We showed that E(M) = kk2 n OR n = = So now, let us...

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We showed that E(M) = kık2 n OR n = ñ = So now, let us say we repeat the experiment R-times independently and generate R measurements M₁, M2, M2, ...MR. We report kık2 E(M) k₁k2 R 1/2 Σ2₁ 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 ₁M₁, i.e. we want to say with probability 1 - 6, ₁ M₁ € 1 Σ 1 Σ Μ ε [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 ñ. Can we argue some cases where estimation is really hard. 3 points We showed that E(M) = kık2 n OR n = ñ = So now, let us say we repeat the experiment R-times independently and generate R measurements M₁, M2, M2, ...MR. We report kık2 E(M) k₁k2 R 1/2 Σ2₁ 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 ₁M₁, i.e. we want to say with probability 1 - 6, ₁ M₁ € 1 Σ 1 Σ Μ ε [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 ñ. Can we argue some cases where estimation is really hard. 3 points

Answer rating: 100% (QA)

Based on the formula given it seems that the quantity represents the mean of the quantity kik2 over R independent measurements M1 M2 MR Specifically i... View the full answer