Question

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.

Answer 1

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).

Step-by-step 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.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 375, \pi = \frac{30}{375} = 0.08$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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).