20 points Making single sampling plan and calculating the Average Outgoing
Examination test 08. June 2007
by M. KLINCSIK & Cs. SÁRVÁRI
The producer and the consumer agreed in the following single sampling plan.
(i) When the fraction defective of the lot is smaller than p0=0,03 then the consumer can be rejected it maximum
(ii) When the fraction defective is greater than p1=0,07 the consumer would like to accept it maximum in β=4% .
(a) Find the sample size n and the allowable number c of defective items in it using the c2 distribution!
(b) Draw the graphs of the first and second type of errors varying the sample size between the smallest and gr
(c) Draw the OC-curve belongs to the smallest n1 sample size!
(d) Draw the AOQ(p) Average Outgoing Quality function for the smallest n1 sample size for the rectifying sche
(e) On the the basis of exercise (d) give the AOQL Average Outgoing Quality Limit!
nd calculating the Average Outgoing Quality
hen the consumer can be rejected it maximum with α=3.5% ;
mer would like to accept it maximum in β=4% .
e items in it using the c2 distribution! 6 points
g the sample size between the smallest and greatest values! 4 points
4 points
smallest n1 sample size for the rectifying scheme! 4 points
oing Quality Limit! 2 points
Totally 20 points
20 points Making single sampling plan and calculating the Average Outg
Examination test 08. June 2007
The producer and the consumer agreed in the following single sampling plan.
(i) When the fraction defective of the lot is smaller than p0=0,03 then the consumer can be rejected
(ii) When the fraction defective is greater than p1=0,07 the consumer would like to accept it maximu
2
(a) Find the sample size n and the allowable number c of defective items in it using the c distributi
(b) Draw the graphs of the first and second type of errors varying the sample size between the smallest and gr
(c) Draw the OC-curve belongs to the smallest n1 sample size!
(d) Draw the AOQ(p) Average Outgoing Quality function for the smallest n1 sample size for the rectifying sche
(e) On the the basis of exercise (d) give the AOQL Average Outgoing Quality Limit!
SOLUTION
(a) Agreement between the producer and the consumer
Producer side Consumer side
Acceptable Quality Level p0 =AQL= 0.03 Lot Tolerance Percent Defective
Producer's Risk α = 0.035 Consumer's Risk (Type II error)
(Type I error) 1-α= 0.965
Calculate the p1/p0 quotient
p1/p0= 2.333333
Make the following table
Number of Degree of
Ratios
defective freedom χ2(1-a;2(c+1)) χ2(b;2(c+1))
r(c)
c 2*(c+1)
12 26 14.55117795 39.88912356 2.741299
13 28 16.05485584 42.36993129 2.639073
14 30 17.57614409 44.83355892 2.550819
15 32 19.11329468 47.28172663 2.473761
16 34 20.66483003 49.71588534 2.405821
17 36 22.22948894 52.13727227 2.34541
18 38 23.80618385 54.54695278 2.291293
19 40 25.39396913 56.94585138 2.242495
20 42 26.99201769 59.33477731 2.198234
p1
Find the value c such that the following inequality holds r (c 1) r (c )
p0
c= 18
Find the smallest value of the sample size n using the following formulas b ;2 c 1
2
12
n1 n n2
n1= 389.6211 2 p1
n2= 396.7697
calculating the Average Outgoing Quality
03 then the consumer can be rejected it maximum with α=3.5% ;
nsumer would like to accept it maximum in β=4% .
2
ective items in it using the c distribution! 6 points
e sample size between the smallest and greatest values! 4 points
4 points
lest n1 sample size for the rectifying scheme! 4 points
g Quality Limit! 2 points
Totally 20 points
Consumer side
erance Percent Defective p1= LTPD= 0.07
mer's Risk (Type II error) β = 0.04
Formulas
χ2 (α, sz) = INVERZ.KHI(alfa; degree of freedom)
b ;2 c 1
2
r (c )
12a ;2 c 1
b ;2 c 1
2
12a ;2 c 1
n1 n n2
2 p1 2 p0
20 points Making single sampling plan and calculating the Average Outgoing Q
Examination test 08. June 2007
The producer and the consumer agreed in the following single sampling plan.
(i) When the fraction defective of the lot is smaller than p0=0,03 then the consumer can be rejected it maximum
(ii) When the fraction defective is greater than p1=0,07 the consumer would like to accept it maximum in β=4% .
2
(a) Find the sample size n and the allowable number c of defective items in it using the c distribution!
(b) Draw the graphs of the first and second type of errors varying the sample size between the smallest and g
(c) Draw the OC-curve belongs to the smallest n1 sample size!
(d) Draw the AOQ(p) Average Outgoing Quality function for the smallest n1 sample size for the rectifying scheme!
(e) On the the basis of exercise (d) give the AOQL Average Outgoing Quality Limit!
