Statistics Assignment: Data Interpretation and Statistical Analysis
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Homework Assignment
AI Summary
This statistics assignment provides a comprehensive analysis of various statistical concepts. The assignment begins with an analysis of stock prices using stem and leaf displays, relative frequency histograms, and bar charts. It then delves into measures of central tendency and dispersion, including mean, median, standard deviation, and box plots, applied to financial data. The assignment further explores probability, analyzing data related to student disciplines and ATAR scores, as well as rainfall patterns. Finally, it covers regression analysis, examining confidence intervals and their ability to predict bankruptcy. The assignment utilizes different statistical tools and techniques to interpret data and draw meaningful conclusions, providing a solid foundation in statistical analysis.

Running Head: STATISTICS ASSIGNMENT
Statistics Assignment
Name of the Student
Name of the University
Author Note
Statistics Assignment
Name of the Student
Name of the University
Author Note
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1STATISTICS ASSIGNMENT
Table of Contents
Answer 1..........................................................................................................................................2
Answer 2..........................................................................................................................................4
Answer 3..........................................................................................................................................6
Answer 4..........................................................................................................................................7
Answer 5..........................................................................................................................................8
Table of Contents
Answer 1..........................................................................................................................................2
Answer 2..........................................................................................................................................4
Answer 3..........................................................................................................................................6
Answer 4..........................................................................................................................................7
Answer 5..........................................................................................................................................8

2STATISTICS ASSIGNMENT
Answer 1
(a) All the quarterly opening price values for the two companies CWN (Crown
Resorts Limited) and TAH (Tabcorp Holdings Limited) have been listed and the stem
and leaf display with the stem values in the middle and the leaf values for CWN and
TAH on either side of the stem values are given in table 1.1.
(b) The relative frequency histogram for CWN and frequency polygon for TAH is
shown in figure 1.1.
0-
2.00 2.00-
4.00 4.00-
6.00 6.00-
8.00 8.00-
10.00 10.00
-
12.00
12.00
-
14.00
14.00
-
16.00
more
than
16.00
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Relative Frequency of opening share prices of CWN and
TAH
CWN
TAH
Class Width
Share Prices
Figure 1.1: Opening share price values of CWN and TAH
Answer 1
(a) All the quarterly opening price values for the two companies CWN (Crown
Resorts Limited) and TAH (Tabcorp Holdings Limited) have been listed and the stem
and leaf display with the stem values in the middle and the leaf values for CWN and
TAH on either side of the stem values are given in table 1.1.
(b) The relative frequency histogram for CWN and frequency polygon for TAH is
shown in figure 1.1.
0-
2.00 2.00-
4.00 4.00-
6.00 6.00-
8.00 8.00-
10.00 10.00
-
12.00
12.00
-
14.00
14.00
-
16.00
more
than
16.00
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Relative Frequency of opening share prices of CWN and
TAH
CWN
TAH
Class Width
Share Prices
Figure 1.1: Opening share price values of CWN and TAH
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3STATISTICS ASSIGNMENT
(c) A bar chart showing the market capitals in 2016 of 5 companies listed in ASX
that deals with leisure and entertainment is given in figure 1.2. All the companies
considered here is over AUD500 million in market capital.
ALL CWN FLT REA TTS
0
2,000,000,000
4,000,000,000
6,000,000,000
8,000,000,000
10,000,000,000
12,000,000,000
14,000,000,000
16,000,000,000
Market Cap
Market Cap
Company Code
Market Capital Value
Figure 1.2: Market capital values of 5 Australian Entertainment Companies
(d) Based on the research of the two companies CWN and TAH, it can be said that
the share prices of CWN is a lot higher than that of TAH. It can also be seen that the
share prices of TAH has been decreasing continually. Thus, it would be better to invest in
CWN.
(c) A bar chart showing the market capitals in 2016 of 5 companies listed in ASX
that deals with leisure and entertainment is given in figure 1.2. All the companies
considered here is over AUD500 million in market capital.
ALL CWN FLT REA TTS
0
2,000,000,000
4,000,000,000
6,000,000,000
8,000,000,000
10,000,000,000
12,000,000,000
14,000,000,000
16,000,000,000
Market Cap
Market Cap
Company Code
Market Capital Value
Figure 1.2: Market capital values of 5 Australian Entertainment Companies
(d) Based on the research of the two companies CWN and TAH, it can be said that
the share prices of CWN is a lot higher than that of TAH. It can also be seen that the
share prices of TAH has been decreasing continually. Thus, it would be better to invest in
CWN.
