Respuesta :
Answer:
a. (0.313,0.447)
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].
Researchers investigating a new drug selected a random sample of 200 people who are taking the drug. Of those selected, 76 indicated they were experiencing side effects from the drug.
This means that [tex]n = 200, \pi = \frac{76}{200} = 0.38[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.38 - 1.96\sqrt{\frac{0.38*0.62}{200}} = 0.313[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.38 + 1.96\sqrt{\frac{0.38*0.62}{200}} = 0.447[/tex]
The correct answer is given by option A.
