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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?

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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.

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{(\pi(1-\pi))/(n)}

In which

z is the z-score that has a p-value of
1 - (\alpha)/(2).

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

This means that
n = 380, \pi = (342)/(380) = 0.9

99% confidence level

So
\alpha = 0.01, z is the value of Z that has a p-value of
1 - (0.01)/(2) = 0.995, so
Z = 2.575.

The lower limit of this interval is:


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

The upper limit of this interval is:


\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.9 + 2.575\sqrt{(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.

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