[Robk]

Let X be your earn per tournament, n the number of tournaments you've played, s your sample standard deviation. Then your 100*(1-a)% confidence interval is approximately

X +/- z(a/2)*s*(1/SQRT(n))

where z(a/2) is the value which has an area of a/2 to its right on a standard normal curve.

For example you've played 150 tournaments and your total result is 16 per tournament, with sample standard deviation 60. Then letting a = .05, your 100*(1-.05) = 95% confidence interval is

16 +/- z(.025)*60*(1/12.25) = 16 +/- 1.96*60*(1/12.25)

= 16 +/- 9.6 = (6.4, 25.6)

Normal tables abound on the web, but here are a few common z values

z(.1) = 1.282
z(.05) = 1.645
z(.025) = 1.96
z(.005) = 3.09

As usual, smart posters please tell me if I messed up.