A STATISTICAL ANALYSIS
AND REVENUE MAXIMIZITON
IN A BEVERAGE FIRM
MURAT ÇAL
TABLE OF CONTENTS
A.
INTRODUCTION ................................................................................................................................ 3
B.
PREVIOUS WORK ............................................................................................................................ 3
C.
ANALYZING THE FACTORS ......................................................................................................... 4
D.
MAXIMIZING THE EXPECTED REVENUE ON THE NEXT MONTH ................................ 14
E.
CONCLUSION ................................................................................................................................. 15
BIBLIOGRAPHY ..................................................................................................................................... 15
A. INTRODUCTION
In recent years, beverage industry increasingly takes a large portion of the whole food
and beverage industry. Beverage production and sales are especially important for remaining
competitive.
As we begin with our survey in the way leading to the profitability of a cherry juice firm,
we need to analyze the sales in the light of four main factors: Seasonality, price of the juice firm,
market price of juice and apricot juice market price. In the second part, we determine the
optimal price in order to maximize our profit. Since the factor X, which describes the market
price of juice, has a probability mass function, we need to pursue a way of finding the expected
value of market price and then for each value that can presumably offered by the department, we
are to compute our expected profits, among which we will then determine our optimal price. Our
assumptions and system judgments within this study are given as follows:
-We create our own data which is reasonable based on a beverage firm’s way of business.
-We use Minitab for our analyses.
B. PREVIOUS WORK
In lots of statistical studies Minitab is used as an analysis tool. A study methodologically
similar to our study was performed by Leishman (2001), where four key mechanisms of the seed
size/number trade-off models to assess their relevance to a general understanding of plant
community structure are examined. Mechanism 1 is about size of the seeds, Mechanism 2
measures the trade-off between number and size, Mechanism 3 is about competition of seedlings,
whereas Mechanism 4 looks for invasion potential of smaller seeds against big ones. No
evidence is found that these mechanisms are not of significant importance.
Baker and Badar (2015) perform a study to determine whether brands and price
parameters affect warranty failure rates significantly. Four years and one million individual sets
of data over 150,000 warranty failures are used in the study. It is indicated that different brands
in the same price range have the similar failure rates, whereas different price ranges within the
same brand have different failure rates.
McLaughlin and Lesser’s study (1987) emphasizes the importance of statistical inference
in potential sales in that it statistically determines the factors in determining demand elasticities
with the help of store scanning, which enables producers to make efficient decision in market
place selection. Another study is made by Huntington (1993) to determine the ways to optimize
box office revenue, where it compares the strategy of selling all the tickets at a single price to the
strategy of selling different seats on different price scales. To the opposite of what has been
expected, more than half of the theaters operate at a single price and do not take visibility of
scene into consideration.
Another trending industry is cigarette, on which Glantz (1993) works to assess tobacco
consumption, pricing and industry revenues depending on California’s Proposition 99 tax
increase on tobacco. This tax increase reduced the rate by 802 million packs, meaning a tripling
rate of falling. Smoking significantly has stopped in California, whereas cigarette producers have
continued to generate more revenues due to price increase.
C. ANALYZING THE FACTORS
To analyze the effects on our linear model resulted from the four factors entitled by the
firm, we need to construct dummy variables for seasonality and give them binary values to
describe their absence or existence. Our factors, each having the pilot X would then be listed as
follows:
Market
Price
X1
Price
of Apricot
Cherry Juice Price
X2
X3
Spring
Summer
Autumn
X4
X5
X6
For winter, we give each X4, X5 and X6 the value 0 to make sure of its existence. To
apply the analysis, first we let our software Minitab use all the factors and determine an initial
equation:
Tableau 1
Regression Analysis: total bottle versus market price, cherry juice, ...
The regression equation is
total bottles sold = - 34867 + 11538 market price - 8347 juice firm price
+ 170 apricot market price - 70 spring - 55 summer - 25 autumn
Predictor
Constant
Coef
-34867
SE Coef
3901
T
-8.94
P
0.000
market price
11537.7
205.0
56.27 0.000
juice firm price -8346.6
252.9 -33.00 0.000
apricot market price
169.7
483.3
0.35 0.728
spring
-69.6
592.2
-0.12 0.907
summer
-55.3
616.3
-0.09 0.929
autumn
-25.4
632.2
-0.04 0.968
S = 1188.79
R-Sq = 99.5%
R-Sq(adj) = 99.4%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
6 8263146690 1377191115 974.51 0.000
Residual Error 28
39570015
1413215
Total
34 830271670
The p value in the ANOVA table in tableau 1 refers to the significance test of regression.
