Accounting Decision Support Tools: Probability, Sales & Hypothesis

This assignment solution provides a detailed analysis of accounting decision support tools, focusing on probability, expected value, hypothesis testing, and control limits. The solution includes calculations for expected value with a practical example, analysis of sales data including average daily sales, probability of selling loaves, variance, and standard deviation. It further explores probability calculations related to machine production and rework rates. The assignment also covers hypothesis testing with z-value calculation and interpretation, concluding whether to reject the null hypothesis based on the given data. Finally, control limits (UCL and LCL) are computed at a 95% confidence level for different sample sizes. Desklib provides this and other solved assignments for students' reference.

ACCOUNTING DECISION SUPPORT TOOLS
Student’s Name
Class
Date

2
Question 1
a. Expected value
Expected value is an average return someone can expect from a series of action or
experiment. It measures the expected result when someone do some random experiment or make
a selection from a set of random values.
Expected value is calculated by multiplying the probability of an event occurring by the total
number of occurrences.1 For example, when you have 10 questions of multiple choices of A, B,
C, D and E. When all the choices is C from all the questions, the probability of passing is 0.20.
Therefore, p ( x )=0.20∧occurences is10 then expected value E ( x )=0.20× 10=2.
b. Sales of loaves
1. Tables
Sales
unit
(x)
Number of
days
p(x) Exp
value
E(x)
More
than
Less
than
0 5 0.05 0 0 2.9 0 0
1 10 0.1 0.1 0.1 2.9 0.81 0.081
2 25 0.25 0.5 0.6 2.8 2.25 0.5625
3 25 0.25 0.75 1.35 2.3 5.0625 1.265625
4 20 0.2 0.8 2.15 1.55 10.24 2.048
5 15 0.15 0.75 2.9 0.75 18.0625 2.709375
Total 100 0.85 2.9 7.1 13.2
0.65 Variance 6.6665
Std
deviation
2.581956622
1 Hancock, Volante, and Lau, "Quantifying Qualitative Probabilities,” 155-159.
[ x−E ( x ) ]2
[ x−E ( x ) ]2
p( x )

3
Sales unit (x) Number of days p(x) Exp value E(x) More than Less than
0 5 =B2/B8 =A2*C2 =D2 =D2+D3+D4+D5+D6+D7=(A2-D2)*(A2-D2) =G2*C2
1 15 =B3/B8 =A3*C3 =E2+D3 =F2-D2 =(A3-D3)*(A3-D3) =G3*C3
2 25 =B4/B8 =A4*C4 =E3+D4 =F3-D3 =(A4-D4)*(A4-D4) =G4*C4
3 25 =B5/B8 =A5*C5 =E4+D5 =F4-D4 =(A5-D5)*(A5-D5) =G5*C5
4 20 =B6/B8 =A6*C6 =E5+D6 =F5-D5 =(A6-D6)*(A6-D6) =G6*C6
5 10 =B7/B8 =A7*C7 =E6+D7 =F6-D6 =(A7-D7)*(A7-D7) =G7*C7
Total 100 =C4+C5+C6+C7 =D2+D3+D4+D5+D6+D7=E2+E3+E4+E5+E6+E7=F2+F3+F4+F5+F6+F7
=C5+C4+C3+C2 Variance =H2+H3+H4+H5+H6+H7
Std deviation =SQRT(H9)
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2. The average daily sales is 2.9. The average daily sales is the accumulated expectation of
all the occurrences.
3. Probability of selling two or more loaves 0.85. The probability of selling two or more we
get by adding the individual probability from 2 sales to 5 sales.
4. The probability of selling three or less loaves is 0.65. The probability of selling 3 or less
we get by adding individual probability from 3 sales back to sales 0.
5. Variance of distribution is 6.6665. We get the variance by adding the values in the
column 8
6. Standard deviation 2.5820. The standard deviation is the square root of variance.
c. Four machines
Machine W X Y Z All machines
Grade
Satisfactory 3,200 800 2,400 1,600 8,000
Rework 600 150 450 300 1,500
Scrap 200 50 150 100 500
All grades 4,000 1,000 3,000 2,000 10,000
1. The probability of the part being produced by Machine W and should be reworked when
selected randomly is;
600
10000 = 6
100 =0.06
Here we find 0.06 by taking parts under Machine W and need to be reworked then divide by
the total number of all parts.
2. The probability of the part being produced by Machine Z and is not satisfactory when
randomly selected is;

