Miller Motor Co. inventory sampling problem.
The following formula will calculate the required sample size. You can memorize this
the night before the test.
(𝒁𝜷 +𝒁𝜶⁄ )𝒔𝒙
𝟐
̅̅̅̅̅
𝑻𝑴
= √𝒏
𝒁𝜷 => 𝜷 𝑟𝑖𝑠𝑘, Risk of Incorrect Acceptance
𝒁𝜶/𝟐 => 𝜶 𝑟𝑖𝑠𝑘, Risk of Incorrect Rejection
Now, let’s try to learn something about sampling.
The purpose of this assignment is to demonstrate how statistical theory helps auditors
manage detection risk. This exercise uses mean-per-unit (MPU) sampling because MPU
was covered in from your statistics class. MPU sampling also most clearly illustrates the
relationship between sample size and risk. Accounting firms all use software incorporating
sophisticated sampling techniques such as: probability proportionate to size, dollar-unit
sampling and stratified sampling. These methods reduce audit cost because they typically
require smaller sample sizes than MPU sampling. However, the underlying theory is more
complicated. This assignment utilizes basic confidence intervals and hypothesis tests
covered in introductory statistics classes.
A secondary objective is to familiarize you with “audit documentation” or “work
papers.”
Miller Motor Co.’s “year-end physical inventory worksheet” has two sections. The
top section includes 1,703 different parts or SKU (stock keeping unit) numbers. The second
section includes the 4 different models of new vehicles in their inventory. The auditors chose
to audit the parts inventory separately from the vehicle inventory. Auditing the two inventories
separately is a simple form of stratification. Including the 5 tail lamp bulbs costing $13.26 in
the same pool as the 3 Impala LTZs costing $86,199.00 would result in a very large standard
deviation (and variance). The formula at the top of the page demonstrates that for any level of
risk the required sample size increases as the standard deviation (variance) increases.
Although this exercise does not involve any further stratification, notice that there is a
great deal of variance in the parts prices. Auditors might be able to further reduce the
required sample size by separating parts that cost more than $1,000 from parts costing less
than $1,000. Accounting firms have software to determine the optimal strata which will
minimize the required sample size for inventory observations.
Designing a sampling plan requires careful consideration. Review the two examples
for new vehicles on the back of the “year-end physical inventory worksheet.” We shouldn’t
assume that grouping items will always reduce the standard deviation. Notice that the
standard deviation is higher when the cars are grouped by model than when each vehicle is
counted as an individual item. Auditing requires judgment.
Review the “year-end physical inventory worksheet” for the parts inventory
The attached work sheet shows the results for our sample of parts inventory. From the
population of 1,703 SKU lines, we randomly selected 387 SKU lines to test for both the
quantity and the cost of the items. We tested the quantity by counting the actual number of
items on actually hand and comparing this with the quantity in the accounting records. We
1
tested the cost by comparing the cost of the items on purchase invoices with the cost in the
accounting records.
We detected two types of errors in our sample. There were instances where the
quantity on hand differed from the quantity in the client’s accounting records. For instance, on
line 3 the auditor counted 239 units of SKU 1851974 during their year-end physical inventory
observation even though the client reports having 240 units. We also detected an instance
where the cost according to the purchase invoice differed from the cost included in the
inventory balance. On line 986, the client’s accounting records indicate SKU 6730291 cost
$132.92 per unit but the purchase invoice revealed these items actually cost $123.92.
We audited the quantity and cost for the 387 items in our sample. The extended cost
was calculated for each SKU item by multiplying the quantity times the cost. For example the
extended cost for line 1, SKU 4470312 would be 5 x $2.65 = $13.26. The sum of the
extended balances for the 387 part lines in the sample came to $441,567.00.
Auditing standards require us compare our projected sample results to the
population. First, we project our sample results to the population to calculate our best
estimate of the balance. Then we compare our best estimate of the balance with the reported
balance to determine the projected overstatement error. A balance is overstated when the
reported book value exceeds our estimate of the balance. Remember that auditing standards
do not require balances to be exact. Auditing standards require auditors to “obtain reasonable
assurance about whether the financial statements are FREE OF MATERIAL MISSTATEMENT.”
Auditing standards require auditors obtain “sufficient, appropriate evidence” the financial
statements “PRESENT FAIRLY, IN ALL MATERIAL RESPECTS.”
