Formula for standard deviation

Question

So I am studying for a final exam and I know, or thought I knew, the formula for standard deviation. I quadruple checked in my notes and the slides for the lecture and see that the formula for the standard deviation when $\sigma$ is unknown is:

$\sqrt((\Sigma d^2 - nd^-) / n - 1)$

But for some reason on the final review solution sheet our professor somehow and some reason replaced $n$ in the numerator with $1/n$ and I have no clue why. So the formula is now:

$\sqrt((\Sigma d^2 - (1/n)d^-) / n - 1)$

My question is, how come we are able to use $1/n$ instead of $n$ and on the test, how can I tell to use $1/n$ as compared to $n$?

Thanks

Answer 1

Yes your professor has a typo. The standard deviation for paired samples is usually written as follows:

$$\sigma = \sqrt{\frac{\sum_{i=1}^n(d_i-\bar d)^2}{n-1}}$$

What you professor is doing is using the trick (or arithmetic if you want to think of it that way) that we can rewrite the numerator as

$$\sum_{i=1}^n(d_i-\bar d)^2 = \sum_{i=1}^nd_i^2-n\bar d^2$$

And thus, the formula for the standard deviation should be

$$\sigma = \sqrt{\frac{\sum_{i=1}^n(d_i-\bar d)^2}{n-1}}= \sqrt{\frac{\sum_{i=1}^n(d_i)^2-n\bar d^2}{n-1}}$$

And finally, it should be $n$ and not $\frac{1}{n}$.

Answer 2

Sample variance is defined as \begin{align} &\sqrt{\frac{1}{n-1}\sum_{i=1}^n\left(d_i-\overline{d}\right)^2}\\ &=\sqrt{\frac{1}{n-1}\sum_{i=1}^n\left(d_i^2+\overline{d}^2-2d_i\overline{d}\right)} \\ &=\sqrt{\frac{1}{n-1}\left[\sum_{i=1}^nd_i^2+\sum_{i=1}^n\overline{d}^2-\sum_{i=1}^n2d_i\overline{d}\right]}\\ &=\sqrt{\frac{1}{n-1}\left[\sum_{i=1}^nd_i^2+n\overline{d}^2-2\overline{d}\sum_{i=1}^nd_i\right]}\\ &\overset{(a)}{=}\sqrt{\frac{1}{n-1}\left[\sum_{i=1}^nd_i^2+n\overline{d}^2-2n\overline{d}^2\right]}\\ &=\sqrt{\frac{1}{n-1}\left[\sum_{i=1}^nd_i^2-n\overline{d}^2\right]}\\ \end{align} $(a)$ follows since sample mean $\overline{d}=\frac{1}{n}\sum_{i=1}^nd_i$. Hence it is $n\overline{d}$ not $(1/n) \overline{d}$