Why is the confidence interval considered a random interval?

I've been reading a lot on confidence intervals lately and I keep seeing statements such as: "A 95% confidence interval is a random interval that contains the true parameter 95% of the time" or "A confidence interval is a random variable because x-bar (its center) is a random variable."

Why is the confidence interval considered random? If it's truly random then why bother with confidence intervals at all? Am I missing something here?

• maybe section 3 of this can help: stats.stackexchange.com/questions/167972/… – user83346 Aug 2 '16 at 6:22
• I think that this question conflates "random" with "unknowable." Probability and statistics are about studying and quantifying randomness. We can't know what the next die roll will show, but we can make assessments of which outcomes are more or less likely, and hence derive confidence intervals. – Sycorax Aug 3 '16 at 1:29

Why is the confidence interval considered random?

You just stated why in your question! You quoted this:

"A confidence interval is a random variable because x-bar (its center) is a random variable."

(In this case, it's presumably an interval for the mean, but the reasoning carries over to other confidence intervals.)

The sample mean is a statistic -- a quantity you calculate from the sample. Because random samples from some population are, well, random, things calculated from them are also going to be random.

Consider: If you drew a second sample from the same population would you have the same observations?

Would the sample mean be the same in both samples? Would the sample standard deviation be the same in both samples? The largest observation? The lower quartile?

No, they vary from sample to sample; indeed they're also random.

A confidence interval is also based on the random sample, so it, too, is a statistic (e.g. define it in terms of its endpoints) and it, too, is random.

If it's truly random then why bother with confidence intervals at all?
Am I missing something here?

Well presumably you'd like to use the data to calculate your interval. After all, it's the thing we have that tells us something about the population we drew the sample from.

If you're using the data - a random sample of your population - then useful quantities you calculate from it will also be random, including confidence intervals.

Random doesn't mean "ignores your data" -- for example a sample mean tells us about our population mean, and our sample standard deviation can be used to help us work out how far the sample mean will tend to be from the population mean.

In fact, we rely on the randomness - we exploit it to get the best possible use of information from our sample. Without random sampling, our intervals wouldn't necessarily tell us much of anything.

[You might like to ponder whether there might be a way to get an interval for a population quantity that is simultaneously reasonably informative and not random.]

Several tentative approximations:

1. Random variables are not random. They are deterministic functions from the outcome to the real line, $X: \Omega \rightarrow \mathbb R$. So you run a random experiment (the experiment, say tossing a coin, is random in the sense that we don't have a formula to return the outcome a priori), and get an outcome; run it again, and get another outcome. Soon you have a sample, and you happen to be interested in a parameter, say the proportion of heads, $p$: you are mapping something like $\small \{H,T,H,H,H,T,H,T,T,T\}$ to the interval $[0,1]$ to get an estimate of the parameter $p$ based on your sample, using the simple formula, $\frac{\text{no.heads}}{\text{total}}$, a deterministic formula. You may label this estimate, $\hat p$.
2. Confidence interval: From this point estimate, you can calculate the CI with some formula, such as, $\hat p\,\pm\,1.96\,\sqrt{\frac{\hat p\,(1-\hat p)}{n}}$. Again deterministic, meaning (crazy nomenclature), a random variable... or two: one for the lower bound, and the other for the upper bound. So effectively you have unfolded the point estimate into two point estimates, based on some underling distributional assumptions (normal approximation), completely unrelated to the specific realization that your sample represents.
3. This interval can contain $p$ or not. Again, think about the point estimate - it can fall very far from the true parameter, $p$, and affect the CI accordingly.
4. But there is one saving grace, which is at the same time a painful yoga position: If you were to repeat this process time and time again, and get many $\hat p$ estimates with their respective confidence intervals, the true parameter $p$ would be contained in $95\%$ of them.

The confidence interval does not tell you that with $95\%$ probability the true proportion is contained between its bounds, which is mind boggling. It is, instead nothing more than "an elaboration" on the sample based on things like the CLT. As such it is "random" (wink, wink).

If you want the probability that the parameter $p$ is contained within an certain interval, you have to change party affiliation, and look up credible intervals under the apparently more satisfying Bayesian paradigm.

• – user83346 Aug 3 '16 at 6:22
• Yes, I had read your post previously. Very useful. – Antoni Parellada Aug 3 '16 at 6:24