You plan to conduct a survey to find what proportion of the workforce has two or more jobs. You decide on the 98% confidence level and a margin of error of 2%. A pilot survey reveals that 7 of the 50 sampled hold two or more jobs.
1. How many in the workforce should be interviewed to meet your requirement?

1635 people

Step-by-step explanation:

The proportion of people with two or more jobs in the sample  is  

7/50 = 0.14

The z-value that gives us a 98% confidence level is 2.33 (this means that 98% of the area under the normal curve N(0;1) is between -2.33 and 2.33)

If we use Simple Random sampling, the size of the sample should be

where z is the z-value which gives us the desired confidence level, p the proportion and e the error. In our case

z = 2.33

p = 0.14

e = 2% = 0.02

So, the sample size should be

rounding up to the nearest integer.

5050 i think

step-by-step explanation: