Statistics — Foundations of Inference

Sampling & Estimation
From Sample to Population

You can never measure everyone — but a well-chosen sample, analysed with the right tools, tells you a remarkable amount about the whole. Understand how samples behave, why the Central Limit Theorem is the most powerful result in statistics, and how to build trustworthy estimates of population parameters.

CLTKey Theorem
10Worked Examples
3Estimator Properties
10Quiz Questions
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§ 01Population vs Sample

Statistical inference begins with a fundamental distinction: the population is every individual we care about, and the sample is the subset we actually measure. Almost every real-world study works with a sample, because measuring an entire population is impossible, too expensive, or destructive.

A parameter is a numerical summary of a population (e.g. the true mean μ or proportion p). A statistic is the corresponding numerical summary computed from a sample (e.g. x̄ or p̂). Because samples vary, so do statistics — and understanding that variability is the entire point of sampling theory.

QuantityPopulation ParameterSample Statistic
Mean μ — fixed, unknown = (1/n)∑xi — varies by sample
Variance σ² — fixed, unknown = ∑(xi−x̄)²/(n−1) — varies
Standard Deviation σ s
Proportion p — fixed, unknown = x/n — varies by sample
Size N (often infinite or very large) n (what we choose)
The Core Problem We want to know μ (or p or σ), but we can only observe x̄ (or p̂ or s). The question is: how reliably does the sample statistic reflect the population parameter, and how do we quantify the uncertainty?

§ 02Random Sampling and Sampling Methods

For a sample statistic to be a useful estimate of a population parameter, the sample must be collected in a way that gives every member of the population a fair chance of being selected. Non-random samples introduce bias — a systematic tendency to over- or under-represent certain parts of the population — which no amount of statistical analysis can correct.

Simple Random Sample (SRS)
Every subset of size n equally likely The gold standard. Each individual is chosen independently with equal probability. Requires a complete list (sampling frame) of the population. Forms the basis of most statistical theory.
Stratified Sampling
SRS within each subgroup Divide the population into homogeneous strata (e.g. age groups), then take an SRS from each stratum. More precise than SRS when strata differ substantially. Guarantees representation.
Cluster Sampling
Sample whole groups randomly Divide the population into clusters (e.g. schools), randomly select clusters, then sample everyone in selected clusters. Practical and cost-effective for geographically spread populations.
Convenience Sampling
NOT random — avoid Selecting whoever is easiest to reach (e.g. surveying friends, online volunteers). Highly prone to selection bias. Results cannot be generalised to the population with any statistical validity.
Example 01 Identifying bias in a sampling design

A school wants to estimate the average number of hours students spend on homework per week. They survey students who stay late in the library. Identify the problem.

Problem

Library-staying students are likely to spend more time on homework than the average student. This is selection bias — the sample systematically over-represents diligent students.

Fix

Use a simple random sample from the complete class register, giving every student an equal chance of being selected regardless of study habits.

Key Point
Bias in collection cannot be undone by statistical analysis. Get the design right first.

§ 03Point Estimation

A point estimator is a statistic — a function of the sample — used to estimate a single unknown population parameter. The number it produces for a particular sample is called a point estimate.

Common Point Estimators

  • x̄ estimates μ (population mean)
  • s² estimates σ² (population variance)
  • p̂ = x/n estimates p (population proportion)
  • Median estimates the population median
  • Sample range estimates population range (poorly)

Estimator vs Estimate

The estimator is the rule or formula (a random variable). The estimate is the specific number computed from one sample.

Example: x̄ = (x1+…+xn)/n is the estimator. For a sample {4, 7, 9, 6}, the estimate is x̄ = 6.5.

Three Desirable Properties of Estimators

Unbiasedness
E[θ̂] = θ On average, across all possible samples, the estimator hits the true parameter exactly. No systematic over- or under-estimation. The sample mean x̄ is unbiased for μ. The sample variance s² = ∑(xᵢ−x̄)²/(n−1) is unbiased for σ² (the divisor n−1 is crucial).
Consistency
θ̂ → θ as n → ∞ As the sample size grows, the estimator converges in probability to the true parameter. Larger samples give more accurate estimates. The sample mean is consistent for μ (by the Law of Large Numbers).
Efficiency
Smallest Var(θ̂) among unbiased estimators Among all unbiased estimators, an efficient estimator has the smallest variance. The sample mean is more efficient than the sample median for estimating μ in a normal population (it has smaller sampling variance).
Example 02 Why divide by n − 1 for the sample variance?

Show why the unbiased sample variance uses n−1 rather than n in the denominator.

