Question

A researcher at a major clinic wishes to estimate the proportion of the adult population of the United States that has sleep deprivation. How large a sample is needed in order to be 95% confident that the sample proportion will not differ from the true proportion by more than 4%?

Answer 1

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