Question

You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable estimate for the population proportion. You would like to be 99.9% confident that you estimate is within 1% of the true population proportion. How large of a sample size is required

Answer 1

sample size should be atleast n= 27069

Explanation:

Given that,

confident level(CI) = 99%= 0.999

desired marginal error=1%= 0.01

note: marginal error = length of CL/2

significant level α = 1 - confident level = 1 - 0.999= 0.001

critical value = Zα/2 = Z(0.001/2) = Z0.0005( value of from z table) = 3.2905267

since we don't have preliminary estimate, p' = 0.5, which is require for maximum value

n = p' × (1 - p')(critical value/desired marginal)²

n= 0.5 × 0.5(3.2905267/0.01)²

n = 27068.91

the value of n has to be an integer = 27069

Answer 2

Answer: 27061

Explanation:

given that Р = 0.5

so

1 - P = 1 - 0.5 = 0.5

ERROR OF MARGIN E = 0.01

SIGNIFICANCE LEVEL α = 1 - confidence level

α = [(1 - (99.9/100)] = 0.0010

α / 2 = 0.001 / 2 = 0.0005

Zα/2 = Z0.0005 = 3.29 (using the Z table )

therefore

Sample size n = ((Zα/2) / E)² * P * (1 - P )

Sample size n = (3.29 / 0.01)² * 0.5 * 0.5

Sample size n = 108,241 * 0.5 * 0.5

Sample size n = 27,060.25 ≈ 27,061