Statistics Assignment on Australian Stock Exchange, Apartment Prices and Employment Type in Australia

This assignment involves analyzing the Australian Stock Exchange, apartment prices in Australian state capital city centres and employment type and occupation in Australia. It includes constructing stem-and-leaf plots, relative frequency histograms, frequency polygons, bar charts, mean, median, first quartile, third quartile, standard deviation, mean absolute deviation, range, probability calculations and box and whisker plots.

Statistics
Student Name:
Instructor Name:
Course Number:
5 September 2018
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Question 1 4 Marks
Visit the Australian Stock Exchange website, www.asx.com.au and from “Prices and research” drop-
down menu, select “Company information”. Type in the ASX code “CSR” (CSR Limited), and find out
details about the company. Also, type in the ASX code “SFR” (Sandfire Resources NL), and find out the
details about that company. Both these companies belong to material sector. Information available in the
ASX website will be inadequate for your purpose, you will need to search the internet for more
information. Your task will be to get the opening prices of CSR and SFR shares for every quarter from
January 2008 to December 2017 (unadjusted prices). If you are retrieving the monthly prices, read the
values in the beginning of every Quarter (January, April, July, October) for every year from 2008 to 2017
(Total 40 observations). To provide you with some guidance as to what the unadjusted prices look like,
two charts accompany this question obtained from ANZ Share Investing, Australia. After you have
researched share prices and financial sector, answer the following questions:
(a) List all the quarterly opening price values in two tables, one for CSR and the other for SFR. Then
construct a stem-and-leaf display with one stem value in the middle, and CSR leaves on the right
side and SFR leaves on the left side. (Must use EXCEL or similar for the plot.) 1 mark
Answer
Quarter CSR SFR Quarter CSR SFR Quarter CSR SFR
Mar-08 3.220 6.833 Sep-11 4.040 7.046 Mar-15 4.010 6.146
Jun-08 4.100 6.659 Dec-11 4.060 6.206 Jun-15 3.890 7.842
Sep-08 3.450 7.200 Mar-12 4.130 6.043 Sep-15 3.900 5.734
Dec-08 3.293 6.367 Jun-12 3.210 5.900 Dec-15 3.920 7.214
Mar-09 2.264 7.313 Sep-12 4.560 5.171 Mar-16 3.950 5.906
Jun-09 2.170 7.966 Dec-12 4.320 7.714 Jun-16 3.950 5.005
Sep-09 2.090 5.332 Mar-13 4.210 7.226 Sep-16 3.980 6.677
Dec-09 3.450 6.901 Jun-13 4.210 5.720 Dec-16 4.200 7.565
Mar-10 3.260 5.386 Sep-13 4.560 6.863 Mar-17 4.210 5.610
Jun-10 3.900 7.892 Dec-13 4.250 6.491 Jun-17 4.450 5.538
Sep-10 3.860 5.331 Mar-14 4.310 5.666 Sep-17 4.480 7.020
Dec-10 3.950 6.327 Jun-14 4.430 6.049 Dec-17 4.490 7.503
Mar-11 3.970 5.309 Sep-14 4.210 6.681
Jun-11 4.010 5.981 Dec-14 4.150 7.943
CSR Stem-and-Leaf Plot
Frequency Stem & Leaf
7.00 Extremes (=<3.29)
2.00 34 . 55
.00 35 .
.00 36 .
.00 37 .
2.00 38 . 69
8.00 39 . 00255578
4.00 40 . 1146
3.00 41 . 035
6.00 42 . 011115
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2.00 43 . 12
4.00 44 . 3589
2.00 45 . 66
Stem width: .10
Each leaf: 1 case(s)
SFR Stem-and-Leaf Plot
Frequency Stem & Leaf
6.00 5 . 013333
8.00 5 . 56677999
7.00 6 . 0012334
6.00 6 . 666889
6.00 7 . 002223
7.00 7 . 5578899
Stem width: 1.00
Each leaf: 1 case(s)
(b) Construct a relative frequency histogram for CSR and a frequency polygon for SFR on the same
graph with equal class widths, the first class being “$0 to less than $1”. Use two different colours for
CSR and SFR. Graph must be done in EXCEL or similar software. 1 mark
Answer
(c) Draw a bar chart of market capitals (or total assets) in 2017 (in million Australian dollars) of 6
companies listed in ASX that trade in similar products or do similar business as CSR or SFR with at
least AUD100 million in market capital. Graphing must be done in EXCEL or with similar software.
1 mark
Answer
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(d) If one wishes to invest in CSR or SFR, 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 (trend, P/E ratio, dividend yield,
debt and Beta)?
1 mark
Answer
The best information to help decide on either of the two stocks would be obtianing their beta. The beta
will tell how volatile th stock is compared to the market prices. Hence this will help in making the right
decision to either invest or not invest based on how volatile the stock is.
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(Note: Use only the original values of share prices and not adjusted values.)
