Transcript Acceptance Sampling for Attributes

SMU
EMIS 7364
NTU
TO-570-N
Statistical Quality Control
Dr. Jerrell T. Stracener,
SAE Fellow
Acceptance Sampling for Attributes
Updated: 4.4.02
1
Decision Risk
• Producer’s Risk –
conforming lots are rejected
• Consumer’s Risk – nonconforming lots are
accepted.
2
Decision Risk
Decision
True Situation
Lot meets
Lot does not
requirements
meet requirements
Accept Lot
no error
Type II error
Reject Lot
Type I error
no error
3
Lot-by-Lot Acceptance Sampling by Attributes
– Single Sampling
4
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.
5
Type A OC Function for Single Sampling Plan
• Sampling Plan Specification
N = Lot Size
n = Sample Size
c = Acceptance number
• D = true number of defectives in the lot
• X = number of defective items in the random
sample.
6
Type A OC Function for Single Sampling Plan
• Probability Distribution of X

X ~ H(N, D, n)

Probability Mass Function of X
  D  N  D 

  
  x  n  x  for x  0,1,..., min n, D 
h x   
 N
 

n

0
, otherwise
7
Type A OC Function for Single Sampling Plan
• OC Function
OCD  PA D 
 P X  c | D 
 D  N  D 
 

c 
x  n  x 


 N
x 0
 
n
8
Example – Single Sampling Plan A
Determine and plot the OC Function for a single
sampling plan specified by
N  50
n 5
c2
9
Example Solution – Single Sampling Plan
OCD  PA D 
 PX  2 | D 
 D  50  D 
 

2 
x  5  x 


 50 
x 0
 
5
D
0
5
10
15
20
25
30
35
40
45
50
PA(D)=OC(D)
1.00
1.00
0.95
0.85
0.69
0.50
0.31
0.15
0.05
0.00
0.00
OC(D)
1.0
0.8
0.6
0.4
0.2
0.0
0
10
20
D
30
40
50
10
Concept
Suppose that the lot size N is large (theoretically
infinite). Under this condition, the distribution of the
number of defectives d in a random sample of n
items is binomial with parameters n and p, where p
is 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 infinite
process, and then to draw random samples of n from
these lots. Sampling from the lot in this manner is
the equivalent of sampling directly from the process.
11
Type B OC Function for Single Sampling Plan
• Sampling Plan Specification
N = Lot Size
n = Sample Size
c = 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.,
D
p
N
• X = number of defective items in the random
sample.
12
Type B OC Function for Single Sampling Plan
• Probability Distribution of X

X ~ B(n, p)

Probability Mass Function of X
n x
n -x






b x   p 1 - p
x
for x  0,1,..., n
13
Acceptance Sampling
The probability of observing exactly x defective
items is:
Px   PX  x 
 n  x nx
  p q
x
n!
nx
x

p 1  p 
x! n  x !
for x = 0, 1, . . ., n
14
OC Function
The probability of acceptance is the probability that
d is less than or equal to c, or:
OCp  PA p 
 PX  c | p 
c
  bx 
x 0
n x
n -x
   p 1 - p 
x 0  x 
c
15
Example – Single Sampling Plan B
Determine and plot the OC Function for a single
sampling plan specified by
N  infinite in size, or at least very large compared to n
n  98
c2
Compare the OC curve for N=500, n=98, c=2.
16
Example Solution – Single Sampling Plan
OCp  PA D 
 PX  2 | p 
 98  x
98- x
  bx     p 1 - p 
x 0
x 0  x 
2
OC(p)
2
1.0
0.8
0.6
0.4
0.2
0.0
0.00
0.02
0.04
0.06
p
0.08
0.10
17
Example
If the lot fraction defective is p = 0.02, n = 89 and
c = 2, then
PA 0.02  Pd  2
2
89!
0.02d 0.9889d

