Respuesta :
Answer:
The 99% confidence interval estimate of the percentage of girls born is (86.04%, 93.96%). Considering the actual percentage of girls born is close to 50%, the percentage increased considerably with this method, which means that it appears effective.
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 the study 380 babies were born, and 342 of them were girls.
This means that [tex]n = 380, \pi = \frac{342}{380} = 0.9[/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}} = 0.9 - 2.575\sqrt{\frac{0.9*0.1}{380}} = 0.8604[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.9 + 2.575\sqrt{\frac{0.9*0.1}{380}} = 0.9396[/tex]
As percentages:
0.8604*100% = 86.04%.
0.9396*100% = 93.96%.
The 99% confidence interval estimate of the percentage of girls born is (86.04%, 93.96%). Considering the actual percentage of girls born is close to 50%, the percentage increased considerably with this method, which means that it appears effective.
