It may be an "easy interpretation," but it doesn't seem like a valid one: "The 95% probability relates to the reliability of the estimation procedure, not to a

It may be an "easy interpretation," but it doesn't seem like a valid one: "The 95% probability relates to the reliability of the estimation procedure, not to a specific calculated interval." - https://en.wikipedia.org/wiki/Confidence_interval#Misunderstandings https://en.wikipedia.org/wiki/Confidence_interval#Misunderst... Similarly, the given example does not seem to represent how statistical confidence intervals work. As just mentioned, the 95% refers to the reliability, not to the probability that a specific interval contains the true parameter value! (In fact, the example is missing which parameter is being estimated.) --- Let's see if we can rework the example to show that the 95% CI is actually well-defined and not arbitrary. To be specific about the problem: - First, the population is given to be the uniform distribution between 0 and 1. - Second, we must choose a parameter to estimate. For simplicity, let's suppose we're estimating the population mean. (Note the true population mean is 0.5.) - Third, each sample is an independently, identically distributed single number from the population (we technically also compute its mean, which is the number itself). --- Here is the challenge: Given an arbitrary sample, can we provide a confidence interval (CI) such that 95% of all CIs computed in this way contain the population mean 0.5? Here is a Python program showing that given a sample mean x, the 95% CI is [x - 0.475, x + 0.475]: from random import random NUM_SAMPLES = 1000000 ERROR = 0.475 POP_MEAN = (1 - 0) / 2 means = [random() for _ in range(NUM_SAMPLES)] num_correct = sum(x - ERROR <= POP_MEAN <= x + ERROR for x in means) print(f'% CIs containing the pop mean: {num_correct / NUM_SAMPLES}') Experimentally, you can see that as NUM_SAMPLES gets larger, the percentage of intervals that contain 0.5 approaches 95%.