0

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?

Funkwecker
  • 2,432
  • 5
  • 24
  • 43
  • 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: https://en.wikipedia.org/wiki/Standard_error – Stefan Mar 06 '18 at 14:09
  • 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 https://stats.stackexchange.com/questions/18603, where I discuss the meaning and interpretation of a standard error. – whuber Mar 06 '18 at 14:37
  • @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. – Funkwecker Mar 06 '18 at 18:57
  • 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. – user5957401 Mar 06 '18 at 19:33

1 Answers1

0

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.

Richter65
  • 139
  • 2
  • 2
    If bias were of any real concern, we could (and would) correct for it. Bias is really beside the point. – whuber Mar 06 '18 at 15:42
  • 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. – Richter65 Mar 06 '18 at 18:35
  • 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. – whuber Mar 06 '18 at 18:59
  • 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". – Richter65 Mar 06 '18 at 19:20
  • @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? – Funkwecker Mar 06 '18 at 19:23
  • Put another way, the OP's question is "does the estimated SEM still tell how far the sample mean is likely to be from the population mean"? The answer is yes, it does. It just provides a slightly different answer than the answer calculated from the actual (and almost always unknown) population parameter. How much different? That depends on the distribution, and also gets into a discussion of bias. – Richter65 Mar 06 '18 at 19:23
  • @Julian: in principle, you can't. Although for most practical purposes, with a large enough sample, the difference becomes small. Granted, it's hard to quantify "small" without knowing what the underlying distribution is, but as an example for a normally-distributed sample of 1000 points the difference is less than 0.1%. (https://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation) – Richter65 Mar 06 '18 at 19:29
  • I question your use of the word bias. Bias typically refers to a systematic error in some direction (i.e. the expectation of the estimator and the true value are different), and I don't think that there is a systematic bias introduced here [(source)](https://en.wikipedia.org/wiki/Variance#Sample_variance). There is noise introduced, but thats different. It is true that if you use `1/n` instead of `1/(n-1)` you get a bias, but that is easily fixed (and indeed not a problem if you use just about any stats package's defaults). – user5957401 Mar 06 '18 at 19:31
  • @user5957401: That is true for variance. But not for standard deviation. The difference may seem trivial, but it does have important implications when dealing with concepts like bias. Those implications are, for example, the basis of the frequentist's objections to Bayesian techniques and the selection of the Bayesian prior. – Richter65 Mar 06 '18 at 19:42
  • You aren't wrong that the correction for SD is different. Thats on me -- I got ahead of myself. However, I think you are very wrong that its not a tractable problem. Wikipedia has an unbiased estimator of [SD](https://en.wikipedia.org/wiki/Standard_deviation#Corrected_sample_standard_deviation) when things are normally distributed -- which sample means are (under weak assumptions). I'll admit though that I would have thought there would be a general solution and there clearly isn't. I'm curious what you mean about the bayes bit though -- could you elaborate? – user5957401 Mar 06 '18 at 19:56
  • I never claimed the bias problem was intractable; I only claimed that it's a function of the underlying distribution. I further asserted that, for most practical purposes, you can ignore the difference between the actual and estimated SEM when you're actually trying to solve a real-world problem. – Richter65 Mar 06 '18 at 20:34
  • Regarding Bayesian priors (getting a little off-topic): suppose you were trying to estimate the value of something you know to be positive (say, the mass of a particle) using Bayseian inference. You use an uninformative prior (https://en.wikipedia.org/wiki/Prior_probability), such as a Heaviside step function, so as not to bias the result, and get the most "likely" mass value (let's call that value $m$) with that prior. But what about the square of the mass? Would that be $m^2$? When you square the prior, it's not longer quite so uninformative, and you've unintentionally introduced a bias. – Richter65 Mar 06 '18 at 20:34