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Questions:
Question 1

(a)List all the quarterly opening price values in two tables, one for CWN and the other for TAH. Then construct a stem-and-leaf display with one stem value in the middle, and CWN leaves on the right side and TAH leaves on the left side. (Must use EXCEL or similar for the plot.)

(b)Construct a relative frequency histogram for CWN and a frequency polygon for TAH on the same graph with equal class widths, the first class being “$0 to less than $2”. Use two different colours for CWN and TAH. Graph must be done in EXCEL or similar software.

(c)Draw a bar chart of market capitals (or total assets) in 2016 (in million Australian dollars) of 5 companies listed in ASX that trade in leisure and entertainment (may include gaming, gambling and cinema) with over AUD500 million in market capital.  Graphing must be done in EXCEL or with similar software.

(d)If one wishes to invest in CWN or TAH, what is the market recommendation (for example, from Morningstar, Fatprophets, InvestSmart, etc.)? If you cannot find the information, what would be your recommendation based on your research of these two companies?

Question 2

(a)only the data provided in the table, do not add or change anything in the table) using the exact position, (n+1)f, where n is the number of observations and f the relevant fraction for the quartile.

(c)Draw a box and whisker plot for the annual dividends of each bank and put them side by side on one graph with the same scale so that the dividends can be compared. (This graph must be done in EXCEL or similar software and cannot be hand-drawn.)  Australia’s financial regulator, Australian Prudential Regulation Authority, or APRA, has increased pressure on the banks on lending in recent times. Please explain what this pressure is about and how it is affecting the banks.

Question 3

(a)Based on the proportion of offers made for different grades, which discipline is the most popular for the best students and what is that proportion?

(b)If an Australian student is selected at random, then from the table above, what is the probability that he or she is studying “Society and Culture” and had an ATAR score of 80.00 or less?

(c)Based on the proportion of offers made, which discipline has the highest proportion of students with the lowest ATAR grades, and what is that proportion?

(d)Which discipline has majority of the students from “No ATAR/Non-Yr 12” background? Please find out the reasons for that through internet searches?

Question 4

(a) The following data collected from the Australian Bureau of Meteorology Website  gives the daily rainfall data (includes all forms of precipitation such as rain, drizzle, hail and snow) for the year 2016 in Adelaide, South Australia. The zero values indicate no rainfall and the left-most column gives the date. Assuming that the weekly rainfall event (number of days in a week with rainfall) follows a Poisson distribution (There are 52 weeks in a year and a week is assumed to start from Monday. The first week starts from 4 January 2016 – you are expected to visit the website and get the daily values which are not given in the table below.

(i) What is the probability that on any given week in a year there would be no rainfall?

(ii) What is the probability that there will be 2 or more days of rainfall in a week?

(b)Assuming that the weekly total amount of rainfall (in mm) from the data provided in part (a) has a normal distribution, compute the mean and standard deviation of weekly totals.

(i) What is the probability that in a given week there will be between 8mm and 16mm of rainfall?

(ii) What is the amount of rainfall if only 12% of the weeks have that amount of rainfall or higher?

Answers:
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

 

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

 

 

 

 

 

 

 

 

 

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.

Figure 1: Comparison of share prices of CWN and TAH

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

Figure 2: Comparison of Market Capital

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%

Part c

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).

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 =

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

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.

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

Part a

The total number of days when no rainfall occurred = 228

Number of weeks in a year = 52

Thus,

Part i

à Thus, the probability that in a week there would be NO rainfall                                                                                    

Part ii

The probability that in a week there will 2 or more days of rainfall

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 mm

The standard deviation of the amount of rainfall in a week = 14.59mm

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 =

The probability of rainfall more than 8 mm =

The z-score of rainfall less than 16 mm =

The probability of rainfall less than 16 mm =

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 :    

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

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:

Figure 4: Normal Probability Plot of Net Profit /Total Assets

Figure 5: Normal Probability Plot of Total Liability/ Total Assets

Figure 6: Normal Probability Plot of Working Capital /Total Assets

Figure 7: Normal Probability Plot of Operating Expenses / Total Liabilities

Figure 8: Normal Probability Plot of Profit on Sales/ Total Sales

Figure 9: Normal Probability Plot of Total Costs/ Total Sales

The above figure shows that the predictors of bankruptcy are not normally distributed.

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.

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

 

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.

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

 

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

References

Abc.net.au [2017] https://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

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