Chapter G, Problem 73EP

a (1)

To determine

Compute the appropriate sample size for the given combination of risk of incorrect acceptance (RIA) and risk of incorrect rejection (RIR).

Explanation of Solution

Classical variable sampling: It is a sampling method in which each record is considered as a sampling unit. Therefore, each record has an equal chance of being selected as the sample.

Sample size is calculated using mean-per-unit estimations is as follows:

$\text{n=}{\left(\frac{\text{N}\text{×}\left[\text{R}\left(\text{IR}\right)\text{+}\text{R}\left(\text{IA}\right)\right]\text{×}\text{SD}}{\text{TM}-\text{EM}}\right)}^{\text{2}}$

Where:

“n” is Sample size.

“N” is Population size (number of transactions)

“R(IR)” is Confidence factor for the risk of incorrect rejection

“R(IA)” is Confidence factor for the risk of incorrect acceptance

“SD” is Standard deviation

“TM” is Tolerable misstatement

“EM” is Expected misstatement

Compute the appropriate sample size:

Given information: N is 2,000; R (IR) is 1.96; R (IA) is 2.33; SD is $30; TM is $32,500 (5% of $650,000); and EM is $22,500.

$\begin{array}{c}\text{n=}{\left(\frac{\text{N}\text{×}\left[\text{R}\left(\text{IR}\right)\text{+}\text{R}\left(\text{IA}\right)\right]\text{×}\text{SD}}{\text{TM}-\text{EM}}\right)}^{\text{2}}\\ \text{n}={\left(\frac{2,000\times \left[1.96+2.33\right]\times \$30}{\$32,500-\$22,500}\right)}^2\\ \text{n}={\left(\frac{257,400}{\$10,000}\right)}^2\\ \text{n}=662.547\text{or 663 items}\end{array}$

Therefore, appropriate sample size is 663 items.

a (2)

To determine

Compute the appropriate sample size for the given combination of risk of incorrect acceptance (RIA) and risk of incorrect rejection (RIR).

Explanation of Solution

Compute the appropriate sample size:

Given information: N is 2,000; R (IR) is 1.65; R (IA) is 2.33; SD is $30; TM is $32,500 (5% of $650,000); and EM is $22,500.

$\begin{array}{c}\text{n=}{\left(\frac{\text{N}\text{×}\left[\text{R}\left(\text{IR}\right)\text{+}\text{R}\left(\text{IA}\right)\right]\text{×}\text{SD}}{\text{TM}-\text{EM}}\right)}^{\text{2}}\\ \text{n}={\left(\frac{2,000\times \left[1.65+2.33\right]\times \$30}{\$32,500-\$22,500}\right)}^2\\ \text{n}={\left(\frac{238,800}{\$10,000}\right)}^2\\ \text{n}=570.25\text{or 571 items}\end{array}$

Therefore, appropriate sample size is 571 items.

a (3)

To determine

Compute the appropriate sample size for the given combination of risk of incorrect acceptance (RIA) and risk of incorrect rejection (RIR).

Explanation of Solution

Compute the appropriate sample size:

Given information: N is 2,000; R (IR) is 1.65; R (IA) is 1.65; SD is $30; TM is $32,500 (5% of $650,000); and EM is $22,500.

$\begin{array}{c}\text{n=}{\left(\frac{\text{N}\text{×}\left[\text{R}\left(\text{IR}\right)\text{+}\text{R}\left(\text{IA}\right)\right]\text{×}\text{SD}}{\text{TM}-\text{EM}}\right)}^{\text{2}}\\ \text{n}={\left(\frac{2,000\times \left[1.65+1.65\right]\times \$30}{\$32,500-\$22,500}\right)}^2\\ \text{n}={\left(\frac{198,000}{\$10,000}\right)}^2\\ \text{n}=392.04\text{or 393 items}\end{array}$

Therefore, appropriate sample size is 393 items.

b)

To determine

Identify the factors that Person JH should consider in establishing the RIA and RIR.

Explanation of Solution

Person JH would decide the risk of incorrect acceptance utilizing the audit risk model and past appraisals of audit risk, risk of material misstatement, and analytical procedure risk.

Person JH would then decide the risk of incorrect rejection dependent on the expenses of growing the sample and choosing extra items. As these costs increment, the risk of incorrect rejection ought to be evaluated at lower level

c)

To determine

Explain the manner that the sample size is influenced by the RIA and RIR based on sub-part a) calculations.

Explanation of Solution

A contrast of above sub-parts “a (1)” and “a (2)” shows that as the risk of incorrect rejection escalates from 5 percent to 10 percent, the sample size declines from 663 items to 571 items. Thus, the risk of incorrect rejection has an inverse relationship with sample size.

A contrast of above sub-parts “a (1)” and “a (2)” shows that as the risk of incorrect rejection escalates from 5 percent to 10 percent A contrast of above sub-parts “a (2)” and “a (3)” , as the risk of incorrect acceptance rises from 1 percent to 5 percent, the sample size falls from 571 items to 393 items. Thus, the risk of incorrect acceptance has an inverse relationship with sample size.

d (1)

To determine

Compute the precision interval for given combination of risk of incorrect acceptance (RIA) and risk of incorrect rejection (RIR).

