Question

The Labor Bureau wants to estimate, at a 90% confidence level, the proportion of all households that receive welfare. A preliminary sample showed that 20.5% of households in this sample receive welfare. The sample size that would limit the margin of error to be within 0.023 of the population proportion is:

Answer #1

Solution :

To estimate the the population proportion at 90% confidence level, the required sample size is given as follows :

Where, E is margin of error, n is sample size, p̂ is sample proportion, q̂ = 1 - p̂ and Z(0.10/2) is critical z-value at 90% confidence level.

Sample proportion of households who receive welfare is, p̂ = 20.5% = 0.205

q̂ = 1 - 0.205 = 0.795

E = 0.023

Using Z-table we get, Z(0.10/2) = 1.645

Hence, required sample size is,

On rounding to nearest integer we get,

n = 834

**The required sample size is 834.**

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