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Sagot :
Answer:
A sample size of 392 is required.
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].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
18% of students struggle in their classes because they don't spend more than 8 hours studying independently outside of a 4-unit class.
This means that [tex]\pi = 0.18[/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].
You would like to be 99% confident that your estimate is within 5% of the true population proportion. How large of a sample size is required?
A sample size of n is required.
n is found when M = 0.05. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.05 = 2.575\sqrt{\frac{0.18*0.82}{n}}[/tex]
[tex]0.05\sqrt{n} = 2.575\sqrt{0.18*0.82}[/tex]
[tex]\sqrt{n} = \frac{2.575\sqrt{0.18*0.82}}{0.05}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.575\sqrt{0.18*0.82}}{0.05})^2[/tex]
[tex]n = 391.5[/tex]
Rounding up:
A sample size of 392 is required.
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