Tests around Bernoulli Population¶

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 \(p_{0}\)

\[ \begin{aligned} H_{0}: p \leq p_{0} \quad \text{versus} \quad p > p_{0} \end{aligned} \]

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 \(k\)

\[ \begin{aligned} P(X \geq k) = \sum_{i=k}^{n} \binom{n}{i} = \sum_{i=k}^{n} p^{i}(1-p)^{n-i} \end{aligned} \]

which is an increasing function in \(p\). Hence, when \(H_{0}\) is true,

\[ \begin{aligned} P(X \geq k) \leq \sum_{i=k}^{n} p_{0}^{i}(1-p_{0})^{n-i} \end{aligned} \]

and we reject when \(X \geq k^{\*}\) depending on the significance level \(\alpha\)

\[ \begin{aligned} k^{\*} = \text{minimum}\quad k \quad \text{where} \quad \sum_{i=k}^{n} p_{0}^{i}(1-p_{0})^{n-i} \leq \alpha \end{aligned} \]

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

\[ \begin{aligned} \text{p-value} &= P(Bin(n, p_{0}) \geq x)\newline &= \sum_{i=x}^{n}p_{0}^{i}(1-p_{0})^{n-i} \end{aligned} \]

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,

\[ \begin{aligned} \frac{X - np_{0}}{\sqrt{np_{0}(1-p_{0})}} \sim \mathcal{N}(0,1) \end{aligned} \]

and criteria discussed in section hold.