Acceptance-sampling scheme

Consider an acceptance-sampling scheme in which a factory takes delivery of a batch of components if a random sample of 400 components contains fewer than M defectives. Otherwise the batch is returned to the supplier. Suppose the factory manager wants to set M so that there is no more than a 10% risk of accepting a batch that has 5% or more defectives. Let P be the sample proportion of defectives. (In the following, use the approximate Normal distribution for P).

a) Suppose that the batch has 5% defectives. What is the value of c for which pr(p> c)=0.10? What value should M have?

p= 5%
q=1-p= 95%

n= 400 (sample size)
s=square root of pq/n= 0.0109 (standard error of proportion)

alpha (a) = 10%
Assuming normal distribution,z value for alpha (a) = 10% is = 1.28155

Acceptance limit (c<p+z*s)

c= 6.40% =0.05+1.28155*0.0109

M=c*n= 25.6 =0r rounded down to 25

b) Suppose the supplier makes a batch in which the true proportion of defectives is only 2%. If the cutoff in (a) is adopted, what is the probability that the batch will be sent back?

c= 6.40%
p= 2%
q=1-p= 98%

n= 400 (sample size)
s=square root of pq/n= 0.007

z=(c-p)/s= 6.285714

Probability value corresponding to z = 6.285714 is = 0.00000002%

Almost zero probability that the batch will be sent back

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