Statistics for Managerial Decisions Assignment

This document contains solutions to Statistics for Managerial Decisions Assignment. It includes stem-and-leaf plot, frequency polygon, histogram, risk-adjusted returns computation, Airbnb listings analysis, probability calculation, summarised rainfall amount data, normal probability plots, and confidence interval computation.

STATISTICS FOR MANGERIAL DECISIONS
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Question 1
Part (a)
The quarterly opening stock prices for RMD and FPH along with the requisite stem-and-leaf
plot is shown as follows.
January April July October January April July October
2009 2.63 2.53 2.52 2.58 2009 2.72 2.55 2.35 2.63
2010 2.93 3.48 3.61 3.43 2010 2.76 2.54 2.57 2.28
2011 3.41 2.91 2.91 2.85 2011 2.33 2.33 2.13 1.99
2012 2.49 3.00 3.06 3.89 2012 1.86 1.75 1.52 1.83
2013 3.96 4.42 4.90 5.70 2013 1.99 2.11 2.89 3.22
2014 5.31 4.81 5.40 5.63 2014 3.59 3.98 4.43 4.65
2015 6.91 9.42 7.30 7.30 2015 5.85 6.42 6.09 6.47
2016 7.40 7.52 8.41 8.48 2016 8.36 8.93 9.65 9.51
2017 8.59 9.39 10.07 9.76 2017 8.19 8.91 10.89 11.72
2018 10.89 12.47 14.18 15.91 2018 12.83 12.28 13.75 13.71
Quaterly Opening Price of Stocks
Res Med Inc Fisher and Paykel Healthcare
Part (b)
The frequency polygon and histogram as required has been illustrated as follows.
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Part (c)
As per the task, six publicly traded healthcare companies have been selected with market
capitalisation in excess of $ 100 million and the following bar chart is obtained from the
same.
Part (d)
In order to opine as to which stock is better from investment perspective, the risk adjusted
returns have to be computed. Beta is deployed as a measure of risk. Considering that three
year beta has been used, hence for returns computation also, the price appreciation during the
last three year ought to be considered.
Three year returns (RMD) = (16.30-8.30)/8.30 = 96.39%
Three year returns (FPH) = (15.45 -9.56)/9.56 = 61.61%
Also, RMD beta is 0.72 as compared to FPH beta of 1.21.
Hence, over the last three years, RMD has the lower risk and superior returns owing to which
this is the preferred bet.
Question 2
Part (a) “Mean, Median, Quartile First and Second”
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Part (b) “Standard deviation, range and mean absolute deviation”
Part (c) Box-and-whisker plot
Part (d)
The listings have been analysed for the above cities at Airbnb. There is limited availability of
one bed apartment for most of the cities and usually bigger apartments are available for rent.
In terms of the price, for Sydney and Melbourne, the price variation in accordance with the
location was very significant. Additionally, the rent also seems to be driven by the underlying
amenities that are being provided. While a close comparison between the rentals given as
sample data and that found on Airbnb is not possible owing to limited information about
location and amenities, the prices in general for most cities was found to be little cheaper than
the sample data.
Question 3
Part (a) “Probability of selecting Canola field in Australia by Tourist”
P =2538678/ (4623527+2538678+371339+955321+11720277)=0.1256
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Part (b) “Probability of selecting Wheat field in NSW by Tourist”
= 9556517/ ((2755310+1201045+403121+483081+9556517)) =0.6637
Part (c) Probability of selecting Barley in South Australia considering that it does not grow
in WA and Tas
Yield of states of Barley and required probability
Part (d) The crop for which the estimates are unreliable in the data provided is grain sorghum
with the state in question being South Australia. This conclusion has been drawn based on the
additional information in notes to the data where it is mentioned that because of the relative
standard error being higher than 50%, the estimate cannot be relied on.
Question 4
Summarised rainfall amount data for 52 weeks
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Part (b) When rainfall follows Normal Distribution
(i) P (Rainfall would be between 10 mm and 15 mm)
Question 5
(a) The normal probability plots (NPP) for the given data i.e. four variables have been
obtained using Excel.
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The key to interpret the normal probability plot is to consider the trend formed by the plot. If
an approximately linear trend is observed from the plot, then it can be concluded that the
normal distribution would be able to fit the underlying data. For the resting blood pressure,
the plot indicates a linear trend with some mild disruption at the extreme points. Hence, it can
be inferred that the normal distribution would be a suitable choice to summarise the above
variable.
The key to interpret the normal probability plot is to consider the trend formed by the plot. If
an approximately linear trend is observed from the plot, then it can be concluded that the
normal distribution would be able to fit the underlying data. For the serum cholesterol the
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plot indicates a linear trend with some mild disruption at the extreme points. Hence, it can be
inferred that the normal distribution would be a suitable choice to summarise the above
variable.
The key to interpret the normal probability plot is to consider the trend formed by the plot. If
an approximately linear trend is observed from the plot, then it can be concluded that the
normal distribution would be able to fit the underlying data. For the maximum heart rate
achieved the plot indicates a linear trend with some mild disruption at the extreme points.
Hence, it can be inferred that the normal distribution would be a suitable choice to summarise
the above variable.
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The key to interpret the normal probability plot is to consider the trend formed by the plot. If
an approximately linear trend is observed from the plot, then it can be concluded that the
normal distribution would be able to fit the underlying data. For the oldpeak variable the plot
indicates a linear trend with some mild disruption at the extreme points. Hence, it can be
inferred that the normal distribution would be a suitable choice to summarise the above
variable.
(b) The confidence interval for the given four variables have been computed using Excel..
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The objective is to identify any of the four variables listed above to identify if the patient is
suffering from heart disease or not. As a result, the only useful variables would be those
where the confidence intervals for the two categories of patient are separate and display no
overlapping. As a result, for such cases based on the underlying value of the variable, it can
be determined with reasonable precision as to whether the patient has heart disease or not.
There are three variables where no overlapping of intervals is observed. These are listed
below.
 Resting blood pressure
 Maximum heart rate achieved
 Oldpeak
It is noteworthy that serum cholesterol cannot be used for differentiating between patients as
some values such as 250 exist in both intervals and hence conclusion cannot be drawn about
the prevalence of heart disease.
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