Application of Statistical Techniques PDF

STATISTICS FOR
MANAGERIAL DECISIONS

TABLE OF CONTENTS
INTRODUCTION...........................................................................................................................1
QUESTION 1...................................................................................................................................1
(a) Listing quarterly opening price (2003-2015) of WPL and WOR with stem and leaf display
.....................................................................................................................................................1
(b) Constructing relative frequency histogram for WPL and frequency polygon for WOR .....2
(c) Drawing bar chart of market capital in 2015 of WPL, WOR, ORG, STO, CTX and SOL...5
(d) Proportion of stock price above $40 for each of WPL and WOR.........................................6
QUESTION 2...................................................................................................................................6
(a) Computation mean, median, first quartile and third quartile of price...................................6
(b) Computation of standard deviation, range and coefficient of variation of every publisher. .6
(c) Drawing box whisker plot for price of each publisher..........................................................7
QUESTION 3...................................................................................................................................9
(a) Probability it gets water from on farm dams or tanks...........................................................9
(b) Probability it gets water from groundwater and is located in Queensland............................9
(c) Probability it gets water from water from rivers, creeks or lakes..........................................9
(d) In New south Wales, which proportion of farm does not take water from either channels of
irrigation or pipelines or river, creeks or lakes.........................................................................10
QUESTION 4.................................................................................................................................10
(a.i) Probability that on any given week in a year would be no rainfall...................................10
(a.ii) Probability that there would be 3 on more days of rainfall in a week..............................10
(b.i) Probability in given week there would be among 6 mm and 12 mm of rainfall...............10
(b.ii) Amount of rainfall if 20% of the weeks have amount of rainfall or higher.....................10
QUESTION 5.................................................................................................................................11
(a) Normal Probability Plots.....................................................................................................11
(b) Constructing 95% confidence interval for every variable...................................................12
(c) Testing hypothesis that more areas in forest burn when temperature is above 25 degree C
than below 25 degree c with 1% level of significance..............................................................14
CONCLUSION..............................................................................................................................15

INTRODUCTION
Statistics is considered as science of good decision making in uncertainty and used in
multiple disciplines like financial analysis, auditing, econometrics, operations and production
which comprises services improvement and marketing research. The present report will provide
interpretation and application of statistical techniques and scenarios. In the same series, it will
imply statistical techniques as tool for decision making in business environment.
QUESTION 1
(a) Listing quarterly opening price (2003-2015) of WPL and WOR with stem and leaf display
The stem and leaf plot is considered as special table where every data value is split into a
stem and a leaf. This organizes data points through place value of the leading digits as every
stems is usually comprises digits in greases common place value of every item of data. The
leaves consist the other digits of every item of data.
WOR Quarter 1 Quarter 2 Quarter 3 Quarter 4
2003 1.81 1.95 2.38 3.36
2004 3.18 3.02 3.28 5.22
2005 5.9 6.86 7.94 9.73
2006 14.26 19.36 19.91 18.02
2007 21.36 27.5 33.9 47.62
2008 38.949 38.493 31.737 14.065
2009 14.197 17.188 24.282 24.837
2010 22.193 25.072 21.77 21.591
2011 26.013 28.553 25.881 26.21
2012 25.674 26.577 24.545 23.209
2013 23.736 21.422 20.726 20.754
2014 15.476 15.815 16.934 12.757
2015 9.05 10.772 8.655 6.134
WPL Quarter 1 Quarter 2 Quarter 3 Quarter 4
2003 11.34 10.67 12.2 13.21
2004 14.3 15.41 16.37 19.22
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2005 20.42 24.29 28.16 34.76
2006 44.26 45.67 41.76 36.65
2007 36.45 38.47 42.25 50.75
2008 44.73 53.95 51.92 40.42
2009 34.15 37.04 44.11 46.04
2010 41.19 44.18 40.48 42.33
2011 40.52 45.54 37.32 35.58
2012 33.29 33.98 32.84 33.48
2013 34.52 37.14 37.17 38.39
2014 36.97 40.35 42.05 39.68
2015 33.9 34.66 35.25 29.28
The stem and leaf plot shows that open price of WOR is low as compared with WPL
shares of open prices. With the aggregate, distribution of both WOR and WPL distribution has
longer tail which is facing towards left side of normality curve and signifies that distribution of
WOR and WPL is skewed at left.
(b) Constructing relative frequency histogram for WPL and frequency polygon for WOR
WPL
2

