Could I get some help with this please!!!!!

Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 600 babies were born, and 330 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 < = ? (Round to three decimal places as needed.)

Answer 1

I think that this website would help you tremendously. 

 

https://onlinecourses.science.psu.edu/stat200/node/48

 

In this particular problem, you need to figure out the necessary components to plug'n'chug into the formula:

 

n = the total number in the sample = 600 

p-hat = the proportion of the sample that were girls = 330/600 = 0.55 

se = standard error = √([p-hat(1-p-hat)]/n) = √([0.55(1-0.55)]/600) = 0.01936491673

z-multiplier = 2.58

 

0.55 ± (2.58)(0.01936491673) = 0.55 ± 0.04996148516

 

This gives you the interval of [0.50003851483,0.59996148516] ≈ [0.50,0.60]

 

We are 99% confident that the true population proportion of female babies born given the clinical trial drug was taken is between 0.50 and 0.60.

 

Therefore, since the confidence interval technically contains 0.50, which is the 50/50 chance of having a girl, then it can be concluded that there is not statistical evidence that the true population proportion of conceiving a girl given the use of this clinical trial drug is not equal to 0.50, so it's not any higher than one's typical likelihood of having a girl.