Chapter 11 Systematic Sampling - IITK

Chapter 11 Systematic Sampling The systematic sampling technique is operationally more convenient than the simple random sampling. It also ensures at the same time that each unit has equal probability of inclusion in the sample. In this method of sampling, the first unit is selected with the help of random numbers and the remaining units are selected automatically according to a predetermined pattern. This method is known as systematic sampling. Suppose the N units in the population are numbered 1 to N in some order. Suppose further that N is expressible as a product of two integers n and k , so that N = nk .

To draw a sample of size n , -

select a random number between 1 and k .

-

Suppose it is i .

-

Select the first unit whose serial number is i .

-

Select every k th unit after i th unit.

-

Sample will contain i, i + k ,1 + 2k ,..., i + (n − 1)k serial number units.

So first unit is selected at random and other units are selected systematically. This systematic sample is called kth systematic sample and k is termed as sampling interval. This is also known as linear systematic sampling. The observations in the systematic sampling are arranged as in the following table: 1

2

3



i



k

1

y1

y2

y3



yi



yk

2

yk +1 

yk + 2 

yk + 3 

 

 

y( n −1) k +1

y( n −1) k + 2

y( n −1) k +3

yk + i 



y( n −1) k +i



y2k  ynk

Systematic sample number Sample composition 

n Probability

1 k

1 k

1 k



1 k



1 k

Sample mean

y1

y2

y3



yi



yk

Sampling Theory| Chapter 11 | Systematic Sampling | Shalabh, IIT Kanpur

Page 1

Example: Let N = 50 and n = 5. So k = 10. Suppose first selected number between 1 and 10 is 3. Then systematic sample consists of units with following serial number 3, 13, 23, 33, 43.

Systematic sampling in two dimensions: Assume that the units in a population are arranged in the form of m rows and each row contains nk units. A sample of size mn is required. Then -

select a pair of random numbers (i, j ) such that i ≤  and j ≤ k .

-

Select the (i, j )th unit, i.e., j th unit in i th row as the first unit.

-

Then the rows to be selected are i, i + , i + 2,..., i + (m − 1)

and columns to be selected are j , j + k , j + 2k ,..., j + (n − 1)k .

-

The points at which the m selected rows and n selected columns intersect determine the position of mn selected units in the sample.

Such a sample is called an aligned sample. Alternative approach to select the sample is independently select n random integers i1 , i2 ,..., in such that each of them is less than or equal to . - Independently select m random integers j1 , j2 ,..., jm such that each of them is less than or equal to k . - The units selected in the sample will have following coordinates: (i1 + r , jr +1 ), (i2 + r , jr +1 + k ), (i3 + r , jr +1 + 2k ),..., (in + r , jr +1 + (n − 1)k ) . Such a sample is called an unaligned sample. -

Under certain conditions, an unaligned sample is often superior to an aligned sample as well as a stratified random sample.

Advantages of systematic sampling: 1.

It is easier to draw a sample and often easier to execute it without mistakes. This is more advantageous when the drawing is done in fields and offices as there may be substantial saving in time.

2.

The cost is low and the selection of units is simple. Much less training is needed for surveyors to collect units through systematic sampling .

3.

The systematic sample is spread more evenly over the population. So no large part will fail to be represented in the sample. The sample is evenly spread and cross section is better. Systematic sampling fails in case of too many blanks.


Page 2

Relation to the cluster sampling The systematic sample can be viewed from the cluster sampling point of view. With n = nk , there are k possible systematic samples. The same population can be viewed as if divided into k large sampling

units, each of which contains n of the original units. The operation of choosing a systematic sample is equivalent to choosing one of the large sampling unit at random which constitutes the whole sample. A systematic sample is thus a simple random sample of one cluster unit from a population of k cluster units.

Estimation of population mean : When N = nk: Let yij : observation

on

the

unit

bearing

the

serial

number

i + ( j − 1)k

in

the

population,

i = 1, 2,..., k , j = 1, 2,..., n.

