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Sagot :
Answer:
6546 students would need to be sampled.
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].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
The dean randomly selects 200 students and finds that 118 of them are receiving financial aid.
This means that [tex]n = 200, \pi = \frac{118}{200} = 0.59[/tex]
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
If the dean wanted to estimate the proportion of all students receiving financial aid to within 1% with 90% reliability, how many students would need to be sampled?
n students would need to be sampled, and n is found when M = 0.01. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.01 = 1.645\sqrt{\frac{0.59*0.41}{n}}[/tex]
[tex]0.01\sqrt{n} = 1.645\sqrt{0.59*0.41}[/tex]
[tex]\sqrt{n} = \frac{1.645\sqrt{0.59*0.41}}{0.01}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.645\sqrt{0.59*0.41}}{0.01})^2[/tex]
[tex]n = 6545.9[/tex]
Rounding up:
6546 students would need to be sampled.
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