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Comments on What is the formula for sample standard deviation of a small sample size?

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What is the formula for sample standard deviation of a small sample size?

+6
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The formula for sample standard deviation is given by:

$$s = \sqrt{\frac{\sum_{i=1}^{i=N} (x_i - \bar{x})^2}{N-1}}$$

Am I right that when the sample size is small ($N<30$), the formula for sample standard deviation becomes:

$$s = t_{N-1, \text{confidence}} \sqrt{\frac{\sum_{i=1}^{i=N} (x_i - \bar{x})^2}{N-1}} \ \ \ \ \ \ \ \ ?$$

Here, $t_{N-1, \text{confidence}}$ is the Student's $t$ coefficient obtained from the tables (the tables are given, for instance, here).


The reason I am asking is the following. I had no doubts whatsoever that I must multiply standard deviation by the Student's $t$ coefficient for a small sample size. And I have been doing it all the time. I used it in a draft for an article. When I was checking the draft, I decided to check this formula. First, I could not remember where I got it from. Second, I searched the books and the Internet only to find out that people use Student's $t$ coefficient to calculate confidence interval, not standard deviation. I tried to derive my formula from the formula for confidence interval and failed. I talked to a fellow student who also thinks standard deviation must be multiplied by Student's $t$ coefficient and also does not remember why.

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From the little that I know ...

  1. If the sample = population (census)
    $\sigma^2 = \displaystyle\frac{1}{N}\sum_{i = 1}^N (x_i - \mu)^2$ where $N$ is the size of the population and $\mu$ is the population mean. The variance is $\sigma^2$ and the standard deviation then is $\sigma$

  2. If the sample is smaller than the population (any study except a census)
    $\sigma^2 = \displaystyle\frac{1}{n - 1}\sum_{i = 1}^n(x_i - \bar x)^2$ where $n$ is the size of the sample and $\bar x$ is the sample mean. This $\sigma^2$ is called the unbiased variance and the standard deviation is $\sigma$.

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2 comment threads

In point (2), just as you replaced $\mu,$ the population mean, with $\overline x,$ the sample mean, I... (1 comment)
Thank you, I know that (except for the term _unbiased variance_. My question was about not just small... (2 comments)
In point (2), just as you replaced $\mu,$ the population mean, with $\overline x,$ the sample mean, I...
Michael Hardy‭ wrote 4 months ago

In point (2), just as you replaced $\mu,$ the population mean, with $\overline x,$ the sample mean, I would have replaced $\sigma^2,$ the population variance, with $s^2.$ And the conclusion that this is an unbiased estimator of the variance relies on the assumption that this is an i.i.d. sample, i.e. with replacement.