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I have this distribution: $\{1, 2, 3, 4\}$, $\sigma = 1.118$ and I wrote down all the possible samplings of $n=2$ and their respective means:

Samplings: $$\{(1,2), (1,3), (1, 4), (2, 3), (2, 4), (3, 4)\}$$ Samplings means (sm): $$\{1.5, 2, 2.5, 2.5, 3, 3.5\}, \sigma_{sm} = 0.645$$

So according to the formula of the Standard Deviation of the Sampling Distribution or Standard Error of the Mean I was expecting to get the same value, but it turns out:

$$\frac{\sigma_\mu}{\sqrt{n}} = \frac{1.118}{\sqrt{2}} = 0.790$$

Trying to understand this result I found out that 0.790 is actually the standard deviation of the permutation of the distribution:

Samplings: $$\{(1,1), (1,2), (1,3), (1, 4), (2,1), (2,2), (2,3), (2, 4), (3,1), (3,2), (3,3), (3, 4), (4,1), (4,2), (4,3), (4, 4)\}$$

Samplings means (sm): $$\{1, 1.5, 2, 2.5, 1.5, 2. 2.5, 3, 2, 2.5, 3, 3.5, 2.5, 3, 3.5, 4\}, \sigma_{sm} = 0.790$$

So now I'm actually more confused than before and actually a bit skeptical of the actual meaning of $SE = \frac{\sigma}{\sqrt{n}}$. If I have a distribution $\{1, 2, 3, 4\}$, I can't draw the sample $(1, 1)$ from it, so why do we use this formula to calculate the standard deviation of the sampling distribution, instead of something like: $$\frac{\sigma}{\sqrt{n+1}} = \frac{1.118}{\sqrt{2+1}} = 0.645$$

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  • $\begingroup$ If you're sampling without replacement you need a finite population correction $\endgroup$ – Glen_b Aug 17 '17 at 0:55
  • $\begingroup$ but does it make any sense to actually sample doing a replacement? I can't get an intuition for that in real life $\endgroup$ – Danilo Souza Morães Aug 17 '17 at 0:59
  • $\begingroup$ The decision to use it is okay but it affects the variance. If you do it, you need the finite population correction. $\endgroup$ – Glen_b Aug 17 '17 at 2:04
  • $\begingroup$ I understand that it affects the variance and I can see how the finite population correction works. What I don't understand is why we would calculate the standard error as if it came from an "infinite" population as there is no such thing. So in the end, why is a "no replacement" formula any good when that's never gonna happen in "real" or "imaginary" life? Why don't we actually use the standard deviation of the sampling distribution as the name states? Why do we use a formula that actually calculates the standard deviation of the permutation of the samplings? $\endgroup$ – Danilo Souza Morães Aug 17 '17 at 2:36
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    $\begingroup$ Infinite populations are perfectly natural for many problems. In many cases every new observed value is a new member of an inexhaustible notional population of possible observed values (that is, much more like a die roll than drawing tickets from a hat and keeping them once drawn) $\endgroup$ – Glen_b Aug 17 '17 at 3:06

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