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I'm reading this book and on page 419 the authors state the following theorem:

I'm really confused. What is the standard deviation? $\sigma$ or $\sigma/\sqrt{n}$?

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  • $\begingroup$ Interesting. My understanding is that the equation, $\frac{\sigma}{\sqrt{n}}$, is the standard error. $\endgroup$
    – Mari153
    Sep 21 '20 at 3:30
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I think the confusion disappears once you realise that the theorem talks about two distributions.

The first distribution is the population, which has a mean $\mu$ and a standard deviation $\sigma$. Let's assume the height of all humans is normal distributed, with $(\mu, \sigma) = (1.5, 0.5)$ in metres.

From that population, you draw one sample of size $n$, and compute its mean -- that's called the sample mean. Say you measure the height of $n=100$ people, and get a mean height for that one sample of 1.57.

Repeat many times: you'll get a distribution of sample means, which is the second distribution -- and that has a mean of $\mu$ and a standard deviation of $\sigma / \sqrt n$.

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Standard deviation of what? It really depends on the random variable you consider. With that being said, let $X_i\stackrel{\text{i.i.d.}}{\sim}(\mu,\sigma^2)$ for $i=1,\ldots,n$. Now $\sigma$ is obviously the standard deviation of the random variables $X_1,\ldots,X_n$, whereas $\frac{\sigma}{\sqrt{n}}$ denotes the standard deviation of the sample mean $\overline{X}=\frac{1}{n}\sum_{i=1}^n X_i$.

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