Question

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? (Round up your answer to the next whole number.)

Answer 1

865 in the workforce should be interviewed to meet your requirements

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the z-score that has a p-value of
`1 - (\alpha)/(2)`.

The margin of error is given by:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

A pilot survey reveals that 5 of the 50 sampled hold two or more jobs.

This means that
`\pi = (5)/(50) = 0.1`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a p-value of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

How many in the workforce should be interviewed to meet your requirements?

Margin of error of 2%, so n for which M = 0.02.

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.02 = 1.96\sqrt{(0.1*0.9)/(n)}`

`0.02√(n) = 1.96√(0.1*0.9)`

`√(n) = (1.96√(0.1*0.9))/(0.02)`

`(√(n))^2 = ((1.96√(0.1*0.9))/(0.02))^2`

`n = 864.4`

Rounding up:

865 in the workforce should be interviewed to meet your requirements