Question

#7. Of 380 randomly selected medical students, 21 said that they planned to work in a rural community. Find a 99% confidence interval for the true proportion of all medical students who plan to work in a rural community.

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 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* (((0.055*0.945) /380 )

E = 0.030

A 99% confidence interval is ,

- E < p < + E

0.055-0.030 < p <0.055+0.030

0.025< p < 0.085

(0.025 , 0.085)