Chap14Rev

69
Introduction to Statistical Q uality Control, 4th Edition Chapter 14 Lot-by-Lot Acceptance Sampling for Attributes

Transcript of Chap14Rev

Page 1: Chap14Rev

Introduction to Statistical Quality Control, 4th Edition

Chapter 14

Lot-by-Lot Acceptance Sampling for Attributes

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Introduction to Statistical Quality Control, 4th Edition

14-1. The Acceptance-Sampling Problem

• Acceptance sampling is concerned with inspection and decision making regarding products.

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Three aspects of sampling

• The purpose of acceptance sampling is to sentence lots, not to estimate the lot quality– Although, some plans do this

• Acceptance sampling is not quality control– Reject or accept lots only– Even if lots are of the same quality, sampling

will accept some lots and reject others

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Three aspects of sampling

• Quality cannot be inspected into the product– Acceptance sampling is an audit tool that

insures that the output of a process conforms to requirements

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14-1. The Acceptance-Sampling Problem

• Three approaches to lot sentencing:

1. Accept with no inspection

2. 100% inspection

3. Acceptance sampling

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14-1. The Acceptance-Sampling Problem

Why Acceptance Sampling and Not 100% Inspection?

• Testing can be destructive• Cost of 100% inspection is high• 100% inspection is not feasible

– Requires too much time– Can be inaccurate

• If vendor has excellent quality history

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14-1. The Acceptance-Sampling Problem

14-1.1 Advantages and Disadvantages of Sampling Advantages• Less expensive• Reduced damage• Reduces the amount of inspection errorDisadvantages• Risk of accepting “bad” lots, rejecting “good” lots• Less information generated• Requires planning and documentation

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14-1. The Acceptance-Sampling Problem

14-1.2 Types of Sampling Plans • There are variables sampling plans and attribute

sampling plans (this chapter is about attributes)

1. Single sampling plan

2. Double-sampling plan

3. Multiple-sampling plan

4. Sequential-sampling

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14-1. The Acceptance-Sampling Problem

14-1.3 Lot Formation

• Considerations before inspection:– Lots should be homogeneous

• Produced by the same machine, same operators, common raw materials, approximately the same time

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14-1. The Acceptance-Sampling Problem

14-1.3 Lot Formation

• Considerations before inspection:– Larger lots more preferable than smaller lots

• More economical

– Lots should be conformable to the materials-handling systems used in both the vendor and consumer facilities.

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14-1. The Acceptance-Sampling Problem

14-1.4 Random Sampling• The units selected for inspection should be

chosen at random.

• If 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.

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Non-randomization

• Pick a unit from the top layer of each box– Uncle Charlie

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Randomization

• Example: Assign a number to each unit in the lot – 1, 2, …, N– Select n unique random numbers from 1, 2, …,

N– The selected numbers constitute the sample

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14-2. Single-Sampling Plans For Attributes

14-2.1 Definition of a Single-Sampling Plan• A single sampling plan is defined by sample size, n, and the

acceptance number c. Say there are N total items in a lot. Choose n of the items at random. If more than c of the items are unacceptable, reject the lot.

• N = lot size

• n = sample size

• c = acceptance number

• d = observed number of defectives

• 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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Example

• N = 10000, n = 89, c = 2– From a lot of 10,000, take a sample of size 89– Observe the number of defectives, d– If d < 2, accept– Otherwise, reject

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14-2. Single-Sampling Plans For Attributes

14-2.2 The OC Curve• 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.

• 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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Example

• See Fig. 14-2 on pg. 683 and discussion following

• If p = .01, Pa = .9397

– See computation at bottom of pg. 683

• See Table 14-2– If p = .02, Pa = .7366 means that 73.66% of

lots will be accepted and 26.34% will be rejected

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Effect of n and c on OC curves

• Fig. 14-3, ideal OC curve– Pa = 1.0 until a level of quality that is

considered ‘bad’ is reached

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Effect of n and c on OC curves

• Fig. 14-4, OC curve for different values of n– By increasing the sample size, we get closer to

the ideal OC curve

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Effect of n and c on OC curves

• Fig. 14-5, OC curve for different values of c– As c is decreased, the OC curve shifts to the left– When c = 0, it is very hard on the vendor

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14-2. Single-Sampling Plans For Attributes

14-2.3 Designing a Single-Sampling Plan with a Specified OC Curve

• Let the probability of acceptance be 1 - for lots with fraction defective p1.

• Let the probability of acceptance be for lots with fraction defective p2.

• Assume binomial sampling is appropriate.– For type B OC curves (from large lots)

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14-2. Single-Sampling Plans For Attributes

14-2.3 Designing a Single-Sampling Plan with a Specified OC Curve

• The sample size n and acceptance number c are the solution to

c

0d

dn2

d2

c

0d

dn1

d1

)p1(p)!dn(!d

!n

)p1(p)!dn(!d

!n1

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14-2. Single-Sampling Plans For Attributes

Example• Consider constructing a sampling plan for which

– p1 = 0.01

= 0.05

– p2 = 0.06

= 0.10

– N = 1000

• Using computer software or a graphical approach (using an appropriate binomial nomograph) it can be shown that the necessary values of n and c are 85 and 2, respectively.

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Using the nomograph

• Figure 14-9– Draw a line from p1 = .01 on the left side to (1–

) = .95 on the right side

– Draw a second line from p2 = .06 on the left to = .10 on the right

– The intersection of the two lines comes close to defining the plan n = 89, c = 2

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Using the nomograph

• Figure 14-9– Since n and c must be integers, this procedure

will actually produce several plans that have OC curves that pass close to the desired plans

– Holding the first line constant, and holding p2, two plans are observed, with values of different than desired, one lower and the other higher

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Using the nomograph

• Figure 14-9– Holding the second line constant, and holding

p1, two plans are observed, with values of different than desired, one lower and the other higher

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Rectifying inspection

• Require corrective action when lots are rejected– 100% screening of rejected lots– Defective items are removed– Affects the outgoing quality

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Rectifying inspection

Inspection activity

Rejected lots

Accepted lots

Incoming lots

Fraction defective p0

Fraction defective

0

Fraction defective

p0

Outgoing lots

Fraction defective p1<p0

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Average outgoing quality

• AOQ is the result of applying rectifying inspection– In a lot of size N, there will be

• n items in the sample that, after inspection, contain no defectives (all of the defectives were replaced)

• N-n items that, if the lot is rejected, also contain no defectives (the balance of the lot was inspected 100%)

• N-n items that, if the lot is accepted, contain p(N-n) defectives

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Average outgoing quality

• AOQ = [Pa p (N-n)]/N

• Example– N = 10000, n = 89, c = 2, p = .01

– Previously determined that Pa = .9397

– AOQ [(.9397)(.01)(10000-89)]/10000– AOQ = .0093

– Since (N-n)/N 1, AOQ ~ Pap

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AOQ curve for rectifying inspection

• See Fig. 14-11 for n = 89, c = 2

• When incoming quality is very good, average fraction defective of outgoing lots is low

• When incoming quality is very poor, average fraction defective of outgoing lots is low

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Average outgoing quality limit

• See Fig. 14-11

• AOQL = .0155– No matter how bad the incoming lots are, the

outgoing quality level will never be worse than 1.55% fraction defective

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Average total inspection

• ATI = n + (1- Pa)(N - n)– N = 10000, n = 89, c = 2, p = .01

– Pa = .9397

– ATI = 89 + (1-.9397)(10000-89) = 687

• See Fig. 14-12– ATI curves for n=89, c=2, N=1000, 5000,

10000

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Double sampling plans

• Defined by four parameters:– n1 = sample size for the first sample

– c1 = acceptance number of the first sample

– n2 = sample size for the second sample

– c2 = acceptance number of the second sample

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Inspect a random sampleof n1 = 50 from the lot

d1 = number of observed defectives

Inspect a random sampleof n2 = 100 from the lot

d2 = number of observed defectives

Acceptthe lot

Acceptthe lot

Rejectthe lot

Rejectthe lot

d1<c1=1 d1>c2=3

1<d1<3

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

n1=50, c1=1

n2=100, c2=3

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Double sampling plans

• Advantages– Can reduce the total amount of inspection– Allows the vendor a second chance

