A medical assistant claims the proportion of people who take a certain medication and develop serious side effects is different than $12 \%$. To test this claim, a random sample of 900 people taking the medication is taken and it is determined 93 of them have experienced serious side effects. The following is the setup for this hypothesis test: \[ \begin{array}{l} H_{0}: p=0.12 \\ H_{a}: p \neq 0.12 \end{array} \] Round $p^{\prime}$ to four decimal places. Find the $p$-value for this hypothesis test for a proportion and round your answer to 3 decimal places. The following table can be utilized which provides areas under the Standard Normal Curve:

Step 1 ：First, we need to calculate the sample proportion $p'$, which is the proportion of people in the sample who have experienced serious side effects. This is given by the formula $p' = \frac{x}{n}$, where $x$ is the number of people who have experienced serious side effects and $n$ is the total number of people in the sample. Substituting the given values, we get $p' = \frac{93}{900} = 0.1033$.

Step 2 ：Next, we need to calculate the standard error of the proportion, which is given by the formula $SE = \sqrt{\frac{p(1-p)}{n}}$, where $p$ is the proportion under the null hypothesis and $n$ is the total number of people in the sample. Substituting the given values, we get $SE = \sqrt{\frac{0.12(1-0.12)}{900}} = 0.0104$.

Step 3 ：Then, we need to calculate the test statistic, which is given by the formula $Z = \frac{p' - p}{SE}$, where $p'$ is the sample proportion, $p$ is the proportion under the null hypothesis, and $SE$ is the standard error of the proportion. Substituting the calculated values, we get $Z = \frac{0.1033 - 0.12}{0.0104} = -1.61$.

Step 4 ：Finally, we need to find the $p$-value for this hypothesis test. The $p$-value is the probability of obtaining a test statistic as extreme as the one calculated, assuming the null hypothesis is true. Since this is a two-tailed test, we need to find the area under the standard normal curve to the left of $-1.61$ and to the right of $1.61$. Using the standard normal table, we find that the area to the left of $-1.61$ is approximately $0.0537$ and the area to the right of $1.61$ is also approximately $0.0537$. Therefore, the $p$-value is $2 \times 0.0537 = 0.107$.

Step 5 ：Since the $p$-value is greater than the significance level of $0.05$, we fail to reject the null hypothesis. This means that there is not enough evidence to support the medical assistant's claim that the proportion of people who take the medication and develop serious side effects is different than $12\%$.

Step 6 ：The final answer is \(\boxed{0.107}\).