Question

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

300

babies were​ born, and

255

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?

less than<

pless than<

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

Answer 1

Solution :

Given that,

n = 300

x = 255

Point estimate = sample proportion = = x / n = 255/300=0.85

1 -   = 1- 0.85 =0.15

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* (((0.85*0.15) /300 )

E = 0.053

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.85-0.053 < p < 0.85+0.053

0.797< p < 0.903