Predicting Bankruptcy: An Empirical Analysis of Financial Ratios

This assignment examines the relationship between bankruptcy and four financial ratios: TL/TA, WC/TA, PS/TS, and TC/TS. The results of an independent sample t-test indicate statistically significant differences between non-bankrupt and bankrupt firms for these ratios. Specifically, there are significant differences in mean values for WC/TA (p=0.001), PS/TS (p=0.000), and TC/TS (p=0.000). The findings suggest that bankruptcy can be predicted from the ratios of TL/TA, WC/TA, PS/TS, and TC/TS.

Running head : STATISTICS
Statistics
Name of the student
Name of the university
Author’s note

1STATISTICS
Table of Contents
Question 1........................................................................................................................................3
Part a............................................................................................................................................3
Part b............................................................................................................................................5
Part c............................................................................................................................................6
Part d............................................................................................................................................7
Question 2........................................................................................................................................7
Part a............................................................................................................................................7
Part b............................................................................................................................................7
Part c............................................................................................................................................8
Part d............................................................................................................................................8
Question 3........................................................................................................................................9
Part a............................................................................................................................................9
Part b............................................................................................................................................9
Part c............................................................................................................................................9
Part d..........................................................................................................................................10
Question 4......................................................................................................................................11
Part a..........................................................................................................................................12
Part i.......................................................................................................................................12

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Part ii......................................................................................................................................12
Part b..........................................................................................................................................12
Part i.......................................................................................................................................13
Part ii......................................................................................................................................13
Question 5......................................................................................................................................14
Part a..........................................................................................................................................14
Part b..........................................................................................................................................17
Part c..........................................................................................................................................17
References......................................................................................................................................22

3STATISTICS
Question 1
Part a
The opening prices of Crown Resorts Limited (CWN) are presented in Table 1:
Table 1: Opening prices of CWN
Year
Month
JAN APR JUL OCT
2008 13.45 10.75 9.44 9
2009 5.91 6.3 7.3 8.8
2010 8.07 8.19 7.66 8.48
2011 8.37 8.34 8.93 7.8
2012 8.09 8.8 8.57 9.04
2013 10.67 12.3 12 15.54
2014 16.85 17.03 15.1 13.75
2015 12.69 13.26 12.61 10.03
2016 12.5 12.3 12.6 13.06
2017 11.58 11.8 12.15
The opening prices of Tabcorp Holdings Limited (TAH) are presented in Table 2:
Table 2: Opening prices of TAH
Year
Month
JAN APR JUL OCT
2008 6.92 6.76 4.68 3.90
2009 3.27 3.16 3.30 3.35
2010 3.25 3.24 2.95 3.27
2011 3.37 3.55 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 3.36 3.30 3.54
2015 4.08 4.79 4.59 4.70
2016 4.71 4.25 4.55 5.01
2017 4.81 4.77 4.35

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The stem and leaf plot of CWN and TAH is presented in table 3. The leaf of CWN is on
the right side and of TAH is on left side. The stem is in the center.
Table 3: Stem and Leaf Plot
TAH Stem CWN
9 9 9 7 6 6 5 2
3 3 3 3 3 2 2 2 2 2 2 1 1 0 3
8 7 7 7 7 6 5 5 2 0 4
0 5 9
9 7 6 3
7 3 6 8
8 0 0 1 3 3 4 5 8 8 9
9 0 0 4
10 0 6 7
11 5 8
12 0 1 3 3 5 6 6 6
13 0 2 4 7
14
15 1 5
16 8 0
17 0

