6.2.3.2. Choosing a Sampling Plan with a given OC Curve (2024)

Number of defectives is approximately binomialIt is instructive to show how the points on this curve are obtained,once we have a sampling plan (\(n,c\)) -later we will demonstratehow a sampling plan (\(n,c\)) is obtained.

We assume that the lot size \(N\)is very large, as compared to the sample size \(n\),so that removing the sample doesn'tsignificantly change the remainder of the lot, no matter how manydefects are in the sample. Then the distribution of the number ofdefectives, \(d\), in a random sample of \(n\)items isapproximately binomial with parameters \(n\) and \(p\),where \(p\)is the fraction of defectives per lot.

The probability of observing exactly \(d\)defectives is given by

The binomialdistribution$$ P_d = f(d) = \frac{n!}{d! (n-d)!} p^d (1-p)^{n-d} \, .$$The probability of acceptance is the probability that \(d\),the number of defectives, is less than or equal to \(c\),the accept number. This means that$$ P_a = P(d \le c) = \sum_{d=0}^c \frac{n!}{d!(n-d)!} p^d(1-p)^{n-d} \, .$$Sample table for \(P_a\), \(P_d\) using the binomial distributionUsing this formula with \(n=52\), \(c=3\), and \(p = 0.01, \, 0.02, \, \ldots, \, 0.12\),we find:

\(P_a\)	\(P_d\)
0.998	0.01
0.980	0.02
0.930	0.03
0.845	0.04
0.739	0.05
0.620	0.06
0.502	0.07
0.394	0.08
0.300	0.09
0.223	0.10
0.162	0.11
0.115	0.12

Solving for (n,c)Equations for calculating a sampling planwith a given OC curveIn order to design a sampling plan with a specified OC curve oneneeds two designated points. Let us design a sampling plan such thatthe probability of acceptance is \(1-\alpha\)for lots with fraction defective \(p_1\)and the probability of acceptance is \(\beta\)for lots with fraction defective \(p_2\).Typical choices for thesepoints are: \(p_1\)is the AQL,\(p_2\)is theLTPDand \(\alpha, \, \beta\)are the Producer'sRisk (Type I error) and Consumer's Risk (Type II error), respectively.

If we are willing to assume that binomial sampling is valid, then thesample size \(n\),and the acceptance number \(c\)are the solution to$$ \begin{eqnarray}1 - \alpha & = & \sum_{d=0}^c \frac{n!}{d!(n-d)!} p_1^d (1-p_1)^{n-d} \\ \beta & = & \sum_{d=0}^c \frac{n!}{d!(n-d)!} p_2^d (1-p_2)^{n-d} \, .\end{eqnarray} $$These two simultaneous equations are nonlinear so there is no simple,direct solution. There are however a number of iterative techniquesavailable that give approximate solutions so that composition of acomputer program poses few problems.

Average Outgoing Quality (AOQ)Calculating AOQsWe can also calculate theAOQfor a (\(n,c\))sampling plan, provided rejected lots are 100 %inspected and defectives are replaced with good parts.

Assume all lots come in with exactly a \(p_0\)proportion of defectives. After screening a rejected lot, the finalfraction defectives will be zero for that lot. However, acceptedlots have fraction defective \(p_0\).Therefore, the outgoing lots from the inspection stations are a mixture of lots with fractions defective \(p_0\)and 0. Assuming the lot size is \(N\),we have.$$ \mbox{AOQ} = \frac{P_a p(N-n)}{N} \, . $$For example, let \(N=10000\), \(n=52\), \(c=3\),and \(p\), the quality of incoming lots, equals 0.03. Now at \(p = 0.03\),we glean from the OC curve table that \(p_a = 0.930\) and$$ \mbox{AOQ} = \frac{(0.930)(0.03)(10000-52)}{10000} = 0.02775 \, . $$

Sample table of AOQ versus \(p\)Setting \(p = 0.01, \, 0.02, \, \ldots, \, 0.12\),we can generate the following table.

AOQ	\(p\)
0.0010	0.01
0.0196	0.02
0.0278	0.03
0.0338	0.04
0.0369	0.05
0.0372	0.06
0.0351	0.07
0.0315	0.08
0.0270	0.09
0.0223	0.10
0.0178	0.11
0.0138	0.12

Sample plot of AOQ versus \(p\)A plot of the AOQversus \(p\)is given below.Interpretation of AOQ plotFrom examining this curve we observe that when the incoming qualityis very good (very small fraction of defectives coming in), then theoutgoing quality is also very good (very small fraction ofdefectives going out). When the incoming lot quality is very bad,most of the lots are rejected and then inspected. The "duds" areeliminated or replaced by good ones, so that the quality of theoutgoing lots, the AOQ,becomes very good. In between these extremes, the AOQrises, reaches a maximum, and then drops.

The maximum ordinate on the AOQcurve represents the worstpossible quality that results from the rectifying inspectionprogram. It is called the average outgoing quality limit,(AOQL).

From the table we see that the \(\mbox{AOQL} = 0.372\) at \(p=0.06\)for the above example.

One final remark: if \(N \gg n\),then the \(\mbox{AOQ} \approx P_a p\).

The AverageTotal Inspection (ATI)Calculating the Average Total InspectionWhat is the total amount of inspection when rejected lots arescreened?

If all lots contain zero defectives, no lot will be rejected.

If all items are defective, all lots will be inspected, and theamount to be inspected is \(N\).

Finally, if the lot quality is \(0 \lt p \lt 1\),the averageamount of inspection per lot will vary between the sample size \(n\),and the lot size \(N\).

Let the quality of the lot be \(p\)and the probability of lot acceptance be \(P_a\),then the ATIper lot is$$ \mbox{ATI} = n + (1 - P_a) (N - n) \, . $$For example, let \(N=10000\), \(n=52\), \(c=3\), and \(p = 0.03\).We know from the OC table that \(P_a = 0.93\).Then \(\mbox{ATI} = 52 + (1-0.930)(10000 - 52) = 753\).(Note that while 0.930 was rounded to threedecimal places, 753 was obtained using more decimal places.)

Sample table of ATI versus \(p\)Setting \(p = 0.01, \, 0.02, \, \ldots, \, 0.14\)generates the following table.

ATI	\(p\)
70	0.01
253	0.02
753	0.03
1584	0.04
2655	0.05
3836	0.06
5007	0.07
6083	0.08
7012	0.09
7779	0.10
8388	0.11
8854	0.12
9201	0.13
9453	0.14

Plot of ATI versus \(p\)A plot of ATIversus \(p\),the Incoming Lot Quality (ILQ)is given below.