Question

The ability to find a job after graduation is very important to GSU students as it is to the students at most colleges and universities. Suppose we take a poll (random sample) of 3653 students classified as Juniors and find that 3005 of them believe that they will find a job immediately after graduation. What is the 99 % confidence interval for the proportion of GSU Juniors who believe that they will, immediately, be employed after graduation. (0.812, 0.833) (0.806, 0.839) (0.81, 0.835) (0.816, 0.829)

Answer 1

(0.806, 0.839)

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)`.

Suppose we take a poll (random sample) of 3653 students classified as Juniors and find that 3005 of them believe that they will find a job immediately after graduation.

This means that
`n = 3653, \pi = (3005)/(3653) = 0.823`

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a pvalue of
`1 - (0.01)/(2) = 0.995`, so
`Z = 2.575`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.823 - 2.575\sqrt{(0.823*0.177)/(3653)} = 0.806`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.823 + 2.575\sqrt{(0.823*0.177)/(3653)} = 0.839`

The answer is (0.806, 0.839)