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Given a normally distributed population with mean $\mu$ and say I constructed a confidence interval of $\mu \pm a$ with a confidence level of 95%. Is the following statement correct?

There is a 0.95 probability/chance that the population mean is captured within this confidence interval?

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In the sense of the Bayesian interpretation of probability, that is actually correct. But it is easy to be misled by that if you don't also understand how frequentists look at it.

In the sense of the frequentist interpretation of probability, it is not correct. Because from the frequentist point of view, you construct one confidence interval, and now the mean is either in that interval or it isn't. Where is the repeatable experiment? The repeatable experiment is the construction of the confidence interval itself. That is, from the frequentist point of view, we "have 95% confidence that the population mean is in the confidence interval" because if we were to take many samples and construct many confidence intervals, 95% of the samples would produce 95% confidence intervals which overlap with $\mu$.

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  • $\begingroup$ Thank you, one follow up question. So if I were to take a 100 samples and constructed a confidence interval for all of them 95 would contain the population mean $\mu$? Or do you mean by many someting like in the limit (as the amount of samples goes to infinity) the proportion of intervals that contain $\mu$ will go to 0.95. $\endgroup$ – Keep_On_Cruising Jul 29 at 18:55
  • $\begingroup$ @Keep_On_Cruising It is in the limit, yes. $\endgroup$ – Ian Jul 29 at 19:08
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You will get different answers for your question depending on philosophical understanding. I like to say that the proposed statement is at best inaccurate because the population mean "exists" but is unknown, therefore there is no randomness to account for (and thus no probability).

I prefer to say the following:

"The confidence interval was produced with a method that captures the mean $\mu$ with $95~\%$ probability."

Note that the probability is associated with the method (say, the random sampling and the formulae) rather than with its actual, numerical output.

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  • $\begingroup$ Oh, I see what you mean! Thanks $\endgroup$ – Keep_On_Cruising Jul 29 at 18:59

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