Question

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?

Answer 1

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.