• Disadvantages– Unless curtailment is used, can lose the

economical advantage– More record keeping is needed

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OC curve

• Pgs. 696-698• Pa = probability of acceptance on the

combined samples• Pa

I = Probability of acceptance on the first sample

• PaII = Probability of acceptance on the

second sample• Pa = Pa

I + PaII

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OC curve

n1=50, c1=1

n2=100, c2=3

• For p = .05

• Compute PaI = .279 as shown on pg. 697

• Then, compute PaII = .010 as shown on pg.

697

• So, Pa = .279 + .010 = .289

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Average sample number

• ASN = n1P1 + (n1 + n2)(1 – P1) = n1 + n2(1 – P1)

where P1 = PaI + Pr

I

See Fig. 14-15

Compares the ASN for n1=60, c1=2, n2=120, c2=3 and the ASN for n=89, c=2The OC curves of the two plans are nearly identical

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Designing a sampling plan

• Grubb’s tables are commonly used– n2 = n1 or n2 = 2n1 and = .05, = .10

– Portion of table for n2 = 2n1 shown on next slide

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Portion of Grubb’s tables

pn1 for pn1 forPlan R=p2/p1 c1 c2 1- = .95 = .10

2 8.07 0 2 0.3 2.423 6.48 1 3 0.6 3.894 5.39 0 3 0.49 2.64

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Example

• Want a double sampling plan with = .05, = .10, p1 = .02, p2 = .12 and n2 = 2n1

• R = p2/p1 = .12/.02 = 6

• Plan 3 comes closest where c1=1 and c2=3

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Introduction to Statistical Quality Control, 4th Edition

Example, cont.

• Determine n1

– Hold – pn1 = .60

– n1 = pn1/p1=60/.02 = 30

– So, n2 = 60

• Summary: n1 = 30, c1 = 1, n2 = 60, c2 = 3

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Introduction to Statistical Quality Control, 4th Edition

Example, cont.

• Another way, hold constant for plan 3– n1 = pn1/p2 = 3.89/.12 = 32.42

– Rounding up, n1 = 33, n2 = 66, c1 = 1, c2 = 3

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Multiple sampling plans

• See pg. 701 for an example– After first sample of n1 = 20, if d1 = 0 accept, if d1 = 3,

reject

– If a decision is made, curtailment can be applied

– If d1 = 1 or 2, take a second sample of n2 = 20, and if the cumulative number of defectives is 1, then accept, or, if the cumulative number of defectives is 4, reject

– This continues until the 5th sample at which time a decision is made

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Multiple sampling plans

• Advantage is that the average sample number may be lower than single- or double-sampling

• Disadvantage is increased complexity

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Sequential sampling plans

• Take samples of size one and continue until a decision is made

• This could continue indefinitely

• In practice, truncation is used

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Sequential sampling plans

• Sequential probability ratio test (SPRT)

• See Fig. 14-16

• See equations on pg. 702

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Example

• We want a sequential-sampling plan for which p1 = .01, = .05, p2 = .06, = .10

• Limit lines are – XA = -1.22 + .028n

– XR = 1.57 + .028n

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Introduction to Statistical Quality Control, 4th Edition

Example, cont.

• Can also use a table to make decisions• See Table 14-3 on pg. 704• Calculations for n = 45

– XA = -1.22 +.028(45) = .04– XR = 1.57 + .028(45) = 2.83– Acceptance and rejection numbers must be

integer• Round down XA to 0 and round up XR to 3

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Example, cont.

