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. Sup

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)

in progress 0
Dâu 2 months 2021-08-28T18:21:08+00:00 1 Answers 26 views 0

Answers ( )

    0
    2021-08-28T18:22:16+00:00

    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)

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )