Introduction To Management Statistics

Published Date: 02 Nov 2017

Introduction

Statistics "is the study of the collection, organization, analysis, interpretation, and presentation of data. It deals with all aspects of this, including the planning of data collection in terms of the design of surveys and experiments" (Dodge, 2006)

This research introduces various statistical terms in the form of problems, which are further calculated to show sequentially how an answer is arrived at.

This research introduces readers to terms such as the Laspeyres and Paache price index, decision tree, time series, regression and linear programming.

Project Background

This project is done as a partial fulfillment of the researcher course of study, and forms an important part of the overall final grade. This research was done not only to edify the researcher on the content of the course, but also to be used as a tool for other readers to understand and get a clearer and more simplified approach to solving statistical problems like those that are in this research. This research may be used as question and answer booklet in the future for student who may have difficulty in this area.

Literature review

Mathematics, economics, statistics and other numerical courses pose as a challenge for many, and the difficulty surrounding these problems are base in not the calculating part of it, but the interpretation part. Many may view statics and advance mathematic of any form as not being a necessity in society’s day to day running’s however, Florence Nightingale (a pioneer nurse, writer and avid statistician) stated "statistic.....the most important science in the whole world: for upon it depends the practical application of every other science and of every art; the science essential to all political and social administration, all education, all organization based upon experience for it only gives the result of our experience" (PSI, 2007)

Statistics "is the study of the collection, organization, analysis, interpretation, and presentation of data. It deals with all aspects of this, including the planning of data collection in terms of the design of surveys and experiments" (Dodge, 2006)

The importance of statistics is evident and cannot be over stated as calculations such as Laspeyres Price Index help economist determine inflation, taxation, value of a dollar, G.D.P among many other economic variations of a base year as appose to another particular year. Price index is the "Measure of change in a set of prices, consisting of a series of numbers arranged so that a comparison of the values for any two periods or places will show the change in prices between periods or the difference in prices between places. Price indexes were first developed to measure changes in the cost of living in order to determine the wage increases necessary to maintain a constant standard of living." (Farlex, 2008)

Objectives

This research aims to:

Introduce advance mathematical terms, symbols and methodologies to readers, so that they may have a better understanding and use of such term in the business environment

Provide a clear understanding of, while demonstrating correct methods for solving Mathematical problems in a simplified way.

Show the importance of how methodologies such as Differentiation, Decision Tree ect. can be applied to economics and daily lives.

Methodology

The completion of this research was mostly done via the primary and secondary sources, as the lecturer and other colleagues played an integral role. Primary day was obtain by conducting short interview with the lecturer for assurance that the research was on the right path as well as, information gained during lecture sessions.

Secondary data was obtained from the use of books such as Qualitative Methods for Business by Butter-Worth, Heineman and Baglear (2004), from website and YouTube short course videos on how to solve such research problems.

Research Findings

Task one

Introduction to management statistics:

Using the information in the table:

Calculate:

Mean

Mode

Observation (X)

Frequency (F)

10

4

15

5

20

9

40

16

60

8

80

5

100

3

Solution:

Observation (X)

Frequency (F)

F(x)

10

4

40

15

5

75

20

9

180

40

16

640

60

8

480

80

5

400

100

3

300

Total= 50

Total= 2115

Mean

Formula: µ= ∑ f (x)

f (n)

Step one: ∑ f (x) = 2115

40+75+180+640+480+400+300=2115

Step two: f (n) = 50

Total sample (n) of "f" is = 4+5+9+16+8+5+3 =50

Step three: 2115 42 = 42 3/10 OR 42.3

50 1

Mode

There is no formula for the mode however, the number of observations that occur the most determine the mode. Therefore:

The observation "40" which occurs "16" times would be the mode.

Task two

Regression;

Calculate the r (correlation) between x and y.

Interpret the value.

