MB0042 – Statistics for Management

Question 1: Distinguish between Classification and Tabulation. Explain the structure and components of a Table with an example.

Answer 1:

Meaning of Classification and Tabulation:

Classification:

Classification is a process of arranging things or data in groups and classes according to their resemblances and affinities.

According to Secrist, “Classification is the process of arranging data into sequences and groups according to their common characteristics or separating them into different but related parts”. Tabulation:

Tabulation follows classification. It is a logical or systematic listing of related data in rows and columns. The row of a table represents the horizontal arrangement of data and column represents the vertical arrangement of data. The presentation of data in tables should be simple, systematic and unambiguous.

Differences between Classification and Tabulation:

Classification TabulationIt is the basis for tabulation. It is the basis for further analysis.It is the basis for simplification. It is the basis for presentation.Data is divided into groups and sub-groups on the basis of similarities and dissimilarities.

Data is listed according to a logical sequence of related characteristics.

Structure and Components of a Table with an example:

Table 1: Percentage of P.G. Employees based on their Age and Departments

(Age in Years)

Source: ………….

1 2

9

4

3

7

5

6

8

10100

1: Table number:

Table number is to identify the table for reference. When there are many tables in an analysis, then table numbers are helpful in identifying the tables.

Tab 2: Title: Title indicates the scope and the nature of contents in a concise form. Title should not be lengthy.

Tab 3 and Tab 4: Captions

Captions are the headings and subheadings describing the data present in the columns.

Tab 5 and Tab 6: Stubs

Stubs are the headings and subheadings of rows.

Tab 7: Body of the table

Body of the table contains numerical information.

Tab 8: Totals

The sub-totals for each separate classification and a general total for all combined classes should be given at the bottom or right side of the figures whose totals are taken. Ruling and spacing separate columns and rows.

Tab 9: Head note

Head note is given below the title of the table to indicate the units of measurement of the data and is enclosed in brackets.

Tab 10: Source note

Source note indicates the source from which data is taken.

Question 2 : (a) Explain Arithmetic mean. (b) The mean wage is Rs. 75 per day, SD wage is Rs. 5 per day for a group of 1000 workers and the same is Rs. 60 and Rs. 4.5 for the other group of 1500 workers.Find the mean and standard deviation for the entire group.

Answer 2:

Arithmetic mean :

Arithmetic mean can be defined as the sum of all values divided by number of values and is represented by X. Arithmetic mean is also called ‘average’. It is most commonly used measures of central tendency. Arithmetic Mean of a series is the value obtained by adding all the observations of a series and dividing this total by the number of observation.

Solution:

Group - 1 Group -2

Mean wages 75 60No. of workers 1000 1500

N1 = 1000 N2 = 1500 X1 = 75 & X2 = 60 σ1 = 5, σ2 = 4.5

Combined mean = X12 = N1 X1 + N1X2

N1 + N2

= 1000 × 75 + 1500 × 60 1000 + 1500 X12 = Rs. 66

Now, Standard deviation d1 = X1 – X12, d2 = X2 – X12

d1 = (75 – 65) = 9 d12 = (9)2 = 81

d2 = (60 – 66) = -6 d2 2 = (-6)2 = 36

σ2 (N1 + N2) = N1(σ2 + d1 2) + N2 (σ2

2 + d22)

2500 σ2 = 1000(52 + 81) + 1500(4.52 + 36)

2500 σ2 = 106000 + 84375

2500 σ2 = 190375

σ2 = 190375 2500

σ2 = 76.15

σ = 8.726

Question 3: Mr. Arun and Mr. Bhandari play a game. If Mr. Arun picks up an even number from1 to 6, Mr. Bhandari will pay him double the amount equal to picked up number. If Mr. Arun picks up an odd number then he has to pay amount equal to double the picked up number. What is Mr. Arun’s expectation?

Answer 3:

Solution :

Let XI be the random variable and P(XI) be its probability. The probabilities are indicated in the table.

Required Values for Calculating Mean and Variance for the Data

No. (XI) P(XI) XI

P(XI) 1 -2 1/6 -2/6 2 4 1/6 4/6 3 -6 1/6 -1 4 8 1/6 8/6 5 -10 1/6 -10/6 6 12 1/6 12/6 Total 1 1

Expectation of Mr. Arun is E(X) = ∑ XI P (XI) = 1.

