Question

7. Of 380 randomly selected medical students, 21 said that they planned to work in a rural community. Find a 95% confidence interval for the true proportion of all medical students who plan to work in a rural community.

A) 0.0251 < p < 0.0854

B) 0.0323 < p < 0.0782

C) 0.0280 < p < 0.0826

D) 0.0360 < p < 0.0745

Answer 1

Solution :

Given that,

n = 380

x = 21

Point estimate = sample proportion = = x / n =21/380=0.055

1 -   = 1- 0.055 =0.945

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.055*0.945) /380 )

= 0.0232

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.055- 0.0232< p < 0.055+0.0232

0.0323 < p < 0.0782

The 95% confidence interval for the population proportion p is :  0.0323 < p < 0.0782