If it is too small, we reject our hypothesis:
H0: The regression is insignificant, β1=β2=β3=β4=β5=β6=0
HA: H0 is false
Although we can say, by looking at the p value which is equal to zero, that we can reject
the null hypothesis and conclude for significance, there are insignificant factors. We determine
these factors by looking at the p values listed in the above table. The hypothesis for each separate
factor is:
H0: β1= 0
HA: H0 is false
This hypothesis is for X1 and β1 is replaced with β2 for X2 and so on. If the p value for
each case is again too small, say, smaller than some given value α, or zero, then we reject H0.
The values corresponding to apricot market price, spring, summer and autumn are 0.728, 0.907,
0.929, 0.968, respectively. Therefore, we reject each hypothesis corresponding to β3, β4, β5 and
β6. Now, in order to eliminate the insignificant variables and retrieve the best equation fitting to
our model we employ the technique Best Subsets Regression which gives us the following table.
Note also that there are several methods like backward elimination or forward selection, among
which we wanted to choose the subsets based one in order to see some combinations and the
values corresponding to them. We will now handle the issue of selecting the best fitted equation
in relation with what these values represent.
Tableau 2
Best Subsets Regression: total bottle versus market price, cherry juice, ...
Response is total bottles sold
m d
a
Vars
1
1
2
2
3
3
4
4
5
5
6
R-Sq
72.5
31.9
99.5
77.7
99.5
99.5
99.5
99.5
99.5
99.5
99.5
R-Sq(adj)
71.6
29.8
99.5
76.3
99.5
99.5
99.5
99.5
99.4
99.4
99.4
Mallows Cp
1585.6
3972.6
-0.8
1280.2
1.0
1.2
3.0
3.0
5.0
5.0
7.0
S
8320.5
13094
1115.6
7603.7
1130.1
1132.9
1148.6
1148.8
1168.1
1168.3
1188.8
r
k
e
t
r
a
k
i
b
e
e
r
p
r
i
c
e
X
p
r
i
c
e
p
r
i
c
e
X
X
X
X
X
X
X
X
X
s
p
r
i
n
g
s
u
m
m
e
r
a
u
t
u
m
n
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X X
X
X
X X X
As we infer from the table, the best equation involves the factors juice firm price and
market price of juice. The combination is underlined and made bold in the table. Yet here, the
question is how and in what sense we select our factors to be involved in our equation. Our first
criteria would be Mallows’ Cp, which is, as also stated in Minitab, the statistic used as an aid in
choosing between competing multiple regression models. Mallows’ Cp compares the precision
and bias of the full model to models with the best subsets of predictors. We want our Cp value to
be as close as much to the k+1, which is the number of factors in the regression model plus one.
By this approach, we would draw the conclusion that we are to include all the six factors in our
model and choose the sixth combination: k is equal to 6, 6+1=7=7=Cp. So they are in maximum
closure. Yet, the factors apricot price, spring, summer and autumn are not included in the model
and we take the underlined, second combination, because after the tests; stepwise, forward
selection and backward elimination, whose results are shown in tableau 3, we see that these
factors are obviously insignificant. We decide this by looking their p values and their absence in
further steps of stepwise regression.
Tableau 3
Stepwise Regression: total bottle versus
market price, cherry juice, ...
Alpha-to-Enter: 0.15
Remove: 0.15
Alpha-to-
Response is total bottles sold on 6
predictors, with N = 35
Step
Constant
market price
T-Value
P-Value
1
-92577
2
-33979
4893
9.32
0.000
11564
67.19
0.000
juice firm price
T-Value
P-Value
Forward selection.
0.25
Response is total bottles sold on 6
predictors, with N = 35
Step
1
2
Constant
-92577 -33979
-8386
-42.47
0.000
S
R-Sq
R-Sq(adj)
Mallows Cp
8321
72.48
71.65
1585.6
Alpha-to-Enter:
market price
T-Value
P-Value
1116
99.52
99.49
-0.8
4893
9.32
0.000
juice firm price
T-Value
P-Value
Stepwise Regression: total bottle versus
market price, cherry juice, ...
