Question

A research group conducted an extensive survey of 2985 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?" 1618 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 3.0% of the population percentage? (Hint: Use p ≈ 0.54 as a preliminary estimate. Round your answer up to the nearest whole number.)
workers

Answer 1

Solution :

Given that,

= 0.54

1 - = 1 - 0.54 = 0.46

margin of error = E = 3.0% = 0.03

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Sample size = n = (Z_/2 / E)² * * (1 - )

= (1.96 / 0.03)² * 0.54 * 0.46

= 1060

Sample size =1060