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I have 974 samples in front of me. I want to test these samples and ideally I should test all 974 samples. However, testing the sample will break the sample and I cannot afford to break all of them. The manual states that if I test (by pulling to see if the sample is strong against the force that I am applying) 14 samples and all 14 samples pass (a certain threshold and don't fail completely) then this results in a 99% statistical confidence level in the results of the tests. What statistical confidence level do I have if I only test 9 samples.

Edit: the range is that the sample has to pass a certain strength-level. This means that there is a lower bound that determines if the sample is good or it fails. There is no upper bound as a sample can be as strong as it wants and it will still pass.

Mr.Happy
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    You talk about a "confidence level" so it *sounds like* you intend that there's a *confidence interval* for some population quantity. What you haven't stated here is what this is intended to be a confidence interval *for* -- and that matters. You don't just have some vague generic "confidence" associated with some sample size, it depends on what you're trying to say about the whole set (presuming the 974 actually represent the population of interest). Are you trying to produce some interval for the proportion of the 974 that would pass that threshold? It will have an upper and a lower limit. – Glen_b Feb 05 '22 at 04:37
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    With 14 in the sample and no failures the rule of three would give an approximate $95\%$ interval on the proportion of passes in the population -- $(1-3/14, 1)$, i.e. the lower limit of the interval is that at least $78.6\%$ of the population pass (ignoring finite population correction, which will be quite small). A $99\%$ interval would be wider than that, down to roughly about $67\%$. see https://en.wikipedia.org/wiki/Rule_of_three_(statistics) and https://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval ... As you see $n=14$ doesn't bound the population proportion all that well – Glen_b Feb 05 '22 at 04:45
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    That $67\%$ is probably not terribly accurate (I haven't checked, it will probably be out by several percentage points) but a more accurate value will still be a lot lower than I expect you realized. At $n=9$, and a $99\%$ interval, it will go considerably lower. Edit: having checked, yes, it was out by a few percentage points but the issue remains. – Glen_b Feb 05 '22 at 04:53

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