PMBA 6312 Problem Set 4: Statistical Analysis Solutions

This document provides a comprehensive solution to PMBA 6312 Problem Set 4, focusing on statistical analysis and probability. The solution addresses several key questions, including probability calculations related to an Arena Football League (AFL) survey, exploring conditional probabilities, and applying statistical concepts to real-world scenarios. The document further delves into time series analysis, comparing moving average and exponential smoothing techniques. Additionally, the solution includes calculations involving the normal distribution, regression analysis, and interpretation of trends, with a focus on predicting future values and assessing the achievement of specific goals. The document is designed to assist students in understanding and solving complex statistical problems, providing detailed explanations and calculations for each question. References to relevant sources are also included, enhancing the solution's credibility and providing additional avenues for further exploration of the concepts.

PROBLEM SET 1
PMBA 6312. PROBLEM SET 4
Name
Course Number
Date
Faculty Name

PROBLEM SET 2
PMBA 6312. PROBLEM SET 4
Question 17
Arena football league (AFL) games and fans. The survey ad 989 respondents; 759 males and 230
females. 196 had attended multiple AFL games and of them 177 were male.
a) Probability of a randomly selected fan having attended multiple games
probability of attending multiple games= 196
989 =0.198
b) Probability of attending multiples games and being male
probability of a male multiple games attender =177
196 =0.903
c) The probability of having attended multiple games and being male
This is the union of the probabilities in section a and b
Probability of attending multiple games∧being male=0.198∗0.903=0.179
d) Given that a randomly selected fan is a male, what is the probability that he had attended
multiple games
probability of a randomly selected male having attended multiple games= probability of being male∗probabili
759
989∗0.179=0.137
e) The probability of a randomly selected fan is male or has attended multiple games
probability of a selecting a male∨has attended multiple games= 759
989 + 0.198=0.965
Question 23
An oil company purchased an option on land in Alaska. Preliminary studies assigned the following
studies
P ( highquality oil )=0.50
P(medium quality oil)=0.20
P(no oil)=0.30
a) Probability of finding oil
The probability of either high or medium quality oil make-up of the chance of finding oil

PROBLEM SET 3
probability of finding oil= p ( high quality ) + p ( mediumquality )
0.5+0.2=0.7
b) After 200 feet of drilling, soil test was performed and the probabilities of finding oil changed
The soil test shows that the probability of finding high quality oil reduces significantly from 0.5 to
0.2. The revised probabilities are as shown below
P ( highquality oil )=0.20
P(medium quality oil)=0.20
P ( mixed qualities of oil ) 1−0.2−0.2−0.3=0.3
P(no oil)=0.30
Question 25
547,038 of the juniors who took ACT achieved a mean score of 530 with a standard deviation of 123
on the mathematics portion. Assuming normal distribution.
a) Probability of a high school junior who takes the score will score at least 610 on the
mathematics.
Z= 610−530
123 =0.6504
The probability is 1−0.74215=0.2578
b) The probability that a high school junior will score at least 460 marks on the mathematics
portion
Z= 460−530
123 =−0.5691
The probability is 1−0.2843=0.7157 ( Reddy & Rukmangadachari, 2015)
c) The probability of scoring between 460 and 550 marks.
460−530
123 < Z < 550−530
123 =−0.569< Z <0.1626
0.564−0.284=0.28
d) How high does a student must score to be in the top 10% of the class?
0.1= X −530
123

PROBLEM SET 4
X =542.3
A student would have to score at least 542 marks on mathematics.
Question 11
a) Time series plot
1 2 3 4 5 6 7 8 9 10 11 12
77
78
79
80
81
82
83
84
85
86
shipments percentages time series
Upward trends are observed in the months of January to march, May to July and September to
November. On the other months, downward trends are observed.
b) Compare 3 month moving average and exponential smoothing with .2 damping factor
Moving average Exponential
Predicted Errors Predicted Errors
#N/A #N/A #N/A #N/A
#N/A #N/A 80 #N/A
82 #N/A 81.6 #N/A

PROBLEM SET 5
83 #N/A 83.52 #N/A
83.33333 1.170628 83.104 1.828515
83.33333 0.430331 83.0208 1.419063
84 0.720082 83.80416 0.642923
84.33333 0.720082 84.76083 0.894368
83.66667 1.13855 84.15217 0.994607
83 0.981307 82.43043 1.487808
83 1.122167 82.88609 1.358319
83.33333 0.608581 83.77722 1.437246
Moving average has smaller errors for the predicted values, hence the best method.
1 2 3 4 5 6 7 8 9 10 11 12
0
10
20
30
40
50
60
70
80
90
Exponential Smoothing
Actual
Forecast
Data Point
Value
c) Prediction for the next month
1 2 3 4 5 6 7 8 9 10 11 12
0
20
40
60
80
100
Moving Average
Actual
Forecast
Data Point
Value

PROBLEM SET 6
month shipments
1 80
2 82
3 84 82
4 83 83
5 83 83.33333
6 84 83.33333
7 85 84
8 84 84.33333
9 82 83.66667
10 83 83
11 84 83
12 83 83.33333
83.5
83
The predicted value of the next one month is 83.5%
Question 21
a) Time series plot
1 2 3 4 5 6 7 8 9 10 11
15
16
17
18
19
20
21
22
23
24
Percentage of Adult smokers
There is an overall decreasing trend from the first year to the 11th years. However, at year 7, in increase
was observed for a period of 1 year.
b) Simple linear regression

PROBLEM SET 7
1 2 3 4 5 6 7 8 9 10 11
15
16
17
18
19
20
21
22
23
24
f(x) = − 0.336363636363636 x + 22.8090909090909
R² = 0.858953444597816
Percentage of Adult smokers
c) Achieving 12% of adult smokers
A minimum of 12% adult smokers has not been achieved. The linear model is on target to meet the OSH
goal (Ryan, 2013).
12=−0.3364 ( x )+ 22.809
x= 12−22.809
−0.3364 =32.13
It will take approximately 23 years for OSH objective of attaining at least 12% of adult smokers
(Shumway, 2000).
References
Ryan, A. G. (2013). Solutions Manual to Accompany Introduction to Linear Regression Analysis, 5th
Edition. John Wiley & Sons.
Shumway, R. H. (2000). Time series analysis and its applications. Springer.