Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 345 babies were​ born, and 276 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?

Answer 1

Answer with explanation:

Given : A clinical trial tests a method designed to increase the probability of conceiving a girl.

Also, in a normal case the probability of conceiving a girl is 0.5.

Let p be the proportion of girls born.

Then , we have

`H_0:p=0.50\\\\H_1:p>0.50`

Sample size : n= 345

The sample proportion of girls born =
`(276)/(345)=0.8`

Critical value :
`z_(\alpha/2)=z_(0.005)=2.576`

The confidence interval for population proportion is given by :_

`p\ \pm z_(\alpha/2)\sqrt{(p(1-p))/(n)}\\\\=0.8\pm(2.576)\sqrt{(0.8(0.2))/(345)}\\\\\approx0.8\pm0.055\\\\=(0.745,0.855)=(74.5\%,85.5\%)`

Since the confidence interval do not contains 50% that means there is increase in the probability of conceiving a girl, then it can be concluded that we have statistical evidence that the the clinical trial method increase the probability of conceiving a girl.