Standard Deviation of Population from Sample.

Question

I guess I am a little confused. I am doing t test statistics in my class. I think I know but I would love some insight.

I am trying to get the estimated standard deviation for the population.

These are not the scores from my sample but lets uses these as examples. $$1\ \ \ 5\ \ \ 3\ \ \ 5\ \ \ 6\ \ \ 7\ \ \ 2\ \ \ 3\ \ \ 4\ \ \ 1$$ What is my process? I have an equation for the estimated population variance is that the same thing? $\sigma^2$ is that the correct symbol? The equation is $E(x-\bar X)^2 / (n-1) = \sigma^2$ is that correct?

Answer 1

As I understand your question, you're interested in the formulas for variance and standard deviation.

Being calculated over the whole population, the variance is $\sigma^2=E[(X-\mu)^2]$. This also can be writen as $\sigma^2=\frac{1}{n}\displaystyle\sum_{i=1}^{n}(x_i - \overline{x})^2$. When calculated over a sample of population this formula introduces a bias (but is still applicable as it tends to be true as the sample size tends to the population size).

The unbiased sample variance is $s^2=\frac{1}{n - 1}\displaystyle\sum_{i=1}^{n}(x_i-\overline{x})^2$. This also can be writen as $s^2=\frac{n}{n-1}E[(X-\mu)^2]$. Now $s$ is so called corrected sample standard deviation (but not unbiased). The true unbiased sample standard deviation is harder to calculate because it depends on the distribution of $X$, look here for details.

Answer 2

In order to find the estimated standard deviation of the population, you would simply take the sum of squares divided by degrees of freedom (n-1) and that is the estimated population variance. In order to get the estimated SD, just take the square root of the variance.