Miller Motor Co. inventory sampling problem.
The purpose of this assignment is to demonstrate how statistical theory helps auditors manage
audit risk. This exercise uses mean-per-unit sampling because the relationship between sample size and
risk is most clearly illustrated using mean-per-unit sampling. Accounting firms have software which
uses 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 that mean-per-unit sampling. However, the underlying theory is more complicated. This
assignment utilizes the basic confidence intervals and hypothesis tests that were covered in your
statistics class.
Miller Motor Co.’s physical inventory worksheet has two sections. The top section includes
1,703 different parts or SKU numbers. MMC also has 4 different models of vehicles in inventory. The
auditors chose to separate the parts inventory from the vehicle inventory. This is a very simple form of
stratification. If the 5 tail lamp bulbs costing $13.26 were in the same pool as the 3 Impala LTZs which
cost $86,199.00 it would result in a very large variance (and standard deviation). As you work this
exercise you will observe that the sample size required to reduce risk to a given level increases as the
variance increases.
Although this exercise does not involve further stratification, notice that there is a great deal of
variance in the parts prices. Auditors might be able to reduce the required sample size by separating
parts that cost more than $1,000 from parts costing less than $1,000. In fact, accounting firms have
software that determines the optimal strata to minimize the required sample size for inventory
observations.
Designing a sampling plan requires careful consideration. Notice that the new vehicle inventory
will result in lower variance if each vehicle is treated as a sampling unit rather than grouping the
vehicles by SKU.
Hypothesis Test - review
Let’s test the null hypothesis (Ho:) that the unknown true balance of the inventory is less than
$1,765,560. Statistics uses sample means so we divide $1,765,560 by 1,703 and will use $1,036.74 in
our hypothesis test. The power of statistics is in the rejection of the null hypothesis. So if we can reject
HO: then we can accept the alternate hypothesis (HA:) which states that the true mean must be greater
than $1,036.74 ($1,765,000). Later we will discuss how to determine the appropriate critical value but
for now we will arbitrarily set the critical value of the hypothesis test equal to $1,865,560 ($1,765,560
+ $100,000). We again divide these values by 1,703 which results in $1,036.74 + $58.72 = $1,095.46.
The hypothesis test will state
𝐻𝑂 ∶ 𝜇 < $1,036.74
𝑜𝑟 (
𝐻𝐴 ∶ 𝜇 > $1,036.74
𝑜𝑟 (
$1,765,560
1,703
$1,765,560
)
)
1,703
$1,865,560
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 = $1,095.46 𝑜𝑟 (
1,703
)
We will take a sample from Miller Motor Co.’s
inventory and if our sample mean exceeds the Critical Value,
we will reject HO: and accept the HA: the true value is greater than $1,036.74. We conclude that the
1
true mean is greater than $1,036.74. Because the sample mean is so much greater than the $1,036.74
(the hypothetical mean) it is very unlikely the sample mean came from a population whose mean is
less than $1,036.74. The normal distribution enables auditors to quantify the risk of reaching an
incorrect conclusion. If inventory is materially overstated, we can limit the risk of concluding
inventory is not materially overstated to 10% by setting the critical value 1.282 standard deviations to
the right of the hypothetical mean.
The standard deviation of the distribution of sample means is referred to as the standard error of
the means which is a function of the standard deviation of the population from which the sample was
𝑠𝑥
selected and the size of the sample, 𝑠𝑒 = √𝑛
. This relationship allows us to determine the sample size
necessary to reduce risk to the desired level. The best estimate of the standard deviation of the
population from Miller Motor Co.’s inventory worksheet, is $861.52.The critical value in the example
is $1,095.46, so we can determine the required sample size.
𝑠
𝑥
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 = 𝜇𝑜 + 𝑍𝛽 √𝑛
𝐶𝑉 = $1,036.74 + 1.282
$861.52
√𝑛
$1,095.46 − $1,036.74 = 1.282
$58.72 =
√𝑛 =
$1,104.47
√𝑛
$1,104.47
$58.72
$861.52
√𝑛
1.282
= 18.809
n = 353.7 the required sample size must always be rounded up i.e. 354
If the sample mean of $1,141 from our random sample of 354 SKU lines is greater than
$1,095.46 we will reject HO: and accept the HA: and conclude that the actual mean is greater than
$1,036.74. There is less than a 10% probability of obtaining a sample mean greater than $1,095.46
from a population with a mean of $1,036.74. This does assume the standard deviation is $861.52.
