In a survey, the planning value for the population proportion is p* = 0.16. How large a sample should be taken to provide a 90% confidence interval with a margin of error of .04?

Answer 1

n = σ²z²/ME²

For a 90% confidence interval, as this is two-sided, we need Φ(z) = .95. From Excel, this is norm.s.inv(.95) = 1.64485362695147

σ² = p(1-p) = .16(1 - .16) = .1344

ME = .04

σ²z²/ME² = .1344 * 1.64485362695147² /.04² = 227.265650144014, which we round up to 228

Answer 2

The simple answer would be 227, but since you want to know HOW to get the answer not find out the actual answer you need to know the formula used .

{available from standard statistical texts and sources}

n = N*X / (X + N – 1),

where,

X = Z_α/2² ­*p*(1-p) / MOE^2.

α = error rate (1 - deg of confidence), MOE = margin of error, p = proportion, Z = Z score critical (for α/2) and N represents the population size.

( note: where population size is very large the formula simplifies to n = Z_α/2² ­*p*(1-p) / MOE^2. )

or you can cheat and just go to this location for an on line calculator

https://select-statistics.co.uk/calculators/sample-size-calculator-population-proportion/