# The problem

My professor made a statement today which I really hated. We were chatting about confidence intervals. Suppose we have some normally distributed population p ~ N(10, 1) with a mean of 10 and a standard deviation of 1. Let X be a size-100 random sample from p. Suppose we compute the 95% confidence interval I = (u, v) for the mean of p. The question is: what is the probability that mean(p) is inside I?

Let’s look at an example:

library(BSDA)    # For Z test

set.seed(8)
p <- data.frame(val=rnorm(1000000, 10, 1))
mean(p$val) sd(p$val)
N <- 100    # Sample size
X <- sample(p\$val, N, replace=F)
z.test(X, sigma.x=1, mu=10, conf.level=0.90)    # 90% CI, I1
z.test(X, sigma.x=1, mu=10, conf.level=0.95)    # 95% CI, I2


This outputs the following (truncated) text describing two z-Tests, one with a 90% confidence interval and one with a 95% confidence interval.

 Population mean: 9.99916568117066
 Population std:  9.99916568117066

One-sample z-Test

data:  X
z = 1.8306, p-value = 0.06716
alternative hypothesis: true mean is not equal to 10
90 percent confidence interval:
10.01857 10.34754
sample estimates:
mean of x
10.18306

One-sample z-Test

data:  X
z = 1.8306, p-value = 0.06716
alternative hypothesis: true mean is not equal to 10
95 percent confidence interval:
9.987062 10.379055
sample estimates:
mean of x
10.18306


For simplicity, here is the population mean and the confidence intervals:

 Population mean: 9.99916568117066

90 percent confidence interval:
10.01857 10.34754

95 percent confidence interval:
9.987062 10.379055


So Prof asks “What’s the probability that the population mean is in the 90% confidence interval?” and everyone says “90%”. Prof is all “lulz nah it’s 0”, and we’re like “well yeah, obviously, but you know what we mean..” and that’s when he drops one on us: he says that a 90% confidence interval doesn’t have a 90% chance of containing the population mean. It has either a 0% chance or 100% chance.