Acceptance sampling: OC curves, single and double sampling

Why it matters

When a lot of 2000 bought-out bolts arrives at the stores, inspecting every one is slow and costly, and for destructive tests impossible. Acceptance sampling decides whether to accept or reject the whole lot from a small random sample, with known risks to both the supplier and the buyer. The OC curve is the tool that makes those risks visible and lets two companies agree on a fair plan.

Key ideas

What sampling does and does not do. Acceptance sampling is a decision rule for lots; it does not control or improve the process that made them. It is used for incoming material, between departments and for destructive tests, and is gradually replaced by supplier SPC and certification as trust grows.

Single sampling plan (N, n, c). From a lot of N items take a random sample of n. Count the defectives d. Accept the lot if d ≤ c (the acceptance number); otherwise reject it.

Probability of acceptance. If the lot fraction defective is p, the number of defectives in the sample follows:

• the hypergeometric distribution exactly (finite lot, sampling without replacement);
• the binomial distribution when n is small compared with N (n/N < about 0.1);
• the Poisson distribution with mean np when n is large and p small (commonly n ≥ 20, p ≤ 0.1). This is the usual GATE approximation and is what standard Pa tables use. Pa is the probability of d ≤ c: a sum of terms from 0 to c, not just the term for d = c.

The OC curve. A plot of Pa (y-axis) against lot quality p (x-axis). The ideal curve is a vertical step: accept every lot better than the agreed quality, reject every worse lot. Real curves are S-shaped. Effects of the parameters:

• Increasing n (with c/n fixed) makes the curve steeper, i.e. more discriminating.
• Increasing c (with n fixed) moves the curve to the right: the plan becomes more lenient.
• c = 0 plans give a curve that drops sharply from p = 0; they look strict but reject many reasonably good lots.
• The lot size N has only a small effect provided n/N is small; the sample size, not the sampling percentage, governs protection.

Points on the OC curve.

• AQL (acceptable quality level) p₁: the worst quality the consumer regards as satisfactory as a process average. The producer's risk α = 1 − Pa(p₁) is the probability of rejecting such a good lot (typically 5 %; a Type I error).
• LTPD (lot tolerance percent defective) or RQL p₂: the poorest quality the consumer will tolerate in an individual lot. The consumer's risk β = Pa(p₂) is the probability of accepting such a bad lot (typically 10 %; a Type II error). A plan is designed so that its OC curve passes close to (p₁, 1 − α) and (p₂, β).

Rectifying inspection, AOQ and AOQL. If rejected lots are 100 % inspected and all defectives replaced, the outgoing quality improves. The average outgoing quality AOQ rises with p at first, reaches a maximum, then falls (bad lots are mostly rejected and screened). The maximum is the AOQL, the worst average outgoing quality the plan can deliver. The ATI (average total inspection) counts the inspection effort.

Double sampling (n₁, c₁, n₂, c₂). Take a first sample n₁ with d₁ defectives:

• accept if d₁ ≤ c₁; reject if d₁ > c₂;
• if c₁ < d₁ ≤ c₂, take a second sample n₂ with d₂ defectives and accept if d₁ + d₂ ≤ c₂, otherwise reject. Very good and very bad lots are decided on the smaller first sample, so the average sample number (ASN) is usually lower than for an equivalent single plan; the price is more complex administration and variable workload. Multiple and sequential plans extend the idea further. Standard plans are tabulated in IS 2500 / ISO 2859 (by AQL and lot size) and in Dodge–Romig tables (by LTPD or AOQL); take plan values from these tables, not from memory.

Formulas

Pa = Σ (d = 0 to c) C(n, d) · pᵈ · (1 − p)ⁿ⁻ᵈ

• Binomial Pa for a single plan. n = sample size; c = acceptance number; p = lot fraction defective; C(n, d) = n! / [d!(n − d)!].

Pa ≈ Σ (d = 0 to c) e^(−λ) · λᵈ / d!, with λ = n·p

• Poisson approximation.

α = 1 − Pa(AQL), β = Pa(LTPD)

AOQ = Pa · p · (N − n) / N (≈ Pa · p when N ≫ n)

• Average outgoing quality (fraction defective) with rectifying inspection. N = lot size.