SOLUTION (b)
Producer side Consumer side
Acceptable Quality Level p0 =AQL= 0.03 Lot Tolerance Percent Defective
Producer's Risk α = 0.035 Consumer's Risk (Type II error)
(Type I error) 1–α= 0.965
Enabled maximum number of defects in the sample
c= 18
Trends of the errors when the sample size increase Formulas for e
sample Type I Max of type Type II Max of type
size n error I error error II error Type I. error =P( Reject good lot) ≤
388 0.02700 0.035 0.03664 0.04
389 0.02762 0.035 0.03561 0.04 c
n
390 0.02825 0.035 0.03460 0.04 1 P d c | p0 1
391 0.02888 0.035 0.03362 0.04 k 0 k
392 0.02953 0.035 0.03266 0.04
393 0.03019 0.035 0.03173 0.04 Type II. error =P( Accept bad lot) ≤
394 0.03086 0.035 0.03082 0.04
395 0.03154 0.035 0.02993 0.04 c
n
P d c | p1 1 p1
n
396 0.03223 0.035 0.02907 0.04
k 0 k
397 0.03294 0.035 0.02822 0.04
First and second type of errors
0.045
Probability (errors)
0.040
0.035
0.030
0.025
Probability (errors
0.025
0.020
388 389 390 391 392 393 394 395 396 397
Type I error
Type II error
Sample size max of type I error
max of type II error
nd calculating the Average Outgoing Quality
hen the consumer can be rejected it maximum with α=3.5% ;
mer would like to accept it maximum in β=4% .
2
ms in it using the c distribution! 6 points
ying the sample size between the smallest and greatest values! 4 points
4 points
t n1 sample size for the rectifying scheme! 4 points
2 points
Totally 20 points
Consumer side
t Tolerance Percent Defective p1= LTPD= 0.07
onsumer's Risk (Type II error) β = 0.04
Formulas for errors
c , p0 and p1 are constants
Type I. error =P( Reject good lot) ≤ a
1– BINOM.ELOSZLÁS(c,n, p0,IGAZ)
c
n
1 P d c | p0 1 1 p0 p0 a
nk k
k 0 k
increase as n increase
Type II. error =P( Accept bad lot) ≤ b
BINOM.ELOSZLÁS(c,n, p1,IGAZ)
c
n
P d c | p1 1 p1 p1 b
nk k
k 0 k
decrease as n increase
20 points Making single sampling plan and calculating the Average Outg
Examination test 08. June 2007
The producer and the consumer agreed in the following single sampling plan.
(i) When the fraction defective of the lot is smaller than p0=0,03 then the consumer can be rejected it maxim
(ii) When the fraction defective is greater than p1=0,07 the consumer would like to accept it maximum in β=4
(a) Find the sample size n and the allowable number c of defective items in it using the c2 distribution!
(b) Draw the graphs of the first and second type of errors varying the sample size between the smallest and greatest va
(c) Draw the OC-curve belongs to the smallest n1 sample size!
(d) Draw the AOQ(p) Average Outgoing Quality function for the smallest n1 sample size for the rectifying
(e) On the the basis of exercise (d) give the AOQL Average Outgoing Quality Limit!
SOLUTION
(c) n= 390
c= 18
Calculation of the Opera
Pa OC ( p) binom.eloszlás(c, n, p,
Lot Probabilty Average
fraction of Outgoing
Differences
defective acceptance Quality OC-
(p) (Pa=OC(p)) (AOQ(p))
0.000 1.00000000 0.0000000
1.00
Probability of acceptance
0.005 1.00000000 0.0050000 -0.00500
0.010 0.99999997 0.0100000 -0.00500 0.90
0.015 0.99998992 0.0149998 -0.00500 0.80
0.020 0.99958732 0.0199917 -0.00499 0.70
0.025 0.99499920 0.0248750 -0.00488 0.60
0.030 0.97175308 0.0291526 -0.00428 0.50
0.035 0.90493662 0.0316728 -0.00252 max value 0.40
0.040 0.77855937 0.0311424 0.00053 0.30
0.045 0.60527609 0.0272374 0.00390 0.20
0.050 0.42119502 0.0210598 0.00618 0.10
0.055 0.26232811 0.0144280 0.00663 0.00
0.060 0.14697729 0.0088186 0.00561 0.00
0.065 0.07462204 0.0048504 0.00397
0.070 0.03460153 0.0024221 0.00243 Lot fraction defe
0.075 0.01476434 0.0011073 0.00131
0.080 0.00583758 0.0004670 0.00064
(d ) and (e )
Calculation of Average Outgoing Quality as the fuction of p
Average Outgoing Quality for sampling plan n=390 and c=18
0.035
Outgoing Quality
0.030
0.025
Average Outgoing Quality
0.025
0.020
0.015
0.010
0.005
0.000
0.00 0.01 0.02 0.03 0.04 0.05 0.06
Lot percent defective = Incoming Quality Average Outgoing Quality
From the table we see that the AOQL = 0,0316727815 at p = .035 for the sampling plan n=3
ulating the Average Outgoing Quality
sumer can be rejected it maximum with α=3.5% ;
ke to accept it maximum in β=4% .
e c2 distribution! 6 points
ween the smallest and greatest values! 4 points
4 points
1 sample size for the rectifying scheme! 4 points
2 points
Totally 20 points
Calculation of the Operating Characteristic Curve
c
n
OC ( p) binom.eloszlás(c, n, p, IGAZ ) p d 1 p
nd
d 0 d
OC-curve for values n=390, c=18
0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08
Lot fraction defective
Probability of
acceptance
AOQ( p) Pa p
390 and c=18
The maximal value of
the AOQ curve
represents the worst
possible quality that
results from the
rectifying inspection
program.
The maximal value of
the AOQ curve
represents the worst
possible quality that
results from the
rectifying inspection
program.
0.07 0.08
Average Outgoing Quality
5 for the sampling plan n=390 and c=18.