Confidence Interval - review
Now, let’s create an interval around the book value,
$1,965,560 plus and minus $100,000, or $1,865,560 to $2,065,560.
This interval is similar to a confidence interval except that it is
centered on the book value rather than being centered on a sample
mean. Statistics uses means so these values are divided by 1,703
and converted to averages: $1,154.17 ± $58.72 or $1,095.45 - $1,212.89.
In this case, the risk is that we conclude the book value of the inventory is incorrect when it is
actually correct. Assume, for a minute, the book value is correct and we want to limit the risk of a
Type I or α error to 20%. In other words, we want to be 80% confident we won’t conclude the
inventory’s book value is incorrect when it is correct. In order to achieve 80% confidence the normal
2
distribution indicates that the lower and upper limits of the interval would be ± 1.282 standard
deviations to the left and to the right of the average book value.
𝑠
𝑥
Again, the standard error of the means equals 𝑠𝑒 = √𝑛
. The interval would be
𝑠
𝑥
𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 = 𝑎𝑣𝑒 𝐵𝑉 ∓ 𝑍 √𝑛
. The best estimate of the standard deviation of the population, at this time,
is again $861.52. In this case, the interval is ± $58.72 or $117.44.These relationships allow us to solve
for the required sample size (n).
𝑠
𝑠𝑥
𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 = ̅̅̅̅
𝐵𝑉 + 𝑍𝛼/2 √𝑛
− ( ̅̅̅̅
𝐵𝑉 − 𝑍𝛼/2
𝑥
𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 = ̅̅̅̅
𝐵𝑉 ± 𝑍𝛼/2 √𝑛
𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 = 𝑍𝛼/2
𝑠𝑥
√𝑛
+ 𝑍𝛼/2
$117.44 = 2 ∗ (1.282
√𝑛 = 1.282
$861.52
$58.72
n = 353.78
)
𝑠𝑥
√𝑛
$861.52
√𝑛
𝑠𝑥
√𝑛
)
$58.72 = 1.282
$861.52
√𝑛
√n = 18.809
always rounded up i.e. n = 354
Applying what we know about Hypothesis Tests and Confidence Intervals
The risk of incorrectly concluding the book value of the inventory is
incorrect when it is correct is a Type I error or α risk. This is the risk of
incorrect rejection. The risk of incorrectly concluding inventory is fairly
presented when the book value is materially overstated is a Type II error or
β risk. This is the risk of incorrect acceptance.
In the previous examples the critical value of the hypothesis test and
the lower limit of the interval were arbitrarily set to equal to the midpoint
between the hypothetical mean and the book value. This will result in Alpha
risk that is twice as large as Beta risk because the hypothesis test is a one-tail
test while confidence intervals have two tails.
conclusion / opinion
unmodified
inventory
adverse
Type I
fairly
presented
inventory
correct
materially
Beta
overstated
error
Alpha
error
Type II
correct
For planning purposes assume that performance materiality (tolerable error) for the parts
inventory has been set as $200,000, which is approximately 10% of the $1,965,560 balance. Statistical
sampling cannot prove that the balance is correct. But 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.” Auditors are
typically concerned that asset accounts, such as inventory, are overstated. If performance materiality
(tolerable error) 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 $1,036.74 ($1,765,560 / 1,703) and the average
book value $1,154.17 ($1,965,560 / 1,703) is $117.44 ($200,000 / 1,703). This will remain the same
regardless of how we allocate risk between Alpha Risk and Beta Risk. The interval is comprised of two
3
elements: the upper portion of the hypothesis test and the lower tail of the confidence interval.
Together these two elements comprise the performance materiality (tolerable error).
