Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study

360360

babies were​ born, and

288288

of them were girls. Use the sample data to construct a

9999​%

confidence interval estimate of the percentage of girls born. Based on the​ result, does the method appear to be​ effective?

nothingless than<

pless than<nothing

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

Answer 1

Solution :

Given that,

n = 360

x = 288

Point estimate = sample proportion = = x / n = 288/360=0.8

1 -   = 1-0.8 =0.2

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576* (((0.8*0.2) / 360)

E = 0.054

A 99% confidence interval proportion p is ,

- E < p < + E

0.8- 0.054 < p < 0.8 +0.054

0.746< p < 0.854