Where does the $\sqrt{N}$ come from in standard error of the mean formula?? When calculating z scores for multiple samples and want to describe the standard deviation of those sample means I know the formula is z = $\frac{(x - \mu)}{\frac{\sigma}{\sqrt{N}}}$ where N is our sample size. Intuitively why does dividing by $\sqrt{N}$ make sense??
Where does the $\sqrt{N}$ come from in standard error of the mean formula??
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$\begingroup$
probability
statistics
standard-deviation
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0$z$ should be $\frac{\bar{X}-\mu}{\sqrt{\sigma}/N}$ in your context, i.e. the standardization of the random variable which corresponds to the arithmetic mean. you standardize a variable by subtracting its expectation and dividing through the standard deviation. In the special case where your variable of interest is $\bar{X}$ (and you are given an iid sample), you have $E(X) = \mu $ and $\sqrt{Var(\bar{X})} = \sigma/\sqrt{N}$, hence the formular. Note: when standardizing a single variable $X$ you have $\frac{X-E(X)}{\sigma}$ for its standardized value. – 2017-02-23
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0It is straightforward to show formulaically but that does not provide much insight. Intuitively -- because the variance of an average of independent random variables is the average of the variances. – 2017-02-23
1 Answers
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The intuitive part is that the average of $n$ values is less variable than any single observation from the population.
Suppose you are sampling from a population that has variance $\sigma^2.$ That is $V(X_i) = \sigma^2.$ However, $V(\bar X) = \sigma^2/n.$ So the variance of $\bar X$ does decrease as $n$ increases, which matches intuition.
Then it follows that the "standard error of the mean" is $$SD(\bar X) = \sqrt{V(\bar X)} = \sqrt{\sigma^2/n} = \sigma/\sqrt{n},$$ as in @LarryB's link.
It is not stretching intuition to understand that means are less variable than individual observations. Suppose you are trying to estimate the average weight of melons in a crate; it contains melons of very different sizes. On any one draw from the crate we might get a huge one or a tiny one. But if we draw a dozen melons, it seems likely we will get ones of various sizes and their mean weight will be closer to the mean weight for the crate.
But my intuition does not tell me that the exact relationship must be $V(\bar X) = \sigma^2/n$ for the variance or $SD = \sigma/\sqrt{n}$ for the standard error. For that, I need to do the math. Smaller, yes intuitive; divide exactly by $\sqrt{n},$ no.
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1The really interesting underlying fact here is that variances of independent quantities add. That's not obvious at all, even though it is easy enough to do the math. – 2017-02-23