Tools › Quality › Control › Acceptance Sampling › Example
Montgomery Section 15.3: double sampling plan n1 = 50, c1 = 1, r1 = 3, n2 = 100, c2 = 3
| Tool | Acceptance Sampling |
| Source | Montgomery, D. C. (2013). Introduction to Statistical Quality Control, 7th ed. Wiley, Hoboken, NJ. Section 15.3. |
| Level | Undergraduate |
Problem
A double sampling plan takes a first sample of 50, accepting on 1 or fewer defectives and rejecting on 3 or more; otherwise a second sample of 100 is taken and the lot is accepted if the combined defectives do not exceed 3. For a lot fraction defective of 0.05, find the probability of acceptance and the average sample number.
Inputs to Enter
- Analysis 'Evaluate a double sampling plan'; n1 50, c1 1, r1 3, n2 100, c2 3 (the default values of those fields)
- OC model binomial; read the row p = 0.05 of the OC table (or set the LTPD to 5 %)
Answers
- Pa^I = P(d1 ≤ 1 | 50, 0.05) = 0.279; P(d1 = 2) = 0.2611
- Pa^II = P(d1 = 2) P(d2 ≤ 1 | 100, 0.05) = 0.0097
- Pa = 0.279 + 0.0097 = 0.289
- ASN = 50 + 100 × 0.2611 = 76.11
Notes
Switch the Analysis select to the double plan; the plan's fields already hold these values. Not the default selection.