Confidence Intervals · Superchallenge problem

Instead of $\hat p \pm z^*\sqrt{\hat p(1-\hat p)/n}$, a confidence interval for a proportion can be found by collecting every $p$ that the data do not reject: all $p$ with $|\hat p - p| \le z^*\sqrt{p(1-p)/n}$ (the standard error uses $p$, not $\hat p$). With $z^* = 2$, $n = 16$ trials and 8 successes, find the UPPER end of this interval in exact form.