Respuesta :
Answer:
The 99% confidence interval is (0.0787, 0.1613).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence interval 1-\alpha, 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].
For this problem, we have that:
The survey included 411 adults between the ages of 55 and 64, and 50 of them said that they had used online dating. This means that [tex]n = 411, \pi = \frac{50}{411} = 0.12[/tex]
The 99% confidence interval is
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)}{411}} = 0.12 - 2.575\sqrt{\frac{0.12*0.88}{411}} = 0.0787[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{411}} = 0.12 + 2.575\sqrt{\frac{0.12*0.88}{411}} = 0.1613[/tex]
The 99% confidence interval is (0.0787, 0.1613).
