Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 580 babies were born, and 319 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?
______ < p < ______


Does the method appear to be effective?

Answer 1

Answer:

0.4967<p<0.6033

Not effective

Step-by-step explanation:

given that a  clinical trial tests a method designed to increase the probability of conceiving a girl.

In the study 580 babies were born, and 319 of them were girls

Sample proportion p= $\frac{319}{580} =0.55$

$q=1-p =0.45$

Standard error of proportion p = $\sqrt{\frac{pq}{n} } =0.0207$

Margin of error for 99%= Z critical * std error=$2.58*0.0207\\=0.0533$

(Since sample size is large Z value is used)

Confidence interval 99%

=$(0.55-0.0533, 0.55+0.0533)\\= (0.4967, 0.6033)$

0.4967<p<0.6033

This method appears to be not effective as the confidence interval 99% contains 0.50 indicating probability for girls is still 0.50