Download ECO220Y Sampling Distributions: Central Limit Theorem

ECO220Y
Sampling Distributions:
Central Limit Theorem
Readings: Section 9.1 (pages 296-301, 305-307), 9.3, 9.4
Fall 2010
Lecture 18
(Fall 2010)
Sampling Distributions
Lecture 18
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Sampling Distribution of Sample Mean
The sample mean X̄ is a random variable that is determined by the
particular items picked for the sample.
If X̄ is a random variable, we can find it’s distribution and parameters
of that distribution (we did it analytically in Lecture 17).
We know that population has distribution with mean µX and variance
σX2 .
Sample is a sum of random variables, each of which is distributed
with µX and σX2 .
The sample mean, X̄ , is the sum of these random variables, divided
n
by the sample size n, or X̄ = X1 +X2 +...+X
.
n
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Sampling Distributions
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Parameters of Sampling Distribution of Sample Mean
Use Laws of Expectation and Variance to find parameters of sampling
distribution of sample mean:
h
i
n
E [X̄ ] = E X1 +X2 +...+X
= n1 E [X1 + X2 + . . . + Xn ] =
n
1
n
(E [X1 ] + E [X2 ] + . . . + E [Xn ]) = n1 (µx + µx + . . . + µx ) = n1 nµx = µx
V [X̄ ] = V
1
n2
h
X1 +X2 +...+Xn
n
i
=
(V [X1 ] + V [X2 ] + . . . + V [Xn ]) =
1
V [X1
n2
+ X2 + . . . + Xn ] =
1
(σX2 +σX2 +. . .+σX2 )
n2
=
1
nσX2
n2
=
σX2
n
What must be true about the relationship between Xi0 s?
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Sampling Distributions
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Parameters of Sampling Distribution of Sample Mean
Sample mean X̄ is a random variable.
Distributed with parameters:
1
Mean: µX̄ = µX
2
Variance: σX̄2 =
σX2
n
3
St. dev.: σX̄ =
σX
√
n
Standard deviation of the sampling distribution is called
standard error.
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Shape of Sampling Distribution of Sample Mean
Use Central Limit Theorem to learn about the shape of sampling
distribution of sample mean.
Central Limit Theorem (CLT) implies that, no matter what the
underlying distribution, the sample mean will tend to a normal
distribution.
CLT states that the sum of n independent, identically distributed
random variables approaches a normal distribution as n increases.
Recall that the sample mean is just a sum of n independent,
identically distributed random variables divided by n. −→ Conditions
σx
for the CLT are satisfied −→ X̄ ∼ N(µx , √
)
n
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Sampling Distributions
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Sufficiently Large Sample Size
Rule of thumb:
Sample size at least 30 (n ≥ 30)
Normal approximation improves as n rises
n < 30 sufficient for modest departures from normal
If population is normal, then n ≥ 1 is sufficient
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Sampling Distributions
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Which one is a sampling distribution?
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What does sample size
affect?
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What happens to the standard
error as sample size goes up?
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Example
North American adults watch TV on average 6 hours per day with a
standard deviation of 1.5 hours.
If we randomly sample 5 adults, what is the chance that on average
they watch TV more than 7 hours a day?
Notation: P(X̄ > 7kµx = 6, σx = 1.5, n = 5) =?
Is it statistically possible that the average time watching television by
a random sample of 16 adults is less than 4.5 hours?
Notation: P(X̄ < 5kµx = 6, σx = 1.5, n = 16) =?
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Sampling Distributions
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TV Example
µx̄ = µx = 6
(a)
σx̄ =
σx
√
n
=
1.5
√
5
(b)
σx̄ =
σx
√
n
=
1.5
√
16
(Fall 2010)
= 0.67
= 0.375
Sampling Distributions
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TV Example
P(X̄ > 7) = P
=P Z >
7−6
0.67
X̄ −µx̄
σx̄
>
7−µx̄
σx̄
= P(Z > 1.49) = 1 − P(Z < 1.49)
= 1 − 0.9319 = 0.0681
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TV Example
P(X̄ < 4.5) = P
=P Z <
4.5−6
0.375
X̄ −µx̄
σx̄
<
4.5−µx̄
σx̄
= P(Z < −4)
Do we need table to find P(Z < −4)?
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Sampling Distributions
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Comparing Means of Two Populations
Suppose we want to compare the average salary of male and female
employees in the same job in the same industry.
Population one is the women, and population two is the men.
Population one has mean salary µ1 and standard deviation σ1 .
Population two has mean salary µ2 and standard deviation σ2 .
A random sample of size n1 from population one and n2 from
population two gives sample means of X¯1 and X¯2 and standard errors
σX¯1 and σX¯2
We are interested in µ2 − µ1 = 0?
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Comparing Means of Two Populations
We need to know the sampling distribution of difference between two
sample means.
Sample means are random variables, difference between two sample
means is a linear combination of two random variables.
A linear combination of independent normal random variables yields a
normal variable.
We can standardize the difference!
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Mean and Variance of Sampling Distribution of Difference
σ22
σ12
X̄2 ∼ N µ2 , n2
X̄1 ∼ N µ1 , n1
E [X̄1 − X̄2 ] = E [X̄1 ] − E [X̄2 ] = µ1 − µ2
V [X̄1 − X̄2 ] = 12 V [X̄1 ] + (−1)2 V [X̄2 ] =
q 2
σ
X̄1 − X̄2 ∼ N µ1 − µ2 , n11 +
σ22
n2
σ12
n1
+
σ22
n2
What is assumed about the relationship between two sample means?
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Example
The manager of a restaurant believes that waitresses who introduce
themselves by telling customers their names will get larger tips than those
who don’t. Specifically, she claims that the average tip for the former is
18% whereas that for the later is only 15% with a standard deviation 3%.
What is the chance that the average tip in a random sample of 10
waitresses who introduce themselves will exceed that of the random
sample of 10 waitresses who don’t?
Notation: P X¯1 − X¯2 > 0 =?
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Example
Mean = E [X̄1 − X̄2 ] = 18 − 15 = 3
Standard Error =
q
σ12
n1
+
σ22
n2
=
q
9
10
9
10
+
= 1.34
X̄1 − X̄2 ∼ N(3, 1.34)
P(X̄1 − X̄2 > 0) = P Z >
0−3
1.34
= P(Z > −2.24)
= 1 − P(Z < −2.24) = 1 − 0.0125 = 0.9875
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Example
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Example
What is the chance that in a random sample of those who do not
introduce themselves, the average tip will exceed that of those who
tell the customers their names?
X̄1 − X̄2 ∼ N(3, 1.34)
P(X̄1 − X̄2 < 0) =?
P(X̄1 − X̄2 < 0) = P Z <
(Fall 2010)
0−3
1.34
= P(Z < −2.24) = 0.0125
Sampling Distributions
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