Suppose the drawn random number is i ≤ k . Sample consists of i th column (in earlier table). Consider the sample mean given by

ysy= y= i

1 n ∑ yij n j =1

as an estimator of the population mean given by Y = =

1 k n ∑∑ yij nk=i 1 =j 1 1 k ∑ yi . k i =1

Probability of selecting i th column as systematic sample =

1 . k

So ( ysy ) = E

1 k = ∑ yi Y . k i =1

Thus ysy is an unbiased estimator of Y .

Further, ( ysy ) = Var

1 k ( yi − Y ) 2 . ∑ k i =1


Page 3

Consider

( N − 1)= S2

k

n

∑∑ ( y

=i 1 =j 1 k

ij

− Y )2

n

∑∑ ( yij − yi ) + ( yi − Y ) 

=

2

=i 1 =j 1 k

n

k

∑∑ ( yij − yi )2 + n∑ ( yi − Y )2

=

=i 1 =j 1

=i 1

k

2 = k (n − 1) S wsy + n∑ ( yi − Y ) 2 i =1

where 2 = S wsy

k n 1 ∑∑ ( yij − yi )2 k (n − 1)=i 1 =j 1

is the variation among the units that lie within the same systematic sample . Thus N − 1 2 k (n − 1) 2 S − S wsy N N N − 1 2 (n − 1) 2 S − S wsy = N n ↓ ↓

Var = ( ysy )

Variation as a whole

Pooled within variation of the k systematic sample

with N = nk . This expression indicates that when the within variation is large, then Var ( yi ) becomes smaller. Thus higher heterogeneity makes the estimator more efficient and higher heterogeneity is well expected in systematic sample.

Alternative form of variance: 1 k ( ysy ) ( yi − Y ) 2 Var = ∑ k i =1  1 k 1 n = ∑  ∑ yij − Y  k i 1= = n j 1  =

2

 1 k  n ( yij − Y )  2 ∑ ∑ kn=i 1 = j 1 

n n  n  2 y Y − + ( ) ( yij − Y )( yi − Y )  ∑ ∑ ∑  ∑ ij i= 1  j= 1 j ( ≠ )= 1 = 1  k n n  1  2 nk S ( 1) ( yij − Y )( yi − Y )  . = − + ∑ ∑ ∑  2 kn  i = 1 j ( ≠ )= 1 = 1 

=

1 kn 2

k


Page 4

The intraclass correlation between the pairs of units that are in the same systematic sample is

ρw =

E ( yij − Y )( yi − Y ) E ( yij − Y )

2

;

−

1 ≤ ρw ≤ 1 nk − 1

k n n 1 ∑ ∑ ∑ ( yij − Y )( yi − Y ) nk (n − 1) i = 1 j ( ≠  ) = 1  = 1 . =  nk − 1  2  S  nk 

So substituting k

n

n

∑ ∑ ∑(y i = 1 j ( ≠ )= 1 = 1

ij

− Y )( yi − Y ) = (n − 1)(nk − 1) ρ w S 2

in Var ( yi ) gives nk − 1 S 2 Var (= ysy ) [1 + ρ w (n − 1)] nk n N −1 S 2 = [1 + ρ w (n − 1)]. N n

Comparison with SRSWOR: For a SRSWOR sample of size n , N −n 2 S Nn nk − n 2 = S Nn k −1 2 = S . N

Var ( ySRS ) =

Since

N −1 2 n −1 2 S − S wsy N n N = nk

= Var ( ysy )

 k −1 N −1  2 n −1 2 − Var ( ySRS ) − Var ( ysy ) = S wsy  S + N  n  N n −1 2 = ( S wsy − S 2 ). n Thus ysy is -

2 more efficient than ySRS when S wsy > S2 .

-

2 < S2. less efficient than ySRS when S wsy

-

2 equally efficient as ySRS when S wsy = S 2.


