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While calculating the standard error, we take the standard deviation of the sampling distribution.

But for calculating the standard deviation we must consider a 'mean' to take difference about.

In Wikipedia its given that the 'mean' is the population mean while I came across a lecture which said it to be sampling distribution mean.

So for standard error do we take the population mean or the sampling distribution mean as the 'mean'?

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    $\begingroup$ When you calculate the standard deviation of the sampling distribution, you use the mean of the sampling distribution. Usually, mean of the population is not available anyway. $\endgroup$ – Evargalo May 29 '17 at 13:17
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    $\begingroup$ If the sampling distribution you are dealing with is the sampling distribution of the sample mean, then the mean of the sampling distribution is the population mean. $\endgroup$ – Just_to_Answer May 29 '17 at 19:39
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In general, you must have given the dataset with sample size $n$. Now obviously the population standard deviation is $\sqrt{E(X-\mu)^2}$, but when we try to find the sample standard deviation we use $$s=\sqrt{\frac{1}{n-1}\sum_{j=1}^{n}(x_j -\bar{x})^2}$$ which is a good estimator of population standard deviation.

You can not calculate $\sqrt{E(X-\mu)^2}$, since the everything in the expression is unknown. But if you know population mean $s$ changes to $$s=\sqrt{\frac{1}{n}\sum_{j=1}^{n}(x_j -\mu)^2}$$.

(But the second case does not happen in practical field.)

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