Tools › Quality › Control › Acceptance Sampling › Example
Montgomery Chapter 15: OC curve, AOQL and ATI of the single sampling plan n = 89, c = 2
| Tool | Acceptance Sampling |
| Source | Montgomery, D. C. (2013). Introduction to Statistical Quality Control, 7th ed. Wiley, Hoboken, NJ. Table 15.1, Sections 15.2.2 and 15.2.3. |
| Level | Undergraduate |
Problem
Lots are inspected with a sample of 89 and an acceptance number of 2. Tabulate the probability of acceptance against lot fraction defective, state the producer's risk at an AQL of 1 % and the consumer's risk at an LTPD of 6 %, and under rectifying inspection of 10,000-unit lots find the average outgoing quality, its limit and the average total inspection.
Inputs to Enter
- Analysis 'Evaluate a single sampling plan'; n = 89, c = 2 (loaded by default)
- AQL 1.0 %, LTPD 6.0 %, OC model binomial
- Lot size 10,000 for the AOQ / ATI of Section 15.2.3 (the default lot is 2,000)
Answers
- Pa: p = 0.01 → 0.9397, 0.02 → 0.7366, 0.03 → 0.4985, 0.04 → 0.3042, 0.05 → 0.1721, 0.06 → 0.0919, 0.07 → 0.0468, 0.08 → 0.0230, 0.09 → 0.0109, 0.10 → 0.0050 (Table 15.1)
- α = 1 − 0.9397 = 0.0603, β = Pa(0.06) = 0.0919
- AOQ at p = 0.01: 0.9397 × 0.01 × 9911/10000 = 0.0093; ATI = 89 + 0.0603 × 9911 = 687
- AOQL = 0.0155 (book, from the curve; exact maximum 0.01525 at p = 0.0253)
Notes
The default inputs are this plan with the lot of 2,000 from the Z1.4 example; enter 10,000 as the lot size to match the book's AOQ and ATI numbers. The book's Pa(0.05) is 0.1721 (a value of 0.4049 sometimes quoted for this plan is wrong).