Days of the month

Table scale of the product A (X)

Table scale of the product B (Y)

1st day

15

16

4th day

14

13

7th day

10

11

10th day

18

19

15th day

15

13

18th day

20

22

22nd day

16

15

25th day

15

17

27th day

12

16

30th day

13

17

Total = 148

Total = 159

Solution:

x

y

xy

x2

y2

1st day

15

16

240

225

256

4th day

14

13

182

196

169

7th day

10

11

110

100

121

10th day

18

19

342

324

361

15th day

15

13

195

225

169

18th day

20

22

440

400

484

22nd day

16

15

240

256

225

25th day

15

17

255

225

289

27th day

12

16

192

144

256

30th day

13

17

221

169

289

∑=148

∑=159

∑=2417

∑=2264

∑=2619

Formula:

n ∑ Xꜟ Yꜟ – ∑ Xꜟ ∑ Yꜟ

R = r = √ (n ∑ Xꜟ2 ­– (∑Xꜟ )2) ( n ∑Yꜟ 2 – (∑Yꜟ)2 )

Step one: n ∑ Xꜟ Yꜟ – ∑ Xꜟ ∑ Yꜟ

10 (2417) – (148) (159) =

24170 – 23532 = 638

Step two: √ (n ∑ Xꜟ2 ­– (∑Xꜟ )2) ( n ∑Yꜟ 2 – (∑Yꜟ)2 )

(10 (2264) – (148)2 ) ( 10 (2619) ­– (159)2 ) =

(22640-21904) (26190 – 25281) =

(736) (909) =

669024

√ 669024 = 817.939

Step three: 638

817.939

= 0.78

ii. Interpret the value

The value r ≈ 0.78, indicate that X has a strong positive relationship to Y.

Task three

Calculate the Lapeyres and Paasche price indices for the following data.

Take 2005 as the base year.

Liter of Beer

Liter of Whiskey

Liter of Wine

Year

Price

Qty

Price

Qty

Price

Qty

2005

0.95

200

19.80

10

10.50

36

2006

0.99

150

20.39

12

11.15

48

2007

1.05

120

20.99

11

12.35

60

i. Laspeyres Price Index (LPI) = ∑ qo pn

∑ qo po

Step one: ∑ q05 p05

∑ q05 p05

(200×0.95) + (10×19.80) + (36×10.50)

(200×0.95) + (10×19.80) + (36×10.50)

= 190+198+378

190+198+378

= 766

766 = 1 ×100 =100

Step two: ∑ q05 p06

∑ q05 p05

= (200×0.99) + (10×20.39) + (36×11.15)

(200×0.95) + (10×19.80) + (36×10.50)

= 198+203.9+401.4

190+198+378

= 803.3

766

= 1.049

= 1.049×100 = 104.9

Step three: ∑ q05 p07

∑ q05 p05

= (200×1.05) + (10×20.99) + (36×12.35)

(200×0.95) + (10×19.80) + (36×10.50)

= 210+209.9+444.6

190+198+378

= 864.5

766

= 1.129

= 1.129×100 = 112.9

ii. Paasche Price Index (PPI) = ∑ qn pn

∑ qn po

Step one: ∑ q05 p05

∑ q05 p05

(200×0.95) + (10×19.80) + (36×10.50)

(200×0.95) + (10×19.80) + (36×10.50)

= 190+198+378

190+198+378

= 766

766

= 1 ×100 = 100

Step two: ∑ q06 p06

∑ q06 p05

= (150×0.99) + (12×20.39) + (48×11.15)

(150×0.95) + (12×19.80) + (48×10.50)

= 148.5 + 244.68 + 535.2

142.5 + 237.6 + 504

= 928.38

884.1

= 1.050

= 1.050×100 = 105.0

Step two: ∑ q07 p07

∑ q07 p05

= (120×1.05) + (11×20.99) + (60×12.35)

(120×0.95) + (11×19.80) + (60×10.50)

=126 + 230.89 + 741

114 + 217.8 + 630

= 1,097.89

961.8

=1.141

= 1.141×100

= 114.1

Task four

Time series the number of rate captured in a grain stor is summarized below. Use simple exponential smoothing with alpha = 0.2 and alpha = 0.7 to forcast the number of rates that will be caught in week 7. GUIDE

Week 1.

216

Week 2.

224

Week 3.

217

Week 4.

233

Week 5.

245

Week 6.

229

Solution:

T

Y

F (É‘= 0.7)

F (É‘= 0.2)

Week 1.

216

216

216

Week 2.

224

216

216

Week 3.

217

221.6

217.6

Week 4.

233

218.4

217.5

Week 5.

245

228.6

220.6

Week 6.

229

240.1

225.5

Week 7.

