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Sagot :
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 [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 zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
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 [tex]n = 3653, \pi = \frac{3005}{3653} = 0.823[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue 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}} = 0.823 - 2.575\sqrt{\frac{0.823*0.177}{3653}} = 0.806[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.823 + 2.575\sqrt{\frac{0.823*0.177}{3653}} = 0.839[/tex]
The answer is (0.806, 0.839)
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