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

Question

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!

Answer 1

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.

Answer 2

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 $