Business Statistics: Analyzing Data for Improved Decision-Making

Desklib provides past papers and solved assignments for students. This report explores statistical methods for business decision-making.

Statistics for business
1

Table of Contents
Introduction......................................................................................................................................3
Question 1........................................................................................................................................4
Question 2........................................................................................................................................8
Question 3......................................................................................................................................12
Question 4......................................................................................................................................17
Conclusion.....................................................................................................................................22
References......................................................................................................................................23
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Introduction
In this report, the importance of the statistics will be discussed for a company. The report will be
divided into four parts. In the first part of the report, it will discuss the analysis of the two
variable will be made in which various interpretation will be made. For the second part, it will be
discussing the test analysis which will include the hypothesis on which analysis will be made. In
the third part of the report, it will compute all the data by using mean, mode, and median various
techniques will be shown through it. And in the last part of the report, it will be showing the
various calculation for profitability and how it could help the organisation to make an analysis.
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Question 1
A.
Calculation of covariance between x and y
Years of
experience
Salary in
1000
x y
5 20
3 23
7 15
9 11
2 27
4 21
6 17
8 14
Formula for covariance = 1/n ∑
i=0
n
( xi−x ) ( yi− y )
n−1
= -11.125
4

B.
Reason for a negative covariance
The covariance is used to calculate the overall variation of the two random elements from their
assumed outcome, as it just shows the direction in which analysis is going. If the element's
variances are related to each other than outcome are positive covariance. And if the variables are
not related to each other than outcome are negative covariance. In negative covariance, the two
variables always tend to move in the opposite direction.
C.
Calculation of coefficient of correlation, the relation between x and y
Years of
experience
Salary in
1000
x y
5 20
3 23
7 15
9 11
2 27
4 21
6 17
8 14
Formula of coefficient of correlation = r =n ( ∑ xy )−∑ x−∑ y / √ {¿
= -0.991093772
The coefficient of correlation is used to compute the relationship between two variables. The
variables shown in the question results in the negative outcome which is -0.99 approx. which is
near to no correlation or no relationship. In the above calculation, the two variables that are taken
for analysis are the year of experience and the salary, the analysis is near to 0 but still, its shows
a negative correlation which means that the two variables are related to each other but to some
extent.
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D.
Reason for negative correlation
If the two variables are negative in nature, then it shows inverse relation. The data represented in
the scattered graph represent inverse relation as the two variables are away from each other. This
shows that the outcome is not relevant to the analysis.
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Question 2
A.
Reason to test the claim and analyse the actual percentage is not 10%. The actual null and
alternative hypothesis –
Null hypothesis – 10% of the users of a certain sinus drug experience drowsiness.
Alternative hypothesis – 10% of the users of a certain sinus drug does not experience drowsiness.
B.
If the significance level is at 5% what is the analysis
t-Test: Two-Sample Assuming Equal
Variances
Variable
1
Variabl
e 2
Mean 0.1 0.09
Variance #DIV/0! #DIV/0!
Observations 1 1
Pooled Variance 65535
Hypothesized Mean Difference 0
df 0
t Stat 65535
P(T<=t) one-tail #NUM!
t Critical one-tail #NUM!
P(T<=t) two-tail #NUM!
t Critical two-tail #NUM!
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C.
Construct a 95% confidence interval estimate of the population proportion of the users for the
drug who are experiencing drowsiness –
Anova: Single
Factor
SUMMARY
Groups Count Sum Averag
e
Varianc
e
Column 1 1 0.1 0.1 #DIV/0!
Column 2 1 0.09 0.09 #DIV/0!
ANOVA
Source of
Variation
SS df MS F P-
value
F crit
Between Groups 5E-05 1 5E-05 65535 #NUM! #NUM!
Within Groups 0 0 65535
Total 5E-05 1
D.
How to use a confidence interval to test the hypothesis
The confidence level is a range of data, which concludes the group of unknown data for analysis.
If a random sample is concluded for many times then it shows a certain amount of percentage in
the interval of confidence which includes meaning population.
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Question 3
A.
Calculation of mean, median, and mode
730 730 730 930 700 570
690 1,030 740 620 720 670
560 740 650 660 850 930
600 620 760 690 710 500
730 800 820 840 720 700
Mean 725
Median 715
Mode 730
Mean is calculated to analyse the average number of output.
Median is calculated to analyse the middle of the raw data as an output
The mode is computed to analyse the number of times that particular raw data is occurring.
B
Measure of central tendency
The measure of central tendency is a brief of the statistical methods which are shown from the
set of the raw data. This computes the values that fall in the distribution table. The main methods
used are mean, median and mode.
C
Calculate the standard deviation
730 730 730 930 700 570
690 1,030 740 620 720 670
560 740 650 660 850 930
9

600 620 760 690 710 500
730 800 820 840 720 700
Std.
Dev
120.307
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D
Are the outliers or unusual data values
The outliers is a data that is different from the set of data available for analysis. The outlier data
shows that the data is inappropriate or data occurred for analysis is not usable as it can affect the
rest of the data. For the given set of data, the outlier will be considered as 500.
730 730 730 930 700 570
690 1,030 740 620 720 670
560 740 650 660 850 930
600 620 760 690 710 500
730 800 820 840 720 700
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E
Using Empirical rule, do you think data could be from a normal population
730 730 730 930 700 570
690 1,030 740 620 720 670
560 740 650 660 850 930
600 620 760 690 710 500
730 800 820 840 720 700
Mean 725
Std.Dev 120.3079
Results
Approx. 99.7% of the data lies between ± 3 SD, or between 364.0763 and 1085.9237
Approx. 95% of the data lies between ± 2 SD, or between 484.38419999999996 and
965.6158
Approx. 68% of the data lies between ± 1 SD, or between 604.6921 and 845.3079
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Question 4
A
Calculate profitability A and O
Service A = 60%
On-time rate = 80%
P (A & O) = P (A)*P (O)
= 60%*80%
= (60/100)*(80/100)
= (0.6)*(0.8)
= 0.48
B
Calculate the profitability that was delivered on time
 P (A U O1) = P (A) + P (O1) + P (A*O1)
= 60% + 80% + 60%*80%
= (60/100) + (80/100) + (60/100)*(80/100)
= (0.6) + (0.8) + (0.6*0.8)
= 1.88
 P (B U O2) = P (B) + P (O2) + P (B*O2)
= 30% + 60% + 30%*60%
= (30/100) + (60/100) + (30/100)*(60/100)
= (0.3) + (0.6) + (0.3*0.6)
= 1.38
 P (C U O3) = P (C) + P (O3) + P (C*O3)
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