Question

Suppose we take a poll (random sample) of 3923 students classified as Juniors and find that 3196 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.

Answer 1

The 99% confidence interval for the proportion of GSU Juniors who believe that they will, immediately, be employed after graduation is (0.7987, 0.8307).

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 z-score that has a p-value of
`1 - (\alpha)/(2)`.

Suppose we take a poll (random sample) of 3923 students classified as Juniors and find that 3196 of them believe that they will find a job immediately after graduation.

This means that
`n = 3923, \pi = (3196)/(3923) = 0.8147`

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a p-value 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)} = 0.8147 - 2.575\sqrt{(0.8147*0.1853)/(3923)} = 0.7987`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.8147 + 2.575\sqrt{(0.8147*0.1853)/(3923)} = 0.8307`

The 99% confidence interval for the proportion of GSU Juniors who believe that they will, immediately, be employed after graduation is (0.7987, 0.8307).