205k views
1 vote
Of 375 randomly selected students, 30 said that they planned to work in a rural community. Find 95% confidence interval for the true proportion of all medical students who plan to work in a rural community.

User Allen Kim
by
8.2k points

1 Answer

1 vote

Answer:

The 95% confidence interval for the true proportion of all medical students who plan to work in a rural community is (0.0525, 0.1075).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
\pi, and a confidence level of
1-\alpha, we have the following confidence interval of proportions.


\pi \pm z\sqrt{(\pi(1-\pi))/(n)}

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

For this problem, we have that:


n = 375, \pi = (30)/(375) = 0.08

95% confidence level

So
\alpha = 0.05, z is the value of Z that has a pvalue of
1 - (0.05)/(2) = 0.975, so
Z = 1.96.

The lower limit of this interval is:


\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.08 - 1.96\sqrt{(0.08*0.92)/(375)} = 0.0525

The upper limit of this interval is:


\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.08 + 1.96\sqrt{(0.08*0.92)/(375)} = 0.1075

The 95% confidence interval for the true proportion of all medical students who plan to work in a rural community is (0.0525, 0.1075).

User Zikes
by
7.9k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories