1

SMUEMIS 7364

NTUTO-570-N

Acceptance Sampling for AttributesUpdated: 4.4.02

Statistical Quality ControlDr. Jerrell T. Stracener, SAE Fellow

2

Decision Risk

• Producer’s Risk – conforming lots are rejected

• Consumer’s Risk – nonconforming lots are accepted.

3

Decision Risk

True SituationLot meets Lot does not

Decision requirements meet requirements

Accept Lot no error Type II error

Reject Lot Type I error no error

4

Lot-by-Lot Acceptance Sampling by Attributes– Single Sampling

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Single Sampling

N = lot size = the number of items in the lot from which the sample is to be drawn.

n = sample size = the number of items drawn at random from the lot

c = the maximum allowable number of defective items in the sample. More than c defectives in the sample will cause rejection of the lot.

6

Type A OC Function for Single Sampling Plan

• Sampling Plan SpecificationN = Lot Sizen = Sample Sizec = Acceptance number

• D = true number of defectives in the lot

• X = number of defective items in the random sample.

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Type A OC Function for Single Sampling Plan

• Probability Distribution of X

X ~ H(N, D, n)

Probability Mass Function of X

otherwise , 0

Dn,min0,1,...,for x

n

N

xn

DN

x

D

xh

8

Type A OC Function for Single Sampling Plan

• OC Function

DOC

c

x

A

DcXP

DP

0

n

N

xn

DN

x

D

|

9

Example – Single Sampling Plan A

Determine and plot the OC Function for a single sampling plan specified by

2c

5n

50N

10

Example Solution – Single Sampling Plan

DOC

2

0x

A

5

50

x5

D50

x

D

D|2XP

DP

0.0

0.2

0.4

0.6

0.8

1.0

0 10 20 30 40 50

OC(D)

D

D PA(D)=OC(D)0 1.005 1.0010 0.9515 0.8520 0.6925 0.5030 0.3135 0.1540 0.0545 0.0050 0.00

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Concept

Suppose that the lot size N is large (theoreticallyinfinite). Under this condition, the distribution of thenumber of defectives d in a random sample of nitems is binomial with parameters n and p, where pis the fraction of defective items in the lot. An equivalent way to conceptualize this is to draw lots of N items at random from a theoretically infiniteprocess, and then to draw random samples of n fromthese lots. Sampling from the lot in this manner isthe equivalent of sampling directly from the process.

12

Type B OC Function for Single Sampling Plan

• Sampling Plan SpecificationN = Lot Sizen = Sample Sizec = Acceptance number

• N is infinite, or at least much larger than n

• D = true number of defective items in the lot

• p = proportion of the population that is defective, i.e.,

• X = number of defective items in the random sample.

N

Dp

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Type B OC Function for Single Sampling Plan

• Probability Distribution of X

X ~ B(n, p)

Probability Mass Function of X

n0,1,...,for x p-1px

nxb x-nx

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Acceptance Sampling

The probability of observing exactly x defective items is:

for x = 0, 1, . . ., n

xnx

xnx

p1p!xnx!

n!

qpx

n

xXP

xP

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OC Function

The probability of acceptance is the probability that d is less than or equal to c, or:

pOC

c

0x

x-nx

c

0x

A

p-1px

n

xb

p|cXP

pP

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Example – Single Sampling Plan B

Determine and plot the OC Function for a single sampling plan specified by

Compare the OC curve for N=500, n=98, c=2.

2c

89n

n tocompared large least veryat or size,in infiniteN

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Example Solution – Single Sampling Plan

pOC

2

0x

2

0

x-98x

A

p-1px

98xb

p|2XP

DP

x

OC(p)

p

0.0

0.2

0.4

0.6

0.8

1.0

0.00 0.02 0.04 0.06 0.08 0.10

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Example

If the lot fraction defective is p = 0.02, n = 89 and c = 2, then

2dP0.02PA

7366.0

98.002.0!782!

89!

98.002.0!881!

89!98.002.0

!980!

89!

98.002.0!d98d!

89!

782

881980

d98d2

0

d

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Example

The OC curve is developed by evaluating PA(p) for various values of p. The following table displays the calculated value of several points on the curve.

The OC curve shows the discriminatory power of thesampling plan. For example, in the sampling plan n = 98, c = 2, if the lots are 2% defective, theprobability of acceptance is approximately 0.74. Thismeans that if 100 lots from a process thatmanufactures 2% defective product are submittedto this sampling plan, we will expect, in the long run, to accept 74 of the lots and reject 26 of them.

