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Sagot :
Answer:
865 in the workforce should be interviewed to meet your requirements
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is given by:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
A pilot survey reveals that 5 of the 50 sampled hold two or more jobs.
This means that [tex]\pi = \frac{5}{50} = 0.1[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
How many in the workforce should be interviewed to meet your requirements?
Margin of error of 2%, so n for which M = 0.02.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 1.96\sqrt{\frac{0.1*0.9}{n}}[/tex]
[tex]0.02\sqrt{n} = 1.96\sqrt{0.1*0.9}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.1*0.9}}{0.02}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.1*0.9}}{0.02})^2[/tex]
[tex]n = 864.4[/tex]
Rounding up:
865 in the workforce should be interviewed to meet your requirements
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