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4STATISTICS ASSIGNMENT
Answer 2
(a) The mean, median, first quartile and third quartile of annual dividends for each of
the banks are given in table 2.1.
Table 2.1: Centrality measures of Annual Dividends
(b) The Standard Deviation, Range and Coefficient of Variation for the dividends of
each of the given banks are given in table 2.2.
Table 2.2: Measures of dispersion of Annual Dividends
(c) The box and whisker plot to show the dividends of each of the banks are given in
figure 2.1.
Answer 2
(a) The mean, median, first quartile and third quartile of annual dividends for each of
the banks are given in table 2.1.
Table 2.1: Centrality measures of Annual Dividends
(b) The Standard Deviation, Range and Coefficient of Variation for the dividends of
each of the given banks are given in table 2.2.
Table 2.2: Measures of dispersion of Annual Dividends
(c) The box and whisker plot to show the dividends of each of the banks are given in
figure 2.1.

5STATISTICS ASSIGNMENT
CBA NAB ANZ WBC
0
0.5
1
1.5
2
2.5
3
3.5
4
Boxplot
Bank Codes
Dividend Values
Figure 2.1: Boxplot of Annual Dividends of the Banks
(d) Australia’s financial regulator, Australian Prudential Regulation Authority, or
APRA, has increased pressure on the banks on lending in recent times. Thus, by lending
more money, the dividends earned by the banks will increase. The people burrowing
money will have to return the money along with the interest. Thus, more lending of
money indicates more income of the banks. Thus, to increase the dividends of the banks,
this pressure is made.
CBA NAB ANZ WBC
0
0.5
1
1.5
2
2.5
3
3.5
4
Boxplot
Bank Codes
Dividend Values
Figure 2.1: Boxplot of Annual Dividends of the Banks
(d) Australia’s financial regulator, Australian Prudential Regulation Authority, or
APRA, has increased pressure on the banks on lending in recent times. Thus, by lending
more money, the dividends earned by the banks will increase. The people burrowing
money will have to return the money along with the interest. Thus, more lending of
money indicates more income of the banks. Thus, to increase the dividends of the banks,
this pressure is made.
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6STATISTICS ASSIGNMENT
Answer 3
(a) The discipline which is the most popular for the best students is Engineering and
related Technologies. The proportion of best students taking up that course is 0.30.
(b) If an Australian student is selected at random, the probability that he or she is
studying “Society and Culture” and had an ATAR score of 80.00 or less is ((5030 + 3806
+ 2807 + 2814) / 221060) = 0.065.
(c) Based on the proportion of offers made, the discipline which has the highest
proportion of students with the lowest ATAR grades is Education. The required
proportion is 0.073.
(d) The discipline which has the majority of students from “No ATAR/Non-Yr 12”
background is Health.
Answer 3
(a) The discipline which is the most popular for the best students is Engineering and
related Technologies. The proportion of best students taking up that course is 0.30.
(b) If an Australian student is selected at random, the probability that he or she is
studying “Society and Culture” and had an ATAR score of 80.00 or less is ((5030 + 3806
+ 2807 + 2814) / 221060) = 0.065.
(c) Based on the proportion of offers made, the discipline which has the highest
proportion of students with the lowest ATAR grades is Education. The required
proportion is 0.073.
(d) The discipline which has the majority of students from “No ATAR/Non-Yr 12”
background is Health.
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7STATISTICS ASSIGNMENT
Answer 4
(a) (i) It can be seen from the given data that out of the 52 weeks of a year, there are 5
weeks which do not have any amount of rainfall. Thus, the probability that in a given
week there will be no rainfall is (5/52) = 0.096.
(ii) There were 33 weeks in 2016 in which two or more days of rainfall has taken
place. Thus, the probability that there will be two or more days of rainfall in a week is
(33/52) = 0.635.
(b) The mean of the weekly total amount of rainfall is 12.035 and the standard
deviation of the weekly total amount of rainfall is 14.73.
(i) The probability that there will be less than 8 mm rainfall is given by:
P (z) = P ((8 – 12.035) / 14.73) = 0.3921.
The probability that there will be less than 16 mm rainfall is given by:
P (z) = P ((16 – 12.035) / 14.73) = 0.6061.
The probability that there will be rainfall between 8 mm and 16 mm is given by (0.0601 –
0.3921) = 0.2140.
(ii) 12 percent of the weeks have 0.224 mm rainfall or higher.
