We showed that E(M) = kk2 n OR n = = So now, let us say we repeat the experiment R-times independently and generate R measurements M, M2, M2, ...MR. We report kk2 E(M) kk2 R 1/2 2 Mi i=1 R as our estimator of n. The goal is to find a high probability confidence interval for n. A good idea is to find confidence interval for M, i.e. we want to say with probability 1 - 6, M 1 1 [E(M) a, E(M) + a], here a is a function of 8 (uncertainty), variance Var(M), and of course R (more repetitions are sharper). R R i=1 R - - Can you find an interesting a (use Chebyshev). It could be of the form contant x std (where std is the standard deviation or square root of some variance) 15 points - Comment on how you will use the formula to find the number of repetitions to get to a particular fraction f of error, compared to E(M), with a probability of failure less than 0.05. 2 points - - How does the above translate into an interval for . Can we argue some cases where estimation is really hard. 3 points