Question

Suppose the mayor of Los Angeles, Mr. Lucas wants an estimate of the proportion of the population who support his current policy towards the homeless. The mayor wants the estimate to be within 5% of the true proportion. Assume a 95% level of confidence. The mayor's political advisors estimated the proportion supporting the current policy to be 0.56. How large of a sample is required? How large of a sample would be necessary if no estimate were available for the proportion supporting current policy? Why is there a difference between (1) and (2)

Answer 1

Solution,

Given that,

a) = 0.56

1 - = 1 - 0.56 = 0.44

margin of error = E = 0.05

At 95% confidence level

= 1 - 95%

= 1 - 0.95 =0.05

/2 = 0.025

Z_/2 = Z_0.025 = 1.96   

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

= (1.96 / 0.05 )² * 0.56 * 0.44

= 378.62

sample size = n = 379

b) =  1 - =   0.5

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

= (1.96 / 0.05)² * 0.5 * 0.5

= 384.16

sample size = n = 385

c) Having an estimate of the population proportion reduces the minimum sample size is needed .