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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.)

1 Answer

4 votes

Answer:

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

User Djhoese
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