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Part b
The relative frequencies of the opening prices of CWN and TAH are presented in table 4
Table 4: Relative Frequencies of CWN and TAH
Range
Relative Frequency
CWN TAH
$0 to < $2 0.00
$2 to < $4 0.64
$4 to < $6 0.03 0.31
$6 to < $8 0.10 0.05
$8 to < $10 0.33
$10 to < $12 0.15
$12 to < $14 0.28
$ 14 to < $16 0.05
$16 to < $18 0.05
The relative frequency histogram of CWN and polygon of TAH is shown in figure 1.
$0 to
>$2 $2 to
>$4 $4 to
>$6 $6 to
>$8 $8 to
>$10 $10 to
>$12 $12 to
>$14 $14 to
>$16 $16 to >
$18
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Comparison of share prices CWN and TAH
CWN
TAH
Price Range
Relative Frequency
Figure 1: Comparison of share prices of CWN and TAH

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Part c
The market capital of five organization was obtained from investsmart website
(investsmart.com.au, 2017). Table 5 presents a comparison of market capital of the five
organizations.
Table 5: Comparison of Market Capital
Name of the organization MKT CAP (M)
Aristocrat Leisure (ALL) 13614
Crown Resorts Limited (CWN) 8087
The star entertainment group (SGR) 4285
Tabcorp (TAH) 3475
Tatts Group (TTS) 5801
Aristocrat Leisure
(ALL) Crown Resorts
Limited (CWN) The star
entertainment
group (SGR)
Tabcorp (TAH) Tatts Group (TTS)
0
2000
4000
6000
8000
10000
12000
14000
16000
Comparison of Market Capitals
Name of Organization
Market Capital
Figure 2: Comparison of Market Capital

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Part d
In the report by investsmart (2017) when we compare the P/E % and Dividend yield of
CWN and TAH, we find that the two ratios of CWN is higher than TAH’s. Thus, it would be
wiser to invest in CWN than to invest in shares of TAH.
Question 2
Part a
Table 6 presents the statistical comparison of the annual dividends of the four banks.
Table 6: Comparison of Annual Dividends
Name of Bank
CBA NAB ANZ WBC
Mean 3.10 2.31 2.06 2.04
Median 3.05 2.36 2.05 2.04
1st Quartile 2.49 2.19 1.91 1.91
3rd Quartile 3.73 2.45 2.17 2.24
Part b
Table 7 presents some statistical comparisons of the annual dividends of the four banks.
Table 7: Comparison of Annual Dividends
CBA NAB ANZ WBC
Standard Deviation 0.79 0.23 0.36 0.37
Range 2.23 0.79 1.21 1.30
Coefficient of Variation 25% 10% 17% 18%

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Part c
CBA NAB ANZ WBC
1.00
1.50
2.00
2.50
3.00
3.50
4.00
Comparison of Annual Divdends of Banks
Name of Bank
Annual Dividend
Figure 3: Comparison of Annual dividends of Banks
From figure 3 it can be seen that the annual dividend of CBA is the highest. The annual
dividends of ANZ and WBC are very similar. In addition the dividend of WBC bank is left
skewed.
Part d
Australian Prudential Regulation Authority (APRA) in 2017 has commenced the process
to crackdown on home loans. APRA has asked banks to restrict interest-only home loans to be
restricted at 30to40% level. Banks across Australia have accepted the new directions of APRA.
This, they (banks) say would boost the profits of the banks (abc.net.au, 2017).

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Question 3
Part a
 The best students have an ATAR of 90.05 or more.
 30.0% of all students of “Engineering and Related Technologies” have more than 90.05 or
more ATAR.
 The number of students = 4203.
 The total number of students offered in 2016 = 14027.
 Thus the proportion of students offered = 4203
14207 =0.2996
Part b
 The total number of student’s admitted in 2016 = 221060
 The total number of students admitted in “Society and Culture” = 46841
 The number of students having 80.05 to 90.00 = 6043
 The number of students having 90.05 or more = 8180
 Thus the number of students having more than 80.05 = 6043 + 8180 = 14223
 Hence, number of students having less than 80.05 = 46841 - 14223 = 32681
 Hence, a randomly selected Australian Student studying in “Society and Culture” and has
ATAR score of less than 80.05 = 32681
221060 =0.148

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Part c
 The lowest ATAR grade is 50.00 or less
 Education course has the highest proportion of students with ATAR grade less than or equal
to 50.00
 The proportion of students = 0.073
Part d
The maximum number of students with “No ATAR/Non-Yr 12” is in the field of health
education. According to Macquarie (2017) ATAR score of a student is just of the ways to
measure a student. Macquarie also uses other methods to determine the suitability of a candidate.
Thus the other methods are found suitable by a large number of students to take admission.

