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Chapter 3: Simple Random Sampling and Systematic
Sampling
Simple random sampling and systematic sampling provide the foundation for almost all of the more complex sampling designs that are based on probability sampling. They are also usually the easiest designs to implement. These two designs highlight a trade-off inherent in all sampling designs: do we select sample units at random to minimize the risk of introducing biases into the sample or do we select sample units systematically to ensure that sample units are well- distributed throughout the population?
Both designs involve selecting n sample units from the N units 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 advance knowledge about the target population
Disadvantages:
- Imprecise relative to other designs if the population is heterogeneous
- More expensive to implement 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 population 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 finite populations (i.e., those where every potential sampling unit can be identified in advance), this can involve generating pseudorandom numbers with a computer. For small finite 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 ( μˆ or y ) which has an unbiased estimator:
n
y
n
i
∑ i
μˆ= =^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 μˆ is:
n
s N
N n (μ)
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 as is a function of the variation inherent in the population ( σ^2 ) and the size of the sample ( n ).
Standard error of μˆ is:
n
s N
N n SE(μ)
2 ⎟ ⎠
ˆ = ⎛^ −.
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. Simply, as the amount of information we know about the population through sampling increases, the remaining uncertainty decreases. Therefore, the correction factor reflects the proportion of the population that remains unknown. Consequently, as the number of sampling units measured ( n ) approaches the total number of sampling units in the population ( N ), the finite population correction factor approaches zero, so the amount of uncertainty in the estimate also approaches zero.
When the sample size n is small relative to the population size N , the fraction of the population being sampled n / N also is small, therefore the correction factor has little effect on the variance of the estimator (Fig. 2 - FPC.xls). If the finite population correction factor is ignored, which is what
FPC with N = 100
0
1
0 20 40 60 80 100 n
FPC
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 size 10, where 4 entities have a value of 1 and 6 entities have a value of 0 (e.g., 1 = presence of a trait, 0 = absence of a trait). The proportion of entities in the sample with the trait is 4/10 or 0.40 which is also equal to the sample mean, which = 0. ([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 = μˆ= =^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: 1
2
−
n
p( p) N
N n n
s N
N n (p )
va ˆr ˆ.
Standard error of p ˆ^ is: 1
2
−
n
p( p) N
N n n
s N
N n SE(p )
Example. (to be added)
Determining Sample Sizes
How many sample units should we measure from the population so that we have confidence that parameters have been estimated adequately?
Determining how many sample units ( n ) to measure requires that we establish the degree of precision that we require for the estimate we wish to generate; we denote a quantity B , the desired bound of the error of estimation, which we define as the half-width of the confidence interval we want to result around the estimate we will generate from the proposed sampling effort.
To establish the number of sample units to measure to estimate the population mean μ at a desired level of precision B with simple random sampling, we set Z × SE( y ) (the formula for a
confidence interval) equal to B and solve this expression for n. We use Z to denote the upper α/ point of the standard normal distribution for simplicity (although we could use the Student’s t distribution), where α is the same value we used to establish the width of a confidence interval, the rate at which we are willing to tolerate Type I errors.
We set ⎟
n
σ N
N n B Z
2 and solve for n:
2
2 2 0
0
B
Z
n
n N
n
If we anticipate n to be small relative to N , we can ignore the population correction factor and use only the formula for n 0 to gauge sample size.
Example : Estimate the average amount of money μ for a hospital’s accounts receivable. Note, however, that no prior information exists with which to estimate population variance σ^2 but we know that most receivables lie within a range of about $100 and there are N = 1000 accounts. How many samples are needed to estimate μ with a bound on the error of estimation B = $3 with 95% confidence (α = 0.05, Z = 1.96) using simple random sampling?
Although it is ideal to have data with which to estimate σ^2 , 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 into the formula above:
2
2 2
n =
Therefore, about 218 samples are needed to estimate μ with a bound on the error of estimation B = $3.
To establish the number of sample units to measure to estimate the population total τ at a desired
level of precision B with simple random sampling, we set (^) ⎟⎟ ⎠
n
B Z NN n
σ^2
( ) and solve for n :
2
2 2 2 0
0
B
N Z
n
n N
n
And as with establishing n for the population mean, 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 d = 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:
2
2 2 2 0 =^ ≈
n.
To adjust for the size of the finite population:
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.
Estimating the Population Mean
The population mean ( μ ) is estimated with: n
y
n
i
∑ i
μˆ= =^1
The population variance ( σ^2 ) is estimated with: 1
1
2 2 −
= n
y y s
n
i
( i )
Variance of the estimate μˆ is:
n
s N
N n^2 ⎟ ⎠
vaˆr( μˆ )=⎛^ −.
Standard error of μˆ^ is:
n
s N
N n SE
2 ⎟ ⎠
Estimating the Population Total
The population total τ is estimated with: ∑
=
n
i
yi n
N
N
1
Variance of the estimate τˆ is: ⎟
N
N n n
s N N
2
vaˆr(τˆ )^2 var( μˆ)^2.
Standard error of τˆ is: ⎟
N
N n n
s N
2
vaˆr( τˆ )^2
Estimating the Population Proportion
The population proportion ( p ) is estimated with the sample proportion ( p ˆ ) which has an unbiased
estimator:
n
y p
n
i
∑ i
ˆ = μˆ= =^1.
Because we are estimating a dichotomous proportion, the population variance σ^2 is again computed with a binomial which is the proportion of the population with the trait ( p ) times the proportion without 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: 1
2
−
n
p p N
N n n
s N
N n p
va ˆr(ˆ).
Examples. (to be added)
