Financial Analysis of Tabcorp Holdings

This assignment delves into the financial performance of Tabcorp Holdings Limited, a leading Australian wagering operator. It analyzes key financial ratios such as profitability, liquidity, and solvency to assess the company's financial health. Additionally, it conducts a SWOT analysis to identify Tabcorp's strengths, weaknesses, opportunities, and threats. Finally, the assignment provides insights into the future outlook of Tabcorp based on its financial performance and industry trends.

Running head: BUSINESS STATISTICS
Business Statistics
Name of the Student:
Name of the Student:
Author’s Note:

2BUSINESS STATISTICS
Table of Contents
Answer 1..............................................................................................................................4
Part 1a..............................................................................................................................4
Part 1b..............................................................................................................................5
Part 1c..............................................................................................................................6
Part 1d..............................................................................................................................7
Answer 2..............................................................................................................................8
Part 2a..............................................................................................................................8
Part 2b..............................................................................................................................8
Part 2c..............................................................................................................................9
Part 2d..............................................................................................................................9
Answer 3............................................................................................................................10
Part 3a............................................................................................................................10
Part 3b............................................................................................................................10
Part 3c............................................................................................................................10
Part 3d............................................................................................................................10
Answer 4............................................................................................................................12
Part 4ai...........................................................................................................................14
Part 4aii..........................................................................................................................14
Part 4bi...........................................................................................................................14

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Part 4bii..........................................................................................................................14
Answer 5............................................................................................................................15
Part 5a............................................................................................................................15
Part 5b............................................................................................................................17
Part 5c............................................................................................................................18
References..........................................................................................................................20

4BUSINESS STATISTICS
Answer 1
Part 1a
Opening prices for Quarterly data
Year January April July October
2008 13.45 10.5 9.44 9
2009 5.91 6.3 7.3 8.8
2010 8.07 8.08 7.66 8.48
2011 8.37 8.13 8.93 7.8
2012 8.09 8.8 8.57 9.04
2013 1.67 12.3 12 15.54
2014 16.85 16.72 15.1 13.75
2015 12.69 13.3 12.61 10.03
2016 12.5 12.35 12.6 13.06
2017 11.58 11.8 12.15 -
CWN Quarterly opening prices 2008-2017
Year January April July October
2008 6.92 6.4 4.68 3.9
2009 3.27 3.03 3.3 3.35
2010 3.25 3.25 2.95 3.27
2011 3.37 3.53 3.21 2.53
2012 2.68 2.70 2.91 2.68
2013 3.00 3.18 2.95 3.20
2014 3.57 3.33 3.30 3.54
2015 4.08 4.70 4.59 4.7
2016 4.71 4.22 4.55 5.01
2017 4.81 4.77 4.35 -
TAH Quarterly opening prices 2008-2017
The stem and leaf for the two companies can be given as:
Leaves for TAH Stem Leaves for CWN

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0 6 6 7 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9
1 0 1 1 2 2 2 2 2 2 3 3 3 3 3 3 3 4 5 6 7 7
9 9 7 7 7 5 2
9 6 5 5 4 3 3 3 3 3 3 3 3 2 2 2 2 0 0 0 3
8 8 7 7 7 7 6 6 4 2 1 4
0 5
9 4 6
Stem and Leaf Display
Part 1b
CWN plotted as relative frequency histogram
TAH plotted as relative frequency polygon
 Frequency/ Total Frequency = Relative Frequency
The following interval will be starting from $0 to less than $2
Opening prices CWN (Relative Frequency) TAH (Relative Frequency)
$0 to less than $2 0.000 0
$2 to less than $4 0.000 0.641
$4 to less than $6 0.000 0.308
$6 to less than $8 0.128 0.051
$8 to less than $10 0.333 0
$10 to less than $12 0.128 0
$12 to less than $14 0.308 0
$14 to less than $16 0.051 0
$16 to less than $18 0.051 0
CWN and TAH and their Relative Frequencies

