6: II 7: III
3.2 The M-Method
R>library(AcceptanceSampling)
R>find.plan(PRP=c(0.01, 0.95), CRP=c(0.06, 0.10), type="normal", s.type="unknown")
3.1.3 Upper and Lower Specification Limits 3.1.3.1 Standard Deviation Known
When there is an upper (USL) and a lower specification limit (LSL), Schilling[84] proposed a simple procedure that may be used to determine if two separate single specification limit plans may be used. The procedure is as follows:
1. CalculateZp∗= (LSL−U SL)/2σ
2. Calculatep∗=pnorm(Zp∗) the area under the standard normal density to the left ofZp∗.
3. If 2p∗ ≥ RQL reject the lot, because even if the distribution is centered between the specification limits, the a proportion outside the specification limits will be too high.
4. If 2p∗ ≤AQLuse two single specification sampling plans (i.e., one for LSL and one for USL).
5. IfAQL≤2p∗ ≤RQLthen use the M-Method for upper and lower speci- fication limits that is described inSection 3.2
3.1.3.2 Standard Deviation Unknown
When there is an upper (USL) and a lower specification limit (LSL), and the standard deviation is unknown, use the M-Method described inSection 3.2.
deviation known, the uniform minimum variance unbiased estimate of the proportion defective is:
PL=Z ∞ QL
√1
2πe−t2/2dt, (3.8)
or the area under the standard normal distribution to the right of QL = ZLq n
n−1
,
whereZL= (x−LSL)/σ. The maximum allowable proportion defective is:
M =Z ∞ k√ n
n−1
√1
2πe−t2/2dt, (3.9) or the area under the standard normal distribution to the right of kq n
n−1, wherekis the acceptance constant used in the k-method.
Example 3To illustrate, reconsider Example 1 from Mitra[72]. The sam- ple size wasn= 10, and the acceptance constant wask= 1.6094. The lower specification limit wasLSL= 100, and the known standard deviation wasσ= 8. In the sample of 10,x= 110. Therefore,QL= 110−8100q
109
= 1.3176.
The estimated proportion defective and the R command to evaluate it is:
PL=Z ∞ 1.3176
√1
2πe−t2/2dt=1-pnorm(1.3176)= 0.0938.
The maximum allowable proportion defective and the R command to evaluate it is:
M =Z ∞ 1.6094√
109
√1
2πe−t2/2dt=1-pnorm(1.6094*sqrt(10/9))= 0.0449. Since 0.0938 > 0.0449 (or PL > M), reject the lot. This is the same conclusion reached with the k-method shown in Example 1. When there is only one specification limit the results of the k-method and the M-method will always agree.
M and PL can be calculated with the R function pnorm or alternatively with theMPnandEPnfunctions in the R packageAQLSchemesas shown in the code below.
R>M<-pnorm(1.6094*sqrt(10/9),lower.tail=F)
R>PL<-pnorm(((110-100)/8)*sqrt(10/9),lower.tail=F) R>M
[1] 0.04489973 R>PL
[1] 0.09381616 library(AQLSchemes)
R>PL<-EPn(sided="one",stype="known",LSL=100,sigma=8,xbar=110, n=10)
R>PL
[1] 0.09381616
R>M<-MPn(k=1.6094,n=10,stype="known") R>M
[1] 0.04489973
3.2.1.2 Standard Deviation Unknown
When the standard deviation is unknown, the symmetric standardized Beta distribution is used instead of the standard normal distribution in calculating the uniform minimum variance unbiased estimate of the proportion defective.
The standardized Beta CDF is defined as Bx(a, b) = Γ(a+b)
Γ(a)Γ(b) Z x
0
νa−1(1−ν)b−1dν, (3.10) where 0≤x≤1,a >0, andb >0. The density function is symmetric when a=b. The estimate of the proportion defective is then
ˆ
pL=Bx(a, b), (3.11)
wherea=b= n2 −1, and x= max
0, .5−.5ZL
√ n n−1
,
and the sample standard deviationsis substituted forσin the formula for ZL =x−LSL
s .
When the standard deviation is unknown, the maximum allowable propor- tion defective is calculated as:
M =BBM
n−2 2 ,n−2
2
, (3.12)
where
BM =.5 1−k
√n n−1
,
and k is the acceptance constant. The Beta CDF function can be evaluated with R using theBx(a, b) =pbeta(x,a,b)function.
Example 4 To illustrate, reconsider Example 2 from Montgomery[74].
The sample size wasn= 42, and the acceptance constant wask= 1.905285.
The lower specification limit wasLSL = 225. In the sample of 42,x= 255, and the sample standard deviation wass= 15. Therefore,
ZL=
255−225 15
= 2.0.
The estimated proportion defective and the R command to evaluate it is:
ˆ
pL=Bx(a, b) =pbeta(.3419322,20,20)= 0.02069563, wherea=b= 422 −1 = 20, and
x= max
0, .5−.5(2.0) √
42 42−1
= 0.3419332
The maximum allowable proportion defective and the R command to evaluate it is:
M =BBM
42−2
2 ,42−2 2
=pbeta(.3494188,20,20)= 0.02630455, where
BM =.5
1−1.905285 √
42 42−1
= 0.3494188.
