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Sagot :
Answer:
A sample of 595 should be obtained.
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]
Estimate of 0.55
This means that [tex]\pi = 0.55[/tex]
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 sample size should be obtained if she wishes the estimate to be within a margin of error of 0.04?
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{0.55*0.45}{n}}[/tex]
[tex]0.04\sqrt{n} = 1.96\sqrt{0.55*0.45}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.55*0.45}}{0.04}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.55*0.45}}{0.04})^2[/tex]
[tex]n = 594.2[/tex]
Rounding up,
A sample of 595 should be obtained.
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