Quantitative Statistical Methods Test Essay
 Words: 940
 Category: Database
 Pages: 4
Get Full Essay
Get access to this section to get all the help you need with your essay and educational goals.
Get AccessQuantitative 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 ) 
Midpoint ( x ) 
Accumulative Frequency ( F ) 
fx 

7079 
 
3 
74.5 
3 
223.5 
5550.25 
16650.7 
8089 
I?I?I? 
6 
84.5 
9 
507 
7140.25 
42841.5 
9099 
I?I?I?I?I?I? 
11 
94.5 
20 
1039.5 
8930.25 
98232.7 
100109 
I?I?I?I?I?I? 
10 
104.5 
30 
1045 
10920.25 
109202.5 
110119 
I?I?I? 
7 
114.5 
37 
801.5 
13110.25 
91771.75 
120129 
 
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. Itâ€™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 wellordered on each root – difficult work. Itâ€™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’sbreadth 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
Itâ€™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 
 Using NPV analysis, select the better investing
 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:
 Plot spread diagram of Y against X
 Find the least squares regression equation of production costs on end product, and plot the line on the diagram
 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
=10.5836.095
=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 ( â€˜000 ) 
Electricity ( â€˜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
=10.279
=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 doesnâ€™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 )