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INTRODUCTION TO ACCEPTANCE
SAMPLING PLAN
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HISTORY
Acceptance sampling plan was applied by the US
military to the testing of bullet during World War ll. If
every bullet was tested in advance, no bullets
would be left to ship, since testing was required for
firing. If on the other hand, none were tested,malfunction might occur in the battle field with
potential disastrous result.
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WHY ACCEPTANC SAMPLING
The are several situations when 100% inspection is not
practical:
When testing is destructive, otherwise all the products will be
lost. When inspection cost is very high.
When many similar products are
to be tested.
When efforts required for testing
is very high.
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When time and technology limitations are very high.
When lot size is very large.
Suppliers quality history is good enough to justify less than
100% inspection.
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ADVANTAGES
It is economical.
It requires less time and less effort.
It requires less personnel, etc.
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DISADVANTAGES
There is always high risk for both the producer and
the customer.
Requires expertise in statistical aspects.
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SOME DEFINITIONS RELATED TO
SAMPLING PLAN
Acceptance Quality Level (AQL): this is the
poorest quality level of the suppliers process that
the customer would consider to be acceptable as a
process average.
Lot Tolerance Percentage Defect (LTPD): This is
the poorest quality that a consumer is willing to
tolerate in an individual lot.
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Customers Risk: this is the probability of
accepting a bad lot.
Producers Risk: this is the probability of rejecting
a good lot.
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SINGLE SAMPLING PLAN
A single sampling plan is the simplest and shortest
plan. Here decision is based on single trial. A
sample size (n) is drawn from a batch or lot (N). If
the number of non conforming units is less or equal
to a predetermined number(c),then the lot is accepted, otherwise rejected.
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Let lot size, N=1000, sample size=30, acceptance number,
c=3.
This means that the batch to be inspected contains 1000pieces of garments. 30 pieces garments are randomly
drawn from the batch and inspected. If the number of
nonconforming units if found less than or equal to 3 pieces,
then entire batch is accepted, if the number of
nonconforming units is more than 3 pieces, then the entirebatch is rejected.
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OPERATING CHARACTERISTIC CURVEOC curve
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PERMUTATIONAND COMBINATION
A permutationis any arrangement
ofr objects selected from n
possible objects. The order of
arrangement is important in
permutations.
EXAMPLE
Suppose that in addition toselecting the group, he must also
rank each of the players in that
starting lineup according to their
ability.
A combination is the number of ways to
choose r objects from a group ofn
objects without regard to order.
EXAMPLE
There are 12 players on the Carolina
Forest High School basketball team.Coach Thompson must pick five players
among the twelve on the team to
comprise the starting lineup. How many
different groups are possible?
792)!512(!5
!12512
C040,95)!512(
!12512
P
5-12
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PROBABILITYOFACCEPTANCE
SINGLESAMPLINGPLAN
Suppose,
N = 1000
n = 30
c = 2
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PROBABILITYOFACCEPTANCEWHENACTUAL
NONCONFORMANCE 1% OR 0.01
P(0) = 30C0x (0.01)0x (1-0.01)30-0
=0.7397
P(1) = 30C1x (0.01)1x (1-0.01)30-1
=0.22415
P(2) = 30C2x (0.01)2x (1-0.01)30-2
=.03283
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Total probability of acceptance when actual non
conformance 1% or 0.01
Pa (0.01) = P(0) + P(1) + P(2)
= 0.7397 + 0.22415 + 0.03283
= 0.99668= 99.668 %
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Total probability of acceptance when actual non
conformance 15% or 0.15
Pa (0.15) = P(0) + P(1) + P(2)
= 0.00763 + 0.04039 + 0.10337
= 0.15139= 15.139%
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OC CURVE SINGLE SAMPLING PLAN4
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This curve plots the probability of accepting the lot
(Y-axis) versus the lot fraction nonconforming or
percent defectives (X-axis).
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DOUBLE SAMPLING PLAN
Lot Size = N
Sample size in the 1st trial = n1
Sample size in the 2nd trial = n2
Acceptance no. for 1st
trial = c12nd Acceptance no.= c2
No. of defects for 1st sample = D1
No. of defects for 2nd sample = D2
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In the first trial, sample n1 is taken from the lot N. If the
defects D1 is less or equal to acceptance no c1, then
the entire lot is accepted, without requiring the second
trial. If D1 is greater than acceptance no c2, then the
entire lot is rejected without requiring the 2nd trial. Incase D1 is greater than c1 but less or equal to c2,
then the 2nd sample size n2 is taken. If D1+D2 is less
than or equal to c2, then the lot is accepted otherwise
rejected.
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Suppose, N=3000, n1=40, n2 = 80, c1=1, c2=4
A lot contains 3000 units of garments, in the 1st trial 40units are randomly drawn. If the defective unit is less or
equal to 1, lot is accepted, without requiring the 2nd trial.
If the no defective units are greater than 4 then the
entire lot is rejected without requiring the 2nd
trial. If thedefective units are 2 or 3 or 4 then 2nd
sample 80 pieces are taken. If the total defective units
(1st + 2nd ) is less or equal to
4 then the lot is accepted, otherwise rejected.
