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

Answer:

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).

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 z-score that has a p-value of $1 - \frac{\alpha}{2}$.

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 $n = 3923, \pi = \frac{3196}{3923} = 0.8147$

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.  

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8147 - 2.575\sqrt{\frac{0.8147*0.1853}{3923}} = 0.7987$

The upper limit of this interval is:

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

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).