1. Quantitative Methods
ANOVA

2. Analysis of variance
Analysis of variance is a technique that can be used to decide if the means of two or more populations are equal.
Let us consider k populations with
means μ1, μ2, …, μk and
standard deviation σ1, σ2, …, σk respectively.
We want to test the hypothesis
H0: μ1 = μ2 = … = μk against an alternative H1 that population means are not all equal.

3. Analysis of variance
The data for analysis of variance are obtained by taking a sample from each population and computing the sample mean and variance for each sample.
For the ith population (i = 1,2, …, k),
Let ni be the sample size,
x¯i be the sample mean and
Si2 be the sample variance based on the divisor ni -1.

4. Test Statistics The test statistic is F = MST / MSE where n1 n2 nk k k
MSE = ∑ (xi – x¯1)2 + ∑ (xi – x ¯ 2)2…+ ∑ (xi – x ¯ k)2 /( n1 + n2 + … + nk – k)
i = i = i = 1
MST = [ n1 (x ¯ 1 – x = )2 + n2 (x ¯ 2 – x =)2 + … + nk (x ¯ k – x =)2 ] / (k – 1)
k k
x= is mean of sample means = ∑ ni x ¯ i / ∑ ni
i = i = 1
MSE =(n1-1)S21 +(n2-1)S22…+(nk-1)S2k /( n1 + n2 + … + nk – k)

5. Test Statistics
Under the assumption that Ho is true, the test statistic F follows
F – distribution with
(k – 1)and ( n – k )degrees of freedom,
Where
k represents number of groups or populations and
n is total sample size k
n = ∑ ni
i = 1

6. Procedure to Test Equality of Means
The procedure to test the hypothesis
H0: μ1=μ2=…= μk is as follows:
Making use of the sample data compute the value of test statistic F.
Specify the level of significance.
Find out the critical value from the F – table at k – 1 and n – k degrees of freedom and at the specified level of significance.
Reject Ho if the computed value of test statistics F is greater than the critical value. Otherwise accept it.

7. One Way ANOVA: Completely Randomized Experimental Design
ANOVA Table
Sources of error df SS MS F
Trtments k-1 SST MST=SST/(k-1) MST/MSE
Error n-k SSE MSE=SSE/(n-k)
Totals n-1 TSS
Where
SST= [ n1 (x ¯ 1 – x = )2 + n2 (x ¯ 2 – x =)2 + … + nk (x ¯ k – x =)2 ]
n n nk
SSE = ∑ (xi – x¯1)2 + ∑ (xi – x ¯ 2)2…+ ∑ (xi – x ¯ k)2
i = i = i = 1
= (n1-1)S21 +(n2-1)S22… (nk-1)S2k
Varsha Varde 7 7

8. TOTAL SUM OF SQUARES
= SUM OFSQUARES DUE TO TREATMENT + SUM OF SQUARES DUE TO ERROR
= n
TSS = ∑ (xi – x = )2
i = 1

9. Assumptions
There are three basic assumptions that must be satisfied before analysis of variance can be used:
(a) The samples are independent random samples,
(b) The samples are drawn from normal population,
(c) The populations have equal variances
σ21 = σ22 …, σ2k.
However, the test holds good for samples drawn from population which are not highly skewed and for which variances are approximately equal.