WOR
Quarter WOR WPL
01/01/03 1.81 11.34
01/04/03 1.95 10.67
01/07/03 2.38 12.2
01/10/03 3.36 13.21
01/01/04 3.18 14.3
01/04/04 3.02 15.41
01/07/04 3.28 16.37
01/10/04 5.22 19.22
01/01/05 5.9 20.42
01/04/05 6.86 24.29
01/07/05 7.94 28.16
01/10/05 9.73 34.76
01/01/06 14.26 44.26
01/04/06 19.36 45.67
01/07/06 19.91 41.76
01/10/06 18.02 36.65
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01/01/07 21.36 36.45
01/04/07 27.5 38.47
01/07/07 33.9 42.25
01/10/07 47.62 50.75
01/01/08 38.949 44.73
01/04/08 38.493 53.95
01/07/08 31.737 51.92
01/10/08 14.065 40.42
01/01/09 14.197 34.15
01/04/09 17.188 37.04
01/07/09 24.282 44.11
01/10/09 24.837 46.04
01/01/10 22.193 41.19
01/04/10 25.072 44.18
01/07/10 21.77 40.48
01/10/10 21.591 42.33
01/01/11 26.013 40.52
01/04/11 28.553 45.54
01/07/11 25.881 37.32
01/10/11 26.21 35.58
01/01/12 25.674 33.29
01/04/12 26.577 33.98
01/07/12 24.545 32.84
01/10/12 23.209 33.48
01/01/13 23.736 34.52
01/04/13 21.422 37.14
01/07/13 20.726 37.17
01/10/13 20.754 38.39
01/01/14 15.476 36.97
01/04/14 15.815 40.35
01/07/14 16.934 42.05
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01/10/14 12.757 39.68
01/01/15 9.05 33.9
01/04/15 10.772 34.66
01/07/15 8.655 35.25
01/10/15 6.134 29.28
WOR
n 52
mean 18.0736
median 19.635
std. dev. 10.59
minimum 1.81
maximum 47.62
WPL
n 52
mean 34.78962
median 37.005
std. dev. 10.83959
minimum 10.67
maximum 53.95
5

(c) Drawing bar chart of market capital in 2015 of WPL, WOR, ORG, STO, CTX and SOL
Companies Market capital (AUD in million)
WPL 32629
WOR 5301
ORG 33367
STO 21926
CTX 5105
SOL 4253
(d) Proportion of stock price above $40 for each of WPL and WOR
Proportion of stock price above $40
WPL 49/50 0.98
6

WOR 35/50 0.7
QUESTION 2
(a) Computation mean, median, first quartile and third quartile of price
Pearson
education
John wiley and
Sons Cengage Learning
Oxford
university press
McGraw Hill
Education
Mean 156.67 128.67 182.14 83.02 202.09
median 138.32 114.94 181.12 77.89 199.21
Quartile
1 106.94 66.48 130.08 59.69 144.94
Quartile
3 207.34 198.46 222.20 97.94 215.83
(b) Computation of standard deviation, range and coefficient of variation of every publisher
Pearson
education
John wiley and
Sons Cengage Learning
Oxford
university press
McGraw Hill
Education
Standard
deviation 72.13 87.71 56.40 36.99 70.89
Max 293.33 240.08 262.44 154.99 342.59
Min 62.53 15.79 94.86 38.40 112.68
Range 230.80 224.29 167.58 116.59 229.91
Coefficie
nt of
variation 46.04 68.17 30.97 44.55 35.08
(c) Drawing box whisker plot for price of each publisher
Pearson education
7

John wiley and Sons
Cengage Learning
8

Oxford university press
McGraw Hill Education
9

QUESTION 3
(a) Probability it gets water from on farm dams or tanks
Total sources of agriculture water 9779.7
Used from on farm tanks or dams 1163.9
Probability (water used from on farm tanks or
dams) 1163.9 / 9779.7
0.1190
(b) Probability it gets water from groundwater and is located in Queensland
Queensland groundwater 631.1
Total sources of agriculture water 9779.7
Probability (Groundwater and located in
Queensland) 631.1 / 9779.7
0.0645
(c) Probability it gets water from water from rivers, creeks or lakes
Total 6174.3
Taken from Rivers, creeks or lakes 1414.4
10

