6 a) What is analysis of variance? What are the assumptions of this technique? b) Three samples below have been obtained from normal populations with equal variances. Test the hypothesis at 5% level that the population means are equal. A B C 8 7 12 10 5 9 7 10 13 14 9 12 11 9 14 [The table value of F at 5% level of significance for 1 = 2 and 2 = 12 is 3.88] a) Meaning and Assumptions b) Formulas/Calculation/Solution to the problem
Answer:- Analysis of Variance (ANOVA) is useful in such situations as comparing the mileage achieved by five different brands of gasoline, testing which of four different training methods produce the fastest learning record, or comparing the first-year earnings of the graduates of half a dozen different business schools. In each of these cases, we would compare the means of more than two samples. Hence, in most of the fields, such as agriculture, medical, finance, banking, insurance, education, etc., the concept of ANOVA is used.
In statistical terms, the difference between two statistical data is known as variance. When two data are compared for any practical purpose, their difference is studied through the techniques of ANOVA. With the analysis of variance technique, we can test the null hypothesis and the alternative hypothesis.
Null hypothesis, ‘H0’: All sample means are equal. Alternate Hypothesis, ‘H1’: All sample means are not equal or at least one of the samples means differ.
Assumptions for study of ANOVA
The underlying assumptions for the study of ANOVA are:
- Each of the samples is a simple random sample
- Population from which the samples are selected are normally distributed
- Each of the samples is independent of the other samples
- Each of the population has the same variation and identical means
- The effect of various components are additive.
- b) Three samples below have been obtained from normal populations with equal variances. Test the hypothesis at 5% level that the population means are equal.
[The table value of F at 5% level of significance for V1 = 2 and V2 = 12 is 3.88]
Solution: – let Ho: there is no significant different in the means of three samples
| X1 | X2 | X3 |
| 8
10 7 14 11 |
7
5 10 9 9 |
12
9 13 12 14 |
| x1=50 | x2=40 | x3=60 |
T=Sum of all observations = 150
Correction Factor = T2/N = 150*2/15=1500
SST(Total sum of the Squares ) = sum of Squares of all observations – T2/N
=(8*2+7*2+12*2+10*2+…..+14*2) – 1500=1600-1500=100
Sum of the Square of error between the columns (samples):
Sum of the square of error between the columns (Samples);
SSE=SST-SSC=100-40=60
Variance between samples:
Variance within the samples:
The degree of freedom = (k-1, n-k) = (2.12)
[K is the number of columns and n is the total number of observations]
ANOVA Table