𝑎𝑢𝑑𝑖𝑡𝑒𝑑 𝑏𝑎𝑙𝑎𝑛𝑐𝑒
$441,567
𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑒𝑑 𝑏𝑎𝑙𝑎𝑛𝑐𝑒 = 𝑁 × (
)
1,703 × (
) = $1,943,123
𝑛
387
𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑒𝑑 𝑒𝑟𝑟𝑜𝑟 = 𝑏𝑜𝑜𝑘 𝑏𝑎𝑙𝑎𝑛𝑐𝑒 − 𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑒𝑑 𝑎𝑢𝑑𝑖𝑡 𝑏𝑎𝑙𝑎𝑛𝑐𝑒
𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑒𝑑 𝑜𝑣𝑒𝑟𝑠𝑡𝑎𝑡𝑒𝑚𝑒𝑛𝑡 𝑒𝑟𝑟𝑜𝑟 𝑜𝑓
$22,437.03 = $1,965,560.03, −$1,943,123.00
The work papers state that tolerable misstatement (TM) for the parts inventory is
$200,000. But simply calculating a projected error that is less than tolerable misstatement
does not necessarily mean there is an acceptably low level of detection risk. Auditing
standards caution that “AS THE PROJECTED MISSTATEMENT APPROACHES OR EXCEEDS TOLERABLE
MISSTATEMENT, THE MORE LIKELY THAT ACTUAL MISSTATEMENT IN THE POPULATION EXCEEDS
TOLERABLE MISSTATEMENT” (AU-C 530.27).
Auditing standards do not require the use of statistical sampling. We might compare
our $22,437 projected misstatement to the $200,000 tolerable misstatement and conclude
that there is an acceptably low risk any overstatement would exceed $200,000 because the
projected error is quite a bit smaller than the tolerable misstatement. Or we might use
methods learned in statistics classes to quantify the risk.
“AS THE PROJECTED MISSTATEMENT APPROACHES OR EXCEEDS TOLERABLE MISSTATEMENT, THE
MORE LIKELY THAT ACTUAL MISSTATEMENT IN THE POPULATION EXCEEDS TOLERABLE
MISSTATEMENT” (AU-C 530.27). How do we determine when the projected error
is too
close to tolerable misstatement? Statistics provides methods to estimate the probability or
likelihood that the actual misstatement exceeds tolerable misstatement.
2
(A)
The normal distribution indicates that if the projected error
is 1.645 standard deviations less than the tolerable misstatement there
is a 5% probability inventory is materially overstated. If the projected
error 1.282 standard deviations less than TM there is a 10% probability
inventory is materially overstated. If 0.842 standard deviations less than
TM there is a 20% probability it is materially overstated. As the projected
error gets larger and approaches the TM the likelihood that the
inventory balance is materially overstated increases.
The parts inventory has a $1,965,560 ($1,154.18) book balance
and tolerable misstatement is $200,000 ($117.44). So, the parts inventory
would be materially overstated if the actual balance is less than
$1,765,560 ($1,036.74) ($1,965,560 - $200,000 TM).
(B)
The normal distribution indicates when the sample mean is
1.645 standard deviations greater than $1,036.74 there is a 5% probability that inventory is
materially overstated. If is 1.282 standard deviations greater than $1,036.74 there is a 10%
probability. If is 0.84 standard deviations greater than $1,036.74 there is a 20% probability.
As the sample mean gets smaller and approaches $1,036.74 it becomes more likely that the
balance is materially overstated.
You should reflect carefully on paragraphs (A) and (B) to make sure you
understand that they say the same thing.
Hypothesis Test – a review
If the parts inventory has a $1,965,560 ($1,154.18) book balance and tolerable
misstatement is $200,000 ($117.44) then the parts inventory would be materially overstated if
the actual balance is less than $1,765,560 ($1,036.74). We can test the null hypothesis (Ho:)
that the unknown true mean of the inventory is less than $1,036.74 ($1,765,560). The power of
statistics is in the rejection of the null hypothesis. So if we reject HO: then we can accept the
alternate hypothesis (HA:) which states that the true mean must be greater than
HA: > $1,036.74 ($1,765,000).
