MDP 308

Quality Management

Lecture #10

Acceptance Sampling

Today’s lecture

What is acceptance sampling?

Selecting items randomly for inspection

Problems with sampling

Single sampling plan and OC curves

Double sampling plan

Rectified sampling plan and AOQL

Using military standards for acceptance sampling

What is acceptance sampling?

Acceptance sampling is a method used to accept or reject a lot (batch) of products based on a random sample of the lot.

The purpose of acceptance sampling is to sentence lots (accept or reject) rather than to estimate the quality of a lot.

Acceptance sampling plans do not improve quality. The nature of sampling is such that acceptance sampling will accept some lots and reject others even though they are of the same quality.

The most effective use of acceptance sampling is as an auditing tool to help ensure that the output of a process meets requirements.

Problems with sampling

The sampling rules in any formal sample plans are based on probability. But application of probability predicts the acceptance of lots with substandard quality.

Example: A lot containing 100 items with a probability of an item to be defective of 5% (this is called the quality level of the lot). A sampling plan is decided such that 20 items will be drawn randomly, if the number of defective items is found to be zero or 1, the lot will be accepted; otherwise, the lot will be rejected. In this example, number of defective items = 0.05 x 100 = 5

The probability of accepting the lot =

Prob.{zero defective items are drawn from the 20-item sample}

+ Prob.{only one defective item is drawn out of the 20-item sample}

Problems with sampling

Using Hypergeometric distribution (sampling without

replacement), we have N = 100, n = 20, r = 100x5% = 5

The probability of accepting the lot =

Prob. {zero defective items are drawn from the 20-item

sample} + Prob.{only one defective item is drawn out of

the 20-item sample}

= +

= 0.42 + 0.319 = 0.739

5 100 5

0 20 0

100

20

−

−

5 100 5

1 20 1

100

20

−

−

Problems with sampling

Using Binomial distribution as an approximation (sampling

with replacement), we have N = 100, n = 20, p = 0.05

The probability of accepting the lot =

Prob. {zero defective items are drawn from the 20-item

sample} + Prob.{only one defective item is drawn out of

the 20-item sample}

= +

= 0.36 + 0.377 = 0.736

0200 )05.01(05.00

20−−

1201 )05.01(05.01

20−−

Acceptance sampling risks As mentioned, acceptance sampling can reject “good” lots and

accept “bad” lots. More formally:

Producers risk refers to the probability of rejecting a good lot. In order to calculate this probability there must be a numerical definition as to what constitutes “good” AQL (Acceptable Quality Level) - the numerical definition of a good lot.

The ANSI/ASQC standard describes AQL as “the maximum percentage or proportion of nonconforming items or number of nonconformities in a batch that can be considered satisfactory as a process average”

Consumers Risk refers to the probability of accepting a bad lot where: LTPD (Lot Tolerance Percent Defective) - the numerical definition of a

bad lot described by the ANSI/ASQC standard as “the percentage or proportion of nonconforming items or non-comformities in a batch for which the customer wishes the probability of acceptance to be a specified low value.

Selecting items randomly for inspection

It is extremely important to have random samples in any

acceptance procedure to help in avoiding systematic and

random errors.

Sampling plans must be devised to fit the packaging from the

incoming pieces.

Example: Consider an item delivered in 100 boxes, each box

contains 4 rows of 9 pieces each. The sampling plan needs to

decide randomly which box to draw from, then which row and

then finally which box in the row.

…………

100boxes

1 2 3

4 5 67 8 9

123

4

Selecting items randomly for inspection

Using the random digits generation function in the calculator, we can get three random digits from which we need only two to decide which box to draw from. The digits that represent the boxes are from 00 to 99.

Then another random digits are generated and one digit is used to decide the row number in the selected box. Here we need one digit from 0 to 3.

Finally, another random digit from 0 to 8 is generated to decide which piece will be drawn from the selected row.

It is important to show that if a random digit is found to be outside the defined range, it must be neglected and another random digits are generated until the selected digit lies inside the defined range.

Single (one-stage) sampling plan

From a lot of N items, select n items randomly

Let X be the number of defective items found in the

sample

If X > c, the lot is rejected

If X c, the lot is accepted

The probability of accepting the lot is calculated using

binomial distribution as follows:

𝑃𝑎 = 𝑃𝑟𝑜𝑏. 𝑋 ≤ 𝑐 =

𝑘=0

𝑐𝑛𝑘

𝑝𝑘(1 − 𝑝)𝑛−𝑘

Where p is the probability that an item is defective.

