# why is the population standard deviation approximated as the sample standard deviation [closed]

This question addresses calculating a p value from the mean and standard deviation statistics of a sample. I understand that the -general- philosophy is to divide the sample standard deviation by the root of the population size to get the standard deviation of the sampling distribution, using the assumption that the standard deviation of the sample is roughly equal to the standard deviation of the hypothetical larger population. Then one calculates a z-score from the number of standard deviations of the sampling distribution to calculate what percent of the time the observed result would have occurred by random chance. I understand that the particular formula for the sampling distribution standard deviation depends on the particular statistic, say difference of means is a different formula.

The texts and videos that I've looked at use language like "the sample standard deviation is the best number we have available to estimate the population standard deviation." I just don't find that explanation satisfying.

This approach hinges on the validity of estimating the standard deviation of the entire population as being approximately equal to the standard deviation of the representative sample. However, we don't make the same assumption that the mean of the population is the approximately equal to the mean of the sample. At some level, it feels like the final result of significance or non-significance is only self-validating or checking for self-consistency of an assumption that is baked into the methodology.

So to restate, why is the sample standard deviation a good approximation of the populatoin standard deviation, but the sample mean is not a good approximation of the population mean? I found online an equation for standard deviation of the sampling distribution of standard deviations:

standard error of standard deviation = .71 sample standard deviation / root N.

Does the relative narrowness of the standard error compared to standard deviation play a role in justifying the approximation?

Thank you

## closed as off-topic by Dilip Sarwate, StubbornAtom, Daryl, postmortes, StammeringMathematicianJun 16 at 16:28

• This question does not appear to be about math within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

• If you do not get an answer here, maybe try stats.stackexchange.com – GEdgar Jun 15 at 23:46
• – 高田航 Jun 16 at 0:43
• "So to restate, why is the sample standard deviation a good approximation of the populatoin standard deviation, but the sample mean is not a good approximation of the population mean?" Why do you think the sample mean is not a good approximation? It is the best available. For both the mean and the variance, the means of the sample estimates are equal to the population mean and variance. – herb steinberg Jun 16 at 1:19
• Sample statistics are often used as estimators of parameters, and there are ways of comparing estimators to see which are "better." One way is bias, another the variance of the estimator - which you allude to at the end of your question. You want an estimator to have low variance so that you can expect it close to the parameter in practice. Sample mean and (corrected) variance will be the MVUEs of a normal population.Another way of comparing estimators, "maximum likelihood," seeks to construct estimators that maximize the likelihood of the observed statistics. – runway44 Jun 16 at 1:33
• I'm voting to close this question as off-topic because the OP has followed @GEdgar's recommendation and posted the same question on stats.SE where it has already received an answer, and the OP has not deleted this one as he should. – Dilip Sarwate Jun 16 at 1:42