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krm8_ism_i - Supplement I Acceptance Sampling Plans...

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Supplement I Acceptance Sampling Plans PROBLEMS 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 I.1 where c = 4, ( 29 0.996 P x c = . Alpha ( 29 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 I.1 where c = 4, ( 29 0.100 P x c = . Beta = 0.100, or 10 percent. 2. a. A variety of sampling plans will satisfy the requirements. One such plan is n = 280, c = 6. To find Alpha: ( 29 280 0.01 2.8 np = × = . Then ( 29 0.976 P x c = and Alpha ( 29 1 0.976 0.024 = - = , or 2.4 percent. To find Beta, np = (280 × 0.04) = 11.2. Then ( 29 0.071 P x c = and Beta = 0.071, or 7.1 percent. Another plan is n = 232, c = 5. In this case, Alpha = 3.1% 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.
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2 l SUPPLEMENT I b. The OC curve is: c. The maximum value of the AOQ over various possible values of the proportion defective is 0.0122. The AOQL is therefore 1.22 percent defective when the lot proportion defective is 0.02. Here we have used n = 280 and c = 6. ( 29 ( 29 a p P N n AOQ N - = Proportion Defective ( p ) np a P ( N n / N ) AOQ 0.000 0.0 1.000 0.907 0.0000 0.005 1.4 0.999 0.907 0.0045 0.010 2.8 0.976 0.907 0.0089 0.015 4.2 0.867 0.907 0.0118 0.020 5.6 0.670 0.907 0.0122 0.025 7.0 0.450 0.907 0.0102 0.030 8.4 0.267 0.907 0.0073 0.035 9.8 0.143 0.907 0.0045 0.040 11.2 0.071 0.907 0.0026 0.071 0.976 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 Pr ob ab ilit y of Ac ce pt an ce , (P a)
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Acceptance Sampling Plans l SUPPLEMENT I l 3 3. SunSilk Shampoo Company. AQL = 5/500, or 1%; LTPD = 5%. a. Again, a variety of sampling plans will satisfy the requirements. One plan is n = 160, c = 4. To find Alpha: np = (160 × 0.01) = 1.6. Then ( 29 0.976 P x c = and Alpha = (1 – 0.976) = 0.024, or 2.4 percent. To find Beta, np = (160 × 0.05) = 8.0. Then ( 29 0.100 P x c = and Beta = 0.100, or 10.0 percent. Another plan is n = 134, c = 3. Alpha = 4.7 percent and Beta = 9.9 percent. b. np = 160(0.03) = 4.80. From Table I.1, a P = 0.476. There is a 52.4 percent probability that a shipment with 3 percent defectives will be rejected.
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