Acceptance Sampling
• Acceptance sampling is a method used to accept or
reject product based on a random sample of the
product.
• The purpose of acceptance sampling is to sentence
lots (accept or reject) rather than to estimate the
quality of a lot.
• Acceptance sampling plans do not improve quality.
The nature of sampling is such that acceptance
sampling will accept some lots and reject others
even though they are of the same quality.
• The most effective use of acceptance sampling is as
an auditing tool to help ensure that the output of a
process meets requirements.
• As mentioned acceptance sampling can reject
“good” lots and accept “bad” lots. More formally:
Producers risk refers to the probability of rejecting a
good lot. In order to calculate this probability there
must be a numerical definition as to what
constitutes “good”
– AQL (Acceptable Quality Level) - the numerical
definition of a good lot. The ANSI/ASQC standard
describes AQL as “the maximum percentage or
proportion of nonconforming items or number of
nonconformities in a batch that can be considered
satisfactory as a process average”
• Consumers Risk refers to the probability of
accepting a bad lot where:
– LTPD (Lot Tolerance Percent Defective) - the
numerical definition of a bad lot described by the
ANSI/ASQC standard as “the percentage or
proportion of nonconforming items or
noncomformities in a batch for which the customer
wishes the probability of acceptance to be a specified
low value.
• The Operating Characteristic Curve is typically
used to represent the four parameters (Producers
Risk, Consumers Risk, AQL and LTPD) of the
sampling plan as shown below where the P on the x
axis represents the percent defective in the lot:
• Note: if the sample is less than 20 units the
binomial distribution is used to build the OC Curve
otherwise the Poisson distribution is used.
OC Curve
1.2
Producers Risk
Probability of Acceptance
1
0.8
0.6
Consumers Risk
0.4
0.2
0
AQL
LTPD
P
Designing Sampling Plans
• Operationally, three values need to be determined
before a sampling plan can be implemented (for
single sampling plans):
– N = the number of units in the lot
– n = the number of units in the sample
– c = the maximum number of nonconforming units in
the sample for which the lot will be accepted.
• While there is not a straightforward way of
determining these values directly given desired
values of the parameters, tables have been
developed. Below is an excerpt of one of these
tables.
Excerpt From a Sampling Plan Table with Producers Risk = 0.05 and Consumers Risk = 0.10
C
LTPD/AQL
n(AQL)
n(LTPD)
0
44.89
0.052
2.334
1
10.946
.355
3.886
2
6.509
.818
5.324
3
4.89
1.366
6.68
4
4.057
1.97
7.992
5
3.549
2.613
9.274
6
3.206
3.286
10.535
7
2.957
3.981
11.772
8
2.768
4.695
12.996
9
2.618
5.426
14.205
Three alternatives for specifying
sampling plans
• Producers Risk and AQL specified
• Consumers Risk and LTPD specified
• All four parameters specified
• For the first two cases we simply choose the
acceptance number and divide the appropriate
column by the associated parameter to get the
sample size.
– Example 1: Given a producers risk of .05 and an
AQL of .015 determine a sampling plan
• c = 1: n = .355/.015 ~ 24
• c = 4: n = 1.97/.015 ~ 131
– Example 2: Given a consumers risk of .1 and a
LTPD of .08 determine a sampling plan
• c = 0: n = 2.334/.08 ~ 29
• c = 5: n = 9.274/.08 ~ 116
• Note: you can check these with the OC curve
posted on my web site
All four parameters specified
• When all four parameters are specified we must
first find a value close to the ratio LTPD/AQL in
the table. Then values of n and c are found.
• Example: Given producers risk of .05, consumers
risk of .1, LTPD 4.5%, and AQL of 1% find a
sampling plan.
– Since 4.5/1 = 4.5 is between c= 3 and c = 4. Using
the n(AQL) column the sample sizes suggested are
137 and 197 respectively. Note using this column
will ensure a producers risk of .05. Using the
n(LTPD) column will ensure a consumers risk of .1
Double Sampling
• In an effort to reduce the amount of inspection
double (or multiple) sampling is used. Whether or
not the sampling effort will be reduced depends on
the defective proportions of incoming lots.
Typically, four parameters are specified:
n1
c1
n2
c2
=
=
=
=
number of units in the first sample
acceptance number for the first sample
number of units in the second sample
acceptance number for both samples
Procedure
• A double sampling plan proceeds as follows:
A random sample of size n1 is drawn from the lot.
If the number of defective units (say d1 )  c1 the
lot is accepted.
If d1 c2 the lot is rejected.
If neither of these conditions are satisfied a second
sample of size n2 is drawn from the lot.
If the number of defectives in the combined
samples (d1 + d2) > c2 the lot is rejected. If not the
lot is accepted.
Example
• Consider a double sampling plan with n1 = 50; c1 =
1; n2 = 100; c2 = 3. What is the probability of
accepting a lot with 5% defective?
• The probability of accepting the lot, say Pa, is the
sum of the probabilities of accepting the lot after
the first and second samples (say P1 and P2)
• To calculate P1 we first must determine the
expected number defective in the sample (50*.05)
we can then use the Poisson function to determine
the probability of finding 1 or fewer defects in the
sample, e.g., Poisson(1,2.5,true) = .287
• To calculate P2 we need to determine first the
scenarios that would lead to requiring a second
sample. Then for each scenario we must determine
the probability of acceptance. P2 will then be the
some of the probability of acceptance for each
scenario.
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Example continued
For this example, there are only two possible scenarios
that would lead to requiring a second sample:
Scenario 1: 2 nonconforming items in the first sample
Scenario2: 3 nonconforming items in the first sample.
The probability associated with these two scenarios is
found as follows:
P(scenario 1) = Poisson(2,2.5,false) = .257
P(scenario 2) = Poisson(3,2.5,false) = .214
Under scenario 1 the lot will be accepted if there are
either 1 or 0 nonconforming items in the second sample.
The expected number of nonconforming items in the
second sample .05(100) = 5
the probability is then: Poisson(1,5,true) = .04
therefore the probability of acceptance under scenario 1
is (.257)*(.04) = .01
Under scenario 2 we will accept the lot only if there are 0
nonconforming items in the second sample, i.e.,
Poisson(0,5,false) = .007
therefore the probability of acceptance under scenario 2
is (.214)*(.007) = .0015
so the probability of acceptance on the second sample is
.01 + .0015 = .0115
So the overall probability of acceptance is (probability of
acceptance on first sample + probability of accepting
after 2 samples) = .287 + .0115 = .299
OC curve for double sampling
p
0.11
0.11
0.1
0.09
0.08
0.07
0.07
0.06
0.05
0.04
0.03
0.03
0.02
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.01
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
Pa
OC Chart - Double Sampling Plan
P1
Pa = P1 + P2
P(rejection 1st)
Average Sample Number
• Since the goal of double sampling is to reduce the
inspection effort, the average number of units
inspected is of interest. This is easily calculated as
follows
• Let PI = the probability that a lot disposition
decision is made on the first sample, i.e.,
(probability that the lot is rejected on the first
sample + probability that the lot is accepted on the
first sample)
• Then the average number of items is:
n1(PI) +(n1 + n2)(1 - PI)
= n1 + n2(1 - PI)
The average sample number curve
can be derived from the data on
the OC chart
ASN Curve
100
80
60
40
20
p
0.12
0.11
0.11
0.1
0.09
0.09
0.08
0.07
0.06
0.06
0.05
0.04
0.04
0.03
0.02
0.02
0.01
0
0
Average sample number
120