Source: Principles of Statistics I, Texas A&M / Tamhane & Dunlop; Mason, Gunst & Hess; Somerville (1958)

Tags: prediction interval, tolerance interval, normal PI, exponential PI, normal TI, exponential lower tolerance bound, distribution-free tolerance interval, engineering tolerance, natural tolerance, statistical tolerance

TL;DR

Prediction intervals forecast the value of the next observation (wider than CIs because they account for both estimation uncertainty and individual variability). Tolerance intervals estimate a region that captures a specified proportion of the population with a given confidence level, and come in parametric (normal, exponential) and distribution-free flavours. Distribution-free TIs are valid for any continuous distribution but are wider than parametric ones when the parametric model is correct.

Key Terms

Prediction interval (PI)

An interval that captures the next randomly selected observation Y_n+1 with probability 1 − α. It is wider than a CI for μ because it includes both the estimation error and the natural variability of a single observation.

Engineering tolerance interval

Fixed specification limits (S_L, S_U) set by the manufacturer. If a product measurement falls outside this range, the product is rejected. This is not a statistical interval in the usual sense.

Natural tolerance interval

When the population distribution F is fully known (no unknown parameters), the interval (Q(u₁), Q(u₂)) with u₁ + (1 − u₂) = α captures exactly 100(1 − α)% of the population.

Statistical tolerance interval (TI)

A 100(P, γ)% TI is an interval (L_P,γ, U_P,γ) such that we are 100γ% confident it contains at least 100P% of the population. P is the population proportion to be captured; γ is the confidence that the interval achieves this capture.

K_P,γ factor

The multiplier used in normal-based TIs: (Ŷ − K_P,γS, Ŷ + K_P,γS). Values are obtained from tables or numerical approximation.

Core Content

Prediction Intervals

PI for a Normal Population

Setting: Y₁, …, Y_n iid N(μ, σ²). We want to predict Y_n+1.

The point predictor is Ŷ_n+1 = Ŷ (replace μ with its MLE, replace Z_n+1 with E[Z_n+1] = 0).

Pivot: (Y_n+1 − Ŷ) / [S√((n+1)/n)]

This has a t-distribution with df = n − 1 because:

• Y_n+1 − Ŷ ~ N(0, σ²(1 + 1/n)) since Y_n+1 and Ŷ are independent

• Y_n+1 − Ŷ is independent of S

• The ratio N(0,1) / √(χ²/df) gives the t-distribution

100(1 − α)% PI for Y_n+1:

Ŷ ± t_{α/2, n−1} · S · √(1 + 1/n)

Compare to the CI for μ: Ŷ ± t_{α/2, n−1} · S · √(1/n). The PI replaces √(1/n) with √(1 + 1/n), reflecting the additional uncertainty of predicting a single observation.

As n → ∞, the PI approaches Ŷ ± t_α/2 · S, which remains non-trivially wide because of the irreducible variability of a single observation. A CI, by contrast, shrinks to zero width.

PI for an Exponential Population

Setting: W₁, …, W_n iid Exp(β). Predict W_n+1.

Pivot: W_n+1 / W̄

This has an F-distribution with df = (2, 2n), since 2W_n+1/β ~ χ²(2) and 2nW̄/β ~ χ²(2n), both independent.

100(1 − α)% PI for W_n+1:

(W̄ · F_{1−α/2, 2, 2n} , W̄ · F_{α/2, 2, 2n})

where F_α/2 = qf(α/2, 2, 2n) and F_1−α/2 = qf(1 − α/2, 2, 2n) in R.

Tolerance Intervals

Three Types of "Tolerance"

1. Engineering tolerance: Fixed specification limits (S_L, S_U). Not statistical. The statistical question is simply: what proportion of output meets spec?

1. Natural tolerance: When F is completely known. For N(μ, σ²) with known μ and σ, the 100(1 − α)% natural tolerance interval is μ ± σZ_α/2.

   • Example: μ = 20, σ = 3 → 99% natural TI = 20 ± 3(2.576) = (12.27, 27.74)

1. Statistical tolerance interval: When F has unknown parameters that must be estimated from data. This is the type we focus on.

TI for a Normal Population

Case 1: μ and σ known. This is just the natural tolerance interval: (μ − σZ_(1−P)/2, μ + σZ_(1−P)/2) captures exactly 100P% of the population with 100% certainty (γ = 1).

Case 2: μ and σ unknown. Let Y₁, …, Y_n iid N(μ, σ²).

Two-sided 100(P, γ)% TI:

Ŷ ± K_P,γ · S

where K_P,γ is obtained from tables (e.g., Mason, Gunst & Hess). K_P,γ accounts for the uncertainty in estimating both μ and σ. It is always larger than Z_(1+P)/2 (the value if parameters were known).

One-sided tolerance bounds:

• Lower bound: L*_P,γ = Ŷ − K*_P,γ · S (at least 100P% of population exceeds this)

• Upper bound: U*_P,γ = Ŷ + K*_P,γ · S (at least 100P% of population falls below this)

K* values differ from the two-sided K values and have their own tables.

Applications:

• Warranty: "95% confident that 99% of light bulbs last at least L*_.99,.95 hours."

