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Sagot :
Using the z-distribution, it is found that the needed sample sizes are:
a) 242
b) 1842
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:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
99% confidence level, hence[tex]\alpha = 0.99[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.99}{2} = 0.995[/tex], so [tex]z = 2.575[/tex].
Item a:
The estimate is:
[tex]\pi = 0.223 - 0.189 = 0.034[/tex]
The sample size is n for which M = 0.03, then:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 2.575\sqrt{\frac{0.034(0.966)}{n}}[/tex]
[tex]0.03\sqrt{n} = 2.575\sqrt{0.034(0.966)}[/tex]
[tex]\sqrt{n} = \frac{2.575\sqrt{0.034(0.966)}}{0.03}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{2.575\sqrt{0.034(0.966)}}{0.03}\right)^2[/tex]
[tex]n = 241.97[/tex]
Rounding up, a sample of 242 is needed.
Item b:
No prior estimates, hence [tex]\pi = 0.5[/tex] is used.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 2.575\sqrt{\frac{0.5(0.5)}{n}}[/tex]
[tex]0.03\sqrt{n} = 2.575\sqrt{0.5(0.5)}[/tex]
[tex]\sqrt{n} = \frac{2.575\sqrt{0.5(0.5)}}{0.03}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{2.575\sqrt{0.5(0.5)}}{0.03}\right)^2[/tex]
[tex]n = 1841.8[/tex]
Rounding up, a sample of 1842 is needed.
For more on the z-distribution, you can check https://brainly.com/question/25404151
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