You plan to conduct a survey to find what proportion of the workforce has two or more jobs. You decide on the 95% confidence level and a margin of error of 2%. A pilot survey reveals that 5 of the 50 sampled hold two or more jobs. How many in the workforce should be interviewed to meet your requirements

865

Step-by-step explanation:

We have that in 95% confidence level the value of z has a value of 1.96. This can be confirmed in the attached image of the normal distribution.

Now we have the following formula:

n = [z / E] ^ 2 * (p * q)

where n is the sample size, which is what we want to calculate, "E" is the error that is 2% or 0.02. "p" is the probability they give us, 5 out of 50, is the same as 1 out of 10, that is 0.1. "q" is the complement of p, that is, 1 - 0.1 = 0.9, that is, the value of q is 0.9.

Replacing these values we are left with:

n = [1.96 / 0.02] ^ 2 * [(0.1) * (0.9)]

n = 864.36

865 by rounding to the largest number.

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step-by-step explanati∵∵∴∴on: