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

Answer:

(0.806, 0.839)

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 $1 - \frac{\alpha}{2}$.

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 $n = 3653, \pi = \frac{3005}{3653} = 0.823$

99% confidence level

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

The lower limit of this interval is:

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

The upper limit of this interval is:

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

The answer is (0.806, 0.839)