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Quantitative Techniques-II
Notes O E 2
x 2 = (26.6)
E
Where O is the observed frequency, E is the expected frequency.
D.F = 10 - 1 = 9
Table value at 5% for 9 degree of freedom = 16.91
Since calculated value = 26.6 greater than table value of 19.19, null hypothesis rejected at 5%
level of significance.
Conclusion: The accident occurring are not uniform over a 10-week period.
Task What hypothesis, test and procedure would you use in the following situation?
1. An automobile company has manufacturing facility at two different geographical
locations. Each location manufactures two-wheelers of a different model. The customer
wants to know if the mileage given by both the models is the same or not. Samples
of 45 numbers may be taken for this purpose.
2. A company has 22 sales executives. They underwent a training programme. The test
must evaluate whether the sales performance is unchanged or improved after the
training programme.
3. A company has three categories of managers:
a. With professional qualifications but without work experience.
b. With professional qualifications accompanied by work experience.
c. Without professional qualifications but with work experience.
Self Assessment
Fill in the blanks:
1. There are two types of tests: Small Sample Tests and ………………….
2. With the help of ……………. test, we will come to know whether two or more attributes
are associated or not.
13.2.3 ANOVA
(a) ANOVA: It is a statistical technique. It is used to test the equality of three or more sample
means. Based on the means, inference is drawn whether samples belongs to same
population or not.
(b) Conditions for using ANOVA:
(1) Data should be quantitative in nature.
(2) Data normally distributed.
(3) Samples drawn from a population follows random variation.
(c) ANOVA can be discussed in two parts:
(1) One-way classification
(2) Two and three-way classification.
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