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Simple Random Sampling and Systematic Sampling
Simple random sampling and systematic sampling provide the foundation for almost all of the more complex sampling designs based on probability sampling. They are also usually the easiest designs to implement. These two designs highlight a trade‐offs inherent in selecting a sampling design: to select sample units at random to minimize the risk of introducing biases into the sample or to select samples systematically to ensure that sample units are well‐distributed throughout the population. Both designs involve selecting n sample units from the N units available in the population and can be implemented with or without replacement.
Simple Random Sampling
When the population of interest is relatively homogeneous then simple random sampling works well, which means it provides estimates that are unbiased and have high precision. When little is known about a population in advance, such as in a pilot study, simple random sampling is a common design choice. Advantages: Easy to implement Requires little knowledge of the population in advance Disadvantages: Imprecise relative to other designs if the population is heterogeneous More expensive than other designs if entities are clumped and the cost to travel among units is appreciable How it is implemented: Select n sample units at random from N available in the population All units within the sampling universe must have the same probability of being selected, therefore each and every sample of size n drawn from the population has an equal chance of being selected. There are many strategies available for selecting a random sample. For large populations, this often involves generating pseudorandom numbers with a computer and for small populations it might involve using a table of random numbers or even writing a unique identifier for every sample unit in the population on a scrap of paper, placing those numbers in a jar, shaking it, then selecting n scraps of paper from the jar blindly. The approach used for selecting the sample matters
little provided there are no constraints on how the sample units are selected and all units have an equal chance of being selected.
Estimating the Population Mean
The population mean ( μ ) is the true average number of entities per sample unit and is estimated with
the sample mean ( ˆ^ or y ) which has an unbiased estimator:
n
y
n i
i
ˆ ^1
where y (^) i is the value from each unit in the sample and n is the number of units in the sample. The population variance ( σ^2 ) is estimated with the sample variance ( s^2 ) which has an unbiased estimator:
1 2 2
n
y y
s
n i i
Variance of the estimate ˆ^ is:
n
s
N
N n^2
va ˆr(ˆ)
The standard error of the estimate is the square root of variance of the estimate, which as always is the standard deviation of the sampling distribution of the estimate. Standard error is a useful gauge of how precisely a parameter has been estimated.
Standard error of ˆ^ is:
n
s
N
N n
SE
2
The quantity
N
N n
is the finite population correction factor which adjusts variance of the estimator (not variance of the population which does not change with n ) to reflect the amount of information that is known about the population through the sample. Practically, the correction factor reflects the proportion of the population that remains unknown. Therefore, as the sample size n approaches the population size N , the finite population correction factor approaches zero, so the amount of variation associated with the estimate also approaches zero. When the sample size n is large relative to the population size N , the fraction of the population being sampled n / N is small, so the correction factor has little effect on the estimate of variance (Fig. 2 ‐ FPC.xls). If the finite population correction factor is ignored, including those cases where N is unknown, FPC with N = 100 0
1 0 20 40 60 80 100 n FPC
The estimated variance of the sample mean: var( )
y .
The estimated standard error of the mean is: 58 06. 7 62..
An estimate of the total number of caribou in the area is: 286 25 9333(. ) 7417
An estimate of variance of the estimated total is: var( ) 286 ( 58 0576. ) 4 748 879, ,
2
The estimated standard error of the total is: 4 748 879, , 2179
Estimating a Population Proportion
If there is interest in the composition of a population, we could use a simple random sample to estimate the proportion of the population p that is composed of elements with a particular trait, such as the proportion of plants that flower in a given year, the proportion of juvenile animals captured, the proportion of females in estrus, and so on. We will consider only classifications that follow binomial trials which means that either an element in the population has the trait of interest (flowering) or not (not flowering) although extending this idea to more complex settings is straightforward. In the case of simple random sampling, the population proportion follows the mean exactly; that is, p = μ. If this idea is new to you, convince yourself by working through an example. Say we generate a sample of 10 elements, where 4 have a value of 1 and 6 have a value of 0 (1 = presence of a trait, 0 = absence of a trait). The proportion of the sample with the trait is 4/10 or 0.40 and so is the arithmetic mean, which = 0.40 ([1+1+1+1+0+0+0+0+0+0]/10 = 4/10). Cosmic.