S
R-Sq
R-Sq(adj)
Mallows Cp
11564
67.19
0.000
-8386
-42.47
0.000
8321
72.48
71.65
1585.6
Stepwise Regression: total bottle versus market price, cherry juice, ...
Backward elimination. Alpha-to-Remove: 0.05
Response is total bottles sold on 6 predictors, with N = 35
Step
1
2
3
4
5
Constant
-34867 -34831 -34912 -34955 -33979
market price
T-Value
P-Value
juice firm price
T-Value
P-Value
11538
56.27
0.000
11540
58.69
0.000
11536
61.18
0.000
11537
62.27
0.000
11564
67.19
0.000
-8347
-8350
-8345
-8345
-8386
-33.00 -35.48 -37.04 -37.65 -42.47
0.000
0.000
0.000
0.000
0.000
apricot market price
T-Value
P-Value
170
169
182
184
0.35
0.36
0.42
0.43
0.728
0.725
0.680
0.671
spring
T-Value
P-Value
-70
-0.12
0.907
-57
-0.12
0.909
summer
T-Value
P-Value
-55
-0.09
0.929
-42
-0.08
0.935
autumn
T-Value
P-Value
-25
-0.04
0.968
S
R-Sq
R-Sq(adj)
Mallows Cp
1189
99.52
99.42
7.0
1168
99.52
99.44
5.0
-42
-0.09
0.926
1149
99.52
99.46
3.0
1130
99.52
99.48
1.0
1116
99.52
99.49
-0.8
Now, our resulting equation is as displayed in tableau 4:
Tableau 4 : Regression Analysis: total bottle versus market price, cherry juice
The regression equation is
total bottles sold = - 33979 + 11564 market price - 8386 juice firm price
1116
99.52
99.49
-0.8
Predictor
Constant
Coef
SE Coef
T
P
-33979
2553
-13.31
0.000
market price
11563.7
172.1
67.19 0.000
juice firm price -8386.4
197.5 -42.47 0.000
S = 1115.62
R-Sq = 99.5%
R-Sq(adj) = 99.5%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
2 8262889160 4131444580 3319.47 0.000
Residual Error 32
39827545
1244611
Total
34 8302716706
In a linear regression, the following are to be assumed in determining our model: Residuals
are normally distributed with mean equal to zero and variance σ2 and they are independent. To
check whether our model is following to our assumptions, we are to observe the following
graphs:
Graph 1
Probability Plot of RESI1
Normal
99
Mean
StDev
N
AD
P-Value
95
90
Percent
80
70
60
50
40
30
20
10
5
1
-3000
-2000
-1000
0
1000
RESI1
2000
Note RESI1 describes residuals stemming from tableau 4.
Graph Set 1
3000
4000
1.663076E-12
1082
35
0.708
0.059
Residual Plots for total bottles sold
Normal Probability Plot
Versus Fits
Standardized Residual
99
Percent
90
50
10
1
-2
0
2
Standardized Residual
3
2
1
0
-1
4
20000
40000
60000
Fitted Value
Histogram
Standardized Residual
Versus Order
Frequency
8
6
4
2
0
-1
0
1
2
Standardized Residual
80000
3
2
1
0
-1
3
1
5
10
15
20
25
Observation Order
30
35
By looking at the graph 1, we see that p value is somehow critical: its value would
determine the normality of residuals according to the value of α. The value is 0.059 and values of
α smaller than this value would imply normality for these residuals. Furthermore in graph set 1
we look at the standard residual values plotted against the fitted values and observe a behavior,
which is not desired for the sake of independence. It seems like the graph of x 2, therefore it
makes us choose our response as the square root of the current one. We could also choose other
roots of our response, accordingly. Now, we are to find a model for our new response, which is
found in tableau 5:
Tableau 5
Regression Analysis: tot bot sold versus market price, cherry juice
The regression equation is
tot bot sold^1/2 = 23.2 + 24.6 market price - 17.2 juice firm price
Predictor
Constant
market price
juice firm price
Coef SE Coef
T
P
23.226
1.321
17.58 0.000
24.6205
0.0891
276.37 0.000
-17.2317
0.1022 -168.59 0.000
S = 0.577440
R-Sq = 100.0%
Analysis of Variance
Source
DF
SS
R-Sq(adj) = 100.0%
MS
F
P
Regression
Residual Error
Total
2
32
34
39420
11
39430
19710
0
59110.93
0.000
In tableau 5, we see that F value is a tremendous one, improving the significance of our
model. P values are zero, which also states there is no insignificant factor in the model. Further,
the stepwise regression gives the same result with more concrete R-Sq and R-Sq(adj) values and
also the corresponding Mallows’ Cp:
Step1
17.0
75.94
75.21
26230.0
S
R-Sq
R-Sq(adj)
Mallows Cp
Step2
0.577
99.97
99.97
0.5
Graph Set 2
Residual Plots for tot bot sold^1/2
Normal Probability Plot
Versus Fits
Standardized Residual
99
Percent
90
50
10
1
-2
-1
0
1
Standardized Residual
2
2
1
0
-1
-2
150
200
Histogram
Standardized Residual
Frequency
6
4
2
-2
-1
0
1
Standardized Residual
300
Versus Order
8
0
250
Fitted Value
2
2
1
0
-1
-2
1
5
10
15
20
25
Observation Order
30
35
In graph set 2, we see that standardized residuals show no behavior, they are like
sprinkled, as desired. In graph 2, we conclude for the normality of residuals also by looking
at the p value equaling 0.875. Their variances are changing between the values -2 and +2,
which is also desired.
