The "normal" standard error of the mean (SEM) is the population standard deviation divided by the square root of the sample size. Wikipedia states that the SEM is an estimate of how far the sample mean is likely to be from the population mean.

In practice you don't know about the population standard deviation and use the sample standard deviation instead. The sample standard deviation, however, is only an estimate of the population standard deviation with some unknown error... Despite this unknown error, does the estimated SEM still tell how for the sample mean is likely to be from the population mean?

  • $\begingroup$ You divide the population standard deviation $\sigma$ (and not the population mean) by the square root of $N$. And then in the second paragraph first sentence it should be standard deviation (and not mean) for population and sample when calculating the SEM. See also here: en.wikipedia.org/wiki/Standard_error $\endgroup$ – Stefan Mar 6 '18 at 14:09
  • $\begingroup$ Although it's unclear what you really mean by "not include the population standard deviation," much (if not all) of this question may already have been answered at stats.stackexchange.com/questions/18603, where I discuss the meaning and interpretation of a standard error. $\endgroup$ – whuber Mar 6 '18 at 14:37
  • $\begingroup$ @whuber Thanks. I rephrased the question and hope it is more clear now. Unfortunately, my knowledge in statistics is not sufficient to evaluate if my question is already answered since the answer covers a lot more than just SEM. $\endgroup$ – Julian Mar 6 '18 at 18:57
  • $\begingroup$ The "proper" frequentist response ought to be along the lines that the SEM does not say anything about the population mean, but rather about the sample mean. Specifically about (as you say) how far from the population mean we might expect the sample mean to be. Whether or not that reversal of your question matters at all is left as an exercise for the reader. $\endgroup$ – user5957401 Mar 6 '18 at 19:33

The answer is yes: both the actual standard deviation of the mean $\sigma_{\bar{x}}$ (calculated from the actual population standard deviation $\sigma$) and the estimated standard deviation of the mean $s_{\bar{x}}$ (calculated from the estimated population standard deviation $s$) tell you, in a sense, how far the sample mean is likely to be from the estimated mean.

What's the difference, then? It turns out that the estimated standard deviation $s_{\bar{x}}$ is a biased estimator, and tends to underestimate the actual standard deviation.

  • 2
    $\begingroup$ If bias were of any real concern, we could (and would) correct for it. Bias is really beside the point. $\endgroup$ – whuber Mar 6 '18 at 15:42
  • $\begingroup$ Yes, bias is beside the point, except that it's the only practical difference between $s_{\bar{x}}$ and $\sigma_{\bar{x}}$. If you ignore bias and look at things qualitatively, then the estimated SEM and actual SEM are the same thing. $\endgroup$ – Richter65 Mar 6 '18 at 18:35
  • $\begingroup$ No, that's not true: as the OP points out, one is an estimate and the other is a population parameter. You seem to be trying to dismiss the essential aspect of the question. $\endgroup$ – whuber Mar 6 '18 at 18:59
  • $\begingroup$ I think you're misunderstanding the question. True, one is an estimate and the other is a population parameter. But the OP's question is: what difference does that make? The answer to that question involves bias. But if you ignore bias, and look at it qualitatively, the answer then becomes "not much". $\endgroup$ – Richter65 Mar 6 '18 at 19:20
  • $\begingroup$ @Richter65 But how do you know that you can savely ignore the bias? Wouldn't it be helpful or even necessary to quantify the bias? $\endgroup$ – Julian Mar 6 '18 at 19:23

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