Question

Refer to the previous exercise. The researchers wanted a sufficiently large sample to be able to estimate the probability of preferring the new analgesic to within 0.08, with confidence 0.95. If the true probability is 0.75, how large a sample is needed to achieve this accuracy

Answer 1

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 = $\frac{Z_{0.05} \sqrt{p(1-p)} }{\sqrt{n} }$

$0.08 = \frac{2 X \sqrt{0.75(1-0.75)} }{\sqrt{n} }$

Cross multiplication , we get

$\sqrt{n} = \frac{2 X \sqrt{0.75(1-0.75)} }{0.08 }$

√n = $\frac{2 X0.4330}{0.08} = 10.825$

squaring on both sides , we get

n = 117.07

Final answer:-

The large  sample n = 117.07