Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 380 babies were​ born, and 342 of them were girls. Use the sample data to construct a 99​% confidence interval estimate of the percentage of girls born. Based on the​ result, does the method appear to be​ effective?

Answer 1

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 $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

In the study 380 babies were​ born, and 342 of them were girls.

This means that $n = 380, \pi = \frac{342}{380} = 0.9$

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.  

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.9 - 2.575\sqrt{\frac{0.9*0.1}{380}} = 0.8604$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.9 + 2.575\sqrt{\frac{0.9*0.1}{380}} = 0.9396$

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.