Intuition

The deviations (xi − x̄) are computed around the sample mean x̄, not the true mean μ. Since x̄ is chosen to minimise ∑(xi−x̄)², the sum is always smaller than it would be around μ. Dividing by n−1 corrects for this underestimation.

Formal Result
E[∑(xi−x̄)²] = (n−1)σ²

Therefore E[s²] = E[∑(xi−x̄)² / (n−1)] = σ²  ✓

Key Takeaway
Dividing by n gives a biased (too small) estimate of σ². Always use n−1 for s².
Example 03 Computing point estimates from a sample

A sample of n = 6 battery lifetimes (hours): 48, 52, 55, 47, 51, 53. Compute x̄, s², and s.

Mean
x̄ = (48+52+55+47+51+53)/6 = 306/6 = 51 hours
Deviations
(48−51)²+(52−51)²+(55−51)²+(47−51)²+(51−51)²+(53−51)² = 9 + 1 + 16 + 16 + 0 + 4 = 46
Variance & SD
s² = 46/(6−1) = 46/5 = 9.2    s = √9.2 ≈ 3.03 hours
Estimates
x̄ = 51 h  (estimates μ)    s² = 9.2 h²  (estimates σ²)    s ≈ 3.03 h  (estimates σ)

§ 04The Sampling Distribution of x̄

If we drew many different samples of size n from the same population and computed x̄ each time, those values would form a distribution — the sampling distribution of x̄. This distribution describes how x̄ varies across samples, and its properties are the foundation of all inferential statistics.

Sampling Distribution of the Sample Mean
E[x̄] = μ    (mean of x̄ equals population mean)
Var(x̄) = σ²/n    (variance decreases as n grows)
SE(x̄) = σ/√n    (standard error of the mean)
These results hold regardless of the population distribution, for any sample size n.

The standard error SE = σ/√n is perhaps the single most important quantity in sampling theory. It measures the typical distance between x̄ and μ. Notice it decreases as n increases — larger samples produce more precise estimates, but with diminishing returns: doubling precision requires quadrupling the sample size.

Example 04 Computing the standard error and its dependence on n

A population has μ = 80 and σ = 20. Find SE(x̄) for n = 4, 16, 25, 100.

Formula
SE = σ/√n = 20/√n
Results
n=4: SE = 20/2 = 10    n=16: SE = 20/4 = 5 n=25: SE = 20/5 = 4    n=100: SE = 20/10 = 2
Observation
To halve the SE from 10 to 5, n must quadruple (4 → 16). Precision improves with √n, not n.
Example 05 Exact sampling distribution (normal population)

If the population is normal: X ∼ N(120, 15²), find the distribution of x̄ for n = 9. What is P(x̄ > 125)?

Distribution
x̄ ∼ N(μ, σ²/n) = N(120, 225/9) = N(120, 25) So SE = 5
Probability
P(x̄ > 125) = P(Z > (125−120)/5) = P(Z > 1.00) = 1 − Φ(1.00) = 1 − 0.8413 = 0.1587
Result
There is a 15.87% chance that a sample of n = 9 gives x̄ > 125.

§ 05The Central Limit Theorem

When the population is normal, x̄ is exactly normal for any n. But what if the population is skewed, discrete, or has an unusual shape? The Central Limit Theorem (CLT) provides a remarkable answer.

The Central Limit Theorem
If X1, X2, …, Xn are i.i.d. with mean μ and finite variance σ², then as n → ∞:

x̄  ∼approx  N(μ, σ²/n)

Equivalently:   Z = (x̄ − μ) / (σ/√n)  →  N(0,1)
The approximation is typically excellent for n ≥ 30, and often for much smaller n if the population is not highly skewed.

The CLT is the reason the normal distribution appears everywhere in statistics. It means that regardless of how a population is distributed — whether it is skewed, bimodal, uniform, or exponential — sample means from large enough samples behave approximately normally. This is what makes it possible to construct confidence intervals and perform hypothesis tests without knowing the population distribution.

Why the CLT Is So Powerful Every probability result about x̄ derived from the normal distribution is applicable to any population, as long as n is large enough. Polling, quality control, clinical trials, and financial risk models all depend on this single theorem.
Example 06 CLT applied to a skewed population

Customer service call durations follow an exponential distribution with mean μ = 4 minutes and σ = 4 minutes (exponential: σ = μ). A sample of n = 64 calls is taken. Find P(x̄ < 3.5).

