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For a normal distribution with a mean of X, and a standard deviation of SD, the 5% lower limit of the population is computed as X-1.645*SD. Meaning, 5% of the population will not reach that level. If I have a sample set of 10.85, 10.32, 10.11, 10.96, and 10.47, that set has a mean of 10.52 and a standard deviation of 0.3573. How do I create a confidence limit based on this data. Meaning how do I say that I am 90% sure that the 5% lower limit of the population will exceed Y.

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  • $\begingroup$ Are you after tolerance intervals? $\endgroup$ – Glen_b -Reinstate Monica Apr 16 '15 at 10:46
  • $\begingroup$ One thought is to use Chebyshev inequality with plug-in estimates of the mean and SD and use the delta method to calculate 90% CIs for the threshold. This relies on data being symmetric or right skewed which can be assessed from the sample. Another approach is to use the empirical distribution to find the 5-th percentile and calculate CIs from Donsker's theorem. $\endgroup$ – AdamO Mar 9 '18 at 16:09
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In the big picture, assuming the example you provide 'is' your problem and not merely an example, you can't draw meaningful inferences from a sample of 5.

Moreover, you seem to misunderstand the meaning of confidence intervals. They are really a measure of precision. They do not mean that you are 90% or 95% certain that the true value falls within the specified range. They are are frequentist statistic and, as such, they mean that were you to collect every possible sample consisting of 5 elements from the population, in 95% of those samples the confidence interval would contain the population parameter. Of course, the sample statistic (the mean in this case) would be different in a lot of those samples and thus so too would the confidence interval.

You have only 1 sample and the corresponding confidence interval, that interval either does or does not contain the population parameter, you don't know which. There is no "probability" associated with it (that would reflect a bayesian approach to the question).

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