Assignment 2: Statistical Techniques for Business Environment

This document provides details about Assessment item 2- Assignment 2 for Statistical Techniques for Business Environment. It includes information about the assessment criteria, format, and plagiarism policy. The assignment includes questions related to Australian Stock Exchange, stem-and-leaf display, frequency histogram, bar chart, mean, standard deviation, and box and whisker plot. The output is in JSON format.

Assessment details for all students
Assessment item 2—Assignment 2
Due date: 6:00pm, Wednesday Week 10 ASSESSMENT
Weighting: 20%
2Format: Submit one file online as .doc, .docx, .rtf or .pdf
Assessment criteria
 This assignment must be typed, word-processed or clearly hand-written (but plots and graphs
must be done using EXCEL or equivalent software),and submitted online as a single file
through Moodle.Important note:The file size must not be over 100MB.
 MicrosoftExcel allows students to cut and paste information easily into Microsoft Word
documents. Word also allows the use of Microsoft Equation Editor to produce all necessary
formulae (use of these are recommended).
 It is expected that Excel would be used to assist in statistical calculations for questions in this
assignment.Where Excel is used, use copy function, “Snipping tool” or similar to cut and paste
relevant parts of the spreadsheet to verify that you have done the work. (In that case there is no
need to write the equations.)
 For those questions where Excelis not used to do the computations, all formulae and working
must be included to obtain full marks.
 Only one file will be accepted in any of the formats mentioned above. No zipped file or any other
file extension will be accepted.
 There will be late submission penalty for submissions beyond the deadline unless prior approval
is obtained from the Unit Coordinator through the extension system in Moodle. Under no
circumstances any submission that is late beyond 14 days from the deadline of Friday of Week 10
will be marked, or get any score other than zero.
Assignment markers will be looking for answers which
 demonstrate the student’s ability to interpret and apply the statistical techniques in the
scenarios and
 use statistical techniques as decision making tools in the business environment.
Full marks will not be awarded to answers which simply demonstrate statistical procedures without
comment, interpretation or discussion (as directed in the questions).
Plagiarism
CQU values academic honesty. Consequently, plagiarism will not be tolerated in assessment items. This
assignment must be completed by each student individually.
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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 theASX code “NAB” (National Australia Bank), and
find out details about the company. Also, type in the ASX code “WBC” (Westpac Banking Corporation),
and find out the details about that company. 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 NAB and WBCshares for every quarter from January 2009 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 2009 to 2017 (Total 36 observations). After
you have researched share prices, answer the following questions:
(a) List all the quarterly opening price values in two tables, one for NAB and the other for WBC. Then
construct a stem-and-leaf display with one stem value in the middle, and NABleaves on the right side
and WBC leaves on the left side. (Must use EXCEL or similar for the plot
NAB Quarterly Opening Price WBC Quarterly Opening Price
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Quarter WBC Open
Oct-17 32.13
Jul-17 30.56
Apr-17 35
Jan-17 32.71
Oct-16 29.78
Jul-16 29.41
Apr-16 29.85
Jan-16 33.4
Oct-15 29.825
Jul-15 32.083
Apr-15 38.895
Jan-15 32.819
Oct-14 31.456
Jul-14 33.694
Apr-14 34.36
Jan-14 32.272
Oct-13 32.614
Jul-13 28.459
Apr-13 30.539
Jan-13 25.823
Oct-12 24.597
Jul-12 21.167
Apr-12 21.839
Jan-12 19.971
Oct-11 19.654
Jul-11 21.958
Apr-11 24.083
Jan-11 22.146
Oct-10 23.154
Jul-10 20.732
Apr-10 27.494
Jan-10 25.062
Oct-09 25.902
Jul-09 19.634
Apr-09 18.843
Jan-09 16.698
Stem and Leaf Display
9 9 9 8 6 1 9 9
9 9 8 7 5 5 4 4 3 2 1 1 1 0 2 0 1 1 2 2 2 3 3 4 4 4 5 5 5 6 7 8 8 9 9 9
8 5 4 3 3 2 2 2 2 2 2 1 0 0 3 0 0 1 1 1 2 2 3 3 3 6
WBC Leaf NAB Leaf
Stem
Stem and Leaf Display shows the price of both the banks are more concentrated on 20 to 30 range
basket.
(b) Construct a relative frequency histogram for NAB and a frequency polygon for WBC on the same
graph with equal class widths, the first class being “$14 to less than $16”. Use two different colours
for NAB and WBC. Graph must be done in EXCEL or similar software.
