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alexdok [17]
3 years ago
9

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.
Mathematics
1 answer:
exis [7]3 years ago
3 0

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).

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Step-by-step explanation:

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