Question

For a 95% confidence interval what would the sample size need to be as it's most conservative to have a 2% margin of error

Answer 1

Answer:

The size of the sample 'n' = 2401

Step-by-step explanation:

Step:1

Given that the margin of error = 2 % = 0.02

The margin of error is determined by

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

we know that the proportion

               $\sqrt{p(1-p)} < \frac{1}{2}$

Step:2

The margin of error is determined by

              $0.02 = \frac{1.96( \frac{1}{2} ) }{\sqrt{n} }$

              $\sqrt{n} = \frac{1.96}{2 X 0.02}$

              √n  = 49

Squaring on both sides, we get

                n = 49 × 49

                n = 2401

Final answer:-

The size of the sample 'n' = 2401