Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 340 babies were​born, and 306 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?

​A. No, the proportion of girls is not significantly different from 0.5.
B. Yes, the proportion of girls is significantly different from 0.5

Answer 1

Answer:

B. Yes, the proportion of girls is significantly different from 0.5

Step-by-step explanation:

Given that a  clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 340 babies were​born, and 306 of them were girls.

Sample proportion = $\frac{306}{340} \\=0.9$

Std error of p = $\sqrt{\frac{pq}{n} } \\=0.01627$

For sample size large, we find that t distribution almost coincides with z distribution

Hence 99% critical value = 2.58

Margin of error = 2.58*std error = 0.0420

Confidence interval = $(0.9-0.0420, 0.9+0.0420)$

=$(0.858, 0.942)$

0.858<p<0.942

Yes, this method is effective becuse lower bound itself is 0.858 i.e. there is 85.8%chance for getting girls in birth

B. Yes, the proportion of girls is significantly different from 0.5