Acceptance Sampling Plans Supplement G AQL LTPD Copyright ©2013 Pearson Education, Inc. publishing as Prentice HallG- 01

Acceptance Sampling PlansSupplement G

a

AQL LTPD

Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall G- 01

What is Acceptance Sampling?

Acceptance Sampling

An inspection procedure used to determine whether to accept or reject a specific quantity of material.

G- 02Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

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

Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall G - 03

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.


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


Sampling Plans

• Single-Sampling Plan

• Double-Sampling Plan

• Sequential Sampling Plan


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.Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall G - 07

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

Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall G -08

Sequential-Sampling Plan

• 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


Reject

Continue sampling

Accept

8 –

7 –

6 –

5 –

4 –

3 –

2 –

1 –

0 –

Cumulative sample size

| | | | | | |

10 20 30 40 50 60 70

Num

ber o

f def

ectiv

esSequential Sampling Chart


Operating Characteristic Curve

• 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.


Operating Characteristic Curve

a

Ideal OC curve

Typical OC curve

1.0

AQL LTPD

Prob

abili

ty o

f acc

epta

nce

Proportion defective


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.


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 = 0.878. Repeat this process for a range of p values.

• The following table contains the remaining values for the OC curve.


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 = 1.000 – 0.878 = 0.122

0.02 1.2 0.6630.03 1.8 0.4630.04 2.4 0.3080.05 3.0 0.1990.06 (LTPD) 3.6 0.126 = 0.1260.07 4.2 0.0780.08 4.8 0.0480.09 5.4 0.0290.10 6.0 0.017

Example G.1

0.878

0.663

0.463

0.308

0.1990.126 0.078

0.048 0.0290.017

(AQL) (LTPD)

1.0 –

0.9 –

0.8 –

0.7 –

0.6 –

0.5 –

0.4 –

0.3 –

0.2 –

0.1 –

0.0 – | | | | | | | | | |1 2 3 4 5 6 7 8 9 10

Proportion defective (hundredths)

Prob

abili

ty o

f acc

epta

nce

= 0.126

= 0.122


Cumulative Poisson Probabilities


Cumulative Poisson Probabilities


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)

p =np =Pa =

=

0.035.79 (use 5.8 in table)0.9650.035 (or 1.0 – 0.965)

Finding (probability of accepting LTPD quality)

p =np =Pa =

=

0.0815.440.100.10


Application G.1


Application G.1

1.0 –

0.8 –

0.6 –

0.4 –

0.2 –

0.0 –| | | | | | | | | | |0 2 4 6 8 10

Prob

abili

ty o

f acc

epta

nce

Percentage defective

= 0.035

= 0.10


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

nProducer’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



1.0 –

0.9 –

0.8 –

0.7 –

0.6 –

0.5 –

0.4 –

0.3 –

0.2 –

0.1 –

0.0 – | | | | | | | | | |1 2 3 4 5 6 7 8 9 10

(AQL) (LTPD)


Prob

abili

ty o

f acc

epta

nce

n = 60, c = 1

n = 80, c = 1

n = 100, c = 1

n = 120, c = 1



• Acceptance level effect– Increasing c while holding n constant

decreases the producer’s risk and increases the consumer’s risk

cProducer’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



1.0 –

0.9 –

0.8 –

0.7 –

0.6 –

0.5 –

0.4 –

0.3 –

0.2 –

0.1 –

0.0 – | | | | | | | | | |1 2 3 4 5 6 7 8 9 10

(AQL) (LTPD)


Prob

abili

ty o

f acc

epta

nce

n = 60, c = 1

n = 60, c = 2

n = 60, c = 3n = 60, c = 4


Acceptance Sampling Plan Data

AQL Based LTPD BasedAcceptance

NumberExpected

DefectivesSample

SizeExpected

DefectivesSample

Size

0 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

5 2.6256 263 9.2647 154

6 3.2838 328 10.5139 175

7 3.9794 398 11.7726 196

8 4.6936 469 12.9903 217

9 5.4237 542 14.2042 237

10 6.1635 616 15.4036 257


Noise King Sampling Plan: c = 3 and n = 111 are best

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.

N

nNPp a AOQ

wherep = true proportion defective of the lotPa = probability of accepting the lotN = lot sizen = sample size


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 0.603 + 0.558)/2 = 0.580.Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

G - 28

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 = (0.603 + 0.558)/2

0.04 4.40 0.359

0.05 5.50 0.202 = (0.213 + 0.191)/2

0.06 6.60 0.105

0.07 7.70 0.052 = (0.055 + 0.048)/2

0.08 8.80 0.024


Example G.2

Step 2: Calculate the AOQ for each value of p.

For p = 0.01: 0.01(0.974)(1000 – 110)/1000 = 0.0087For p = 0.02: 0.02(0.819)(1000 – 110)/1000 = 0.0146For p = 0.03: 0.03(0.581)(1000 – 110)/1000 = 0.0155For p = 0.04: 0.04(0.359)(1000 – 110)/1000 = 0.0128For p = 0.05: 0.05(0.202)(1000 – 110)/1000 = 0.0090For p = 0.06: 0.06(0.105)(1000 – 110)/1000 = 0.0056For p = 0.07: 0.07(0.052)(1000 – 110)/1000 = 0.0032For p = 0.08: 0.08(0.024)(1000 – 110)/1000 = 0.0017


Example G.2

Step 3: Identify the largest AOQ value, which is the estimate of the AOQL.

In this example, the AOQL is 0.0155 at p = 0.03.

AOQL1.6 –

1.2 –

0.8 –

0.4 –

0 –| | | | | | | |1 2 3 4 5 6 7 8

Defectives in lot (percent)

Aver

age

outg

oing

qua

lity

(per

cent

)


Application G.2

• Demonstrate the model for computing AOQ• Management has selected the following parameters:

AQL = 0.01 = 0.05LTPD = 0.06 = 0.10

n = 100 c = 3

What is the AOQ if p = 0.05 and N = 3000?

p =np =Pa =

AOQ =

0.05100(0.05) = 50.265

300029002650050 .. = 0.0128


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.

Solved Problem

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?


Solved Problema. 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,

np =

np =

180(0.03) = 5.4 = 0.049

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.


Solved Problemb. 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

0.011.8

1.000

0.023.6

0.996

0.03 (AQL) 5.4

0.951 = 1 – 0.951 = 0.049

0.047.2

0.810

0.059.0

0.587

0.0610.8

0.363

0.0712.6

0.194

0.08 (LTPD) 14.4

0.092 = 0.092

0.0916.2

0.039

0.1018.0

0.015


Solved Problem

c. According to the table, the probability of accepting a lot with 5 percent defectives is 0.587. Therefore, the probability that a lot with 5 percent defects will be rejected is 0.413, or 1 – 0.587

= 0.092

= 0.0491.0 —

0.9 —

0.8 —

0.7 —

0.6 —

0.5 —

0.4 —

0.3 —

0.2 —

0.1 —

0 — | | | | | | | | | |1 2 3 4 5 6 7 8 9 10

Proportion defective (hundredths)(p)

Prob

abili

ty o

f acc

epta

nce

(Pa)

(AQL) (LTPD)

0.996

0.9510.810

0.587

0.363

0.1940.092

0.0390.015

1.000



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Acceptance Sampling Plans Supplement G AQL LTPD Copyright ©2013 Pearson Education, Inc. publishing as Prentice HallG- 01