Answer:
b) 74
The sample size that needs to be taken if the desired margin of error is 5 or less is 74
Step-by-step explanation:
Explanation:-
Given population variance σ² = 484
σ = √484
σ = 22
The level of significance ∝=0.95
The z-score of 0.95 level of significance = 1.96
Given Margin of error = 5
we know that the margin of error is determined by
[tex]M.E = \frac{Z_{\alpha }S.D }{\sqrt{n} }[/tex]
cross multiplication, we get
[tex]\sqrt{n} = \frac{Z_{\alpha }S.D }{M.E}[/tex]
[tex]\sqrt{n} = \frac{1.96 X22 }{5} = 8.624[/tex]
Squaring on both sides, we get
n = (8.624) ²
n = 74.37 ≅74
Conclusion:-
The sample size that needs to be taken if the desired margin of error is 5 or less is 74
