Of 314 randomly selected medical students,25 said that they planned to work in a rural community.Construct a 95% confidence interval for the percentage of all medical students who plan to work in a rural community.
A)(4.40%,11.5%)
B)(4.97%,11.0%)
C)(5.45%,10.5%)
D)(4.03%,11.9%)
E)(7.02%,8.90%)

Question

Quizplus · Accepted Answer

We can use the formula for a confidence interval for a proportion:
CI = p ± z*√(p(1-p)/n)
where p is the proportion of medical students who plan to work in a rural community, z is the critical value for a 95% confidence interval (which is 1.96), and n is the sample size (which is 314).

Plugging in the values we have:
CI = 25/314 ± 1.96*√((25/314)*(1-25/314)/314)
CI = 0.0796 ± 0.0283
CI = (0.0513, 0.1080)

Multiplying by 100 to get the percentage:
CI = (5.13%, 10.80%)

The closest answer choice is B, which is (4.97%, 11.0%).

Of 314 Randomly Selected Medical Students,25 Said That They Planned