d  0 d!89  d !
89!
89!
89
0
0.021 0.9888
0.02 0.98 

1!88!
0!89 !
89!
87
2
0.02 0.98

2!87 !
 0.7366
18
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 the
sampling plan. For example, in the sampling plan
n = 98, c = 2, if the lots are 2% defective, the
probability of acceptance is approximately 0.74. This
means that if 100 lots from a process that
manufactures 2% defective product are submitted
to this sampling plan, we will expect, in the long
run, to accept 74 of the lots and reject 26 of them.
19
Example
Probabilities of Acceptance for the Single-Sampling
plan n = 89, c = 2
Fraction
Defective, p
0.005
0.010
0.020
0.030
0.040
0.050
0.060
0.070
0.080
0.090
Probability of
Acceptance, Pa (p)
0.9897
0.9397
0.7366
0.4985
0.3042
0.1721
0.0919
0.0468
0.0230
0.0109
20
If N = 500, n =98 & c=2,
PA D  OCD  PX  2 | D 
 D  500  D 
 

2 
x  98  x 


 500 
d 0


 98 
For comparison, select p=0.02. Then
10  490 
D  N p  5000.02  10, and
 

2 
x  98  x 

PA 10  OCD   PX  2 | 10  
 500 
d 0


 98 
 0.2753  0.3049  0.1976
 0.7778
21
Probability of acceptance, Pa
Example
1
0.8
0.6
0.4
0.2
0
0
0.02
0.04
0.06
0.08 -p
0
10
20
30
40
-D
Lot fraction defective, p
22
Single Sample Test Plan Design
• Probability Distribution of X
P0 = specified fraction defective
P1 = minimum acceptable fraction defective
a = producer’s risk
b = consumer’s risk
23
Test Procedure
To test
vs
H0: p = p0
H1: p = p1
at the a  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
24
Single Sampling Plan
x
number
of defective
c
items
.
.
.
2
1
0
reject
accept
0
n
n = sample size
x = number of defective items
c = maximum number of defective items for
acceptance
25
Operating Characteristic (OC) Function
OC  p   PA  p 
 P X  c | p 
 n x
n x
    p 1  p 
x 0  x 
c
Note that:
OCp1   β
and
OCp0   1  α
26
OC Curve
OC  p   PA  p 
1
1-a
b
0
p0
p1
1
p
27
Test Plan Design
To determine a single sample test plan for testing H0:
p = p0, specify values of p0, p1, a, and b such that
PA(p0) = 1 - a and PA(p1) = b, then find the values of
n and c that satisfy the following equations.
 n x
n x
  p0 1  po   1  a

x 0  x 
c
and
 n x
n x
  p1 1  p1   b

x 0  x 
c
28
Lot-by-Lot Acceptance Sampling by Attributes
– Double Sampling
29
Double Sampling
A double-sampling plan is a procedure in which,
under certain circumstances, a second sample is
required before the lot can be sentenced. A doublesampling plan is defined by four parameters:
n1 = sample size on the first sample
c1 = acceptance number of the first sample
n2 = sample size on the second sample
c2 = acceptance number for both samples
30
Single Sampling Plan
x
number
of defective
items
c2
.
.
.
c1+1
Reject
Reject
Continue
Accept
c1
.
.
.
2
1
0
Accept
0 1 2
...
n1
...
n1+n2
31
Double Sampling - Advantages
The principal advantage of a double-sampling plan
with respect to single sampling is that it may reduce
the total amount of required inspection. Suppose that
the first sample taken under a double-sampling plan
is smaller than the sample that would be required
using a single-sampling plan that offers the consumer
the same protection. In all cases, then, in which a lot
is accepted or rejected on the first sample, the cost
of 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 second
sample (called curtailment of the second sample).
32
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.
33
Double Sampling - The OC Curve
The performance of a double-sampling plan can be
conveniently summarized by means of its operatingcharacteristic (OC) curve. A double-sampling plan
has a primary OC curve that gives the probability of
acceptance as a function of lot of process quality. It
also has supplementary OC curves that show the
probability of acceptance as a function of lot
acceptance and rejection on the first sample.
34
Double Sampling
Inspect a random sample of
n1 = 50 from the lot
d1 = number of observed defectives
Accept
the
lot
d1  c1 = 1
d1 > c1 = 3
Reject
the
lot
Inspect a random sample of
n2 = 100 from the lot
d2 = number of observed defectives
Accept
the
lot
d1 + d2  c2 = 3
d1 + d2 > c2 = 3
Operation of the double-sampling plan with
n1 = 50, c1 = 1, n2 = 100, c2 = 3
Reject
the
lot
35
Example

