Join IDNLearn.com and start getting the answers you've been searching for. Join our platform to receive prompt and accurate responses from experienced professionals in various fields.
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
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com is committed to your satisfaction. Thank you for visiting, and see you next time for more helpful answers.