1. E(X) = -2 × 1/6 = -2/6

2. E(X) = 4 × 1/6 = 4/6

3. E(X) = -6 × 1/6 = -1

4. E(X) = 8 × 1/6 = 8/6

5. E(X) = -10 × 1/6 = -10/6

6. E(X) = 12 × 1/6 = 12/6.

Question 4: The probability that an employee will get an occupational disease is 20%. In a firm having five employees, what is the probability that:

i) None of the employees get the disease

ii) Exactly two will get the disease.

iii) More than four will contract the disease.

Solution 4 :

Given that :

P = 20/ 100 = 0.2

q = 1 – 0.2 = 0.8

n = 5

Therefore, by binomial distribution, P(X = x) = 5 Cx (0.8)5 – x (0.2)x

i) The probability that none of the employees get the disease is given by :

P(X = 0) = (0.8)5 = 0.3277

Therefore, the probability that none of the employees get the disease is 0.3277.

ii) The probability that exactly two employees will get the disease is given by:

P(X = 2)= 5C2 (0.8)3(0.2)2 = (10)(0.512)(0.04) = 0.2048.

Therefore, the probability that exactly two employees will get the disease is 0.2048.

iii) The probability that more than four employees will get the disease is given by:

P(X > 4) = P(X = 5) = (0.2)5 = 0.00032.

Therefore, the probability that more than four employees will get the disease is 0.00032.

Question 5: Microsoft estimated that out of 10,000 potential software buyers, 35% wait to purchase the new OS Windows Vista, until an upgrade has been released. After and advertising campaign to reassure the public was released, Microsoft surveyed 3000 buyers and found 950 who are still skeptical. At 5% level of significance, can the company conclude that the population of skeptical people had decreased?

Solution 5 : The procedure is explained in the following steps:

1. Null hypothesis Ho : p = 0.35

Alternate hypothesis H1 : p < 0.35

2. Level of significance α = 0.05 = Ztab = -1.645 and R : Z < -1.645

3. Test statistics

Z = (P ^ - P) √(pq)√ (N - n) n N – 1

4. Given p^ = 950/ 3000 = 19/60 = 0.317, p = 1 - p = 1 – 0.35 = 0.65,

N = 10,000 , n = 3000

√(pq) √(N – n ) = √(0.35 × 0.65)√(10,000 - 3000) n N – 1 3000 10,000 – 1 = 0.0073

Zcal = (0.317 - 0.35) 0.0073 = -4.52

5. Conclusion : Since Zcal (-4.52) < Ztab (-1.645)and is in the rejection region, HO is rejected. At 5% level of significance, we conclude that the proportion of skeptical people has significantly decreased.

Question 6: Explain Chi-square test and the conditions for applying chi-square test.

Answer 6:

Chi – Square test:

The Chi-square test is one of the most commonly used non-parametric tests in statistical work. The Greek Letter X2 is used to denote this test. X2 describe the magnitude of discrepancy between the observed and the expected frequencies. The value of X2 is calculated as:

Where, O1, O2, O3….On are the observed frequencies and E1, E2, E3…En are the corresponding expected or theoretical frequencies.

Uses of Chi-Square test :

The X2 test is used broadly to:

Test goodness of fit for one way classification or for one variable only.

Test independence or interaction for more than one row or column in the form of a contingency table concerning several attributes.

Test population variance X2 through confidence intervals suggested by X2 test.

Characteristics of Chi – square test:

The X2 test is based on frequencies and not on parameters.

It is a non parametric test where no parameters regarding the rigidity of populations are required.

Additive property is also found in X2 test.

While testing whether the observed frequencies of certain outcomes fits with expected frequencies defined by a theoretical distribution, the X2 value defined here follows X2

distribution:

Conditions for applying the Chi – Square:

1. The frequencies used in Chi-Square test must be absolute and not in relative terms.

2. The total number of observations collected for this test must be large.

3. Each of the observations which make up the sample of this test must be independent of each other.

4. As test is X2 based wholly on sample data, no assumption is made concerning the population distribution. In other words, it is a non parametric-test.

5. X2 test is wholly dependent on degrees of freedom. As the degrees of freedom increase, the Chi-Square distribution curve becomes symmetrical.

6. The expected frequency of any item or cell must not be less than 5, the frequencies of adjacent items or cells should be polled together in order to make it more than 5.

7. The data should be expressed in original units for convenience of comparison and the given distribution should not be replaced by relative frequencies or proportions.

8. This test is used only for drawing inferences through test of the hypothesis, so it cannot be used for estimation of parameter value.