Business Decision Analysis Assignment

This document is an individual problem assignment for the course BUSI 2013 Business Decision Analysis at Yorkville University. It includes a question about forecasting and regression analysis, with step-by-step solutions and calculations. The assignment is due on April 19th, 2019.

Kwaku Tweneboa Kodua
Yorkville University
BUSI 2013 Business Decision Analysis
Individual Problem 2
Professor: Aniket Mahanti
April 19th, 2019

Question 1
Wee
k
1 2 3 4 5 6
Valu
e
18 13 16 11 17 14
It can be solved using excel table as below.
Wee
k
Valu
e
Forecas
t Error
Forec
ast
Error
Absolut
e value
for
forecas
t error
Squar
ed
Forec
ast
Error
Percent
age
Error
Absolut
e Value
of
Percent
age
Error
1 18
2 13 18 -5 5 25 -38.4615 38.4615
3 16 13 3 3 9 18.7500 18.7500
4 11 16 -5 5 25 -45.4545 45.4545
5 17 11 6 6 36 35.2941 35.2941
6 14 17 -3 3 9 -21.4286 21.4286
Totals -4 22 104
-
51.3005
159.388
8
Alternatively, we can let the F denotes the Forecast for t period, At to denote
the actual value in t period, and Et to denote the forecast error in t period.
Now the Naïve forecast for the value in t weeks be given by
Ft=At-1

F2=A1=18
E2=A2-F1=13-18=-5
F3=A2=13
E3=A3-F2=16-13=3
F4=A3=16
E4=A4-F3=11-16=-5
F5 =A4 =11
E5 =A5-F4=17- 11=6
F6=A5 =17
E6=A6 -F5=14-17=-3
Part a
Mean Absolute Error (MAE) =∑
i=2
6 |Ei|
5 = (5+ 3+5+6+3 )
5 =4.40
Part b
Mean Squared Error (MSE) =∑
i=2
6 ( Ei)2
5 ¿
[52 +32+52+62 +32 ] /5=20.80
Part c
Mean Absolute Percentage Error (MAPE) =100 % /5 ∑
i=2
6
¿ Et
At ∨¿

¿ 100/5 ∑
i=2
6
[|−5
13 |+| 3
16 |+|−5
11 |+| 6
17|+|−3
14 |]=31.88 %
Part d
To find the forecast for week 7, we will use the formula for linear
regression stated below where b will represent the forecast for week 7.
y=a+bx
Where b= n∑ xy−∑ x ∑ y
n ¿ ¿
Wee
k (x)
Valu
e (y) xy x^2 y^2
1 18 18 1 324
2 13 26 4 169
3 16 48 9 256
4 11 44 16 121
5 17 85 25 289
6 14 84 36 196
∑x=2
1
∑y=8
9
∑xy=3
05
∑x^2
=91
∑y^2=1
355
b= 6 ( 305 )−(21)( 89)
6 ( 91 )−(21)2
b=1830− ( 1869 )
546−441 =−39
105 =−0.3714
y= ∑ y
n = 89
6 =14.83

x=∑ x
n =21
6 =3.5
a= y−bx=14.83− (−0.3714 ) ( 3.5 )=16.13
Therefore, the forecast for week 7 is shown by the formula below;
y=16.13−0.3714 x
Question 2.
Wee
k
Valu
e
Foreca
st
foreca
st
error
Absolu
te
Error
of
Foreca
st
Error
Squar
ed
Foreca
st
Error
Percenta
ge Error
Absolute
Value of
Percenta
ge Error
1 18
2 13 18.000 -5.000 5.000 25.000 -38.462 38.462
3 16 15.500 0.500 0.500 0.250 3.125 3.125
4 11 15.667 -4.667 4.667 21.778 -42.424 42.424
5 17 14.667 2.333 2.333 5.444 13.725 13.725
6 14 15.000 -1.000 1.000 1.000 -7.143 7.143
TOTAL
S -7.833 13.500 53.472 -71.178 104.879
a) mean squared error
MSE=(∑ Squared Forecast Error)/n=(53.472/5)=10.694
b) mean absolute percentage error
MAE=(∑Absolute Error of Forecast Error)/n=(13.500/5)=2.700
c)
week 7= (sum of Forecast
Values/n)=(18+13+16+11+17+14)/6= 14.833

