It is estimated that the proportion of union members in favor of the new contract is between 0.3 and 0.5. The sponsors of the proposition would like to survey union members and get a 95 percent confidence interval for the true proportion of people in favor of the new contract. They want to restrict the error level to 5 percentage points (5%). Given the likelihood of the survey being delivered to a union member is 0.7 (incidence rate), and the response rate is 0.5, how many surveys should they send out? Please show your calculation.
Answer:
Given,
p = (0.3 + 0.5)/2
= 0.4
Here at 95% CI, z value is 1.96
Margin of error E = 5% = 0.05
consider,
sample size n = p(1-p)*(z/E)^2
substitute values
= 0.4(1-0.4)*(1.96/0.05)^2
= 368.7936
sample size n = 369
Number of surveys they send out = n/(incidence rate * response rate)
substitute values
= 369/(0.7*0.5)
= 1054.285714
= 1055
It is estimated that the proportion of union members in favor of the new contract is...