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(2) Sampling Distribution & Central Limit
Theorem
Prof. Mohammad Almahmeed
2
Sampling Distribution
And
Central Limit Theorem
Sampling distribution
of the sample mean
 If we sample a number of samples (say k samples where k
is very large number) each of size n, from a normally
distributed population with mean  and standard deviation
.
 And compute the mean of each of these samples.
 We will have different sample mean for each sample:
x1 , x 2 , . . . , xk
 All of these means estimate the same unknown population
mean .
 These means are values of a random variable
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QMIS 220
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(2) Sampling Distribution & Central Limit
Theorem
Prof. Mohammad Almahmeed
Sampling distribution
of the sample mean
 From mathematical statistics one could prove that this random
variable follows the normal distribution with mean equals to 
2
(the
population mean) and variance equals to 
.
Population
n
x1
=?
N(, σ)
n
x2
n
sample # 1
sample # 2

n
xk
sample # k
X is a random variable

σ 
X ~ N μ,

n

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Sampling distribution
of the sample mean
In other words, if x is the mean of a sample of size n
taken from a normally distributed population with mean
() and standard deviation (σ) [i.e. X~N(,σ)] . Then X
is a random variable that follows the normal
distribution with mean  and standard deviation
σ
. i.e. :
n
σ 

X~N  μ ,

n 

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QMIS 220
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(2) Sampling Distribution & Central Limit
Theorem
Prof. Mohammad Almahmeed
Central Limit Theorem
If X is a random variable that follows any distribution
(known or unknown) but with mean (μ) and Standard
deviation (σ). If X is the mean of a sample of size n (n
large i.e n > 30). Then the distribution of X will approach
the normal distribution with mean μ and Standard


deviation n . (
is known as the standard error of the mean)
n
That is
X . N (  ,
 

X ~ Approach
  N   ,

n


n
)
 

 N  ,

i.e. Distribution of ( X ) n

n

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Central Limit Theorem
Population
n large:
Sample
(n>30)
X ~ ? (μ,σ)
X
X
n
X .~ N(,

)
n
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QMIS 220
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(2) Sampling Distribution & Central Limit
Theorem
Prof. Mohammad Almahmeed
Central Limit Theorem
 Distribution of ( X ) :
(1) If
X ~ N (  ,  )  sample n (large / small)
 

 X ~ N  ,

n

(2)
IF
X ~ ? ( , ) 
small
sample
 X ~ ? u nknown
( 3) IF X ~ ? (  ,  )  large sample
 

.
 X ~ N  ,

n

(C . L.T .)
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Sampling distribution of the sample mean
and Central Limit Theorem
Example - 1
X is a random variable that follows the normal distribution with mean 80 and
standard deviation equal to 10. X is the mean of a random sample of size 15
taken from that population. Find:
(1) P(75 < X < 92) =
(2) P(78 < X <83) =
75 - 80
92  80
Z
)
10
10
 P(-0.5  Z  1.2)
 P(0  Z  1.2)  P(0  Z  0.5)
 0.3849  0.1915  0.5764
(1) P(75  X  92)  P(
78 - 80
83  80
Z
)
10/ 15
10/ 15
 P(-0.77  Z  1.16)
(2) P(78  X  83)  P(
 P(0  Z  1.16)  P(0  Z  0.77)
 0.3770  0.2794  0.6534
QMIS 220
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4
(2) Sampling Distribution & Central Limit
Theorem
Prof. Mohammad Almahmeed
Sampling distribution of the sample mean
and Central Limit Theorem
Example - 2
For the same previous example if the distribution of X is unknown. And we took a
sample of size 40. find:
(1) P(75 < X < 92) =
(2) P(78 < X <83) =
(1) we can not find the required probability since the distribution of X is
unknown.
(2) Since the sample size is large enough we will apply the central limit theorem
to have:
78 - 80
83  80
Z
P(78  X  83)  P(
)
10/ 40
10/ 40
 P(-1.26  Z  1.90)
 P(0  Z  1.90)  P(0  Z  1.26)
 0.4713  0.3962  0.8675
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Sampling distribution of the sample mean
and Central Limit Theorem
Example - 3
If the standard error of the mean for a sample of 36 is 15. In order to
decrease the standard error of the mean to 5 what should be the value
of n (the sample size)?
the standard error of the mean is  x 

