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?

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 , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the z-score that has a p-value of .

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

This means that

99% confidence level

So , z is the value of Z that has a p-value of , so .  

The lower limit of this interval is:

The upper limit of this interval is:

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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step-by-step explanation:

the correct answer is 1: 2

step-by-step explanation:

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