Cross Validated Asked on November 16, 2021
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)^2sigma^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.
Answered by Kieran on November 16, 2021
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