Page 5

Also, the relative efficiency of ysy relative to ySRS is RE =

Var ( ySRS ) Var ( ysy ) N −n 2 S Nn

=

N −1 2 S [1 + ρ w (n − 1) ] Nn  N −n  1 =   N − 1 1 + ρ w (n − 1)   n(k − 1)  1  ; (nk − 1) 1 + ρ w (n − 1) 

−

1 ≤ ρ ≤ 1. nk − 1

Thus ysy is -

more efficient than ySRS when ρ w < −

-

less efficient than ySRS when ρ w > −

-

equally efficient as ySRS when ρ w = −

1 nk − 1

1 nk − 1

1 . nk − 1

Comparison with stratified sampling: The systematic sample can also be viewed as if arising as a stratified sample. If population of N = nk units is divided into n strata and suppose one unit is randomly drawn from each of the strata. Then we get a stratified sample of size n . In doing so, just

consider each row of the following arrangement as a

stratum.

1

2

3



i



k

1

y1

y2

y3



yi



yk

2

yk +1 

yk + 2 

yk + 3 

 

 

y( n −1) k +1

y( n −1) k + 2

y( n −1) k +3

yk + i 



y( n −1) k +i



y2k  ynk

Systematic sample number Sample composition 

n Probability

1 k

1 k

1 k



1 k



1 k

Sample mean

y1

y2

y3



yi



yk


Page 6

Recall that in case of stratified sampling with k strata, the stratum mean

yst =

1 N

k

∑N j =1

j

yj

is an unbiased estimator of population mean.

Considering the set up of stratified sample in the set up of systematic sample, we have -

Number of strata = n

-

Size of strata = k (row size)

-

Sample size to be drawn from each stratum = 1

and yst becomes yst = =

1 n ∑ ky j nk j =1 1 n ∑ yj n j =1

Var ( yst ) =

1 n2

n

∑Var ( y ) j =1

j

N −n 2 1 n k −1 2  S j  using Var ( ySRS ) S  = 2 ∑ n j =1 k .1 Nn   =

k −1 n 2 ∑Sj kn 2 j =1

k −1 2 S wst nk N −n 2 S wst = Nn =

where = S 2j

1 k ∑ ( yij − y j )2 k − 1 i =1

is the mean sum of squares of units in the j th stratum. 2 = S wst

k n 1 n 2 1 = S ( yij − y j ) 2 ∑ ∑∑ j n=j 1 n(k − 1)=i 1 =j 1

is the mean sum of squares within strata (or rows). The variance of systematic sample mean is


Page 7

1 k ( yi − Y ) 2 ∑ k i =1

( ysy ) Var =

 1 k 1 n 1 n y yj  = − ∑ ∑  ∑ ij k i 1= n j1  = n j 1 =  1 k  n = 2 ∑  ∑ ( yij − y j )  n k=i 1 = j 1  =

2

2

k n n  1  k n 2 ( ) ( )( ) y y y y y y − + − − ∑∑ ∑∑∑  .   ij j ij j i n 2 k  =i 1 =j 1 =i 1 j ≠ = 1 

Now we simplify and express this expression in terms of intraclass correlation coefficient. The intraclass correlation coefficient between the pairs of deviations of units which lie along the same row measured from their stratum means is defined as

ρ wst =

E ( yij − Y )( yi − Y ) E ( yij − Y ) 2

k n n 1 ∑∑∑ ( yij − y j )( yi − y ) nk (n − 1)=i 1 j ≠= 1 = 1 k n ( yij − y j ) 2 ∑∑ nk=i 1 =j 1 k

=

n

n

∑∑∑

=i 1

j ≠= 1

( yij − y j )( yi − y )

2 ( N − 1)(n − 1) S wst

So 1 2 2 ( N − n) S wst  + ( N − n)(n − 1) ρ wst S wst 2  nk N −n 2 S wst [1 + (n − 1) ρ wst ] . (using N nk ) = = Nn Var ( y= sy )

Thus Var ( ysy ) − Var ( yst ) =

N −n 2 (n − 1) ρ wst S wst Nn

and the relative efficiency of systematic sampling relative to equivalent stratified sampling is given by

= RE

Var ( yst ) 1 = . Var ( ysy ) 1 + (n − 1) ρ wst

So the systematic sampling is -

more efficient than the corresponding equivalent stratified sample when ρ wst > 0 .