232.33

226.2

Time series formula: F t+1 = Ft + ɑ (Yt – Ft) with 0 ≤ ɑ ≤ 1

F t+1 = ɑ Yt + (1– ɑ) Ft

É‘=0.7

Step one: F 0+1= ɑYt + (1–ɑ) F1

F1 = 0.7Y1 + (1 – 0.7) F1

= 0.7 (216) + 0.3 (216)

=151.2 + 64.8 =216

Step three: F 2+1= ɑYt+(1–ɑ) F3

F3 = 0.7 Y3 + (1­– 0.7) F3

F3 = 0.7 (224) + (1 – 0.7) 216

= 156.8 + 0.3 (216)

= 156.8 + 64.8 = 221.6

Step five: F 4+1= ɑYt + (1–ɑ) F5

F5 = 0.7 Y5 + (1­– 0.7) F5

F5 = 0.7 (233) + (1– 0.7) 218.4

= 163.1 + 0.3 (218.4)

= 163.1 + 65.52 = 228.6

Step seven: F 6+1= ɑYt+(1–ɑ)F7

F7 = 0.7 Y7 + (1­– 0.7) F7

F7 = 0.7 (229) + (1– 0.7) 240.1

=160.3 + 0.3 (240.1)

=160.3 + 72.03 = 232.33

Step two: F 1+1= ɑYt + (1–ɑ) F2

F2 = 0.7 Y2 + (1­– 0.7) F2

= 0.7 (216) + 0.3 (216)

= 151.2 + 64.8 = 216

Step four: F 3+1= ɑYt + (1–ɑ) F4

F4 = 0.7Y4 + (1–0.7) F4

F4 = 0.7 (217) + (1 – 0.7) 221.6

=151.9 + 0.3 (221.6)

= 151.9 + 66.48=218.4

Step six: F 5+1= ɑYt + (1–ɑ) F6

F6 = 0.7 Y6 + (1­– 0.7) F6

F6 = 0.7 (245) + 0.3 (228.6)

= 171.5 + 68.58

=240.1

É‘=0.2

Step one: F 0+1 = ɑYt + (1–ɑ) F1

F1 = 0.2 Y1 + (1­– 0.2) F1

F1 = 0.2 (216) + (1– 0.2) 216

= 43.2 + 0.8 (216)

= 43.2 + 172.8= 216

Step three: F 2+1 = ɑYt + (1–ɑ) F3

F3 = 0.2 Y3 + (1­– 0.2) F2

F3 = 0.2 (224) + (1 –0.2) 216

= 44.8 + 0.8 (216)

= 44.8 + 172.8 = 217.6

Step five: F 4+1 = ɑYt + (1–ɑ) F5

F5 = 0.2 Y5 + (1­– 0.2) F5

F5 = 0.2 (233) + (1 –0.2) 217.5

= 46.6 + 0.8 (217.5)

= 46.6 + 174 = 220.6

Step six: F 6+1 = ɑYt + (1–ɑ) F7

F7 = 0.2 Y7 + (1­– 0.2) F7

F7 = 0.2 (229) + (1 –0.2) 225.5

= 45.8 + 0.8 (225.5)

= 45.8 + 180.4 = 226.2

Step two: F 1+1 = ɑYt + (1–ɑ) F2

F2 = 0.2 Y2 + (1­– 0.2) F2

F2 = 0.2 (216) + (1– 0.2) 216

= 43.2 + 0.8 (216)

= 43.2 + 172.8 = 216

Step four: F 3+1 = ɑYt + (1–ɑ) F4

F4 = 0.2 Y4 + (1­– 0.2) F4

F4 = 0.2 (217) + (1 – 0.2) 217.6

= 43.4 + 0.8 (217.6)

= 43.4 + 174.08 = 217.5

Step six: F 5+1 = ɑYt + (1–ɑ) F6

F6 = 0.2 Y6 + (1­– 0.2) F6

F6 = 0.2 (245) + (1 –0.2) 220.6

= 49 + 0.8 (220.6)

= 49 + 176.5 = 225.5

Task five

Inferential statistics

A cruise ship was interested in the typical duration each client spent in the breakfast buffet. The entry and exist times of 30 cruisers was noted.

i. Calculate and approximate 99% confidence interval for the mean breakfast time.