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Example

Probabilities of Acceptance for the Single-Samplingplan n = 89, c = 2

Fraction Probability of Defective, p Acceptance, Pa (p)0.005 0.98970.010 0.93970.020 0.73660.030 0.49850.040 0.30420.050 0.17210.060 0.09190.070 0.04680.080 0.02300.090 0.0109

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If N = 500, n =98 & c=2,

For comparison, select p=0.02. Then

DOCDPA

2

0d

98

500

98

500

D|2XP

x

D

x

D

10|2XPDOC10P

and 10,0.02500ND

A

p

7778.0

1976.03049.02753.0

98

500

98

490102

0d

xx

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Example

0

0.2

0.4

0.6

0.8

1

0 0.02 0.04 0.06 0.08

Lot fraction defective, p

Prob

abili

ty o

f acc

epta

nce,

Pa

10 200 30 40 -D

-p

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Single Sample Test Plan Design

• Probability Distribution of X

P0 = specified fraction defective

P1 = minimum acceptable fraction defective

= producer’s risk

= consumer’s risk

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Test Procedure

To testH0: p = p0

vs H1: p = p1

at the 100% level of significance,

• Obtain a random sample of size n

• Inspect the n items and determine the number, X, that are defective

• Reject H0 if x > c, otherwise accept H0

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Single Sampling Plan

accept

rejectnumberof defectiveitems

c...2

10

n

x

n = sample sizex = number of defective itemsc = maximum number of defective items for acceptance

0

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Operating Characteristic (OC) Function

Note that:

and

pPpOC A

pcXP |

xnxc

x

ppx

n

1

0

α1pOC 0

βpOC 1

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OC Curve

pPpOC A

p1 1

0

1 -

p

1

p0

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Test Plan Design

To determine a single sample test plan for testing H0: p = p0, specify values of p0, p1, , and such that PA(p0) = 1 - and PA(p1) = , then find the values of n and c that satisfy the following equations.

and

110

0

xno

xc

x

ppx

n

xnxc

x

ppx

n11

0

1

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Lot-by-Lot Acceptance Sampling by Attributes– Double Sampling

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Double Sampling

A double-sampling plan is a procedure in which, under certain circumstances, a second sample isrequired before the lot can be sentenced. A double-sampling plan is defined by four parameters:

n1 = sample size on the first samplec1 = acceptance number of the first samplen2 = sample size on the second samplec2 = acceptance number for both samples

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Single Sampling Plan

numberof defectiveitems

c1

210

Accept

Reject

n1

x

0

Accept

Reject

n1+n2

Continue

.

.

.

c1+1

.

.

.

.

.

.

c2

. . .1 2 . . .

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Double Sampling - Advantages

The principal advantage of a double-sampling plan with respect to single sampling is that it may reducethe total amount of required inspection. Suppose thatthe first sample taken under a double-sampling planis smaller than the sample that would be required using a single-sampling plan that offers the consumerthe same protection. In all cases, then, in which a lotis accepted or rejected on the first sample, the costof inspection will be lower for double sampling than it would be for single sampling. It is also possible to reject a lot without complete inspection of the secondsample (called curtailment of the second sample).

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Double Sampling - Disadvantages

Double sampling has two potential disadvantages:

1. Unless curtailment is used on the second sample, under some circumstances double sampling may require more total inspection than would be required in a single-sampling plan that offers the same protection.

2. Double-sampling is administratively more complex, which may increase the opportunity for the occurrence of inspection errors. Furthermore, there may be problems in storing and handling raw materials or component parts for which one sample has been taken, but that are awaiting a second sample before a final lot dispositioning decision can be made.

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Double Sampling - The OC Curve

The performance of a double-sampling plan can be conveniently summarized by means of its operating-characteristic (OC) curve. A double-sampling planhas a primary OC curve that gives the probability ofacceptance as a function of lot of process quality. It also has supplementary OC curves that show the probability of acceptance as a function of lotacceptance and rejection on the first sample.

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Double Sampling

Operation of the double-sampling plan with n1 = 50, c1 = 1, n2 = 100, c2 = 3

Inspect a random sample of n1 = 50 from the lot

d1 = number of observed defectives

Inspect a random sample of n2 = 100 from the lot

d2 = number of observed defectives

Acceptthelot

Acceptthelot

Rejectthelot

Rejectthelot

d1 c1 = 1 d1 > c1 = 3

d1 + d2 c2 = 3 d1 + d2 > c2 = 3

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Example

If denotes the probability of acceptance on the combined samples, and and denote theprobability of acceptance on the first and secondsamples, respectively, then

is just the probability that we will observed1 c1 = 1 defectives out of a random sample of n1 = 50 items. Thus

pPaIaP

pPpPpP IIa

Iaa

IIaP

pP Ia

11

1

d50d1

0d 11

Ia p1p

!d50!d

50!pP

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Example

If p = 0.05 is the fraction defective in the incominglot, then

To obtain the probability of acceptance on the secondsample, we must list the number of ways the secondsample can be obtained. A second sample is drawnonly if there are two or three defectives on the firstsample - that is, if c1 < d1 c2.

279.095.00.05!d50!d

50!05.0P 11

1

d50d1

0d 11

Ia

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Example

1. d1 = 2 and d2 = 0 or 1; that is, we find two defectives on the first sample and one or less defectives on the second sample. The probability of this is:

P(d1 = 2, d2 1) = P(d1 = 2) x P(d2 1)

= (0.261)(0.037)

= 0.009

22

2

d100d1

0d 22

482 0.950.05!d100!d

100!0.950.05

2!48!

50!

