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) $0.0323

Question

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) $0.0323 < p < 0.0782$
B) $0.0360 < p < 0.0745$
C) $0.0251 < p < 0.0854$
D) $0.0280 < p < 0.0826$

Quizplus · Accepted Answer

To find a 95% confidence interval for the proportion, we use the formula: $\hat{p} \pm z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p} = \frac{21}{380}$, $n = 380$, and $z$ is the z-score for a 95% confidence level, which is approximately 1.96. Plugging in the values, we get the interval $0.0323 < p < 0.0782$.

Of 380 Randomly Selected Medical Students, 21 Said That They 0.0323<p<0.07820.0323 < p < 0.07820.0323<p<0.0782

Of 380 Randomly Selected Medical Students, 21 Said That They $0.0323 < p < 0.0782$