𝐻𝑂 ∶ 𝜇 < $1,036.74
𝑜𝑟 (
𝐻𝐴 ∶ 𝜇 > $1,036.74
𝑜𝑟 (
$1,765,560
1,703
$1,765,560
1,703
)
)
The auditor’s “year-end physical inventory worksheet” shows the sample mean ( ) is
$1,141.00 for the 387 items which were sampled. Logically, the more by which exceeds
Ho: μ<$1,036.74, the lower the risk the actual mean of the parts inventory is less than
$1,036.74. Conversely, as the amount by which exceeds Ho: μ<$1,036.74 decreases and
approaches $1,036.74 then the risk that the mean might actually be less than $1,036.74
increases
3
Table 1
projected value and projected error
sample
mean
N
projected
value
book
balance
1,154.18
1,141.00
1,094.09
1,060.18
1,036.74
1,703 1,965,568.54
1,943,123.00
1,863,235.27
1,805,486.54
1,765,559.71
1,965,560.03
projected
overstatement
-8.51
22,437.03
102,324.76
160,073.49
200,000.33
Reflect on the above table. Make sure you understand how each of the
projected values was calculated and how each projected overstatement error was calculated.
Make sure you understand the logic underlying each of the following statements:
If
If
If
If
If
is $1,154 there is almost no risk the book balance is overstated by $200,000
is $1,141 there is a small risk the book balance is overstated by $200,000
is $1,094 the risk the book balance is overstated by $200,000 increases
is $1,060 the risk the book balance is overstated by $200,000 continues to increase
is $1,036 there is a 50% risk the book balance is overstated by $200,000
Let’s return to the hypothesis test. Later we will discuss how to determine the
appropriate sample size and Critical Value but for now we will arbitrarily set the
Critical Value half way between the book value $1,965,560 ($1,154.18) and the hypothesis
test $1,765,560 (Ho: μ = $1,036.74).
We divide the $1,865,560 midpoint by 1,703 which results in $1,095.46.
𝐻𝑂 ∶ 𝜇 < $1,036.74
𝑜𝑟 (
𝐻𝐴 ∶ 𝜇 > $1,036.74
𝑜𝑟 (
$1,765,560
1,703
$1,765,560
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 (𝐂𝐕) = $1,095.46 𝑜𝑟 (
)
)
1,703
$1,865,560
1,703
)
Whenever is greater than $1,095.46 the
projected balance will be greater than $1,865,560 and
the projected error will be less than $100,000 ($58.72).
Whenever is less than $1,095.46, the projected
balance will be less than $1,865,560 and the
projected error will be more than $100,000 ($58.72).
Make sure you understand why the risk that the parts
inventory is materially overstated is greater when is
less than $1,095.46 than when is more than
$1,095.46.
The audited balances for the 387 items selected from the parts inventory have a
sample mean of $1,141.00. Because exceeds the $1,095.46 Critical Value, we reject Ho:
and accept the HA: that the true balance is greater than $1,036.74 ($1,765,560).
4
We conclude that the true mean is greater than $1,036.74 ($1,765,560). Because the
sample mean is so much greater than the 𝑯𝑶 ∶ 𝝁 < $𝟏, 𝟎𝟑𝟔. 𝟕𝟒 it is very unlikely the sample
came from a population with a mean that is less than $1,036.74 ($1,765,560).
Just for Fun
From your statistics class, you might remember that we use the standard error of the
means to evaluate sample results. The standard error the means is the standard deviation for
the distribution of sample means and is calculated by dividing the standard deviation for the
𝑠𝑥
population by the square root of the sample size, 𝑠𝑒 = √𝑛
.
Table 2
difference in std errors converted to probabilities
sample
mean
hypoth
mean
Ho:
1,154.18 1,036.74
If
If
If
If
If
estimate
of
error
n
117.44 387
sample
std dev
880.00
std
error
differ in
std errors probability
"z"
2.63
0.004
1,141.00
1,094.09
104.26
57.35
44.73
2.33
1.28
0.010
0.100
1,060.18
1,036.74
23.44
-0.01
0.52
0.00
0.300
0.500
is $1,154 there is less than a 0.01% risk the book balance is overstated by $117.44 ($200,000/1,703)
is $1,141 there is a 0.01% risk the book balance is overstated by $200,000
is $1,094 there is a 0.10% risk the book balance is overstated by $200,000
is $1,060 there is a 30% risk the book balance is overstated by $200,000
is $1,036 there is a 50% risk the book balance is overstated by $200,000
Calculating the Critical Value
The normal distribution allows us to quantify the risk of reaching an incorrect
conclusion. If inventory is materially overstated, the risk of concluding inventory is not
materially overstated can be limited to 10% by setting the Critical Value 1.282 standard
deviations to the right of the hypothetical mean.
