STAT2009 Statistics for Managerial Decision Assignment Sample
Question 1
Visit the Australian Stock Exchange website, https://www2.asx.com.au. Type in the ASX code “COL” (Coles Group Limited), and find out details about the company. Also, type in the ASX code “WOW” (Woolworths Group Limited), 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 COL and WOW shares for every quarter from January 2013 to December 2021 (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 2013 to 2021 (Total 36 observations). After you have researched share prices, answer the following questions for University Assignment Help -
(a) List all the quarterly opening price values in two tables, one for COL and the other for WOW. Then construct a stem-and-leaf display with one stem value in the middle, and COL leaves on the right side and WOW leaves on the left side. (Must use EXCEL or similar for the plot.) 1 mark
(b) Construct a relative frequency histogram for COL and a frequency polygon for WOW on the same graph with equal class widths”. Use two different colours for COL and WOW. Graph must be done in EXCEL or similar software. 1 mark
(c) Draw a bar chart of market capitals (or total assets) in 2021 (in million Australian dollars) of 5 companies listed in ASX that trade in similar products or do similar business as COL or WOW with at least AUD500 million in market capital. Graphs must be done in EXCEL or with similar software.
(d) If one wishes to invest in COL or WOW, 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)?
Question 2
The table below lists the retail turnover by industry group between January 2018 and December 2018.
Retail turnover by industry group between Jan 2018 and Dec 2018 (in $ millions)
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From the information provided in the table above
(a) Calculate the mean and standard deviation for each industry group.
(b) Calculate the Minimum, Q1, Median, Q3 and Maximum values for each industry group.
(c) Draw a box and whisker plot for the retail turnover of each industry group 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)
(d) Discuss the retail turnover and the trends of each industry group.
Question 3
The Table below is taken from the Australian Bureau of Statistics (ABS) website. It provides data on households use of the internet from Household Use of Information Technology Survey, Australia.
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From the information provided in the table above,
(a) What is the probability that a randomly selected household lives in Tasmania?
(b) What is the probability that a randomly selected household lives in South Australia and accesses the internet for banking purpose?
(c) Given that a household lives in Western Australia, what is the probability that a randomly selected household accesses the internet for purchasing goods or services? What is the probability that a randomly selected household has accesses the internet for social networking or health services?
Question 4
Melbourne city council found that the average time of a regional train arrival at the Southern Cross station is normally distributed, with a mean of 31 minutes and a standard deviation of 2 minutes.
What value does the upper 5% of the time exceed?
A marketing company is conducting an experiment in which participants are to taste one of two different brands of tea. Their task is to correctly identify the brand tasted. A random sample of 300 participants is taken and it is assumed that the participants have no ability to distinguish between the two brands.
(i) What is the probability that the sample percentage of correct identifications is greater than 1
(ii) What is the probability that the sample will have between 70% and 80% of the identifications correct?
(iii) It is recorded that the length of action movies shown in a local cinema varies uniformly between 112 minutes and 135 minutes. What is the probability that the length of an action movie will be less than 121 minutes?
Question 5
A local university found that sixty four percent of business management students attend their lecture online. If a random sample of 300 students is selected, what is the approximate probability that at least 125 students actually attended their lecture online? Use normal approximation of the binomial distribution.
A shopping centre wants to estimate the mean time that shoppers spend at the shopping centre. Find the 95% confidence interval estimate of the mean given the following random sample (in minutes) of 124, 62, 80, 105, 115, 131, 84, 121, 92, 74.
Tourism Australia found that 172 out of 450 tourists postponed their travel plans after they had confirmed their travel dates. At 0.01 level of significance, is there evidence that the population proportion of tourists who confirmed their travel plans and postponed their travel dates is less than 0.50?
Answers-
Question 1
a) Quarterly opening price values one for Coles Group Limited and other one is for Woolworths Group Limited
Here, it is noteworthy that the company Coles Group Limited (COL) was listed on Nov 22, 2018 and thus, there is no quarterly opening price data before this period.
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The stem and leaf display with STEM in the middle and COL leaves on right side and WOW leaves on the left side
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b) The requisite relative frequency histogram and frequency polygon
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c) The relevant total asset bar chart for five companies is as follows.
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d) In order to recommend the appropriate stock investment between Woolworths and Coles, the following information obtained from Yahoo Finance is useful.
The data above highlights that P/E ratio for Woolworths is 27.57 as compared to 21.90 for Coles. This might indicate that the market expects more growth from Woolworths as compared to Coles. Additionally, the risk associated with Woolworths is also marginally lower than Coles based on the comparison of 5 year beta. Hence, Woolworths seems to be a more favourable investment choice.
