Problem DC 10 1 Page 547 Original Problem

Problem DC 10 -1, Page 547 (Original Problem) When Marge Simpson, PA, audited the Candle Company inventory, a random sample of inventory types was chosen for physical observation and price testing. The sample size was 80 different types of candles and candle-making inventory. The entire inventory contained 1, 740 types, and the amount in the inventory control account was $166, 000. Simpson had already decided that a misstatement of as much as $6, 000 in the account would not be material. The audit work revealed the following eight errors in the sample of 80. Book Value (a) Audit value Error Amount (b) $600. 00 $622. 00 $(22. 00) (0. 037) 15. 50 14. 50 1. 00 0. 065 65. 25 31. 50 33. 75 0. 517 83. 44 53. 45 29. 99 0. 359 16. 78 15. 63 1. 15 0. 069 78. 33 12. 50 65. 83 0. 840 13. 33 14. 22 (0. 89) (0. 067) 93. 87 39. 87 54. 00 0. 575 $966. 50 $803. 67 $162. 83 Percent misstatement for the whole sample: 162. 83/966. 50 = 17%. Could use 20% for Slide 51 % Misstatement b/a Sampling 1

Assume the sample was chosen using Monetary Unit Sampling, and the sampling risk is 10%. Simpson also assumes that the Misstatement Assumption for zero misstatements is 50% for both over and under misstatements. Is the inventory materially misstated? Total Population = Sample Size = 166, 000. 00 80 ARACR = 10. 00% Upper Misstatement Unit Error Assumption 50. 00% Lower Misstatement Unit Error Assumption 50. 00% Materiality This part of the population is not in the sample. Thus zero misstatements have been found here because these items have not been examined. The misstatements of the remaining part of the population are not known. Thus a misstatement assumption has to be made. E. g. they are 50% misstated. See next slide. 6, 000 Sample The misstatements of the sample, as percentages, are known. See previous slide 29. Sampling 2

Could use 20% based on actual deviations. But still would be material Overstatements Number of Misstatements (1) Upper Precision Limit Portion (2) Recorded Value (3) Misstatement Unit Error Assumption (4) Bound Portion 2 x 3 x 4 0 0. 029 166, 000 0. 500 $2, 407 1 0. 019 166, 000 0. 840 2649 2 0. 018 166, 000 0. 575 1718 3 0. 016 166, 000 0. 517 1373 4 0. 016 166, 000 0. 359 954 5 0. 015 166, 000 0. 069 172 6 0. 015 166, 000 0. 065 162 Overstatements Upper Precision Limit Initial Misstatement Bound 0. 128 $9, 435 Sampling 3

Sample size ACTUAL NUMBER OF DEVIATIONS FOUND 0 1 2 3 4 5 6 7 8 9 10 10 PERCENT RISK OF OVER RELIANCE (RIA or Beta Risk) 20 25 10. 9 8. 8 18. 1 14. 7 30 35 40 45 50 55 60 65 70 75 80 90 100 125 150 200 300 400 500 7. 4 12. 4 6. 4 10. 7 5. 6 9. 4 5. 0 8. 4 4. 6 7. 6 4. 2 6. 9 3. 8 6. 4 3. 5 5. 9 3. 3 5. 5 3. 1 5. 1 2. 9 4. 8 1. 9 2. 6 4. 3 1. 8 2. 3 3. 9 1. 9 3. 1 1. 6 2. 6 1. 2 2. 0 0. 8 1. 3 0. 6 1. 0 0. 5 0. 8 24. 5 20. 0 30. 5 24. 9 36. 1 29. 5 41. 5 34. 0 46. 8 38. 4 51. 9 42. 6 56. 8 46. 8 61. 6 50. 8 66. 2 54. 8 16. 8 14. 5 12. 8 11. 4 10. 3 9. 4 8. 7 8. 0 7. 5 7. 0 6. 6 5. 9 1. 6 5. 3 4. 3 3. 6 2. 7 1. 8 1. 4 1. 1 21. 0 18. 2 16. 0 14. 3 12. 9 11. 8 10. 0 9. 3 8. 7 8. 2 7. 3 1. 6 6. 6 5. 3 4. 4 3. 4 2. 3 1. 7 1. 4 24. 9 21. 6 19. 0 17. 0 15. 4 14. 1 12. 9 12. 0 11. 1 10. 4 9. 8 8. 7 1. 5 7. 9 6. 3 5. 3 4. 0 2. 7 2. 0 1. 6 28. 8 24. 9 22. 0 19. 7 17. 8 16. 3 15. 0 13. 9 12. 1 11. 3 10. 1 1. 9 9. 1 7. 3 6. 1 4. 6 3. 1 2. 4 1. 9 32. 5 28. 2 24. 9 22. 3 20. 2 18. 4 16. 9 15. 7 14. 6 13. 7 12. 8 11. 5 10. 3 8. 3 7. 0 5. 3 3. 5 2. 7 2. 1 36. 2 31. 4 27. 7 24. 8 22. 5 20. 5 18. 9 17. 5 16. 3 15. 2 14. 3 12. 8 11. 5 9. 3 7. 8 5. 9 3. 0 2. 4 39. 7 34. 5 30. 5 27. 3 24. 7 22. 6 20. 8 19. 3 18. 0 16. 8 15. 8 14. 1 12. 7 10. 2 8. 6 6. 5 4. 3 3. 3 2. 6 43. 2 37. 6 33. 2 29. 8 27. 0 24. 6 22. 7 21. 0 19. 6 18. 3 17. 2 15. 4 13. 9 11. 2 9. 4 7. 1 4. 7 3. 6 2. 9 46. 7 40. 6 35. 9 32. 2 29. 2 26. 7 24. 6 22. 8 21. 2 19. 8 18. 7 16. 7 15. 0 12. 1 10. 1 7. 6 5. 1 3. 9 3. 1 The Upper and Lower Bounds must be determined. What do these mean? Obtain the Precision Limits from this table – 10% RIA Sampling 4