𝑠𝑥
𝑠𝑥
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 = 𝜇𝑜 + 𝑍𝛽 √𝑛
𝐿𝑜𝑤𝑒𝑟 𝐿𝑖𝑚𝑖𝑡 = ̅̅̅̅
𝐵𝑉 − 𝑍𝛼/2 √𝑛
Setting the critical value equal to the lower limit of the interval will result in the following
equation. Notice that the right side of third equation equals the performance materiality (tolerable
error) divided by the number of lines in the parts inventory or the average performance materiality
(tolerable error).
𝑠𝑥
𝑠𝑥
𝜇𝑜 + 𝑍𝛽 √𝑛
= ̅̅̅̅
𝐵𝑉 − 𝑍𝛼/2 √𝑛
𝑠𝑥
𝑠𝑥
𝜇𝑜 + 𝑍𝛽 √𝑛
= ̅̅̅̅
𝐵𝑉 − 𝑍𝛼/2 √𝑛
𝑠𝑥
(𝑍𝛼/2 + 𝑍𝛽 ) √𝑛
= ̅̅̅̅
𝐵𝑉 − 𝜇𝑜
(𝑍𝛼/2 +𝑍𝛽 )𝑠𝑥
̅̅̅̅
𝑇𝐸
n = 353.78
= √𝑛
$861.52
√n = 18.809
√𝑛 = (1.282 + 1.282) $117.44
we must always round up i.e. n = 354
Although the interval between the hypothetical mean and the average book value will remain
the same ($117.44), the required sample size will change if we change the allocation of risk between
Alpha Risk and Beta Risk. For instance, the sample size will increase if we limit the risk of incorrect
acceptance (Beta Risk) to 5% and the risk of incorrect rejection (Alpha Risk) to 30%.
𝛽 = .05 => 𝑍𝛽 = 1.645
(𝑍𝛼/2 +𝑍𝛽 )𝑠𝑥
̅̅̅̅
𝑇𝐸
n = 386.81
= √𝑛
𝛼 = .30 => 𝑍𝛼/2 = 1.036
√𝑛 = (1.036 + 1.645)
$861.52
√n = 19.667
$117.44
we must always round up i.e. n = 387
Also notice that the critical value will change because it is now 1.645 standard deviations to the
right of the hypothesized mean.
𝑠
𝑥
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 = 𝜇𝑜 + 𝑍𝛽 √𝑛
𝐻𝑜 ∶ 𝜇 > $1,036.74
𝐶𝑉 = $1,036.74 + 1.645
$861.52
√387
𝐶𝑉 = $1,036.74 + 72.04 𝑜𝑟 $1,108.78
Later, during the audit we will select and audit a random sample of 387 inventory line items. If
everything goes exactly as planned and the sample mean of the audited values exceeds $1,108.78, we
will accept the hypothesis that the parts inventory balance is not materially overstated.
Observation of the Physical Inventory
The attached work sheet shows the results of our sample. We found two types of errors in our
sample. We observed instances where the actual quantity on hand differed from the quantity in the
4
client’s accounting records. We also observed one instance where the cost per the invoice was different
than the cost used to determine the inventory balance. The sum of the extended audited balances for
the 387 part lines in the sample came to $441,567.00. The sample mean of $1,141.00 lies in the region
where we accept the alternate hypothesis. However, the standard deviation of the sample exceeded the
standard deviation used to calculate the original critical value.
Our conclusion on the parts inventory should be based on the best available evidence. We can
re-calculate the critical value using the standard deviation
from the sample.
𝐻𝑜 ∶ 𝜇 > $1,036.74
𝑠𝑥
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 = 𝜇𝑜 + 𝑍𝛽 √𝑛
$880.00
𝐶𝑉 = $1,036.74 + 1.645
√387
$1,036.74 + 73.59 𝑜𝑟 $1,110.33
𝐶𝑉 =
Because the standard deviation of the sample is larger than the standard deviation used to
determine the required sample size, the critical value is further to the right. In this case, the sample still
lies in the acceptance region.
What if the Sample Mean is less than the Critical Value?