Question3
To answer this question, we will consider the calculated forecast
errors in Exercise 1 and 2
Let’s use a table to make a good comparison
Exercise 1 Exercise 2
MAE 4.40 2.700
MSE 20.80 10.694
MAPE 31.88 20.976
Exercise 2 gives a more accurate forecast than method used in
exercise 1. Exercise 2 uses the average of all values.
Question 11.
Part a
A time series plot generated using excel. The graph obtained shows
that the plotted data appear to follow a horizontal pattern.

1 2 3 4 5 6 7 8 9 10 11 12
77
78
79
80
81
82
83
84
85
86
Time Series Plot
Time (Months)
Shipment Percentage
Part b.
The
Mean Squared Error (MSE) =∑
i=3
12 ( Ei)2
9 = 11.11
9 =1.2344
Now, we plot the moving average
Mont
h
shipme
nt
Forec
ast
(MA3)
Foreca
st
Error
Absolu
te
Value
Foreca
st
Error
Squared
Forecast
Error
1 80
2 82
3 84
4 83 82.00 1.00 1.00 1.00
5 83 83.00 0.00 0.00 0.00
6 84 83.33 0.67 0.67 0.44
7 85 83.33 1.67 1.67 2.78
8 84 84.00 0.00 0.00 0.00
9 82 84.33 -2.33 2.33 5.44
10 83 83.67 -0.67 0.67 0.44
11 84 83.00 1.00 1.00 1.00
12 83 83.00 0.00 0.00 0.00
Sum 1.33 7.33 11.11

1 2 3 4 5 6 7 8 9 10 11 12
77
78
79
80
81
82
83
84
85
86
Moving average plot
shipment ForeCast(MA3)
Time (month)
Shipment Percentage
We will now check the exponential Smoothing Forecast.
Let us use this Ft =αA (t−1)+ ( 1−α ) F (t−1) formula to generate a table
Where Ft=Forecast demand for t month
α =Is the exponential smoothing constant
At-1=The previous average percentage
Ft-1=previous period forecast demand
Mont
h
shipme
nt
Forec
ast
Forec
ast
Error
Absolute
focus Error
Squared
Forecas
t error
1 80 0
2 82 80.000 2.000 2.000 4.000
3 84 80.400 3.600 3.600 12.960
4 83 81.120 1.880 1.880 3.534
5 83 81.496 1.504 1.504 2.262
6 84 81.797 2.203 2.203 4.854
7 85 82.237 2.763 2.763 7.632
8 84 82.790 1.210 1.210 1.464
9 82 83.032 -1.032 1.032 1.065
10 83 82.826 0.174 0.174 0.030

11 84 82.860 1.140 1.140 1.299
12 83 83.088 -0.088 0.088 0.008
Sum
15.35
3 17.594 39.108
The Mean squared error for the exponential demand (MSE) =
(39.108/11) =3.555
Plotting the exponential smoothing on a marker plot.
1 2 3 4 5 6 7 8 9 10 11 12
77
78
79
80
81
82
83
84
85
86
Exponential Smmothing gragh
shipment Forecast
Time(Month)
% of Shipment
Part c
To get the forecast for month 13, we would average the moving averages
= (83+84+83)/3=83.33
Therefore, the forecast for 13th month is 83.33%
Question 19.
t 1 2 3 4 5 6 7

Y
t
12
0
11
0
10
0
9
6
9
4
9
2
8
8
Part a
A scatter plot for the above data is as shown.
1 2 3 4 5 6 7
0
20
40
60
80
100
120
140
A scatter plot plot
Yt
t
Yt
The data has a horizontal trend.
Part b
We will apply linear regression analysis technique here.
We will start by creating a table as shown below:
t Yt tYt t^2 (Yt)^2
1 120 120 1 14400
2 110 220 4 48400
3 100 300 9 90000

