# On the variance of the sample mean from finite population

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Journal of Scientific Research

Volume XXXIV No. 2 October 2005

ON THE VARIANCE OF THE SAMPLE MEAN
FROM FINITE POPULATION
Syed Shakir Ali Ghazali1*, Ghausia Masood Gilani**, Muhammad Hussain Tahir***
*Department of Statistics, Govt. S.E. College, Bahawalpur.
**Institute of Statistics, Punjab University, Lahore
***Department of Statistics, Islamia University, Bahawalpur

ABSTRACT
Analternate proof of the variance of the sample mean in case of simple random
sampling without replacement (SRSWOR) is obtained. This proof is very simple
and avoids the use of expectation.
Key words: Sample mean, simple random sampling, variance, without replacement

sampling.
1. INTRODUCTION
Simple random sampling from a finite population
has

attracted

much

of

APPROACH I:

theresearchers and

Barnett (2002, p.32-35) has given the proof of the

practitioners working in surveys. It is the simplest,

variance of mean as follows:

most preferable and widely used probability

If a sample of size n is drawn from a finite

sampling technique. The variance or standard error

population of size N having y1 , y1 , L , y N units,

of the sampling distribution ofmean serves as a

N
n

then there are K =   distinct samples each


basis for efficiency comparison with other sampling
methods like stratified random sampling, systematic

−1

N
having the same probability   .
n


sampling, cluster sampling etc. The variance of
sample mean in case of SRSWOR has been

Let y i , i = 1, 2, L , n be the ith chosen member, thendiscussed by Hansen et al. (1953), Murthy (1967),

the probability for obtaining this ordered sequence

Sukhatme and Sukhatme (1970), Cochran (1977),
Jessen (1978), Singh and Chaudhary (1986),

is

But the probability for obtaining any particular set

(1998), Govindarajulu (1999), Barnett (2002) and

of n distinct population members(irrespective of

Sampath (2005). In literature, three different

−1

approaches for the variance of sample mean in

the order) is just

simple random sampling without replacement
(SRSWOR) are available, which are given below:
1

( N − n)!
1
1
1
1
.
L
=
.
.
N N −1 N − 2 N − n +1
N!

For proof and correspondence: S. S. A. Ghazali

19

n! ( N − n)!  N 
=  .
n
N!
 Journal of Scientific Research

Volume XXXIV No. 2 October 2005

Then the estimator of Y based on simple random

Consider

sample of size n is

y=

n( y − Y ) = ( y1 − Y ) + ( y 2 − Y ) + L + ( y n − Y )

1n
∑ yi ,
n i =1

But

[

which is an unbiased estimator of Y .

E ( y1 − Y ) 2 + L + ( y n − Y ) 2

Now for variance
1
Var ( y ) = Var 
n

=

1
n

2n

i =1

=

yi 

2

i

2

i =1

i

j)

E ( y i2 ) =

1
N

( y1 − Y )( y 2 − Y )

n(n − 1) 
=
+ ( y1 − Y )( y 3 − Y )

N ( N − 1) 

+ L + ( y N −1 − Y )( y N − Y )

=Y ,

i =1
N

∑Y

i

2

and

units in the sample and population respectively. The

N

∑Y Y
i

j

(i ≠ j )

sum on the left contains

i< j

We getVar ( y i ) =

(1. 5)

Here the sum of products extends over all pair of

i =1

2
E( yi y j ) =
N ( N − 1)

(1.4)

( y1 − Y )( y 2 − Y ) + ( y1 − Y )( y 3 − Y )
E

+ L + ( y n −1 − Y )( y n − Y )

(1.1)

i< j

N

i

]

n

∑ Var ( y ) + n ∑ Cov ( y , y
∑Y

n
( y1 − Y ) 2 + L + ( y N − Y ) 2
N

Also

n

Using the results on Barnett (2002, p34)that
1
E( yi ) =
N

[

]

(1.3)

( N − 1) 2
S2
S and Cov ( y i , y j ) = −
N
N

right contains

(1. 2)

n(n − 1)
and that on the
2

N ( N − 1)
terms.
2

Now squaring (1.3) and averaging over all simple

Hence by using (1.2) in (1.1), we have

random samples. We get by using (1.4) and (1.5),

1  n ( N − 1) S 2 n (n − 1) S 2 

Var ( y ) = 2 

N
N
...