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Sagot :
Using the z-distribution, it is found that the smallest sample size required to obtain the desired margin of error is of 4330.
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:
- The estimate of the proportion is of [tex]\pi = 0.8[/tex].
- The desired margin of error is of M = 0.01.
- 90% confidence level, hence[tex]\alpha = 0.9[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.9}{2} = 0.95[/tex], so [tex]z = 1.645[/tex].
Then, we solve for n to find the minimum sample size.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.01 = 1.645\sqrt{\frac{0.8(0.2)}{n}}[/tex]
[tex]0.01\sqrt{n} = 1.645\sqrt{0.8(0.2)}[/tex]
[tex]\sqrt{n} = \frac{1.645\sqrt{0.8(0.2)}}{0.01}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.645\sqrt{0.8(0.2)}}{0.01}\right)^2[/tex]
[tex]n = 4329.6[/tex]
Rounding up, the smallest sample size required to obtain the desired margin of error is of 4330.
To learn more about the z-distribution, you can take a look at https://brainly.com/question/12517818
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