Quantitative Statistical Methods Test Essay

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Quantitative Statistical Methods


Table of Contentss

Question 12

Question 26

Question 38

Question 410

Bibliography10

Question 1

The undermentioned informations are the tonss obtained by 40 kids in a Quantitative Methods trial

100 92 96 91 81 110 70 119 94 88 104 115 97 108 93 116 86 89 78 108 127 91 114 84 101 121 95 105 112 102 124 92 76 105 90 81 109 116 96 107

a. Draw the root secret plan

B. Find average, Q1 and Q3

c. Draw the box secret plan

d. Find the mean and standard divergence for the tonss

e. Discussion the restriction of utilizing stem secret plans and box secret plans

Answer

a ) The Stem Plot

Root

Leafs

7

0,6,8

8

1,1,4,6,8,9

9

0,1,1,2,2,3,4,5,6,6,7

10

0, 1, 2, 4,5,5,7,8,8,9

11

0,2,4,5,6,6,9

12

1,4,7

B )

Time interval

Tally

Frequency ( degree Fahrenheit )

Mid-point ( x )

Accumulative Frequency ( F )

fx

70-79

|||

3

74.5

3

223.5

5550.25

16650.7

80-89

|I?|I?|I?||

6

84.5

9

507

7140.25

42841.5

90-99

|I?|I?|I?||I?|I?|I?||

11

94.5

20

1039.5

8930.25

98232.7

100-109

|I?|I?|I?||I?|I?|I?|

10

104.5

30

1045

10920.25

109202.5

110-119

|I?|I?|I?|||

7

114.5

37

801.5

13110.25

91771.75

120-129

|||

3

124.5

40

373.5

15500.25

46500.75

40

597

3990

61151.5

405199.9

N=40 ( even )

=

=90+

=90+

=90+10

=100

=90+

=90+

=90+

=90+

=

=

90.90

=100+

=100+

=100+

=100+10

110

degree Celsius ) The Box Plot is given below

vitamin D ) Mean and the standard divergence

Mean,=

=

=99.75

S.D =

=

=

=

=13.41

=134.1

vitamin E ) Here is the Restrictions of Stem secret plans and Box Plots:

Root and Leaf Plots

Stem.and Leaf Plot.is.a.way.of.briefing.a.set.of.data.dignified.on.an.interlude.scale. It.is.often.used.inexperimental.data.analysis.to.illustrate.the.key.features.of.the.circulation.of.data.in.an appropriate.and.easy.way. Stem.plots.record.data.values.in.rows, and.easily.can.be.made.into.a histogram. Huge.data.sets.can.be.billeted.by.splitting.stems.

( Purplemath, 2014 ) ( Statistics Canada, 2014 ) ( Math is for merriment, 2014 )

Restrictions of utilizing Stem Plots

Besides a batch of advantages there has besides some Restrictions of utilizing Stem and Leaf Plots.

Root secret plan is non visually appealing. Does non easy indicate steps of centrality for

Large informations sets. Root and Leaf Plots can non be used for categorical informations and Difficult with big figure of observations. It’s really hard if there are a batch of roots with no foliages. And sometimes outlier might be a job. Preferably have need of foliages to be well-ordered on each root – difficult work. It’s tricky to utilize ( though non impossible ) if denary stem/leaf are non the optimum solution.

( Purplemath, 2014 ) ( Statistics Canada, 2014 ) ( Math is for merriment, 2014 )

Box Plot

Box Plot is a one sort of graph which used to expose the array of quantitative informations. It is besides known as Box and Whisker Plot.

Box Plot fragmented the information set into quartiles. It is often used in analyzing informations analysis. It is a sort of graph used to demo the signifier of the distribution, its cardinal value, and variableness. A box secret plan is specially utile for meaning whether a distribution is tilted and whether there are any infrequent observations ( outliers ) in the information set. Box and hair’s-breadth secret plans are besides really helpful when immense sum of readings are convoluted and when two or more informations sets are being associated.

Restrictions of utilizing Box Plot

It’s non as visually attractive as other graphs, & A ; exact values non retained.

( Purplemath, 2014 ) ( Statistics Canada, 2014 ) ( Math is for merriment, 2014 )

Question 2

A company is sing puting in a undertaking it has two options ( Project X and Y ) . You have been called to urge to better of the two options. The information in the below is available for your analysis.

Undertaking Ten

Initial investing = ?27000

Cost of Capital =16 %

Expected net hard currency flows for 5 old ages:

Year

1

2

3

4

5

Net hard currency flows

1780

2750

4200

4570

8600

Undertaking Y

Initial investing = ?48000

Cash of Capital =10 %

Expected net hard currency flows for 5 old ages:

Year

1

2

3

4

5

Net hard currency flows

18700

19512

20900

22978

23838

  1. Using NPV analysis, select the better investing
  2. What are the other factor that need to be considered in your concluding determination?

