U21110 Quantitative Methods for Accountants: Coursework Analysis

This assignment provides a comprehensive analysis of AirCampania's seat occupancy data using time series analysis, specifically focusing on additive and multiplicative models to determine seasonal indices and forecast future occupancy. It also includes an analysis of product marketing strategies, evaluating expected profit for different product options (Pippo and Papperino) and making recommendations based on profitability. Furthermore, the assignment delves into probability and statistical independence, utilizing a Venn diagram to analyze work-related injuries and determine if they are statistically independent. The solutions use Microsoft Excel for data analysis and graphical representations. The assignment explores the use of moving averages to smooth data, calculate seasonal indices, and compare different forecasting models to determine the most appropriate method for the given scenario. Finally, it evaluates the expected profit of two different products and assess the statistical independence of two events related to work-related injuries.

Part 1A Solutions
A. The aim of the moving average is to smooth out data i.e. to get rid of the irregularity component and
seasonality components. Since AirCampania’s cycle is 4 quarters, we can take a moving average of 4
periods to smooth out the data, hence identify a trend for its seat occupancy. The graph below shows
the centered moving average plotted with the original data. The trend line tells us that the seat
occupancy appears to be increasing at a slow rate each year.
Part 1B Solutions
Seasonal indices for seat occupancy using Additive model
Yt= St + It +Tt+ Ct
Yt -(Tt+ Ct) = St +It
Unadjusted seasonal index = average (S+I) for each quarter
∑ i seasonal index for quarter i=0
Adjusted seasonal index=unadjusted seasonal index –mean of all Unadjusted seasonal index
Yt
t Year Quarter Seat Occupancy (%) MA(4) CMA(4) St,+It St
1 2014 1 22 -6.047
2 2 34 6.391
3 3 19 25.3 24.8 -5.75 -2.578
4 4 26 24.3 25.0 1.00 2.234
5 2015 1 18 25.8 26.8 -8.75 -6.047
6 2 40 27.8 28.4 11.63 6.391
7 3 27 29.0 30.0 -3.00 -2.578
8 4 31 31.0 29.5 1.50 2.234

9 2016 1 26 28.0 28.4 -2.38 -6.047
10 2 28 28.8 29.3 -1.25 6.391
11 3 30 29.8 29.1 0.88 -2.578
12 4 35 28.5 29.5 5.50 2.234
13 2017 1 21 30.5 30.0 -9.00 -6.047
14 2 36 29.5 28.9 7.13 6.391
15 3 26 28.3 28.8 -2.75 -2.578
16 4 30 29.3 29.4 0.63 2.234
17 2018 1 25 29.5 29.4 -4.38 -6.047
18 2 37 29.3 29.3 7.75 6.391
19 3 25 29.3 -2.578
20 4 30 2.234
Quarter
Seasonal indices, St Unadjusted Seasonal
indices St
Adjusted Seasonal indices
St
1 =average(-8.75,-2.38,-9,-4.38) -6.125 -6.047
2 =average(11.63,-1.25,7.13,7.75) 6.313 6.391
3 =average(-5.75,-3,0.88,-2.75) -2.656 -2.578
4 =average(1,1.50,5.50,0.63) 2.156 2.234
Sum -0.313 0.000
Quarter Seasonal indices
1
January-March
-6.047
2
April-June
6.391
3
July-September
-2.578
4
October-December
2.234
Part 1C Solutions
Multiplicative model vs Additive Model
t
Quarte
r
Seat Occupancy
(%)
MA(4
)
CMA(4
) St,+It St
Deseasonalize
d Tt
Forecas
t
ABS
Error
1 1 22
-
6.047 28.047
26.69
6 21 4.70
2 2 34 6.391 27.609
26.86
4 33 7.14
3 3 19 25.3 24.8 -5.75
-
2.578 21.578
27.03
3 24 8.03

