Question

Use the given data to find the minimum sample size required to estimate the population proportion. margin of error: 0.045; confidence level: 96%; and unknown

Answer 1

The formula to find the minimum sample size while testing population proportion is

n = $p (1-p) (\frac{z(\alpha/2)}{ME} ) ^{2}$

Where p is sample proportion. If there is no information about sample proportion then we take it as 0.5

ME = Margin of error = 0.045

z (α/2) = Critical z score value

This is z score such that P(Z < - z (α/2)) = α/2 and P(Z > z (α/2)) = α/2

α = 1 -confidence level = 1- (96/100) = 0.04

α/2 = 0.04/2 = 0.02

So we have to find z score such that

P(Z < -z (α/2)) = 0.02 and P(Z > z (α/2)) = 0.02

Using z score table to find probability exactly or close to 0.02

Probability value 0.0202 is closest probability value and the corresponding z score is -2.05

It means P(Z < -2.05) =0.02 and P(Z > 2.05) = 0.02

So we have z (α/2) = 2.05

Using all the given values into formula to find sample size

n = 0.5 * (1-0.5) (2.05/0.045)^2

= 0.25 * (45.5556)^2

= 0.25 * 2075.313

n = 518.82

Round the sample size to nearest integer we get

n =519

The minimum sample size is 519