Question 1 (a)Quarterly opening price for Res Med Inc and Fisher and Paykel Healthcare Stem and leaf display (b)Graph Relative frequency histogram: RMD Frequency polygon: FPH 2
0 < 22 < 44 < 66 < 86 < 1010 < 1212 < 1414 < 16 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 50.00 Relative Frequency Histogram and Frequency Polygon Res Med Inc Fisher and Paykel Healthcare Stock Price ($) Relative Frequency (c)Bar chart for market capitals as on December 31, 2018 is shown below. (d)The relevant information about the two stocks to facilitate comparison is shown below. P/E Ratio (FPH) = 62.57 Beta (FPH) = 1.21 3
Dividend Yield (FPH) = 1.22% P/E Ratio (RMD) = 52.82 Beta (RMD) = 0.72 Dividend Yield (RMD) = 1.28% Considering the above data and the fact that both the stocks have shown significant gains in the recent times, the preferable stock for investment would be RMD considering that the beta is lower which highlights lower risk and volatility in the stock. Also, the dividend yield is much higher while the P/E multiple is lower than FPH which provides more comfort. Question 2 (a)Mean, median, first and third quartile of weekly rents (b)Standard deviation, mean absolute deviation and range (c)Box and whisker for weekly rents 4
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(d)It is quite evident that there is significant difference in the average advertised weekly rents for 1 bed 1 bath apartment in various Australian cities. As expected highest weekly rents are observed for Sydney while the lowest is observed for Perth. With regards to Melbourne also, the average weekly rent is about $ 35 lower than the average mean rent for apartment in Sydney. The given rates are significantly higher than the corresponding weekly rents available at the current places through Airbnb. In most of the locations such as Sydney and Melbourne, the listings are available are lower rent values then the above mean. Question 3 (a)Probability that the a randomly picked crop field would be canola field in Australia (Hillier, 2016) P(Canolafield)=2538678 (4623527+2538678+371339+955321+11720277)=0.1256 (b)Probability that the a randomly picked crop field would be Wheat field in NSW P(Wheatfield)=9556517 (2755310+1201045+403121+483081+9556517)=0.6637 (c)YieldofBarley(Aust.)=Production Area=12920461 4623527=2.795 YieldofBarley(NSW)=Production Area=2755310 1008269=2.733 YieldofBarley(VIC)=Production Area=3100466 954176=3.249 YieldofBarley(QID)=Production Area=413023 141960=2.909 YieldofBarley(SA)=Production Area=2744507 922787=2.974 5
Probability that barley grown in SA¿2.974 2.733+3.249+2.909+2.974=0.251 (d)The grain sorghum estimates for South Australia (SA) would be considered as unreliable because of the underlying standard error being more than 50%. Question 4 (a)Poisson distribution (i)Probability that on any week there is no rainfall Number of days in a year on which rainfall was incurred = 170 Mean number of days on which rainfall was incurred = 170/52 = 3.3077 No rainfall means X = 0 Now, P(X=0∨λ=3.3077)=3.30770×e−3.30779 0! P(X=0∨λ=3.3077)=0.0366 (ii)Probability that there will be 3 days or more than 3 days of rainfall in a week Rainfall 3 days or more than 3 days means X >= 3 Now, P(X≥3∨λ=3.3077)=1−3.30773×e−3.30779 3!=1−0.5786 P(X≥3∨λ=3.3077)=0.4214 (b)Normal distribution Mean amount of rainfall = 83.75 mm per week Standard deviation of amount of rainfall = 152.74 mm 6
Standarderror=Standarddeviation √(52)=21.18 (i)Probability that in a given week there is rainfall between 10 mm and 50 mm of rainfall (Fick, 2015) Z1=X1−mean Standarderror=10−83.75 21.18=−3.48 P(Z1<−3.48)=0.000249 Z2=X2−mean Standarderror=50−83.75 21.18=−1.59 P(Z2<−1.59)=0.055526 Hence, P(10<X<50)=P(Z2<−1.59)−P(Z1<−3.48) P(10<X<50)=0.055526−0.000249=0.055277 Hence, there is 0.05527probability that in a given week there is rainfall between 10 mm and 50 mm of rainfall. (i)Amount of rainfall when only 12% of weeks have that amount of rainfall or higher P(rainfall>x)=0.12 The z value for 12% can be found by using NORMSINV () excel function(Harmon, 2016). z=NORMSINV(0.12)=−1.17499 Let the required amount of rainfall be X mm and hence, z=X−mean Standarderror −1.17499=X−83.75 21.18 X=58.86mm 7
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Therefore, the amount of rainfall when only 12% of weeks have that amount of rainfall or higher would be 58.86 mm. Question 5 (a)Normality plot of the variables is shown below. 80100120140160180200220 -4 -3 -2 -1 0 1 2 3 4 Normal Probability Plot Resting Blood Pressure Resting Blood Pressure Zj For the above variable, the distribution can be assumed as normal since there is a broad linear pattern that is observed in the above plot(Hillier, 2016). 8
100150200250300350400450500550600 -4 -3 -2 -1 0 1 2 3 4 Normal Probability Plot Serum Cholestoral in mg/dl Serum cholestoral in mg/dl Zj For the above variable, the distribution can be assumed as normal since there is a broad linear pattern that is observed especially if an outlier on the higher end is ignored. 6080100120140160180200220 -4 -3 -2 -1 0 1 2 3 4 Normal Probability Plot Maximum Heart Rate Achieved Maximum heart rate achieved Zj For the above variable, the distribution can be assumed as normal since there is a broad linear pattern that is observed in the above plot. 9
01234567 -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 For the above variable, the distribution can be assumed as normal since there is a broad linear pattern that is observed especially if couple of outliers on the upper and lower end are ignored (Fehr and Grossman, 2013). (b)90% confidence interval for the variables for the patients with heart disease and without heart disease Absence heart diseasePresence heart disease 10
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Significant difference between the patients with and without heart disease would be observed for those variables where there is no overlapping of the 90% confidence interval determined above (Medhi, 2016). These variables are as indicated below. 1)Resting blood pressure 2)Oldpeak 3)Maximum heart rate achieved 12
References Fehr, F. H.and Grossman, G. (2013).An introduction to sets, probability and hypothesis testing.3rd ed. Ohio: Heath. Flick, U. (2015)Introducing research methodology: A beginner's guide to doing a research project.4th ed. New York: Sage Publications. Harmon, M. (2016)Hypothesis Testing in Excel - The Excel Statistical Master.7th ed. Florida: Mark Harmon. Hillier, F. (2016)Introduction to Operations Research.6th ed.New York: McGraw Hill Publications. Medhi, J. (2016)Statistical Methods: An Introductory Text. 4th ed. Sydney: New Age International. 13