CLT Setup

n = 64 ≥ 30, so by the CLT: x̄ ∼ N(4, 4²/64) = N(4, 0.25)

SE = 4/√64 = 4/8 = 0.5
Standardise
Z = (3.5 − 4) / 0.5 = −0.5/0.5 = −1.00
Probability
P(x̄ < 3.5) = P(Z < −1.00) = 1 − Φ(1.00) = 0.1587
Result
Despite the exponential (skewed) population, P(x̄ < 3.5) ≈ 15.87% via the CLT.
Example 07 CLT applied to a sum of random variables

A factory machine produces parts with weight μ = 200 g and σ = 8 g. A box holds n = 36 parts. What is P(total weight > 7300 g)?

Reformulate

Total weight T = ∑Xi = n×x̄. So P(T > 7300) = P(x̄ > 7300/36) = P(x̄ > 202.78).

SE
SE = 8/√36 = 8/6 ≈ 1.333
Standardise
Z = (202.78 − 200)/1.333 = 2.78/1.333 ≈ 2.08
Result
P(T > 7300) = P(Z > 2.08) = 1 − 0.9812 = 0.0188 ≈ 1.88%

§ 06Sampling Distribution of the Proportion p̂

When the parameter of interest is a population proportion p, the sample proportion p̂ = X/n (where X is the count of successes) is the natural estimator. By the CLT applied to a Bernoulli population, p̂ has an approximately normal sampling distribution for large samples.

Sampling Distribution of p̂
E[p̂] = p    (unbiased)
Var(p̂) = p(1−p)/n
SE(p̂) = √(p(1−p)/n)
p̂  ∼approx  N(p, p(1−p)/n)
Rule of thumb: approximation is good when np ≥ 10 and n(1−p) ≥ 10.
Example 08 Sampling distribution of a proportion

In a large city, 35% of residents recycle regularly (p = 0.35). A survey samples n = 200 residents. Find the mean and standard error of p̂, and find P(p̂ > 0.40).

Check
np = 200×0.35 = 70 ≥ 10 ✓    n(1−p) = 200×0.65 = 130 ≥ 10 ✓
Mean & SE
E[p̂] = 0.35    SE = √(0.35×0.65/200) = √(0.001138) ≈ 0.03373
Probability
Z = (0.40 − 0.35)/0.03373 ≈ 1.483 P(p̂ > 0.40) = P(Z > 1.483) = 1 − 0.9310 = 0.0690
Result
There is about a 6.9% chance that a random sample of 200 shows more than 40% recycling.

§ 07The Law of Large Numbers

The CLT tells us about the shape of the sampling distribution. The Law of Large Numbers (LLN) tells us about where it centres as n grows. There are two versions, but both say the same thing intuitively: averages of large samples converge to the true mean.

Weak LLN

For any ε > 0:

P(|x̄ − μ| > ε) → 0 as n → ∞

The sample mean converges in probability to μ. For any fixed tolerance, the probability of being outside it goes to zero.

Strong LLN

P(x̄ → μ as n → ∞) = 1

The sample mean converges almost surely to μ. With probability 1, every sequence of sample means eventually stays arbitrarily close to μ. Stronger than the Weak LLN.

The LLN underpins the consistency property of x̄ and justifies long-run frequency interpretations of probability. It is also why the insurance industry works: individual claims are random, but the average claim over millions of policies converges reliably to the expected value.

Example 09 LLN and the CLT working together — sample size determination

A quality engineer wants x̄ to be within 1.5 units of μ with 95% probability. The population has σ = 10. How large must n be?

Set Up

By CLT, x̄ ∼ N(μ, σ²/n). We want P(|x̄ − μ| ≤ 1.5) = 0.95.

P(|Z| ≤ 1.5/SE) = 0.95  ⇒  1.5/SE = 1.96
Solve for n
SE = 1.5/1.96 ≈ 0.7653  ⇒  σ/√n = 0.7653 10/√n = 0.7653  ⇒  √n = 10/0.7653 ≈ 13.07  ⇒  n ≈ 170.8
Required Sample Size
n = 171  (always round up to guarantee the required precision)

§ 08Unknown σ and the t-Distribution

In the CLT results above, we assumed σ is known. In practice it almost never is, so we substitute the sample standard deviation s. This introduces extra uncertainty, and the standardised sample mean no longer follows the standard normal — it follows a t-distribution with n−1 degrees of freedom.

Studentised Sample Mean
T = (x̄ − μ) / (s/√n)  ∼  tn−1
The t-distribution has heavier tails than the normal — reflecting extra uncertainty from estimating σ. As n → ∞, tn−1 → N(0,1).

The t-distribution was discovered by William Gosset (writing as “Student”) while working at Guinness Brewery, where he needed to make inferences from small samples of barley. It is now the standard distribution used in confidence intervals and hypothesis tests when σ is estimated.

