Respuesta :
Answer:
The 99% confidence interval for the true population proportion of adults with children is (0.6367, 0.7433).
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 z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
In a sample of 500 adults, 345 had children.
This means that [tex]n = 500, \pi = \frac{345}{500} = 0.69[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a p-value 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}} = 0.69 - 2.575\sqrt{\frac{0.69*0.31}{500}} = 0.6367[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.69 + 2.575\sqrt{\frac{0.69*0.31}{500}} = 0.7433[/tex]
The 99% confidence interval for the true population proportion of adults with children is (0.6367, 0.7433).
