Download 06 SAMPLING DISTRIBUTIONS

Quantitative Methods
Varsha Varde
Sampling Distributions
• Concept
• The Central Limit Theorem
• The Sampling Distribution of the Sample
Mean
• The Sampling Distribution of the Sample
Proportion
• The Sampling Distribution of the Difference
Between Two Sample Means
• The Sampling Distribution of the Difference
Between Two Sample Proportions
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Sample
Population
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A number describing a population
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A number describing a sample
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Random Sample
Every unit in the population has
an equal probability of being
included in the sample
Common Sense Thing #1
A random sample should
represent the population well, so
sample statistics from a random
sample should provide reasonable
estimates of population
parameters
Common Sense Thing #2
All sample statistics have some
error in estimating population
parameters
Common Sense Thing #3
If repeated samples are taken
from a population and the same
statistic (e.g. mean) is calculated
from each sample, the statistics
will vary, that is, they will have a
distribution
The probability distribution of a statistic is
called its
sampling distribution.
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Mean and Standard Deviation
of X
mean =  x  
and

standard deviation = x 
n
Distribution of X when
sampling from a normal
distribution
X has a normal distribution with
mean =  x  
and

standard deviation =  x 
n
The Central Limit Theorem (CLT
• Whatever be the probability distribution
of the population from which sample is
drawn the probability distribution of
sample mean will approximately normal
if size of sample is large ;generally
larger than 30
• Strictly speaking CLT was derived only
for sample means but it applies also for
sample totals and sample proportions
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Central Limit Theorem
If the sample size (n) is large enough,
has a normal distribution with
mean =  x  
and

standard deviation =  x 
n
regardless of the population
distribution
X
n  30
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Does X have a normal
distribution?
Is the population normal?
Yes
No
Is n  30?
X is normal
Yes
X is considered to be
normal
Varsha Varde
No
X may or may not be
considered normal
(We need more info)
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The Sampling Distribution of the
Sample Mean
• Suppose X is a random variable with
mean µ and standard deviation  .
• (i) What is the the mean (expected value)
and standard deviation of sample mean
x¯?
• Answer: μx ¯ = E(x ¯ ) = µ
σ x ¯ = S.D.(x¯ ) =  /√n
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• (ii)What is the sampling distribution of the
sample mean x ¯?
• Answer: x¯ has a normal distribution with
mean µ and standard deviation  /√n ,
• Equivalently
• Z =(x ¯ − μx¯ )/σ x ¯
• Z = (x¯ − µ ) /  /√n
• provided n is large (i.e. n ≥ 30)
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Example.
• Consider a population, X, with mean μ = 4
and standard deviation σ = 3.
• A sample of size 36 is to be selected.
• (i) What is the mean and standard
deviation of X¯?
• (ii) Find P(4 <X ¯ <5),
• (iii) Find P(X ¯ >3.5), (exercise)
• (iv) Find P(3.5 ≤ X ¯ ≤ 4.5). (exercise)
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The Sampling Distribution of the
Sample Proportion
• Suppose the distribution of X is binomial
with parameters n and p.
• (ii) What is the the mean (expected value)
and standard deviation of sample
proportion p ¯ ?
• Answer:μP ¯ ˆ = E(p ¯ ) = p
σ P ¯ = S.E.(p ¯ ) = √pq/n
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• What is the sampling distribution of the
sample proportion p ¯?
• Answer: p ¯ has a normal distribution with
mean p and standard deviation √pq /n ,
• Equivalently
• Z =(p ¯ − μ P ¯ )/σ P ¯
• Z = (p ¯ − p) / √pq/n
• provided n is large (i.e. np ≥ 5, and nq ≥ 5)
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Example.
• It is claimed that at least 30% of all adults favour
brand A versus brand B.
• To test this theory a sample n = 400 is selected.
Suppose 130 individuals indicated preference for
brand A.
• DATA SUMMARY: n = 400, x = 130, p = .30,
p ¯ = 130/400 = .325
• (i) Find the mean and standard deviation of the
sample proportion p ¯.
• Answer:
• μp ¯ = p = .30
• σp ¯ =√pq/n = .023
• (ii) Find Prob( p ¯ > 0.30)
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Comparing Two Sample Means
• E(X ¯ 1 − X ¯ 2) = μ1 − μ2
• σX ¯ 1−X ¯ 2 =√(σ21/n1 +σ22/n2)
( X ¯ 1 − X ¯ 2) − (μ1 − μ2)
• Z = ---------------------------√ σ12/n1+ σ22/n2
• provided n1, n2 ≥ 30.
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Comparing Two Sample Proportions
• E( P ¯ 1 - P ¯ 2) = p1 - p2
• σ2P ¯ 1- P ¯ 2 = p1q1/n1+p2q2/n2
•
(P ¯ 1 - P ¯ 2) - (p1 - p2)
• Z= -------------------------√ p1q1/n1+ p2q2/n2
provided n1 and n2 are large.
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