An experiment is conducted to determine whether there is a difference among the

Question

An experiment is conducted to determine whether there is a difference among the mean increases in growth produced by five strains (A, B, C, D and E) of growth hormones for plants. The experimental material consists of 20 cuttings of a shrub (all of equal weight), with four cuttings randomly assigned to each of the five different strains. The increases in weight for each cutting along with the sample mean and sample standard deviation of each group are given in the table below.

It is also given that the overall mean = 17.15.

Compute the following:
(a) SSTR=  
(b) SSE =  
(c) MSTR =  
(d) MSE =  
(e) F =

A B C D E Plant 1 14 27 20 12 15 Plant 2 14 20 23 18 10 Plant 3 15 22 24 13 10 Plant 4 10 19 25 17 15 Mean 13.25 22 23 15 12.5 Standard Dev. 2.2174 3.5590 2.1602 2.9439 2.8868

Explanation / Answer

Step 1                          
   Null Hypothesis Ho : µ1 =µ2 =µ3 =µ4 =µ5           
   Alternative Hypothesis : µ1 µ2 µ3 µ4 µ5           
Step 2                          
   Degrees of freedom between = k - 1 = 5 - 1 = 4                      
   Degrees of freedom Within = n - k = 20 - 5 = 15                      
                          
   Degrees of freedom Total F( k-1,n - k,) at 0.05 is = F Crit = 3.056                      
                          
Step 3                          
   Grand Mean = G / N = 13.25+22+23+15+12.5 / 5 = 17.15                      
    SST = ( Xi - GrandMean)^2 = (14-17.15)^2 + (14-17.15)^2 + (15-17.15)^2 + ……..& so on = 514.55                      
   SS Within = (Xi - Mean of Xi ) ^2 =,(14-13.25)^2 + (14-13.25)^2 + (15-13.25)^2 + ……..& so on = 92.75                      
                          
   SS Between = SST - SS Within = 514.55 - 92.75 = 421.8                      
Step 4                          
   Mean Square Between = SS Between / df Between = 421.8/4 = 105.45                      
   Mean Square Within = SS Within / df Within = 92.75/15 = 6.183                      
                          
Step 5                          
   F Cal = MS Between / Ms Within = 105.45/6.183 = 17.054                      
   We got |F cal| = 17.054 & |F Crit| =3.056                      
                          
MAKE DECISION                          
   Hence Value of |F cal| > |F Crit|and Here We Reject Ho                      

Treatments ONE WAY ANOVA Mean = X /n A 14 14 15 10 13.25 B 27 20 22 19 22 C 20 23 24 25 23 D 12 18 13 17 15 E 15 10 10 15 12.5

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