So, in this case we have
\begin{equation*}
\sigma = \sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.2 \cdot 0.8}{100}} = 0.04
\end{equation*}
and the z-statistic (using the normal distribution) for the sample statistic of p = 0.27 is
\begin{equation*}
z = \frac{0.27 - 0.2}{0.04} = 1.75.
\end{equation*}
Remember, the alternate hypothesis has two tails so to determine the P value we need to determine from the normal distribution
\begin{equation*}
P(Z \gt 1.75) + P(Z \lt -1.75)
\end{equation*}
and find that this has probability approximately 0.0392 + 0.0392 = 0.0784. However, this P value is greater than our significance level \(\alpha = 0.05\) so we cannot reject the null hypothesis at the 5 percent significance level. However, if we had chosen initially to use a 10 percent significance level then we would have rejected the null hypothesis and accepted the alternate.