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Quarter NAB Open
Oct-17 31.54
Jul-17 29.61
Apr-17 33.17
Jan-17 30.7
Oct-16 28
Jul-16 25.63
Apr-16 25.9
Jan-16 29.01
Oct-15 28.955
Jul-15 32.252
Apr-15 36.522
Jan-15 31.836
Oct-14 30.552
Jul-14 31.189
Apr-14 33.699
Jan-14 33.129
Oct-13 32.701
Jul-13 27.929
Apr-13 29.336
Jan-13 23.784
Oct-12 24.24
Jul-12 22.539
Apr-12 23.527
Jan-12 22.415
Oct-11 20.78
Jul-11 24.288
Apr-11 24.554
Jan-11 22.814
Oct-10 24.269
Jul-10 21.883
Apr-10 26.047
Jan-10 25.951
Oct-09 29.088
Jul-09 21.008
Apr-09 19.126
Jan-09 19.915

14-16 16-18 18-20 20-22 22-24 24-26 26-28 28-30 30-32 32-34 36-38
0
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9
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1
NAB Frequency and WBC Frequency Polygon Distribution
(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 NAB or WBC with
at least AUD50 million in market capital. Graphs must be done in EXCEL or with similar software.
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Company Ticker MarketCap(inBillions)
COMMONWEALTH BANK OF AUSTRALIA. CBA 129.63
AUSTRALIA AND NEW ZEALAND BANKING GROUP LIMITED ANZ 74.8
BENDIGO AND ADELAIDE BANK LIMITED BEN 5.41
CYBG PLC CYB 4.84
BANK OF QUEENSLAND LIMITED. BOQ 4.17
AUSTRALIAN FINANCE GROUP LTD AFG 0.27

CBA ANZ BEN CYB BOQ AFG
0
20
40
60
80
100
120
140 129.63
74.8
5.41 4.84 4.17 0.27
Australian Other banking Companies MarketCap
MarketCap(inBillions)
(d) If one wishes to invest in NAB or WBC, 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)?
Description NBA WBC
P/E Ratio 12.3 11.01
Divident Yield 7.98% 7.19%
Total Debt to Total Equity 3.36 2.82
Beta 1.22 1.46
Price $24.91 $26.2
As we can see that both the banks details are quite good with low P/E ratio and high dividend yield
but the beta for WBC is more volatile than NBA therefore it is better to choose NBA to invest in.
Question 2 4 Marks
The table below lists the new motor vehicles sales between January 2016 and December 2016 in 8
Australian states.
The complete data are available on the website
http://www.abs.gov.au/AUSSTATS/abs@.nsf/DetailsPage/9314.0December%202017?OpenDocument.
New motor vehicles sales by State between Jan 2016 and Dec 2017
Month State
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NSW VIC QLD SA WA TAS NT ACT
Jan-2016 28392 22426 17193 5208 7861 1298 626 1509
Feb-2016 32633 26782 18711 5988 8725 1319 851 1528
Mar-2016 35963 27716 21470 6087 9054 1515 1023 1550
Apr-2016 29391 23963 17753 5174 7601 1313 969 1563
May-2016 32867 26125 19565 5949 7935 1503 1180 1560
Jun-2016 43449 34262 27270 7613 10749 1853 1240 1542
Jul-2016 30219 25517 18445 5210 8234 1389 831 1524
Aug-2016 31788 27096 18062 5785 8144 1743 881 1519
Sep-2016 34424 28766 19702 6551 8668 2074 880 1531
Oct-2016 31850 27546 16870 5450 7580 1937 713 1552
Nov-2016 32955 28991 18733 6259 7657 1903 808 1570
Dec-2016 33632 27065 19252 6464 8026 1918 734 1573
From the information provided in the table above,
(a) Calculate the mean and standard deviation for each state.
(b) Calculate the Minimum, Q1, Median, Q3 and Maximum values for each state.
NSW VIC QLD SA WA TAS NT ACT
Mean
33130.2
5
27187.9
2
19418.8
3
5978.16
7
8352.83
3
1647.08
3
893.541
7
1543.41
7
Media
n 32750 27080.5 18722 5968.5 8085 1629 880.5 1546
Min 28392 22426 16870 5174 7580 1298 626 1509
Max 43449 34262 27270 7613 10749 2074 1240 1573
Q1
31395.7
5 25973
17984.7
5 5390 7810 1371.5 847.5 1527
Q3 33830 27978.5
19599.2
5 6310.25 8682.25 1906.75 936 1560.75
STD
3874.42
6
2913.56
1
2762.80
7
705.478
9
889.271
1
286.471
1
129.165
2
21.1249
9
(c) Draw a box and whisker plot for the number of new vehicles sales of each state and put them side by
side on one graph with the same scale for comparison.(This graph must be done in EXCEL or
similar software and cannot be hand-drawn.)
