Suppose we have a set of $n$ samples and we want to test how many of them satisfy a property (or equivalently, success). Let $p$ be the fraction of population satisfying he property and we want to check if this equals $p_{0}$ \begin{align} H_{0}: p \leq p_{0} \quad \text{versus} \quad p > p_{0} \end{align}
i.e., we reject this batch if the size of sample not satisfying the property (defective) is more than some predefined quantity/significance $p_{0}$.
We reject when the defectives in the sample ($X$) are more than a threshold $k$ \begin{align} P(X \geq k) = \sum_{i=k}^{n} \binom{n}{i} = \sum_{i=k}^{n} p^{i}(1-p)^{n-i} \end{align}
which is an increasing function in $p$. Hence, when $H_{0}$ is true, \begin{align} P(X \geq k) \leq \sum_{i=k}^{n} p_{0}^{i}(1-p_{0})^{n-i} \end{align}
and we reject when $X \geq k^{*}$ depending on the significance level $\alpha$ \begin{align} k^{*} = \text{minimum}\quad k \quad \text{where} \quad \sum_{i=k}^{n} p_{0}^{i}(1-p_{0})^{n-i} \leq \alpha \end{align} because there can be multiple $k$ which satisfy the above equation, and we want to reject $H_{0}$ as soon as the number of defectives in sample $X$ is more than the minimum $k$.
The test can also be done using p-value \begin{align} \text{p-value} &= P(Bin(n, p_{0}) \geq x)\newline &= \sum_{i=x}^{n}p_{0}^{i}(1-p_{0})^{n-i} \end{align}
where $x$ is the count of defects in the sample. We reject $H_{0}$ at any $\alpha >$ p-value since in that situation the number of defects required will be much less than $x$.
For large $n$, $X$ will behave like a normal distribution and when $H_{0}$ is true, \begin{align} \frac{X - np_{0}}{\sqrt{np_{0}(1-p_{0})}} \sim \mathcal{N}(0,1) \end{align}
and criteria discussed in section hold.