11STATISTICS
Question 4
The weekly data for Adelaide Airport, number 23034 is presented in the following table:
Table 8: Rainfall data for Adelaide Airport, Number 230304
week Jan Feb Mar Apr May Jun July Aug Sep Oct Nov Dec
1st 0 0 0 0.4 3.6 3.4 9.2 1.4 1.8 0.2 0
2nd 10.8 0 0 0 0 0.4 1 0 0 0 0
3rd 3 0 0 0 2.4 0 0.6 0.4 11 0 0
4th 0 0 0 0 5.6 0 0 0 1 9.4 0 0
5th 0 0 0 0 0 0.4 32.8 0 0 0 0 4.6
6th 0 0 0 4.4 0 6.8 1 0 0 0 0 0
7th 0 0 11.6 0 0 2 3 0 0 0 0 0
8th 0 0 0 0 1 4.8 0 0 2.4 0 0 7.8
9th 0 0 0 0 0.2 5.8 0 0.4 18.4 0 0 2.4
10th 0 0 19.6 0 11.8 0.6 7.4 2.4 0 1 0 0
11th 0 0 0.6 0.2 0.2 0.4 0 9.2 0.2 0.6 0 0
12th 1.2 0 0 0 1.2 0.2 3 0 0 0.6 13.2 0
13th 0 0 0 0 1.2 0 4.6 0 4.8 0 12.2 0
14th 0 0 0 0 0 0 0.2 0 7.6 0 2.2 0.2
15th 0 0 0 0 0 0 0 0 9.8 0 0.4 0
16th 0 0.6 0 0 0.2 0.6 0 0 0 0.8 0 0
17th 0 0 0 0 0 3.6 0 0.8 0 11.6 0 0.4
18th 0 0 9 0 0 0 0 0 3 8 0 0
19th 0 0 0 0 0 0 0 11.8 0.4 2.4 0 0
20th 1 0 0 0 0 0.6 1 4.2 0.6 0 0 0.8
21st 0 0 0 0.6 0 3.8 0 0.6 2.2 11.2 0 0
22nd 6.6 0 0 0 0 12.2 0.2 0 0 0 1.6 0
23rd 8.8 0 0 0 3.8 2.6 3.4 0 0 0 0.4 0
24th 0 0.4 0 0 0.4 14 0.8 0 1.6 0 0 0
25th 0 0 0.8 0 4.2 0 8 0 1.6 0 0 0
26th 0 0 0 0 22.8 0 8.6 0.6 0 0.4 0 0.8
27th 0 0 0 0 1.8 5.4 14.4 0 0 0 0 4.4
28th 0 0 0 0.2 8 0 0 0 0 0 0 53.2
29th 6.2 0 0 2.6 0 0 0.2 2.4 38 0 0 1.2
30th 12.2 0 0 0.6 2.2 0 5.8 8.8 0 0 0
31st 0.2 0 0.8 3 0 0.8 0

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Part a
 The total number of days when no rainfall occurred = 228
 Number of weeks in a year = 52
 Thus, λ= Number of days of No Rainfall
Number of Weeks = Σx
n =228
52 =4.385
Part i
 Thus, the probability that in a week there would be NO rainfall¿ λ0 e−λ
0 !
¿ 4.3850 e−4.385
0 ! =0.0125
Part ii
 The probability that in a week there will 2 or more days of rainfall =
1− λ0 e− λ
0 ! – λ1 e−λ
1! =1−0.0125− 4.3851 e−4.385
1!
¿ 1−0.0125−0.0546=0.9329
 Hence, the probability that in a week would be 2 or more days of rainfall = 0.9329
Part b
The total amount of Rainfall at Adelaide Airport = 649.2 mm
Hence, the average rainfall in a week ¿ 649.2
52 =12.48mm
The standard deviation of the amount of rainfall in a week = 14.59mm

