An auditor wants to estimate what proportion of a bank’s commercial loan files are incomplete. The auditor wants to be within 10% 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.
Here is one formula you might use for this problem:
n = [(z-value)^2 * p * q]/E^2
With your data:
n = [(1.96)^2 * .5 * .5]/.10^2
I'll let you finish the calculation (round to the next highest whole number).
Note: n = sample size needed; .5 for best estimate of the population proportion; .5 for q, which is (1 - p). E = maximum error, which is 10% or .10 from the problem. Z-value is found using a z-table (for 95%, the value is 1.96).
I hope this helps.
To estimate the required sample size for the auditor, we can use the formula for sample size calculation for proportions. The formula is:
n = (Z^2 * p * (1-p)) / E^2
Where:
n is the required sample size
Z is the z-score corresponding to the desired confidence level
p is the estimated population proportion
E is the desired margin of error as a proportion
In this case, the auditor wants to use a 95% confidence level, which corresponds to a z-score of approximately 1.96. The auditor wants to be within 10% of the true proportion, so the desired margin of error (E) is 0.10.
Since no estimate of the proportion is available, we can use 0.5 (50%) as the estimated population proportion (p).
Now let's plug these values into the formula and calculate the required sample size (n):
n = (1.96^2 * 0.5 * (1-0.5)) / 0.10^2
n = (3.8416 * 0.25) / 0.01
n = 0.9604 / 0.01
n ≈ 96.04
Since the sample size should be a whole number, we need to round up to the nearest whole number. Therefore, the auditor should sample at least 97 files in order to estimate the proportion of incomplete loan files with a 95% confidence level and a desired margin of error of 10%.