--- title: "Process capability and acceptance sampling" subject: "Metrology" chapter: 15 key: "semester_3/metrology:process-capability-and-sampling-inspection" source: https://www.bctengineer.com/q/semester_3/metrology/process-capability-and-sampling-inspection site: BCT Engineering --- # Process capability and acceptance sampling Metrology · Chapter 15: Quality Control Management - Asked in: year not given, 3+5 marks - Video: [Acceptance Sampling and ANSI Z1.4 Sampling Procedure Explained](https://www.youtube.com/watch?v=cEY4bN9ko7A) — CQE Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/metrology/process-capability-and-sampling-inspection - Whole chapter: https://www.bctengineer.com/md/semester_3/metrology/chapter-15 ## Question (a) Write short notes on acceptance sampling (single and double sampling plans). (b) A shaft diameter is specified as $25.00 \pm 0.06$ mm. The process is in control with mean 25.01 mm and standard deviation 0.018 mm. Calculate $C_p$ and $C_{pk}$, the percentage of shafts outside the limits (assume normal distribution), and comment. ## Short answer Acceptance sampling inspects a random sample to decide lot acceptance. Single plan: sample n, accept if defectives ≤ c. Double plan uses two stages for less inspection. For the shaft: Cp = (25.06-24.94)/(6*0.018) = 1.11, Cpk = min((25.06-25.01)/(3*0.018), (25.01-24.94)/(3*0.018)) = 0.926. Outside ≈ 0.28%; off-centre mean lowers Cpk. ## Answer #### (a) Acceptance sampling In **acceptance sampling** a random sample is drawn from a lot and inspected; the lot is accepted or rejected from the result. It is used when 100 % inspection is too costly, too slow or destructive. - **Single sampling plan:** sample size $n$ and acceptance number $c$. If the number of defectives found is $\le c$, the lot is accepted; otherwise it is rejected. Simple and easy to administer. - **Double sampling plan:** a first small sample $n_1$ is taken. If defectives $\le c_1$ accept; if $\ge c_2$ reject; if in between, take a second sample $n_2$ and decide on the combined count. It needs on average less inspection and gives a second chance to good lots. The **OC (operating characteristic) curve** shows the probability of accepting a lot against its fraction defective. Producer's risk $\alpha$ (rejecting a good lot at AQL) and consumer's risk $\beta$ (accepting a bad lot at LTPD) are fixed by choice of $n$ and $c$. #### (b) Process capability Specification limits: $USL = 25.06$, $LSL = 24.94$ (tolerance $= 0.12$ mm). $$ C_p = \frac{USL - LSL}{6\sigma} = \frac{0.12}{6(0.018)} = 1.11 $$ $$ C_{pu} = \frac{USL - \mu}{3\sigma} = \frac{25.06-25.01}{0.054} = 0.926, \qquad C_{pl} = \frac{\mu - LSL}{3\sigma} = \frac{25.01-24.94}{0.054} = 1.296 $$ $$ C_{pk} = \min(C_{pu}, C_{pl}) = 0.926 $$ Fraction outside the limits: $$ z_{upper} = \frac{25.06 - 25.01}{0.018} = +2.78, \qquad z_{lower} = \frac{24.94 - 25.01}{0.018} = -3.89 $$ - Above $USL$: $P(z > 2.78) = 0.00274$ - Below $LSL$: $P(z < -3.89) = 0.00005$ - Total $\approx 0.00279$, i.e. **0.28 %** (about 2800 parts per million) **Comment:** $C_p > 1$ means the process spread fits the tolerance, but $C_{pk} < C_p$ shows that the process is off-centre (mean 0.01 mm above the middle of the tolerance). Re-centring on 25.00 mm would give $C_{pk} = C_p = 1.11$ and a reject rate of about 0.09 %. A $C_{pk}$ of 1.33 or more is usually required. **Answer: $C_p$ = 1.11, $C_{pk}$ = 0.926; about 0.28 % of shafts out of limits.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In a single sampling plan with sample size n and acceptance number c, when is the lot accepted? A. When no defective is found at all B. When defectives exceed c C. When the number of defectives found is c or fewer D. After taking a second sample Answer: C. Defectives <= c means accept; otherwise reject. (Key point: Single sampling) 2. What does an OC curve show and what risks relate to it? A. Probability of acceptance versus fraction defective; producer's risk alpha at AQL and consumer's risk beta at LTPD B. Number of defectives versus time; only producer's risk C. Cost against sample size; no risks D. Process mean against time; consumer's risk only Answer: A. The plan (n, c) fixes alpha (rejecting a good lot) and beta (accepting a bad lot). (Key point: OC curve) 3. For 25.00 +/- 0.06 mm with sigma = 0.018 mm, Cp is: A. 0.926 B. 1.11 (0.12 / 0.108) C. 2.22 D. 1.30 Answer: B. Cp = (USL - LSL)/(6 sigma) = 0.12/0.108 = 1.11. (Key point: Cp) 4. With mean 25.01, why is Cpk = 0.926 less than Cp? A. Cpk = max(Cpu, Cpl) and sigma is large B. The process is in perfect centre C. Cpk ignores the mean D. Cpk = min(Cpu, Cpl) and the process is off-centre toward the upper limit Answer: D. Cpu = 0.05/0.054 = 0.926 is the smaller value; re-centring at 25.00 would give Cpk = Cp = 1.11. 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