4
300+100
10000 = 400
10000 = 4
100 =0.04
We find 0.04 by dividing the number of parts under machine Z and are scrapped and need
rework by the total number of all parts
3. The probability of the part being produced by Machine Y and should be scrapped when
selected randomly is;
150
10000 = 15
1000 =0.015
We find 0.015 by dividing the parts which needs to be scrapped and is under machine Y by
the total number of all parts.
4. The probability of parts need to be reworked is;
1500
10000 = 15
100 =0.15
We get 0.15 by dividing all the parts which needs to be reworked by the total number of all parts
5. The probability a part needs to be scrapped given it produced by machine W is;
200
10000 = 2
100 =0.02
Here we divide the number of parts scrapped under machine W by the total number of all the
parts.
d. The average sales of apples is 4000 with a standard deviation of 500.
1. The probability of sales will be greater than 4250 apples
z= x−x՟
σ

5
z= 4250−4000
500 = 250
500 =0.5
From the z-distribution table, z ( 0.5 ) =0.6915 .therefore, the probability of selling greater
than 4250 apples is;
p=1−0.6915=0.3085
2. The probability of sales will be less than 3600 apples.
z= x−x՟
σ
z= 3600−4000
500 =−400
500 =−0.8
From the z-distribution table, z (−0.8 )=0.2919 .therefore, the probability of selling less
than 3600 apples is;
p=0.2919
3. The probability of sales will be less than 4500 apples.
z= x−x՟
σ
z= 4500−4000
500 = 500
500 =1.0
From the z-distribution table, z (1.0 )=0.8413 .therefore, the probability of selling less than
4500 apples is;
p=0.8413
Question 2
1. The average Australian population is 24,995,335. The statistics of 2019 in the internet put
Australia with that population.
2. Average age to die in Australia for both men and women is 82.5 years. From the statistics
of 2017, the mean death age in Australia is 82.5 years.
3. The percentage of working population in Australia is 53% this is according to the
internet.

6
Question 3
a. Control limits
1. UCL and LCL at 95% confidence level, with 64 observations, average of 20 hours and
standard of deviation of 5.23
Upper Control Limits UCL=μ X՟+3 σ X՟
Lower Control Limits LCL=μX՟−3 σ X՟
Where μX՟=μ∧σ X՟= σ
√ n
μX՟=20 , σ =5∧n=64
UCL=20+ 5
√64
UCL=20+0.625
∴ UCL=20.625
LCL=20− 5
√ 64
LCL=20−0.625
∴ LCL=19.375
2. UCL and LCL at 95% confidence level, with 9 observations, average of 20 hours and
standard of deviation of 5.
Upper Control Limits UCL=μ X՟+3 σ X՟
Lower Control Limits LCL=μX՟−3 σ X՟
2 Muñoz-Rosas et al., “Empirical Comparisons,” 453-464.
3 Oprime et al., "X-bar control chart design with asymmetric control limits and triple
sampling." 1-14.

7
Where μX՟=μ∧σ X՟= σ
√ n
μX՟=20 , σ =5∧n=9
UCL=20+ 5
√9
UCL=20+1.6667
∴ UCL=21.6667
LCL=20− 5
√9
LCL=20−1.6667
∴ LCL=18.3333
b. Hypothesis testing
1. Hypotheses
H0 : μ=42
H1 : μ ≠ 42
2. Z-value456
4 Zhu and Bradic, “Linear Hypothesis Testing," 1583-1600.
5. Pyle and Solanko, "The Composition and Interests of Russia's Business Lobbies,” 19-41.
6. Griffin, Lont and Yuan, "Agency problems and audit fees: further tests of the free cash
flow hypothesis." 321-350.
5
6

8
z= x՟−μ
σ
√ n
z= 45−42
10.8
√ 49
z= 3
10.8
7
z= 3
1.5429
z=1.9445
3. Sketch
Rejected region
1.9445

9
4. From the z-distribution table, z (1.94) =97.38%. 100-97.38=2.62. 0.0262<0.05 we
therefore we reject the null hypothesis and conclude that the average age of employees at
Sturt Ltd is not 42.78
7 Brent, "Confidence intervals for variance components in unbalanced one-way random
effects model using non-normal distributions."
8. Braga et al., "The Odd Log–Logistic Normal Distribution,” 311-335.
8

10
Bibliography
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"The Odd Log–Logistic Normal Distribution: Theory and Applications in Analysis of
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Muñoz-Rosas, J. F., Álvarez-Verdejo, E., Pérez-Aróstegui, M. N., Gutiérrez-Gutiérrez, L.
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