Question

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.05 for the estimation of a population proportion? Assume that past data are not available for developing a planning value for p*. (Round your answer up to the nearest whole number.)

Answer 1

Answer:

The sample size 'n' = 384

Step-by-step explanation:

Step(i):-

Given data the margin of error = 0.05

The margin of error of the estimation of a population proportion is determined by

$M.E = \frac{Z_{\frac{\alpha }{2} }\sqrt{p(1-p} }{\sqrt{n} }$

Step(ii):-

In data not given sample proportion 'p' but

we know that $\sqrt{p(1-p} \leq \frac{1}{2}$

$0.05 = \frac{1.96 X \frac{1}{2} }{\sqrt{n} }$

cross multiplication, we get

0.05 ×√n = 0.98

$\sqrt{n} = \frac{0.98}{0.05} = 19.6$

squaring on both sides, we get

n = 384.16≅ 384

Final answer:-

The sample size 'n' = 384