Assume the sum of the 387 lines that were audited came to $423,412.83; the standard deviation
of the sample was $880.00 and the sample mean was $1,094.09. 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 $1,863,235.27
$423,412.83
( 1,703 ×
𝑜𝑟 1,703 × $1,094.09 ). The projected discrepancy is $102,324.73
387
($1,965,560 - $1,863,235) which is less than performance materiality (tolerable error).
The sample does not provide sufficient evidence to state that the parts inventory is
not materially overstated with 95% confidence. However, the sample 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
($1,094.09−$1,036.74)√387
880
= 𝑍𝛽 =
𝑠𝑥
𝑋̅ = 𝜇𝑜 + 𝑍𝛽 √𝑛
(𝑋̅ −𝜇𝑜 )√𝑛
𝑠𝑥
= 𝑍𝛽
1.282 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛𝑠
The sample mean is 1.282 standard deviations to the right of the hypothetical mean. The table
for one-tail tests indicates that there is a 10% probability a sample mean would be 1.282 standard
deviations to the right of the actual mean. We would be 90% confident that the parts inventory is not
materially overstated. In some situations you will audit additional lines of inventory. In other
5
situations, the partner or manager may be satisfied that the sample results in combination with the
results of other audit tests provide sufficient evidence.
6
Work Paper Lead Sheet
Miller Motor Co
Inventory Lead Sheet
performed by:
date:
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
John
1/01/13
balance
per
audit
7
Work Paper 1
Miller Motor Co.
Inventory
Parts inventory
performed by:
date:
John
1/01/13
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
Performance
Materiality:
For parts inventory, performance materiality (tolerable error) 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, 2013, 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.
8
Work Paper 2
Miller Motor Co.
Inventory
Parts inventory
performed by:
date:
John
1/01/13
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
Performance
Materiality:
For parts inventory, performance materiality (tolerable error) has been set at $200,000
Procedure:
Sample size calculation
Book
= $1,965,560.03
μ ≥ 1,965,560.03 – 200,000.00
performance materiality = $200,000 Ho: μ ≥ 1,154.175 – 117.44 perf. mat. Is $200,000 / 1,703
std dev =
861.52
Ho: μ ≥ 1,036.74
risk of incorrect rejection
α = 0.30
Zα/2 = 1.036
risk of incorrect acceptance
β = 0.05
Zβ = 1.645
CV + Zβ* Sx/√n = LL – Zα/2* Sx/√n
Zβ* Sx/√n = CV- LL – Zα/2* Sx/√n
Zα/2* Sx/√n + Zβ* Sx/√n = CV- LL
1.036*861.52/√n + 1.645*861.52/√n = 117.44
2.681*861.52 /√n
= 117.44
2.681*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
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
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.
Evaluation of Sample Results
Sample Mean $1,141.00 is greater than CV of $1,110.33
9
Work Paper 2 (alternative presentation)
Miller Motor Co.
Inventory
Parts inventory
performed by:
date:
John
1/01/13
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
Performance
Materiality:
For parts inventory, performance materiality (tolerable error) has been set at $200,000
Procedure:
Sample size calculation
Book
= $1,965,560.03
μ ≥ 1,965,560.03 – 200,000.00
performance materiality = $200,000
Ho: μ ≥ 1,154.175 – 117.44 perf. mat. is $200,000 / 1,703
std dev =
861.52
Ho: μ ≥ 1,036.74
risk of incorrect rejection
α = 0.30
Zα/2 = 1.036
risk of incorrect acceptance
β = 0.05
Zβ = 1.645
CV + Zβ* Sx/√n = LL - Zα/2* Sx/√n
Zβ* Sx/√n = CV- LL - Zα/2* Sx/√n
Zα/2* Sx/√n + Zβ* Sx/√n = CV- LL
1.036*861.52/√n + 1.645*861.52/√n = 117.44
2.681*861.52 /√n
= 117.44
2.681*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=
Projected
to population
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
387
73.59
86.77
1703
125,323.77
147,760.80
Since the projected error plus the allowance is $147,760.80 we can conclude that the total
misstatement in the account is less than the $200,000 performance materiality (tolerable error)
with less than 5% risk of incorrect acceptance.
10
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
11
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
12