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 17.5% of households in this sample receive welfare. The sample size that would limit the margin of error to be within 0.028 of the population proportion is:

Answer 1

Solution :

Given that,

= 0.175

1 - = 1 - 0.175 = 0.825

margin of error = E = 0.028

Z/2 = 1.645

sample size = n = (Z / 2 / E )2 * * (1 - )

= (1.645 / 0.028)2 * 0.175 * 0.825

= 498

sample size = 498