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?

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.

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

In which

z is the zscore that has a pvalue of .

For this problem, we have that:

99% confidence level

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

The lower limit of this interval is:

The upper limit of this interval is:

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.

50.6

step-by-step explanation:

the answer is 1 becuz if it were 1.0 that would rip apart the entire so ya. : )

~potato.

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