Question

You plan to conduct a survey to find what proportion of the workforce has two or more jobs. You decide on the 99% confidence level and a margin of error of 2%. A pilot survey reveals that 8 of the 50 sampled hold two or more jobs. (Use z Distribution Table.)

How many in the workforce should be interviewed to meet your requirements? (Round z-score to 2 decimal places. Round up your answer to the next whole number.)

Number of persons to be interviewed:

Answer 1

Solution :

Given that,

= 8 / 50 = 0.16

1 - = 1 - 0.16 = 0.84

Margin of error = E = 2% = 0.02

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.58

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

= (2.58 / 0.02)² * 0.16 * 0.84

= 2236.55 = 2237

Number of persons to be interviewed : 2237