Chapter 9, Problem 95CRE

Why Is $n-1$ in the Sample Standard Deviation? Why do we calculate $s$ by dividing by $n-1$, rather than just $n$ ?

$s^2=\frac{{\displaystyle \sum {\left(x-\bar{x}\right)}^2}}{n-1}$

TI-84 output

The reason is that if we divide by $n-1$, then $s^2$ is an unbiased estimator of ${\sigma }^2$, the population variance.

We want to show that $s^2$ is an unbiased estimator of ${\sigma }^2$, sigma squared. The mathematical proof that this is true is beyond the scope of an introductory statistics course, but we can use an example to demonstrate that it is.

First we will use a very small population that consists only of these three numbers: 1, 2, and 5.

You can determine that the population standard deviation, $\sigma$, for this population is $1.699673$ (or about $1.70$ ), as shown in the TI-84 output. So the population variance, sigma squared, ${\sigma }^2$, is $2.888889$ (or about $2.89$ ).

Now take all possible samples, with replacement, of size 2 from the population, and find the sample variance, $s^2$, for each sample. This process is started for you in the table. Average these sample variances $\left(s^2\right)$, and you should get approximately $2.88889$. If you do, then you have demonstrated that $s^2$ is an unbiased estimator of ${\sigma }^2$, sigma squared.

Show your work by filling in the accompanying table and show the average of $s^2$.