Respuesta :
Answer:
The large sample n = 117.07
Step-by-step explanation:
Step(i):-
Given that the estimate error (M.E) = 0.08
The proportion (p) = 0.75
q =1-p = 1- 0.75 =0.25
Level of significance = 0.05
Z₀.₀₅ = 1.96≅ 2
Step(ii):-
The Marginal error is determined by
M.E = [tex]\frac{Z_{0.05} \sqrt{p(1-p)} }{\sqrt{n} }[/tex]
[tex]0.08 = \frac{2 X \sqrt{0.75(1-0.75)} }{\sqrt{n} }[/tex]
Cross multiplication , we get
[tex]\sqrt{n} = \frac{2 X \sqrt{0.75(1-0.75)} }{0.08 }[/tex]
√n = [tex]\frac{2 X0.4330}{0.08} = 10.825[/tex]
squaring on both sides , we get
n = 117.07
Final answer:-
The large sample n = 117.07
