# Acceptance sampling

I have this question.

"Units of different lengths are produced. If a unit has length greater than 10 metres it is defective. Lengths are assumed to be normally distributed, $$N(\mu,0.01)$$ - that is, the variance is 0.01.

The batch is accepted if the sample mean, of size $$n$$ units, is less than $$k$$ .

Find $$n$$ and $$k$$ such that a batch with 0.1% defectives is accepted with probability 90% and a batch with 1% defectives is accepted with probability 5%. Ans: k=9.73, n=15."

(1) I tried using this formula to handle the variation in $$\mu$$.

$$n$$ = $$(Z_\alpha + Z_\beta)^2{\sigma}^2$$/$$(\Delta_\mu)^2$$

You get this by getting two equations using the probabilities of a Type I or Type II error and eliminating the sample mean between these.

(2) Another way is to use the Poisson distribution, noting that $$\lambda = np$$, where $$p_1 = 0.01$$ and $$p_2 = 0.1$$. This leads to selecting a sample of about 50, on checking the ratio $$p_1/p_2$$ through appropriate tables. If there is more than one defective unit the batch is rejected. But, this does not take account of the prior knowledge about $$\mu$$.

Assume $$\mu_1 \gt \mu_0$$ and that the sample mean, $$\bar X$$, is at the critical value.Then,

$$\bar X = \mu_0+ \sigma Z_\alpha/\sqrt n$$

$$\bar X = \mu_1- \sigma Z_\beta/\sqrt n$$

$$\implies n = (Z_\alpha + Z_\beta)^2\sigma^2/(\mu_1-\mu_0)^2$$

Units over 10 metres are defective and lengths are $$N(\mu,\sigma^2)$$.

So,

$$10-\mu_0 = \sigma Z_{p_0}$$

$$10-\mu_1 = \sigma Z_{p_1}$$

$$\implies \mu_1 - \mu_0 = \sigma (Z_{p_0} - Z_{p_1})$$

$$\implies n = (Z_\alpha + Z_\beta)^2/(Z_{p_0}-Z_{p_1})^2$$

$$Z_a$$ is defined by $$P(Z \le Z_a) = 1- a$$

where $$0 \le a \le 1$$ and $$Z$$ is $$N(0,1)$$ random variable.

Note that $$Z_{1-\beta} = -Z_\beta$$

$$p_0 = 0.001, p_1 = 0.01, \alpha = 0.10, \beta = 0.05, Z_{p_0} = 3.09, Z_{p_1} = 2.32, Z_\alpha = 1.28,$$ $$Z_\beta = 1.64$$

So, $$n \ge (1.28+1.64)^2/(3.09-2.32)^2$$

$$\implies n = 15$$.

Either $$\mu_0$$ or $$\mu_1$$ can be calculated and then the first or second equation above gives the critical value of $$\bar X = 9.73$$

Note that the sample size, $$n$$, does not depend on $$\sigma$$, which need not be known, but the critical value of $$\bar X$$ depends on a value for $$\sigma$$ being known, or estimated.