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Sagot :
Answer:
The minimum sample size required is 461.
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]
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].
An interval estimate of the proportion p with a margin of error of 0.06. What is the minimum sample size required?
The minimum sample size required is n, which is found when M = 0.06.
We don't have an estimate for the true proportion, which means that we use [tex]\pi = 0.5[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.06 = 2.575\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.06\sqrt{n} = 2.575*0.5[/tex]
[tex]\sqrt{n} = \frac{2.575*0.5}{0.06}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.575*0.5}{0.06})^2[/tex]
[tex]n = 460.5[/tex]
Rounding up
The minimum sample size required is 461.
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