1. Acceptance Sampling Plans Supplement G
AQL LTPD
Acceptance Sampling Plans
Supplement G
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G- 01

2. What is Acceptance Sampling?
An inspection procedure used to determine whether to accept or reject a specific quantity of material.
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3. Acceptance Sampling Plan Decisions
Basic procedure
Take random sample
Accept or reject, based on results
Producer, or seller, is the origin of the material or service
Consumer, or buyer, is the destination of the material or service
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4. Quality and Risk Decisions
Acceptable quality level (AQL)
The quality level desired by the consumer
Producer’s risk ()
The risk that the sampling plan will fail to verify an acceptable lot’s quality and, thus, reject it (Type 1 Error)
Most often the producer’s risk is set at 0.05, or 5 percent.
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5. Quality and Risk Decisions
Lot tolerance proportion defective (LTPD)
The worst level the customer can tolerate
Consumer’s risk, ( )
The probability of accepting a lot with LTPD quality (Type II Error)
A common value for the consumer’s risk is 0.10, or 10 percent
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Sampling Plans
Single-Sampling Plan
Double-Sampling Plan
Sequential Sampling Plan
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Single Sampling Plan
Single Sampling Plan
A decision to accept or reject a lot based on the results of one random sample from the lot.
States the sample size, n, and the acceptable number of defectives, c
If the quality characteristic of the sample passes the test (defects ≤ c), accept the lot.
If the sample fails (defects > c) there may be complete inspection of the lot or the entire lot is rejected.
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Double-Sampling Plan
Double-Sampling Plan
A plan in which management specifies two sample sizes, (n1 and n2), and two acceptance numbers (c1 and c2)
Take a random sample of relatively small size n1, from a large lot
If the sample passes the test (≤ c1), accept the lot
If the sample fails (> c2), the entire lot is rejected
If the sample is between c1 and c2, then take a larger second random sample, n2
If the combined number of defects ≤ c2 accept the lot, otherwise reject
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9. Sequential-Sampling Plan
A plan in which the consumer randomly selects items from the lot and inspects them one-by-one.
Results are compared to sequential-sampling chart
Chart guides decision to reject, accept, or continue sampling, based on cumulative results
Average number of items inspected (ANI) is generally lower with sequential sampling
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10. Sequential Sampling Chart
8 –
7 –
6 –
5 –
4 –
3 –
2 –
1 –
0 –
Cumulative sample size
| | | | | | |
Number of defectives
Reject
Continue sampling
Accept
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11. Operating Characteristic Curve
A graph that describes how well a sampling plan discriminates between good and bad
Select sample size n and acceptance number c to achieve the level of performance specified by the AQL, , LTPD, and 
The OC curve shows the probability of accepting a lot Pa, as a dependent function of p, the true proportion of defectives in the lot.
For every possible combination of n and c, there exists a unique operating characteristic curve.
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12. Operating Characteristic Curve
1.0
AQL LTPD
Probability of acceptance
Proportion defective a
Ideal OC curve
Typical OC curve 
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Example G.1
The Noise King Muffler Shop, a high-volume installer of replacement exhaust muffler systems, just received a shipment of 1,000 mufflers.
The sampling plan for inspecting these mufflers calls for a sample size n = 60 and an acceptance number c = 1.
The contract with the muffler manufacturer calls for an AQL of 1 defective muffler per 100 and an LTPD of 6 defective mufflers per 100.
Calculate the OC curve for this plan, and determine the producer’s risk and the consumer’s risk for the plan.
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Example G.1
Let p = 0.01.
Multiply n by p to get 60(0.01) = 0.60.
Locate 0.60 in Table G.1. The probability of acceptance = Repeat this process for a range of p values.
The following table contains the remaining values for the OC curve.