-

less efficient than the corresponding equivalent stratified sample when ρ wst < 0

-

equally efficient than the corresponding equivalent stratified sample when ρ wst = 0.


Page 8

Comparison of systematic sampling, stratified sampling and SRS with population with linear trend: We assume that the values of units in the population increase according to linear trend.

So the values of successive units in the population increase in accordance with a linear model so that yi =+ a bi, i = 1, 2,..., N .

Now we determine the variances of ySRS , ysy and yst under this linear trend.

Under SRSWOR V ( ySRS ) =

N −n 2 S . Nn

Here N = nk 1 N ∑i N i =1 1 N ( N + 1) = a+b N 2 N +1 = a+b 2

Y= a + b

1 N ∑ ( yi − Y )2 N − 1 i =1

= S2 =

N + 1 1 N  a + bi − a − b ∑  N − 1 i =1  2 

b2 N  N + 1  = ∑  i − 2  N − 1 i =1  =

2

2

2 b2  N 2  N +1   i N − ∑    N − 1  i =1  2  

b 2  N ( N + 1)(2 N + 1) N ( N + 1) 2  −  N − 1  6 4  = b2

N ( N + 1) 12

nk − n 2 nk (nk + 1) b 12 nk .n 2 b (k − 1)(nk + 1). = 12

Var ( ySRS ) =


Page 9

Under systematic sampling Earlier yij denoted the value of study variable with the j th unit in the i th systematic sample. Now yij represents the value of [i + ( j − 1)k ] unit of the population, so th

yij =a + b [i + ( j − 1)k ] , i =1, 2,..., k ; j =1, 2,..., n. ysy = yi Var = ( ysy ) yi =

1 k ( yi − Y ) 2 ∑ k i =1

1 n ∑ yij n j =1 1 n ∑ a + b {i + ( j − 1)k} n j =1 

=

 n −1  a + bi + k = 2  

 nk + 1   n −1  ( yi − Y= ) ∑ a + b  i + k −a −b ∑ 2  2   =i 1 =i 1  k

k

2

2

k  k +1  = b2 ∑  i −  2  i =1  2  k 2 k +1 k   k +1  2 2 =b  ∑ i + k  − ∑ i  2 i 1   2  =  i 1 = 2

 k (k + 1)(2k + 1)  k + 1 2 k (k + 1)  = b2  +   − (k + 1) 6 2   2   b2 k (k 2 − 1) = 12

1 b2 k (k 2 − 1) k 12 b2 2 = (k − 1). 12

= Var ( ysy )


Page 10

Under stratified sampling yij =a + b [i + ( j − 1)k ] , i =1, 2,..., k , j =1, 2,..., n yst =

k

1 N

∑N y i =1

i

N −n 2 k −1 2 = S wst S wst Nn nk

= Var ( yst )

1 n 2 ∑Sj n j =1

2 where S wst =

=

i

k n 1 ∑∑ ( yij − y j )2 n(k − 1)=i 1 =j 1

k n  1  k +1  = a + b {i + ( j − 1)k} − a − b  + ( j − 1)k  ∑∑  n(k − 1)=i 1 =j 1   2  k n b2  k +1 = ∑∑ i −  2  n(k − 1)=i 1 =j 1 

2

2

b 2 nk (k 2 − 1) 12 n(k − 1) k (k + 1) = b2 12 =

k − 1 2 k (k + 1) b 12 nk b2  k 2 − 1  =   12  n 

Var ( yst ) =

If k is large, so that

1 is negligible, then comparing Var ( yst ), Var ( ysy ) and V ( ySRS ), k

Var ( yst ) :