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Part i
 The probability that the rainfall would be between 8mm and 16mm is P(8<x<16)
 The z-score of rainfall more than 8mm = 8−12.48
14.59 =−4.48
14.59 =−0.3071
 The probability of rainfall more than 8 mm = P ( z> 8 )=0.3794
The z-score of rainfall less than 16 mm = 16−12.48
14.59 = 3.52
14.59 =0.2413
The probability of rainfall less than 16 mm = P ( z<16 ) =0.5953
 Hence, the probability P(8<x<16) = 0.3794+0.5953 = 0.9747
Part ii
 For measuring the rainfall more than 12% we have to use the right tailed test.
 The z-score for 12% = 1.1750
 The equation can be depicted as : 1.1750= x−12.48
14.59
 Solving for “x” in the above equation: X = 29.6231
 Thus if 12% of the weeks have an average of 12.48 mm rainfall then the amount of rainfall =
29.6231mm

14STATISTICS
Question 5
Part a
The normal probability plot of NP/TA, TL/TA, WC/TA, OE/TL, PS/TS and TC/TS can
be represented as:
-4 -3 -2 -1 0 1 2 3 4
-500
-400
-300
-200
-100
0
100
probability plot NP/TA
z-value
Observed Values
Figure 4: Normal Probability Plot of Net Profit /Total Assets
-4 -3 -2 -1 0 1 2 3 4
0
10
20
30
40
50
60
70
80
TL/TA
Figure 5: Normal Probability Plot of Total Liability/ Total Assets

15STATISTICS
-4 -3 -2 -1 0 1 2 3 4
-80
-70
-60
-50
-40
-30
-20
-10
0
10
WC/TA
Figure 6: Normal Probability Plot of Working Capital /Total Assets
-4 -3 -2 -1 0 1 2 3 4
-100
0
100
200
300
400
500
600
700
800
OE/TL
Figure 7: Normal Probability Plot of Operating Expenses / Total Liabilities

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-4 -3 -2 -1 0 1 2 3 4
-16
-14
-12
-10
-8
-6
-4
-2
0
2
PS/TS
Figure 8: Normal Probability Plot of Profit on Sales/ Total Sales
-4 -3 -2 -1 0 1 2 3 4
-5
0
5
10
15
20
25
TC/TS
Figure 9: Normal Probability Plot of Total Costs/ Total Sales
The above figure shows that the predictors of bankruptcy are not normally distributed.

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Part b
Table 9 presents the 95% confidence interval of bankruptcy predictors.
Table 9: 95% Confidence Interval for bankruptcy predictors
Bankruptcy Predictors Lower Limit Upper Limit
NP/TA -1.5825 0.4282
TL/TA 0.5226 1.0947
WC/TA -0.3720 0.2861
OE/TL 0.0050 10.6323
PS/TS -0.2264 0.0619
TC/TS 0.8356 1.2220
Part c
Table 8: Test for Hypothesis NP/TA
t-Test: Two-Sample Assuming Unequal Variances
Non-Bankrupt Bankrupt
Mean 0.07 -1.37
Variance 0.02 529.64
Observations 501 408
Hypothesized Mean Difference 0
df 407
t Stat 1.265
P(T<=t) one-tail 0.103
t Critical one-tail 1.649
P(T<=t) two-tail 0.207
t Critical two-tail 1.966
The test for hypothesis finds that there is statistically no significant difference between
Non-Bankrupt (0.070) and Bankrupt (-1.372) for NP/TA, t(407) = 1.265, p = 0.207 at 0.05 level
of significance.

