SMU MBA ASSIGNMENTS

MB0040 — Statistics for Management

MB0040 — Statistics for Management

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 Ã¥X = 60 Ã¥Y = 60 Ã¥X2 = 880 Ã¥Y2 = 904 Ã¥XY = 840

Hence, Karl Pearson’s correlation coefficient is 0.70.