4 4 26 24.3 25.0 1.00 2.234 23.766
27.20
2 29 1.20
5 1 18 25.8 26.8 -8.75
-
6.047 24.047
27.37
1 21 9.37
6 2 40 27.8 28.4
11.6
3 6.391 33.609
27.54
0 34 12.46
7 3 27 29.0 30.0 -3.00
-
2.578 29.578
27.70
9 25 0.71
8 4 31 31.0 29.5 1.50 2.234 28.766
27.87
8 30 3.12
9 1 26 28.0 28.4 -2.38
-
6.047 32.047
28.04
7 22 2.05
1
0 2 28 28.8 29.3 -1.25 6.391 21.609
28.21
6 35 0.22
1
1 3 30 29.8 29.1 0.88
-
2.578 32.578
28.38
4 26 1.62
1
2 4 35 28.5 29.5 5.50 2.234 32.766
28.55
3 31 6.45
1
3 1 21 30.5 30.0 -9.00
-
6.047 27.047
28.72
2 23 7.72
1
4 2 36 29.5 28.9 7.13 6.391 29.609
28.89
1 35 7.11
1
5 3 26 28.3 28.8 -2.75
-
2.578 28.578
29.06
0 26 3.06
1
6 4 30 29.3 29.4 0.63 2.234 27.766
29.22
9 31 0.77
1
7 1 25 29.5 29.4 -4.38
-
6.047 31.047
29.39
8 23 4.40
1
8 2 37 29.3 29.3 7.75 6.391 30.609
29.56
7 36 7.43
1
9 3 25 29.3
-
2.578 27.578
29.73
6 27 4.74
2
0 4 30 2.234 27.766
29.90
4 32 0.10
The Mean Absolute Deviation (MAD) of the Errors using Additive model is calculated as 4.619.
The Mean Absolute Deviation of the Errors using multiplicative model is given as 5.
Based on the Mean Absolute Deviation (MAD) criteria, a moving average with MAD of 4.619 is to be
preferred over a moving average with a MAD of 5.
Therefore, for this scenario, the additive model is more appropriate than the multiplicative model as
its MAD is smaller.

Part 1D Solutions
Forecast 2019
Yt = St +It +Tt +Ct
Yt -(Tt +Ct) = St +It
Deseasonalized value= Yt- St
Trend estimate=b0+b1t
Forecast= T+C+S
t Year Q Occupancy (%) St Tt Forecast
21 2019 1 -6.047 30.073 24
22 2 6.391 30.242 37
23 3 -2.578 30.411 28
24 4 2.234 30.580 33
2019
Quarter
Forecast Seat
Occupancy (%)
1
January-March 24
2
April-June 37
3
July-September 28
4
October-December 33
Part 2-1 solutions
A. Approach- Expected Profit
NPV = ∑
i=1
5
P( i)∗Profit (i)
NPV = 160*0.1+120*0.25+700*0.20+200*0.15+-300*0.20+-800*0.10
=16+30+140+30-60-80
=76
Since the Expected Profit is greater than zero (positive), it should market Pippo

B. Approach - Expected Profit
= ∑
i=1
5
P( i)∗Profit (i)
= 85*0.1+60*0.25+45*0.20+25*0.15+15*0.20+10*0.10
=8.5+15+9+3.75+3+1
=40.25
The company should market Papperino.
The reason being even though Papperino’s Expected Profit is greater than zero (hence profitable), its
value is less than Pippo’s Expected Profit. This suggests that marketing Pippo will be more profitable
than the Papperino product.
Part 2-2 Solutions
A. Venn diagram
A= work-related injuries, occurred on Fridays
B= work-related injuries, occurred in the last hour of a day’s shift
AB= work-related injuries, occurred on Fridays and in the last hour of a day’s shift
B. The probability that a work-related injury that occurs on Friday does not occur in the last hour of the
day’s shift is 0.22. This is shown as below
P(AB’)=P(A)-P(AB)=0.25-0.03=0.22
C. No , the two events, work-related injuries that occurred on Fridays and work-related injuries that
occurred in the last hour of a day’s shift, are not statistically independent. If the events were
P(A)=0.25
P(AB)=0.03
P(B)=0.18
P(AB’)=0.22

independent, then the probabilities P(AB) would be equal i.e P(AB)= P(A)*P(B)= 0.18*0.25=0.045.
However, we have been given that P(AB) is 0.03. Therefore the condition of conditional probability for
independent event has not been met.