The “year-end physical inventory worksheet” indicates the best estimate of the
standard deviation is $880.00. In order to limit the risk of incorrectly concluding the inventory
balance is not materially overstated when the true mean is less than $1,036.74 ($1,765,560)
the Critical Value in the example is $1,094.09. We calculated the Critical Value by adding
1.282 standard deviations to our hypothetical mean.
𝑠
𝑥
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 = 𝜇𝑜 + 𝑍𝛽 √𝑛
𝐶𝑉 = $1,036.74 + 1.282
$880.00
√387
= $1,094.09
Continuing with the example, the of $1,141 exceeds the $1,094.09 CV. So, we can
reject Ho: with a 10% level of β risk, accept the HA: and conclude that the actual mean is
5
greater than $1,036.74 ($1,765,560). There is less than a 10% probability we would obtain
greater than $1,094.09 from a population with a mean of $1,036.74. This assumes a sample
size of 387 and a standard deviation of $880.00.
If we wanted to limit the risk of concluding inventory is not materially overstated to 5%
we would set the Critical Value 1.645 standard deviations to the right of the hypothetical
mean and the Critical Value would be $1,110.33.
𝑠
𝑥
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 = 𝜇𝑜 + 𝑍𝛽 √𝑛
𝐶𝑉 = $1,036.74 + 1.645
$880.00
√387
= $1,110.33
The of $1,141 in our example still exceeds the $1,110.33 Critical Value. So, we can
reject Ho: with a 5% level of β risk, accept the HA: and conclude that the actual mean is
greater than $1,036.74 ($1.750.560). There is less than a 5% probability we would obtain
greater than $1,110.33 from a population with a mean of $1,036.74 (again assuming n = 387
and std. dev. = $880.00).
How did we determine the sample size of n =387?
So far we have discussed the risk of incorrectly concluding
that the book value of the inventory is fairly presented when it is
materially overstated. In your statistics class, this was the risk of
incorrect acceptance: Type II error or β risk. There is also the risk of
incorrectly concluding the inventory value of is incorrect when it is
correct. This is the risk of incorrect rejection: Type I error or α risk.
conclusion / opinion
unmodified
inventory
adverse
Type I
fairly
presented
inventory
correct
materially
Beta
overstated
error
Alpha
error
Type II
correct
Typically, hypothesis tests are associated with Type II / β risk
and confidence intervals with Type I / α risk.
Remember that confidence and risk are complements; they add up to one. If we are
70% confident then there is 30% risk of error.
Confidence Interval - review
We can create an interval around the book value,
$1,965,560 ( $1,154.18). This interval is similar to a
confidence interval except that a confidence interval would
be centered on a sample mean rather than the book value.
Type II
β risk
z score
1 tail
0.010
0.025
0.050
0.075
0.100
0.150
0.200
0.300
Type I
α risk
2 tail
2.326
1.960
1.645
1.440
1.282
1.036
0.840
0.524
0.02
0.05
0.10
0.15
0.20
0.30
0.40
0.60
While we are more
concerned about concluding
inventory is fairly presented if the balance is materially
overstated, we are also concerned that we might conclude
inventory is incorrect when it is actually correct. Let’s
assume the book value is correct and we want to limit the risk
of a Type I ( α ) error to 30%. We want to be 70% confident
we won’t conclude the inventory book value is incorrect when
it is correct. In order to achieve 70% confidence, the normal
distribution indicates that the lower and upper limits of the
interval would be ± 1.036 standard deviations to the left and
to the right of the average book value.
6
𝑠
𝑥
Again, the standard error of the means equals 𝑠𝑒 = √𝑛
. The interval would be
𝑠
𝑥
𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝐵𝑉 ∓ 𝑍 √𝑛
. The best estimate of the standard deviation of the population,
at this time, is again $880.00 and our sample size is again n = 387. In this case, the interval
turns out to be ± $46.36 or a total of $92.69.
The interval in this example would be
𝑠
̅̅̅̅ ± 𝑍𝛼/2 𝑥
𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 = 𝐵𝑉
√𝑛
̅̅̅̅ − 𝑍𝛼
𝐋𝐨𝐰𝐞𝐫 𝐋𝐢𝐦𝐢𝐭 = 𝐵𝑉
2
𝐿𝐿 = $1,154.18 − 1.036
𝑠𝑥
̅̅̅̅ + 𝑍𝛼/2
𝑼𝒑𝒑𝒆𝒓 𝑳𝒊𝒎𝒊𝒕 = 𝐵𝑉
√𝑛
880
√387
= $1,107.84
𝑈𝐿 = $1,154.18 + 1.036
𝑠𝑥
√𝑛
880
√387
= $1,200.52
Planning the Inventory Audit and Determining the Sample Size
We need to determine the sample size in the planning stages of the audit, prior to the
Jan. 1st physical inventory observation. The work papers indicate tolerable misstatement for
the parts inventory has been set as $200,000, which is approximately 10% of the $1,965,560
balance. Auditing standards require auditors to “obtain reasonable assurance about whether
the financial statements are FREE OF MATERIAL MISSTATEMENT.” Auditing standards require
auditors obtain “sufficient, appropriate evidence” the financial statements “ PRESENT FAIRLY, IN
ALL MATERIAL RESPECTS.”
