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Jackknife confidence interval for the median

In this exercise, we will calculate the jackknife 95% CI for a non-standard estimator. Here, we will look at the median. Keep in mind that the variance of a jackknife estimator is n-1 times the variance of the individual jackknife sample estimates where n is the number of observations in the original sample.

Returning to the wrench factory, you are now interested in estimating the median length of the wrenches along with a 95% CI to ensure that the wrenches are within tolerance.

Let's revisit the code from the previous exercise, but this time in the context of median lengths. By the end of this exercise, you will have a much better idea of how to use jackknife resampling to calculate confidence intervals for non-standard estimators.

This exercise is part of the course

Statistical Simulation in Python

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Exercise instructions

  • Append the median length of each jackknife sample to median_lengths.
  • Calculate the mean of the jackknife estimate of median_length and assign to jk_median_length.
  • Calculate the upper 95% confidence interval jk_upper_ci and lower 95% confidence intervals of the median jk_lower_ci using 1.96*np.sqrt(jk_var).

Hands-on interactive exercise

Have a go at this exercise by completing this sample code.

# Leave one observation out to get the jackknife sample and store the median length
median_lengths = []
for i in range(n):
    jk_sample = wrench_lengths[index != i]
    median_lengths.append(____)

median_lengths = np.array(median_lengths)

# Calculate jackknife estimate and it's variance
jk_median_length = ____
jk_var = (n-1)*np.var(median_lengths)

# Assuming normality, calculate lower and upper 95% confidence intervals
jk_lower_ci = jk_median_length - ____
jk_upper_ci = jk_median_length + ____
print("Jackknife 95% CI lower = {}, upper = {}".format(jk_lower_ci, jk_upper_ci))
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