Applied Quantitative Methods - Desklib

This article covers topics like summary statistics, graphs, measures of variability and association, linear regression, probability and more. It includes solved assignments, essays, and dissertations on Applied Quantitative Methods. The article also explains the difference between population and sample, standard deviation, interquartile range, correlation coefficient, regression equation, coefficient of determination, and probability.

Applied Quantitative Methods
Name:
Institution:
23rd May 2018
1

HA1011 – Applied Quantitative Methods
Group Assignment
Attempt all the questions (8x2.5 = 20 Marks)
Question 1 of 8
HINT: We cover this in Lecture 1 (Summary Statistics and Graphs)
Data were collected on the number of passengers at each train station in Melbourne. The numbers
for the weekday peak time, 7am to 9:29am, are given below.
456 1189 410 318 648 399 382 248 379 1240 2268 272
267 1113 733 262 682 906 338 1750 530 1584 7729 323
1311 1632 1606 982 878 169 583 548 429 658 344 2630
538 494 1946 268 435 862 866 579 1359 1022 1618 1021
401 1181 1178 637 2830 1000 2958 962 697 401 1442 1115
Tasks:
a. Construct a frequency distribution using 10 classes, stating the Frequency, Relative
Frequency, Cumulative Relative Frequency and Class Midpoint
Solution
Mid-point Frequency Relative
frequency
Cumulative relative
frequency
100-299 199.5 6 0.098 0.098
300-499 399.5 14 0.230 0.328
500-999 749.5 17 0.279 0.607
1000-1999 1499.5 18 0.295 0.902
2000-2999 2499.5 4 0.066 0.967
3000-3999 3499.5 0 0.000 0.967
4000-4999 4499.5 0 0.000 0.967
5000-5999 5499.5 0 0.000 0.967
6000-6999 6499.5 1 0.016 0.984
7000-7999 7499.5 1 0.016 1.000
b. Using (a), construct a histogram. (You can draw it neatly by hand or use Excel)
Solution
2

c. Based upon the raw data (NOT the Frequency Distribution), what is the mean, median and
mode? (Hint – first sort your data. This is usually much easier using Excel.)
Solution
number of passengers
Mean 1033.433
Median 715
Mode 401
Question 2 of 8
HINT: We cover this in Lecture 2 (Measures of Variability and Association)
You are the manager of the supermarket on the ground floor below Holmes. You are wondering if
there is a relation between the number of students attending class at Holmes each day, and the
amount of chocolate bars sold. That is, do you sell more chocolate bars when there are a lot of
Holmes students around, and less when Holmes is quiet? If there is a relationship, you might want to
keep less chocolate bars in stock when Holmes is closed over the upcoming holiday. With the help of
the campus manager, you have compiled the following list covering 7 weeks:
Weekly attendance Number of chocolate bars sold
472 6 916
413 5 884
503 7 223
612 8 158
399 6 014
538 7 209
455 6 214
Tasks:
a. Is above a population or a sample? Explain the difference.
3

Solution
This is a sample since it is covering a portion of the period. A population covers the entire
period.
b. Calculate the standard deviation of the weekly attendance. Show your workings. (Hint –
remember to use the correct formula based upon your answer in (a).)
Solution
Stdev= √var
var=∑ ( xi−x )2
n−1
x
472 -13 158.041
413 -72 5122.469
503 18 339.612
612 127 16238.041
399 -86 7322.469
538 53 2854.612
455 -30 874.469
3392 0 32910
var=∑ ( xi−x ) 2
n−1 =32910
7−1 = 32910
6 =5485
Stdev= √var= √5485=74.0608
c. Calculate the Inter Quartile Range (IQR) of the chocolate bars sold. When is the IQR more
useful than the standard deviation? (Give an example based upon number of chocolate bars
sold.)
Solution
IQR=Q1 −Q3
Q1=2 nd term=413
Q3=6 th term=538
IQR=Q1 −Q3=538−413=125
d. Calculate the correlation coefficient. Using the problem we started with, interpret the
correlation coefficient. (Hint – you are the supermarket manager. What does the correlation
coefficient tell you? What would you do based upon this information?)
Solution
r = n ( ∑ xy )− ( ∑ x )( ∑ y )
√ [ n∑ x2− ( ∑ x )
2
] [ n∑ y2− ( ∑ y ) 2
] = 7∗23425707−3392∗47618
√ ( 7∗1676576−33922 ) ( 7∗327928878−476182 ) =0.968
The correlation coefficient is 0.968; this means that a very strong positive linear relationship
exists between weekly attendance and number of chocolates sold. Based on this
information, as a manager I would try to see on ways of increasing the weekly attendance
through attractive events in order to increase on sales.
Question 3 of 8
4

