Question

Resume fraud. In 2002 the Veritas Software company found out that its chief financial officer did not actually have the MBA he had listed on his resume. They fired him, and the value of the company's stock dropped 19%. 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. 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?

Answer 1

They would have to do 407 such checks.

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 of:

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

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
`\pi = 0.25`

98% confidence level

So
`\alpha = 0.02`, z is the value of Z that has a p-value of
`1 - (0.02)/(2) = 0.99`, so
`Z = 2.327`.

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

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

`0.05 = 2.327\sqrt{(0.25*0.75)/(n)}`

`0.05√(n) = 2.327√(0.25*0.75)`

`√(n) = (2.327√(0.25*0.75))/(0.05)`

`(√(n))^2 = ((2.327√(0.25*0.75))/(0.05))^2`

`n = 406.12`

Rounding up:

They would have to do 407 such checks.