43

35

36

25

30

35

42

28

18

21

39

43

34

38

27

34

38

41

19

44

34

39

19

36

29

33

24

40

31

18

SUM =973

Step one: finding the mean

µ= ∑ x

n

= 973

30

= 32.433

Step two: finding the variance

S2 = ∑ (χ i ­­– µ) 2

n – 1

χ

χ – µ

(χ – µ)2

43

10.57

111.7

35

2.57

6.6

39

6.57

43.2

34

1.57

2.5

34

1.57

2.5

33

0.57

0.3

35

2.57

6.6

42

9.57

91.6

43

10.57

111.7

38

5.57

31.0

39

6.57

43.1

24

-8.43

71.1

36

3.57

12.7

28

-4.43

19.6

34

1.57

2.5

41

8.57

73.4

19

-13.43

180.4

40

7.57

57.3

25

-7.43

55.2

18

-14.43

208.2

38

5.57

31.0

19

-13.43

180.4

36

3.57

12.7

31

-1.43

2.0

30

-2.43

5.9

21

-11.43

130.6

27

-5.43

29.4

44

11.57

133.9

29

-3.43

11.8

18

-14.43

208.2

∑ = 973

∑ = 1877.1

S2 = ∑ (χ i ­­– µ) 2

n – 1

= 1877.1

30 – 1

= 1877.1

29

= 64.73

S2 = 64.73

√ S2 = √ 64.73

S = 8.05

Step three: calculating the 99% CI.

Formula: (χ ± t )

= 32.43 + (2.576) 8.05

√ 30

= 32.43 + (2.576) (1.47)

= 32.43 + 3.79

= 36.2

OR

= 32.43­–3.79

= 28.6

Thus the 99% CI = (36.2, 28.6)

Task six

Linear programming

Minimize; cost = 9x+3y

Subject to the following constraints;

Constraints 1: Y ≥ 5

Constraints 2: 6X + 7Y ≥ 210

Constraints 3: 7X + 15Y ≥ 525

Constraints 4: 5X + 28Y ≤ 700

Constraints 5: X ≥ 0, Y ≥ 0

Which of the constraints are binding and which are non-binding.

In solving the problem let’s assume X = 0 and Y = 0 to solve the inequalities, so as to find the coordinates for x and y on the graph

Constraints 2: 6X + 7Y ≥ 210

Constraints 3: 7X + 15Y ≥ 525

6x + 7(0) = 210

6x + 0 = 210

6x = 210

6 6

x = 35

6(0) + 7y =210

0 + 7y = 210

7y = 210

7 7

y = 30

coordinates are (x, y) (35,30)

7(0) + 15y = 525

0 + 15y = 525

15y = 525

15 15

y = 35

7x + 15(0) = 525

7x + 0 = 525

7x = 525

7 7

x= 75

coordinates are (x, y) (35, 75)

Constraints 4: 5X + 28Y ≤ 700

5x + 28(0) = 700

5x + 0 = 700

5x = 700

5 5

x = 140

5(0) + 28y = 700

0 + 28y = 700

28y = 700

28 28

y = 25

coordinates are (x, y) (140,25)

Plotting the points on a graph:

The shaded area is the feasible region.

Minimize; cost = 9x+3y

Using coordinates at points A, B, C and D from the graph in the optimizing function, minimized cost will be determined.

The Coordinates are:

A: (7.37, 23.68) B: (34.71, 18.8)

C: (64.29, 5) D: (29.17, 5)

Now:

A: 9 (7.37) + 3(23.68) B: 9(34.71) + 3(18.8)

66.33+71.04 312.39+56.4

=137.37 =368.79

C: 9(64.29) + 3(5) D: 9(29.17) 3(5)

578.61+ 15 262.53+15

=593.61 =277.53

Therefore cost is minimized at point A.

Constraints one (y≥5) and three (7x+15y≥525) are non-binding because they are redundant and does not affect solution of the optimizing function, while constraints four (5x+28y ≤700) and two (6x+7y≥210) are binding as any change in these constraints will cause the solution of the optimizing function to change.

Task Seven

What is a decision tree?

Decision trees find use in a wide range of disciplines. It is applied in medical and cognitive science, engineering, economics among many other theoretical and math base disciplines. "A decision tree is a decision support tool that uses a tree-like graph or model of decisions and their possible consequences, including chance event outcomes, resource costs, and utility. It is one way to display an algorithm. an algorithm is a step-by-step procedure for calculations. Algorithms are used for calculation, data processing, and automated reasoning." (Shaw, 1995)

According to Rafael Olivas in the book "decision trees: A primer for decision making professionals" A decision tree "is a method you can use to help make good choices, especially decisions that involve high costs and risks. Decision trees use a graphic approach to compare competing alternatives and assign values to those alternatives by combining uncertainties, costs, and pay offs into specific numerical values." (Olivas, 2007).

