Respuesta :
Answer:
They would have to do 407 such checks.
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 of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
Kroll, Inc., a firm that specializes in investigating such matters, said that they believe as many as 25% of back ground checks might reveal false information.
This means that [tex]\pi = 0.25[/tex]
98% confidence level
So [tex]\alpha = 0.02[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.327[/tex].
How many such random checks would they have to do to esti mate the true percentage of people who misrepresent their backgrounds to within ±5% with 98% confidence?
This is n for which M = 0.05. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.05 = 2.327\sqrt{\frac{0.25*0.75}{n}}[/tex]
[tex]0.05\sqrt{n} = 2.327\sqrt{0.25*0.75}[/tex]
[tex]\sqrt{n} = \frac{2.327\sqrt{0.25*0.75}}{0.05}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.327\sqrt{0.25*0.75}}{0.05})^2[/tex]
[tex]n = 406.12[/tex]
Rounding up:
They would have to do 407 such checks.
