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A proportion or percentage must be between 0 and 1, therefore, a confidence interval for a proportion should also be between 0 and 1.

But is it possible that if the "10-successes-and-10-failures" rule of thumb isn't respected, the confidence interval is so large that it goes outside of the normal boundaries?

Is that part of why this rule exists?

Or should the confidence interval always be within 0-1 and if it isn't, it means that you have made a calculation error?

Maxime Dupré
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    It depends on how you calculate your CI. There are various approximations in use which may give nonsensical interval endpoints outside [0,1], especially if you have few trials and/or a large preponderance of successes or failures. We have a couple of threads on this here. – Stephan Kolassa Jan 17 '23 at 17:54
  • I see. I'm new to this, so I wasn't even aware that there are multiple ways to calculate a CI. I calculate my CI using what I suppose is the standard error of the proportion: SEP = sqrt{p * (1 - p) / n} and then Sample proportion +/- 1.96 * SEP (assumes a 95% confidence). – Maxime Dupré Jan 17 '23 at 18:07
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    There's no point to extending a confidence interval beyond the range $[0,1]$ of the parameter it is intended to cover. If you did have a confidence interval procedure that yielded values beyond this range, you could replace it by clamping its result to that range (convert negative values to $0,$ values exceeding $1$ to $1$) and (a) its coverage would be the same and (b) its average length could not increase. Therefore the modified procedure would dominate the original one, making the original inadmissible and not worth considering. – whuber Jan 17 '23 at 18:07
  • @whuber Got it. This is beyond the scope of my current knowledge, but I understand that you can simply clamp the result to a normal proportion range. I was simply wondering if this kind of confidence interval was possible and if it is, if it is the result of a "low sample size" (or more accurately a sample size that doesn't have enough failures/successes). Assuming that I'm only using basic techniques (the formula I used for standard error of the proportion, normal distribution, etc). – Maxime Dupré Jan 17 '23 at 18:12
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    It's definitely possible. For instance, your formula for the lower confidence limit (LCL) yields a negative value whenever $p^2$ is less than $Z^2p(1-p)/n$ with $Z=1.96.$ This inequality is equivalent to $0\lt p \lt Z^2/(n+Z^2),$ showing there is always a possibility of obtaining a negative LCL with your formula, no matter what $Z$ and $n$ might be. For a better approach when $p$ could be very small, search our site for Binomial confidence intervals. – whuber Jan 17 '23 at 18:18
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    Many similar: https://stats.stackexchange.com/questions/493537/does-an-impossible-value-on-a-confidence-interval-for-a-probability-distribution, https://stats.stackexchange.com/questions/55598/what-should-i-do-when-a-confidence-interval-includes-an-impossible-range-of-valu, https://stats.stackexchange.com/questions/507336/confidence-intervals-of-positive-parameters-reaching-into-negatives, https://stats.stackexchange.com/questions/123779/confidence-interval-violating-physical-boundaries, https://stats.stackexchange.com/questions/21854/interpreting-a-negative-confidence-limit-for-a-proportion – kjetil b halvorsen Jan 17 '23 at 18:46

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