Since 0.02069563 < 0.02630455 (or ˆpL< M), accept the lot. This is the same conclusion reached with the k-method shown in Example 2.
The calculation of the estimated proportion defective, ˆpL, and the max- imum allowable proportion defective,M, can be simplified using theEPn() and MPn() functions in the R package AQLSchemes as shown in the R code below.
R>library(AQLSchemes)
R>PL<-EPn(sided="one",stype="unknown",LSL=225,xbar=255, s=15,n=42)
R>PL
[1] 0.02069563
R>M<-MPn(k=1.905285,stype="unknown",n=42) R>M
[1] 0.02630455
3.2.2 Upper Specification Limit 3.2.2.1 Standard Deviation Known
When there is an upper specification limit and the standard deviation is known, the acceptance criterion changes fromPL< M toPU < M, where
PU =Z ∞ QU
√1
2πe−t2/2dt, (3.13) or the area under the standard normal distribution to the right of QU = ZU
q n
n−1
andZU = (U SL−x)/σ.
3.2.2.2 Standard Deviation Unknown
When there is an upper specification limit and the standard deviation un- known, the acceptance criterion is ˆpU < M where
ˆ
pU =Bx(a, b), (3.14)
a=b= n 2 −1, x= max
0, .5−.5ZU √
n n−1
,
ZU = U SL−x s ,
andM is the same as that defined in Equation 3.12.
3.2.3 Upper and Lower Specification Limit 3.2.3.1 Standard Deviation Known
When there are both upper and lower specification limits and the standard deviation is known, the acceptance criterion becomes: accept if
P = (PL+PU)< M, (3.15) wherePL is defined in Equation 3.8, PU is defined in Equation 3.13, and M is defined in Equation 3.9.
3.2.3.2 Standard Deviation Unknown
When the standard deviation is unknown, the acceptance criterion becomes:
accept if
ˆ
p= (ˆpL+ ˆpU)< M, (3.16) where ˆpL is defined in Equation 3.11, ˆpU is defined in Equation 3.14 and M is defined in Equation 3.12.
Example 5 Reconsider the variables sampling plans whose OC curves were shown inFigure 3.3 where σ was known, and Figure 3.5 where σ was unknown. Suppose the upper specification limit was U SL = 100, and the lower specification limit wasLSL= 90.
When the standard deviation was known to be σ = 2.0, the sample size wasn= 21, and the acceptance constantk= 1.967411, as indicated inFigure 3.3. Ifxwas determined to be 96.68 after taking a sample of 21, then
QU =(100−96.68) 2.0
r21
20 = 1.701, and
PU =Z ∞ 1.701
√1
2πe−t2/2dt= 0.04447. QL=
(96.68−90) 2.0
r21
20 = 3.4225, and
PL=Z ∞ 3.4225
√1
2πe−t2/2dt= 0.00031. M =Z ∞
1.967411√
2120
√1
2πe−t2/2dt= 0.0219.
Therefore,P = (PL+PU) = 0.0448>0.0219 =M, and the decision would be to reject the lot.
P andM can again be calculated using theEPn()andMPn()functions as shown below.
R>library(AQLSchemes) R># sigma known
R>P<-EPn(sided="two",stype="known",sigma=2,LSL=90, USL=100,xbar=96.68,n=21)
R>P
[1] 0.04478233
R>M<-MPn(k=1.967411,stype="known",n=21) R>M
[1] 0.02190018
When the standard deviation was unknown (as inFigure 3.5), the sample size was n = 63, and the acceptance constant k = 1.97403, as indicated in Figure 3.5. If x was determined to be 97.006, and the sample standard deviation was 1.9783 after taking a sample of 63, then
ˆ
pU =Bx(a, b) = 0.06407, (3.17) where
a=b= 63
2 −1 = 30.5, x= max
0, .5−.5QU √
63 63−1
= 0.4031, and
QU = 100−97.006
1.9783 = 1.51342.
ˆ
pL=Bx(a, b) = 0.000095, (3.18) where
a=b= 63
2 −1 = 30.5, x= max
0, .5−.5QL √
63 63−1
= 0.2733, and
QL=97.006−90
1.9783 = 3.541. M =BBM
63−2 2 ,63−2
2
= 0.02284, (3.19) whereBM =.5
1−1.9740363√−631
= 0.37364.
Therefore, ˆp= (ˆpL+ˆpU) = 0.06416>0.02284 =M, and again the decision would be to reject the lot.
ˆ
pandM can again be calculated using theEPn()andMPn()functions as shown below.
R>library(AQLSchemes) R># sigma unknown
R>P<-EPn(sided="two",stype="unknown",LSL=90,USL=100, xbar=97.006,s=1.9783,n=63)
R>P
[1] 0.06416326
R>M<-MPn(k=1.97403,stype="unknown",n=63) R>M
[1] 0.02284391