Auditors are typically concerned that asset accounts are overstated. If tolerable
misstatement is $200,000, we need evidence that the actual balance of the parts inventory is
greater than $1,765,560.
The interval between the hypothetical mean Ho: μ >$1,036.74 ($1,765,560) and the
average book value $1,154.17 ($1,965,560) is $117.44 ($200,000). This will remain the same
regardless of how we allocate Alpha Risk and Beta Risk. The interval is comprised of two
elements: the upper portion of the hypothesis test and the lower tail of the confidence
interval. Together these two intervals comprise the tolerable misstatement and are captured
by 𝑍𝛼/2 𝑎𝑛𝑑 𝑍𝛽 in the sample size formula.
𝑠
𝑠
̅̅̅̅ − 𝑍𝛼/2 𝑥
𝐿𝑜𝑤𝑒𝑟 𝐿𝑖𝑚𝑖𝑡 = 𝐵𝑉
√𝑛
𝑥
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 = 𝜇𝑜 + 𝑍𝛽 √𝑛
The formula to calculate the minimum sample size at the very beginning to this paper
was determined by setting the Critical Value equal to the Lower Limit. Because we are
working with averages, Tolerable Misstatement must be divided by the 1,703 lines in the
parts inventory, ̅̅̅̅̅
𝑇𝑀.
𝐶𝑉 = 𝐿𝐿
OR
𝑠𝑥
̅̅̅̅ − 𝜇𝑜
(𝑍𝛼/2 + 𝑍𝛽 ) √𝑛
= 𝐵𝑉
(𝑍𝛼/2 +𝑍𝛽 )𝑠𝑥
̅̅̅̅̅
𝑇𝑀
𝑠𝑥
𝑠𝑥
𝜇𝑜 + 𝑍𝛽 √𝑛
= ̅̅̅̅
𝐵𝑉 − 𝑍𝛼/2 √𝑛
̅̅̅̅ − 𝜇𝑜 = 𝑇𝑀
̅̅̅̅̅
𝐵𝑉
SO
𝑠𝑥
̅̅̅̅̅
(𝑍𝛼/2 + 𝑍𝛽 ) √𝑛
= 𝑇𝑀
= √𝑛
7
Beta Risk and Alpha Risk can be controlled with the number of standard deviations
(𝑍𝛼/2 + 𝑍𝛽 ) in the formula. In order to limit Beta Risk to 10% and Alpha Risk to 15%, 𝑍𝛼/2 +
𝑍𝛽 would be 1.282 and 1.440 respectively. The minimum sample size would be 399.
𝛽 = .10 => 𝑍𝛽 = 1.282
(𝑍𝛼/2 +𝑍𝛽 )𝑠𝑥
̅̅̅̅̅
𝑇𝑀
n = 398.7
= √𝑛
𝛼 = .15 => 𝑍𝛼/2 = 1.440
√𝑛 = (1.282 + 1.440)
$861.52
$117.44
√n = 19.968
we must always round up i.e. n = 399
In the example work papers Beta Risk and Alpha Risk were set at .05% and 30%.
Although the interval between the hypothetical mean and the average book value remains the
same ($117.44), the required sample size will change as Alpha Risk and Beta Risk change.
In the case, the sample size will be the 387 which has been used throughout the paper.
√𝑛 =
(𝑍𝛼/2 +𝑍𝛽 )𝑠𝑥
̅̅̅̅̅
𝑇𝑀
n = 386.8
$861.52
√𝑛 = (1.645 + 1.036) $117.44
√n = 19.667
we must always round up i.e. n = 387
Take a moment to notice that the Critical Value
will be different when Beta Risk is 10% than when it is
5%. That is because the number of standard
deviations assigned to 𝑍𝛽 will change. In order to
reduce Beta Risk more standard deviations must be
included in the portion of the distribution to the left of
the Critical Value.