Understatements Number of Misstatements (1) Upper Precision Limit Portion (2) Recorded Value (3) Misstatement Unit Error Assumption (4) Bound Portion 2 x 3 x 4 0 0. 029 166, 000 0. 500 $2, 407 1 0. 019 166, 000 0. 067 211 2 0. 018 166, 000 0. 037 111 Understatements Upper Precision Limit Initial Misstatement Bound 0. 066 $2, 729 Sampling 5

Offsetting Adjustments Number of Misstatements Misstatement Unit Error Assumption (a) Sample Size (b) Recorded Population (c) Point Estimate a(c/b) Initial Overstatement Bounds $9, 435 Understatement Misstatements 1 0. 067 80 166, 000 139 (139) 2 0. 037 80 166, 000 77 (77) Total 0. 104 80 166, 000 216 (216) Adjusted Overstatement Bound 9, 219 Initial Understatement Bound 2, 729 Overstatement Misstatements 1 0. 840 80 166, 000 1743 (1743) 2 0. 575 80 166, 000 1193 (1193) 3 0. 517 80 166, 000 1073 (1073) 4 0. 359 80 166, 000 745 (745) 5 0. 069 80 166, 000 143 (143) 6 0. 065 80 166, 000 135 (135) Total 2. 425 5032 (5032) Adjusted Understatement Bound -2303 As the Upper Bound is greater than the materiality of $6, 000, there may be a material misstatement Sampling 6

The Decision Rule Tolerable misstatement ($6, 000) $6, 000 Reject $2, 303 $9, 219 LMB UMB Problem Since materiality is $6, 000, there may be material misstatement. Sampling 7

Problem EP 10 -5, Page 546 – Point Estimate Assume the results shown below were obtained from a stratified sample. Required : Apply the ratio calculation method (Point Estimate) to each stratum to calculate the projected likely misstatement (PLM). What is the PLM for the entire sample? SAMPLE RESULTS Stratum Population Size Recorded Amount Sample Misstatement Amount PLM 1 6 $100, 000 $(600) 2 80 75, 068 23 21, 700 (274) (948) 3 168 75, 008 22 9, 476 (66) (522) 4 342 75, 412 22 4, 692 (88) (1, 414) 5 910 74, 512 23 1, 973 23 869 1, 506 $400, 000 96 $137, 841 $(1, 005) (2, 615) Total 1: (600)/100, 000 x 100, 000 = (600) 2: (274)/21, 700 x 75, 068 = (948) 3: (66)/9, 476 x 75, 008 = (522) 4: (88)/4, 692 x 75, 412 = (1, 414) 5: 23/1, 973 x 74, 512 = 869 Cannot do this: (1, 005)/137, 841 x 400, 000 = (2916) NOT CORRECT! Sampling 8

Problem DC 10 -1, Page 547 – The Difference Method Sample size is 80 Population consists of 1, 740 types Book Value (a) Audit value Error Amount (b) % Misstatement b/a $600. 00 $622. 00 $(22. 00) (0. 037) 15. 50 14. 50 1. 00 0. 065 65. 25 31. 50 33. 75 0. 517 83. 44 53. 45 29. 99 0. 359 16. 78 15. 63 1. 15 0. 069 78. 33 12. 50 65. 83 0. 840 13. 33 14. 22 (0. 89) (0. 067) 93. 87 39. 87 54. 00 0. 575 $966. 50 $803. 67 $162. 83 Projected likely misstatement = ((162. 83)/80)X 1740 = (3541. 56). Remember that TM = $6, 000 This is a little over half of what is considered material (tolerable) and would likely be acceptable Sampling 9