Question 2 4 Marks
The table below lists the prices of apartments with 2 bedrooms and 2 bathrooms sold between January
and July of 2018 in Australian state capital city centres (Sydney NSW 2000, Melbourne VIC 3000,
Brisbane QLD 4000, Adelaide SA 5000, Perth WA 6000, and Hobart TAS 7000). The data are available
in the website www.realestate.com.au/sold and the prices are given in thousand Australian dollars.
Consider the information as sample data since some selling prices were undisclosed.
Apartments (2 bed, 2 bath) sold between Jan and Jul 2018 in Australian state capitals.
Capital city Prices Sold (in thousand AUD)
Sydney NSW 2000 3680, 1825, 1110, 1190, 1850, 1165, 940, 1399, 1075, 1445, 1900, 2000,
2550, 2350, 1810, 1265, 2300, 1550, 2380
Melbourne VIC 3000
643, 1320, 600, 615, 895, 700, 610, 801,830, 600, 550, 645, 690, 733, 610,
608, 525, 700, 534, 625, 770, 1325, 1142, 650, 585, 680, 770, 675, 650,
580, 570, 1667, 790, 690, 870, 515, 821, 825, 560, 470, 621
Brisbane QLD 4000
750, 452, 502, 542, 620, 490, 455, 600, 625, 527, 820, 650, 550, 498, 355,
997, 498, 450, 420, 1010, 345, 445, 760, 485, 412, 435, 505, 730, 900, 650,
510, 827, 380, 358, 375, 600, 595, 615
Adelaide SA 5000 620, 595, 360, 530, 610, 1415, 485, 691, 458, 578, 820, 600, 725, 615
Perth WA 6000 650, 535, 440, 365, 445, 440, 480, 440, 460, 490, 430, 480, 470, 545, 628,
815, 520, 490
Hobart TAS 7000 615, 893, 637, 730, 810
From the information provided in the table above,
(a) Compute the mean, median, first quartile, and third quartile of sold prices for each city using the
exact position, (n+1)f, where n is the number of observations and f the relevant fraction for the
quartile.
1 mark (Question 2 continued next page)
Answer
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City Mean Median 1st
quartile
3rd
quartile
Sydney NSW
2000
1778.11 1810.00 1190.00 2300.00
Melbourne VIC
3000
733.17 650.00 600.00 780.00
Brisbane QLD
4000
572.05 518.50 450.00 650.00
Adelaide SA
5000
650.14 605.00 530.00 691.00
Perth WA 6000 506.83 480.00 440.00 490.00
Hobart TAS
7000
737.00 730.00 626.00 851.50
(Question 2 continued)
(b) Compute the standard deviation, mean absolute deviation and range for each city. 1 mark
Answer
City Standard
Deviation
MAD Range
Sydney NSW
2000 674.21
511.99 2740.00
Melbourne VIC
3000 240.41
160.72 1197.00
Brisbane QLD
4000 173.08
136.64 665.00
Adelaide SA
5000 247.83
150.06 1055.00
Perth WA 6000 103.29 72.44 450.00
Hobart TAS 7000 116.83 91.60 278.00
(c) Draw a box and whisker plot for the sold prices of each city and put them side by side on one graph
with the same scale so that the sold prices for different cities can be compared. (This graph must be
done in EXCEL or similar software and cannot be hand-drawn.) 1mark
Answer
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(d) Write a paragraph on 2 bed, 2 bath apartment prices of each city, number of apartments sold, and
recent trends in apartment prices in the Australian capital cities after doing some research. 1 mark
Answer
The above results shows that Sydney city has the highest average prices for the 2-bedroom.
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Question 3 4 Marks
The Table below is taken from the Australian Bureau of Statistics (ABS) website
http://www.abs.gov.au/AUSSTATS/abs@.nsf/DetailsPage/6105.0July%202014?OpenDocument in
Employee type Excel sheet Table 6. It provides data on employment type and occupation in
Australia.
Answer parts (a), (b) and (c) of the following based on the table above:
(a) What is the probability that an employee in Australia selected at random will be a
Professional?
1 mark
Answer
P ( Professional ) = x
n =1878.9+684.2
7964+ 3609.8 = 2563.1
11573.8 =0.221 5
(b) What is the probability that an employee in Australia selected at random will be a Male and a
Sales worker? 1 mark
Answer
P ( Male∧Sales )=P ( Male )∗P (Sales)
P ( Male )= 5179.4+1090.4
7964+3609.8 = 6269.8
11573.8 =0.5417
P ( Sales ) = 440.2+ 632.9
7964 +3609.8 = 1073.1
11573.8 =0.0927
P ( Male∧Sales )=P ( Male )∗P ( Sales )=0.5417∗0.0927=0.0502
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(c) Given that a female employee is working part-time, what is the probability that she belongs
to the category of Clerical and administrative workers?