HINT: We cover this in Lecture 3 (Linear Regression)
(We are using the same data set we used in Question 2)
You are the manager of the supermarket on the ground floor below Holmes. You are wondering if
there is a relation between the number of students attending class at Holmes each day, and the
amount of chocolate bars sold. That is, do you sell more chocolate bars when there are a lot of
Holmes students around, and less when Holmes is quiet? If there is a relationship, you might want to
keep less chocolate bars in stock when Holmes is closed over the upcoming holiday. With the help of
the campus manager, you have compiled the following list covering 7 weeks:
Weekly attendance Number of chocolate bars sold
472 6 916
413 5 884
503 7 223
612 8 158
399 6 014
538 7 209
455 6 214
Tasks:
a. Calculate AND interpret the Regression Equation. You are welcome to use Excel to check
your calculations, but you must first do them by hand. Show your workings.
Solution
a= (∑ y ) (∑ x2 )− (∑ x ) (∑ xy )
n ∑ x2− (∑ x )2
b= n ( ∑ xy )− (∑ x ) (∑ y )
n∑ x2− ( ∑ x )2
x y x^2 xy
472 6916.0
0
222784 3264352
413 5884.0
0
170569 2430092
503 7223.0
0
253009 3633169
612 8158.0
0
374544 4992696
399 6014.0
0
159201 2399586
538 7209.0
0
289444 3878442
455 6214.0
0
207025 2827370
∑ x
=339
2
∑ y=
47618
∑ x2=1
676576
∑ x y=23
425707
5

a= ( ∑ y ) ( ∑ x2 ) − ( ∑ x ) ( ∑ xy )
n ∑ x2− ( ∑ x ) 2 = 47618∗1676576−3392∗23425707
7∗1676576−33922 =1628.689
b= n ( ∑ xy )− (∑ x ) (∑ y )
n∑ x2− ( ∑ x )2 =7∗23425707−3392∗47618
7∗1676576−33922 =10.6772
Weekly attendance affects number of chocolate bar sold. Weekly attendance is the independent
variable while number of chocolate bar sold is the dependent variable. From the above calculations
we observe that the regression model is;
y=1628.689+10.6772 x
The coefficient of the weekly attendance is 10.6772; this means that a unit increase in the weekly
attendance would result to an increase in the number of Holmes sold by 10.6772. If the number of
weekly attendance increases by 10 then number of Holmes sold would increase by 106.772. It also
means that a unit decrease in the weekly attendance would result to a decrease in the number of
Holmes sold by 10.6772. If the number of weekly attendance decreases by 10 then number of
Holmes sold would also decrease by 106.772.
Using Excel We obtained;
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.967993
R Square 0.93701
Adjusted R Square 0.924412
Standard Error 224.5952
Observations 7
ANOVA
df SS MS F
Significanc
e F
Regression 1 3751817 3751817 74.37736 0.000346
Residual 5 252215 50442.99
Total 6 4004032
Coefficient
s
Standard
Error t Stat P-value Lower 95% Upper 95%
Intercept 1628.689 605.9 2.688049 0.0434 71.1734 3186.205
x 10.67723 1.238051 8.624231 0.000346 7.494724 13.85974
b. Calculate AND interpret the Coefficient of Determination.
Solution
6

