Respuesta :
Answer: n = 600.25 ≈ 601 ( sample size)
the minimum sample size of the adults in United State who have sleep deprivation is 601
Step-by-step explanation:
Given that;
the confidence level = 95% = 0.05
the Margin of error E = 4% = 0.04
∴ level of significance ∝ = 1 - C.L
∝ = 1 - (95/100)
= 0.05
Z∝/2 = 0.05/2 = 0.025
Z-critical value at 95% confidence level is OR level of significance at 0.025 = 1.96 (z-table)
Now to calculate the sample size, we say
n = (Z(critical) / M.E )² P ( 1 - P)
now we substitute
n = (1.96 / 0.04)² × 0.5 ( 1 - 0.5)
n = 2401 × 0.5 × 0.5
n = 600.25 ≈ 601
Therefore the minimum sample size of the adults in United State who have sleep deprivation is 601
Using the formula for the margin of error, it is found that a sample of 601 is needed.
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]
95% confidence level
So [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 [tex]z = 1.96[/tex].
In this problem:
- There is no previous estimate, hence [tex]\pi = 0.5[/tex] is used.
- The minimum sample size is n for which M = 0.04, hence:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}[/tex]
[tex]0.04\sqrt{n} = 1.96(0.5)[/tex]
[tex]\sqrt{n} = \frac{1.96(0.5)}{0.04}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.96(0.5)}{0.04}\right)^2[/tex]
[tex]n = 600.25[/tex]
Rounding up, a sample of 601 is needed.
A similar problem is given at https://brainly.com/question/25420216
