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Sagot :
Using the z-distribution, as we are working with a proportion, it is found that you should sample 577 households.
What is a confidence interval of proportions?
A confidence interval of proportions is given by:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which:
- [tex]\pi[/tex] is the sample proportion.
- z is the critical value.
- n is the sample size.
The margin of error is given by:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In this problem, we have a 95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so the critical value is z = 1.96.
In this problem, the estimate is of [tex]\pi = 0.6[/tex], and we want a margin of error of M = 0.05, hence we solve for n to find the sample size.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.05 = 1.96\sqrt{\frac{0.6(0.4)}{n}}[/tex]
[tex]0.05\sqrt{n} = 1.96\sqrt{0.6(0.4)}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.6(0.4)}}{0.05}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.6(0.4)}}{0.05}\right)^2[/tex]
[tex]n = 576.24[/tex]
Rounding up, you should sample 577 households.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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