Statistics for Managerial Decisions Assignment - Analysis & Reports

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Homework Assignment
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This assignment provides a comprehensive analysis of statistical concepts applied to managerial decisions. It includes analysis of stock data, rental prices, and agricultural production, using techniques such as stem-leaf displays, histograms, box plots, and probability calculations. The solution covers topics like P/E ratios, dividend income, volatility, and the application of Poisson and normal distributions. It also addresses confidence intervals and normality probability plots to assess data characteristics and make informed decisions. The assignment references key statistical methods and tools, providing a practical application of statistical principles to real-world scenarios. The document is well-structured, presenting the data, calculations, and interpretations in a clear and concise manner, demonstrating the application of statistical tools for managerial decision-making, with references to relevant literature to support the analysis.
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STATISTICS FOR MANAGERIAL DECISIONS
STUDENT ID:
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Question 1
(a) All the quarterly opening prices of two stocks for the given period are shown below
(Flick, 2015).
Table 1 RMD (ResMed Inc)
Table 2 FPH (Fisher and Paykel Healthcare)
Stem-leaf display
Right leaf: RMD and Left leaf: FPH
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(a) Graphical display of histogram and polygon
(b) The requisite graph indicating the market capitalisation of six ASX listed companies with
market cap in excess of AUD 100 million is shown as follows.
Relevant date for the above is December 31, 2018.
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(c) As no analyst recommendation on the two stocks could be found, thus, key parameters
regarding the two stocks have been obtained from Yahoo Finance and summarised in a
tabular format.
Considering the above information, RMD stock seems more suitable for investment that FPH.
This is because the P/E ratio associated with RMD is significantly lower than FPH which
provides comfort in valuation. Also, RMD has a higher dividend income besides having
lower volatility as captured by the beta.
Question 2
(a) Requisite numerical summary of weekly rents of one bedroom apartment in given
location (Flick, 2015)
(a) Requisite numerical summary of weekly rents of one-bedroom apartment in given
location
(b) Box & Whisker Plot
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(c) The box plot along with the relevant descriptive statistics clearly highlight that the rent
demanded in various cities show significant difference. This can be best illustrated by
comparing the mean rent across Perth and Sydney as the latter is about twice the former.
The rent for apartments seem particularly high in case of Sydney which enjoys on average
a premium in excess of $ 250 over the next highest mean rent city of Melbourne. The
search of similar apartments on Airbnb for the above cities clearly hints hat the above
averages seem to be skewed on the highest end especially for Sydney (Hillier, 2016).
Question 3
(a) P (Picked canola field crop in Australia)
=2538678/ (4623527+2538678+371339+955321+11720277)=0.1256 or 12.56%
(b) P (Picked wheat field crop in NSW)
= 9556517/ ((2755310+1201045+403121+483081+9556517))=0.6637 or 66.37%
(c) Formula for yield
Yield = Production/Area
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(Australia) (NSW) (VIC) (QID) (SA)
¿ 12920461
4623527
¿ 2.795
¿ 2755310
1008269
¿ 2.733
¿ 3100466
954176
¿ 3.249
¿ 413023
141960 =2.909 ¿ 2744507
922787 =2.974
(d) The estimate of grain sorghum production in the state of South Australia is most
unrealistic as the standard error associated with this data is in excess of 50% as per the
ABS data from the given link.
Question 4
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(a) Rainfall distribution: “Poisson distribution”
(i) Probability of no rainfall in week
Days of rainfall = 170 and Mean λ = 170
52 = 3.3077
P(X = 0 | λ= 3.3077) = 3.30770 × e3.30779
0! = 0.0366
(ii) Probability that rainfall experienced for 3 days orgreater than 3 days in week
(b) Rainfall distribution: “Normal distribution”
Standard error = Standard deviation
52 = 21.18
(i) Probability that amount of rainfall would fall between 10 mm and 50 mm (Harmon,
2015)
(i) Assuming that the expected rainfall amount will be X millimetre.
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Question 5
(a) The required normality probability plot
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0 1 2 3 4 5 6 7
-4
-3
-2
-1
0
1
2
3
4
Normal Probability Plot
Old Peak= ST Depression Induced by Exercise Relative
to Rest
Old peak =ST Depression induced by exercise relative to rest
Zj
The given variable can be assumed to be normal distribution is a linear trend is exhibited in
the respective normal plot. In the given case, there are four variables whose normal
probability plots have been provided above. For all these, the broad trend in the plot is
observed as linear if the few outliers are ignored (Fehr & Grossman, 2016).
(b) 90% confidence interval
Comment: This variable does differ for the two sets of patients as the respective confidence
interval does not overlap as is apparent from the above output.
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Comment: This variable does not differ for the two sets of patients as the respective
confidence intervals do overlap as is apparent from the above output.
Comment: This variable does differ for the two sets of patients as the respective confidence
interval does not overlap as is apparent from the above output.
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