Answer :
Answer:
The best point estimate of the population proportion p is 0.0251.
The 99% confidence interval for the proportion of adverse reactions is (0.0191, 0.0311).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
In clinical trials, among 4547 patients treated with the drug, 114 developed the adverse reaction of nausea.
This means that [tex]n = 4547, \pi = \frac{114}{4547} = 0.0251[/tex]
The best point estimate of the population proportion p is 0.0251.
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.0251 - 2.575\sqrt{\frac{0.0251*0.9749}{4547}} = 0.0191[/tex]
The upper limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.0251 + 2.575\sqrt{\frac{0.0251*0.9749}{4547}} = 0.0311[/tex]
The 99% confidence interval for the proportion of adverse reactions is (0.0191, 0.0311).