Supplement
G Acceptance Sampling Plans
PROBLEMS
Operating Characteristic Curves
1. Alpha is the producer’s risk. To find Alpha, p is set equal to the acceptable quality
level (AQL = 0.5%). Then np = (200 × 0.005) = 1.00. From Table G.1 where c = 4,
P ( x ≤ c ) = 0.996 . Alpha = (1 − 0.996 ) = 0.004 , or 0.4 percent.
Beta is the consumer’s risk. To find Beta, p is set equal to the lot tolerance proportion
defective. (LTPD = 4%). Then np = (200 × 0.04) = 8.0. From Table G.1 where c = 4,
P ( x ≤ c ) = 0.100 . Beta = 0.100, or 10 percent.
2. Hospital Supply.
a. p = 0.0010, n = 350.
np = (350)(0.0010) = 0.35.
From Table G.1: when c = 1, Pa = 0.951,
Therefore, the producer’s risk is (1 – 0.951) = 0.049, or 4.9 percent.
b. np = (350)(0.0017) = 0.60.
From Table G.1: Pa = 0.878 .
Therefore, there is a 12.2 percent (100% – 87.8%) probability that a shipment
with a proportion defective equal to 0.17 percent will be returned.
c. If Alpha is required to be less than 5 percent, Pa ≥ 0.95 (100% − 5%) . Given
np = (350 × 0.0017) = 0.6, then, from Table G.1, when c = 1, Pa = 0.878 , and
when c = 2, Pa = 0.977 . c should have been set equal to 2.
At np = 0.35 and c = 2, Pa = 0.994 and the producer’s risk is 1 − 0.994 = 0.006 .
3. Electronic components.
np = 1500(20/5000) = 6.0.
Use Table G.1. At np = 6.0 and c = 3, Pa = 0.151 and therefore Beta, the consumer’s
risk, is = 0.151. Therefore, the consumer would want c = 3.
If the AQL = 10/5000 = 0.002
np = 1500(0.002) = 3.0.
G-1
Copyright © 2016 Pearson Education, Inc.
, G-2 l SUPPLEMENT G Acceptance Sampling Plans
At np = 3.0 and c = 3, Pa = 0.647 and Alpha, the producer’s risk, is = 0.353.
Increasing the sample would reduce the risk to both parties.
Selecting a Single-Sampling Plan
4. a. A variety of sampling plans will satisfy the requirements.
One such plan is n = 280, c = 6.
To find Alpha:
np = ( 280 × 0.01) = 2.8 . Then P ( x ≤ c ) = 0.976 and Alpha
= (1 − 0.976 ) = 0.024 , or 2.4 percent.
To find Beta, np = (280 × 0.04) = 11.2. Then P ( x ≤ c ) = 0.071 and Beta = 0.071,
or 7.1 percent.
Another plan is n = 232, c = 5. In this case, Alpha = 3.1 percent and Beta = 10
percent. The tradeoff between this plan and the other one is that you have a
smaller sample size but the producer’s and consumer risks increase.
Copyright © 2016 Pearson Education, Inc.
G Acceptance Sampling Plans
PROBLEMS
Operating Characteristic Curves
1. Alpha is the producer’s risk. To find Alpha, p is set equal to the acceptable quality
level (AQL = 0.5%). Then np = (200 × 0.005) = 1.00. From Table G.1 where c = 4,
P ( x ≤ c ) = 0.996 . Alpha = (1 − 0.996 ) = 0.004 , or 0.4 percent.
Beta is the consumer’s risk. To find Beta, p is set equal to the lot tolerance proportion
defective. (LTPD = 4%). Then np = (200 × 0.04) = 8.0. From Table G.1 where c = 4,
P ( x ≤ c ) = 0.100 . Beta = 0.100, or 10 percent.
2. Hospital Supply.
a. p = 0.0010, n = 350.
np = (350)(0.0010) = 0.35.
From Table G.1: when c = 1, Pa = 0.951,
Therefore, the producer’s risk is (1 – 0.951) = 0.049, or 4.9 percent.
b. np = (350)(0.0017) = 0.60.
From Table G.1: Pa = 0.878 .
Therefore, there is a 12.2 percent (100% – 87.8%) probability that a shipment
with a proportion defective equal to 0.17 percent will be returned.
c. If Alpha is required to be less than 5 percent, Pa ≥ 0.95 (100% − 5%) . Given
np = (350 × 0.0017) = 0.6, then, from Table G.1, when c = 1, Pa = 0.878 , and
when c = 2, Pa = 0.977 . c should have been set equal to 2.
At np = 0.35 and c = 2, Pa = 0.994 and the producer’s risk is 1 − 0.994 = 0.006 .
3. Electronic components.
np = 1500(20/5000) = 6.0.
Use Table G.1. At np = 6.0 and c = 3, Pa = 0.151 and therefore Beta, the consumer’s
risk, is = 0.151. Therefore, the consumer would want c = 3.
If the AQL = 10/5000 = 0.002
np = 1500(0.002) = 3.0.
G-1
Copyright © 2016 Pearson Education, Inc.
, G-2 l SUPPLEMENT G Acceptance Sampling Plans
At np = 3.0 and c = 3, Pa = 0.647 and Alpha, the producer’s risk, is = 0.353.
Increasing the sample would reduce the risk to both parties.
Selecting a Single-Sampling Plan
4. a. A variety of sampling plans will satisfy the requirements.
One such plan is n = 280, c = 6.
To find Alpha:
np = ( 280 × 0.01) = 2.8 . Then P ( x ≤ c ) = 0.976 and Alpha
= (1 − 0.976 ) = 0.024 , or 2.4 percent.
To find Beta, np = (280 × 0.04) = 11.2. Then P ( x ≤ c ) = 0.071 and Beta = 0.071,
or 7.1 percent.
Another plan is n = 232, c = 5. In this case, Alpha = 3.1 percent and Beta = 10
percent. The tradeoff between this plan and the other one is that you have a
smaller sample size but the producer’s and consumer risks increase.
Copyright © 2016 Pearson Education, Inc.
