Sampling Theory
Probability sampling designs
In probability sampling every unit has a known, non-zero chance of selection, which is what makes the sampling error measurable.
| Design | How units are chosen | Suits |
|---|---|---|
| Simple random (SRS) | every sample of size n equally likely; by lottery or random numbers | homogeneous population with a full list |
| Stratified | divide into homogeneous strata, draw SRS from each | heterogeneous population |
| Systematic | random start, then every k-th unit, k = N/n | ordered lists, quick field work |
| Cluster | choose whole groups (villages, blocks) and observe every unit in them | no list of units, but a list of groups |
| Multistage | sample at stages: districts → villages → households | large national surveys |
| Multiphase (double) | a large first-phase sample gives cheap information; a sub-sample gets the detailed schedule | when an auxiliary variable is cheap to measure |
In SRS without replacement there are ᴺCₙ possible samples (for N = 10, n = 3, ¹⁰C₃ = 120); with replacement, Nⁿ. In stratified sampling, proportional allocation takes nₕ = n·Nₕ/N, so strata of 600, 300 and 100 with n = 50 get 30, 15 and 5; Neyman (optimum) allocation takes nₕ ∝ NₕSₕ, giving more units to larger and more variable strata.
Non-probability sampling
When units are chosen by judgement or convenience rather than by chance, the chance of selection is unknown and the sampling error cannot be measured.
- Purposive (judgement): the investigator picks units he believes are typical.
- Convenience: whoever is easiest to reach — passers-by, volunteers.
- Quota: each interviewer must fill fixed quotas (so many women, so many over 60), choosing the people himself — stratified in design, non-random in selection.
Sampling distribution and sample size
Statement only, as the syllabus asks: if samples of size n are drawn from a population with mean μ and SD σ, the sample mean x̄ has mean μ and standard error σ/√n (with replacement or an infinite population). Without replacement from a finite population of N, the SE is multiplied by the finite population correction √((N − n)/(N − 1)). By the central limit theorem, x̄ is approximately normal for large n whatever the shape of the population. The SE of a sample proportion is √(PQ/n).
Sample size. To estimate a mean within a margin E with confidence given by z, take n = (zσ/E)²; for a proportion, n = z²PQ/E². With σ = 10, E = 2 and 95% confidence (z = 1.96), n = (1.96 × 10/2)² = 96.04, so 97 units — always round up. Halving the margin quadruples the sample, because n grows with 1/E².
Exam traps and a quick revision
- Parameter = population, fixed; statistic = sample, varies.
- A census has no sampling error but can have large non-sampling error.
- Systematic: k = N/n; proportional allocation: nₕ = nNₕ/N; Neyman: nₕ ∝ NₕSₕ.
- Purposive, convenience and quota are non-probability designs.
- SE(x̄) = σ/√n; n = (zσ/E)², rounded up.
Key takeaways
- A parameter describes the population; a statistic, computed from a sample, estimates it.
- Sampling error comes from observing a part; non-sampling error from everything else and exists even in a census.
- Strata are internally alike and all sampled; clusters are internally varied and sampled whole.
- Purposive, convenience and quota sampling cannot measure their own sampling error.
- SE(x̄) = σ/√n, times √((N − n)/(N − 1)) for a finite population without replacement; n = (zσ/E)².
Practice questions (10)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A numerical characteristic of a population, such as its mean, is called a:
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Answer: C — parameter
Population quantities are parameters; the same quantities computed from a sample are statistics, used as estimates of the parameters.Which type of error can occur in a complete census?
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Answer: A — non-sampling error only
A census observes every unit, so there is no sampling error; but non-response, wrong answers and processing mistakes still occur, and are often larger than in a well-run sample.How many different simple random samples of size 3 can be drawn without replacement from a population of 10 units?
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Answer: B — 120
Without replacement the order does not matter, so the count is ¹⁰C₃ = 120. 1000 = 10³ is the count with replacement, and 720 counts ordered selections.A systematic sample of 50 is to be drawn from a list of 1,000 units. The sampling interval k is:
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Answer: B — 20
k = N/n = 1000/50 = 20: choose a random start between 1 and 20, then every 20th unit.A population has three strata of sizes 600, 300 and 100. With proportional allocation, how many units of a sample of 50 go to the second stratum?
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Answer: B — 15
nₕ = n·Nₕ/N = 50 × 300/1000 = 15. The three strata get 30, 15 and 5, adding to 50.Which of the following is a non-probability sampling method?
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Answer: A — quota sampling
In quota sampling the interviewer chooses the people who fill each quota, so selection chances are unknown. The other three give every unit a known chance.A population has standard deviation 12. The standard error of the mean of a random sample of 36 is:
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Answer: B — 2
SE = σ/√n = 12/√36 = 12/6 = 2. Dividing by n instead of √n gives 1/3, and quoting σ itself (12) ignores the averaging that the sample mean does.A survey first selects districts, then villages within the selected districts, then households within the selected villages. This design is:
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Answer: A — multistage sampling
Sampling successively smaller units at each stage is multistage. Multiphase sampling keeps the same units and collects more information from a sub-sample of them.In sampling without replacement from a finite population, the factor √((N − n)/(N − 1)) applied to the standard error is called the:
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Answer: B — finite population correction
It corrects σ/√n for the shrinking pool when units are not replaced; it is close to 1 when n is a small fraction of N. The sampling fraction is n/N itself.To estimate a population mean within ±2 units with 95% confidence (z = 1.96), when σ = 10, the minimum sample size is:
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Answer: A — 97
n = (zσ/E)² = (1.96 × 10/2)² = 9.8² = 96.04. A sample must be a whole number at least this large, so 97 — rounding down to 96 misses the required precision.