I have a set of directions in degrees in which I want to find the average and the distribution error. At the end of the day, I want to say something like:
"Given these 100 directions, the average is 5 $\pm$ 2$^{\circ}$"
There is one complications being that I have a 90$^{\circ}$ ambiguity of direction such that a N-S and E-W line of direction should be treated as equal in the averaging process. I have gotten around this by multiplying all my angles by 4 and then doing the vector averaging:
Let $J = 4$ be the multiplier and $\theta$ be a vector of directions in radians with length $N$. So,
$ \phi_n = J\theta_n $
$\bar{c} = \frac{1}{N}\sum_{n=1}^N($cos$(\phi_n))$ is the average cosines for all $n$.
$\bar{s} = \frac{1}{N}\sum_{n=1}^N($sin$(\phi_n)) $ is the average sines for all $n$.
$\bar{R} = \sqrt{\bar{c}^2+\bar{s}^2}$ is the resultant.
$\bar{\theta} = \frac{1}{J}$tan$^{-1}(\bar{s}/\bar{c})$ is the average direction in radians.
Now, how do I get an interval of plus-minus?
I have found the following equation for standard deviation:
$\sigma = \frac{1}{J}\sqrt{-2 ln ( \bar{R})}$ which gives a result in radians.
However, this number is always very large, even when I have little scatter in my data. Am I making a mistake somewhere or is radial standard deviation a poor measure of confidence intervals?
Any help is appreciated.
Thanks