# The confidence interval can be expressed in terms of samples (or repeated samples): "Were this procedure to be repeated on multiple samples, the calculated confidence interval (which would differ for each sample) would encompass the true population parameter 90% of the time."[1] Note that this does not refer to repeated measurement of the same sample, but repeated sampling.

Aleah Avery 2022-11-18 Answered
According to frequentists, why can't probabilistic statements be made about population paramemters?
The confidence interval can be expressed in terms of samples (or repeated samples): "Were this procedure to be repeated on multiple samples, the calculated confidence interval (which would differ for each sample) would encompass the true population parameter 90% of the time."[1] Note that this does not refer to repeated measurement of the same sample, but repeated sampling.
And:
The confidence interval can be expressed in terms of a single sample: "There is a 90% probability that the calculated confidence interval from some future experiment encompasses the true value of the population parameter." Note this is a probability statement about the confidence interval, not the population parameter.
And:
A 95% confidence interval does not mean that for a given realised interval calculated from sample data there is a 95% probability the population parameter lies within the interval, nor that there is a 95% probability that the interval covers the population parameter.[11] Once an experiment is done and an interval calculated, this interval either covers the parameter value or it does not; it is no longer a matter of probability.
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## Answers (1)

Eva Cochran
Answered 2022-11-19 Author has 14 answers
Step 1
Suppose that you want to model the random behaviour of a certain population. Then you have to associate to the population a density function f (that is, you choose a "normal distribution", "exponential distribution", etc.), and a parametre $\theta$ (that is, if for example your density is a normal, then $\theta$ can be the population mean or the variance, etc.).
Suppose that you have decided which f you want, that is, the distribution for your population. The goal now is to estimate $\theta$. In frequentist statistics, $\theta$ is an unknown contant to be discovered. That is why we speak about confidence and not about probability.
Example: imagine I want to model the height of the people in England. I associate to it the normal distribution, so f is the density function of a normal. Now I want to estimate $\mu =\text{population mean}$. One takes a sample ${X}_{1},\dots ,{X}_{n}$ of heights and uses the fact that
$\frac{\overline{{X}_{n}}-\mu }{{s}_{n}/\sqrt{n}}\sim {t}_{n-1}.$
One computes a and b so that
$P\left(a<\frac{\overline{{X}_{n}}-\mu }{{s}_{n}/\sqrt{n}}
that is,
$P\left(\overline{{X}_{n}}-a\cdot {s}_{n}/\sqrt{n}<\mu <\overline{{X}_{n}}-b\cdot {s}_{n}/\sqrt{n}\right)=0.95.$
Step 2
Here it makes sense to speak about probability because $\overline{{X}_{n}}$ is a random variable. Now, what you do is to substitute $\overline{{X}_{n}}$ (random variable) by the sample mean $\overline{{x}_{n}}$ (constant value), and your confidence interval is
$I=\left[\overline{{x}_{n}}-a\cdot {s}_{n}/\sqrt{n},\overline{{x}_{n}}+b\cdot {s}_{n}/\sqrt{n}\right].$
The parametre $\mu$ is a constant, so either it belongs to I or not (you do not have probability here). But you have a lot of confidence that it will belong to I.
Remark: opposite to frequentist statistics, one may use bayesian statistics, which assumes that the parametre $\theta$ is a random variable, with a probability distribution to be discovered. In this case one speaks about credible regions (probabilities) and not confidence intervals (confidence).
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