r = n ( ∑ xy )− ( ∑ x )( ∑ y )
√ [ n∑ x2− ( ∑ x )
2
] [ n∑ y2− ( ∑ y ) 2
] = 7∗23425707−3392∗47618
√ ( 7∗1676576−33922 ) ( 7∗327928878−476182 ) =0.968
Thus the coefficient of determination r2 is 0.937; this means that 93.7% of the variation in
the dependent variable (number of Holmes sold) is explained by the number of weekly
attendances.
Question 4 of 8
HINT: We cover this in Lecture 4 (Probability)
You are the manager of the Holmes Hounds Big Bash League cricket team. Some of your players are
recruited in-house (that is, from the Holmes students) and some are bribed to come over from other
teams. You have 2 coaches. One believes in scientific training in computerised gyms, and the other in
“grassroots” training such as practising at the local park with the neighbourhood kids or swimming
and surfing at Main Beach for 2 hours in the mornings for fitness. The table below was compiled:
Scientific training Grassroots training
Recruited from Holmes
students
35 92
External recruitment 54 12
Tasks (show all your workings):
a. What is the probability that a randomly chosen player will be from Holmes OR receiving
Grassroots training?
Solution
P ( Holmes∨Grassroot training )=P ( Holmes ) +P ( grassroot training ) −P ( Holmes∧Grassroot training )
P ( Holmes ) = 127
193 =0.658
P(Grassroot training)=104
193 =0.5389
P ( Holmes∧Grassroot training )= 92
193 =0.4767
P ( Holmes∨Grassroot training )=0.658+ 0.5389−0.4767=0.7202
b. What is the probability that a randomly selected player will be External AND be in scientific
training?
Solution
P ( External∧Scientific training )= 54
193 =0.2798
c. Given that a player is from Holmes, what is the probability that he is in scientific training?
Solution
P ( Scientific training|Holmes ) = P(Scientific∧Holmes)
P( Holmes)
7

P ( Scientific∧Holmes )= 35
193 =0.18135
P ( Holmes ) = 127
193 =0.658
P ( Scientific training|Holmes )= 0.18135
0.658 =0.2756
d. Is training independent from recruitment? Show your calculations and then explain in your
own words what it means.
Solution
If A and B are independent events, the probability of this event happening
can be calculated as shown below:
P ( A ∩B )=P ( A ) × P(B)
P ( Holmes∧Grassroot training )= 92
193 =0.4767
P ( Holmes ) = 127
193 =0.658
P(Grassroot training)=104
193 =0.5389
If the two events are independent then we would expect;
P ( Holmes∧Grassroot training )=P ( Holmes ) × P(Grassroot training)
P ( Holmes ) × P ( Grassroot training )=0.658∗0.5389=0.354596
P ( Holmes ) × P ( Grassroot training ) ≠ P ( Holmes∧Grassroot training )
Hence from the above, we conclude that training is not independent from recruitment
Question 5 of 8
HINT: We cover this in Lecture 5 (Bayes’ Rule)
A company is considering launching one of 3 new products: product X, Product Y or Product Z, for its
existing market. Prior market research suggest that this market is made up of 4 consumer segments:
segment A, representing 55% of consumers, is primarily interested in the functionality of products;
segment B, representing 30% of consumers, is extremely price sensitive; and segment C representing
10% of consumers is primarily interested in the appearance and style of products. The final 5% of the
customers (segment D) are fashion conscious and only buy products endorsed by celebrities.
To be more certain about which product to launch and how it will be received by each segment,
market research is conducted. It reveals the following new information.
 The probability that a person from segment A prefers Product X is 20%
 The probability that a person from segment B prefers product X is 35%
 The probability that a person from segment C prefers Product X is 60%
 The probability that a person from segment C prefers Product X is 90%
Tasks (show your workings):
A. The company would like to know the probably that a consumer comes from segment A if it is
known that this consumer prefers Product X over Product Y and Product Z.
Solution
8

