21
Feb 17

## The pearls of AP Statistics 37

### Confidence interval: attach probability or not attach?

I am reading "5 Steps to a 5 AP Statistics, 2010-2011 Edition" by Duane Hinders (sorry, I don't have the latest edition). The tip at the bottom of p.200 says:

For the exam, be VERY, VERY clear on the discussion above. Many students
seem to think that we can attach a probability to our interpretation of a confidence
interval. We cannot.

This is one of those misconceptions that travel from book to book. Below I show how it may have arisen.

### Confidence interval derivation

The intuition behind the confidence interval and the confidence interval derivation using z score have been given here. To make the discussion close to Duane Hinders, I show the confidence interval derivation using the t statistic. Let $X_1,...,X_n$ be a sample of independent observations from a normal population, $\mu$ the population mean and $s$ the standard error. Skipping the intuition, let's go directly to the t statistic

(1) $t=\frac{\bar{X}-\mu}{s/\sqrt{n}}$.

At the 95% confidence level, from statistical tables find the critical value $t_{cr,0.95}$ of the t statistic such that

$P(-t_{cr,0.95}

Plug here (1) to get

(2) $P(-t_{cr,0.95}<\frac{\bar{X}-\mu}{s/\sqrt{n}}

Using equivalent transformations of inequalities (multiplying them by $s/\sqrt{n}$ and adding $\mu$ to all sides) we rewrite (2) as

(3) $P(\mu-t_{cr,0.95}\frac{s}{\sqrt{n}}<\bar{X}<\mu+t_{cr,0.95}\frac{s}{\sqrt{n}})=0.95.$

Thus, we have proved

Statement 1. The interval $\mu\pm t_{cr,0.95}\frac{s}{\sqrt{n}}$ contains the values of the sample mean with probability 95%.

The left-side inequality in (3) is equivalent to $\mu<\bar{X}+t_{cr,0.95}\frac{s}{\sqrt{n}}$ and the right-side one is equivalent to $\bar{X}-t_{cr,0.95}\frac{s}{\sqrt{n}}<\mu$. Combining these two inequalities, we see that (3) can be equivalently written as

(4) $P(\bar{X}-t_{cr,0.95}\frac{s}{\sqrt{n}}<\mu<\bar{X}+t_{cr,0.95}\frac{s}{\sqrt{n}})=0.95.$

So, we have

Statement 2. The interval $\bar{X}\pm t_{cr,0.95}\frac{s}{\sqrt{n}}$ contains the population mean with probability 95%.

### Source of the misconception

In (3), the variable in the middle ($\bar{X}$) is random, and the statement that it belongs to some interval is naturally probabilistic. People not familiar with the above derivation don't understand how a statement that the population mean (which is a constant) belongs to some interval can be probabilistic. It's the interval ends that are random in (4) (the sample mean and standard error are both random), that's why there is probability! Statements 1 and 2 are equivalent!

In all statistical derivations random variables are ex-ante (before the event). No book says that but that's the way it is. An estimate is an ex-post (after the event) value of an estimator. An estimate is, of course, a number and not a random variable. Ex-ante, a confidence interval always has a probability. Ex-post, the fact that an estimate belongs to some interval is deterministic (has probability either 0 or 1) and it doesn't make sense to talk about 95%.

Since confidence levels are always strictly between 0 and 100%, students should keep in mind that we deal with ex-ante variables.