Use z (Normal)
σ known, any n; or n ≥ 30 When the population SD is known, or when n is large enough that s ≈ σ reliably, the z-distribution gives a valid and slightly tighter result.
Use t
σ unknown, any n When σ must be estimated by s. Always valid for normal populations. For non-normal populations, the CLT ensures approximate validity when n ≥ 30.
Small n, non-normal
Caution required With n < 30 and an unknown, non-normal population, both z and t approximations can be unreliable. Consider non-parametric methods or bootstrap confidence intervals.
Example 10 Probability under the t-distribution

A sample of n = 12 measurements gives x̄ = 47.3 and s = 5.8. The population mean is claimed to be μ = 50. Compute the t-statistic and find the p-value for a two-tailed test.

t-Statistic
t = (x̄ − μ) / (s/√n) = (47.3 − 50)/(5.8/√12) = −2.7 / (5.8/3.464) = −2.7 / 1.675 ≈ −1.612
Degrees of Freedom
ν = n−1 = 11
p-value

From t-tables: P(|t11| > 1.612) ≈ 0.136    (two-tailed)

Conclusion
p ≈ 0.136 > 0.05 → insufficient evidence to reject μ = 50 at the 5% level.

§ 09Common Mistakes and Misconceptions

Confusing SE with SD
SE = σ/√n ≠ σ The standard deviation σ (or s) describes spread in the population (or sample). The standard error SE describes how much x̄ varies between samples. They differ by a factor of √n.
Using n instead of n−1
s² = ∑(xᵢ−x̄)² / (n−1) Dividing by n gives the biased estimator. Most software uses n−1 by default (called the “corrected” or “unbiased” sample variance). Always check which formula your calculator uses.
Over-relying on n ≥ 30
Rule of thumb, not a law The “n ≥ 30 is enough for CLT” rule is a starting point. Heavily skewed or thick-tailed populations may need n ≥ 100 or more for good normal approximation. Always check residuals.
Larger sample = no bias
n ↑ reduces variance, not bias A larger biased sample is still biased. The LLN only works when E[x̄] = μ, i.e., the estimator is unbiased. Convenience samples remain biased regardless of size.
Using z when t is needed
Unknown σ → always use t Replacing s with z is anti-conservative: it understates the tails and gives too-narrow intervals. Use tn−1 whenever σ is estimated from data.
Interpreting SE as uncertainty about individuals
SE concerns x̄, not individuals The SE of the mean is NOT the spread of individual data. It only measures how precisely we have estimated the population mean. An individual prediction requires the full SD, not the SE.

§ 10Summary — The Inferential Chain

Sampling and estimation form the bridge between raw data and population knowledge. Every subsequent technique — confidence intervals, hypothesis tests, regression — rests on the ideas introduced here.

The Key Results at a Glance
x̄ is unbiased, consistent, and efficient for μ
s² (with n−1) is unbiased for σ²
SE(x̄) = σ/√n  →  precision improves with √n
CLT: x̄ ≈ N(μ, σ²/n) for large n, any population
Unknown σ: use T = (x̄−μ)/(s/√n) ∼ tn−1
Where to Go Next The sampling distribution of x̄ derived here is the direct input to confidence intervals (which give a range of plausible values for μ) and hypothesis tests (which formally assess a specific claim about μ). Master the CLT and standard error, and the rest of inferential statistics follows naturally.

Continue the Statistics Series

§ 11Quiz — Test Your Understanding

Ten questions on sampling, estimation, and the Central Limit Theorem. Immediate feedback on every answer.

Sampling & Estimation Quiz

Question 1 of 10
The standard error of the sample mean is…
Question 2 of 10
An estimator θ̂ is unbiased if…
Question 3 of 10
A population has σ = 24. To halve the standard error of x̄ compared to n = 16, what sample size is needed?
Question 4 of 10
The Central Limit Theorem states that, for large n, the sampling distribution of x̄ is approximately…
Question 5 of 10
The unbiased sample variance uses which denominator?
Question 6 of 10
X ∼ N(50, 100). For samples of n = 25, what is P(x̄ > 54)?
Question 7 of 10
Which sampling method is most prone to selection bias?
Question 8 of 10
When σ is unknown and n = 20, the standardised sample mean T = (x̄−μ)/(s/√n) follows a…
Question 9 of 10
The Law of Large Numbers guarantees that as n → ∞, x̄ converges to…
Question 10 of 10
A population proportion is p = 0.6. For n = 100, the standard error of p̂ is…
0 out of 10

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