The Poisson distribution chart

The operating characteristic (OC) curve

Probability of acceptance

(Pa)

Percent defective (100p)

1.0

Drawing an OC curve To draw an OC curve, we need first to identify two parameters:

n: sample size

c: limit on the number of allowable defective items in a sample, beyond which the sample will not be accepted.

The steps of constructing OC curves are:

1. Select p values from 0 to an upper limit suitable to the selected n and c values (maximum is 0.99)

2. multiply each value by n and enter it in the table

3. make a percent column for p and mark it 100p

4. Start at the pn value on the Poisson distribution graph and go to the curve with the corresponding c value to get Pa. Or if the sample size is less than 20, use the cumulative binomial distribution formula for calculating Pa as stated earlier.

5. Record the Pa value on the table.

Note: if the sample size (n) is less than 20 units the binomial distribution is used to build the OC Curve otherwise the Poisson distribution is used as sufficient approximation.

Example of drawing an OC curve

Example: Draw an OC curve with (n = 200 and c = 6)

We first construct the following table using the procedure

described in the previous slide:

p np 100p Pa

0 0 0 1.00

0.01 2 1 1.00

0.015 3 1.5 0.97

0.02 4 2 0.89

0.025 5 2.5 0.76

0.03 6 3 0.61

0.04 8 4 0.31

0.05 10 5 0.13

0.06 12 6 0.05

0.07 14 7 0.01

0.08 16 8 0.00

0.09 18 9 0.00

Example of drawing an OC curve (cont.)

On the OC curve we the following terms are defined:

AQL: Acceptable Quality Level

IQL: Indifference Quality Level

LTPD: Lot Tolerance Percent Defective

Indecisive zone = LTPD - AQL

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

1.10

0 1 2 3 4 5 6 7 8 9 10

100p

Pa

AQL IQL RQL

Indecisive zone

Producer and consumer risks

0

0.2

0.4

0.6

0.8

1

1.2

Pro

ba

bil

ity o

f A

cce

pta

nce

100P

OC Curve

LTPDAQL

Producers Risk

Consumers Risk

Hypothesis testing in acceptance sampling

The null and alternative hypotheses of the single-sampling plan

can be stated as follows:

H1: p > p0

H0: p = p0

Where p is the true percentage defective in the population (lot)

p0 is the rejection limit = c / N

The probability of type I error (-error) is the probability that

the lot will be rejected while its true percent defective is

actually less than or equal p0. This is what is called producer

risk.

The probability of type II error (-error) is the probability that

the lot will be accepted while its true percent defective is

actually greater than p0. This is what is called consumer risk.

Producer and consumer risks

0

0.2

0.4

0.6

0.8

1

1.2P

rob

ab

ilit

y o

f A

cce

pta

nce

100P

OC Curve

LTPDAQL

Producers Risk

Consumers Risk

Producer risk = 1 – Probability of acceptance at the AQL

Consumer risk = Probability of acceptance at the LTPD

The effect of n and c on the OC curve

Pa

100p

1.0

n increases

Pa

100p

1.0

c increases

The effect of n and c on the OC curve

0

0.2

0.4

0.6

0.8

1

1.2

0 2 4 6 8 10

100p

Pa

Designing sampling plans

Operationally, three values need to be determined

before a sampling plan can be implemented (for single

sampling plans):

N = the number of units in the lot

n = the number of units in the sample

c = the maximum number of nonconforming units in the

sample for which the lot will be accepted.

Designing sampling plans

While there is not a straightforward way of determining

these values directly given desired values of the

parameters, tables have been developed. Below is an

excerpt of one of these tables.

Excerpt From a Sampling Plan Table with Producers Risk = 0.05 and Consumers Risk = 0.10

c LTPD/AQL n(AQL) n(LTPD)

0 44.89 0.052 2.334

1 10.946 .355 3.886

2 6.509 .818 5.324

3 4.89 1.366 6.68

4 4.057 1.97 7.992

5 3.549 2.613 9.274

6 3.206 3.286 10.535

7 2.957 3.981 11.772

8 2.768 4.695 12.996

9 2.618 5.426 14.205

Designing sampling plans

Pa

100p

1.0

The ideal OC curve

Given the choice from a set of OC curves, select the one that has the lowest c/n

ratio and smallest LTPD-AQL difference

Three alternatives for specifying sampling

plans

1. Producers Risk and AQL specified

2. Consumers Risk and LTPD specified

3. All four parameters specified

First two alternatives of sampling plan

preparation

For the first two cases we simply choose the acceptance

number and divide the appropriate column by the

associated parameter to get the sample size.