Also take a moment to notice that we used the
standard deviation from our sample results to
calculate the Critical Value in our hypothesis test.
However, we did not know the standard deviation of the sample at time we were planning the
audit and needed to determine the minimum sample size. We used the standard deviation of
the recorded values to determine the sample size because that was the best estimate we had
at that time.
𝑠
𝑥
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 = 𝜇𝑜 + 𝑍𝛽 √𝑛
CV = $1,036.74 + 1.645
$861.52
CV = $1,036.74 + 1.645
$880.00
√387
√387
= $1,108.78
Beta Risk = 5% with $861.52 std dev
= $1,110.33
Beta Risk = 5% with $880.00 std dev
Think!
If you want to be conservative when you determine the Critical Value for your
hypothesis test would you use:
The standard deviation of the values recorded in the book balance,
The standard deviation of the items in the sample,
The smaller of the std. dev. of the recorded values or the sample, or
The larger of the std. dev. of the recorded values of the sample?
8
What if the Sample Mean is less than the Critical Value?
Assume the sum of the 387 lines that were audited came to $410,289.70; the standard
deviation of the sample was $880.00 and the sample mean was $1,060.18. The sample
mean would lie in the region where we accept Ho, or more correctly we fail to reject Ho.
However, the sample mean is larger than the hypothetical mean of $1,036.74. The projected
balance would be $1,805,486.54. The projected error is $160,073.49, which is less than
tolerable misstatement.
In this case, the sample would not provide sufficient evidence to state that the parts
inventory is not materially overstated with 95% confidence. However, the sample also does
not provide evidence that the parts inventory is materially overstated. We can modify the
equation used to calculate the Critical Value and determine the level of confidence the
sample results do provide.
𝑠
𝑥
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 = 𝜇𝑜 + 𝑍𝛽 √𝑛
($1,060.18−$1,036.74)√387
880
= 𝑍𝛽 =
𝑠𝑥
𝑋̅ = 𝜇𝑜 + 𝑍𝛽 √𝑛
(𝑋̅ −𝜇𝑜 )√𝑛
𝑠𝑥
= 𝑍𝛽
0.524 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛𝑠
A sample mean of $1,060.18 is 0.524 standard deviations to the right of $1,036.74.
The table for one-tail tests indicates for a population with a mean of $1,036.74 there would be
a 30% probability of a obtaining a sample mean 0.524 standard deviations to the right of
$1,036.74. We would be 70% confident that the parts inventory is not materially overstated.
In some situations the partner may want you to audit additional lines of inventory to obtain
addition assurance. In other situations, the partner might be satisfied that the sample results
when they are combined with the results of other audit tests.
9
10
Work Paper Lead Sheet
Miller Motor Co
Inventory Lead Sheet
performed by: John
date:
1/01/15
balance
per
G.L.
15101
15201
15301
15401
15501
parts
new cars
new trucks
used cars
used trucks
1,965,560.03
231,824.00
866,948.12
185,558.53
479,862.11
Total Inventory
3,729,752.79
balance
per
audit
11
Work Paper 1
Miller Motor Co.
Inventory
Parts inventory
performed by: John
date:
1/01/15
Nature of test: Substantive Test of Details of Account Balances
Objective:
The objective of this procedure is to determine if the parts inventory account
is materially overstated.
Assertion(s):
Existence, Valuation and Allocation
Tolerable
Misstatement: For parts inventory, tolerable misstatement has been set at $200,000
Procedure:
DC & H, LLP selected a random sample of 387 entries from MMC’s
inventory worksheet. On Jan. 1, 2015, we observed MMC’s physical
inventory observation.
For each SKU item selected in our sample, we counted the number of items
in MMC’s parts warehouse and compared our count with the quantity on
MMC’s parts department physical inventory worksheet. For those items
where there was a discrepancy between our count and the quantity reported
by MMC, we re-counted those items accompanied by MMC’s parts inventory
supervisory.
For each SKU item selected in our sample, we obtained the most recent
invoice to determine the appropriate cost (MMC uses FIFO). In some cases
the quantity on hand for an SKU number exceeded the quantity purchased
on the most recent invoice. In those cases we obtained sufficient previous
invoices to account for the quantity on hand. We compared our cost with the
cost on MMC’s parts department physical inventory worksheet. We reviewed
all SKU numbers for which we found a discrepancy between the cost on the
most recent invoice and the cost reported on MMC’s inventory work sheet
with the parts inventory supervisory.
