ACCEPTANCE SAMPLING

Some approximations:

Hyp(N, n, p)

Bin(n, p) Po( )

n>10 p10

p+Nn

Ex. Suppose we have a large lot. To control the quality we pick 10 units randomly. If at most one of them is defect then the lot is accepted otherwise it is rejected. The fraction defective is p. Calculate the acceptance probability for 6 different values of p and use these to draw the corresponding OC Curve.

Solution: Since the lot is big we approximate the number of defective units in the sample with a binomial distribution.

Pa = 91100 )p1(p110)p1(p

010

+

We calculate the acceptance probability for p = 1%, 5%, 10%, 15%, 20% and 30%.

p = 0.01 Pa(0.01) = 91100 99.001.011099.001.0

010

+

0.996

p = 0.05 Pa(0.05) = 91100 95.005.011095.005.0

010

+

0.914

p = 0.10 Pa(0.10) = 91100 90.010.011090.010.0

010

+

0.736

p = 0.15 Pa(0.15) = 91100 85.015.011085.015.0

010

+

0.544

p = 0.20 Pa(0.20) = 91100 80.020.011080.020.0

010

+

0.376

p = 0.30 Pa(0.30) = 91100 70.030.0110

70.030.00

10

+

0.149

0.2 0.4 0.6 0.8 1

0.2

0.4

0.6

0.8

1

The slope will be different for different situations.

Pa

p

Double sampling plans

Ex. Suppose that a lot contains 1000 units. We have decided to use the following double sampling plan:

1) Pick 30 units randomly.

- if all are correct then accept the lot, - if three or more are defective then reject the lot, - if one or two units are defective go to point 2.

2) 60 new units are selected at random.

- if the number of defectives totally in both samples add up to at most two then accept the lot,

- if the number of defectives totally in both samples add up to three or more then reject the lot.

Suppose that the lot contains 2% defective units. How big is the acceptance probability?

Solution: In this example the acceptance probability is quite as easy to calculate exact. Since the conditions for both a binomial approximation (n/N = 30/1000 0.03 < 0.1) and a Poisson approximation are fulfilled we can use the approximation that feels easiest.

In the solution that follows the binomial approximation is used:

n1 = 30, n2 = 60, c1 = 0, c2 = 2, r1 = r2 = 3

Let di denote the number of defectives in sample i.

This means that Sample 1 if d1 = 0 accept the lot, if d1 3 reject the lot, if d1 = 1 or 2 take a new sample.

Sample 2 if d1 + d2 2 accept the lot if d1 + d2 3 reject the lot

The we will accept the lot in the following situations:

Sample 1 Sample 2 d1=0 d1=1 d2=0 d1=1 d2=1 d1=2 d2=0

This gives the acceptance probability:

Pa = P(d1=0) + P(d1=1 d2=0) + P(d1=1 d2=1) + P(d1=2 d2=0)

The probabilities that contain the intersection between the number of defectives in sample 1 and sample 2 are solved by using conditional probabilities.

Accept the lot in sample 1: P(d1 = 0) = 300 98.002.0030

0.5455

Accept the lot in sample 2 when d1 = 1:

In sample 1: Take 30 observations

P(d1 = 1) = 291 98.002.0130

0.28736

In sample 2: Take 60 observations

P(d2=0 | d1=1) = 600 98.002.0060

0.2976

P(d2=1 | d1=1) = 591 98.002.0160

0.3644

Accept the lot is sample 2 when d1 = 2:

In sample 1: Take 30 observations

P(d1 = 2) = 282 98.002.0230

0.0988

In sample 2: Take 60 observations

P(d2=0 | d1=2) = 600 98.002.0060

0.2976

These calculations are put together:

Pa = P(d1=0) + P(d2=0 | d1=1) P(d1=1) + P(d2=1 | d1=1) P(d1=1) + + P(d2=0 | d1=2) P(d1=2) =

= 0.5455 + 0.2976 0.3340 + 0.3644 0.3340 + 0.2976 0.0988 0.7960

Double sampling plans with the OC curve going through the points (p1, 1 ) and (p2, ) where = 5% and = 10%

n1 = n2

Sampling- plan no

1

2pp

Acceptance number

Approximate value of pn1 when Pa =

Approx. value of ASN(p)/n1 for p95

1c

2c

0.95

0.50

0.10

1 11.90 0 1 0.21 1.00 2.50 1.170 2 7.54 1 2 0.52 1.82 3.92 1.081 3 6.79 0 2 0.43 1.42 2.96 1.340 4 5.39 1 3 0.76 2.11 4.11 1.169 5 4.65 2 4 1.16 2.90 5.39 1.105 6 4.25 1 4 1.04 2.50 4.42 1.274 7 3.88 2 5 1.43 3.20 5.55 1.170

2n1 = n2

Sampling- plan no

1

2pp

Acceptance number

Approximate value of pn1 when Pa =

Approx. value of ASN(p)/n1 for p95

1c

2c

0.95

0.50

0.10

1 14.50 0 1 0.16 0.84 2.32 1.273 2 8.07 0 2 0.30 1.07 2.42 1.511 3 6.48 1 3 0.60 1.80 3.89 1.238 4 5.39 0 3 0.49 1.35 2.64 1.771 5 5.09 1 4 0.77 1.97 3.92 1.359 6 4.31 0 4 0.68 1.64 2.93 1.985 7 4.19 1 5 0.96 2.18 4.02 1.498

Ex. In a factory you buy large lots of bolts. When the lots arrive to the factory the quality is controlled using the following double sampling plan.