Answer

Undertaking Ten

Initial investing = 27000

Cost of capital=16 %

Ten

Year

Cash flow

Discount factor ( 16 % )

Present value ( CF?DF )

0

-27000

1

– 27000

1

1780

0.862068965

1534.48

2

2750

0.743162901

2043.69

3

4200

0.640657673

2690.76

4

4570

0.552291097

2523.97

5

8600

0.476113015

4094.57

Total=

-5100

NPV=-14,112.53

Undertaking Y

Initial investing = 48000

Cost of capital= 10 %

Yttrium

Year

Cash flow

Discount factor ( 10 % )

Present value ( CF?DF )

0

-48000

1

-48000

1

18700

0.909090909

17000

2

19512

O.82644624

16125.61

3

20900

0.7513148

15702.47

4

22978

0.683013455

15694.28

5

23838

0.620921323

14801.52

Total=

57928

NPV=17,223.88

( a ) Using NPV analysis the better investing is Project

B ) To do the concluding determination I have to see the Discount Factor.

Question 3

A leather industry has derived the following information on production costs ( ?Y ) and units of end product ( X ) for the last 12 months:

Yttrium

16

6.4

6.5

16

12.7

12.5

9.3

6.3

10.5

9

9.8

12

Ten

29

4

7

35

28

23

16

8

21

14

12

26

You are required to:

  1. Plot spread diagram of Y against X
  2. Find the least squares regression equation of production costs on end product, and plot the line on the diagram
  3. Predict production cost for the following month if it is able to bring forth 40 units of end product and discourse the likely dependability of this production

Answer

a ) Scatter diagram of Y against X

B ) The least squares regression equation of production costs on end product, and line on the diagram:

ten

Y

ten Y

29

16

464

841

256

4

6.4

25.6

16

40.96

7

6.5

45.5

49

42.25

35

16

560

1225

256

28

12.7

355.6

784

161.29

23

12.5

287.5

529

156.25

16

9.3

148.8

256

86.49

8

6.3

50.4

64

39.69

21

10.5

220.5

441

110.25

14

9

126

196

81

12

9.8

117.6

144

96.04

26

12

312

676

144

?x=223

?y=127

?xy=2713.5

?=5221

?=1470.22

The least squares regression equation

b=

=

=

=

=0.328

a =-b

=-0.328

=10.583-6.095

=4.487

=4.49

Y=a+bx

=4.49+0.328x

X=0, Y=4.49+0.328?0=4.49

X=10, Y=4.49+0.328?10=7.77

X=20, Y=4.49+0.328?20=11.05

Question 4

The gas and electricity gross revenues between 2002 and 2009 by the public supply system are shown below:

Year

Oil and gas ( ‘000 )

Electricity ( ‘000 )

2002

27

19

2003

31

18

2004

40

21

2005

46

27

2006

53

30

2007

65

41

2008

67

29

2009

66

27

a ) Calculate the coefficient of correlativity between oil & A ; gas and electricity gross revenues

( 1 ) The merchandise minute correlativity and coefficient

( 2 ) Spearman rank correlativity and coefficient

B ) Discuss the significance of both these steps.

Answer

a ) ( 1 )

Year

Oil & A ; gas

ten

Electricity

Y

Ten Y

2002

27

19

513

729

361

2003

31

18

558

961

324

2004

40

21

840

1600

441

2005

46

27

1242

2116

729

2006

53

30

1590

2809

900

2007

65

41

2665

4225

1681

2008

67

29

1943

4489

841

2009

66

27

1782

4356

729

?x=395

?y=212

?xy=11133

?=21285

?=6006

R =

=

=

=

=

= 0.80

The merchandise minute correlativity and coefficient is Strong.

( 2 )

Year

Oil & A ; Gas

Electricity

Ro

Rhenium

vitamin D

2002

27

19

8

6

2

4

2003

31

18

7

7

0

0

2004

40

21

6

5

1

1

2005

46

27

5

4.5

0.5

0.25

2006

53

30

4

2

2

4

2007

65

41

3

1

2

4

2008

67

29

1

3

-2

4

2009

66

27

2

4.5

2.5

6.25

n=8

395

212

?

P=1-

=1-

=1-

=1-

=1-

=1-0.279

=0.720

=0.72

The Spearman Rank correlativity between oil & A ; gas and electricity is strong.

B )Significance of utilizing

The other steps of correlativity are parametric within the significance is upon possible relationship of a parameterized signifier, such as a additive relationship. This both two steps are much easier to utilize than other and since it doesn’t affair in which manner we rank the information, go uping or falling. It is less sensitive to bias due to the consequence of outliers. It can be used to cut down the weight of outliers. It does non necessitate of premise of normalcy. When the interval between informations points are debatable, it is advisable to analyze the rankings instead than the existent value.

( Laerd Statistics, 2014 )

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