Master of
Business Administration – MBA Semester 1
MB0040 –
Statistics for Management – 4 Credits
(Book ID: B1129)
Assignment
Set – 2 (60 Marks)
Q1). Explain the following
terms with respect to Statistics: (i) Sample, (ii) Variable, (iii) Population
Ans:
Universe or Population
Statistical
survey or enquiries deal with studying various characteristics of unit
belonging to a group. The group consisting of all the units is called Universe
or Population. The figure 7.1 illustrates the population.
Sample
Sample
is a finite subset of a population. A sample is drawn from a population to
estimate the characteristics of the population. Sampling is a tool which
enables us to draw conclusions about the characteristics of the population. The
figure 7.3 illustrates the population and sample.
Q2). What are the types of
classification of data?
Ans:
Types of classification
The
important types of classification are:
Geographical
classification
Data
classified according to region is geographical classification.
Chronological
classification
Data
classified according to the time of its occurrence is called chronological
classification.
Conditional
classification
Classification
of data done according to certain conditions is called conditional classification.
Qualitative
classification
Classification
of data that is immeasurable is called qualitative classification. For example,
sex of a person, marital status, color and others.
Quantitative
classification
Classification
of data that is measurable either in discrete or continuous form is called
quantitative classification.
Statistical
Series
Data
is arranged logically according to size or time of occurrence or some other
measurable or non-measurable characteristics.
Q3). Find the (i)
arithmetic mean and (ii) range if the following data: 15, 17, 22, 21, 19,
26,
20
Ans: The arithmetic mean is given by:
Therefore, the arithmetic mean is 20.
Q4). Suppose two houses in
a thousand catch fire in a year and there are 2000 houses in a village. What is the probability that:
i)
None of the houses catches fire
ii)
At least one house catches fire
Ans:
Given the probability of a house catching fire is:
and
Therefore,
the required probabilities are calculated as follows:
i.
The probability that none
catches fire is given by:
Therefore,
the probability that none of the houses catches fire is 0.01832.
ii.
The probability that at least
one catches fire is given by:
Therefore,
the probability that at least one house catches fire is 0.98168.
Q5). (i) what are the characteristics of Chi-square test?
Characteristics of Chi-Square test
The following are the
characteristics of Chi-Square test (c2 test).
· The c2 test is based on
frequencies and not on parameters
· It is a non-parametric test
where no parameters regarding the rigidity of population of populations are
required
· Additive property is also found
in c2 test
· The c2 test is useful to test
the hypothesis about the independence of attributes
· The c2 test can be used in complex
contingency tables
· The c2 test is very widely used
for research purposes in behavioral and social sciences including business
research
· It is defined as:
where, ‘O’ is the observed frequency and
‘E’ is the expected frequency.
(ii) The table below gives the production
in three shifts and the number of defective goods that turned out in three
weeks. Test at 5% level of significance whether weeks and shifts are
independent.
Shift
|
1 Week
|
2 Week
|
3 Week
|
Total
|
|
|
|
|
|
I
|
15
|
5
|
20
|
40
|
II
|
20
|
10
|
20
|
50
|
III
|
25
|
15
|
20
|
60
|
Total
|
60
|
30
|
60
|
150
|
Solution: The table below displays the
observed and expected values required to calculate c2.
Observed Value (O)
|
Expected Value (E)
|
(O – E)2
|
|
15
|
40 x 60 /150 = 16
|
1
|
0.0625
|
20
|
50 x 60/150 = 20
|
0
|
0
|
25
|
60 x 60/150 = 24
|
1
|
0.0417
|
5
|
40 x 30/150 = 8
|
9
|
1.125
|
10
|
50 x 30/150 = 10
|
0
|
0
|
15
|
60 x 30/150 = 12
|
9
|
0.75
|
20
|
40 x 60/150 = 16
|
16
|
1
|
20
|
50 x 60 /150 = 20
|
0
|
0
|
20
|
60 x 60/150 = 24
|
16
|
0.6667
|
|
|
c2
|
3.6459
|
|
|
|
|
The steps followed to calculate c2 are described below.
1. Null hypothesis ‘Ho’: The week and shifts are
independent
Alternate hypothesis ‘HA’: The week and shifts are
dependent
2. Level of Significance is 5% and D.O.F (3 – 1) (3 –
1) = 4
3. Test Statistics
4. Test c2cal = 3.6459
5. Conclusion: Since c2cal (3.6459) < c2tab (9.49),
‘Ho’ is accepted. Hence, the attributes ‘week’ and ‘shifts’ are independent.
Q6). Find Karl
Pearson’s correlation co-efficient for the data given in the below table:
X
|
20
|
16
|
12
|
8
|
4
|
Y
|
22
|
14
|
4
|
12
|
8
|
Solution:
X
|
Y
|
X2
|
Y2
|
XY
|
|
|
|
|
|
20
|
22
|
400
|
484
|
440
|
16
|
14
|
256
|
196
|
224
|
12
|
4
|
144
|
16
|
48
|
8
|
12
|
64
|
144
|
96
|
4
|
8
|
16
|
64
|
32
|
åX = 60
|
åY = 60
|
åX2 = 880
|
åY2 = 904
|
åXY = 840
|
Hence, Karl Pearson’s correlation
coefficient is 0.70.
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