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A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 335335 babies were​ born, and 268268 of them were girls. Use the sample data to construct a 9999​% confidence interval estimate of the percentage of girls born. Based on the​ result, does the method appear to be​ effective?

User Allen Chan
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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.

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

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

For this problem, we have that:


n = 335, \pi = (268)/(335) = 0.8

99% confidence level

So
\alpha = 0.01, z is the value of Z that has a pvalue 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)} = 0.8 - 2.575\sqrt{(0.8*0.2)/(335)} = 0.7437

The upper limit of this interval is:


\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.8 + 2.575\sqrt{(0.8*0.2)/(335)} = 0.8563

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.

User Danivovich
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