This document provides solutions for Applied Quantitative Methods Assignment 2. It includes topics such as frequency distribution, mean, median, mode, standard deviation, correlation coefficient, regression line, correlation of determination, probability, binomial distribution, Poisson distribution, and more.
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APPLIED QUANTITATIVE METHODS ASSIGNMENT 2 Student Name [Pick the date]
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Question 1 (a)A frequency distribution (Total number of passengers at train station: Melbourne) has been obtained from the given data and presented in the tabular manner. (a)Frequency histogram (through excel) has been obtained for the above data and pasted below. (b)The mean, median and mode for raw data Data in Ascending order – Sorted 1
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Mean Sample size (n) = 60 Total of the sample data points = 58594 Mean of data = 58594/60 = 976.57 Median Median=½{(n/2)value+((n/2)+1)value}=½{(30thvalue+31stvalue}=½ (733+862=797.5 Mode Maximum occurred number is 401 and thus, it would be the mode of the data. Question 2 (a)The data that has been given in the question clearly highlights that it corresponds to onlyafewoftheweeks.Thepopulationdatawouldcorrespondtoallthe observations and thereby would have given data for all the 52 weeks present in a year. However, the data provided only provide information about some selective weeks and thereby would be declared as sample (Medhi, 2016). (b)Calculation of standard deviation:weekly attendance Mean of sample (= 3392 / 7= 484.57 Standarddeviationofsample=ට1 N−1σ(x−xത)2=ට1 7−1×(32909.714)=74.06 (c)Calculation of Inter Quartile Range:Number of chocolate bars sold 3
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Data in Ascending order – Sorted First quartile = 25thpercentile = {25*(7+1)/100}thterm = 2ndvalue = 6014 Third quartile = 75thpercentile = {75*(7+1)/100}thterm = 6th value = 7223 Inter quartile range =Third quartile – First quartile = 7223-6014 = 1209 It is noteworthy that both inter-quartile range (IQR) and standard deviation are both indicators of dispersion in the given data. When the data is symmetric and does not have outliers, then standard deviation is considered to be a superior measure. However, in cases where the data is asymmetric or skewed and has outliers present, then standard deviation would not be a suitable choice. In such circumstances, IQR would be a superior measure of dispersion as it is immune to the outliers and does not get impacted by the same (Eriksson and Kovalainen, 2015). (d)Calculation of correlation coefficient (r) has been carried out based on the table indicated below. 4
Interpretation:The correlation coefficient has been derived as 0.968. The magnitude of this correlation coefficient is quite high and almost equal to the maximum value of 1 which would imply that the magnitude of the relationship between weekly attendance and sale of chocolate bar is quite strong. The value of the coefficient is positive which is an indication that higher weekly attendance would potentially leadto higher chocolate bar sales (Hastie, Tibshirani and Friedman,2014). Question 3 (a)Regression line Computation: 5
Regression equation y = 1628.689 +10.67 x Number of chocolate bars sold = 1628.689 + (10.67* Weekly attendance) In order to interpret the above regression equation, there are two key aspects that need to be considered namely the slope coefficient (10.67) and intercept (1628.689). The intercept coefficient represents the chocolate bars sale that would be expected when the the weekly attendance is zero. In this case, it would be 1629. Also, the slope coeffficient highlights the expected change in the dependent variable (chocolate bars sold) produced by a unit change in the independent variable (i.e. weekly attendance). Clearly, a change in the weekly attendance by 1 student would result in change in chocolate bars sale by 10.67 units. The change for the two variables would take place in the same direction (Taylor and Cihon, 2017). (b)Correlationof determination (R2) Rsquare= ( Interpretation: The above value highlights the ability of the given simple regression model to acocunt for 93.7% of the changes that are visible in the chocolate bar sales on a weekly basis (Lind, Marchal and Wathen, 2014). Question 4 Data Given (a)Probability(Grassroots and Holmes) 6
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=(35+92+12)/(35+92+54+12)=0.7202 (b)Probability(Scientific and External) =54/(35+92+54+12)=0.2798 (c)Probability(Scientific and Holmes) =35/(35+92)=0.2756 (d)Training and Recruitment are independent or not Event 1 and 2 will be classified as independent when P(A)* P(B) = P (A AND B) P(A)=Player is external=(54+12)/(35+92+54+12)=0.3412 P(B)=Player is receiving scientific training=(35+54)/(35+92+54+12)=0.4611 P(A). P(P)=0.3412*0.4611=0.1573 P(AANDB)=P(Player is External AND receiving scientific training)=0.2576 P(A)* P(B) = P (A AND B) (not satisfied) and thus, training and Recruitment are not independent events. Question 5 (a)Probability (segment A | Product x) 7
SegmentProbability A0.55 B0.30 C0.10 D0.05 Probability for product X and all the four segments SegmentProbability P(X|A)0.20 P(X|B)0.35 P(X|C)0.60 P(X|D)0.90 Required probability P(A│X)=P(A). P(X|A)/P(X) =( (0.55)*(0.2))/[{(0.55)*(0.2))}+{(0.35)*(0.30)}+{(0.60)*(0.10)}+{(0.90)*(0.05)}] =0.3537 (b)P (Select product X) ={ሺ0.55ሻ∗ሺ0.2ሻ)}+{(0.35)∗(0.30)}+{(0.60)∗(0.10)}+{ሺ0.90ሻ∗(0.05)}=0.320 Question 6 8
(a)Probability that 2 or lower than 2 customers will purchase Binomial distribution Given information (b)Probability that in 2 minutes, 9 customers would enter in the shop. Poisson distribution 9
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Question 7 μ=$1.1 million =$1100000 σ=$385,000 x=$2 million =$2000000 (a)Probability (Apartment sales > $2 million) (b)Probability ($1 million < sales< $1.1 million) 10
P (1000000<x<1100000)=P((1000000-1100000)/385000<(x-μ)/σ<(1100000- 1100000)/385000)=P(-0.26<z<0) From z tableP(z<-0.26)=0.3975 P(z<0)=0.5 P(1000000<x<1100000)=0.5-0.3975=0.1025 Question 8 (a)Inordertodetermineiftheunderlyingdatacanbeconsideredasnormally distributed or not, consideration needs to be given the central limits theorem. This theorem states that if the number of observations is adequately large i.e. (greater than 30), then it would be fair to assume that the underlying sample takes a normal distribution irrespective of the original distribution being normal(Taylor and Cihon, 2017). (b)Assistant called 45 investors and out of which 11 investors show willingness to invest in new fund. Proportion of agreed investors = 11 /45 = 0.24 Standard error for proportion of sample =√(((1-p)p)/n)=√(0.24(1-0.24)/45)=0.064 Probability(p>30%)=P(z>(0.3-0.24)/0.064)=P(z>0.87) P(z>0.87)=0.192 Probability(p>30%)=0.192 11
References Eriksson, P. and Kovalainen, A. (2015)Quantitative methods in business research. 3rd ed. London: Sage Publications. Hastie, T., Tibshirani, R. and Friedman, J. (2014)The Elements of Statistical Learning. 4th ed.New York: Springer Publications. Lind, A.D., Marchal, G.W. and Wathen, A.S. (2012)Statistical Techniques in Business and Economics.15th ed. New York : McGraw-Hill/Irwin. Medhi, J. (2016)Statistical Methods: An Introductory Text. 4th ed. Sydney: New Age International. Taylor, K. J. andCihon, C. (2017)Statistical Techniques for Data Analysis. 2nd ed. Melbourne: CRC Press. 12