Respuesta :
Answer:
The 99% confidence interval estimate of the percentage of girls born is (74.37%, 85.63%).
Usually, 50% of the babies are girls. This confidence interval gives values considerably higher than that, so the method to increase the probability of conceiving a girl appears to be very effective.
Step-by-step explanation:
Confidence Interval for the proportion:
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].
For this problem, we have that:
[tex]n = 335, \pi = \frac{268}{335} = 0.8[/tex]
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}} = 0.8 - 2.575\sqrt{\frac{0.8*0.2}{335}} = 0.7437[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8 + 2.575\sqrt{\frac{0.8*0.2}{335}} = 0.8563[/tex]
For the percentage:
Multiplying the proportions by 100.
The 99% confidence interval estimate of the percentage of girls born is (74.37%, 85.63%).
Usually, 50% of the babies are girls. This confidence interval gives values considerably higher than that, so the method to increase the probability of conceiving a girl appears to be very effective.
