Quantitative Methods ANOVA

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.

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.

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 = 1 i = 1 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 = 1 i = 1 MSE =(n1-1)S21 +(n2-1)S22…+(nk-1)S2k /( n1 + n2 + … + nk – k)

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

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.

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 ] n1 n2 nk SSE = ∑ (xi – x¯1)2 + ∑ (xi – x ¯ 2)2…+ ∑ (xi – x ¯ k)2 i = 1 i = 1 i = 1 = (n1-1)S21 +(n2-1)S22…..... +(nk-1)S2k Varsha Varde 7 7

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

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.