Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 532 babies were​ born, and 266 of them were girls. Use the sample data to construct a 99​% confidence interval estimate of the percentage of girls born.

__________<p<______________

​(Round to three decimal places as​ needed.)

Answer 1

Solution :

Given that,

n = 532

x = 266

= x / n = 266 / 532 = 0.500

1 - = 1 - 0.500 = 0.500

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 * (((0.500 * 0.500) / 532)

= 0.056

A 99% confidence interval for population proportion p is ,

- E < P < + E

0.500 - 0.056 < p < 0.500 + 0.056

0.444 < p < 0.556