Audit Sampling: Confidence Interval, Test, and Materiality
An audit partner is evaluating the accounts receivable balance of a retailer. The population contains 2,400 invoices with a recorded total of $1,850,000. A simple random sample of 64 invoices is selected without replacement. The audit team finds an average overstatement of $18.50 per invoice and a sample standard deviation of $12.00. The client's materiality threshold for receivables is a $40,000 total overstatement. The partner wants a 95% confidence interval for the population mean and total misstatement, a formal test at the 5% level of whether the true mean overstatement exceeds $15.00, and advice on whether the evidence supports concluding that the population is materially misstated. A junior auditor suggests ignoring the finite population correction because the sample is “large”.
1. Construct the 95% confidence interval for the mean overstatement and for the total overstatement, with and without the finite population correction.
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Standard error without FPC = s/√n = 12/√64 = 12/8 = 1.500. The FPC factor is √[(N − n)/(N − 1)] = √[(2,400 − 64)/(2,400 − 1)] = √(2,336/2,399) = 0.9868. Adjusted SE = 1.500 × 0.9868 = 1.480.
Using t with 63 degrees of freedom, t(0.025,63) ≈ 1.998. Margin = 1.998 × 1.480 = 2.957. Mean CI = 18.50 ± 2.96, i.e. $15.54 to $21.46. Total misstatement CI = 2,400 × the mean limits = $37,296 to $51,504.
Without FPC, margin = 1.998 × 1.500 = 2.997; mean CI $15.50 to $21.50; total CI $37,200 to $51,600. The FPC narrows the interval only slightly because the sample is 2.7% of the population. Ignoring it is conservative but not materially misleading here; the junior auditor's shortcut is acceptable only because the sampling fraction is small.
2. Test whether the true mean overstatement exceeds $15.00, then test whether it exceeds the materiality threshold. What should the partner conclude?
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Test 1: H0: μ = 15 versus H1: μ > 15. t = (18.50 − 15.00)/1.480 = 2.365 with 63 df. The 5% critical value is about 1.669; p ≈ 0.011. Reject H0. There is significant evidence that the mean overstatement exceeds $15 per invoice.
Materiality threshold in mean terms = $40,000/2,400 = $16.667. Test 2: H0: μ = 16.667 versus H1: μ > 16.667. t = (18.50 − 16.667)/1.480 = 1.238; p ≈ 0.11. Fail to reject at the 5% level. The 95% CI lower limit of $15.54 is below $16.667, so the data are also consistent with a total misstatement below $40,000.
The evidence shows a real overstatement problem, but it does not establish material misstatement at the 95% confidence level. The partner should not conclude that the population is materially misstated; further sampling or a larger sample would be needed. This is a Type II error risk: failing to detect material misstatement if it exists.
3. How large a sample is needed for a 95% margin of error of about $2.00 per invoice, allowing for the finite population correction?
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First ignore the FPC: n0 = (zσ/E)^2. Using z = 1.96, σ ≈ 12, and E = 2: n0 = (1.96 × 12/2)^2 = (11.76)^2 = 138.3, so 139.
Apply the FPC: n = n0 / [1 + (n0 − 1)/N] = 139 / [1 + 138/2,400] = 139 / 1.0575 = 131.4, so round up to 132 invoices.
Check: with n = 132, SE = 12/√132 × √[(2,400 − 132)/(2,399)] = 1.0445 × 0.9723 = 1.016; the margin is about 1.98, close to $2.00. A sample of 132 is sufficient, but the auditor should allow for the fact that σ was estimated from the pilot sample; if the population is more variable, a larger sample is prudent.