Question 2
Mean and standard deviation for the retail turnover by industry groups in ($ millions)
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Formula view
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Minimum, first quartile (Q1), median, third quartile(Q3) and maximum values
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Formula view
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The box and whisker display
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d) From the part (a) output, it is apparent that household goods have the highest average retail turnover for the given period. Further, the respective ranges of retail turnover and the box plot clearly highlight that across the various segments, retail turnover shows a significant variation. Additionally, boxplot confirms presence of outliers in the retail turnover data for almost all categories. The analysis clearly implies that retail turnover varies across the industry groups and also the months.
Question 3
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Question 4
Let the required value is denoted by X.
Given data
Normal distribution
Mean (μ)= 31 minutes
Standard deviation (σ)= 2 minutes
95th percentile is indication of the upper 5% of time therefore, the corresponding z value from normal table comes out to be 1.645.
Formula for z score (X-μ)/(σ )=Z
Putting the values ((X-31)/2)=1.645
X=(1.645*2)+31
X=34.29 min
34.29 minutes is the value that upper 5% of the time exceeds.
(i) Let proportion of participants who have correctly identified the brand is 0.5. This means Probability of success (p) is also 0.50. Also, Total number of trials (N) would be 300 participants.
Binomial distribution
μ= (n*p) = (0.50*300) = 150
σ = √(Np(1-p) )= √(300*0.5*(1-0.5) )=5√3
“Probability that sample percent of correct identification greater than 55 is basically 55% of 300 which is 165.
Now, probability that correct identification of brand more than 165 is calculated using normal distribution.
=P(X>165)
= P((x-μ)/(σ )>(165-150)/(5√3))
=P(Z>1.732)
=1-P(Z<1.732) (From standard normal table)
=0.0416
Probability that sample will have between 70% and 80% of correct identification
Here, 70% indicates X1=70% of sample size = (0.70*300) = 210,
Also, 80% indicatesX2=80% of sample size = (0.80*300) = 240
=P(210<X<240)
= P((210-150)/(5√3)<Z<(240-150)/(5√3))
=P(6.92<Z<10.39)(From standard normal table)
=0
In this case, length of an action movies varies uniformly between 112 minutes and 135 minutes
Probability (Lengthof an action Lower Than 121 minutes)
=(121-112)/(135-112)
= 0.3913
Question 5
64% of Business Management students have attended their lecture online
Random sample (N) = 300 students
Proportion mean (p) = 0.640
Proportion (At least 125Business Managementstudents attended their lectureonline)
p'=125/300 = 0.4167
Normal approximation of binomial distribution
P(p>0.4167)
=P(Z>(p'-p)/√(p(1-p)/N))
=P(Z>((0.4167-0.64)/√((0.64*(1-0.64))/300)))
=P(Z>-8.0576)(From standard normal table)
=1
The 95% confidence interval
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N = 10
Degree of freedom (N-1) = 9
Here, population standard deviation is not given and hence, t critical value would be used to determine the confidence interval.
T(critical) for 95% confidence and 9 degree of freedom = 2.2622 (t table)
LL = Mean-t*((Standard deviation)/√(Sample size ))
LL=98.80-2.2622*(23.70/√10)
LL=81.85
Now,
UL = Mean+t*((Standard deviation)/√(Sample size ))
UL=98.80+2.2622*(23.70/√10)
UL=115.75
95% confidence interval = [81.85 115.75]
It can be concluded with 95% confidence that population mean time in shopping at shopping center will fall between above interval.
Claim to test: Proportion of the tourists who have confirmed their travel plans and then postponed their travel dates is lower than 0.50.
One sample proportion Z hypothesis test
H_0: p≥0.5
H_a: p<0.5
Here total number of tourists who have confirmed their travel plans and then postponed their travel dates= 172
Sample number N = 450
Proportion p'=172/450 = 0.3822
Test statistics
z statistics=(p'-p)/√(p(1-p)/n)
z statistics= (0.3822-0.5)/√(0.5(1-0.5)/450)
z statistics =-4.9978
Level of significance = 1% or 0.01
Left tailed P value = 0.00
Observation: P value <Level of significance
Decision: Reject Ho and accept Ha
Finally, the claim of the Tourism Australia is correct that “Proportion of the tourists who have confirmed their travel plans and then postponed their travel dates is lower than 0.50.”
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