1 mark
Answer
P ( C|F ) = P(C∧F)
P(F )
P ( C∧F ) =P ( C )∗P(F )
P ( C ) =1093.1+568.3
7964+3609.8 = 1661.4
11573.8 =0.1435
P ( F ) =2784.6+2519.4
7964 +3609.8 = 5304
11573.8 =0.4583
P ( C∧F )=P ( C )∗P ( F )=0.1435∗0.4583=0.0658
P ( C|F )= P (C∧F )
P ( F ) = 0.0658
0.4583 =0.1435
(d) Visit the ABS website (Employment type, Table 6) and determine the ratio for total persons
in 2013 between Owner managers of incorporated enterprises to Owner managers of
unincorporated enterprises. 1 mark
Answer
Owner managers of incorporated enterprises = 782.6
Owner managers of unincorporated enterprises = 1156.2
Ratio= 782.6
1156.2 =0.6769
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Question 4 4 Marks
(a) The following data collected from the Australian Bureau of Meteorology Website
(http://www.bom.gov.au/climate/data/?ref=ftr) gives the daily rainfall data (includes all forms of
precipitation such as rain, drizzle, hail and snow) for the year 2017 in Hobart, Tasmania. 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 2 January 2017 – you are expected to visit the website and get the daily values which are
not given in the table below. Part of the 52nd week may run into 2018.):
(i) What is the probability that on any given week in a year there would be no rainfall? 1 mark
Answer
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P ( Norainfall on any given week )=31
52 =0.5962
(ii) What is the probability that there will be 3 or more days of rainfall in a week? 1 mark
Answer
P ( ¿ does not rain∈a day ) =217
365 =0.5945
P ( 3∨more days of rainfall∈ a week )=P(2∨less days of no rainfall∈a week)
P ( 2∨less days of no rainfall∈a week )=P ( 2 days of no rainfall ∈a week ) +P (1 day of no rainfall∈a week )+ P ( 0
P ( 2 days of no rainfall∈a week ) = 7 !
5 !2 ! 0.59452 (1−0.5945)5=0.0814
P ( 1day of no rainfall∈ a week )= 7 !
6 !1! 0.59451 (1−0.5945)6 =0.0185
P ( 0 days of no rainfall ∈a week )= 7 !
7 ! 0 ! 0.59450 (1−0.5945)7=0.0018
P ( 3∨more days of rainfall∈ a week ) =0.0814 +0.0185+0.0018=0.1017
(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 3mm and 9mm of rainfall?
1 mark
Answer
Mean=np=7∗( 1−0.5945 ) =2.8385
Standard deviation= √ npq= √ 7∗0.5945∗0.4055= √ 1.68748825=1.2990
P(3<x<9)
Z−score 1=3−2.8385
1.2990 =0.1243
Z−score 2= 9−2.8385
1.2990 =4.7432
P( Z< 4.7432)=1.0000
P( Z> 0.1243)=0.4505
The required probability is thus;
1.0000−0.4505=0.5495
Thus the probability that in a given week there will be between 3mm and 9mm of rainfall is 0.5495.
(ii) What is the amount of rainfall if only 15% of the weeks have that amount of rainfall or
higher? 1 mark
Answer
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P(X>x) = 0.15
Z= x −2.8385
1.2990 =1.03643
x−2.8385=1.03643∗1.2990=1.3463
x=1.3463+2.8385=4.1848
Thus 4.1848 is the amount of rainfall if only 15% of the weeks have that amount of rainfall or higher.
Question 5 4 Marks
Download Absenteeism at work data from the UCI machine learning data repository
(https://archive.ics.uci.edu/ml/datasets/Absenteeism+at+work). The dataset is about unapproved
absenteeism from work in hours (given in the last column of the table, column U). From the data
provided, answer the questions below.
(a) Test for normality of the following variables using normal probability plot:
Transportation expense, Distance from Residence to Work, Service time, Age, and Body mass
index. 2 marks
Answer
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The above plots clearly shows that the five variables are not normally distributed or rather they
don’t come from a normally distributed population.
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(b) Construct a 90% confidence interval for each of the variables in part (a) which can be considered
as normally distributed separating the data between Absenteeism time in hours for less than 10
and Absenteeism time in hours for 10 or more. In other words, there will be two confidence
interval constructions for each normally distributed variable – one for all data with less than 10
in the last column and the other for 10 or higher value in the last column. After all the
confidence intervals have been constructed, identify the variable(s) which are likely to influence
absenteeism from work (i.e., identify the variables where the confidence intervals do not
overlap). 2 marks
Answer
One-Sample Statistics
N Mean Std. Deviation Std. Error Mean
Transportation expense 740 221.3297 66.95222 2.46121
Distance from Residence to
Work
740 29.6311 14.83679 .54541
Service time 740 12.5541 4.38487 .16119
Age 740 36.4500 6.47877 .23816
Body mass index 740 26.6770 4.28545 .15754
One-Sample Test
Test Value = 0
t df Sig. (2-tailed) Mean
Difference
90% Confidence Interval of the
Difference
Lower Upper
Transportation expense 89.927 739 .000 221.32973 217.2763 225.3831
Distance from Residence
to Work
54.328 739 .000 29.63108 28.7328 30.5293
Service time 77.883 739 .000 12.55405 12.2886 12.8195
Age 153.046 739 .000 36.45000 36.0578 36.8422
Body mass index 169.339 739 .000 26.67703 26.4176 26.9365
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