Var ( ysy ) :

Var ( ySRS )

or

k 2 −1 n

:

k 2 −1

:

(k − 1)(1 + nk )

or

k +1 n

:

k +1

:

nk + 1

or

k +1 : n(k + 1)

k +1 k +1

:

nk + 1 k +1

1

:

n



1 n

Thus Var ( yst ) : Var ( ysy ) : Var ( ySRS )

::

1 : 1 : n n

So stratified sampling is best for linearly trended population. Next best is systematic sampling. Sampling Theory| Chapter 11 | Systematic Sampling | Shalabh, IIT Kanpur

Page 11

Estimation of variance: As such there is only one cluster, so variance in principle, cannot be estimated. Some approximations have been suggested. 1.

Treat systematic sample as if it were a random sample. In this case, an estimate of variance is 1 1  2 (y = Var sy )  −  swc  n nk  1 n −1 2 where swc ( yi + jk − yi ) 2 . = ∑ n − 1 j =0

This estimator under-estimates the true variance. 2.

Use of successive differences of the values gives the estimate of variance as 2

n −1 1 1  1 (y ) = − Var ∑ ( yi + jk − yi +( j +1) k ) . sy    n nk  2(n − 1) j =0

This estimator is a biased estimator of true variance. 3.

Use the balanced difference of y1 , y2 ,..., yn to get the estimate of variance as n−2 yi + 2   yi  ( y ) = 1 − 1  1 − + Var y ∑ sy i + 1   2   n nk  5(n − 2) i  2 or

2

2

n−4 y  1  yi  ( y ) = 1 − 1  − yi +1 + yi + 2 − yi +3 + i + 4  . Var ∑ sy    2   n nk  15(n − 4) i  2

4.

The interpenetrating subsamples can be utilized by dividing the sample into C groups each of size y=

n . Then the group means are y1 , y2 ,..., yc . Now find c 1 c ∑ yt c t =1

(y ) = Var sy

c 1 ( yt − y ) 2 . ∑ c(c − 1) t =1


Page 12

Systematic sampling when N ≠ nk . When N is not expressible as nk then suppose N can be expressed as N= nk + p; p < k .

Then consider the following sample mean as an estimator of population mean

 1 n +1  n + 1 ∑ yij j =1  ysy= y= i n 1  ∑ yij  n j =1

if i ≤ p if i > p.

In this case  1  p  1 n +1  n  1 n yij  + ∑  ∑ yij   ∑  ∑ k  i = 1  n + 1 j= 1 1 p +1  n j =  i=  

= E ( yi )

≠ Y. So ysy is a biased estimator of Y .

An unbiased estimator of Y is ysy* = =

k N

∑y

ij

j

k Ci N

where Ci = nyi is the total of values of the i th column. k E (Ci ) N k 1 k = . ∑ Ci N k i =1

E ( ysy* ) =

=Y

Var ( ysy* ) =

k 2  k − 1  *2   Sc N2  k  2

where Sc*2 =

1 k  NY  ∑  nyi −  . k − 1 i =1  k 

Now we consider another procedure which is opted when N ≠ nk . [Reference: Theory of Sample Surveys, A.K. Gupta, D.G. Kabe, 2011, World Scientific Publishing Co.]


Page 13

When population size N is not expressible as the product of n and k , then let N = nq + r. Then take the sampling interval as  q k = q + 1 

n 2. n if r > 2

if r ≤

M  M Let   denotes the largest integer contained in . g g * If = k q= ( q or q + 1) , then the

 N  N  N   *  with probability  *  + 1 −  *   q  q  q  number of units expected in sample =    N  + 1 with probability  N  −  N  .  *   *  *   q  q   q 

If q = q* , then we get  r  r r  n +   with probability   + 1 −    q q q . n* =  n +  r  + 1 with probability  r  −  r        q   q  q 

