Process capability and acceptance sampling

(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.