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:
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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