Submitted by: Submitted by kelliann
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Pages: 5
Category: Business and Industry
Date Submitted: 11/24/2011 12:50 PM
Audit Case 9.6B
1. A)
Customer Number Total
03-5308 644,329
03-8275 617,895
03-7802 514,105
01-5661 506,261
01-6827 482,657
02-7398 478,170
04-9069 468,557
02-0149 467,119
05-3038 460,106
05-2034 407,565
B)
Sample Size = (Sampling Population Book Value) x Confidence Factor
(Tolerable-Expected Misstatement)
Confidence Factor= 1.6 because we assume a moderate level of assessment of risk of material misstatement and a moderate desired level of confidence. (Based on Garrett and Schulzke’s “Confidence Factors for Nonstatistical Sampling” table)
Expected Misstatement= By adding the 10 accounts that were used in direct testing and subtracting them from the book value of 12,881,551, we derived a new sampling book population of 7,834,787. The level of expected misstatement was found by taking the total accounts receivable balance totaling $7,834,787 and multiplying if by .05% because in prior years’ audit aggregate misstatements of less than .05% of the accounts receivable balance were discovered via customer confirmation testing. As auditors, we use prior year experience to estimate expected misstatement. Therefore 7,834,787 x .005 = 39,173.94.
Sample Size = 7,834,787 x 1.6
400,000 – 39,173.94
Sample Size = 34.74 so 35
C)
Determine Sample Size
Given: Tolerable misstatement (TM) $600,000
Risk of incorrect acceptance (RIA) .05
Expected Misstatement (EM) $39,173.94
Book Value of Population (BV) $7,834,787
Calculate:
BV x Reliability Factor
a. Sample Size (n) = ────────────────────────────
TM - (EM x Expansion...