The standard error for this test is
\begin{equation*}
\sigma_e = \frac{0.08}{\sqrt{20}} \approx 0.01789
\end{equation*}
and using the t-distribution with n-1 = 19 degrees of freedom yields a t-statistic of
\begin{equation*}
t = \frac{\overline{x} - 16}{\sigma_e} = \frac{16.05-16}{0.01789} \approx 2.795.
\end{equation*}
To compute the p-value,
\begin{equation*}
P(t \gt 2.795) \approx 0.0058.
\end{equation*}
Since this p-value is less than our significance level \(\alpha = 0.05\) (by a lot) then you can reject the null hypothesis and accept the alternate. It is safe therefore to say that customers can expect at least 16 ounces! However, note that some folks will still be stiffed since the standard deviation of 0.08 certainly means that some bottles have less than 16.05-0.08 ounces of beverage.