ATI = n + (1 − Pa)(N − n)

• Average total inspection per lot, single sampling with rectification.

Pa = P(d₁ ≤ c₁) + Σ (over c₁ < d₁ ≤ c₂) P(d₁) · P(d₂ ≤ c₂ − d₁)

• Double sampling probability of acceptance.

ASN = n₁ + n₂ · P(c₁ < d₁ ≤ c₂)

• Average sample number for double sampling (with no curtailment).

Worked examples

Example 1 (standard): single sampling with rectification Given: N = 2000, n = 50, c = 2. Incoming lot quality p = 0.02. Find Pa (binomial), AOQ and ATI.

1. P(0) = 0.98⁵⁰ = 0.3642.
2. P(1) = 50 × 0.02 × 0.98⁴⁹ = 0.3716.
3. P(2) = C(50, 2) × 0.02² × 0.98⁴⁸ = 1225 × 0.0004 × 0.3792 = 0.1858.
4. Pa = 0.3642 + 0.3716 + 0.1858 = 0.9216.
5. AOQ = Pa · p · (N − n)/N = 0.9216 × 0.02 × 1950/2000 = 0.0180 (1.80 %).
6. ATI = 50 + (1 − 0.9216) × 1950 = 50 + 152.9 = 203 items per lot.

Pa = 0.922, AOQ = 1.80 % defective, ATI ≈ 203 items per lot. (The Poisson approximation with λ = 1 gives Pa = 0.920, very close.)

Example 2 (GATE level): double sampling Given: plan n₁ = 50, c₁ = 1, n₂ = 100, c₂ = 3; lot quality p = 0.02. Use the Poisson approximation. Find Pa and ASN.

1. First sample: λ₁ = 50 × 0.02 = 1.0. P(0) = e⁻¹ = 0.3679, P(1) = 0.3679, P(2) = 0.1839, P(3) = 0.0613.
2. Accept on first sample: P(d₁ ≤ 1) = 0.3679 + 0.3679 = 0.7358.
3. Second sample needed if d₁ = 2 or 3. λ₂ = 100 × 0.02 = 2.0: P(0) = e⁻² = 0.1353, P(1) = 0.2707, so P(d₂ ≤ 1) = 0.4060.
4. Accept on second sample: P(d₁ = 2)·P(d₂ ≤ 1) + P(d₁ = 3)·P(d₂ = 0) = 0.1839 × 0.4060 + 0.0613 × 0.1353 = 0.0747 + 0.0083 = 0.0830.
5. Pa = 0.7358 + 0.0830 = 0.8187.
6. Probability a second sample is needed = 0.1839 + 0.0613 = 0.2453; ASN = 50 + 100 × 0.2453 = 74.5.

Pa ≈ 0.819; average sample number ≈ 74.5 items (against 150 if both samples were always taken).

Common mistakes

• Computing only the term P(d = c) instead of the cumulative sum from 0 to c.
• Swapping producer's and consumer's risk: α is rejecting a good lot (at AQL), β is accepting a bad lot (at LTPD).
• Believing a fixed percentage sample (say 10 % of every lot) gives equal protection; protection depends on n, not n/N.
• In double sampling, forgetting that the second-stage acceptance uses the cumulative count d₁ + d₂.
• Using the Poisson approximation with λ = p instead of λ = n·p.

For GATE PI

Expect Pa calculations for small single plans (binomial or Poisson with given e⁻λ values), identifying α and β from an OC curve or plan, the effect of n and c on the OC curve, AOQ and AOQL concepts, and occasionally a double-sampling Pa. Practise cumulative Poisson sums quickly and keep four decimal places until the end.

Quick check

1. For n = 20, c = 0 and p = 0.03, what is Pa?
2. Which risk is evaluated at the AQL?
3. What happens to the OC curve when c is increased with n fixed?
4. In a double plan with n₁ = 30, n₂ = 60, the second sample is needed with probability 0.2. What is the ASN?

Answers: 1. 0.97²⁰ = 0.544. 2. Producer's risk α. 3. It shifts right (more lenient). 4. 42.