A factory produce bottles of milk. In means, each bottles contains $\mu$ liters of milk with variance $\sigma ^2$ ($\mu$ and $\sigma ^2$ are unknown). We take a sample $\{x_1,...,x_n\}$ of size $n$. Let $\mu_s$ be the means and $\sigma _s^2$ be the variance of the sample. In my lectures, they say that $$\mathbb P\left\{\mu\in \left[\mu_s-1,96\frac{\bar{\sigma}_s }{\sqrt n},\mu_s+1,96\frac{\bar \sigma_s }{\sqrt n}\right]\right\}=0,95,$$ where ${\bar \sigma_s}^2 :=\frac{n}{n-1}\sigma _s^2$. I asked my TA (Teacher Assistant) why taking $\bar \sigma_s $ instead of $\sigma _s$, but he couldn't really answer. I saw in my lecture that they mentioned as a remark : "We use $\bar \sigma_s $ instead of $\sigma _s$, because $\bar \sigma_s $ is a better estimates of $\sigma $".
My Questions :
In which sense is it a better estimates ?
Why is it a better estimates ?
It's a better estimate in the sense that it's unbiased. (In more advanced statistics, sometimes we are willing to tolerate some bias is this sufficiently reduces variance.)
The Wikipedia link shared by Henry demonstrates the unbiasedness. (It might be a good exercise to try showing it yourself.)