I have 10 iid r.v. with Bernoulli distribution with $X_{i} = 1$ for a positive result. I'm given $\sum_{i=1}^{10} X_i= 1$ and need to find a two-sided 99% confidence interval for $\theta$.

So $\alpha = 0.01$ and $z_{\alpha/2} = 2.575$ based on the standard normal distribution. The variance is given $\sigma^{2} = \hat{\theta}(1-\hat{\theta})$, with $\theta = 1/10$ from the given information. The confidence interval is: $$ \left(\hat{\theta} - z_{\alpha/2}\frac{\sigma}{\sqrt{n}}, \hat{\theta} + z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\right) $$ so my confidence interval is $(-0.143, 0.343)$. Have I messed up a calculation somewhere? I'm not sure what this confidence interval tells me ... perhaps that the sample size is too small to really get any useful information on $\theta$?

  • 3
    $\begingroup$ Your interval is generated assuming $\hat \theta$ is normally distributed. In this situation it's not remotely normal, so the interval isn't useful. Read about binomial intervals $\endgroup$
    – Glen_b
    Commented Nov 30, 2013 at 9:36
  • 1
    $\begingroup$ Beside Glen_b's comment: It's perhaps the aim of this exercise to underline the fact that normal approximations are not very good for small samples of (possibly) asymmetrically distributed random variables. Note that your confidence interval is equivalent to (0, 0.343), since $\theta$ can't be negative. $\endgroup$
    – Michael M
    Commented Nov 30, 2013 at 10:09
  • $\begingroup$ A similar question: stats.stackexchange.com/questions/123779/… $\endgroup$ Commented May 5, 2019 at 13:10

1 Answer 1


When the procedure you have used to calculate a confidence interval gives an interval including impossible values, that is an indication of problems with the method. In your case, you have used a normal (central limit theorem-based) CI with so few observations that the approximation is invalid. You can test that easily in R, say:

Bernoulli loglikelihood function

We have plotted the loglikelihodd function for your case. If this is (close to) quadratic, the normal approximation will be good. That is clearly not the case here!

As @Glen_b says in a comment, you need to read up on binomial confidence intervals, see for instance Wikipedia or Binomial confidence interval estimation - why is it not symmetric?.

R code used for the plot:

make_loglik <- function(n, x) {
     function(p) dbinom(x, n, p, log=TRUE)  

loglik <- make_loglik(10, 1)

plot(loglik, from=0, to=1, xlab="p", col="blue", main="log likelihood function\nBernoulli, n=10, x=1")

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