QUESTION $23-1$ POINT .
A medical researcher claims that the proportion of people taking a certain medication that develop serious side effects is $12 \%$. To test this claim, a random sample of 900 people taking the medication is taken and it is determined that 93 people have experienced serious side effects.
The following is the setup for this hypothesis test:
\[
\begin{array}{l}
H_{0}: p=0.12 \\
H_{a}: p \neq 0.12
\end{array}
\]
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:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline$z$ & 0.00 & 0.01 & 0.02 & 0.03 & 0.04 & 0.05 & 0.06 & 0.07 & 0.08 \\
\hline-1.8 & 0.036 & 0.035 & 0.034 & 0.034 & 0.033 & 0.032 & 0.031 & 0.031 & 0.030 \\
\hline-1.7 & 0.045 & 0.044 & 0.043 & 0.042 & 0.041 & 0.040 & 0.039 & 0.038 & 0.038 \\
\hline-1.6 & 0.055 & 0.054 & 0.053 & 0.052 & 0.051 & 0.049 & 0.048 & 0.047 & 0.046 \\
\hline-1.5 & 0.067 & 0.066 & 0.064 & 0.063 & 0.062 & 0.061 & 0.059 & 0.058 & 0.057 \\
\hline-1.4 & 0.081 & 0.079 & 0.078 & 0.076 & 0.075 & 0.074 & 0.072 & 0.071 & 0.069 \\
\hline
\end{tabular}
Provide your answer below:
P-value=
Answer
The p-value for this hypothesis test for a proportion is \(\boxed{0.124}\).
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 \(\boxed{0.124}\).
