113 Acceptance Sampling by Attributes Introduction Acceptance sampling is concerned with inspection and decision making regarding products. Three aspects of sampling are important: o Involves random sampling of an entire “lot” o Accept and Reject Lots (does not achieve quality improvement) o “Lot sentencing” o Audit tool Three approaches to lot sentencing: o Accept with no inspection o 100% inspection o Acceptance sampling Reasons for Acceptance Sampling, not 100% inspection o Testing is destructive o Cost of 100% inspection is high o 100% inspection is not feasible (require too much time) o If vendor has excellent quality history
Transcript
113
Acceptance Sampling by Attributes
Introduction
Acceptance sampling is concerned with inspection and decision
making regarding products.
Three aspects of sampling are important:
o Involves random sampling of an entire “lot”
o Accept and Reject Lots (does not achieve quality
improvement)
o “Lot sentencing”
o Audit tool
Three approaches to lot sentencing:
o Accept with no inspection
o 100% inspection
o Acceptance sampling
Reasons for Acceptance Sampling, not 100% inspection
o Testing is destructive
o Cost of 100% inspection is high
o 100% inspection is not feasible (require too much time)
o If vendor has excellent quality history
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Advantages of Sampling
o Less expensive
o Reduced damage
o Reduces the amount of inspection error
Disadvantages of Sampling
o Risk of accepting “bad” lots, rejecting “good” lots.
o Less information generated
o Requires planning and documentation
Types of Sampling Plans (attribute sampling plans)
o Single sampling plan
o Double-sampling plan
o Multiple-sampling plan
o Sequential-sampling
Lot Formation
Considerations before inspection
o Lots should be homogeneous
o Larger lots more preferable than smaller lots
o Lots should be conformable to the materials-handling
systems used in both the vendor and consumer facilities.
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Random Sampling
The units selected for inspection should be chosen at random.
Random samples are not used, bias can be introduced.
If any judgment methods are used to select the sample, the statistical
basis of the acceptance-sampling procedure is lost.
Definition of a Single-Sampling Plan
A single sampling plan is defined by sample size, n, and the
acceptance number c.
o N = lot size
o n = sample size
o c = acceptance number
o d = observed number of defectives
N total items in a lot. Choose n of the items at random. If c or less
number of items are defective, accept the lot.
The acceptance or rejection of the lot is based on the results from a
single sample - thus a single-sampling plan
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The OC Curves
The operating-characteristic (OC) curve measures the performance of
an acceptance-sampling plan.
The OC curve plots the probability of accepting the lot versus the lot
fraction defective.
c
d
dnda pp
dnd
ncdPP
0
)1()!(!
!}{
Example. If the lot fraction defective is 01.0p , n=89 and c=2, then
9397.0
)99.0()01.0(!87!2
!89)99.0()01.0(
!88!1
!89)99.0()01.0(
!89!0
!89
)99.0()01.0()!89(!
!89}2{
872881890
2
0
89
d
dda dd
dPP
The OC curve shows the probability that a lot submitted with a
certain fraction defective will be either accepted or rejected.
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Effect of OC curves
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Type-A OC curves – based on Hypergeometric distribution
Type-B OC curves – based on binomial distribution
Other aspects of OC Curve Behavior:
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AQL and LTPD
Acceptable quality level (AQL, 1p ) - poorest level of quality for the
vendor’s process that the consumer would consider to be acceptable
as a process average. The probability of such a process not being
accepted is the Producer’s risk.
Lot tolerance percent defective (LTPD, 2p ) – poorest level of quality
that the consumer is willing to accept in an individual lot. The
probability that a lot with lower quality level is accepted is the
Consumer’s risk. Also called rejectable quality level (RQL) or
limiting quality level (LQL)
AQL and LTPD can be used for the design of sampling plans
Designing a Single-Sampling Plan with a Specified OC Curve
Let the probability of acceptance be 1 for lots with fraction
defective 1p
Let the probability of acceptance be for lots with fraction defective
2p .
Assume binomial sampling (with type-B OC curves).
The sample size n and acceptance number c are the solution to
Example. For N=10,000, n = 89, c = 2 and p =0.01, we have aP =0.9397.