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Example G.1
Values for the Operating Characteristic Curve with n = 60 and c = 1
Proportion Defective (p) np
Probability of c or Less Defects (Pa)
Comments
0.01 (AQL)
0.6
0.878
 = – = 0.122
0.02
1.2
0.663
0.03
1.8
0.463
0.04
2.4
0.308
0.05
3.0
0.199
0.06 (LTPD)
3.6
0.126
 = 0.126
0.07
4.2
0.078
0.08
4.8
0.048
0.09
5.4
0.029
0.10
6.0
0.017
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Example G.1
1.0 –
0.9 –
0.8 –
0.7 –
0.6 –
0.5 –
0.4 –
0.3 –
0.2 –
0.1 –
0.0 –
| | | | | | | | | |
Proportion defective (hundredths)
Probability of acceptance
0.878
0.663
0.463
0.308
0.199
0.126
0.078
0.048
0.029
0.017
 = 0.122
 = 0.126
(AQL) (LTPD)
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17. Cumulative Poisson Probabilities
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18. Cumulative Poisson Probabilities
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Application G.1
A sampling plan is being evaluated where c = 10 and n = 193. If AQL = 0.03 and LTPD = 0.08, what are the producer’s risk and consumer’s risk for the plan? Draw the OC curve.
Finding  (probability of rejecting AQL quality)
Finding  (probability of accepting LTPD quality)
p =
np =
Pa =
 =
0.03
5.79 (use 5.8 in table)
0.965
0.035 (or 1.0 – 0.965)
p =
np =
Pa =
 =
0.08
15.44
0.10
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Application G.1
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Application G.1
1.0 –
0.8 –
0.6 –
0.4 –
0.2 –
0.0 –
| | | | | | | | | | |
Probability of acceptance
Percentage defective
 = 0.035
 = 0.10
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22. Explaining Changes in the OC Curve
Sample size effect
Increasing n while holding c constant increases the producer’s risk and reduces the consumer’s risk n
Producer’s Risk
(p = AQL)
Consumer’s Risk
(p = LTPD) 60
0.122
0.126 80
0.191
0.048
100
0.264
0.017
120
0.332
0.006
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23. Explaining Changes in the OC Curve
1.0 –
0.9 –
0.8 –
0.7 –
0.6 –
0.5 –
0.4 –
0.3 –
0.2 –
0.1 –
0.0 –
| | | | | | | | | |
(AQL) (LTPD)
Proportion defective (hundredths)
Probability of acceptance
n = 60, c = 1
n = 80, c = 1
n = 100, c = 1
n = 120, c = 1
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24. Explaining Changes in the OC Curve
Acceptance level effect
Increasing c while holding n constant decreases the producer’s risk and increases the consumer’s risk c
Producer’s Risk
(p = AQL)
Consumer’s Risk
(p = LTPD) 1
0.122
0.126 2
0.023
0.303 3
0.003
0.515 4
0.000
0.706
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25. Explaining Changes in the OC Curve
1.0 –
0.9 –
0.8 –
0.7 –
0.6 –
0.5 –
0.4 –
0.3 –
0.2 –
0.1 –
0.0 –
| | | | | | | | | |
(AQL) (LTPD)
Proportion defective (hundredths)
Probability of acceptance
n = 60, c = 1
n = 60, c = 2
n = 60, c = 3
n = 60, c = 4
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26. Acceptance Sampling Plan Data
Noise King Sampling Plan: c = 3 and n = 111 are best
AQL Based
LTPD Based
Acceptance Number
Expected Defectives
Sample Size
0.0509 5
2.2996 38 1
0.3552 36
3.8875 65 2
0.8112 81
5.3217 89 3
1.3675
137
6.6697
111 4
1.9680
197
7.9894
133
2.6256
263
9.2647
154 6
3.2838
328
175 7
3.9794
398
196 8
4.6936
469
217 9
5.4237
542
237 10
6.1635
616
257
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27. Average Outgoing Quality
AOQ
The expressed proportion of defects that a plan will allow to pass.
Rectified inspection
The assumption that all defective items in the lot will be replaced if the lot is rejected and that any defective items in the sample will be replaced if the lot is accepted.
where
p = true proportion defective of the lot
Pa = probability of accepting the lot
N = lot size
n = sample size
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Example G.2
Suppose that Noise King is using rectified inspection for its single-sampling plan.
Calculate the average outgoing quality limit for a plan with n = 110, c = 3, and N = 1,000.
Use the following steps to estimate the AOQL for this sampling plan:
Step 1:
Determine the probabilities of acceptance for the desired values of p.
