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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 population 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

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    $\begingroup$ "why is the sample standard deviation a good approximation of the population standard deviation, but the sample mean is not a good approximation of the population mean?" Where have you seen it asserted that the sample mean is not a good approximation of the population mean? It is often used as an estimator of a population mean. $\endgroup$ – Michael Hardy Jun 16 at 1:16
  • $\begingroup$ I've wondered the same thing. We may consider the sample mean the best available estimate of the population mean. But we know it's not perfect, so we want to quantify our uncertainty with a p-value or CI. So we do so by estimating the standard error, with the very same data we don't entirely "trust". $\endgroup$ – user3391564 Jun 16 at 1:31
  • $\begingroup$ @Michael When a class shifts from teaching a z test with population standard deviation known to a t test using the sample standard deviation, sometimes unfortunate language like this is used that creates a seemingly illogical inconsistency between how the mean and standard deviation are described and used. The treatment of the sd as a nuisance parameter when testing a mean seems to further the confusion. Some of the handwaving done to transition from the ztest to the ttest without getting into technical details likely leads to confusion as well. $\endgroup$ – jsk Jun 16 at 1:55
  • $\begingroup$ “I just don’t find that satisfying”. It is not satisfying, indeed. The problem is that the text you’re using is not advanced enough to formally justify/prove the statement. You will find proofs for statements like those in mathematical statistics textbooks. Check, for example, Statistical Inference by Casella & Berger. $\endgroup$ – Victor S. Jun 16 at 4:24
  • $\begingroup$ You wrote "However, we don't make the same assumption that the mean of the population is the approximately equal to the mean of the sample." That is simply not true. If you could tell us what caused you to think we do not use the sample mean to estimate the population mean, maybe your question could be understood. $\endgroup$ – Michael Hardy Jun 16 at 16:25
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No matter the shape of the population distribution, the Central Limit Theorem holds that the sampling distribution of the sample means (and sample variances) approaches a normal distribution as the sample size $T$ (number of observations or data points) gets larger, especially for sample sizes over 30. As you take more samples, the sample distribution graph will look more like a normal distribution, and so on average, the sample moments will be the population moments as $T\uparrow$. (If you add up the means or standard deviations from all of your samples and find the average, that average will be close to the actual moments for the population.)

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  • $\begingroup$ You wrote "as $T\uparrow$", but you never said what "$T$" is. $\qquad$ $\endgroup$ – Michael Hardy Jun 16 at 16:28

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