• Safety: "99% confident that 99.5% of paint containers have at most U*_.995,.99 units of lead."

R approximation (use only when P ≥ 0.95, γ ≥ 0.95, and n > 50):

Chi = qchisq(1 - G, n - 1)
z = qnorm((1 + P) / 2)
K2Side = sqrt(((n-1)*(n+1)*z^2) / (n * Chi))

Sample table values for 100(P, γ)% two-sided TI (K_P,γ):

Lower Tolerance Bound for Exponential Distribution

Setting: T₁, …, T_n iid Exp(β). Want a warranty value W such that at least 100P% of devices survive past W.

Case 1: β known. W_P = −β log(P).

• Example: β = 2000, P = 0.98 → W_.98 = −2000 log(0.98) = 40.4 hours.

Case 2: β unknown. Estimate with β̂ = T̄.

100(P, γ)% lower tolerance bound:

W_P,γ = −β̂ · [2n / χ²_{1−γ, 2n}] · log(P)

where χ²_{1−γ, 2n} = qchisq(γ, 2n) in R.

Derivation sketch: A(W_P,γ) ≥ P ⟺ e^−W/β ≥ P ⟺ (β̂/β)K_P,γ ≤ 1 ⟺ 2nT̄/β ≤ 2n/K_P,γ. Since 2nT̄/β ~ χ²(2n), set the right side equal to χ²_{1−γ, 2n}.

Example: n = 10, P = 0.90, γ = 0.95. K_P,γ = 2n / χ²_{0.05, 20} = 20 / 31.41 = 0.637. W_.9,.95 = −T̄ · 0.637 · log(0.9) = 0.067T̄.

If β were known, the bound would be −β log(0.9) = 0.105β, which is larger. The statistical bound is smaller because it accounts for estimation uncertainty.

Distribution-Free Tolerance Intervals

Setting: Y₁, …, Y_n iid with continuous cdf F. No parametric assumption.

100(P, γ)% TI: (Y_(r), Y_(n−s+1))

where r + s = m, and m is determined from the condition:

P[Y ≤ m − 1] ≤ 1 − γ, where Y ~ Bin(n, 1 − P)

R code to find m:

ans = pbinom(r-1, n, 1-P)  # for r = 1, 2, ..., n
# Find largest m such that ans[m] ≤ 1 − γ

One-sided bounds: Take r = 0 for an upper bound (0, Y_(n−m+1)) or s = 0 for a lower bound (Y_(m), ∞).

Example: n = 130, P = 0.90, γ = 0.95. From table/R: m = 8. Taking r = 4, s = 4, the TI is (Y₍₄₎, Y₍₁₂₇₎). True coverage γ = 0.9544.

Distribution-Free vs Parametric TI Width Comparison

For 100(0.9, 0.95)% TIs under a truly normal population:

Even at n = 100, the distribution-free TI is about 10% wider. This is the price of making no distributional assumptions.

Why It Matters / Exam Flags

⚠️ A PI is always wider than a CI for the same parameter (mean). The extra width reflects individual variability, not just estimation error.

⚠️ Tolerance intervals have two parameters (P and γ). Do not confuse them with CIs, which have only α.

⚠️ For the exponential lower tolerance bound, the chi-square df is 2n, not n − 1.

⚠️ Distribution-free TIs require larger sample sizes than parametric ones. At n = 100, they are still 10% wider under normality.

⚠️ The K_P,γ factor for a normal TI is not Z_(1+P)/2. It is always larger because it accounts for unknown μ and σ.

⚠️ As n → ∞, the PI for a normal population does not shrink to zero width. It converges to μ ± Z_α/2σ (the natural tolerance of a single observation).

Practice Q&A

Q: For n = 25 observations from a N(μ, σ²) population with Ŷ = 50 and S = 4, compute a 95% PI for the next observation.

A: t_{0.025, 24} = 2.064. PI = 50 ± 2.064 · 4 · √(1 + 1/25) = 50 ± 2.064 · 4 · 1.0198 = 50 ± 8.42 = (41.58, 58.42).

Q: Explain the difference between a 95% CI for μ and a (P=0.95, γ=0.95) tolerance interval.

A: The CI estimates the single number μ with 95% confidence. The TI estimates an interval that captures at least 95% of all individual population values, with 95% confidence that this capture is achieved. The TI is much wider because it must cover the spread of individuals, not just the location of the mean.

Q: For n = 10 exponential failure times with T̄ = 500, compute a (P=0.90, γ=0.95) lower tolerance bound.

A: W_P,γ = −T̄ · [2n / χ²_{0.05, 20}] · log(0.90) = −500 · [20/31.41] · (−0.10536) = 500 · 0.6366 · 0.10536 = 33.5 hours.

Q: Why is the distribution-free tolerance interval wider than the normal-based tolerance interval when the data really are normal?

A: The distribution-free TI must be valid for every continuous distribution, so it cannot exploit the specific shape of the normal. It uses order statistics placed conservatively, while the normal TI uses the efficient estimators Ŷ and S together with the known Gaussian structure.

Related Terms / Search Tags

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