Explanation of Solution

Precision interval: It is an interval of sample assessments that pedals the audit team’s contact to the risk of incorrect acceptance and risk of incorrect rejection

It is calculated by using following formula:

$\text{Precision interval = Sample estimate }\pm \text{ Precision}$

Precision is calculated as follows:

$\text{Precision = N }\times \text{ R}\left(\text{IR}\right)\times \left(\frac{\text{SD}}{\sqrt{\text{n}}}\right)$

Compute the precision interval:

Given information: Sample estimate is $660,000

Precision is $5,429(working note 1)

$\begin{array}{c}\text{Precision interval = Sample estimate }\pm \text{ Precision}\\ \text{=}\text{\$660,000}\pm \text{\$5,429}\\ \text{=}\text{\$654,571}\text{to \$665,429}\end{array}$

Therefore, precision interval is $654,571 to $665,429.

Working note 1: Compute the precision:

Given information: N is 2,000; R (IA) is 2.33; SD is $30; and n is 663.

$\begin{array}{c}\text{Precision = N }\times \text{ R}\left(\text{IR}\right)\times \left(\frac{\text{SD}}{\sqrt{\text{n}}}\right)\\ =2,000\times 2.33\times \left(\frac{\$30}{\sqrt{663}}\right)\\ =2,000\times 2.33\times 1.16504854368\\ =\$5,429\left(\text{rounded off}\right)\end{array}$

d (2)

To determine

Compute the precision interval for given combination of risk of incorrect acceptance (RIA) and risk of incorrect rejection (RIR).

Explanation of Solution

Compute the precision interval:

Given information: Sample estimate is $660,000

Precision is $5,850(working note 2)

$\begin{array}{c}\text{Precision interval = Sample estimate }\pm \text{ Precision}\\ \text{=}\text{\$660,000}\pm \text{\$5,850}\\ \text{=}\text{\$654,150}\text{to \$665,850}\end{array}$

Therefore, precision interval is $654,150 to $665,850.

Working note 2: Compute the precision:

Given information: N is 2,000; R (IA) is 2.33; SD is $30; and n is 571.

$\begin{array}{c}\text{Precision = N }\times \text{ R}\left(\text{IR}\right)\times \left(\frac{\text{SD}}{\sqrt{\text{n}}}\right)\\ =2,000\times 2.33\times \left(\frac{\$30}{\sqrt{571}}\right)\\ =2,000\times 2.33\times 1.25546125646\\ =\$5,850\left(\text{rounded off}\right)\end{array}$

d (3)

To determine

Compute the precision interval for given combination of risk of incorrect acceptance (RIA) and risk of incorrect rejection (RIR).

Explanation of Solution

Compute the precision interval:

Given information: Sample estimate is $660,000

Precision is $4,994(working note 3)

$\begin{array}{c}\text{Precision interval = Sample estimate }\pm \text{ Precision}\\ \text{=}\text{\$660,000}\pm \text{\$4,994}\\ \text{=}\text{\$655,006}\text{to \$664,994}\end{array}$

Therefore, precision interval is $655,006 to $664,994.

Working note 1: Compute the precision:

Given information: N is 2,000; R (IA) is 1.65; SD is $30; and n is 393.

$\begin{array}{c}\text{Precision = N }\times \text{ R}\left(\text{IR}\right)\times \left(\frac{\text{SD}}{\sqrt{\text{n}}}\right)\\ =2,000\times 1.65\times \left(\frac{\$30}{\sqrt{393}}\right)\\ =2,000\times 1.65\times 1.51330039719\\ =\$4,994\left(\text{rounded off}\right)\end{array}$

d)

To determine

Explain Person JH’s conclusion regarding Company M’s accounts receivables.

Explanation of Solution

Company M’s accounts receivable were reported at $650,000, Person JH would conclude that in all the three of given cases the accounts receivable were not misstated. It would be made as the maximum variances between the recorded balance and the utmost limit from the precision interval do not go beyond tolerable misstatement.

e)

To determine

Explain the manner that the precision interval is influenced by the RIA and RIR based on sub-part d) calculations.

Explanation of Solution

A contrast of above sub-parts “d (1)” and “d (2)” shows that as the risk of incorrect rejection rises from 5 percent to 10 percent, the precision escalates (mainly from the outcome of the risk of incorrect rejection on sample size). The upper level of precision results in a broader precision interval, affecting it more similarly that Person JH will reject the account balance as not properly stated.

A contrast of above sub-parts “d (2)” and “d (3)” shows that as the risk of incorrect acceptance escalates from 1 percent to 5 percent, the precision falls. The lower level of precision outcomes in a thinner precision interval, affecting it more similarly that Person JH will accept the account balance as fairly stated.