Chapter 15
Basic Acceptance Sampling Procedure
From : Book of Statistical Quality
Control by D. C. Montgomery
Prof . Ma.Teodora E.Gutierrez
THE ACCEPTANCE-SAMPLING PROBLEM I
►
lot-by-lot acceptance-sampling plans for attributes
acceptance sampling:
►
concerned with inspection and decision making
regarding product
THE ACCEPTANCE-SAMPLING(™K®Ø)
PROBLEMII
A typical application of acceptance sampling
►
a shipment of product from a supplier
►
Product: a component; raw material used in the
company’s manufacturing process
►
A sample taken from the lot: quality characteristic
of the units in the sample is inspected
►
A decisionis made regarding lot disposition o r
c a l l e d lot sentencing
)
)
accept: put into production
reject: returned to the supplier or to some lot
disposition action
THE ACCEPTANCE-SAMPLING(™K®Ø)
PROBLEM III
Three aspects of sampling:
1.sentence lots,not to estimate the lot quality
2.simply accepts or rejects lots:
Even if all lots are of the same quality, sampling will accept
some lots and reject others
3.an audit( 100%) tool to ensure that the output of a
process conforms to requirements
THE ACCEPTANCE-SAMPLING(™K®Ø) PROBLEMIV
Three approaches to lot sentencing:
1.accept with no inspection
2.100% inspection :
)
inspect every item, removing all defective units
)
Situation: extremely criticalcomponent
3.acceptance sampling
THE ACCEPTANCE-SAMPLING(™K®Ø) PROBLEMV
Acceptance sampling is useful in the following situations:
►
Test is destructive(%û¾)
►
The cost of 100% inspection is extremely high
►
100% inspection is not technologically feasible or ð ¾
►
many item to be inspected or the inspection error rate is
high than 100% inspection
►
the supplier has an excellent quality history
►
potentially serious product liability(.k) risks
ADVANTAGES AND DISADVANTAGES OF SAMPLINGI
Advantage: comparing with 100%inspection
►
less inspection
►
reduced damage
►
applicable to destructive testing
►
fewer personnel are involved
►
reduces the amount of inspection error
►
the simple return of defectives often provides a stronger
motivation to the supplier for quality improvements
ADVANTAGES AND DISADVANTAGES OF SAMPLINGII
Disadvantage:
►
risks of accepting "bad" lots and rejecting "good" lots
►
less information
►
Acceptance sampling requires planning and
documentation of the acceptance-sampling procedure
TYPES OF SAMPLING PLANSI
Variables: Chap. 16
Attributes: Chap. 15
1. single-sampling plan:
(sample size, acceptance number)=(n,
c) if #{defective}>c ⇒ reject thelot
2. double-sampling plan
3. multiple-sampling plan: the extension is
sequential sampling
LOT FORMATION
►
Lots should be homogeneous: the same machine, the
same operators, from common raw materials, at
approximate the same time
►
Larger lots are preferred over smaller ones
►
Lots should be conformable(ʼ¾) tothe
materials-handling systems used in both the supplier and
consumer facilities('n)
SINGLE-SAMPLING PLAN FOR ATTRIBUTESI
►
N : size of a lot; (Ex: N = 10, 000)
►
n: the sample size; (Ex: n = 89)
►
c: the acceptance number; (Ex: c = 2)
►
d: the number of observed defective items
d ≤ c ⇒ acceptes the lot
Operating characteristic(OC) curve
SINGLE-SAMPLING PLAN FOR ATTRIBUTESII
►
