Chapter G, Problem 70EP

a.

To determine

Describe the factors that Mr. S should consider in establishing acceptable levels of the risk of incorrect acceptance and the risk of incorrect rejection.

Explanation of Solution

The risk of incorrect acceptance is determined using the audit risk model based on prior assessments of audit risk, analytical procedures risk and risk of material misstatement.

If the initial results indicate that the account balance is materially misstated, the risk of incorrect rejection is determined based on cost to Mr. S which expands the sample. Mr. S will typically assess the risk of incorrect rejection at lower or higher levels, as the cost of expanding the sample increases or decreases.

b.

To determine

Explain the advantages and disadvantages of Mr. S’s establishing lower levels of the risk of incorrect acceptance and the risk of incorrect rejection.

Explanation of Solution

Establishing lower levels of the risk of incorrect acceptance and the risk of incorrect rejection’s advantages are which the sample of Mr. S will have a likelihood of representing the population.

Thus, Mr. S’s sample evaluation allows Mr. S to make correct decision with respect to Mr. W’s accounts receivable.

On the other hand, the disadvantage is that Mr. S has to select and evaluate a larger sample.

c.

To determine

Calculate the sample size, if Mr. S establishes levels of the risk of incorrect acceptance and the risk of incorrect rejection of 5 percent.

Explanation of Solution

The sample size: The sample size signifies the number of items that the audit team inspects within a population of interest. Four major aspects that impact the sample size in an “attributes sampling application” are as follows:

• “Tolerable rate of deviation”.
• “Sampling risk-risk of overreliance, or risk of assessing control risk too low.
• “Expected population deviation rate”.
• “Population size”.

Step 1: Calculate the tolerable misstatement.

$\begin{array}{c}\text{Tolerable Misstatement}=\left(\text{6\%}\times \text{Accounts receivable}\right)\\ =0.06\times \$2,000,000\\ =\$120,000\end{array}$

Step 2: Calculate the expected misstatement.

$\begin{array}{c}\text{Expected Misstatement}=\left(4\%\times \text{Accounts receivable}\right)\\ =0.04\times \$2,000,000\\ =\$80,000\end{array}$

Step 3: Calculate the sample size.

$\begin{array}{c}\text{n}={\left(\frac{N\times [R\left(IR\right)+R\left(IA\right)\times SD}{TE–EE}\right)}^2\\ \text{n}={\left(\frac{1,250\times \left[1.96+1.65\right]\times \$100}{\$120,000–\$80,000}\right)}^2\\ \text{n}=127.26\text{ or }128\text{ items}\end{array}$

Notes:

N= number of individual customer accounts.

R (IA) = Risk of incorrect rejection.

R (IR) = Risk of incorrect acceptance.

SD = Standard Deviation.

TE= Tolerable Misstatement.

EE = Expected Misstatement.

Therefore, the sample size is 128 times.

d1.

To determine

Calculate the sample size, if Risk of incorrect acceptance of 5 percent, risk of incorrect rejection of 10 percent.

Explanation of Solution

Calculate the sample size.

$\begin{array}{l}\text{n}={\left(\frac{N\times [R\left(IR\right)+R\left(IA\right)\times SD}{TE–EE}\right)}^2\\ \text{n}={\left(\frac{1,250\times \left[1.65+1.65\right]\times \$100}{\$120,000–\$80,000}\right)}^2\\ \text{n}=106.35\text{ or }107\text{ items}\end{array}$

Notes:

N= number of individual customer accounts.

R (IA) = Risk of incorrect rejection.

R (IR) = Risk of incorrect acceptance.

SD = Standard Deviation.

TE= Tolerable Misstatement.

EE = Expected Misstatement.

Therefore, the sample size is 107 times.

d2.

To determine

Calculate the sample size, if Risk of incorrect acceptance of 10 percent, risk of incorrect rejection of 5 percent.

Explanation of Solution

Calculate the sample size.

$\begin{array}{l}\text{n}={\left(\frac{N\times [R\left(IR\right)+R\left(IA\right)\times SD}{TE–EE}\right)}^2\\ \text{n}={\left(\frac{ 1,250\times \left[1.96+1.28\right]\times \$100}{\$120,000–\$80,000}\right)}^2\\ \text{n}=102.52\text{ or }103\text{ items}\end{array}$

Notes:

N= number of individual customer accounts.

R (IA) = Risk of incorrect rejection.

R (IR) = Risk of incorrect acceptance.

SD = Standard Deviation.

TE= Tolerable Misstatement.

EE = Expected Misstatement.

Therefore, the sample size is 103 times.

d3.

To determine

Calculate the sample size, if Risk of incorrect acceptance of 10 percent, risk of incorrect rejection of 10 percent.

Explanation of Solution

Calculate the sample size.

$\begin{array}{l}\text{n}={\left(\frac{N\times [R\left(IR\right)+R\left(IA\right)\times SD}{TE–EE}\right)}^2\\ \text{n}={\left(\frac{1,250\times \left[1.65+1.28\right]\times \$100}{\$120,000–\$80,000}\right)}^2\\ \text{n}=83.84\text{ or }84\text{ items}\end{array}$

Notes:

N= number of individual customer accounts.

R (IA) = Risk of incorrect rejection.

R (IR) = Risk of incorrect acceptance.

SD = Standard Deviation.

TE= Tolerable Misstatement.

EE = Expected Misstatement.

 Therefore, the sample size is 84 times.

e.

To determine

Explain how the levels of sampling risks affect sample size based on the sample size calculation in requirement (c) and (d).

Explanation of Solution

It shows the inverse relationship between sampling risks and sample size by comparing the sample sizes in requirement (c) and (d). For instance by relating the sample size with 5% risk of incorrect rejection and incorrect acceptance i.e. is 128 times (requirement (c)) with the sample size when the risk of incorrect rejection is 10 percent i.e. 107 times (requirement (d1)) or the risk of incorrect acceptance is 10% i.e. 103 times (requirement (d2)), it specifies when the sampling risk increases then the sample size decreases. This provides Mr. S the smallest sample size when both the sampling risks are increased to 10%.