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Sagot :
Using the formula for the margin of error, it is found that Haley should sample 12,724 soldiers.
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]\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]\frac{1+\alpha}{2}[/tex].
The margin of error is of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
Estimate of the proportion of 64%, hence [tex]\pi = 0.64[/tex].
94% confidence level
So [tex]\alpha = 0.94[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.94}{2} = 0.97[/tex], so [tex]z = 1.88[/tex].
Margin of error of less than 0.008 is wanted, hence, we have to find n when M = 0.008.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.008 = 1.88\sqrt{\frac{0.64(0.36)}{n}}[/tex]
[tex]0.008\sqrt{n} = 1.88\sqrt{0.64(0.36)}[/tex]
[tex]\sqrt{n} = \frac{1.88\sqrt{0.64(0.36)}}{0.008}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.88\sqrt{0.64(0.36)}}{0.008})^2[/tex]
[tex]n = 12723.84[/tex]
Rounding up, Haley should sample 12,724 soldiers.
A similar problem is given at https://brainly.com/question/25404151
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