Let's say you are estimating a population proportion, which you model as binomial. One source of error already is using the normal approximation to the binomial when getting your critical values. But what bothers me more is that the theoretically sound interval uses the true population proportion in computation of the interval width. This is usually approximated by the sample proportion, but doesn't this no longer make the confidence level accurate? (It seems like a pretty Bayesian assumption for a frequentist approach to get away with.)

As a common tactic, I see people use the upper bound for the population variance (by assuming the proportion is $1/2$), and use that to determine their intervals. Is this preferable to using the sample proportion to estimate the population variance? At least in this upper bounding case, we can say with mathematical soundness, that our confidence level is at least $90\%$confident (assuming normal perfectly approximates binomial).

  • 1
    $\begingroup$ Are you asking about the mathematical definition of "90% confidence intervals"? Or are you questioning the application of that definition in the special case of estimating a population proportion? $\endgroup$
    – Lee Mosher
    Commented May 6, 2016 at 17:45

2 Answers 2


Traditional Wald Confidence Interval. You are asking about the 'coverage probability' if traditional (sometimes called 'Wald') confidence interval (CI) for binomial success probability $\pi,$ based on $n$ trials of which $X$ are Successes. One estimates $\pi$ as $p = X/n$. Then a "95%" CI fir $\pi$ is of the form $$ p \pm 1.96\sqrt{\frac{p(1-p)}{n}}.$$

You are correct to be suspicious that the coverage probability may not be 95% as claimed: first, because it is based on the assumption that $\frac{p - \pi}{\sqrt{\pi(1 - \pi)/n}} \sim Norm(0, 1);$ second, because it estimates the standard error $\sqrt{\pi(1-\pi)}/n$ as $\sqrt{p(1-p)/n}.$

Coverage Probability. One can check the coverage probability in specific cases. Suppose $n = 25.$ Then there are $26$ possible CIs of the displayed form depending on possible values $X = 0, 1, \dots, 25.$ Some of these intervals include a particular value of $\pi$ and some do not. For example, if $\pi = 0.30,$ the CIs resulting from $4 \le X \le 12$ cover $\pi = 0.30,$ and the rest do not. Because $P(4 \le X \le 12|\pi = .30) = 0.9593,$ the coverage probability is almost exactly 95% as claimed.

However, if $\pi = 0.31,$ then only the CIs corresponding to $5 \le X \le 12$ include $\pi = 0.31$ and the coverage probability is $P(5 \le X \le 12 |\pi = .31) = 0.9024,$ so the coverage probability is nearer 90% than 95%.

Because there are "lucky" and "unlucky" values of $\pi,$ it seems worthwhile to find coverage probabilities for a sequence of 2000 values of $\pi$ in $(0,1).$ Plotting coverage probability against $\pi$, we see that there are many more 'unlucky' values of $\pi$ than 'lucky' ones. Heavy blue dots show the coverage probabilities for $\pi = .30$ and $ \pi = .31,$ mentioned above.

enter image description here

Agresti 'Plus-4' Interval. One cure (best for the 95% level), is to use $n^+ = n + 4$ and $p^+ = (X + 2)/n^+$ instead of $n$ and $p$ in the displayed formula above. This essentially means we append two imaginary successes and two imaginary failures to our observations. Hence it is sometimes called a 'Plus-4' interval. This idea is due to Agresti, and is based on sound (but somewhat complex) reasoning. Here is a graph of coverage probabilities of such Agresti-style 95% confidence intervals for $n = 25.$

enter image description here

Bayeaian-based Interval. Yet another type of CI for $\pi$ is based on a Bayesian argument in which the prior distribution carries very little information. It is based on taking quantiles .025 and .975 of the distribution $Beta(x +1, n-x +1).$ Evaluation requires software. If $n = 25$ and $X = 5$ Successes, then the a 95% interval of this type is is computed in R statistical software as $(0.09, 0.39).$

 qbeta(c(.025, .975), 5 + 1, 20 + 1)
 ## 0.08974011 0.39350553

The corresponding graph of coverage probabilities for this type of CI is shown below.

enter image description here


  1. Agresti, A.; Coull, B.A.: Approximate is better than "exact" for interval estimation of binomial proportions, The American Statistician, 52:2 (1998), pages 119-126.

  2. Brown, L.D.; Cai, T.T; Dasgupta, A.: Interval estimation for a binomial proportion, Statistical Science, 16:2 (2001), pages 101-133

  • $\begingroup$ Very elucidating, thank you. $\endgroup$ Commented May 7, 2016 at 17:55

Your have correctly identified two major sources of error: sampling error and model error. The former can be made arbitrarily small by taking an increasingly large, random sample; the latter cannot be removed by sampling.

To answer your question, as with most models, they not 100% accurate, so all inference is approximate. You will not know with 100% certainty that your confidence is actually $\geq 90$%.

However, there are several mitigating factors at play here so that statisticians are not unduly bothered by the fact that their models are wrong:

  1. If the sample came from a distribution with finite mean and variance, then we can use the results of the Central Limit Theorem and Berry-Esseen to decide if a normal approximation is appropriate.
  2. If this is not good enough, we can rely on boostrapping using the empirical distribution as the maximum likelihood estimate of the actual distribution. This is possible in part due to the Glivenko-Cantelli Theorem.
  3. We can check the fit of data against a hypothesized distribution using goodness of fit tests. This is not an ideal way to go about things, but you can determine if the observed data are consistent with a hypothesized distribution within certain Type I error. Note: this is really a check for consistency, not truth.

So, we have several ways to check that the "model error" part is small, and we can focus on using mathematical statistics to describe the sampling error.

  • $\begingroup$ Considerable experience shows that bootstrapping the binomial success probability is a bad idea. Bootstrapping is useful for many things, but not so useful here. $\endgroup$
    – BruceET
    Commented May 6, 2016 at 19:23
  • $\begingroup$ @BruceET I was answering OP's general question about how we are confident about confidence. OP was using binomial inference as an example of the general issue that we often don't know if the binomial model is even true, so I was pointing out ways statistician's verify that their model assumptions are not too far off. $\endgroup$
    – user237392
    Commented May 6, 2016 at 19:32
  • $\begingroup$ Thank you for the very good answer. I'm learning more about bootstrapping/goodness of fit now and it seems quite interesting. $\endgroup$ Commented May 7, 2016 at 17:58

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