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Sagot :
Answer:
The minimum sample size we should anticipate using is 601.
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]
We have no history with this characteristic, so we have no idea as to what the proportion might be.
This means that we use [tex]\pi = 0.5[/tex], which is when the largest sample size will be needed.
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
What is the minimum sample size we should anticipate using?
This is n for which M = 0.04. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.04\sqrt{n} = 1.96*0.5[/tex]
[tex]\sqrt{n} = \frac{1.96*0.5}{0.04}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2[/tex]
[tex]n = 600.25[/tex]
Rounding up, 601
The minimum sample size we should anticipate using is 601.
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