Step 1 :Given that the null hypothesis \(H_{0}: p=0.12\) and the alternative hypothesis \(H_{a}: p \neq 0.12\).
Step 2 :A random sample of 900 people taking the medication is taken and it is determined that 93 people have experienced serious side effects.
Step 3 :Calculate the observed proportion \(p_{obs} = \frac{x}{n} = \frac{93}{900} = 0.10333333333333333\).
Step 4 :Calculate the standard error \(se = \sqrt{\frac{p_{null}(1-p_{null})}{n}} = \sqrt{\frac{0.12(1-0.12)}{900}} = 0.010832051206181281\).
Step 5 :Calculate the z-score \(z = \frac{p_{obs} - p_{null}}{se} = \frac{0.10333333333333333 - 0.12}{0.010832051206181281} = -1.5386436372416588\).
Step 6 :Using the standard normal distribution table, find the probability of obtaining a z-score as extreme as, or more extreme than, the observed z-score. This will be our p-value.
Step 7 :The p-value for this hypothesis test for a proportion is \(\boxed{0.124}\).