Why can we use empirical standard deviation when computing mean confidence intervalsI'm reviewing some basic statistics, and I'm asking myself questions on things I used to take for granted when I first saw them years ago. I'm going to state things as I understand them, so there might be a mistake in the following.Consider a random variable X following a distribution of mean μ and standard deviation σ. We measure n samples of X, and observe an empirical mean x ¯ and empirical std s.Because we're observing samples, x ¯ and s are random variables themselves. We should therefore not use x ¯ and μ interchangeably, and the same goes for s and σ. Instead, people compute confidence interval on μ based on x ¯ . By the central limit theorem, if n is large enough, we can say that x ¯ ∼ N ( μ , σ n ).When computing confidence intervals, we usually use σ x ¯ = s n . If all of this is correct, my question is the following: since both s and x ¯ are random variables, why does it seems to be ok to consider s = σ when computing confidence intervals for μ?

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Why can we use empirical standard deviation when computing mean confidence intervalsI&#039;m reviewing some basic statistics, and I&#039;m asking myself questions on things I used to take for granted when I first saw them years ago. I&#039;m going to state things as I understand them, so there might be a mistake in the following.Consider a random variable X following a distribution of mean   μ and standard deviation   σ. We measure n samples of X, and observe an empirical mean             x      ¯       and empirical std s.Because we&#039;re observing samples,             x      ¯       and s are random variables themselves. We should therefore not use             x      ¯       and   μ interchangeably, and the same goes for s and   σ. Instead, people compute confidence interval on μ based on             x      ¯      . By the central limit theorem, if n is large enough, we can say that             x      ¯        ∼  N  (  μ  ,      σ          n        ).When computing confidence intervals, we usually use       σ                            x          ¯                      =      s          n      . If all of this is correct, my question is the following: since both s and             x      ¯       are random variables, why does it seems to be ok to consider   s  =  σ when computing confidence intervals for   μ?

Elias Keller · Accepted Answer

Step 1Nearly everything you wrote is accurate, but the conclusion you draw at the end is based on a misapprehension; namely that the standard error of the mean when the variance is unknown, the quantity   s      /        n  , is equal to the standard deviation of the sampling distribution of the mean, which is asymptotically   σ      /        n  . These are not equal since the former is a statistic and the latter is a function of the unknown parameter   σ.That said, the former estimates the latter, in the same way that             x      ¯       estimates   μ.The source of this confusion comes from the notion that the construction of the confidence interval is based on some &quot;recipe&quot; in which the confidence limits may be expressed in the form            x      ¯        ±      t          n      −      1      ,      α            s          n        ,where       t          n      −      1      ,      α       is some &quot;critical value&quot; that depends on the desired level of confidence and the sample size. While the formula is correct, the origin of this formula isn&#039;t by analogy with the corresponding parameters; that is to say, the interval estimate isn&#039;t constructed from some expression like  μ  ±      t          n      −      1      ,      α            σ          n        .Step 2This is incorrect. The interval estimate comes from inverting a hypothesis test. Specifically, we want to control the Type I error   α so that  Pr      [                            |                                      X            ¯                          −                  μ          0                          |                            s                  /                          n                      &amp;gt;    c                      |                                                    H      0        ]    =  α  ,for the hypothesis      H    0    :  μ  =      μ    0      vs.        H    a    :  μ  ≠      μ    0  and c is some critical value that depends on   α. The statistic on the LHS of the inequality is (without the absolute value) is Student t-distributed when       H    0   is true, thus c is a quantile of the Student t distribution.

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