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$0 to less
than $2 $2 to less
than $4 $4 to less
than $6 $6 to less
than $8 $8 to less
than $10 $10 to less
than $12 $12 to less
than $14 $14 to less
than $16 $16 to less
than $18
0.000
0.050
0.100
0.150
0.200
0.250
0.300
0.350
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
CWN and TAH
CWN TAH
Opening prices
Frequency
CWN and TAH and their Relative Frequencies
Part 1c
The market capital for the five companies under ASX can be given as:
Companies Listed under ASX Market Capitalisation (Over AUD
500M)
Aristrocat Leisure Limited (ALL) 13518
Crown Resorts Limited (CWN) 8039
Fairfax Media Limited (FXJ) 2219
Flight Centre Travel Group Limited (FLT) 4885
Tabcorp Holdings Limited (TAH) 3432
The Star Entertainment Group Limited (SGR) 4293
Companies Listed under ASX and their Market Capitalisation

7BUSINESS STATISTICS
ALL
CWN
FXJ
FLT
TAH
SGR
0 2000 4000 6000 8000 10000 12000 14000 16000
Market Capitalisation for leisure and entertainment
companies in million AUD
Market Capital (AUD Million)
Companies Listed under ASX
Companies Listed under ASX and their Market Capitalisation
Part 1d
As per the data given in InvestSMART (2017) for “peer comparison”, based on Market
Capitalisation, Growth and Dividend Yield, it is better to invest in Crown Resorts (CWN)
because shareholders have received more dividend in 2017 than in Tabcorp (TAH) and is
subjected to receive likewise the same in upcoming years

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Answer 2
Part 2a
The mean, median and quartiles for the prices of each publisher can be given as:
CBA NAB ANZ WBC
Mean $3.10 $2.31 $2.06 $2.04
Q1 2.49 2.1875 1.91 1.907
Median $3.05 $2.36 $2.05 $2.04
Q3 3.7325 2.45075 2.173 2.2355
Min $2.49 $2.19 $1.91 $1.91
Max $3.73 $2.45 $2.17 $2.24
Mean, Quartiles and Median for each bank
As per the data calculated, the annual dividends have been given maximum by CBA as
per the mean and median data followed by NAB. Also, Q1 and Q3 is even higher for CBA.
Part 2b
CBA NAB ANZ WBC
Range (Max-Min value) $2.23 $0.79 $1.21 $1.30
Coefficient of Variation (sd/ x̅ * 100) $25.40 $9.86 $17.38 $18.03
Standard Deviation 0.787482
0.22795367
7 0.358824
0.36718812
9
Each Bank’s Range, SD and Coefficient of Variation
.

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Part 2c
CBA NAB ANZ WBC
0
0.5
1
1.5
2
2.5
3
3.5
4
Box and Whister Plot Banks
Q3-Median
Median-Q1
Q1
Banks
Annual Dividends per share
Box and Whisker Plot for each Bank
Part 2d
APRA has increased the pressure on banks primarily for the reason that these four largest
banks have a high capital surplus above the minimum requirements such that these minimum
requirements would be added with 150 base point’s benchmark to find extra capital. Also, this is
done to enable banks handle adversity in the times of need without endangering its core objective
of financial intermediation for Australian society (White, 2017).

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Answer 3
Part 3a
Discipline being most popular for best students is “Engineering and Related
Technologies”
 Proportion = 4203/ 14027 = 0.299636
Part 3b
P (Society and Culture and ATAR 80 and less) = 5030/ 25262 = 0.1991132
Part 3c
Field of education 50.00 or
less
proportion with lowest ATAR
rate
Natural and Physical Sciences 705 705/ 9723 = 0.072508485
Information Technology 418 418/ 9723 = 0.042990846
Engineering and Related Technologies 289 289/ 9723 = 0.029723336
Architecture and Building 316 316/ 9723 = 0.032500257
Agriculture, Environmental and Related Studies 98 98/ 9723 = 0.010079194
Health 1,568 1568/ 9723 = 0.161267099
Education 1,062 1062/ 9723 = 0.109225548
Management and Commerce 1,742 1742/ 9723 = 0.17916281
Society and Culture 2,814 2814/ 9723 = 0.289416847
Creative Arts 683 683/ 9723 = 0.070245809
Total 9,723
P (Highest proportion of students with lowest ATR grades) = 2814/ 9723 = 0.289416847
(Society and Culture)
Part 3d
Field of education No ATAR/
Non-Yr 12
majority
Natural and Physical Sciences 7,434 7,434/ 91,567 = 0.081186454
Information Technology 2,281 2,281/ 91,567 = 0.024910721

