ANOVA Analysis of Variance or The F distribution

Example A randomized clinical trial carried out to compare the effect of three treatments A, B and C for the reduction of serum cholesterol levels in obese patients. The reduction in serum cholesterol levels after the intervention is shown in the table Reduction in serum cholesterol levels Treatment ATreatment BTreatment c Can we conclude from these data there is a difference in effect of treatments used for the reduction of serum cholesterol level? 2. Which treatment is most effective ?

K = number of groups n = Sample size of single groups(n 1,n 2.n 3 …..) N = Total sample size T = sum of scores in a single group(T 1, T 2, T 3 …..) G = sum of all scores or Grant total

Solution Reduction in serum cholesterol levels Treatment ATreatment BTreatment c nj899 ∑ nj (N) = 26 Tj = = = 218 ∑ Tj (G) = 703 X2X ……+18 2 = … = … = 5547 ∑X² = Tj 2 /nj179 2 /8 = /9 = /9 = ∑Tj²/nj =

ABCTotal nj899∑nj (N) = 26 Tj ∑ Tj (G) = 703 X² ∑X² = Tj²/nj ∑Tj²/nj = Now we will calculate :  Sum of Squares: SS Within, SS Between, SS Total  Degrees of freedom: df Within, df Between,df Total  Mean Squares: MS Within, MS Between  F-Ratio: F = MS Between / MS Within

Sum of Squares: SS Within (SSW), SS Between (SSB), SS Total (SST)  SSB = ∑( T²j/ nj ) – (T²)/N – 703²/26 =  SSW = ∑ X²ij - ∑ (T²j/ nj ) – =  SST = = Degrees of freedom: df Within, df Between, df Total  df Within = N-1=3-1 = 2  df Between = K- N= 26 – 3 = 23  df Total = K- 1 = 26 – 1= 25 Mean Squares: MS Within (MSqW), MS Between (MSqB)  MSqB =SSB/d.f between = /2=  MSqW = SSW/d.f within = /23 =  F-Ratio: F = MS Between / MS Within = / 33.98= 10.03

Analysis of ANOVA can be presented in a table Source of Variation Degrees of freedom Sum of squares(SSq) Mean of squares(MSq)Test Statistic(F) P- valu e Between the Groups K -1 = 3 -1 =2 SSB = MSqB =SSB/(K -1) /2= MSqB MSqW = Within the Groups N – K 26 – 3=23 SSW = MSqw= SSW/(N – k) /23 = TotalN – = 25 SST =

Multiple comparisons between treatment Since Ho : null hypothesis is rejected. We are interested to know which pair of treatments means significantly differ. H01=µ1 =µ2 (X1 – X2) – (µ1- µ2) Test statistic = √Error mean square(1/n1+1/n2 ) TreatmentMeannComparisonTS A22.388A & B B34.009A & C C24.229B & C

Statistical decision : Student t distribution table value for α = 0.05 and d.f. 23 is There is no significantly different from the treatment A & C, where as treatment B is significantly different from the treatment A & C Conclusion : since the average reduction in serum cholesterol levels is highest in treatment B and which is significantly different from the treatment A and C. Hence we conclude that treatment B is most effective for the reduction of serum cholesterol levels.