To determine E carefully, note that from the central limit theorem
\begin{equation*}
\frac{Y-np}{\sqrt{np(1-p)}} = \frac{\tilde{p} - p}{\sqrt{p(1-p)/n}}
\end{equation*}
is approximately standard normal for large n. Presuming that \(\tilde{p} \approx p\) and replacing the unknown p terms on the bottom with \(\tilde{p}\) gives
\begin{equation*}
z = \frac{\tilde{p} - p}{\sqrt{\tilde{p}(1-\tilde{p})/n}}
\end{equation*}
where z is a standard normal distribution variable. So, using the central limit theorem and the standard normal distribution, you can find the value \(z_{ \alpha/2}\) where
\begin{equation*}
P( -z_{ \alpha/2} \lt z \lt z_{ \alpha/2}) = 1 - \alpha
\end{equation*}
\begin{equation*}
P( -z_{ \alpha/2} \lt \frac{\tilde{p} - p}{\sqrt{\tilde{p}(1-\tilde{p})/n}} \lt z_{ \alpha/2}) = 1 - \alpha
\end{equation*}
or by rearranging the inside inequality
\begin{equation*}
P( \tilde{p} - z_{ \alpha/2}\sqrt{\tilde{p}(1-\tilde{p})/n} \lt p \lt \tilde{p} + z_{ \alpha/2}\sqrt{\tilde{p}(1-\tilde{p})/n}) = 1 - \alpha.
\end{equation*}
Setting \(E = z_{ \alpha/2}\sqrt{\tilde{p}(1-\tilde{p})/n}\) gives a way to determine a confidence interval centered on \(\tilde{p} = \frac{Y}{n}\) for p with "confidence level" \(1-\alpha\text{.}\)