statistics assignment

Assignment on Statistics for management

Ans 1. Meaning of classification:

Classification is a process of arranging things or data in groups or 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”.

According to Stockton and Clark, “The process of grouping large number of individual facts and observations, on the basis of similarity among the items is called Classification”.

Meaning of 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.

The objectives of tabulation are to:

a) Simplify complex datab) Highlight important characteristicsc) Present data in minimum spaced) Facilities comparisone) Bring out trends and tendenciesf) Facilitate further analysis

Difference between Classification and Tabulation:

Classification TabulationIt is the basis for tabulation It is the basis for further analysisIt is the basis for simplification It is the basis for presentationData 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

The Structure and components of a table are as follows:

Table: 1 Percentage of PG employees based on their age and department

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1 2

9


(Age in Years)

Source:

…………….

Tab 1: Table NumberTab 2: TitleTab 3 & 4: CaptionTab 5 & 6: Subheadings (Stubs)Tab 7: Body of the tableTab 8: TotalsTab 9: Head noteTab 10: Source of note

Ans 2. Arithmetic mean is defined as the sum of all values divided by number of values and is represented by ....Arithmetic mean is also called ‘average’. It is most commonly used in 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 observations.

There are two types of Arithmetic Mean:

a) Simple Arithmetic Mean – Arithmetic mean is simply sometimes referred as ‘Mean’. For example: mean income, mean expenses, mean marks, etc. Unlike other averages, mean has to be computed by considering each and ever observation in the series. Hence, the mean cannot be found by either inspection or observation of items.

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35

7

4

10

8

9

Departments Age20-40 40 and above

Accounts 2.658 1.348Finance 2.359 1.267Personal 3.168 1.648

Production 4.251 2.159Marketing 1.459 4.359

Total 13.895 10.781


Simple Arithmetic Mean is equal to the sum of values of the variables divided by their number of observations.

b) Weighted Arithmetic Mean – Weighted Arithmetic mean is computed by considering the relative importance of each of the values to the total value. The Arithmetic mean gives equal importance to all the items of distribution. In certain cases, relative importance of items is not the same. To give relative importance, weightage may be given to variables depending on cases. Thus, weightage represents the relative importance of the items.

Solution:We have by data, X1 =75, σ1 = 5, N1 = 1000

X2=60, σ2 = 4.5, N1 = 1500Combined Mean:

N1 X1 + N2 X2 1000*75+ 1500*60X12 = = = 66

N1 + N2 1000 + 1500

Combined Standard deviation: d1 = X1 –X12 = (75-66) = 9 Therefore, d1 2 = 81d2 = X1 – X12 = (60-66) = -6 Therefore, d2 2 = 36

(N1 +N2 ) σ2 = N1 (σ12 + d1

2) + N2 (σ22 + d2

2)(1000+1500) σ2 = 1000 (52 +81) + 1500 (4.52 + 36)

2500 σ2 = 190375σ2 = 76.15

Therefore, σ = 8.73Ans 3. Solution:

Let Xi be the random variable and P (Xi) be its probability. The probabilities and indicated in the below given table:

Required Values for calculating Mean and Variance for the data

No. (Xi) P (Xi)Xi P(Xi)

1 -2 1/6 -2/62 4 1/6 4/63 -6 1/6 -6/64 8 1/6 8/6

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5 -10 1/6 -10/66 12 1/6 12/6

Total 1 1

Therefore, Expectation of Mr. Arun is E (X) = ∑ Xi P ( Xi) = 1

Ans 4. Solution:

Given that:

p= 20100

=0.2

Therefore, q = 1 - 0.2 = 0.8

N = 5

Therefore, by binomial distribution, P (X = x) = 5 C x (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 = 0.3277Therefore, 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) =5 C2 (0.8)3 (0.2)2 = (10) (0.512) (0.04) = 0.2048Therefore, the probability that exactly two employees will get the disease is 0.2048

III. The probability that more than four employees will get the diseases is given by: P (X > 4) = P(X = 5) = (0.2)5 = 0.00032Therefore, the probability that more than four employees will get the disease is 0.00032

Ans 5. Solution:

The procedure is explained in the following steps:

1. Null hypothesis Ho : p = 0.35Alternate hypothesis H1 : p < 0.35

2. Level of significance α = 0.05 ❑⇒ Ztab = -1.645 and R:z < -1.6453. Test statistics

Z= (ˆp−p)

√( pqn )√(N−nN−1 )

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4. Given ˆp = 950/3000 = 19/60 = 0.317, p = 0.35, q = 1-p = 1- p = 1- 0.35 = 0.65, N = 10,000, n = 3000

√( pqn )√( N−nN−1 )=√(0.35∗0.65

3000 )√( 10000−300010000−1 )=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 sceptical people has significantly decreased.

Ans 6. 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:

( Oi – Ei )2 ( O1 – E1 )2 ( O2 – E2 )2 ( O3 – E3 )2 ( On – En )2

x2 = Ʃ = + + +….+ Ei E1 E2 E3 En

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

Conditions for applying the Chi-Square testThe following are the conditions for using the Chi-Square test:

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 x2 test is 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.

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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 be replaced by relative frequencies or proportions.

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

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