It follows that the population proportion ( p ) is estimated with the sample proportion ( p ˆ ) which has an
unbiased estimator:
n
y
p
n i
i
ˆ ˆ ^1.
Because we are dealing with dichotomous proportions (sample unit does or does not have the trait), the population variance σ^2 is computed based on variance for a binomial which is the proportion of the population with the trait ( p ) times the proportion that does not have that trait (1 – p ) or p (1 – p ). The
estimate of the population variance s 2 is: p ˆ^ ( 1 p ˆ).
Variance of the estimate p ˆ is:
vaˆr(ˆ)
2
n
p p
N
N n
n
s
N
N n
p.
Standard error of p ˆ is:
2
n
p p
N
N n
n
s
N
N n
SE p.
Systematic Sampling
Occasionally, selecting sample units at random can introduce logistical challenges that preclude collecting data efficiently. If the chance of introducing a bias is low or if ideal dispersion of sample units in the population is a higher priority that a strictly random sample, then it might be appropriate to choose samples non‐randomly. Like simple random sampling, systematic sampling is a type of probability sampling where each element in the population has a known and equal probability of being selected. The probabilistic framework is maintained through selection of one or more random starting points. Although sometimes more convenient, systematic sampling provides less protection against introducing biases in the sample compared to random sampling. Estimators for systematic sampling and simple random sampling are identical; only the method of sample selected differs. Therefore, systematic sampling is used to simplify the process of selecting a sample or to ensure ideal dispersion of sample units throughout the population. Advantages: Easy to implement Maximum dispersion of sample units throughout the population Requires minimum knowledge of the population Disadvantages: Less protection from possible biases Can be imprecise and inefficient relative to other designs if the population being sampled is heterogeneous How it is implemented: Choose a starting point at random Select samples at uniform intervals thereafter
1 ‐in‐k systematic sample
Most commonly, a systematic sample is obtained by randomly selecting 1 unit from the first k units in the population and every k th^ element thereafter. This approach is called a 1 ‐in‐k systematic sample with a random start. To choose k so than a sample of appropriate size is selected, calculate: k = Number of units in population / Number of sample units required For example, if we plan to choose 40 plots from a field of 400 plots, k = 400/40 = 10, so this design would be a 1 ‐in‐ 10 systematic sample. The example in the figure is a 1 ‐in‐ 8 sample drawn from a population of N = 300; this yields n = 28. Note that the sample size drawn will vary and depends on the location of the first unit drawn.
Variance of the estimate p ˆ^ is:
vaˆr(ˆ)
2
n
p p
N
N n
n
s
N
N n
p.
How Many Samples?
Optimal allocation is an approach to maximize sampling efficiency; that is to provide high precision for a given amount of sampling effort. A different question is how many samples should we take from the population? First, establish the degree of precision required, B , the bound the error of estimation, which is the half‐ width of the confidence interval we wish to attain from sampling. Determine the sample size n required
by setting Z × SE( y ) equal to B and solving this expression for n.