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Table 9: Test for Hypothesis TL/TA
t-Test: Two-Sample Assuming Unequal Variances
Non-Bankrupt Bankrupt
Mean 0.47 1.22
Variance 0.10 18.45
Observations 501 408
Hypothesized Mean Difference 0
df 410
t Stat -3.495
P(T<=t) one-tail 0.000
t Critical one-tail 1.649
P(T<=t) two-tail 0.001
t Critical two-tail 1.966
The test for hypothesis finds that there are statistically significant differences between
Non-Bankrupt (0.474) and Bankrupt (1.22) for TL/TA, t(410) = -3.495, p = 0.001 at 0.05 level of
significance.
Table 10: Test for Hypothesis WC/TA
t-Test: Two-Sample Assuming Unequal Variances
Non-Bankrupt Bankrupt
Mean 0.24 -0.39
Variance 0.10 13.63
Observations 501 408
Hypothesized Mean Difference 0
df 412
t Stat 3.389
P(T<=t) one-tail 0.000
t Critical one-tail 1.649
P(T<=t) two-tail 0.001
t Critical two-tail 1.966

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The test for hypothesis finds that there are statistically significant difference between
Non-Bankrupt (0.24) and Bankrupt (-0.39) for WC/TA, t(412) = 3.389, p = 0.001 at 0.05 level of
significance.
Table 11: Test for Hypothesis OE/TL
t-Test: Two-Sample Assuming Unequal Variances
Non-Bankrupt Bankrupt
Mean 3.68 7.32
Variance 34.46 2268.05
Observations 501 408
Hypothesized Mean Difference 0
df 417
t Stat -1.537
P(T<=t) one-tail 0.063
t Critical one-tail 1.649
P(T<=t) two-tail 0.125
t Critical two-tail 1.966
The test for hypothesis finds that there is statistically no significant difference between
Non-Bankrupt (3.68) and Bankrupt (7.32) for OE/TL, t(417) = -1.537, p = 0.125 at 0.05 level of
significance.

20STATISTICS
Table 12: Test for Hypothesis PS/TS
t-Test: Two-Sample Assuming Unequal Variances
Non-Bankrupt Bankrupt
Mean 0.05 -0.25
Variance 0.03 1.10
Observations 501 408
Hypothesized Mean Difference 0
df 423
t Stat 5.767
P(T<=t) one-tail 0.000
t Critical one-tail 1.648
P(T<=t) two-tail 0.000
t Critical two-tail 1.966
The test for hypothesis finds that there are statistically significant difference between
Non-Bankrupt (0.05) and Bankrupt (-0.25) for PS/TS, t(423) = 5.767, p = 0.000 at 0.05 level of
significance.
Table 13: Test for Hypothesis TC/TS
t-Test: Two-Sample Assuming Unequal Variances
Non-Bankrupt Bankrupt
Mean 0.91 1.17
Variance 0.03 1.49
Observations 501 408
Hypothesized Mean Difference 0
df 418
t Stat -4.194
P(T<=t) one-tail 0.000
t Critical one-tail 1.649
P(T<=t) two-tail 0.000
t Critical two-tail 1.966

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The test for hypothesis finds that there are statistically significant difference between
Non-Bankrupt (0.914) and Bankrupt (1.170) for PS/TS, t(907) = 0.000, p = 0.000 at 0.05 level of
significance.
Thus from the independent sample t-test we find that there are statistically significant
differences between the factors TL/TA, WC/TA, PS/TS and TC/TS. Hence bankruptcy can be
predicted from the ratios of TL/TA, WC/TA, PS/TS and TC/TS

22STATISTICS
References
Abc.net.au [2017] http://www.abc.net.au/news/2017-04-11/why-tougher-apra-rules-do-not-
worry-banks/8421804
Investsmart.com.au [20147] https://www.investsmart.com.au/shares/asx-tah/tabcorp-holdings-
limited
Mq.edu.au [2017] https://www.mq.edu.au/study/find-a-course/undergraduate/macquarie-entry