1
$\begingroup$

I'm having a hard time verbalizing the difference between these two concepts. How does this sound?

Standard deviation: perform a measurement; there is a 68% chance that that measurement will be within one standard deviation of the sample mean. Qualitatively, measures scatter/variation/dispersion of the data. It does not change with sample size (assuming large N).

Standard deviation of the mean (or standard error): perform a series of measurements; there is a 68% chance that that sample mean (or average) will be within one “standard deviation of the mean” of the population mean. Decreases with increasing sample size.

$\endgroup$
  • $\begingroup$ Not quite right. The standard error description is closer. For that one, is better to say standard error = "standard deviation of the sample mean". In contrast, for your standard deviation description, you should say "population mean". The other issue is that your 68% number assumes a normal distribution. For the standard error this is OK if the sample size is "large". However, for the standard deviation the 68% number does not generally apply. $\endgroup$ – GeoMatt22 Sep 9 '16 at 5:15
  • $\begingroup$ @GeoMatt22 thanks. yes, for my situation, normal dist. is assumed. I am not sure why in st. dev. description it would be population mean. I take a bunch of data that have normal dist. The mean of that data is the sample mean, right? The population mean would be the mean of the entire population, no? $\endgroup$ – user5419 Sep 9 '16 at 5:27
  • $\begingroup$ For a normal distribution, the measurement will be within 1 standard deviation of the population mean 68% of the time. Because of the standard error, your sample mean could be further away. $\endgroup$ – GeoMatt22 Sep 9 '16 at 5:33
  • 1
    $\begingroup$ Just to add that it may help to think of them as both standard deviations but of different distributions. $\endgroup$ – mdewey Sep 9 '16 at 8:35
  • $\begingroup$ @geomatt if we rely on sample quantities it's not just the error in the mean you need to worry about; the error on the standard deviation is also there. $\endgroup$ – Glen_b Sep 10 '16 at 3:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.