• When is the first opportunity to accept?– -1.22 + .028n > 0

• n > 43.57 44

• When is the first opportunity to reject?– For n = 1, 1.57 +.028 = 1.598 > 1– For n = 2, 1.57 + .056 = 1.626 < 2 2

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14-4. Military Standard 105E (ANSI/ASQC Z1.4 ISO 2859)

14-4.1 Description of the Standard• Developed during World War II• MIL STD 105E is the most widely used

acceptance-sampling system for attributes• Gone through four revisions since 1950.• MIL STD 105E is a collection of sampling

schemes making it an acceptance-sampling system

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14-4. Military Standard 105E (ANSI/ASQC Z1.4 ISO 2859) 14-4.1 Description of the Standard• Three types of sampling are provided for:

1. Single2. Double3. Multiple

• Provisions for each type of sampling plan include

1. Normal inspection2. Tightened inspection3. Reduced inspection

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Introduction to Statistical Quality Control, 4th Edition

14-4. Military Standard 105E (ANSI/ASQC Z1.4 ISO 2859)

14-4.1 Description of the Standard• The acceptable quality level (AQL) is a primary focal

point of the standard• 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 for the standard provided are used to determine

the appropriate sampling scheme.

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14-4. Military Standard 105E (ANSI/ASQC Z1.4 ISO 2859)

14-4.1 Description of the Standard• Switching Rules

– Normal to tightened– Tightened to normal– Normal to reduced– Reduced to normal– Discontinuance of inspection

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14-4. Military Standard 105E (ANSI/ASQC Z1.4 ISO 2859) 14-4.2 Procedure1. Choose the AQL2. Choose the inspection level3. Determine the lot size4. Find the appropriate sample size code letter from Table

14-45. 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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14-4. Military Standard 105E (ANSI/ASQC Z1.4 ISO 2859)

Example• Suppose a product is submitted in lots of size

N = 2000. The AQL is 0.65%. Say we wanted to generate normal single-sampling plans.

• For lots of size 2000, (and general inspection level II) Table 14-4 indicates that the appropriate sample size code letter is K.

• From Table 14-5 for single-sampling plans under normal inspection, the normal inspection plan is n = 125, c = 2.

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Introduction to Statistical Quality Control, 4th Edition

14-4. Military Standard 105E (ANSI/ASQC Z1.4 ISO 2859)

14-4.3 Discussion• There are several points about the standard that

should be emphasized:1. MIL STD 105E is AQL-oriented2. The sample sizes selected for use in MIL STD 105E

are limited3. The sample sizes are related to the lot sizes.4. Switching rules from normal to tightened and from

tightened to normal are subject to some criticism.5. A common abuse of the standard is failure to use the

switching rules at all.

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Normal TightenedReduced

2 out of 5 consecutive lots

rejected

5 consecutivelots accepted

O Production steadyO 10 consecutive lots acceptedO Approved by responsible authority

O Lot rejectedO Irregular productionO Lot meets neither accept nor reject criteria

O Other conditions warrant return to normal inspection

“And” conditions

“Or” conditions

Start

10 consecutivelots remain on

tightened inspection

Discontinue

inspection

Switching rules

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OC curves

• See pg. 713 for OC curves for sample size code letter K

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Double sampling

• These are included in the full text of the standard

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14-4. Military Standard 105E (ANSI/ASQC Z1.4 ISO 2859)

14-4.3 Discussion• 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 Plans

• For rectifying inspection

• See Table 14-8 on pg. 717 for an example for AOQL = 3%– Indexed by lot size (N) and process average (p)

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Example

• N = 5000, p = .01

• Want a single sampling plan (w/rectifying inspection) with AOQL = 3%

• Read n = 65, c = 3 from the table

• These plans minimize ATI– Pa = .9957 at p = .01 (determined as previously)

– ATI = 65 + (1 - .9957)(5000 – 65) = 86.22

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Example, cont.

• Also, note that LTPD = 10.3%

• This is the point on the OC curve for which Pa = .10

– That is, this plan provides that 90% of incoming lots that are as bad as 10.3% defective will be rejected

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LTPD plans

• Can also develop a plan for a specified LTPD

• Table 14-9 is for LTPD = 1%

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Example

• N = 5000, p = .25%

• We want a single sampling plan (w/rectifying inspection) with LTPD of 1%

• Find n = 770, c = 4

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Assignment

• Work odd numbered exercises through 14-15

• Understand MIL-STD 105E and Dodge-Romig tables

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End