Question

11. A local health care facility noted that in a sample of 200 patients, 180 were referred to them by the local hospital. Provide a 99% confidence interval for all the patients who are referred to this facility by the hospital. a. 0.9 ± 0.013*0.021 b. 0.9 ± 2.575*0.021 c. 0.9 ± 2.601*0.021 d. 0.9 ± 2.601*0.00045 e. 0.9 ± 2.575*0.00045

Answer 1

Solution :

Given that,

n = 200

x = 180

Point estimate = sample proportion = = x / n = 180/200=0.9

1 -   = 1- 0.9 =0.1

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 ( Using z table )

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

= 2.576* (((0.9*0.1) /200 )

E = 2.576*0.021

A 99% confidence interval for population proportion p is ,

0.9 ± 2.575*0.021