If Pa p denotes the probability of acceptance on the
combined samples, and PaI and PaII denote the
probability of acceptance on the first and second
samples, respectively, then
Pa p   P p   P p 
I
a
II
a
P p is just the probability that we will observe
I
a
d1  c1 = 1 defectives out of a random sample of
n1 = 50 items. Thus
1
50!
50d
P p   
p d1 1  p  1
d1 0 d1!50  d1 !
I
a
36
Example
If p = 0.05 is the fraction defective in the incoming
lot, then
1
50!
d1
50d1
0.05 0.95  0.279
P 0.05  
d1 0 d1!50  d1 !
I
a
To obtain the probability of acceptance on the second
sample, we must list the number of ways the second
sample can be obtained. A second sample is drawn
only if there are two or three defectives on the first
sample - that is, if c1 < d1  c2.
37
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)
1
100!
 50!
2
48 
0.05 0.95  
0.05d2 0.95100d 2

 2!48!
d 2 0 d 2!100  d 2 !
= (0.261)(0.037)
= 0.009
38
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)
50!
3
47 100!
0
100
0.05 0.95
0.05 0.95

3!47!
0!100!
= (0.220)(0.0059)
= 0.001
39
Example
Thus, the probability of acceptance on the second
sample is
P 0.05  Pd1  2, d 2  1  Pd1  3, d 2  0
 0.009  0.001
 0.010
II
a
The probability of acceptance of a lot that has fraction
defective p = 0.05 is therefore
Pa 0.05  PaI 0.05  PaII 0.05
 0.279  0.010
 0.289
40
Double Sampling - The OC Curve
1.0
Probability of
acceptance on
combined sample
Probability, P
0.8
0.6
0.4
Probability of
rejection on first
sample
Probability of
acceptance on
first sample
0.2
0.0
0.00
0.02
0.04
0.06
0.08
0.10
0.12
Lot fraction defective, p
OC Curves for the double-sampling plan with
n1 = 50, c1 = 1, n2 = 100, c2 = 3
41
Rectifying Inspection Programs
42
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.
43
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.
44
Rectifying Inspection
Rejected
lots
Incoming lots
Fraction defective
p0
Fraction
defective
0
Outgoing lots
Fraction defective
p1
Inspection
activity
Accepted
lots
Fraction
defective
p0
45
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.
46
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.
47
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.
48
Average Outgoing Quality (AOQ)
The average fraction defective, called average
outgoing quality is
Pa pN  n 
AOQ 
,
N
Where the lot size is N and that all defectives are
replaces with good units. Then in lots of size N,
we have
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
49
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
Pa pN  n 
N

0.93970.0110,000  89

10,000
 0.0093
AOQ 
That is, the average outgoing quality is at 0.93%
defective.
50
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.
0.0175
0.0150
AOQ
0.0125
0.0100
AOQ
0.0075
0.0050
0.0025
0.0000
0.00
0.01
0.02
0.03
0.04
0.05
fraction defective, p
0.06
0.07
51
Average Total Inspection (ATI)
ATI  n  1  Pa N - n ,
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.
52
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
ATI  n  1  Pa N - n 
 89  1 - 0.9397 10,000  89
 687
Remember this is an average number of units
inspected over many lots with fraction defective
p = 0.01.
53
Determination of optimum quality level, p*
total
cost
cost to
achieve p
cost of
inspection
per lot
cost
0
p*
1
incoming quality level ~p
54