Similarly, if q*= q + 1, then  n−r   (n − r )  n−r  +1−  n −   with probability     (q + 1)   q +1    q +1  * n = n +  n − r  + 1 with probability  n − r  −  (n − r )  .          q + 1   (q + 1)    q + 1  

n Example: Let N = 17 and n = 5. Then q = 3 and r = 2 . Since r < , k = q= 3. 2

Then sample sizes would be  r  r  r = = +1−   n +   5 with probability    q q q n* =   r  r  n +  r  + 1 6 with probability = =  −    q   q  q 

1 3 2 . 3


Page 14

This can be verified from the following example: Systematic sample number

Systematic sample

Probability

1

Y1 , Y4 , Y7 , Y10 , Y13 , Y16

1/3

2

Y4 , Y5 , Y8 , Y11 , Y14 , Y17

1/3

3

Y3 , Y6 , Y9 , Y12 , Y15

1/3

We now prove the following theorem which shows how to obtain an unbiased estimator of the population mean when N ≠ nk . Theorem: In systematic sampling with sampling interval k from a population with size N ≠ nk , an unbiased estimator of the population mean Y is given by k  n'  Yˆ =  ∑ y  N i where i stands for the i th systematic sample, i = 1, 2,..., k and n' denotes the size of i th systematic sample. Proof. Each systematic sample has probability

1 . Hence k

k 1 k  n'  E (Yˆ ) = ∑ .  ∑ y  i =1 k N  i

=

1 N

 n'  ∑ ∑ y . i =1  i k

Now, each unit occurs in only one of the k possible systematic samples. Hence N  n'  Yi , ∑  ∑ y  = ∑ =i 1 = i 1 i k

which on substitution in E (Yˆ ) proves the theorem. When N ≠ nk , the systematic samples are not of the same size and the sample mean is not an unbiased estimator of the population mean. To overcome these disadvantages of systematic sampling when N ≠ nk , circular systematic sampling is proposed. Circular systematic sampling consists of selecting a

random number from 1 to N and then selecting

the unit corresponding to this random number.

Thereafter every k th unit in a cyclical manner is selected till a sample of n units is obtained, k being the nearest integer to

N . n


Page 15

In other words, if i is a number selected at random from 1 to N , then the circular systematic sample consists of units with serial numbers i + jk , if i = jk ≤ N  j 0,1, 2,..., (n − 1). = i + jk − N , if i = jk > N  This sampling scheme ensures equal probability of inclusion in the sample for every unit.

Example: Let N = 14 and n = 5. Then, k = nearest integer to

14 = 3. Let the first number selected at random 5

from 1 to 14 be 7. Then, the circular systematic sample consists of units with serial numbers 7,10,13,

16-14=2,

19-14=5.

This procedure is illustrated diagrammatically in following figure.

1 12 2 13 3

12 4

11 5

6

10 7

9 8


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Theorem: In circular systematic sampling, the sample mean is an unbiased estimator of the population mean. Proof: If i is the number selected at random, then the circular systematic sample mean is 1 n  y = ∑ y , n i  n  where  ∑ y  denotes the total of y values in the i th circular systematic sample, i = 1, 2,..., N . We  i

note here that in circular systematic sampling, there are N circular systematic samples, each having probability

1 of its selection. Hence, N

1 n  1 1 N  n  = × y ∑ n  ∑  N Nn ∑ ∑ y i 1 =i 1 = i i = E( y )

N

Clearly, each unit of the population occurs in n of the N possible circular systematic sample means. Hence, N  n  y n Yi , = ∑  ∑  ∑ 1 =i 1 = i i N

which on substitution in E ( y ) proves the theorem.

What to do when N ≠ nk One of the following possible procedures may be adopted when N ≠ nk . (i)

Drop one unit at random if sample has (n + 1) units.

(ii)

Eliminate some units so that N = nk .

(iii)

Adopt circular systematic sampling scheme.

(iv)

Round off the fractional interval k .


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Chapter 11 Systematic Sampling - IITK

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