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Example

2. d1 = 3 and d2 = 0; that is, we find three defectives on the first sample and no defectives on the second sample. The probability of this is:

P(d1 = 3, d2 0) = P(d1 = 3) x P(d2 = 0)

= (0.220)(0.0059)

= 0.001

1000473 95.00.0500!10!

100!95.005.0

!47!3

!50

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Example

Thus, the probability of acceptance on the secondsample is

The probability of acceptance of a lot that has fractiondefective p = 0.05 is therefore

0d3,dP1d2,dP05.0P 2121IIa

001.0009.0

010.0

05.0P05.0P05.0P IIa

Iaa

010.0279.0

289.0

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Double Sampling - The OC Curve

OC Curves for the double-sampling plan with n1 = 50, c1 = 1, n2 = 100, c2 = 3

0.0

0.2

0.4

0.6

0.8

1.0

0.00 0.02 0.04 0.06 0.08 0.10 0.12

Probability of acceptance on first sample

Probability of acceptance on combined sample

Probability of rejection on first sample

Lot fraction defective, p

Pro

babili

ty, P

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Rectifying Inspection Programs

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Rectifying Inspection Programs

• Acceptance-sampling programs require corrective action when lots are rejected.

• Generally takes the form of 100% inspection or screening of rejected lots, with all discovered defective items either removed for subsequent rework or return to the vendor, or replaced from a stock of known good items. Such sampling programs are called rectifying inspection programs.

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Rectifying Inspection Programs (continued)

• The inspection activity affects the final quality of the outgoing product. Suppose that the incoming lots to the inspection activity have fraction defective, po. Some of these lots will be accepted, and others will be rejected. The rejected lots will be screened, and their final fraction defective will be zero. However, accepted lots have fraction defective p0. Consequently, the outgoing lots from the inspection activity are a mixture of lots with fraction defective p0 and fraction defective zero, so the average fraction defective in the stream of outgoing lots is p1, which is less that p0, and serves to “correct” lot quality.

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Rectifying Inspection

Incoming lotsFraction defective

p0

Inspectionactivity

Rejected lots

Accepted

lots

Outgoing lotsFraction defective

p1

Fraction defective

0

Fraction defective

p0

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Rectifying Inspection Programs (continued)

• Used in situations where the manufacturer wishes to know the average level of quality that is likely to result at a given stage of the manufacturing operations.

• Used either at receiving inspection, in-process inspection of semi-finished products, or at final inspection of finished goods

• The objective of in-plant usage is to give assurance regarding the average quality of material used in the next stage of the manufacturing operations.

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Handling of Rejected Lots

• The best approach is to return rejected lots to the vendor, and require it to perform the screening and rework activities.

Has the psychological effect of making the vendor responsible for poor quality

May exert pressure on the vendor to improve its manufacturing processes or to install better process controls.

• Screening and rework take place at the consumer level because the components or raw materials are required in order to meet production schedules.

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Average Outgoing Quality

• Widely used for the evaluation of a rectifying sampling plan.

• Is the quality in the lot that results from the application of rectifying inspection

• Is the average value of lot quality that would be obtained over a long sequence of lots from a process with fraction defective p.

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Average Outgoing Quality (AOQ)

The average fraction defective, called average outgoing quality is

Where the lot size is N and that all defectives are replaces with good units. Then in lots of size N, we have

,

N

nNpPAOQ a

1. N items in the lot that, after inspection, contain no defectives, because all discovered defectives are replaced

2. N – n items that, if the lots is rejected, contain no defectives

3. N – n items that, if accepted, contain (N-n)p defectives

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Example

Suppose that N = 10,000, n = 89, and c = 2, and that the incoming lots are of quality p = 0.01. Now at p = 0.01, we have Pa = 0.9397, and the AOQ is

That is, the average outgoing quality is at 0.93% defective.

0093.0

10,000

98000,0101.00.9397N

nNpPa

AOQ

51

Example

Average outgoing quality will vary as the fraction defective of the incoming lots varies. The curve that plots average outgoing quality against incoming lot quality is called an AOQ curve.

0.0000

0.0025

0.0050

0.0075

0.0100

0.0125

0.0150

0.0175

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07

AOQ

fraction defective, p

AO

Q

52

Average Total Inspection (ATI)

Another important measure relative to rectifying inspection is the total amount of inspection required by the sampling program. If the lots contain no defective items, no lots will be rejected, and the amount of inspection per lot will be the sample size n. If the items are all defective, every lot will be submitted to 100% inspection, and the amount of inspection per lot will be the lot size N. Of the lot quality is 0 < p < 1, the average amount of inspection per lot will vary between the sample size n and the lot size N.

,n-NP1nATI a

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Average Total Inspection (ATI)

Consider our previous example with N = 10,000, n = 89, and c = 2, and p = 0.01. Then since Pa = 0.9397, we have

Remember this is an average number of units inspected over many lots with fraction defective p = 0.01.

687

89000,100.9397-189

n-NP1n a

ATI

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Determination of optimum quality level, p*

cost

incoming quality level ~p

0 1p*

cost toachieve p

cost ofinspectionper lot

totalcost