Case 9.6

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...