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
[tex]H_0:p=0.50\\\\H_1:p>0.50[/tex]
Sample size : n= 345
The sample proportion of girls born =[tex]\dfrac{276}{345}=0.8[/tex]
Critical value : [tex]z_{\alpha/2}=z_{0.005}=2.576[/tex]
The confidence interval for population proportion is given by :_
[tex]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\%)[/tex]
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.