Then 687)8910000)(9397.01(89))(1( nNPnATI a
AOQL is the maximum point on
the curve
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Double Sampling Plans
Procedure
o 1n = sample size of the first sample
o 1c = acceptance number of the first sample
o 2n = sample size of the second sample
o 2c = acceptance number for both samples
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Example. For the plan with 1n =50, 1c =1, 2n =100, 2c =3, a random sample
of 50 will be taken from the lot. If 11 d , the lot will be accepted. If
31 d , the lot will be rejected. If 21 d or 31 d , the second sample of
100 will be taken. If 321 dd , the lot will be accepted. If 321 dd , the
lot will be rejected.
Remarks
o Possible less inspection
o Second chance
o More complicated
o Less inspection may not be realized
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Multiple Sampling Plans
Similar to double sampling
Possible less inspection
More complicated
May be further discussed later
127
Military Standard 105E (ANSI/ASQC Z1.4, ISO 2859)
Developed during World War II
Widely used acceptance-sampling system for attributes
Gone through four revisions since 1950.
a collection of sampling schemes to make a sampling system
Based on AQL
Description of the Standard
Three types of sampling are provided for:
1. Single
2. Double
3. Multiple
Provisions for each type of sampling plan include
1. Normal inspection
2. Tightened inspection
3. Reduced inspection
The AQL is generally specified in the contract or by the authority
responsible for sampling
Different AQLs may be designated for different types of defects
Defects include critical defects, major defects, and minor defects
Tables are used to determine the appropriate sampling scheme
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Switching Rules
1. Normal to tightened – 2 out of five lots are rejected
2. Tightened to normal – 5 lots are accepted
3. Normal to reduced
10 lots have been accepted under normal inspection
total number of defectives of the 10 lots is less than
given limit
stable production
authorized
4. Reduced to normal
a lot is rejected
procedure terminates without meeting acceptance or
rejection criteria. Accept the lot and change to normal
production is not stable
other conditions
5. Discontinuance of inspection – 10 consecutive lots remain on
tightened inspection. Discontinue and take actions.
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Procedure
1. Choose the AQL
2. Choose the inspection level
3. Determine the lot size
4. Find the appropriate sample size code letter from Table 15-4
5. Determine the appropriate type of sampling plan to use
(single, double, multiple)
6. Enter the appropriate table to find the type of plan to be used.
7. Determine the corresponding normal and reduced inspection
plans to be used when required
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Example
Suppose a product is submitted in lots of size N = 2000. The
AQL is 0.65%. Assume that we want to generate normal single-
sampling plans. For lots of size 2000 and general inspection level II,
Table 15-4 indicates the appropriate sample size code letter is K.
From Table 15-5 for single-sampling plans under normal
inspection, the normal inspection plan is n = 125, c = 2. This means that
we accept the lot if there are 2 or less defective units in a random sample
of 125. We reject the lot if there are 3 or more defective units.
If tightened inspection is to be used after inspecting 5 lots with
normal inspection, then Table 15-6 shows that n = 125, c =1 for
tightened inspection. This means that we accept the lot if there is 1 or 0
defective units in a random sample of 125. We reject the lot if there are
2 or more defective units.
If reduced inspection can be used after accepting 10 consecutive
lots with normal inspection, and all other conditions satisfied, then Table
15-7 shows that in the reduced inspection, the sample size is n =50 Ac=1
and Re=3. This means that:
If there are 1 or less defectives in the sample, we will accept the lot
If there are 3 or more defective units, we will reject the lot and use
normal inspection for inspecting the next lot.
If there are 2 defective units, we will accept the lot and use normal
inspection for inspecting the next lot.
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132
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Discussion
Several points to be emphasized:
MIL STD 105E is AQL-oriented
The sample sizes selected for use in MIL STD 105E are limited
The sample sizes are related to the lot sizes.
Switching rules from normal to tightened and from tightened to
normal are subject to some criticism.
A common abuse of the standard is failure to use the switching rules
at all.
ANSI/ASQC Z1.4 or ISO 2859 is the civilian standard counterpart of
MIL STD 105E.
