University of Sydney · FACULTY OF STATISTICS

QBUS5001 Chap.6 Sampling Distributions, CLT & Estimation

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Chapter 6 of 11 · QBUS5001

Sampling Distributions, CLT & Estimation

Modules 5 and 6 connect samples to populations.

The sampling distribution of the sample mean has mean μ and standard error σ/√n; the Central Limit Theorem makes that distribution approximately Normal for n ≥ 30 regardless of the population shape, while the Law of Large Numbers says x̄ converges to μ as n grows.

From there you build confidence intervals (point estimate ± critical value × standard error) for a mean (Z when σ is known, t when it is not) and for a proportion, and you determine the required sample size for a target margin of error.

The interpretation of a CI — a statement about the procedure, not a probability about a fixed parameter — is a perennial exam point.

In this chapter

What this chapter covers

  • 01

    Sampling distribution of x̄: mean μ, standard error σ/√n

  • 02

    Central Limit Theorem: approximately Normal for n ≥ 30

  • 03

    Law of Large Numbers: x̄ → μ as n → ∞

  • 04

    Sampling distribution of a proportion (np ≥ 5, n(1−p) ≥ 5)

  • 05

    Unbiased and consistent point estimators

  • 06

    CI for a mean: σ known → Z, σ unknown → t(n−1)

  • 07

    CI for a proportion

  • 08

    Sample-size determination for mean and proportion

  • 09

    Correct frequentist interpretation of a confidence interval

Worked example · free

CLT for the sample mean

Q [6 marks]. LED bulbs have a mean lifetime of 12,000 hours with standard deviation 900 hours. A quality-assurance team draws a random sample of n = 144 bulbs. Using the Central Limit Theorem, find the probability that the sample mean lifetime is below 11,880 hours.
  • 1 markCheck the CLT applies: n = 144 ≥ 30, so the sampling distribution of x̄ is approximately Normal.
  • 1 markState the sampling distribution: x̄ ~ N(12000, 900²/144).
  • 1 markCompute the standard error: σ/√n = 900/√144 = 900/12 = 75 hours.
  • 1 markStandardise the cut-off: Z = (11880 − 12000)/75 = −120/75 = −1.60.
  • 1 markFind the lower-tail probability: P(Z < −1.60) = NORM.S.DIST(−1.60, 1) ≈ 0.0548.
  • 1 markConclude: there is about a 5.48% chance the sample mean lifetime falls below 11,880 hours.
SE = 75 hours; Z = −1.60; P(x̄ < 11,880) = NORM.S.DIST(−1.60, 1) ≈ 0.0548 (about 5.48%).
Sia tip — The CLT question is just a Normal probability where the SD is replaced by the standard error σ/√n. Dividing by √n (not n) is the step students most often miss.
Glossary

Key terms

Sampling distribution
The probability distribution of a sample statistic (e.g. x̄) over all possible samples of size n; for the mean it is centred at μ with spread σ/√n.
Standard error of the mean
σ/√n, the standard deviation of the sample mean's sampling distribution; it shrinks as n grows, tightening estimates.
Central Limit Theorem
For n ≥ 30 the sampling distribution of x̄ is approximately Normal regardless of the population's shape, enabling Normal-based inference on non-Normal data.
Law of Large Numbers
As the sample size grows without bound, the sample mean converges to the population mean μ and the standard error tends to zero.
Unbiased estimator
An estimator whose expected value equals the parameter it estimates; the sample mean is unbiased because E[x̄] = μ, and it is also consistent by the LLN.
FAQ

Sampling Distributions, CLT & Estimation FAQ

Should I divide by σ or by σ/√n when standardising a sample mean?

By the standard error σ/√n. A single observation X uses σ; a sample mean x̄ uses σ/√n because averaging reduces variability. Mixing these up is the classic CLT error.

What is the correct interpretation of a 95% confidence interval?

Over repeated sampling, 95% of the intervals constructed this way would contain the true parameter. It is wrong to say there is a 95% probability the parameter lies in this particular interval — the parameter is fixed; it is the interval that is random.

When do I use Z and when t for a confidence interval for a mean?

Use Z (with z(α/2)) when the population standard deviation σ is known. Use t (with t(α/2, n−1)) when σ is unknown and you estimate it with the sample s, which is the usual real-world case.

Study strategy

Exam move

Burn the standard error σ/√n into memory and treat every CLT problem as a Normal probability on x̄. For confidence intervals, memorise the universal template (estimate ± critical × SE) and the Z-vs-t decision (is σ known?), and rehearse the verbal interpretation of a CI word-for-word, because the exam regularly awards a mark purely for the correct frequentist statement.

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