# How to prove that using the Sample Standard Deviation formula is more accurate than the population standard population formula on a sample dataset?

So, for an assignment for a Python class at college I have to demonstrate that the Sample Standard Deviation formula is more accurate than the population standard population formula on a sample data Set.

So the full original data Set is an array of numbers 5,7,8,3,10,21,4,13,1,0,0,9,17

I get a random sub set of those so, for example, it could be [5, 21, 7]

Using the population formula on that set gets a standard deviation of: 7.118052168020874

Using the sample formula on that set gets a standard deviation of: 8.717797887081348

So does this demonstrate the sample formula being more accurate? I feel like you need to know what the answer is beforehand to know which is more accurate for calculating it.

• You can update your answer with MathJaX formatting, the numbers 12345 do not look as good as $12345$ Dec 6, 2020 at 18:51

The sample formula of the st. dev on $$[5;21;7]$$ is not the value you showed but it is $$\approx 8.72$$

The sample formula is more accurate because $$S^2$$ is an unbiased estimator of $$\sigma^2$$ (unbiased estimator of the population variance)

Proof:

$$\mathbb{E}[S^2]=\mathbb{E}\Bigg[\frac{1}{n-1}\Sigma_i(X_i-\overline{X}_n)^2\Bigg]=\frac{1}{n-1}\mathbb{E}[\Sigma_i(X_i-\mu)^2-n(\overline{X}_n-\mu)^2]=$$

$$=\frac{1}{n-1} \left[\sum_i\mathbb{E}(X_i-\mu)^2-n\mathbb{E}(\overline{X}_n-\mu)^2\right] = \frac{1}{n-1}\Bigg(n\sigma^2-n\frac{\sigma^2}{n} \Bigg) = \sigma^2$$

• thanks, i fixed the error on the sample set calculation. And that explanation helps a lot, thank you! Dec 6, 2020 at 19:16
• I down-voted this. It deals with the variance whereas the question asked about the standard deviation, and it ignores the fact that it is not more accurate if "more accurate" is construed as having a smaller mean squared error (and sometimes it is). Dec 6, 2020 at 19:25