However, the values for p = 0.03, 0.05, and 0.07 had to be interpolated because the table does not have them.
For example, Pa for p = 0.03 was estimated by averaging the Pa values for np = 3.2 and np = 3.4, (or )/2 =
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29. Proportion Defective (p) Probability of Acceptance (Pa)
Example G.2
Proportion Defective (p) np
Probability of Acceptance (Pa)
0.01
1.10
0.974
0.02
2.20
0.819
0.03
3.30
0.581
= ( )/2
0.04
4.40
0.359
0.05
5.50
0.202
= ( )/2
0.06
6.60
0.105
0.07
7.70
0.052
= ( )/2
0.08
8.80
0.024
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Example G.2
Step 2: Calculate the AOQ for each value of p.
For p = 0.01: 0.01(0.974)(1000 – 110)/1000 =
For p = 0.02: 0.02(0.819)(1000 – 110)/1000 =
For p = 0.03: 0.03(0.581)(1000 – 110)/1000 =
For p = 0.04: 0.04(0.359)(1000 – 110)/1000 =
For p = 0.05: 0.05(0.202)(1000 – 110)/1000 =
For p = 0.06: 0.06(0.105)(1000 – 110)/1000 =
For p = 0.07: 0.07(0.052)(1000 – 110)/1000 =
For p = 0.08: 0.08(0.024)(1000 – 110)/1000 =
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Example G.2
1.6 –
1.2 –
0.8 –
0.4 –
0 –
| | | | | | | |
Defectives in lot (percent)
Average outgoing quality (percent)
Step 3: Identify the largest AOQ value, which is the estimate of the AOQL.
In this example, the AOQL is at p = 0.03.
AOQL
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Application G.2
Demonstrate the model for computing AOQ
Management has selected the following parameters:
AQL = 0.01  = 0.05
LTPD = 0.06  = 0.10
n = 100 c = 3
What is the AOQ if p = 0.05 and N = 3000?
p =
np =
Pa =
AOQ =
0.05
100(0.05) = 5
0.265 =
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Solved Problem
In a manufacturing facility a feeder process, when operating correctly, has an AQL of 3 percent. A consuming process has a specified LTPD of 8 percent. Management wants no more than a 5 percent producer’s risk and no more than a 10 percent consumer’s risk.
a. Determine the appropriate sample size, n, and the acceptable number of defective items in the sample, c.
b. Calculate values and draw the OC curve for this inspection station.
c. What is the probability that a lot with 5 percent defectives will be rejected?
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Solved Problem
For AQL = 3 percent, LTPD = 8 percent,  = 5 percent, and  = 10 percent, use Table G.1 and trial and error to arrive at a sampling plan. If n = 180 and c = 9,
180(0.03) = 5.4
 = 0.049
np =
np =
180(0.08) = 14.4
 = 0.092
Sampling plans that would also work are n = 200, c = 10; n = 220, c = 11; and n = 240, c = 12.
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35. Proportion Defective (p) Probability of c or Less Defectives (Pa)
Solved Problem
b. The following table contains the data for the OC curve. Table G.1 was used to estimate the probability of acceptance.
Proportion Defective (p) np
Probability of c or Less Defectives (Pa)
Comments
0.01
1.8
1.000
0.02
3.6
0.996
0.03 (AQL)
5.4
0.951
 = 1 – = 0.049
0.04
7.2
0.810
0.05
9.0
0.587
0.06
10.8
0.363
0.07
12.6
0.194
0.08 (LTPD)
14.4
0.092
 = 0.092
0.09
16.2
0.039
0.10
18.0
0.015
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Solved Problem
0.996
0.951
0.810
0.587
0.363
0.194
0.092
0.039
0.015
1.000
c. According to the table, the probability of accepting a lot with 5 percent defectives is Therefore, the probability that a lot with 5 percent defects will be rejected is 0.413, or 1 – 0.587
1.0 —
0.9 —
0.8 —
0.7 —
0.6 —
0.5 —
0.4 —
0.3 —
0.2 —
0.1 —
0 —
| | | | | | | | | |
Proportion defective (hundredths)(p)
Probability of acceptance (Pa)
(AQL) (LTPD)
 = 0.049
 = 0.092
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