Example 1: Given a producers risk of .05 and an AQL of .015

determine a sampling plan

c = 1: n = .355/.015 ~ 24

c = 4: n = 1.97/.015 ~ 131

Example 2: Given a consumers risk of .1 and a LTPD of .08

determine a sampling plan

c = 0: n = 2.334/.08 ~ 29

c = 5: n = 9.274/.08 ~ 116

Sampling plans when All four parameters

specified

When all four parameters are specified we must first find

a value close to the ratio LTPD/AQL in the table. Then

values of n and c are found.

Example: Given producers risk of .05, consumers risk of

.1, LTPD 4.5%, and AQL of 1% find a sampling plan.

Since 4.5/1 = 4.5 is between c= 3 and c = 4. Using the n(AQL)

column the sample sizes suggested are 137 and 197

respectively. Note using this column will ensure a producers

risk of .05. Using the n(LTPD) column will ensure a consumers

risk of .1

Double sampling plans

In an effort to reduce the amount of inspection, double

(or multiple) sampling is used. Whether or not the

sampling effort will be reduced depends on the defective

proportions of incoming lots. Typically, four parameters

are specified:

n1 = number of units in the first sample

c1 = acceptance number for the first sample

n2 = number of units in the second sample

c2 = acceptance number for both samples

Procedure of double sampling plans

A double sampling plan proceeds as follows:

A random sample of size n1 is drawn from the lot.

If the number of defective units (say d1 ) c1 the lot is

accepted.

If d1 c2 the lot is rejected.

If neither of these conditions are satisfied a second

sample of size n2 is drawn from the lot.

If the number of defectives in the combined samples (d1 +

d2) > c2 the lot is rejected. If not the lot is accepted.

Example

Consider a double sampling plan with n1 = 50; c1 = 1; n2 =

100; c2 = 3. What is the probability of accepting a lot with

5% defective?

The probability of accepting the lot, say Pa, is the sum of

the probabilities of accepting the lot after the first and

second samples (say P1 and P2)

Example (cont.)

To calculate P1 we first must determine the expected

number defective in the sample (50*.05)

we can then use the Poisson function to determine the

probability of finding 1 or fewer defects in the sample,

e.g., POISSON.DIST(1,2.5,TRUE) = .287

To calculate P2 we need to determine first the scenarios

that would lead to requiring a second sample. Then for

each scenario we must determine the probability of

acceptance. P2 will then be the sum of the probability of

acceptance for each scenario.

Example (cont.)

For this example, there are only two possible

scenarios that would lead to requiring a second

sample:

Scenario 1: 2 nonconforming items in the first sample

Scenario2: 3 nonconforming items in the first sample.

The probability associated with these two scenarios

is found as follows:

P(scenario 1) = POISSON.DIST(2,2.5,FALSE) = .257

P(scenario 2) = POISSON.DIST(3,2.5,FALSE) = .214

Example (cont.) Under scenario 1 the lot will be accepted if there are either 1 or 0

nonconforming items in the second sample. The expected number of nonconforming items in the second sample .05(100) = 5

the probability is then: POISSON.DIST(1,5,TRUE) = .04

therefore the probability of acceptance under scenario 1 is (.257)*(.04) = .01

Under scenario 2 we will accept the lot only if there are 0 nonconforming items in the second sample, i.e., POISSON.DIST(0,5,FALSE) = .007

therefore the probability of acceptance under scenario 2 is (.214)*(.007) = .0015

so the probability of acceptance on the second sample is .01 + .0015 = .0115

So the overall probability of acceptance is (probability of acceptance on first sample + probability of accepting after 2 samples) = .287 + .0115 = .299

OC curve for double sampling

OC Chart - Double Sampling Plan

0

0.10.2

0.3

0.40.5

0.6

0.7

0.80.9

1

0

0.0

1

0.0

2

0.0

3

0.0

3

0.0

4

0.0

5

0.0

6

0.0

7

0.0

7

0.0

8

0.0

9

0.1

0.1

1

0.1

1p

Pa

0

0.10.2

0.3

0.4

0.50.6

0.7

0.80.9

1

P1

Pa = P1 + P2

P(rejection 1st)

Average Sample Number

Since the goal of double sampling is to reduce the inspection effort, the average number of units inspected is of interest. This is easily calculated as follows

