Content INTRODUCTION Basic Principle of ANOVA Assumptions in ANOVA ANOVA Techniques EXAMPLE OF ANOVA
INTRODUCTION ANOVA stands for Analysis of Variance . It was introduced by Sir R.A Fisher in 1920. It is a statistical method used to analyse the differences between the means of two or more samples or treatments . ANOVA is used to test the differences among different groups of data for homogeneity. The basic principle of ANOVA is to test for differences among the populations by examining amount of variation within each of these samples relative to amount of variation between the samples. ANOVA is an extension of T test as T-test is not suitable to test for significance of differences among more than two sample means
However, conducting multiple t-tests can lead to severe inflation of the Type I error rate. ANOVA can be used to test differences among several means for significance without increasing the Type I error rate.
Assumptions in ANOVA Population distribution is normal Samples are random and independent Homogeneity of sample variance
ANOVA TECHNIQUES Two types:- One-way ANOVA Two-way ANOVA One-way ANOVA can only be used when investigating a single factor and a single dependent variable .
The F-test in One-Way ANOVA
Basics It tests the null hypothesis H₀:There is no significant difference between the means of all groups.(all groups are same) H₀= μ₁ =, μ₂ =, μ₃ =….= μ ĸ Where μ =group mean ,K=no. of group ALTERNATIVE HYPOTHESIS H Α :There are at least two groups means that are statistically significantly different from each other. H A : μ₁≠μ₂≠μ₃≠ …≠ μκ
One way ANOVA (single factor)
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ANOVA TABLE (ONE-WAY)
EXAMPLE
CONT… STEP1: CALCULATE MEAN OF EACH SAMPLE.
CONT… STEP2: CALCULATE GRAND SAMPLE MEAN (mean of sample means)
CONT… STEP3: Calculate sum of square between the samples and sum of square within the samples.
CONT… STEP4: Calculate sum of square (SS) for total variance. SS(total)= SS Between the samples+ SS within the samples. SS(total )= 8+24 =32
CONT… STEP5: Setting up ANOVA Table.
conclusion The above table shows that calculated value of F is less than the table value of 4.26 at 5% level with d.f being v 1 = 2 and v 2 =9 .