P ( Segment A| prefers X ¿= P(Segment A∧Prefers X )
P( P refers X )
P ( Segment A∧Prefers X ) =0.2
P ( Prefers X ) = P ( Segment A∧Prefers X )
P(Segment A ) = 0.2
0.55 =0.3636
P ( Segment A| prefers X ¿= P(Segment A∧Prefers X )
P(Prefers X ) = 0.2
0.3636 =0.55
B. Overall, what is the probability that a random consumer’s first preference is product X?
Solution
P ( Prefers X ) = P ( Segment A∧Prefers X )
P(Segment A ) = 0.2
0.55 =0.3636
Question 6 of 8
HINT: We cover this in Lecture 6
You manage a luxury department store in a busy shopping centre. You have extremely high foot
traffic (people coming through your doors), but you are worried about the low rate of conversion
into sales. That is, most people only seem to look, and few actually buy anything.
You determine that only 1 in 10 customers make a purchase. (Hint: The probability that the
customer will buy is 1/10.)
Tasks (show your workings):
A. During a 1 minute period you counted 8 people entering the store. What is the probability
that only 2 or less of those 8 people will buy anything? (Hint: You have to do this by hand,
showing your workings. Use the formula on slide 11 of lecture 6. But you can always check
your calculations with Excel to make sure they are correct.)
Solution
P ( 2∨less ) =P ( x=2 ) +P ( x=1 ) + P(x=0)
P ( x=2 )= (8
2 )0.12 (1−0.1)8−2=0.1488
P ( x=1 ) = ( 8
1 ) 0.11(1−0.1)8−1=0.3826
P ( x=0 ) =(8
0 )0.10 (1−0.1)8−0 =0.4305
P ( 2∨less ) =0.1488+0.3826+0.4305=0.9619
B. (Task A is worth the full 2 marks. But you can earn a bonus point for doing Task B.)
On average you have 4 people entering your store every minute during the quiet 10-11am
slot. You need at least 6 staff members to help that many customers but usually have 7 staff
on roster during that time slot. The 7th staff member rang to let you know he will be 2
minutes late. What is the probability 9 people will enter the store in the next 2 minutes?
(Hint 1: It is a Poisson distribution. Hint 2: What is the average number of customers
entering every 2 minutes? Remember to show all your workings.)
9

Solution
P ( x )= e−μ μx
x ! =¿
μ=4
P ( x=9 ) =e− 4 49
9 ! =0.0132
Question 7 of 8
HINT: We cover this in Lecture 7
You are an investment manager for a hedge fund. There are currently a lot of rumours going around
about the “hot” property market on the Gold Coast, and some of your investors want you to set up a
fund specialising in Surfers Paradise apartments.
You do some research and discover that the average Surfers Paradise apartment currently sells for
$1.1 million. But there are huge price differences between newer apartments and the older ones left
over from the 1980’s boom. This means prices can vary a lot from apartment to apartment. Based on
sales over the last 12 months, you calculate the standard deviation to be $385 000.
There is an apartment up for auction this Saturday, and you decide to attend the auction.
Tasks (show your workings):
A. Assuming a normal distribution, what is the probability that apartment will sell for over $2
million?
Solution
P ( x>2 m )=P (Z > 2000000−1100000
385000 )=P ( Z >2.3377 )=0.0097
B. What is the probability that the apartment will sell for over $1 million but less than $1.1
million?
Solution
P ( 1m< x <1.1 m ) =¿
P (Z < 1100000−1100000
385000 =0 )
P (Z >1000000−1100000
385000 =−0.2597 )
P (−0.2597< Z<0 )=0.10245
Question 8 of 8
HINT: We cover this in Lecture 8
10

You are an investment manager for a hedge fund. There are currently a lot of rumours going around
about the “hot” property market on the Gold Coast, and some of your investors want you to set up a
fund specialising in Surfers Paradise apartments.
Last Saturday you attended an auction to get “a feel” for the local real estate market. You decide it
might be worth further investigating. You ask one of your interns to take a quick sample of 50
properties that have been sold during the last few months. Your previous research indicated an
average price of $1.1 million but the average price of your assistant’s sample was only $950 000.
However, the standard deviation for her research was the same as yours at $385 000.
Tasks (show your workings):
A. Since the apartments on Surfers Paradise are a mix of cheap older and more expensive new
apartments, you know the distribution is NOT normal. Can you still use a Z-distribution to
test your assistant’s research findings against yours? Why, or why not?
Solution
Yes you can still use normal distribution since the sample size used is relatively large and
using the Central limit Theorem the samples will be considered to be approximately
normally distributed hence z-distribution can be used to test.
B. You have over 2 000 investors in your fund. You and your assistant phone 45 of them to ask
if they are willing to invest more than $1 million (each) to the proposed new fund. Only 11
say that they would, but you need at least 30% of your investors to participate to make the
fund profitable. Based on your sample of 45 investors, what is the probability that 30% of
the investors would be willing to commit $1 million or more to the fund?
Solution
P ( p=0.3 )
Z= p−P
σ
σ = √ P(1−P)
n = √ 0.3(1−0.3)
45 =0.0683
Z= p−P
σ =
11
45 −0.3
0.0683 =−0.8134
P ( Z >−0.8134 )=0.7920
11