15 
5
36
90
n

n
  = 90
90
 n =  
 5
2
n = 324
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QMIS 220
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(2) Sampling Distribution & Central Limit
Theorem
Prof. Mohammad Almahmeed
Sampling distribution
of the Population Proportion (p)
The proportion of any incident in a population is the number of elements in the
population that belong to that incident divided by the total number of elements
in the population. “P” usually denotes this proportion.
P=
The number of elements that has certain character in the population
the population size
This proportion can be estimated from a sample by P̂ where:
pˆ =
The number of elements with that certain character in the sample
sample size
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Sampling distribution
of the Population Proportion (p)
The population of sample proportion has a Binomial distribution. But when the
sample size is large the population of all possible sample proportions has
approximately normal distribution, with mean (
 p̂ ) equals P, and standard
P(1  P )
. For this approximation to be good, the
n
following conditions should be met:
deviation (
 p̂ ) equals to
The sample size (n) is large:
(1) (n * p) > 5
(2) (n * q) > 5 , where q=1-p
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QMIS 220
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(2) Sampling Distribution & Central Limit
Theorem
Prof. Mohammad Almahmeed
The sample size (n) is large when:
(1) (n * p) > 5
(2) (n * q) > 5 , where q=1-p
Sampling distribution
of the Population Proportion (p)
When can we consider n as sufficiently large enough?
n
10
15
45
55
90
105
p
.4
.4
.1
.1
.05
.05
q
.6
.6
.9
.9
.95
.95
np
4
6
4.5
5.5
4.5
5.25
nq
6
9
41.5
49.5
85.5
99.75
Not large enough
Large enough
Not large enough
Large enough
Not large enough
Large enough
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Application of the sampling
distribution of p̂
Example:
Ahmad is a broker in Kuwait stock market, if we know from his record
that 65% of his deals are profitable. Let p̂ be the proportion in a
random sample of 20 of his latest deals. Find the probability that the
value of p̂ will be greater than 70%?
p = 0.65
q = 0.35
P( p̂ > .70) = ?
Conditions:
n=20
n * p = 20 * 0.65 = 13 >5
n * q = 20 * 0.35 = 7 >5
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QMIS 220
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(2) Sampling Distribution & Central Limit
Theorem
Prof. Mohammad Almahmeed
Application of the sampling
distribution of p̂
Example: cont.
 p̂ = 0.65
 p̂ =
and
pq
0.65* 0.35

 0.114  0.1067
n
20
 pˆ   pˆ
p(pˆ  0.70)  p 

  pˆ


0.70 - 0.65
0.1067 

P( Z > 0.469 ) = 0.5 – P( 0 < Z < 0.469) = 0.5 – 0.1808
= 0.3192 = 31.92%
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Estimator and Estimate
A sample Statistics used to estimate a population parameter is
called an Estimator
The value(s) assigned to a population parameter based on the
value of a sample statistics is called an Estimate
To estimate the population mean  we took a random sample of
size 22. The computed value of the sample mean x is 43.7
Here x is the estimator for  and 43.7 is the estimate for it.
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QMIS 220
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(2) Sampling Distribution & Central Limit
Theorem
Prof. Mohammad Almahmeed
Biased and Unbiased Estimator
An Estimator of the population parameter is said to be Unbiased
estimator when the expected value (or the mean) of this estimator is equal to the value of the corresponding population parameter x
(i.e., for the case of the sample mean, if E( ) = 
x
If not, (i.e. E( ) ≠ the Estimator is said to be Biased
.
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QMIS 220
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