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 =$\dfrac{276}{345}=0.8$

Critical value : $z_{\alpha/2}=z_{0.005}=2.576$

The confidence interval for population proportion is given by :_

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

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.