Answer:
<em>The large sample n = 117.07</em>
Step-by-step explanation:
<u><em>Step(i):-</em></u>
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
<u><em>Step(ii):-</em></u>
The Marginal error is determined by
M.E = ![\frac{Z_{0.05} \sqrt{p(1-p)} }{\sqrt{n} }](https://tex.z-dn.net/?f=%5Cfrac%7BZ_%7B0.05%7D%20%5Csqrt%7Bp%281-p%29%7D%20%20%7D%7B%5Csqrt%7Bn%7D%20%7D)
![0.08 = \frac{2 X \sqrt{0.75(1-0.75)} }{\sqrt{n} }](https://tex.z-dn.net/?f=0.08%20%3D%20%5Cfrac%7B2%20X%20%5Csqrt%7B0.75%281-0.75%29%7D%20%20%7D%7B%5Csqrt%7Bn%7D%20%7D)
Cross multiplication , we get
![\sqrt{n} = \frac{2 X \sqrt{0.75(1-0.75)} }{0.08 }](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%20%3D%20%5Cfrac%7B2%20X%20%5Csqrt%7B0.75%281-0.75%29%7D%20%20%7D%7B0.08%20%7D)
√n = ![\frac{2 X0.4330}{0.08} = 10.825](https://tex.z-dn.net/?f=%5Cfrac%7B2%20X0.4330%7D%7B0.08%7D%20%3D%2010.825)
squaring on both sides , we get
n = 117.07
<u><em>Final answer:-</em></u>
<em>The large sample n = 117.07</em>