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I am trying to find a way to affirm with certain confidence that a batch of products have no defects at all.

This question is very similar to this one but I do not know, nor can assume, any percentage of defects, because there can not be any.

How can I find a sample size to be, let's say, 90% confident, that my 1000 products batch has no deffects at all?

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  • $\begingroup$ Do you have some previous historical data which you can use to estimate how many defects you had in the past? The answer will depend a lot on the expected number of defects. If you have a lot of defects then a smaller sample will be enough since you will observe defects faster. If you have very few defects then you will need a larger sample to have the same confidence. $\endgroup$ Jan 11, 2023 at 11:48
  • $\begingroup$ No, I do not have historical data. Our intent is to check for the first time whether this initial batch has defects without a previous history. An approach we tried is the six sigma methodology but there is no apparent way given that we do not have an expected amount of defects. $\endgroup$ Jan 11, 2023 at 12:31
  • $\begingroup$ Then you are proverbially screwed. How can you tell with any confidence if there defects or not if you do not know how frequent they are? Suppose I told you there are 990 defects out of 1000, how big do you think your sample needs to be? What if there is 1 defect out of 1000? This frequency entirely determines the solution to your problem. $\endgroup$ Jan 11, 2023 at 12:37

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Let K be the number of defective products in the 1000 products batch and X the number of defective products in a sample of size n. Assume the observed value of X is 0. X follows a hypergeometric distribution, therefore:

$P(X=0)=\cfrac{{K\choose0}{1000-K\choose n}}{1000\choose n}=\cfrac{\cfrac{(1000-K)!}{(1000-K-n)!n!}}{\cfrac{1000!}{(1000-n)!n!}}$

You have to test the null hypothesis $H_{0}: K>0$ against the alternative $H_{1}: K=0$.

As the null hypothesis is composite, the p-value is the maximum probability under the null hypothesis of X taking a value at least as extreme as the observed value, that is, the maximum probability under the null hypotheiss of X=0. The maximum probability is reached for K=1 (because the probability of getting no defective products in your sample is greater when there are fewer defective products in the batch). Therefore, when the observed value of X is 0, the p-value is:

$pvalue=max_{H_{0}}\{P(X=0)\}=\cfrac{\cfrac{(999)!}{(999-n)!}}{\cfrac{1000!}{(1000-n)!}}=\cfrac{1000-n}{1000}$

You need a sample size n such that, when the observed value of X is 0, the p-value is less than or equal to the significance level $\alpha$. Taking $\alpha$=0.1:

$\cfrac{1000-n}{1000}<=0.1$

$n>=900$

If you find no defective products in a sample of 900 products, you can reject the null hypothesis $H_{0}: K>0$ with significance level 0.1.

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  • $\begingroup$ Thanks! It looks like it is going to be a bit hard to achieve, tho haha. $\endgroup$ Jan 16, 2023 at 16:30

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