75%of the workers support a benefit package. The results are accurate to within + or - 2%, 18 times out of 20. What sample size is needed to obtain an error range of 2% from the statement made 

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The formula I am familiar with for sample size for the confidence interval of a proportion is n = (p hat)(q hat)((z sub c/E)^2 p hat = .75        q hat = 1- .75 = .25 E = .02           confidence level = 18/20 = .9 z sub c at .9 = 1.645 so putting it all together you get n = (.75)(.25)(1.645/.02)^2 = 1268.4  or round up to 1269 Don't know if you use the same symbols or if this is the type of problems you are doing.