Please help with statistics I am desperate!?

Question

An auditor wants to estimate what proportion of a bank’s commercial loan files are incomplete. The auditor wants to be within 4% of the true proportion when using a 95% confidence level. How many files must the auditor sample? No estimate of the proportion is available, so use 0.5 for the population proportion

Answer 1

EDITED.

Aside.

Proof of the formula that is used below.

The confidence interval for estimate a population proportion p is {p(+-)z(α/2)*√[p(1-p)/n]}

To obtain the sample size n we need to make the error bound in the interval equal to B, then we get

z(α/2)*√[p(1-p)/n] = B

i.e., √[p(1-p)/n] = B/[z(α/2)]

Squaring both sides of the equation,

  {√[p(1-p)/n]}^2 = {B/[z(α/2)]}^2

   p(1-p)/n = [B/z(α/2)]^2

   [B/z(α/2)]^2*n = [p(1-p)/n]*n

    [B/z(α/2)]^2*n = p(1-p)

                    n = p(1-p)/ [B/z(α/2)]^2

                    n = p(1-p)*[z(α/2)/B]^2

Let n be the number of files that the auditor must sample, p be the population proportion of a bank's commerical loan files are incomplete, B=Error bound of the true proportion of incompleted files and α be the level of significance.

Given that p=0.5, B=4%=0.04 and

100(1-α)% = 95%

   1-α = 95/100

   1-α = 0.95

    -α = 0.95-1

    -α = -0.05

     α = 0.05

So, using the formula for the sample size for a confidence interval for p is

n= p(1-p)[(zα/2)/B]^2

n = 0.5*(1-0.5)*[z(0.05/2)/0.04]^2

n = 0.5*0.5*(z0.025/0.04)^2

n =0.25* (1.96/0.04)^2

n = 0.25 * (49)^2

n = 0.25 * 2401

n = 600.25

n ≈ 600

Therefore, as shown on above, the auditor must sample a size of n=600 files to determine what proportion of a bank's commercial loan files are incompleted.

Hope this helps.