Population is Normal
� If a population is normal, �� is also
normally distributed, i.e.,
�� ~ ' � , (�/
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Sampling Distribution Properties
(i.e. is unbiased )
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Normal Population Distribution
Normal Sampling Distribution (has the same mean)
xx
x
μμx =
μ
xμ
Ch. 6-22
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Sampling Distribution Properties
� As n increases, decreases!
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Larger sample size
Smaller sample size
x
(continued)
nσ/σx =
μ
Ch. 6-23
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If the Population is not Normal
� We can apply the Central Limit Theorem:
� Even if the population is not normal,
� …sample means from the population will be approximately normal as long as the sample size is large enough.
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Central Limit Theorem
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n↑As the
sample
size gets
large
enough…
the sampling
distribution
becomes
almost normal
regardless of
shape of
population
xCh. 6-25
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If the Population is not Normal
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Population Distribution
Sampling Distribution (becomes normal as n increases)
Central Tendency
Variation
x
x
Larger sample size
Smaller sample size
(continued)
Sampling distribution properties:
μμx =
n
σσx =
xμ
μ
Ch. 6-26
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Z-value for Sampling Distributionof the Mean
� Z-value for the sampling distribution of :
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where = sample mean
= population mean
= standard deviation of ��
X
μ
nσ/
μ)X(Z
−=
X
Ch. 6-27
σ
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Central Limit Theorem
Let �� =�
∑ ��
�� , where {��, �� , … , �} is a
random sample with finite mean and finite
variance.
Define * =+�,-.
/ 0⁄and then,
lim→�
� * < 2 = Φ(2)
where Φ 2 = 5�
�607
-� 8-9::; <2
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Central Limit Theorem
We say that 0 �� − � converges in distribution to a normal with mean
0 and variance (�, which is denoted as
0 �� − � D→ '(0, (�)
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Example
� Suppose a random sample of size n = 36 is
drawn from the population distribution with
mean μ = 8 and standard deviation σ = 3.
� What is the approximated probability that the
sample mean is between 7.8 and 8.2?
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Example
Solution:
� Even though the population is not normally
distributed, we use the central limit theorem to
get an approximated solution
� … the sampling distribution of is
approximately normal
� … with mean = 8
� …and standard deviation
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(continued)
x
xμ
0.536
3
n
σσ x ===
Ch. 6-31
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Example
Solution (continued):
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(continued)
0.31080.4)ZP(-0.4
363
8-8.2
n
- X
363
8-7.8P 8.2) X P(7.8
=<<=
<<=<<�
���
σ
µ
Z7.8 8.2 -0.4 0.4
Sampling Distribution
Standard Normal Distribution .1554
+.1554
Population Distribution
??
??
?????
???Sample Standardize
8μ = 8μX
= 0μz =xX
Ch. 6-32
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Clicker Question 6-3
� Suppose a random sample of size n = 36 is
drawn from the population distribution with
mean μ = 8 and standard deviation σ = 3. What
is the probability that the sample mean is larger
than 8.98?
A). 0.01 B). 0.025 C). 0.05 D). 0.10
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Clicker Question 6-4
� Suppose a random sample of size n = 36 is
drawn from the population distribution with
mean μ = 8 and standard deviation σ = 3.
Which of the following is true?
A). � 8 − 1.96�
�< �� < 8 + 1.96
�
�= 0.90
B). � 8 − 1.96�
�< �� < 8 + 1.96
�
�= 0.95
C). � 8 − 1.96�
�< �� < 8 + 1.96
�
�= 0.99
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Acceptance Intervals
� Let zα/2 be the z-value that leaves area α/2 in the
upper tail of the normal distribution.
� Then,
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ααα −=+<<− 1)σzμσzP(μX/2X/2 X
Ch. 6-35
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Sampling Distributions of Sample Proportions
p = the proportion of the population having some characteristic
� Sample proportion ( ) provides an estimate of p:
� Let ��, ��, … , � be independent Bernouilli random
variables with K �� = L. Then,
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p̂ =number of items in the sample having the characteristic of interest
sample size
p̂
Ch. 6-36
∑=
==n
i
iXn
X1
1 p̂
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� The shape of the average of independent Bernoulli
is approximately normal if n is large
� Standardize to Z from the average of Bernoulli random variable:
Normal Approximation for the average of Bernoulli random var.
p)/np(1
pˆ
)ˆVar(
pˆZ
−
−=
−=
p
p
p
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(continued)
Ch. 5-37
( )
−→−=− ∑
= n
ppNpX
npp
n
i
i
)1(,0
1ˆ
1
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Normal Approximation for the average of Bernoulli random var.
� Bernoulli random variable Xi :
� By Central Limit Theorem,
� Xi =1 with probability p
� Xi =0 with probability 1-p
� By Central Limit Theorem,
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E(Xi) = p Var(X
i) = p(1-p)
Ch. 5-38
5.4
1
n(X
i− E(X
i))
i=1
n
∑ → N 0,Var(Xi)( )
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Sampling Distribution of p
� Normal approximation:
Properties:
and
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(where p = population proportion)
Sampling Distribution
.3
.2
.1
00 . 2 .4 .6 8 1
E(p̂) = p σ p̂
2 = Var p̂( ) =p(1− p)
n
^
P(p̂)
p̂
Ch. 6-39
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Z-Value for Proportions
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Z =p̂ − p
σ p̂
=p̂ − p
p(1− p)
n
Standardize to a Z value with the formula:
Ch. 6-40
p̂
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Example
0.34610.50008461.0
(0)(1.02)
)02.1Z0P(
0.4)/100(0.4)(1
40.045.0Z
0.4)/100(0.4)(1
0.40-40.0P45).0ˆP(0.40
=−=
Φ−Φ=
<<=
−
−≤≤
−=<< p
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
� 40% of all voters support ballot proposition A. What is the probability that between 0.40 and 0.45 fraction of voters indicate support in a sample of n = 100 ?