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Sample n1 is taken
Accept lotIs D1 c 2? Reject lot
Sample n2 is taken
Is D1+D2 > c2 ?
Accept lot
Reject lot
N
N
N
Y
Y
Y
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PROBABILITYOFACCEPTANCE
DOUBLESAMPLINGPLAN
Suppose,
N = 3000
n1 = 40, c1 = 2
n2 = 80, c2 = 4D1 = nonconforming from n1= 40 sample
D2 = nonconforming from n2= 80 sample
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PROBABILITYOFACCEPTANCEIN 1STSAMPLE
WHENACTUALNONCONFORMANCE 5% OR 0.05
P(0) = 40C0x (0.05)0x (1-0.05)40-0
=0.128512
P(1) = 40C1x (0.05)1x (1-0.05)40-1
=0.27055
P(2) = 40C2x (0.05)2x (1-0.04)40-2
=0.27767
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Total probability of acceptance in 1st sample when
actual non conformance 5% or 0.05
P1 = P(0) + P(1) + P(2)
= 0.128512 + 0.27055 + 0.27767
= 0.67673= 67.673%
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PROBABILITYOFACCEPTANCEIN 2ND SAMPLE
WHENACTUALNONCONFORMANCE 5% OR 0.05
P2 (D1=3 & D2=0)
= { 40C3x (0.05)3x (1-0.05)40-3 } X { 80C0x (0.05)
0x (1-0.05)80-0 }
=0.003057
P2 (D1=3 & D2=1) = 40C1x (0.05)1x (1-0.05)40-1
= { 40C3x (0.05)3x (1-0.05)40-3 } X { 80C1x (0.05)
1x (1-0.05)80-1 }
=0.01287
P2 (D1=4 & D2=0) = 40C2x (0.05)2x (1-0.04)40-2
= { 40C4x (0.05)4x (1-0.05)40-4 } X { 80C0x (0.05)
0x (1-0.05)80-0 }
=0.001488
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TOTALPROBABILITYOFACCEPTANCEIN 2ND
SAMPLEWHENACTUALNONCONFORMANCE
5% OR 0.05
P2
= P2 (D1=3 & D2=0) + P2 (D1=3 & D2=1) + P2 (D1=4 & D2=0)
= 0.003057 + 0.01287 + 0.001488
= 0.017415
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TOTALCOMBINEDPROBABILITYOFACCEPTANCE
INCLUDING 1ST & 2ND SAMPLEWHENACTUALNON
CONFORMANCEIS 5% OR 0.05
Pa = P1 + P2
= 0.67673 + 0.017415
= 0.694145
= 69.4145%
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OC CURVE DOUBLE SAMPLING PLAN4
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MULTIPLE SAMPLING PLAN(extension of double sapling plan)
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In double sampling plan, samples are taken
maximum twice, where in multiple sampling plan,
samples are taken many times before a decision is
taken, regarding acceptance or rejection.
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Similar to double sampling plan, an acceptance
number and a rejection number is specified. At any
sample, if cumulative number of defectives is less
than or equal to the acceptance number, the entire
lot is accepted and no more samples are taken or inthe other way to say, the process of sampling is
stopped. At any sample, if the cumulative number of
defectives is more than or equal to rejection
number, the entire lot is rejected and no moresample is taken. The process of taking samples
continues only if the number of defectives is more
then the acceptance number but less than the
rejection number.
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Sample
Number
Sample Size Cumulative
Sample Size
Cumulative
Acceptance
Number
Cumulative
Rejection
Number
Take another sample if
number of cumulative
defectives
1 50 50 1 3 22 50 100
2 4 33 50 150
3 5 4
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SEQUENTIALSAMPLINGPLAN(extension of double sampling plan)
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The acceptance number or rejection number are
not by fixed numbers, rather in terms acceptance
and rejection region.
There are two types of sequecial sampling with
respect to nature of drawing the sample
1. Item by item
2. Group
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For example, if
= 0.05 = probability of type 1 error
= 0.10 = probability of type 2 error
P1 = 0.01 = fraction nonconforming for whichprobability of acceptance is high
P2 = 0.10 = fraction nonconforming for whichprobability of acceptance is low
Then,
X1 = -a1 + b*n = -0.393 + 0.04 * n
X2 = a2 + b*n = 1.205 + 0.04 * n
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Both acceptance number and rejection numbers
must be integers. The acceptance number is the
next integer less than or equal to X1 and the
rejection number is the next number greater than or
equal to X2.
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For, n =1
X1 = -0.899 ~ -1
X2 = 1.245 ~ 2
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Units Inspected Accept the lot ifcumulative
defectives are
Reject the lot ifcumulative
defectives are
1
2 2
3 24 2
5 2
6 2
7 2
8 2
9 2
10 2
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Units Inspected Accept the lot ifcumulative
defectives are
Reject the lot ifcumulative
defectives are
11 2
12 2
13 214 2
15 2
16 2
17 2
18 2
19 2
20 3
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Units Inspected Accept the lot if
cumulative
defectives are
Reject the lot if
cumulative
defectives are
21 3
22 3
23 3
24 3
25 3
26 348 3
49 3
58 4
74 4
83 5
100 5
109 6