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(d) Discuss the number of new vehicles sales of each state, and the trends in new vehicles sales in
different Australian states.
Time series analysis of sales of vehicle by state shows us that there was a huge demand for vehicles
in the Q2 of 2016 which trades back to normal ranges and are still in the nominal level.
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01/01/2016
01/02/2016
01/03/2016
01/04/2016
01/05/2016
01/06/2016
01/07/2016
01/08/2016
01/09/2016
01/10/2016
01/11/2016
01/12/2016
01/01/2017
01/02/2017
01/03/2017
01/04/2017
01/05/2017
01/06/2017
01/07/2017
01/08/2017
01/09/2017
01/10/2017
01/11/2017
01/12/2017
0
5000
10000
15000
20000
25000
30000
35000
40000
45000
50000
NSW
VIC
QLD
SA
WA
TAS
NT
ACT
Question 3 4 Marks
The Table below is taken from the Australian Bureau of Statistics (ABS) website.It provides data on
Australian households with internet access at home from the Household Use of Information Technology
Survey.
From the information provided in the table above,
(a) What is the probability that a randomly selected household lives in Victoria?
probability that a randomly selected household lives in Victoria = TOTAL HH in VIC/Total HH
= 0.251
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(b) What is the probability that a randomly selected household lives in Tasmania and has an internet
access in the year 2010-11?
P(Tasmania and Internet access in 2010-11) = 0.021744
(c) Given that a household has an internet access in 2012-13, what is the probability that a randomly
selected household lives in New South Wales?
Prob that a randomanly selected HH in New south wales 12-13 = 0.309742
(d) What is the probability that a randomly selected household has an internet access in 2010-11 or
2012-13?
probability that a randomly selected household has an internet access in 2010-11 or 2012-13 =
= 0.000149 + 0.000136 = 0.000285
Question 4 4 Marks
(a) In a study conducted by a bus transportation company, the average time of bus arrival at the central
station is normally distributed, with a mean of 73 minutes and a standard deviation of 8 minutes.
What value does the upper 5% of the time exceed?
Z = X-mean/std
= 75-73/8 = 0.25
Prob. 5% UPPER = 0.0483
(b) The Bureau of Meteorology found that the air temperature varies uniformly between 12 degree
Celsius and 25 degree Celsius. What is the probability that the air temperature will be less than 21
degree Celsius?
P(Temp < 21) = d-c/b-a = 21-12/25-12 = 0.69
(c) A marketing company is conducting an experiment in which participants are to taste one of two
different brands of coffee. Their task is to correctly identify the brand tasted. A random sample of
200 participants is taken and it is assumed that the participants have no ability to distinguish
between the two brands.
N = 200
Prob of correct identification p = 0.5
q = 1-p = 0.5
(i) What is the probability that the sample percentage of correct identifications is greater than 65?
z = (p’ - p)/(√(pq/n) = (0.65 - 0.5)/(√(0.5 * 0.5/200) = 4.2426
P(p’ > 0.65) = P(z > 4.2426) = 0
(ii) What is the probability that the sample will have between 50% and 60% of the identifications
correct?
z = (p’ - p)/√(pq/n) = (0.5 - 0.5)/√(0.5 * 0.5/200) = 0 and x = (0.6 - 0.5)/(√(0.5 * 0.5/200) = 2.8284
P(0.5 < p’ < 0.6) = P(0 < z < 2.8284) = 0.4977
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Question 5 4 Marks
(a) A survey conducted by a pharmaceutical company found that the probability of success in the
clinical trial is 0.70. Suppose five hundred clinical trials are performed in the study, what is the
probability of succeeding in the clinical trial between 280 and 355 times? Use normal
approximation of the binomial distribution.
Prob of successful clinical trial = 0.7
Prob of unsuccessful clinical trial = 0.3
P(280<X<355) = P(X=280)+...+P(X=355) = 0.702
(b) A travel agency acts as a selling agent for a major airline company. The agency found that 210 out
of 500 travellers cancelled their airline bookings after they had selected their flight details. At 0.01
level of significance, is there evidence that the population proportion of travellers who select their
flights and cancel their bookings is less than 0.50?
Prob of cancelled bookings = 210/500 = 0.42
(c) An internet service provider wants to estimate the mean amount of time that customers spend
browsing web sites on daily basis. Find the 95% confidence interval estimate of the mean given the
following random sample (in minutes) of 110, 123, 80, 131, 105, 65, 84, 118, 95.
Mean = 110+123+80+131+105+65+84+118+95/9 = 101.22
STD = 21.88
Z = X-mean/std = 110-101.22/21.88 = 0.40
95% confidence interval = 101.22 ± 14.3
= 86.9 to 116
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