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I've come across the post in relevant topics in the confidence interval, I've seen answer said the population mean is not random variable so we can't say 95% probability that CI contains the population mean.

Could someone explain more clearly and avoid credibility for me, please?

My understanding is, given 5% of the p-value, each sample could generate a confidence interval and re-sampling numerous time, consider a collection of their confidence interval, and 95% of such CI contains the population mean, and 5% do not

Analogy: If this is true, why we can't say the CI that we computed, 95% of chance it is the one that contains the population mean. Just like a collection of 100 balls and 95 of which are red. Then we pick a ball blindly, we may infer 95% of our ball is red.

Appreciated for any comments and help

LJNG
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    I have given my answer to your question (+1). However, I hope you will give explicit links to the previous similar questions you mention. – BruceET Jul 13 '20 at 00:00
  • @BruceET https://stats.stackexchange.com/questions/26450/why-does-a-95-confidence-interval-ci-not-imply-a-95-chance-of-containing-the – LJNG Jul 16 '20 at 17:41

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In my view, there is a lot of silliness in various purported interpretations of frequentist confidence intervals. One example of that is the interpretation you quote, "[T]he population mean is not random variable so we can't say 95% probability that CI contains the population mean."

For simplicity, consider the 95% z confidence interval for normal $\mu,$ where $\sigma$ is known: $\bar X \pm 1.96\sigma/\sqrt{n}.$ This comes from the perfectly reasonable statement

$$0.95 = P\left(-1.96 \le \frac{\bar X = \mu}{\sigma/\sqrt{n}} \le 1.96\right)\\ =P\left(\bar X - 1.95\frac{\sigma}{\sqrt{n}}\le \mu\le \bar X + 1.95\frac{\sigma}{\sqrt{n}}\right).$$

The sentence I quoted from your question ignores that $\bar X$ is a random variable. The 95% CI is a reasonable statement that the random interval contains (covers) the unknown $\mu$ with probability 95%. A frequentist interpretation of the probability of this 'coverage event' is that over the long run such an event will be true 95% of the time.

It is unproductive sophistry to say that once we observe $\bar X,$ the 'probability collapses', so that the event is either true or false--no probability about it.

Traditionally, the compromise with hard-core frequentists has been to call this a "confidence" interval, not a "probability" interval. So it is OK to say I have 95% "confidence in" the truth of the interval. (It is best not to try to define what "confidence" means. You may soon get around to admitting it's just a diplomatic synonym of "probability".)

In the same sense, a frequentist would say that "$P(\mathrm{Heads}) = 1/2$" for a fair coin means that over the long run the coin will show Heads nearly half of the time. Few people (even few hard-core frequentists) say it's meaningless to claim a coin is fair because, if you ever toss it and look at the result, the 'probability collapses' and you either have a Head or a Tail--no probability about it.

Note: In a Bayesian setting normal $\mu$ and binomial $p$ are random variables. One begins with a (more or less informative) prior distribution, looks at data, and finds a posterior distribution on $\mu$ or $p.$ From the posterior distribution, one can find a 95% Bayesian posterior probability interval for the parameter. However, details of that approach, which may have some philosophical difficulties of its own, are stories for another day.

BruceET
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