Acceptance sampling

Question

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 $\mu$.

Answer 1

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.