Of 366 randomly selected medical students, 27 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) $0.0513

Question

Of 366 randomly selected medical students, 27 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) $0.0513 < p < 0.0962$
B) $0.0470 < p < 0.101$
C) $0.0419 < p < 0.106$
D) $0.0386 < p < 0.109$

Quizplus · Accepted Answer

The sample proportion is 27/366. Using the formula for a confidence interval for a proportion, the interval is calculated as $ \hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $, where $ \hat{p} = \frac{27}{366} $, $ Z $ is the Z-score for 95% confidence (1.96), and $ n = 366 $. This calculation results in the interval $0.0470 < p < 0.101$.

Of 366 Randomly Selected Medical Students, 27 Said That They 0.0513<p<0.09620.0513 < p < 0.09620.0513<p<0.0962

Of 366 Randomly Selected Medical Students, 27 Said That They $0.0513 < p < 0.0962$