The hypothesis testing for population proportion can be conducted in a similar manner. Here are some examples to depict it.
Example: A supplier of semiconductor wafers claims that of all the wafers he supplies, no more than 10% are defective. A sample of 400 wafers is tested, and 50 of them, or 12.5%, are defective. Can we conclude that the claim is false?
Sol: First of all, the supplier claims that the percentage of defective wafer is less than 10%. Let us denote the proportion of defective wafer by $p$, then the suppliers claim is: $p \leq 0.1$. As the sample size is large, we can apply the central limit theorem and can say that the sample proportion comes from a normal distribution denoted as:
$$\widehat{p} \sim N\bigg(p, \frac{p(1-p)}{n}\bigg)$$
where $p$ is the population proportion and $n = 400$ is the sample size. Our goal is to find that whether the supplier’s claim is false or not. Hence, the null and alternate hypothesis can be defined as:
$$H_0: p \leq 0.1$$
$$H_A: p > 0.1$$
To perform hypothesis test, we assume that null hypothesis is true and hence $p=0.1$. The null distribution comes out to be:
$$\widehat{p} \sim N(0.1, 2.25 \times 10^{-4})$$
The standard deviation and the observed value of $\widehat{p}$ is $\sqrt{2.25 \times 10^{-4}} = 0.015$ and $50 /400 = 0.125$. Hence, the z-score is:
$$z = \frac{0.125 - 0.1}{0.015} = 1.67$$
The corresponding p-value is 0.0475. For a significance level of 5%, we can reject the null hypothesis and say that the supplier’s claim is false.
The only necessary condition to conduct the mentioned test is the fact that sample size must be large, or there should be more than 10 outcomes for both the classes, i.e. $np_0 > 10$ and $n(1-p_0) > 10$.
### Reference :https://www.mheducation.com/highered/product/statistics-engineers-scientists-navidi/M0073401331.html