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Engineering and Related Technologies 3,719 3,719/ 91,567 = 0.040615069
Architecture and Building 2,634 2,634/ 91,567 = 0.028765822
Agriculture, Environmental and Related Studies 1,271 1,271/ 91,567 = 0.013880546
Health 31,341 31,341/ 91,567 =
0.342273963
Education 7,077 7,077/ 91,567 = 0.077287669
Management and Commerce 10,331 10,331/ 91,567 =
0.112824489
Society and Culture 18,161 18,161/ 91,567 =
0.198335645
Creative Arts 6,764 6,764/ 91,567 = 0.073869407
Total 91,567
The maximum number of students from “No ATAR/Non-Yr 12” background is Health.
Generally, health and scirnces are with No ATAR background because they are based on
learning and academic performance to satisfy course prerequisites in their undergraduate course
(Futurestudents.unimelb.edu.au., 2012).

12BUSINESS STATISTICS
Answer 4
Rainfall In Adelaide, South Australia in 2016
Week Number of Rainfall events Total Rainfall Std. Dev
1 0 0 0
2 1 1.2 0.453557368
3 3 16.4 3.732227727
4 3 18.6 4.794043924
5 2 13.8 4.050396806
6 0 0 0
7 1 0.6 0.226778684
8 1 0.4 0.151185789
9 0 0 0
10 3 31.8 7.901657451
11 1 9 3.401680257
12 1 0.8 0.302371578
13 0 0 0
14 1 4.4 1.663043681
15 1 0.2 0.075592895
16 1 0.6 0.226778684
17 3 3.2 0.957178393
18 2 6.6 2.087148887
19 5 14.6 4.31564154
20 1 0.2 0.075592895
21 6 41 7.949603165
22 5 7.8 1.365562853
23 7 20.6 2.794552524
24 2 4.2 1.341640786
25 5 33.2 5.901089488
26 4 11.4 2.127372625

13BUSINESS STATISTICS
27 4 44.2 11.98268592
28 3 7.8 1.893346747
29 4 5.4 1.229788987
30 5 34.2 5.586718945
31 3 10.8 3.39943973
32 3 12 3.415371415
33 4 17.4 4.370899439
34 1 0.6 0.226778684
35 5 11 2.050783822
36 3 21 6.847870715
37 4 25.2 3.981624459
38 5 6.4 0.878310066
39 3 48.6 14.06684721
40 2 20.4 4.998475958
41 4 3 0.423140188
42 4 33.2 5.358748966
43 1 0.4 0.151185789
44 2 1 0.299205297
45 2 25.4 6.203685541
46 2 2.6 0.819988386
47 2 2 0.598410593
48 0 0 0
49 3 14.8 3.067805172
50 2 0.6 0.157359158
51 1 0.8 0.302371578
52 5 59.8 19.75203425
Total 136 649.2 4.992485013
Weekly Rainfall Data for Adelaide in 2016
Days of Rainfall in one year = 136

14BUSINESS STATISTICS
 λ ( Average one week)= 136
52 =2.615
Part 4ai
P (No rainfall in 52 weeks) (X = 0)
 e = 2.6150 x e−2.615
0 ! = 0.0732
Part 4aii
P ( 2 or more days of rainfall) ( X = 0, X = 1)
 P ( 2 or more days of rainfall) = (1- (X = 0) - (X = 1)) = 1 – 2.6150 x e−2.615
0 ! −¿ 2.6151 x e−2.615
1!
 P ( 2 or more days of rainfall) = 1 – (0.0732 + 0.1913) = 1- 0.2645 = 0.7355
Part 4bi
Total Rainfall = 649.2
x̅ = 649.2/ 52 = 12.4846
Std. dev = 4.99
The Z = {(X-μ)/ Standard Deviation}
 P (X>8)= (8 -12.4846)/4.99 = -0.8987; area = 0.1867
 P (X<16)= (16 -12.4846)/4.99 = 0.7045; area = 0.7580
 P (8≤X≤16) = 0.7580 - 0.1867 = 0.5713
Part 4bii
Probability (Rainfall 12% and higher) in which for X can be calculated with the help of Z
score
 12% right tail in normal distribution
 (area = 0.5+0.38 = 0.88) Z = 1.17 , based on this Z = {(X-μ)/ Standard Deviation}
 1.17 = (X – 12.4846)/ 4.99 = 18.3229