Differences include:
1. Terminology “nonconformity”, “nonconformance”, and
“percent nonconforming” is used
2. Switching rules were changed slightly to provide an option for
reduced inspection without the use of limit numbers
3. Several tables that show measures of scheme performance were
introduced
4. A section was added describing proper use of individual
sampling plans when extracted from the system
5. A figure illustrating switching rules was added
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Dodge-Romig Sampling Plans
Based on AOQL or LTPD
Use developed tables
AOQL plans
o Minimize average total inspection
o Rejected lots will be 100% inspected
o Fraction nonconforming is known or can be estimated
o The plan also presents the LTPD values corresponding to
10.0aP on the OC curve of the plan. Or 90% of the lots will
be rejected if its percent defective is higher than the
corresponding LTPD value.
Example. N=5000, p=1%. From Table 15-8, we find that the plan with
AOQL=3% will be n=65, c=3. The corresponding LTPD value at
10.0aP is 10.3%.
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If the incoming quality is indeed at the level of p=1%, we can calculate
or check the corresponding OC curve to see that the acceptance
probability is 9957.0aP for p=1%. Then the average total inspection
will be: 22.86)655000)(9957.01(65))(1( nNPnATI a
LTPD plans
o Provide plans for different LTPD values with lot acceptance
probability of 10%.
Example. N=5000 and incoming percent defective is p=0.25%. We can
find from Table 15-9 that the plan with LTPD=1% will be n=770, c=4.
The corresponding AOQL value for this plan is 0.28%.
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Similarly, we can find that if indeed that the incoming percent defective
is p=0.25%, we can calculate or check the corresponding OC curve to
see that the acceptance probability is 9541.0aP for p=0.25%. Then the
average total inspection will be:
62.961)7705000)(9547.01(770))(1( nNPnATI a
Item-by-Item Sequential Sampling Plans
The procedure is to take one unit from the lot to test at one time and
continue for a number of items. Based on the test results, the entire lot
will be accepted or rejected. In doing so, it may reduce the total number
of items to be tested. This procedure is also indexed on 1p , 2p and .
It calculates the following 2 lines for each sample:
snhX A 1 (acceptance line)
snhX R 2 (rejection line)
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If the total number of nonconforming units is less than or equal to the integer part of AX , we accept the lot. If the total number of nonconforming units is equal to or greater than the integer above RX , we reject the lot. These parameters are calculated by:
kh /1
log1
, kh /1
log2
,)1(
)1(log
21
12
pp
ppk
and
kp
ps /
1
1log
2
1
Example. Assume that 1p =0.01, =0.05 , 2p =0.06 and =0.10, then:
80066.094.001.0
99.006.0log
)1(
)1(log
21
12
pp
ppk
22.180066.0/10.0
95.0log/
1log1
kh
57.180066.0/05.0
90.0log/
1log2
kh
028.080066.0/94.0
99.0log/
1
1log
2
1
kp
ps
Then we have:
snhX A 1 = n028.022.1 (acceptance line) snhX R 2 n028.057.1 (rejection line)
When n = 1, we have:
snhX A 1 = n028.022.1 = 192.1028.022.1 snhX R 2 598.1028.057.1028.057.1 n
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As we just take one sample, these results tell us nothing. When n = 2, we have: snhX A 1 = 164.1056.022.12028.022.1
snhX R 2 626.1056.057.12028.057.1 So it says nothing about accepting the lot but the lot will be rejected if both items are bad. For this plan, the process continues for AX until n = 44 when AX becomes positive. On the other hand, the rejection numbers are shown in Table 15-3.
As the process continues, if there are 1]int[ RX bad ones, reject the lot. If the first 44 are all good ones, accept the lot. Otherwise, continue and stop sampling when you reach 267389 items. The number of 89 corresponds to the single sampling plan for 1p =0.01, =0.05 , 2p =0.06 and =0.10.
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A similar example is to assume that 1p =0.01, =0.05 , 2p =0.10 and =0.10. In this case, we have:
snhX A 1 = n04.0939.0 (acceptance line) snhX R 2 n04.0205.1 (rejection line)
The results for n =1, 2, 3... 26 are tabulated below.
n accept reject n accept reject 1 x x 14 x 2 2 x 2 15 x 2 3 x 2 16 x 3 4 x 2 17 x 3 5 x 2 18 x 3 6 x 2 19 x 3 7 x 2 20 x 3 8 x 2 21 x 3 9 x 2 22 x 3
10 x 2 23 x 3 11 x 2 24 0 3 12 x 2 25 0 3 13 x 2 26 0 3
