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MAST90105 · Methods Of Mathematical Statistics

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Chapter 8 of 10 · MAST90105

Interval Estimation and Pivotal Quantities

A point estimate is a single number; a confidence interval reports a range that traps the parameter with a stated probability, and the chapter shows that one recipe generates every standard CI on the course. Find a pivotal quantity — a function of the data and the parameter whose distribution is known and free of the parameter — bracket it between quantiles of that known distribution, then invert the inequality to isolate the parameter. For a normal mean with known σ the pivot is Z = (X̄ − μ)/(σ/√n); with σ unknown it is the t-statistic with n−1 degrees of freedom; for a variance it is (n−1)S²/σ² ~ χ²; for a proportion the large-sample normal pivot. The chapter also covers large-sample (Wald) intervals built from the asymptotically normal MLE, and distribution-free intervals for percentiles that need no parametric model. The interpretation trap is hammered: confidence is a property of the procedure — 95% of such intervals cover — not a probability about a fixed parameter.

In this chapter

What this chapter covers

  • 018.1 What a confidence interval claims (and the interpretation trap)
  • 028.2 Pivotal quantities — the master recipe
  • 038.3 CI for a normal mean: Z (known σ) and t (unknown σ)
  • 048.4 CI for a variance via the χ² pivot
  • 058.5 Large-sample (Wald) CIs for proportions and the MLE
  • 068.6 Distribution-free CIs for percentiles
Worked example · free

Worked example: a t confidence interval for a normal mean

Q [5 marks]. A sample of n = 16 gives X̄ = 50 and sample standard deviation S = 8. Construct a 95% confidence interval for the population mean μ, assuming normal data. (t0.025,15 = 2.131.)
  • +1Choose the pivot. σ is unknown and the data are normal, so the pivot is T = (X̄ − μ)/(S/√n) ~ tn−1 = t15.
  • +1Standard error. S/√n = 8/√16 = 8/4 = 2.
  • +1Critical value. For 95% with 15 df, t0.025,15 = 2.131 (read off the provided table).
  • +1Invert. μ = X̄ ± t·(S/√n) = 50 ± 2.131(2) = 50 ± 4.262.
  • +1State and interpret. The 95% CI is (45.74, 54.26). ‘95% confident’ means the procedure traps μ in 95% of repeated samples — not that μ has a 95% chance of lying in this particular interval.
The 95% CI for μ is 50 ± 4.26 = (45.74, 54.26). The recipe is universal: pick the pivot, read its quantile off the table, and invert μ = X̄ ± t·(S/√n).
Glossary

Key terms

Confidence interval
A data-based range (L, U) that contains the parameter with a stated coverage probability, e.g. 95%. The coverage is a property of the procedure across repeated samples, not a probability statement about a fixed parameter once the data are observed.
Pivotal quantity
A function of the data and the parameter whose sampling distribution does not depend on the parameter — Z, t, χ² pivots. Bracketing it between known quantiles and inverting produces a confidence interval. The single recipe behind every standard CI.
t-interval for a mean
When σ is unknown and data are normal, X̄ ± tα/2,n−1(S/√n). The t (rather than Z) accounts for estimating σ, and its heavier tails widen the interval, the more so for small n.
Large-sample (Wald) interval
θ̂ ± zα/2·SE(θ̂), built from the asymptotic normality of the MLE with SE from the inverse Fisher information. Used for proportions and general MLEs when n is large enough for the normal approximation.
Distribution-free interval
A confidence interval for a percentile (e.g. the median) built from order statistics and the binomial, with no assumption about the parametric form of the distribution — robust when normality is doubtful.
FAQ

Interval Estimation and Pivotal Quantities FAQ

What does ‘95% confidence’ actually mean?

It is a statement about the procedure, not the parameter. If you repeated the sampling and interval construction many times, about 95% of the resulting intervals would contain the true parameter. For any one observed interval the parameter is either in it or not — there is no 95% probability attached to that fixed interval. This is the most-tested interpretation point in the chapter.

When do I use t instead of z for a mean?

Use t when the population standard deviation σ is unknown and you estimate it with the sample S — which is almost always, given normal (or approximately normal) data. Use z only when σ is known, or as a large-sample approximation where the t and z values nearly coincide. The t has heavier tails, so it gives a slightly wider, more honest interval that accounts for estimating σ.

How does the pivotal-quantity method build every interval?

Find a quantity that involves both the data and the parameter but whose distribution is known and parameter-free — the pivot. Bracket the pivot between the α/2 and 1−α/2 quantiles of its distribution, then algebraically rearrange (invert) the double inequality so the parameter sits alone in the middle. Different pivots (Z, t, χ²) give the mean, variance and proportion intervals from one identical procedure.

Study strategy

Exam move

Learn interval estimation as a single algorithm rather than a list of formulas: identify the pivot for the quantity in question, look up its quantiles on the provided table, and invert to isolate the parameter. Build a short pivot library on your A4 sheet — Z and t for a mean, χ² for a variance, the normal Wald pivot for a proportion — since the table supplies the quantiles but not which pivot to use. Drill the t-interval until the standard-error and critical-value steps are automatic, and rehearse the correct frequentist interpretation of confidence, because a careless ‘95% chance μ is in here’ loses an easy mark.

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