Graph 2
Probability Plot of RESI5
Normal
99
Mean
StDev
N
AD
P-Value
95
90
-2.18441E-13
0.5602
35
0.200
0.875
Percent
80
70
60
50
40
30
20
10
5
1
-1.0
-0.5
0.0
RESI5
0.5
1.0
1.5
Note RESI5 denotes residuals associated with tableau 5.
There is one issue left to be discussed, namely the interaction of factors and their joint
effects on our model. To introduce their joint effect to our software, we simply multiply several
effects with each other and apply linear regression. We note that our main factors, juice firm
price and juice market price are to be included in the model. Because we think of the possibility
of entry of the apricot price into our model, we consider the interaction between apricot price and
juice firm price, and the one between juice market price and apricot price. We do not consider
seasonality here, as when we perform multiplication, we get most of the values equal to zero,
and also the ones not equaling zero deliver no entrance into our model when checked in Minitab.
Now tableau 6 shows the combinatorial regression analysis results associated with the two main
factors and interaction implying factors involving apricot price:
Tableau 6
Regression Analysis: tot bot sold versus market and , cherry and, ...
The regression equation is
tot bot sold^1/2 = 22.9 + 0.011 market and apricot - 0.019 cherry
+ 0.043 apricot squared + 24.6 market price
- 17.2 juice firm price
and apricot
Predictor
Coef SE Coef
T
P
Constant
22.941
2.792
8.22 0.000
market and apricot
0.0110
0.2626
0.04 0.967
cherry and apricot
-0.0189
0.2379
-0.08 0.937
apricot squared
0.0431
0.2530
0.17 0.866
market price
24.5911
0.7552
32.56 0.000
juice firm price -17.1821
0.6765 -25.40 0.000
S = 0.605867
R-Sq = 100.0%
R-Sq(adj) = 100.0%
Analysis of Variance
Source
Regression
Residual Error
Total
DF
5
29
34
SS
39419.5
10.6
39430.2
MS
7883.9
0.4
F
21477.68
P
0.000
As we see in the tableau, the interaction of apricot with the main prediction factors delivers
new insignificant factors. Although the model is significant with a big F value, the p values
corresponding to interaction of market and apricot, cherry and apricot and apricot with itself are
0.967, 0.937 and 0.866, respectively. If the analysis is performed without including the main
factors, then we would get the result shown in tableau 7:
Tableau 7
Regression Analysis: tot bot sold versus market price, cherry , ...
The regression equation is
tot bot sold^1/2 = 45.9 + 23.8 market price - 17.9 juice firm price
+ 0.0248 cherry and market + 0.052 market and apricot
- 0.048 cherry and apricot - 0.022 apricot squared
Predictor
Coef SE Coef
T
P
Constant
45.89
16.38
2.80 0.009
market price
23.7752
0.9383
25.34 0.000
juice firm price -17.8923
0.8317 -21.51 0.000
cherry and market
0.02475 0.01741
1.42 0.166
market and apricot
0.0518
0.2597
0.20 0.843
cherry and apricot
-0.0483
0.2347
-0.21 0.838
apricot squared
-0.0216
0.2528
-0.09 0.933
S = 0.595477
R-Sq = 100.0%
R-Sq(adj) = 100.0%
Analysis of Variance
Source
Regression
Residual Error
Total
DF
6
28
34
SS
39420.3
9.9
39430.2
MS
6570.0
0.4
F
18528.42
P
0.000
Still the interaction factors are insignificant; they become effective only in the absence of
the two main predictors, cherry price and market price of juice. We see that by looking at the
corresponding p values. The interaction between cherry and juice price seems to be better than
others, but this is both due to the single strength of each factor and also the p value is only better,
not acceptable.
