Respuesta :
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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
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 [tex]n = 3923, \pi = \frac{3196}{3923} = 0.8147[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8147 - 2.575\sqrt{\frac{0.8147*0.1853}{3923}} = 0.7987[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8147 + 2.575\sqrt{\frac{0.8147*0.1853}{3923}} = 0.8307[/tex]
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).
