Question

A research group conducted an extensive survey of 3091 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?" 1581 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 2.2% of the population percentage? (Hint: Use p ≈ 0.51 as a preliminary estimate. Round your answer up to the nearest whole number.)
__________ workers

Answer 1

Solution :

Given that,

n = 3091

x = 1581

E = 2.2% = 0.022

Point estimate = sample proportion = = x / n = 1581/3091= 0.5115

1 -    = 1-0.5115 =0.4885

At 90% confidence level the z is

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.1 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )

Margin of error = E = Z/2 * (( * (1 - )) / n)

n = *(1 - )*( Z/2 / E)^2

= 0.5115*0.4885*(1.96/0.022)^2

= 1983.2478

~ 1983 or 1984 (please check)

***please comment if you have any doubts.Happy to help you.Thank you. Please Like.