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I'm reviewing probability and statistics.The textbooks said that

if the sampled population is infinite, then

$$\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$$

I'm curious about how does this result come from. Wikipedia does not tell me much. Is there any provement? How does it come from?

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    $\begingroup$ Why does this question have 3 up-votes????? Correction - why does it have any upvotes? $\endgroup$
    – Epictetus
    Nov 14, 2012 at 9:52
  • $\begingroup$ What is $n$? What is $\sigma$? What is $\bar x$? $\endgroup$
    – martini
    Nov 14, 2012 at 9:56
  • $\begingroup$ $n$ is the number of data, $\sigma$ is the standard deviation of those data, and $\bar{x}$ is the mean. @martini $\endgroup$ Nov 14, 2012 at 11:26

2 Answers 2

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If our population consists of $N$ individuals and $x_i$, $i=1,\ldots,N$, is the variable of interest, then the population mean and population variance are given by $$ \bar{X}=\frac{1}{N}\sum_{i=1}^N x_i,\qquad \sigma^2=\frac{1}{N-1}\sum_{i=1}^N(\,x_i-\bar{X})^2. $$ Suppose a simple random sample $x_1,\ldots,x_n$ of size $n$ is drawn from this population (i.e. every sample of size $n$ has equal probability of being drawn). Then the corresponding sample mean and sample variance is $$ \bar{x}=\frac{1}{n}\sum_{i=1}^n x_i,\qquad \mathrm{var}(\bar{x})=\frac{1}{n-1}\sum_{i=1}^n (x_i-\bar{x})^2. $$ It can be shown that $$ \mathrm{var}(\bar{x})=\frac{N-n}{Nn}\sigma^2=\left(1-\frac{n}{N}\right)\frac{\sigma^2}{n}. $$ Now, let $N\to\infty$ and see what happens.

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Assume we have a group of data, to get the SD of the mean, we can calculate the mean for many subgroups of our data and calculate it's SD. To do this, $\sigma_{\bar{y}}$ can be defined as follows: $$\sigma_{\bar{y}}^2=\frac{\sum\limits_{j=1}^{k}(\bar{y_j}-y_{tm})^2}{k}$$ Where $\bar{y_j}$ is the mean of a finite number of data, $y_{tm}$ is the mean for an infinite number of data, and k is the number of times you took the finite number of data. Since $$\bar{y_j}=\frac{\sum\limits_{i=1}^{n}y_{i}}{n}$$ Substituting that in the first equation we get: $$\sigma_{\bar{y}}^2=\frac{1}{k}\sum\limits_{j=1}^{k}\left(\frac{1}{n}\sum\limits_{i=1}^{n}y_i-y_{tm}\right)^2$$ $$\sigma_{\bar{y}}^2=\frac{1}{kn^2}\sum\limits_{j=1}^{k}\left(\sum\limits_{i=1}^{n}y_i-ny_{tm}\right)^2$$ But $n=\sum_{i=1}^{n}1$, so substituting on the $n$ inside the parentesis: $$\sigma_{\bar{y}}^2=\frac{1}{kn^2}\sum\limits_{j=1}^{k}\left(\sum\limits_{i=1}^{n}y_i-\sum\limits_{i=1}^{n}y_{tm}\right)^2=\frac{1}{kn^2}\sum\limits_{j=1}^{k}\sum\limits_{i=1}^{n}\left(y_i-y_{tm}\right)^2$$Since $\frac{1}{n}\sum_{i=1}^{n}(y_i-y_{tm})^2=\sigma_{j}^2$ because it is the variance of each group of data: $$\sigma_{\bar{y}}^2=\frac{1}{nk}\sum\limits_{j=1}^{k}\sigma_{j}^2$$ Since $\sigma_j^2$ is the variance of a subgroup of the same date, it can be considered the same for all $j$, and at last we get: $$\sigma_{\bar{y}}^2=\frac{\sigma^2}{n}$$ $$\sigma_{\bar{y}}=\frac{\sigma^2}{\sqrt{n}}$$

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