Pick 30 bolts at random. If all are correct then accept the lot. If 3 or more are defective then reject the lot. If the sample consists of one or two defective bolts then you pick another 50 units. If both samples sum up to two or less defectives then the lot is accepted. Otherwise it is rejected.

Draw an OC curve for this sampling plan.

Solution: The sampling plan can be summarized as

n1 = 30, n2 = 50, c1 = 0, c2 = 2 and r1 = r2 = 3.

Acceptance probabilities for the fraction defectives 0.01, 0.02, 0.05, 0.10 and 0.20 are calculated:

p

Acceptance probabilities in sample 1 and sample 2

Pa

0.01 Sample 1:

P(d1 = 0) = 300 99.001.0030

0.7397

Sample 2:

P(d1=1 d2=0) + P(d1=1 d2=1) + + P(d1=2 d2=0) =

= 291500 99.001.0

13099.001.0

050

+

+ 291491 99.001.01

3099.001.01

50

+

+ 282500 99.001.02

3099.001.00

50

0.2240

0.7397 + 0.2240 =

= 0.9637

p = 0.02 Pa 0.827 p = 0.05 Pa 0.329

p = 0.10 Pa 0.048 p = 0.20 Pa 0.001

The OC curve will obtain the following appearance:

0.025 0.05 0.075 0.1 0.125 0.15

0.2

0.4

0.6

0.8

1

Suppose you want to find a simple sampling plan with almost the same OC curve as the double sampling plan. Use the fraction defectives 0.05 and 0.20 to determine such a plan.

Pa

p

Ex. Determine the sequential sampling plan with an OC curve that goes through the points (p1, ) = (0.02, 0.05) och (p2, ) = (0.10, 0.10).

Solution: The constants h1, h2 and s in the rejection line and the acceptance line are calculated. (The value of the constant K will be found in the denominator of all three constants.

( )( )

=

21

12

p1pp1plnK = ))10.01(02.0

)02.01(10.0(ln

= )018.0098.0(ln

We use this expression in the equations for h1, h2 and s:

h1 = K

)1(ln

=

)018.0098.0(ln

)10.0

05.01(ln 1.329

h2 = K

)1(ln

=

)018.0098.0(ln

)05.0

10.01(ln 1.7056

s = K

)p1p1(ln

2

1

=

)018.0098.0(ln

)10.0102.01(ln

0.0503

The acceptance line becomes d1 = -h1 + sn = -1.329 + 0.0503n

The rejection line becomes d2 = h2 + sn = 1.7056 + 0.0503n

If we calculate ASN(p) for different values of the fraction defective , p, then we obtain the following ASN-curve.

p 1 B ASN(p) = 30 + 50 (1 B)

0.01 291 99.001.01

30

+ 282 99.001.0

230

0.2570

30 + 50 0.2570 42.85

0.02 291 98.002.01

30

+ 282 98.002.0

230

0.4328

30 + 50 0.4328 51.64

0.03 291 97.003.01

30

+ 282 97.003.0

230

0.5390

30 + 50 0.5390 56.95

0.05 291 95.005.01

30

+ 282 95.005.0

230

0.5975

30 + 50 0.5957 59.88

0.07 291 93.007.01

30

+ 282 93.007.0

230

0.5354

30 + 50 0.5354 56.77

0.09 291 91.009.01

30

+ 282 91.009.0

230

0.4265

30 + 50 0.4265 51.32

0.12 291 88.012.01

30

+ 282 88.012.0

230

0.3494

30 + 50 0.3494 47.47

0.15 291 85.015.01

30

+ 282 85.015.0

230

0.1438

30 + 50 0.1438 37.19

0.1 0.2 0.3 0.4 0.5

25

30

35

40

45

50

55

60

In an earlier example we saw that the OC-curve for this double sampling plan almost

ASN for the simple sampling plan

ASN for the double sampling plan

p = 0.013 p = 0.114

The values of mx and my are found in the following table.

A Dodge & Roming table

AOQL = )N1

n

1(ym nxp mm =

where mp is the fraction value that gives the maximum value of AOQ(p)

c mx my c mx my

0

1.00

0.3679

11

8.82

7.233 1 1.62 0.8400 12 9.59 7.948 2 2.27 1.371 13 10.37 8.670 3 2.95 1.942 14 11.15 9.398 4 3.64 2.544 15 11.93 10.13 5 4.35 3.168 16 12.72 10.88 6 5.07 3.812 17 13.52 11.62 7 5.80 4.472 18 14.31 12.37 8 6.55 5.146 19 15.12 13.13 9 7.30 5.831 20 15.92 13.89 10 8.05 6.528