# Is the parameter of interest uniformly distributed over the confidence interval?

Let's assume that we are estimating a parameter of the population (e.g., mean height or weight). We've gathered our sample and calculated confidence intervals around the parameter of interest. Are we, therefore, assuming that this parameter follows uniform distribution over the confidence interval? In simple terms, is the parameter equally probable to be anywhere within this confidence interval?

• There are so many posts about confidence intervals on Cross Validated and yet somehow it's hard to find good references and there is no wiki. Perhaps start here and here to learn more about what confidence intervals are. Apr 3, 2022 at 17:48
• Q is addressed pretty directly here link.springer.com/article/10.3758/s13423-015-0947-8 short answer is definitely 'no', long answer is really interesting :-) Apr 4, 2022 at 1:39

Because you are asking about confidence intervals, I will assume these are frequentist interval estimates.

A 90% t CI, based on the sample mean $$\bar X$$ and the sample standard deviation $$S$$ is of the form $$\bar X \pm t^*S/\sqrt{n},$$ where $$t^*$$ cuts probability $$0.05$$ from the upper tail of Student's t distribution with $$n - 1$$ degrees of freedom.

The analogous probability statement is $$P\left(-t^* < \frac{\bar X - \mu}{S/\sqrt{n}} < t^*\right) = 0.9,$$

in which probability is not uniformly distributed in $$(-t^*,t^*).$$

Consequently, you cannot get a 45% CI just by making that interval half as long: $$(-0.5t^*, 0.5t^*).$$

In order to get a 45% CI, you would need to cut probability $$0.275$$ from each tail of $$\mathsf{T}(\nu=99).$$

Suppose you have the fictitious sample below, of size $$n = 100$$ from a normal population with unknown mean $$\mu$$ and $$\sigma,$$ Then a 90% CI for $$\mu$$ would be $$(59.79, 62.16)$$ and a 45% CI would be $$(60.54, 61.40),$$ as computed using R below.

set.seed(2022)
x = rnorm(100, 60, 7)

CI.90= mean(x) + qt(c(.05,.95),99)*sd(x)/10; CI.90
 59.78559 62.15685

CI.45= mean(x) + qt(c(.275,.725),99)*sd(x)/10; CI.45
 60.54292 61.39953


An alternative way to get the 45% CI, would be to use the confidence interval from the t.test procedure in R.

t.test(x, conf.lev=.45)$conf.int  60.54292 61.39953 attr(,"conf.level")  0.45  An incorrect interval $$(60.38, 61.56)$$ that simply makes CI.90half as long, would not be a true 45% CI for $$\mu.$$ CI.u= mean(x) + .5*qt(c(.05,.95),99)*sd(x)/10; CI.u  60.37841 61.56404  • I understand the experiment in R you provided to prove the concept. However, I have two questions: 1. We treat the population parameter, μ, as a r.v. Is that even ok? 2. If μ doesn't follow uniform distribution over the CI, then μ is probably more likely to be in the center of our CI rather than on the edges. That's a layman's intuition. Is it ok? – Glue Apr 4, 2022 at 18:40 • Treating parameter$\mu$as a random variable would make this a Bayesian problem. If it's a Bayesian problem then you need to refer to credible intervals or Bayesian posterior interval estimates. It might be possible to use a nearly uniform prior distribution, but with a reasonable amount of data, the posterior distribution would rarely be uniform. // In my frequentist approach, the distribution on interval$(-t^*, t^*)\$ is a t distribution with heavier density near the center than at the endpoints. Apr 4, 2022 at 19:51

If you are referring to frequentist statistics, in that case, when you repeat an experiment, and construct the 68% confidence level confidence interval each time, then 68% of those intervals should contain the true value. The true parameter is fixed, and does not follow a distribution - the confidence intervals do.

If you are referring to Bayesian statistics and credible intervals, then the posterior distribution of the parameter depends on the prior and on the likelihood.

• In the OP, the sample has already been gathered, so the ci is fixed. Apr 3, 2022 at 15:49