# How to get the population standard deviation from a sample standard deviatoin

I am trying to find the confidence interval, and I need to know the population standard deviation. If I am given the sample standard deviation, how can I get the population one? Thanks!

You need to use $s^{2}$ unbiased estimator for $\sigma^{2}$ population variance which is defined as

$$\sigma^{2} = \frac{1}{N} \sum_{i=1}^{N} (x_{i} - \mu)^{2}$$

The $\sigma^{2}$ computation would require to know $\mu$ true mean which is unknown when working with a sample so again you need an estimator for this which is $\bar x$ sample mean.

The use of $\bar x$ sample mean instead of $\mu$ true meam introduces a bias which tends to zero as the sample numerosity grows so the sample variance estimator uses a corrective term

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

As you can see, with $N \rightarrow \infty$ you get $s^{2} \rightarrow \sigma^{2}$ as

• $\bar x \rightarrow \mu$
• $N-1 \rightarrow N$
• I know this is an old post but it's better me asking here than starting a new one. Where exactly does the $N-1$ come from? Is that in an attempt to reduce the bias of $s^2$? Also, if $\bar{x}$ is an unbiased estimator of $\mu$, how does using it introduce bias? Aug 16, 2017 at 13:31

Generally speaking, if you are taking a sample from a normal distribution and you are trying to obtain a confidence interval for the mean, but you don't know the population standard deviation, the two-sided confidence interval at level of significance $1-\alpha$ is

$$\overline{X} \pm \frac{t_{\alpha/2} S}{\sqrt{n}}$$

where $t_{\alpha/2}$ is the positive number such that $P(T>t_{\alpha/2})=\alpha/2$ when $T$ is a Student's $t$-distributed random variable with $n-1$ degrees of freedom. Here $t_{\alpha/2}$ is a little bit larger than $z_{\alpha/2}$ (which arises from the standard normal distribution), which is what you would use if the population standard deviation were known.