Comprehensive Solution: HA1011 Applied Quantitative Methods Assignment

This document presents a comprehensive solution to an Applied Quantitative Methods assignment, covering various statistical concepts and techniques. The solution includes summary statistics, frequency distribution, histogram construction, and calculations of mean, median, and mode. It addresses the distinction between sample and population data, computes standard deviation and interquartile range, and determines the correlation coefficient between weekly attendance and chocolate bar sales. Furthermore, the solution delves into regression analysis, deriving the least squares regression line and interpreting its coefficients. Probability calculations are performed for training and recruitment scenarios, and binomial and Poisson distributions are applied to customer purchase probabilities. Finally, the document addresses hypothesis testing, including z-test statistics, and concludes with a discussion on investment proportions and standard errors, showcasing a thorough understanding of applied quantitative methods. Desklib offers a wealth of similar solved assignments and resources for students.

APPLIED
QUANTITATIVE
METHODS
HA1011 ASSIGNMENT
Student Name
[Pick the date]

Question 1
(a) Summary statistics
Frequency distribution
(b) Histogram based on frequency distribution table
(c) Mean value (Sorted data from smallest to largest (ascending order))
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Total number of observations = n = 60
Total sum of the given number of observation
Median
N = Even Number
Mode
Maximum occurrence has incurred for the value 401 and therefore, the mode of the data is 401.
Question 2
(a) The decision whether the given data is sample or population is based on the extent of data
made available. In the given scenario, only part of the population data is given since the data
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corresponding to selective 7 weeks is offered and data for all weeks of the year has not been
offered. As a result, this given data is called sample data.
(b) The standard deviation of the sample (s) for ‘ weekly attendance’
(c)The inter quartile range of the sample (IQR) for ‘number of chocolate bars sold’
Sorted data would be required for determination of IQR.
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(d) Dependent variable: Number of chocolate bars sold
Independent variable: Weekly attendance of students
Table to find correlation coefficient
Formula to find correlation coefficient
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Therefore, the correlation coefficient for the dependent and independent variables is 0.968.
Question 3
(a) Regression analysis
Equation of least square regression line
Slope coefficent
Intercept
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Therefore, the least square regression line is y = 1628.689 +10.67 x
Example:
1) Weekly attenance comes out to be zero due to clsoing of Holmes. Here, x = 0.00
The least square regression line
y = 1628.689 +(10.67 * 0) = 1628.689 or 1629 cholcolate bars
2) The increase in the independent variable or student by 10 would imply that the sale of
chocolate bars will increase by 10*10.67 or 1067 units. This is obtained by considering the
interpretation of the regression line slope into consideration.
(b) Formula to find correlation of determination
R^2 = (
Question 4
Scientific training: Recruited from Holmes students = 35
Scientific training: External recruitment = 54
Grassroots training: Recruited from Holmes students = 92
Grassroots training: External recruitment = 12
(a) The requisite probability that a player takes grassroots training and from Holmes =
(35+92+12)/(35+92+54+12) = 139/193 =0.72
(b) The requisite probability that a player takes scientific training and from external =
(54)/(35+92+54+12) = 54/193 =0.28
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(c) The requisite probability that a player takes scientific training and from Holmes
=(35)/(35+92) = 0.276
(d) “Whether events ‘training’ and ‘recruitment’ are classified as independent events or not.
Event 1: Training P (A)
Event 2: Recruitment P (B)
When P(A). P(B) = P (A AND B) then only A and B are classified as independent events.
Here,
Hence, P (A).P(B) = 0.157
Further,
P (A AND B) = 0.257
It is evident from above that condition required for independent events are not satiated and thus,
‘training’ and ‘recruitment’ are not classified as independent events.
Question 5
(a) Probability that customer would choose product x from segment A.
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Probability that customer would choose product x from segment A
(b) Probability that customer would choose product x
Question 6
Assumption
X= Number of customers enter in shop
P = Number of customers who make a purchase from shop
(a) Probability 2 or lower than 2 customers will make a purchase from shop
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In this case, the binomial distribution would be taken into account.
Now,
Therefore,
Therefore, there is 96.19% probability that 2 or lower than 2 customers will make a purchase
from shop.
(b) Probability that in 2 minutes only 9 customers will enter in the shop.
In this case, the Poisson distribution would be taken into account.
Now,
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Here,
Further,
Question 7
Given data and information
(a) Probability that sell would be greater than $2 million
Where,
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