Suppose you use data from a random sample to build a 95% CI around the sample’s mean. Next, suppose you put that sample back into the population. Finally, suppose you get ready to extract a 2^{nd} random sample from the same population, with plans to use the new data to compute just the 2^{nd} sample’s mean. How confident can you be that your 2^{nd} sample’s mean will lie somewhere between the end points of the 1^{st} sample’s 95% CI?

Did you say or think: “95% confident”?

If you did, you’re a bit more confident than you actually should be!

If your 1^{st} sample’s mean were to match perfectly the mean of the population, you *could* be 95% confident that the 2^{nd} sample’s mean would turn out to be “inside” the 1^{st} sample’s 95% CI. That’s because the end points of your CI would coincide with the 2 points in a sampling distribution of the mean that serve to bookend the middle 95% of that distribution’s means. Select a 2^{nd} sample, and its mean would have a 95% chance of landing between those bookends.

Your 1^{st} sample, however, is not likely to have a mean that matches up perfectly with *μ*. This will cause the 1^{st} sample’s 95% CI to be “off-center” in the sampling distribution of means. More than half of the CI will be located on the high (or low) side of that distribution’s midpoint, the true population mean. By having the CI’s end points *not* coincide with the points that bookend the middle 95% of sampling distribution of means, the 95% CI captures less than 95% of those means.

To prove to yourself that a 95% CI based on one sample’s data does not predict, with 95% accuracy, what a 2^{nd} sample’s mean will be like, answer these 2 little questions: (1) How much of a normal distribution lies between the *z-*score points of +1.96 and –1.96? (2) How much of a normal distribution lies between any other pair of *z*-scores that are that same distance (3.92) apart from each other?