Answer 4
(a) (i) It can be seen from the given data that out of the 52 weeks of a year, there are 5
weeks which do not have any amount of rainfall. Thus, the probability that in a given
week there will be no rainfall is (5/52) = 0.096.
(ii) There were 33 weeks in 2016 in which two or more days of rainfall has taken
place. Thus, the probability that there will be two or more days of rainfall in a week is
(33/52) = 0.635.
(b) The mean of the weekly total amount of rainfall is 12.035 and the standard
deviation of the weekly total amount of rainfall is 14.73.
(i) The probability that there will be less than 8 mm rainfall is given by:
P (z) = P ((8 – 12.035) / 14.73) = 0.3921.
The probability that there will be less than 16 mm rainfall is given by:
P (z) = P ((16 – 12.035) / 14.73) = 0.6061.
The probability that there will be rainfall between 8 mm and 16 mm is given by (0.0601 –
0.3921) = 0.2140.
(ii) 12 percent of the weeks have 0.224 mm rainfall or higher.

8STATISTICS ASSIGNMENT
Answer 5
(a) It can be seen from the normal probability plots of all the three variables that none
of the variables are normally distributed.
0.055005500550055 37.018701870187 73.982398239824
-500
-400
-300
-200
-100
0
100
Normal Probability Plot
Series1
Sample Percentile
NP/TA
0.055005500550055 37.018701870187 73.982398239824
0
20
40
60
80
Normal Probability Plot
Series1
Sample Percentile
TL/TA
0.055005500550055 37.018701870187 73.982398239824
-80
-60
-40
-20
0
20
Normal Probability Plot
Series1
Sample Percentile
WC/TA
Answer 5
(a) It can be seen from the normal probability plots of all the three variables that none
of the variables are normally distributed.
0.055005500550055 37.018701870187 73.982398239824
-500
-400
-300
-200
-100
0
100
Normal Probability Plot
Series1
Sample Percentile
NP/TA
0.055005500550055 37.018701870187 73.982398239824
0
20
40
60
80
Normal Probability Plot
Series1
Sample Percentile
TL/TA
0.055005500550055 37.018701870187 73.982398239824
-80
-60
-40
-20
0
20
Normal Probability Plot
Series1
Sample Percentile
WC/TA
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9STATISTICS ASSIGNMENT
0.055005500550055 37.018701870187 73.982398239824-200
0
200
400
600
800
1000
Normal Probability Plot
Series1
Sample Percentile
OE/TL
0.055005500550055 37.018701870187 73.982398239824
-15
-10
-5
0
5
Normal Probability Plot
Series1
Sample Percentile
PS/TS
0.055005500550055 37.018701870187 73.982398239824-5
0
5
10
15
20
25
Normal Probability Plot
Series1
Sample Percentile
TC/TS
(b) The 95 percent confidence intervals have been obtained from the regression
tables. The confidence intervals of each of the variables are given in the following table.
Variable Lower Limit Upper Limit
NP/TA -0.0036 0.0006
TL/TA 0.0107 0.0328
0.055005500550055 37.018701870187 73.982398239824-200
0
200
400
600
800
1000
Normal Probability Plot
Series1
Sample Percentile
OE/TL
0.055005500550055 37.018701870187 73.982398239824
-15
-10
-5
0
5
Normal Probability Plot
Series1
Sample Percentile
PS/TS
0.055005500550055 37.018701870187 73.982398239824-5
0
5
10
15
20
25
Normal Probability Plot
Series1
Sample Percentile
TC/TS
(b) The 95 percent confidence intervals have been obtained from the regression
tables. The confidence intervals of each of the variables are given in the following table.
Variable Lower Limit Upper Limit
NP/TA -0.0036 0.0006
TL/TA 0.0107 0.0328
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10STATISTICS ASSIGNMENT
WC/TA -0.0374 -0.0117
OE/TL 0.0015 0.0035
PS/TS -0.1847 -0.0676
TC/TS 0.0521 0.1288
(c) It can be seen from the regression analysis that all the variables have a very low R
Square value in predicting bankruptcy. Thus, The variables are not a good predictor of
bankruptcy.
WC/TA -0.0374 -0.0117
OE/TL 0.0015 0.0035
PS/TS -0.1847 -0.0676
TC/TS 0.0521 0.1288
(c) It can be seen from the regression analysis that all the variables have a very low R
Square value in predicting bankruptcy. Thus, The variables are not a good predictor of
bankruptcy.
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