1. ANOVA Analysis of Variance or The F distribution

2. 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 204015 192817 213020 25 24 29 283632 243825 184127 4428 1.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 ?

3. 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

4. Solution Reduction in serum cholesterol levels Treatment ATreatment BTreatment c 204015 192817 213020 25 24 29 283632 243825 184127 4428 nj899 ∑ nj (N) = 26 Tj20+19+21+25+24+28+24+18 = 179 40+28+30+25+24+36+38 +41+44+9 = 306 15+17+20+25+29+32+2 5+27+28 = 218 ∑ Tj (G) = 703 X2X2 20 2 +19 2 +21 2 +25 2 ……+18 2 = 4087 40 2 +28 2 +30 2 +…..+44 2 = 10847 152+172+202+….+28 2 = 5547 ∑X² = 20471 Tj 2 /nj179 2 /8 = 4005.13306 2 /9 =10404.00218 2 /9 =5280.44 ∑Tj²/nj = 19689.57

5. ABCTotal nj899∑nj (N) = 26 Tj179306218∑ Tj (G) = 703 X²4087108475547∑X² = 20471 Tj²/nj4005.1310404.005280.44∑Tj²/nj = 19689.57 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

6. Sum of Squares: SS Within (SSW), SS Between (SSB), SS Total (SST)  SSB = ∑( T²j/ nj ) – (T²)/N 19689.57 – 703²/26 = 681.53  SSW = ∑ X²ij - ∑ (T²j/ nj ) 20471 – 19689.57 =781.43  SST = 681.53 + 781.43 = 1462.96 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 = 681.58/2= 340.77  MSqW = SSW/d.f within = 781.43/23 = 33.98  F-Ratio: F = MS Between / MS Within = 340.77/ 33.98= 10.03

7. 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 = 681.53MSqB =SSB/(K -1) 681.53/2= 340.77 MSqB MSqW 340.77 33.98 = 10.03 0.10 Within the Groups N – K 26 – 3=23 SSW = 781.43 MSqw= SSW/(N – k) 781.43/23 = 33.98 TotalN – 1 26- 1= 25 SST = 1462.96

8. 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- 4.10267 B34.009A & C- 0. 64965 C24.229B & C3.559306

9. Statistical decision : Student t distribution table value for α = 0.05 and d.f. 23 is 2.069. 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.