Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 340 babies were​ born, and 289 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?(__)<(__) ​(Round to three decimal places as​ needed.)Does the method appear to be​effective?a)No​,the proportion of girls is not significantly different from 0.5.b)Yes, the proportion of girls is significantly different from 0.5.

Answer 1

The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).

b)Yes, the proportion of girls is significantly different from 0.5.

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence interval
`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:

In the study 340 babies were​ born, and 289 of them were girls. This means that
`n = 340, \pi = (289)/(340) = 0.85`

Use the sample data to construct a 99​% confidence interval estimate of the percentage of girls born.

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))/(340)} = 0.85 - 2.575\sqrt{(0.85*0.15)/(400)} = 0.80`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(340)} = 0.85 + 2.575\sqrt{(0.85*0.15)/(400)} = 0.90`

The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).

Does the method appear to be ​effective?

b)Yes, the proportion of girls is significantly different from 0.5.