Jackknife Standard Error

Problem

Given leave-one-out estimates theta_i, the jackknife standard error is sqrt((n - 1) / n * sum_i (theta_i - mean(theta))^2).

Given the estimates, return the standard error.

Hint

The n - 1 factor compensates for leave-one-out dependence.

case 1

in: [[1,2,3]]

expected: 1.1547005383792515

case 2

in: [[5,5]]

expected: 0

case 3

in: [[0.5,0.6,0.7,0.8]]

expected: 0.19364916731037085

Starter codePython

def jackknife_standard_error(estimates):
    # Your code here
    pass

3 test cases · run your code against them in the editor