Question

Of 254 randomly selected medical students, 25 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.

Answer 1

Solution:
Given in the question
Number of Sample (n) = 254
Number of students that they planned to work in a rural community (X)= 25
Sample proportion p= X/n = 25/254 = 0.0984
Point estimate = 0.0984
95% confidence interval for the true proportion of all medical students who plan to work in a rural community can be calculated as
p +/- Zalpha/2 * sqrt(p*(1-p)/n)
Confidence level = 0.05
level of significance () = 1 - Confidence level = 1- 0.95 = 0.05
/2 = 0.05/2 = 0.025, from Z table we found Zalpha/2 = 1.96
So 95% confidence interval is
0.0984 +/- 1.96*sqrt(0.0984*(1-0.0984)/254)
0.0984 +/- 1.96*0.0187
0.0984 +/- 0.0366
0.062 to 0.135
So we are 95% confident that the true proportion of all medical students who plan to work in a rural community is between 0.062 and 0.135