0
$\begingroup$

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 $\mu$.

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.