Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 340 babies were​ born, and 289 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?(__)<(__) ​(Round to three decimal places as​ needed.)Does the method appear to be​effective?a)No​,the proportion of girls is not significantly different from 0.5.b)Yes, the proportion of girls is significantly different from 0.5.

Answer 1

Answer:

The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).

b)Yes, the proportion of girls is significantly different from 0.5.

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 interval $1-\alpha$, 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:

In the study 340 babies were​ born, and 289 of them were girls. This means that $n = 340, \pi = \frac{289}{340} = 0.85$

Use the sample data to construct a 99​% confidence interval estimate of the percentage of girls born.

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{340}} = 0.85 - 2.575\sqrt{\frac{0.85*0.15}{400}} = 0.80$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{340}} = 0.85 + 2.575\sqrt{\frac{0.85*0.15}{400}} = 0.90$

The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).

Does the method appear to be ​effective?

b)Yes, the proportion of girls is significantly different from 0.5.