2
Sampling Distribution And
Central Limit Theorem
2
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:
All of these means estimate the same unknown population mean .
These means are values of a random variable
x , 1 x , . . . , 2 x k
3
From mathematical statistics one could prove that this random variable follows the normal distribution with mean equals to (the population mean) and variance equals to 2
n
.
Population
sample # 1
sample # 2
sample # k n xk
n x2
n x1
X is a random variable
n μ, σ N
~ X
= ? N(, σ)
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 populationwith 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
n. i.e. :
X ~ N μ , σ n
Sampling distribution
of the sample mean
5
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. (
n
is known as the standard error of the mean) That is
N ( , ) X n
N , n
~ Approach
X
i.e. Distribution of
, n N
)
(X n
6
Central Limit Theorem
Population Sample
n) , ( N
~
X
X ~ ? (μ,σ)
n large:
(n>30)
n XX
.
7
Central Limit Theorem
Distribution of ( X ) :
n N N
X
,
~ X
small) / (large n sample )
, (
~ If ) (1
nknown
?
~ X
sample small ) , (
?
~ IF ) (
u X
2
.) . . ( ,
~ X
sample large ) , (
?
~ IF ) (
T L n C
N X
3
.
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) =
0.5764 0.1915 0.3849
0.5) Z P(0 1.2) Z P(0
1.2) Z P(-0.5
10 ) 80 Z 92 10
80 - P(75 92) X P(75
) 1 (
0.6534 0.2794 0.3770
0.77) Z P(0 1.16) Z P(0
1.16) Z P(-0.77
15) 10/
80 Z 83 15 10/
80 - P(78 83) X P(78
) 2 (
9
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:
0.8675 0.3962
0.4713
1.26) Z P(0 1.90) Z P(0
1.90) Z P(-1.26
40) 10/
80 Z 83
40 10/
80 - P(78 83) X P(78
10
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 n
15 = 90 36
90 90 2
5 = 5 = 324
n n n
11
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 = thepopulation size
population the
in character certain
has that elements of number The
This proportion can be estimated from a sample by Pˆ where:
ˆ The number of elements with that certain character in the sample
p = sample size
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 (ˆp) equals P, and standard deviation ( ˆp) equals to
n ) P 1 ( P
. For this approximation to be good, the following conditions should be met:
The sample size (n) is large:
(1) (n * p) > 5
(2) (n * q) > 5 , where q=1-p
Sampling distribution
of the Population Proportion (p)
13 When can we consider n as sufficiently large enough?
n p q np nq
10 .4 .6 4 6 Not large enough 15 .4 .6 6 9 Large enough 45 .1 .9 4.5 41.5 Not large enough 55 .1 .9 5.5 49.5 Large enough 90 .05 .95 4.5 85.5 Not large enough 105 .05 .95 5.25 99.75 Large enough
(2) (n * q) > 5 , where q=1-p
Sampling distribution
of the Population Proportion (p)
14
Application of the sampling distribution of
Example:
Ahmad is a broker in Kuwait stock market, if we know from his record that 65% of his deals are profitable. Let ˆp be the proportion in a random sample of 20 of his latest deals. Find the probability that the value of ˆp will be greater than 70%?
ˆp
p = 0.65 q = 0.35 n=20 P( ˆp > .70) = ?
Conditions: n * p = 20 * 0.65 = 13 >5 n * q = 20 * 0.35 = 7 >5
15 Example: cont.
ˆp= 0.65 and
ˆp= 0.65*0.35 0.114 0.1067 20
pq
n
ˆ 0.70 - 0.65 0.1067 ˆ
ˆ ˆ
p( 0.70) p p
p
p p
P( Z > 0.469 ) = 0.5 – P( 0 < Z < 0.469) = 0.5 – 0.1808
= 0.3192 = 31.92%
Application of the sampling distribution of ˆp
Estimator and Estimate
A sample Statisticsused 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 xis the estimator for and 43.7 is the estimate for it.
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 (i.e., for the case of the sample mean, if E( ) =
If not, (i.e. E( ) ≠ the Estimator is said to be Biased.
Biased and Unbiased Estimator
17 x
x