Work Paper 2 shows the sample size calculation and the evaluation of the
sample results.
Workpaper 3 shows the sample results.
Conclusion:
Based on the sample results we conclude that parts inventory is not
materially overstated.
12
Work Paper 2
Miller Motor Co.
Inventory
Parts inventory
performed by: John
date:
1/01/15
Nature of test: Substantive Test of Details of Account Balances
Objective:
The objective of this procedure is to determine if the parts inventory account
is materially overstated.
Assertion(s):
Existence, Valuation and Allocation
Tolerable
Misstatement: For parts inventory, tolerable misstatement has been set at $200,000
Procedure:
Book
Tol Mis
Sample size calculation
= $1,965,560.03
= $ 200,000.
std dev =
861.52
risk of incorrect rejection
risk of incorrect acceptance
(𝒁𝜷 +𝒁𝜶⁄ )𝒔𝒙
𝟐
̅̅̅̅̅
𝑻𝑴
̅̅̅̅ = $1,154.18
𝐵𝑉
̅̅̅̅̅
𝑇𝑀 = $ 117.44
μo < 1,965,560.03 – 200,000.00
Ho: μo < 1,154.175 – 117.44
Ho: μo < 1,036.74
α = 0.30
Zα/2 = 1.036
β = 0.05
Zβ = 1.645
= √𝒏
𝟏.𝟔𝟒𝟓+𝟏.𝟎𝟑𝟔 )𝟖𝟔𝟏.𝟓𝟐
𝟏𝟏𝟕.𝟒𝟒
= √𝒏
19.667 = √n
386.8
= n => n = 387 (we always round up)
Hypothesis test using sample results to calculate Critical Value using sample results
𝑥̅ =
𝑠 =
$1,141.00
880.00
CV
= μ + Zβ * Sx/√n
= 1,036.74 + 1.645* 880.00 / √387
= 1,036.74 + 73.586
= 1,110.33
Evaluation of Sample Results
Sample Mean
$1,141.00 is greater than CV of $1,110.33
Since the sample mean of $1,141.00 is greater than the Critical Value of
$1,110.33 we can conclude that the mean of the population from which the
sample was selected is greater than $1,110.33 with less than 5% risk of
incorrect acceptance.
13
line qty
SKU
67.38
79.19
132.92
134.54
176.39
201.02
1,578.71
2,453.83
4,143.35
112.30
131.98
221.53
224.24
293.98
335.04
1,857.31
2,886.86
4,874.53
Sample results n =387
Sample mean
Sample Std Dev
Extended Cost
Standard Deviation of Sampling Lines
2.65
4.38
5.86
26.57
cost
4.42
7.30
9.76
44.28
retail
price
Miller Motor Co.
year-end physical inventory
worksheet
1
5 4470312 Tail lamp bulb
2 40 1118019 Oil filter
3 240 1851974 Spark plug
4 43 3873086 Air intake filter
5 …
…
…
…
…
…
…
…
…
984 21 6202037 Shock
985
2 3680387 Fuel pump diesel
986 11 6730291 Wheel steel
987 19 6037192 Brake pads
988
3 5119378 Water pump
989
4 4030026 Alternator
990 …
…
…
…
…
…
…
…
…
1701
2 2780189 Transmission 4l60e m30
1702
1 1616045 Engine - 4.3L
1703
1 1478027 Engine - 6.0L
14
3,157.43
2,453.83
4,143.35
1,414.98
158.38
1,462.10
2,556.34
529.16
804.10
1,965,560.03
861.52
…
…
…
…
…
…
13.26
175.20
1,405.44
1,142.42
extended
cost
…
…
…
…
…
…
239
42
qty
67.38
5.86
26.57
2.65
441,567.00
1,141.00
880.00
4,143.35
1 4,143.35
603.07
3,157.43
201.02
1,363.12
1,414.98
1,399.58
1,115.86
13.26
2 1,578.71
3
11 123.92
21
5
audit results
extended
cost
cost
15
Work Paper 2 (alternative presentation)
Miller Motor Co.
Inventory
Parts inventory
performed by: John
date:
1/01/15
Nature of test: Substantive Test of Details of Account Balances
Objective:
The objective of this procedure is to determine if the parts inventory account
is materially overstated.
