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
Now we determine the variances of under this linear trend.
Under SRSWOR
Under systematic sampling Earlier yij denoted the value of study variable with the jth unit in the ith systematic sample. Now yij represents the value of [i + ( j− 1)k ]th unit of the population, so
Under stratified sampling
If k is large, so that 1/k is negligible, then comparing
Thus,
So stratified sampling is best for linearly trended population. Next best is systematic sampling.
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
This estimator under-estimates the true variance.
2. Use of successive differences of the values gives the estimate of variance as
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
4. The interpenetrating subsamples can be utilized by dividing the sample into C groups each of
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
In this case
So is a biased estimator of .
An unbiased estimator of is
where is the total of values of the ith column.
Now we consider another procedure which is opted when N ≠ nk.
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
Let denotes the largest integer contained in
number of units expected in sample
If q = q*, then we get
Similarly if = q* + 1, then
Example: Let N = 17 and n = 5. Then q = 3 and r = 2 . Since
Then sample sizes would be
This can be verified from the following example:
Systematic sample number | Systematic sample | Probability |
1 | 1/3 | |
2 | 1/3 | |
3 | 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 is given by
where i stands for the i th systematic sample, i = 1, 2, ..., k andn ' denotes the size of ith systematic sample.
Proof. Each systematic sample has probability1/k. Hence
Now, each unit occurs in only one of the k possible systematic samples. Hence
which on substitution in 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 kth unit in a cyclical manner is selected till a sample of n units is obtained, k being the nearest integer to N/n
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
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 14/5 = 3 Let the first number selected at random 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.
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
where denotes the total of y values in the ith circular systematic sample, i = 1, 2, ...,N . We note here that in circular systematic sampling, there are N circular systematic samples, each having probability 1/N of its selection. Hence,
Clearly, each unit of the population occurs in n of the N possible circular systematic sample means. Hence,
which on substitution in E () 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 .
556 videos|198 docs
|
1. What is systematic sampling? |
2. How is systematic sampling different from random sampling? |
3. What are the advantages of using systematic sampling? |
4. What are the limitations of systematic sampling? |
5. How can systematic sampling be used in CSIR-NET Mathematical Sciences Mathematics? |
556 videos|198 docs
|
|
Explore Courses for Mathematics exam
|