A research group conducted an extensive survey of 3078 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?" 1482 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 1.4% of the population percentage? (Hint: Use p ≈ 0.48 as a preliminary estimate. Round your answer up to the nearest whole number.)

2 answers

Try this formula:

n = [(z-value)^2 * p * q]/E^2
...where z-value is 1.96 which represents 95% confidence using a z-table, p = 0.48, q = 1 - p, and E = .014 (1.4% in decimal form).

I'll let you calculate it from here. Round your answer to the next highest whole number.
0.495 < p < 0.525
Similar Questions
    1. answers icon 1 answer
  1. Which of the following is most likely to need the help of a tax professional?a Salaried workers b Minimum-wage workers c College
    1. answers icon 1 answer
  2. A census is defined as which of the following?A. A collection of data B. A survey conducted with a small group C. A bureau of
    1. answers icon 1 answer
  3. the wages of 80 workers in a small factory is given thus.wage :- 1500 3500 6000 12000 20000 no of workers 8736 10883 11382 10367
    1. answers icon 3 answers
more similar questions