In the light of the information presented above, we determine our final model equation with
corresponding Cp-, p, and F values as follows:
The regression equation is
tot bot sold^1/2 = 23.2 + 24.6 market price - 17.2 juice firm price
Predictor
Constant
market price
juice firm price
Coef SE Coef
T
P
23.226
1.321
17.58 0.000
24.6205
0.0891
276.37 0.000
-17.2317
0.1022 -168.59 0.000
S = 0.577440
R-Sq = 100.0%
Analysis of Variance
Source
Regression
Residual Error
Total
DF
2
32
34
SS
39420
11
39430
R-Sq(adj) = 100.0%
MS
19710
0
F
59110.93
Mallows Cp
=
P
0.000
0.5
D. MAXIMIZING THE EXPECTED REVENUE ON THE NEXT MONTH
In order to find the optimal price, we need to take each value among 32, 33, 34, 35,
36 and by inserting them into our equation, and taking the square of the response, we find our
sales amount. Then multiplying the sales amount with the taken price we calculate our
revenue. Because the average market price has a probability mass function, we apply this
process for each of the values determined for the market price. Let us illustrate this in the
following:
0.3, X  26

P( X )  0.6, X  27
0.1. X  28

P(price)=32: We write the price and first value, which is 26, into our model:
tot bot sold^1/2 = 23.2 + 24.6*26 - 17.2*32= 112.4
Taking the square of it delivers the sales amount: 112.42=12633.76 .The revenue is
calculated as: 12633.76*32= 404280.32. Now, for the partial expected revenue, we multiply it by
0.3: 404280.32*0.3= 121284.09.
As for the market price of 27, we write 27 and 32 into our equation and get the sales
amount:
(tot bot sold^1/2 = 23.2 + 24.6*27 - 17.2*32)2=18769
We get the revenue: 18769*32=600608 and 600608*0.6=360364.8 is the partial expected one.
As for the market price of 28, we get the partial revenue 83566.59
Our expected profit when we determine the price of cherry as 32 would be the sum of
three revenues: 121284.09+360364.8+83566.59=565215.48
If we proceed with the remaining prices as described above, we would get:
P=33: expected revenue= 89724.09+284170.39+68809.49=442703.97
P=34: expected revenue=62056.80+214745.90+55011.46=331814.16
P=35: expected revenue=38814.72+153156.36+42350.00=234321.08
P=36: expected revenue=20530.37+100466.78+31002.62=151999.77
Therefore we need to determine our price as 32, which then maximizes our expected
revenue in the next month. If the firm wants to increase this number by also considering total
cost or production cost, the department may decrease the price of cherry .
E. CONCLUSION
We observed the importance of juice firm price and juice market price in determination of
the factors on the sales amount. Apricot, which has a potential for replacing juice when high
prices of juice are present, played almost no role in computing our sales amount.
For further calculations, we need to be given the costs of production and other costs
associated with distribution or delivery i.e. Here, if the firm wants to increase the expected
revenue by also considering total cost or production cost, the department may decrease the price
of cherry.
BIBLIOGRAPHY
Baker N., Badar M. (2015) “Effect of Price and Brand on Common Platform Appliance
Failure”Journal of Management Research and Innovation 11(4) pp 296-304
Glantz, S. (1993), “Changes in cigarette consumption, prices, and tobacco industry revenues
associated with California's proposition 99” Tob Control 2(4) pp 311-314
Huntington, p. (1993) “Ticket Pricing Policy and Box Office Revenue” Journal of Cultural
Economics 17(1) pp 71-87
Leisman M. (2001) “Does the seed size/number trade-off model determine plant community
structure? An assessment of the model mechanisms and their generality” Synthesizing Ecology
93(2) pp 294-302
McLaughlin E., Lesser W. (1987) “Experimental Price Variability and Consumer Response:
Tracking Potato Sales With Scanners” Journal of Food Distribution Research pp 108-115