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 0.025 of the population proportion is:

Sample size should be atleast 625

Step-by-step explanation:

Given that the  Labor Bureau wants to estimate, at a 90% confidence level, the proportion of all households that receive welfare

Sample proportion = 17.5%

Let n be the sample size

Standard error of sample proportion=

Z critical for 90% = 1.645

Margin of error = 1.645 * std error

Since margin of error<0.025 we have

10/60

step-by-step explanation:

if you divide 10 by 60 the answer is 0.166666 while if you divide or solve the other problems it equals 0.6