Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 290 babies were​ born, and 261 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 between 85.46% and 94.54%.

Usually, babies are equally as likely to be boys or girls. Here, it is desired to increase the probability of conceiving a girl, which is achieved, considering the lower bound of the confidence interval is considerably above 50%.

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 $\pi$, 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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 290, \pi = \frac{261}{290} = 0.9$

99% confidence level

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

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.9 - 2.575\sqrt{\frac{0.9*0.1}{290}} = 0.8546$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.9 + 2.575\sqrt{\frac{0.9*0.1}{290}} = 0.9454$

Percentage:

Proportion multplied by 100.

0.8546*100 = 85.46%

0.9454*100 = 94.54%

The 99​% confidence interval estimate of the percentage of girls born is between 85.46% and 94.54%.

Based on the​ result, does the method appear to be​ effective?

Usually, babies are equally as likely to be boys or girls. Here, it is desired to increase the probability of conceiving a girl, which is achieved, considering the lower bound of the confidence interval is considerably above 50%.