Question

A research group conducted an extensive survey of 2922 wage and salaried workers on issues ranging from relationships with their bosses to household chores. The data were gathered through hour-long telephone interviews with a nationally representative sample. In response to the question, "What does success mean to you?" 1466 responded, "Personal satisfaction from doing a good job." Let p be the population proportion of all wage and salaried workers who would respond the same way to the stated question. How large a sample is needed if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 2.0% of the population percentage? (Hint: Use p ≈ 0.50 as a preliminary estimate. Round your answer up to the nearest whole number.)
workers

Answer 1

Solution,

Given that,

= 0.50

1 - = 1 - 0.50 = 0.50

margin of error = E = 2.0% = 0.02

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025  = 1.96

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

= (1.96 / 0.02)2 * 0.50 * 0.50

= 2401

sample size = n = 2401