Ch. 5-41
0.0490.40)/100-(0.40)(1 = p)/n-p(1 =)ˆVar(
40.0)ˆ(
=
==
p
ppE
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Example
� if P = .4 and n = 100, what is
P(.40 ≤ ≤ .45) ?
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Z.45 1.02
.3461
Standardize
Sampling DistributionStandardized
Normal Distribution
(continued)
Use standard normal table: P(0 ≤ Z ≤ 1.02) = .3461
.40 0p̂
Ch. 6-42
p̂
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Clicker Question 6-5
� 40% of all voters support ballot proposition A.
Let L̂ be the sample fraction of voters who
support proposition A in a sample of n=100.
Which of the following is true?
A). � 0.4 − 1.96 0.049 < L̂ < 0.4 + 1.96 0.049 = 0.90
B). � 0.4 − 1.96 0.049 < L̂ < 0.4 + 1.96 0.049 = 0.95
C). � 0.4 − 1.96 0.049 < L̂ < 0.4 + 1.96 0.049 = 0.99
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Question
� 30% of all voters support ballot proposition A.
Let L̂ be the sample fraction of voters who
support proposition A in a sample of n=100.
What is the value of a?
� 0.3 − P < L̂ < 0.3 + P = 0.95
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Sample Variance
� Let x1, x2, . . . , xn be a random sample from a
population. The sample variance is
� the square root of the sample variance is called
the sample standard deviation
� the sample variance is different for different
random samples from the same population
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∑=
−−
=n
1i
2i
2 )x(x1n
1s
Ch. 6-45
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Sampling Distribution ofSample Variances
� The sampling distribution of s2 has mean σ2
� If the population distribution is normal, then
has a χ2 distribution with n – 1 degrees of freedom.
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22σ)E(s =
2
2
σ
1)s-(n
Ch. 6-46
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Chi-square distribution
� Consider *� for Q = 1, … , R independently drawn
from the standard normal distribution, '(0,1).
Then,
χT� = (*�)�+(*�)�+ ⋯ + (*V)�
� When k = 1,
χ�� = (*)�
where *~' 0,1 .
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 6-47
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Question
� What is the value of b such that
�(χ�� > W) = 0.05?
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The Chi-square Distribution
� The chi-square distribution is a family of distributions, depending on degrees of freedom:
� d.f. = n – 1
� Text Table 7 contains chi-square probabilities
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0 4 8 12 16 20 24 28 0 4 8 12 16 20 24 28 0 4 8 12 16 20 24 28
d.f. = 1 d.f. = 5 d.f. = 15
χχχχ2 χχχχ2χχχχ2
Ch. 6-49
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Degrees of Freedom (df)
Idea: Number of observations that are free to varyafter sample mean has been calculated
Example: Suppose the mean of 3 numbers is 8.0
Let X1 = 7
Let X2 = 8
What is X3?
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
If the mean of these three values is 8.0, then X3 must be 9(i.e., X3 is not free to vary)
Here, n = 3, so degrees of freedom = n – 1 = 3 – 1 = 2
(2 values can be any numbers, but the third is not free to vary for a given mean)
Ch. 6-50
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Chi-square Example
� A commercial freezer must hold a selected
temperature with little variation. Specifications call
for a standard deviation of no more than 4 degrees
(a variance of 16 degrees2).
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� A sample of 14 freezers is to be tested
� Suppose that the population variance
is 16.
� What is the upper limit (K) for the
sample variance such that the
probability of exceeding this limit is
less than 0.05?
Ch. 6-51
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Finding the Chi-square Value
� Use the the chi-square distribution with area 0.05 in the upper tail:
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
probability α = .05
χχχχ213,0.05
χ2
χχχχ213,0.05
= 22.36
= 22.36 (α = .05 and 14 – 1 = 13 d.f.)
2
22
σ
1)s(n −=χ
is chi-square distributed with (n – 1) = 13 degrees of freedom
Ch. 6-52
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Chi-square Example
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0.0522.3616
1)s(nPK)P(s
22 =
>
−=>So:
(continued)
χχχχ213,0.05 = 22.36 (α = .05 and 14 – 1 = 13 d.f.)
22.3616
1)K(n=
−(where n = 14)
so 27.521)(14
)(22.36)(16K =
−=
If s2 from the sample of size n = 14 is greater than 27.52, there is strong evidence to suggest the population variance exceeds 16.
or
Ch. 6-53
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Question
� A commercial freezer must hold a selected
temperature with little variation. Specifications call
for a standard deviation of no more than 2
degrees.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
� A sample of 10 freezers is to be tested
� Suppose that the population variance
is 4.
� What is the upper limit (K) for the
sample variance such that the
probability of exceeding this limit is
less than 0.05?
Ch. 6-54