Let PI = the probability that a lot disposition decision is made on the first sample, i.e., (probability that the lot is rejected on the first sample + probability that the lot is accepted on the first sample)

Then the average number of items is:

n1(PI) +(n1 + n2)(1 - PI)

= n1 + n2(1 - PI)

The average sample number curve can be

derived from the data on the OC chart

ASN Curve

0

20

40

60

80

100

120

0

0.0

1

0.0

2

0.0

2

0.0

3

0.0

4

0.0

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0.0

5

0.0

6

0.0

6

0.0

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0.0

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0.0

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0.0

9

0.1

0.1

1

0.1

1

0.1

2

p

Av

era

ge

sa

mp

le n

um

be

r

Rectified sampling plan

Population

p, N

Sample

n

Reject

Prob. = 1-Pa Accept

Prob. = Pa

Average number of defectives = np < c

replace np defectives

N

n zero defectives p(N-n) defectives

= p(N-n) defectives

Make 100% inspection for N

replace for Np defectives

N Zero defectives

Rectified sampling plan

Expected number of defectives in the lot = (1-Pa)zero + Pa[zero+p(N-

n)] = pPa(N-n)

→ expected percentage defective in the lot = 100 pPa(N-n)/N

The average outgoing quality (AOQ) = pPa(1-n/N)

0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0.018

0.02

0 2 4 6 8 10

p %

AO

Q

Average outgoing quality limit (AOQL)

Military Standard 105E

(MIL STD 105E)

(ANSI/ASQC Z1.4, ISO 2859)

Most widely used acceptance sampling system for

attributes

MIL STD 105E is Acceptance Sampling System

collection of sampling schemes

Can be used with single, double or multiple

sampling plans

Inspection Types

Normal Inspection

Used at start of inspection activity

Tightened Inspection Instituted when vendor’s recent quality history has deteriorated

Acceptance requirements for lots are more stringent

Reduced Inspection Instituted when vendor’s recent quality history has been exceptionally

good

Sample size is usually smaller than under normal inspection

Switching Rules

- Production Steady

- 10 consecutive lots accepted

- Approved by responsible

authority

NormalReduced Tightened

- Lot rejected

- Irregular production

- Lot meets neither accept

nor reject criteria

- Other conditions warrant

return to normal inspection

2 out of 5 consecutive lots

rejected

5 consecutive

lots accepted

10 consecutive lots remain

on tightened inspection

Start

Discontinue

Inspection

AND conditions

OR conditions

Procedure for

MIL STD 105E

STEP 1: Choose AQL MIL STD 105E designed around Acceptable Quality Level,

AQL Recall that the Acceptable Quality Level, AQL, is producer's largest

acceptable fraction defective in process

Typical AQL range: 0.01% AQL 10%

Specified by contract or authority responsible for sampling

STEP 2: Choose inspection level

Level II

Designated as normal

Level I

Requires about one-half the amount of inspection as Level II

Use when less discrimination needed

Level III

Requires about twice as much inspection as Level II

Use when more discrimination needed

S-1, S-2, S-3, S-4

Four special inspection levels used if very small samples necessary

Procedure for

MIL STD 105E

STEP 3–Determine lot size, N

Lot size most likely dictated by vendor

STEP 4: Find sample size code letter

From Table 12-10

Given lot size, N, and Inspection Level, use table to

determine sample size code letters

STEP 5: Determine appropriate type sampling plan

Decide if Single, Double or Multiple sampling plan is to be

used

Procedure for

MIL STD 105E

STEP 6: Find Sample Size, n, and Acceptance Level, c

Given sample size letter code, use Master Tables: 12-11, 12-12,

and 12-13

Find n and c for all three inspection types:

Normal Inspection

Tightened Inspection

Reduced Inspection

Procedure for

MIL STD 105E

Example

Suppose product comes from vendor in lots of size 2000 units. The acceptable quality level is 0.65%. Determine the MIL STD 105E acceptance-sampling system.

Normal Insp. Level

Lot Size

= 2000

Table 12-10

AQL

Plan K

Sample 125 units.

Ac = 2, accept if defects ≤ 2.

Re = 3, reject entire lot if defects ≥ 3.

Table 12-11

AQL

Plan K

Sample 125 units

Ac = 1, accept if defects ≤ 1.

Re = 2, reject entire lot if defects ≥ 2.

Table 12-12

AQL

Plan K

Sample 50 units

Ac = 1, accept if defects ≤ 1.

Re = 3, reject entire lot if defects ≥ 3.

Table 12-13