Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 370 babies were​ born, and 333 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:

CI=[0.8592,0.9402]

Yes, Method appears to be effective.

Step-by-step explanation:

-We first calculate the proportion of girls born:

$\hat p\frac{x}{n}\\\\=\frac{333}{370}\\\\=0.9$

Since np$\geq 10$, we assume normal distribution and calculate the 99% confidence interval as below:

$CI=\hat p\pm z\sqrt{\frac{\hat p(1-\hat p)}{n}}\\\\=0.9\pm2.576\sqrt{\frac{0.9\times 0.1}{370}}\\\\=0.9\pm 0.0402\\\\={0.8598, \ 0.9402]$

Hence, the confidence interval is {0.8598, 0.9402]

-The probability of giving birth to a girl is 0.5 which is less than the lower boundary of the confidence interval, it can be concluded that the method appears to be effective.