Z is a constant that denotes the upper α/2 point of the standard normal distribution where α is the same value used to establish the width of confidence intervals. Population Mean
For simple random sampling, set: B Z
N n
N n
2
solve for n to get: n
n N
n
z
B
0 0 2 2
or n
z
B
N
2 2 2
Note that if n will be small relative to N , the population correction factor can be ignored, and the formula for sample size reduced to n 0. Example: Estimate the average body mass of male freshman μ on campus. Assume that no prior information exists with which to estimate population variance σ^2 but we know that the mass of most male freshmen is within a range of about 100 pounds and there are N = 1000 students. How many samples are needed to estimate μ with a bound on the error of estimation B = 3 pounds using simple random sampling? Although it is best to have data with which to estimate σ^2 , perhaps from a small pilot study, the range is often approximately equal to 4 σ, so one‐fourth of the range might be used as an approximate value of σ:
range
Substituting: n
2 2 2
Therefore, about 218 samples are needed to estimate μ with a bound on the error of estimation B = 3 pounds. Population Total
For simple random sampling, solve for n from: B Z N N ^ n
n
2
n
n N
n
N z
B
0 0 2 2 2
or n
N z
B
N
2 2 2 2
Again, if N is large relative to n , the population correction factor can be ignored, and the formula for sample size reduced to n 0. Example: What sample size is necessary to estimate the caribou population we examined to within B = 2000 animals of the true total with 90% confidence (α = 0.10). Using s 2 = 919 from earlier and Z = 1.645, which is the upper α = 0.10/2 = 0.05 point of the normal
distribution: n 0
2 2 2
To adjust for finite population size: n
How it is implemented: Divide the entire population into non‐overlapping strata Selected a simple random sample from within each strata L = number of strata N (^) i = number of sample units within stratum i N = number of sample units in the population
Estimating the Population Mean
Estimates from stratified random samples are simply the weighted sum of estimates from a series of simple random samples, each generated within a unique stratum. This should be apparent in the estimators below, such as that for the population mean, which is an average of the means from each stratum weighted by the number of sample units measured within each stratum. With only one stratum, stratified random sampling reduces to simple random sampling. The population mean ( μ ) is estimated with:
L i
L L Ni i
N
N N N
N 1
Variance of the estimate ˆ^ is again just a weighted average of estimates from a series of random
samples, although it looks a bit cumbersome:
L i (^) i i i i i L L L L L L
n
s
N
N n
N
n N
s
N
N n
N
n
s
N
N n
N
N 1
2 2 2 1 2 2 1 2 1 1 2 1 1 2 1
va ˆr( ˆ)
Standard error of ˆ^ is:
L i (^) i i i i i
n
s
N
N n
N
N
SE
1 2 2 2 1
Estimating the Population Total
Like the mean, estimating a total for a stratified random sample is a matter of summing individual
estimates of the total from each stratum, Ni ˆ i^.
The population total is estimated with:
L i
N N NL L Ni i
1
Variance of the estimated total ˆ^ is:
L i (^) i i i i i i
n
s
N
N n
N N
1 2 2 2
vaˆr(ˆ) vaˆr( ˆ)
Standard error of ˆ^ is the square root of vaˆr( ˆ ).
Estimating the Population Proportion
Estimating the proportion of the population with a particular trait ( p ) using stratified random sampling involves combining estimates from multiple simple random samples, each generated within a stratum. The population proportion is estimated with the sample proportion:
L i
p N p N p NL pL Nipi
1
Variance of the estimate p ˆ^ is:
L i (^) i i i i i i i L i i i
n
p p
N
N n
N
N
N p
N
p
1 2 2 1 2 2
1 ˆ(^1 ˆ)
vaˆr(ˆ )
vaˆr(ˆ)
Standard error of p ˆ^ is the square root of va ˆr( p ˆ).
Example: Simple example of 12 samples taken from a population of 41 entities.
Stratum ( i ) N i ni y^ si^2
Estimate of the population mean: y ^
Estimate of the population total = 41 × 1.57 = 64.4. Estimated variance of the estimated population mean is:
var( ) ( )
y .
^
Estimated variance of the estimated population total = 412 × 0.192 = 322.8.
Allocating Sampling Effort among Strata
After deciding to use stratify random sampling, we need to decide how to divide sampling effort among different strata; that process is called allocation. When deciding where to expend effort, the question becomes how best to allocate sampling effort among strata so that the sampling process will be the most efficient balance of effort, cost, and precision. Should we allocate the same sampling effort to
sampling. The number of samples selected from each stratum is proportional to the size, variation, as well as the cost ( ci ) of sampling in each stratum. More sampling effort is allocated to larger and more variable strata, and less to strata that are more costly to sample.
k k L k k i i i i
c
N s
c
Ns
n n
1