An importantmeasure of the performance of an
acceptance-sampling plan
►
display(•î) the discriminatory($ïIh¾) power of the
sampling plan
►
the probabilitythat a lot submitted with a certain
fraction(½Iz) defective will be either accepted orrejected
SINGLE-SAMPLING PLAN FOR ATTRIBUTESIII
►
B(n, p)
P{d defectives} =
►
. Σ
n d
p (1 −p)
d
n− d
Probability of acceptance:
Pa= P{d ≤ c} =
Σc . nΣ
pd(1 − p) n−d
d
d=0
Ex: n = 89, c = 2, p = 0.01 ⇒ Pa = P{d ≤ 2} = 0.9397
n = 89, c = 2, p = 0.02 ⇒ Pa = P{d ≤ 2} = 0.74
100 lots: expect to accept 74 of the lots and reject 26 ofthem
SINGLE-SAMPLING PLAN FOR ATTRIBUTESIV
(a) ideal
(b) sample size
(c) c
(a)discriminated perfectly between good and bad lots
(b) c n=constant: the greater the slope of the OC curve, the
greater is the discriminatory power
(c)changing c does not dramatically change the slope of the
OC curve
DESIGNING A SINGLE-SAMPLING PLANI
Type I and Type II errors:
►
H0 : the lot quality is good
Accept H0
Reject H0
►
State of the lot
Good Quality of Lot
Poor Quality of Lot
(H0 is true)
(H0 is false)
Correct
β=Type II error
α=Type I error
Correct
To the design of an acceptance-sampling plan: require that
the OC curve pass through two designated points
DESIGNING A SINGLE-SAMPLING
PLANII
Σ . Σ
c
1 −α =
d=0
n d
p (1 − p1) n−d
d 1
(1 − α = the prob. of acceptance the lots with p1)
Σc . nΣ
β=
pd(1 − p )2 n− d
d 2
d=0
(β = the prob. of acceptance the lots with p2)
DESIGNING A SINGLE-SAMPLING PLANIII
►
1 − α, β: two simultaneous equations are nonlinear
►
there is no simple, direct solution
►
(p1, 1 − α) = (0.01, 0.95), (p2, β) = (0.06, 0.10)
►
Any two points on the OC curve could be used to define
the sampling plan
►
p1 =AQL (producer’s risk), p2 =LTPD (consumer’s risk)
►
AQL: acceptable quality level
►
LTPD: lot tolerance percent defective = RQL(rejectable
quality level)=LQL (limiting quality level)
Designing a Single-Sampling PlanIV
DESIGNING A SINGLE-SAMPLING PLAN
►
►
AQL: acceptable quality level the poorest level of quality for
the supplier process
LTPD: lot tolerance percent defective the poorest level of
quality that the consumer is willing to accept in an
individual lot
Designing a Single-Sampling PlanVI
RECTIFYING INSPECTION I
►
Rectifying inspection:
Average outgoing quality(AOQ)(): the evaluation of a
rectifying sampling plan
AOQ
RECTIFYING
INSPECTION(ƑFL™)II
AOQ =
Pap(N −n)
N
n/N →0
c
Pap
►
n: after inspection, contain no defectives
►
N − n: contain no defectives if the lot is rejected
►
N − n: contain p(N − n) defectives if the lot is
accepted
AOQ:
Rectifying Inspection(ƒfl™)III
DOUBLE-SAMPLING PLANSI
DSP
►
DSP: a procedure; a second sample is
required before the lot can be sentenced.
►
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 of the second
sample
►
di =the number of defectives in the ith sample
Double-Sampling PlansII
DOUBLE-SAMPLING PLANSIII
OC curve:
Pa = PI a+ PII
a
Ex: (n1, n2, c1, c2, p) = (50, 100, 1, 3, 0.05)
►
PaI=
Σ1
d1=0
.
Σ
n
(0.05)d1(0.95)
d1
PaII = P{d =1 2, d ≤ 21
50− d1
= 0.278
}+ P{d =13, d = 0}
2
= P{d1 = 2} · P{d2 ≤ 1} + P{d1 = 3} · P{d2 = 0} =0.0107
Pa = PI a+ PII =
a 0.2897
DOUBLE-SAMPLING PLANSIV
[PaI(N − n ) 1+ P (NaII− n − n )]p
1
2
N
II
ATI = n P1 +aI (n + n
1 )P 2 a + N(1 −P ) a
AOQ =