15BUSINESS STATISTICS
Answer 5
Part 5a
-4 -3 -2 -1 0 1 2 3 4
0
0.2
0.4
0.6
0.8
1
1.2
NP/TA
NP/TA
Normality Plot of NP/TA (Normally distributed)
-4 -3 -2 -1 0 1 2 3 4
0
10
20
30
40
50
60
70
80
TL/TA
TL/TA
Normality Plot of TL/TA (Not normal distribution as extreme values deviate for normality)

16BUSINESS STATISTICS
-4 -3 -2 -1 0 1 2 3 4
-80
-70
-60
-50
-40
-30
-20
-10
0
10
WC/TA
WC/TA
Normality Plot of WC/TA (Not normal distribution as extreme values deviate for normality)
-4 -3 -2 -1 0 1 2 3 4
0
2
4
6
8
10
12
OE/TL
OE/TL
Normality Plot of OE/TL (Not normal distribution as extreme values deviate largely)

17BUSINESS STATISTICS
-4 -3 -2 -1 0 1 2 3 4
-16
-14
-12
-10
-8
-6
-4
-2
0
2
PS/TS
PS/TS
Normality Plot of PC/TS (Approximately normal as with extreme values of outliers)
-4 -3 -2 -1 0 1 2 3 4
-5
0
5
10
15
20
25
TC/TS
TC/TS
Normality Plot of TC/TS (Considered to be normally distributed with deviation from the straight
line)
Part 5b
The 95% CL  CI = ͞ x ± 1.96 * standard error.
NP/TA TL/TA WC/TA OE/TL PS/TS TC/TS
Mean -
0.57713
0.80864
6991
-
0.042955
5.31279
127
-
0.08221
1.02875

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4 62
Standard Deviation
15.4249
895
2.90846
401
2.502224
856
32.2318
854
0.72864
8301
0.83630
672
Standard Error
0.51161
461
0.09646
7662
0.082993
56
1.06906
416
0.02416
7739
0.02773
854
Confidence level = 1.96 *
sample error
1.00408
459
0.18932
5504
0.162881
5
2.09812
394
0.04743
112
0.05443
911
Lower Limit = Mean - CL
-
1.58121
46
0.61932
1487
-
0.205836
737
3.21466
734
-
0.12964
511
0.97431
709
Upper Limit = Mean + CL
0.42695
46
0.99797
2495
0.119926
264
7.41091
521
-
0.03478
287
1.08319
531
Mean, Standard Deviation, Standard Error and Confidence Interval Limits
Part 5c
The Bankrupt condition has been divided into 2 parts – “0 as not bankrupt” and “1 is
bankrupt in a year” and tested for “Working Capital/ Total Assets (WC/TA)”
H0: There is no significant difference between WC/TA of bankrupt and non-bankrupt
companies
H1: There is significant difference between WC/TA of bankrupt and non-bankrupt
companies
To test the statistics,
 T= (x1 – x2)-(μ1 - μ2)/s.
t-Test: Unequal Variances
WC/TA of non Bankrupt WC/TA of Bankrupt
Mean 0.235882756 -0.38535189
Var 0.104893708 13.62622445
N 501 408
H.M.D 0
D.F. 412
t statistics 3.388760997

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P(T<=t) two-tail 0.000769719
t Critical two-tail 1.965738512
Two Samples of WC/TA under Bankrupt condition
The critical t value at 412df and α = 0.05 is 1.96
The calculated t statistics came out to be as 3.388 which is greater than the critical t value
(such as t stat> t critical) at 95% confidence level. Hence, it can be proved that null hypothesis
will be rejected as 3.388 > 1.96 satiating there is significant difference between WC/TA of
bankrupt and non-bankrupt companies.