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 $\theta$?