An example of a tiplical decision tree

Figure 1.

(Olivas, 2007)

Decision trees provide an effective method of Decision Making because they:

Clearly lay out the problem so that all options can be challenged.

Allow us to analyze fully the possible consequences of a decision.

Provide a framework to quantify the values of outcomes and the probabilities of achieving them.

Help us to make the best decisions on the basis of existing information and best guesses.

As with all Decision Making methods, decision tree analysis should be used in conjunction with common sense – decision trees are just one important part of your Decision Making tool kit.

Decision trees are commonly used in operations research, specifically in decision analysis, to help identify a strategy most likely to reach a goal.

Task Eight

Differentiation is "the mathematical process of obtaining the derivative of a function. The derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity." (Anton, et al.,2005) differentiation is primarily a tool used for calculating rates of change.

Differentiation may be used in economics to gain a competitive advantage in a particular market. This may be achieved when the product itself is differentiated, that is: "the process of distinguishing a product or offering from others, to make it more attractive to a particular target market. This involves differentiating it from competitors' products as well as a firm's own product offerings." (Chamberlin, 1993) In this type of differentiation the company is more focus on cost and quality offering of a product. In a marketing sense the main aim of differentiation is to gain a unique place in consumers’ hearts, a position. When this is achieved the product normally does well and the cost is no longer an issue as the quality will worth whatever it is being sold for.

Differentiation may also be used to set prices on goods and service being provided by a company. This may be done by carefully calculating and analyzing the relationship between revenue and cost. "Revenue is the amount of money that is brought into a company by its business activities" while cost is "a monetary valuation of effort, material, resources, time and utilities consumed, risks incurred, and opportunity forgone in production and delivery of a good or service. All expenses are costs, but not all costs (such as those incurred in acquisition of an income-generating asset) are expenses." (dictionary.com, n.d.)

Differentiation helps determine at what price to set goods so that a reasonable profit margin may be obtained.

Recommendation

The successful conclusion of this research lead to the researcher recommending its use as a manual or guide for upcoming students taking said course, as the breakdown of formulas and explanation of techniques are at their simplest, it may be viewed as self explanatory.

The researcher also recommend that this course to taught to student with a background knowledge of statistics or knowledge os some basic statistical terms as research may help, but students need face to face direction from a tutor or lecturer.

Finally the researcher also recommend that in order for students to fully grasp the content of this research much practices must be exercised so that when different situation arise the proper interpretation may be made and the correct application of formula may be applied.

Conclusion

The research introduced various statistical terms in the form of problems, which are further calculated to show sequentially how they are done. Various topic were explored, some of which are Linear Programming, Price Index, Differentiation, Time Series and Regression.

These topics are topics which are not normally used in everyday mathematics however, they are used mostly by engineers and economist to help formulate and draw conclusions to very important researches and decisions to be made.

Reference

Anton, H., Bivens, I. & Davis, S., February 2, 2005. Calculus: Early Transcendentals Single and Multivariable. 8th edition ed. New York: : Wiley.

Btter-Worth, H. a. B., 2004. Qualitative Methods for Business. s.l.:s.n.

Chamberlin, E., 1993. Theory of Monopolistic Competition.. s.l.:s.n.

dictionary.com, b., n.d. [Online]

Available at: http://www.businessdictionary.com/definition/cost.html

Dodge, 2006. The Oxford Dictionary of Statistical Terms. s.l.:s.n.

Farlex, 2008. the free dictionary. [Online]

Available at: http://encyclopedia2.thefreedictionary.com/price+index

[Accessed 2013].

Olivas, R., 2007. Decision Trees: A Primer for Decision Making Professionals. s.l.:s.n.

PSI, C., 2007. Statistics & Research Design:QUALITIVE AND QUANTITATIVE STATISTICS. WHY STUDY STATISTICS IN PSYCHOLOGY?, 2 December, p. 1.

Table of Contents

Title page

Introduction 1

Project background 2

Literature review 3

Objectives 4

Methodology 5

Research findings 6-28

Task one 7-8

Task two 9-11

Task three 12-15

Task four 16-18

Task five 19-21

Task six 22-24

Task seven 25-26

Task eight 27-8

Recommendation 29

Conclusion 30