Sampling distribution: distribution of a statistic across all possible samples from a population.
Key idea: use the sampling distribution to make inferences about a population parameter (mu, p, etc.) from a sample statistic.
Population vs. sample concepts:
Population measures are parameters (e.g., population mean μ, true proportion p).
Sample measures are statistics (e.g., sample mean X̄, sample proportion p̂).
Central Limit Theorem (CLT)
If a random sample is drawn from any population, the sampling distribution of the sample mean X̄ is approximately normal if the sample size is large enough (commonly n ≥ 30).
If the population is already normal, smaller samples can still yield a normal X̄.
CLT enables inference using the standard normal distribution after standardization.
Sampling distribution of the sample mean (X̄)
Mean of the sampling distribution: E[Xˉ]=μ.
Variance of the sampling distribution: Var(Xˉ)=nσ2.
Standard error of the mean: SE(Xˉ)=nσ.
Standardization to Z: Z=σ/nXˉ−μ.
Finite population correction (when population size N is finite): σXˉ=σN−1N−n⋅n1=nσN−1N−n.
Note: often ignored when N is large relative to n.
Inference with X̄
Because X̄ ≈ Normal(μ, σ^2/n) for large n, we can make confidence statements about μ using the standard normal distribution.
Increasing n reduces the SE and tightens the distribution around μ.
Sampling distribution of the sample proportion p̂
For Bernoulli trials with true population proportion p, let X be the number of successes in n trials: X ~ Bin(n, p).
Sample proportion: p^=nX.
Mean and variance of p̂: E[p^]=p,Var(p^)=np(1−p).
Standardization: Z=p(1−p)/np^−p.
Normal approximation to p̂ is valid if the rule of thumb is met: np≥10andn(1−p)≥10.
For X (number of successes), X ~ Bin(n, p) with mean E[X]=np and variance Var(X)=np(1−p).
Bernoulli trials, independence, and sampling with replacement
Bernoulli trial: one trial with two outcomes (success/failure).
Trials must be independent and identically distributed; use sampling with replacement to preserve constant p across trials.
Examples (illustrative calculations)
Chocolate weights example
Given single bar: X ~ Normal(μ=32.2, σ=0.3).
Q1: P(X > 32) = ?
Z = (32 - 32.2)/0.3 = -0.667 ➜ P(X > 32) ≈ 0.75.
Q2: For a pack of n = 4, X̄ ~ Normal(μ=32.2, σ^2/n = 0.3^2/4 = 0.0225).
SE = 0.3/√4 = 0.15; Z = (32 - 32.2)/0.15 = -1.333 ➜ P(X̄ > 32) ≈ 0.91.
intuition: larger samples pull the average toward the true mean and reduce variability.