4 96 384 16 147456
5 94 470 25 220900
6 92 552 36 304704
7 88 616 49 379456
∑t
=
2
8
∑Yt
=7
00
∑tYt
=26
62
∑t^
2=1
40
∑(Yt)^2
=12053
16
b= n∑ ty−∑t ∑ y
n ( ∑ t2 ) −(∑ t)2
¿ 7(2662)−( 28)(700)
7(140)− ( 28 )2
¿ −966
196 =−4.9286
a= ∑ y
n −b ∑ t
n = 700
7 −28
7 b=100−4 b=119.71
Therefore, y=119.71+ ( 4.9286 ¿ t
Part c
The forecast when t=58.
To find this, we will substitute t in the regression equation obtained in
part b, to obtain the absolute values.
Y=119.71+ ( 4.9286 ¿ 58=405.5688
Therefore, the forecast is 405.5688 when t=58
Question 23

Part a
1 2 3 4 5 6 7 8 9 10 11 12 13
0
50
100
150
200
250
300
350
400
450
500
Forecast value, y
Forecast value, y
Time(year)
% of adults
The data obtain has a horizontal trend.
Part b
Yea
r, x
Forecast
value, y xy x^2 y^2
1 41 41 1 1681
2 44.9 89.8 4 2016.01
3 47.1 141.3 9 2218.41
4 45.7 182.8 16 2088.49
5 46.6 233 25 2171.56
6 44.5 267 36 1980.25
7 47.6 333.2 49 2265.76
8 49.8 398.4 64 2480.04
9 48.1 432.9 81 2313.61
10 48.9 489 100 2391.21
11 48.9 537.9 121 2391.21
∑x
=6
6
∑y=472.
1
∑xy=3
146.3
∑x^2
=506
∑y^2=23
997.55

b= n∑ xy −∑ x ∑ y
n ( ∑ x2 )−(∑ x)2
b=11 ¿¿
a= ∑ y
n −b ∑ x
n = 472.1
11 − 66
11 b=42.918−6 b=25.8072
Y=a+bx
Y=25.8072+¿ 2.8518 x
Therefore, the regression equation is Y=25.8072+¿ 2.8518 x
Using the above formula to predict the values of y=12, y=13, y=14 and
y=14 and generate a table
Yea
r, t
Forecast
value
Forecas
t MA(3)
Forecast
Error
Absolute
Value
Forecast error
Squared
forecast
error
1 41
2 44.9
3 47.1
4 45.7 44.33 1.37 1.37 1.87
5 46.6 45.90 0.70 0.70 0.49
6 44.5 46.47 -1.97 1.97 3.87
7 47.6 45.60 2.00 2.00 4.00
8 49.8 46.23 3.57 3.57 12.72
9 48.1 47.30 0.80 0.80 0.64
10 48.9 48.50 0.40 0.40 0.16
11 48.9 48.93 -0.03 0.03 0.00
12 60.0 48.63 11.40 11.40 129.86
13 62.9 52.61 10.27 10.27 105.49
14 65.7 57.27 8.46 8.46 71.62
15 68.6 62.88 5.70 5.70 32.53
Totals 42.67 46.67 363.24
MSE= (363.24)/15= 24.216

Part c
Given x=16
When forecasting for this year, we will use the regression formula
obtained in b and substitute the value of x=16.
Y=25.8072+¿ 2.8518 ( 16 ) =¿71.436
Therefore, the percentage of adults next year when x=16 is 71.436%
Part d
Yes, the trend of equation in part b can equally predict the percentage of
adults participating in a 30-minute exercise.

Reference
Academy, J. (2008). Retrieved from https://www.youtube.com/user/rdjalayer:
https://www.youtube.com/watch?v=gHdYEZA50KE&t=29s
Davis, B. &. (1991). Times Series Thoery and Methods. In Times Series
Thoery and Methods. New York: Springer-Verlag.