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

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).