(a)List all the quarterly opening price values in two tables, one for CWN and the other for TAH. Then construct a stemandleaf 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?
(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 handdrawn.) 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.
(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/NonYr 12” background? Please find out the reasons for that through internet searches?
(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 leftmost 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?
The opening prices of Crown Resorts Limited (CWN) are presented in Table 1:
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:
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.
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 









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
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
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.
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 
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% 
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.
Australian Prudential Regulation Authority (APRA) in 2017 has commenced the process to crackdown on home loans. APRA has asked banks to restrict interestonly 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).
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 =
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
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
The maximum number of students with “No ATAR/NonYr 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.
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 
The total number of days when no rainfall occurred = 228
Number of weeks in a year = 52
Thus,
à Thus, the probability that in a week there would be NO rainfall
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
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
The probability that the rainfall would be between 8mm and 16mm is P(8<x<16)
The zscore of rainfall more than 8mm =
The probability of rainfall more than 8 mm =
The zscore 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
For measuring the rainfall more than 12% we have to use the right tailed test.
The zscore 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
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.
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 
Table 8: Test for Hypothesis NP/TA
tTest: TwoSample Assuming Unequal Variances 


NonBankrupt 
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) onetail 
0.103 

t Critical onetail 
1.649 

P(T<=t) twotail 
0.207 

t Critical twotail 
1.966 

The test for hypothesis finds that there is statistically no significant difference between NonBankrupt (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
tTest: TwoSample Assuming Unequal Variances 


NonBankrupt 
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) onetail 
0.000 

t Critical onetail 
1.649 

P(T<=t) twotail 
0.001 

t Critical twotail 
1.966 

The test for hypothesis finds that there are statistically significant differences between NonBankrupt (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
tTest: TwoSample Assuming Unequal Variances 


NonBankrupt 
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) onetail 
0.000 

t Critical onetail 
1.649 

P(T<=t) twotail 
0.001 

t Critical twotail 
1.966 

The test for hypothesis finds that there are statistically significant difference between NonBankrupt (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
tTest: TwoSample Assuming Unequal Variances 


NonBankrupt 
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) onetail 
0.063 

t Critical onetail 
1.649 

P(T<=t) twotail 
0.125 

t Critical twotail 
1.966 

The test for hypothesis finds that there is statistically no significant difference between NonBankrupt (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
tTest: TwoSample Assuming Unequal Variances 


NonBankrupt 
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) onetail 
0.000 

t Critical onetail 
1.648 

P(T<=t) twotail 
0.000 

t Critical twotail 
1.966 

The test for hypothesis finds that there are statistically significant difference between NonBankrupt (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
tTest: TwoSample Assuming Unequal Variances 


NonBankrupt 
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) onetail 
0.000 

t Critical onetail 
1.649 

P(T<=t) twotail 
0.000 

t Critical twotail 
1.966 

The test for hypothesis finds that there are statistically significant difference between NonBankrupt (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 ttest 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
Abc.net.au [2017] https://www.abc.net.au/news/20170411/whytougheraprarulesdonotworrybanks/8421804
Investsmart.com.au [20147] https://www.investsmart.com.au/shares/asxtah/tabcorpholdingslimited
Mq.edu.au [2017] https://www.mq.edu.au/study/findacourse/undergraduate/macquarieentry
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