Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 650 babies were​ born, and 325 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

Solution:

Given,

n = 650 ....... Sample size

x = 325 .......no. of successes in the sample

Let denotes the sample proportion.

     = x/n   = 325/650 = 0.5

Our aim is to construct 99% confidence interval.

c = 0.99

= 1- c = 1- 0.99 = 0.01

  /2 = 0.01 2 = 0.005 and 1- /2 = 0.995

Search the probability 0.995 in the Z table and see corresponding z value

   = 2.576   

Now , the margin of error is given by

E = /2 *  

= 2.576* [(0.50*0.50)/650]

= 0.051

Now the confidence interval is given by

( - E)   ( + E)

(0.50 - 0.051)   (0.50 + 0.051 )

0.449 0.551  

Required 99% confidence Interval is (0.449 , 0.551 )

Generally , probability of girl is considered to be 0.5

From this confidence interval , we can say that p is not significantly greater than 0.5. Because 0.5 is in the interval and lower limit is less than 0.5.

So , based on the​ result, the method does not appear to be​ effective