Assertion(s):
Existence, Valuation and Allocation
Tolerable
Misstatement: For parts inventory, tolerable misstatement has been set at $200,000
Procedure:
Book
Tol Mis
Sample size calculation
= $1,965,560.03
= $ 200,000
std dev =
861.52
risk of incorrect rejection
risk of incorrect acceptance
(𝒁𝜷 +𝒁𝜶⁄ )𝒔𝒙
𝟐
̅̅̅̅̅
𝑻𝑴
̅̅̅̅ = $1,154.18
𝐵𝑉
̅̅̅̅̅
𝑇𝑀 = $ 117.44
μo < 1,965,560.03 – 200,000.00
Ho: μo < 1,154.175 – 117.44
Ho: μo < 1,036.74
α = 0.30
Zα/2 = 1.036
β = 0.05
Zβ = 1.645
= √𝒏
𝟏.𝟔𝟒𝟓+𝟏.𝟎𝟑𝟔 )𝟖𝟔𝟏.𝟓𝟐
𝟏𝟏𝟕.𝟒𝟒
= √𝒏
19.667 = √n
386.8
= n => n = 387 (we always round up)
Evaluation of Sample Results
n=
Book value
Sample total
441,567.00
387
overstatement
Allowance for
Mean
N=
1,154.18
1703
1,965,560.03
1,141.00
1703
1,943,123.00
22,437.03
13.18
Zβ for 0.05
sampling risk
⁄
880.00
1.645
projected error plus allowance
Projected
to population
387
73.59
86.77
1703
125,323.77
147,760.80
Since the projected error of $22,437 plus the allowance of $125,323 equals $147,760
we can conclude that the total misstatement in the account is less than the $200,000
tolerable misstatement with less than 5% risk of incorrect acceptance.
16
line qty
SKU
67.38
79.19
132.92
134.54
176.39
201.02
1,578.71
2,453.83
4,143.35
112.30
131.98
221.53
224.24
293.98
335.04
1,857.31
2,886.86
4,874.53
Sample results n =387
Sample mean
Sample Std Dev
Extended Cost
Standard Deviation of Sampling Lines
2.65
4.38
5.86
26.57
cost
4.42
7.30
9.76
44.28
retail
price
Miller Motor Co.
year-end physical inventory
worksheet
1
5 4470312 Tail lamp bulb
2 40 1118019 Oil filter
3 240 1851974 Spark plug
4 43 3873086 Air intake filter
5 …
…
…
…
…
…
…
…
…
984 21 6202037 Shock
985
2 3680387 Fuel pump diesel
986 11 6730291 Wheel steel
987 19 6037192 Brake pads
988
3 5119378 Water pump
989
4 4030026 Alternator
990 …
…
…
…
…
…
…
…
…
1701
2 2780189 Transmission 4l60e m30
1702
1 1616045 Engine - 4.3L
1703
1 1478027 Engine - 6.0L
17
3,157.43
2,453.83
4,143.35
1,414.98
158.38
1,462.10
2,556.34
529.16
804.10
1,965,560.03
861.52
…
…
…
…
…
…
13.26
175.20
1,405.44
1,142.42
extended
cost
…
…
…
…
…
…
239
42
qty
67.38
5.86
26.57
2.65
441,567.00
1,141.00
880.00
4,143.35
1 4,143.35
603.07
3,157.43
201.02
1,363.12
1,414.98
1,399.58
1,115.86
13.26
2 1,578.71
3
11 123.92
21
5
audit results
extended
cost
cost
line qty
1
2
3
4
2
2
3
1
retail
price
SKU
S9856 Silverado LT 78.7
M3780 Malibu LTZ
I3772 Impala LTZ
C1851 Corvette Base
26,810.00
26,955.00
29,930.00
48,950.00
cost
24,933.00
25,607.00
28,733.00
44,545.00
Extended Cost
Standard Deviation of Sampling Lines
1
2
3
4
5
6
7
8
1
1
1
1
1
1
1
1
S9856 Silverado LT 78.7
S9856 Silverado LT 78.8
M3780 Malibu LTZ
M3780 Malibu LTZ
I3770 Impala LTZ
I3770 Impala LTZ
I3770 Impala LTZ
C1851 Corvette Base
26,810.00
26,810.00
26,955.00
26,955.00
29,930.00
29,930.00
29,930.00
48,950.00
extended
cost
49,866.00
51,214.00
86,199.00
44,545.00
231,824.00
19,047.53
24,933.00
24,933.00
25,607.00
25,607.00
28,733.00
28,733.00
28,733.00
44,545.00
Extended Cost
231,824.00
Standard Deviation of Sampling6,524.27
Lines
18