Probability (water taken from Rivers, lakes or
creeks for farm located) 1414.3 / 6174.3
0.2291
(d) In New south Wales, which proportion of farm does not take water from either channels of
irrigation or pipelines or river, creeks or lakes
Total 3425.8
Taken from amount of water from New South
Wales 957
Probability (water not taken from Creeks, lakes
or Rivers or pipelines or irrigation) 957 / 3425.8
0.2794
QUESTION 4
(a.i) Probability that on any given week in a year would be no rainfall
In this aspect, Z is 1.51 which is average of overall rainfall where P(X=0)=e-1.51= 0.221
(a.ii) Probability that there would be 3 on more days of rainfall in a week
In this context, Alpha = 78.60/52 = 1.51 (Average of weekly rainfall)
X Probability Outcome
0 e-1.51 * (1.51)0/ 0! 0.221
1 e-1.51 * (1.51)1/ 1! 0.3334
<2 0.221 + 0.3334 0.5544
>=2 1 – P(X < 2)= 1 – 0.5544 0.4457
Thus, probability that there would be 2 or more days of rainfall in particular week is
articulated as 0.4457.
(b.i) Probability in given week there would be among 6 mm and 12 mm of rainfall
Mean weekly aggregate 10.58
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amount of rainfall
Standard deviation 14.61
P (-1.3136 < Z < 0.0972)
P (-infinity < Z < 0.0972) – P(-Infinity < Z < -0.3136)
0.5387 – 0.3769
0.1618
------------1
(b.ii) Amount of rainfall if 20% of the weeks have amount of rainfall or higher
With context to refer normal table, it is observed that
P (Z < 0.8416) = 0.80-----------2
With consideration of 1 and 2,
0.8416
A = 12.29
Thus, it could be concluded that for rainfall 12.29 mm, only 20% of the weeks have
significant amount of rainfall or higher.
QUESTION 5
(a) Normal Probability Plots
It follows normal distribution.
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It follows normal distribution.
It follows normal distribution.
(b) Constructing 95% confidence interval for every variable
The specified formula for 95% confidence interval for mean is stated below:
Thus, table value of corresponding z to significance level of 5% and is stated as 1.96.
Henceforth, margin of error, E is calculated with application of table value of t with standard
deviation and size of sample. The below table is replicating working of confidence interval of
955 for particular mean values.
M 21.13
Z 1.96
Sm √(6.672/140) = 0.56
μ M ± Z(sM)
μ 21.13 ± 1.96*0.56
μ 21.13 ± 1.1049
M (95%) 21.13
CI 20.0251, 22.2349
13

Temperature: The confidence interval of 95% for temperature 0 Celsius is (20.251,
22.2349). It shows that when there is extraction of more sample through sample population, there
is probability of 95% which is 95 out of 100 times, then the true mean temperature 0 Celsius
would underlie within particular interval.
Relative humidity
M 47.07
Z 1.96
Sm
√(17.188192/140)
= 1.45
μ M ± Z(sM)
μ 47.07 ± 1.96*1.45
μ 47.07 ± 2.8472
M (95%) 47.07
CI 44.2228, 49.9172
Relative humidity: The confidence interval of 95% with context to relative humidity is
(44.2228, 49.9172). It reflects that , with extraction of more sample through sample population
then there is a probability of 95% then true mean Relative humidity would lie within particular
interval.
Wind
M 4.228
Z 1.96
Sm √(1.772662/140) = 0.15
μ M ± Z(sM)
μ 4.228 ± 1.96*0.15
μ 4.228 ± 0.29364
M
(95%) 4.228
CI 3.93436, 4.52164
Wind (km/hr): The confidence interval of 95% for Wind (km/hr) is (3.93436, 4.52164).
This replicates that when more number of samples are calculated through sample population,
then there is a 95% chance (95 out of 100 times), the true mean Relative humidity (5) would
underlie within particular interval.
Area
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M 17.88864
Z 1.96
Sm √(70.882/140) = 5.99
μ M ± Z(sM)
μ 17.88864 ± 1.96*5.99
μ 17.88864 ± 11.7410728
M
(95%) 17.88864
CI 6.1475, 29.6297
Area burned (ha): The confidence interval of 95% for particular area burned (ha) is
(6.1475, 29.6297). It gives indication when more number of samples are extracted through similar
population, then there is 95% probability and true mean Area burned (ha) would underneath
within this stated interval.
(c) Testing hypothesis that more areas in forest burn when temperature is above 25 degree C than
below 25 degree c with 1% level of significance
With context to identify that more areas in the burnt forest when temperature is more than
250 Celsius than when it is lower than 250 Celsius, with performing two mean z test. The
alternative and null hypothesis is stated below:
Null Hypothesis: H0: There is no significant difference in forest area burned when temperature is
above 250 Celsius and below 250 Celsius.
Alternative hypothesis: H1: There is significant difference in forest area burned when
temperature is above 250 Celsius and below 250 Celsius.
Level of significance: 0.05
The test statistic is: Z = X1 – X2s12n1 + s22n2
Area burned (ha) > 25 degree Area burned (ha)<=25 degree
Average 29.0431
St dev 117.2704
Sample size 42
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Z test
Hypothesized variance 0
Significance level 0.05
Area burned (ha)> 25 degree
Sample size 42
Mean 29.04
St dev 117.2704
Area burned (ha)< 25 degree
Sample size 98
Mean 13.108
St dev 36.2119
Mean difference 1.6449
Standard error 18.4612
Z test 0.8632
Upper tail test
Upper Critical value 1.6449
P value 0.194
The value for z test is 0.8632 and its corresponding p value is 0.1940 > 0.05 It shows
that probability of rejecting null hypothesis is cancelled, Henceforth, it could not be concluded
that more areas in forest burn when temperature is above 250 Celsius compares to below 250
Celsius.
CONCLUSION
From the above report it could be concluded that statistics play very important role in
process of decision making of managers with business perspective. It has shown importance of
hypothesis testing and normality curve with confidence interval of 95%.
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