Question

An experiment was conducted see the effect on fertilizer on production of Mameys. H0 : fertilizers have no impact on mamey production. H1 : fertilizers have an impact on mamey production. There are four treatment (a=4); three types of fertilizer and the control. Each treatment has four replicates (n=4). The number of mameys produced is given in the table below.

A. Using these data complete a 1-Way ANOVA table.

F1: 1, 2, 6, 11

F2: 2, 4, 2, 4

F3: 12, 4, 2, 6

Con: 3, 3, 1, 1

B. What is the proportion of explained variance for this treatment?

Answer 1

    We use the Excel Data Analysis Tool for ANOVA single factor with repetition                  
   The null and alternative hypotheses are :                  
A)   Ho :  Fertilizers have no impact on mamey production                  
H1 :  Fertilizers have an impact on mamey production                  

SUMMARY

ANOVA

    p-value = 0.3288              
                  
    Let α = 0.05       ...5% level of significance          
   0.3288 > 0.05              
   that is, p-value > 0.05              
   Hence, we Do Not Reject Ho              
                  
   Conclusion :              
   At 5% level of significance, there does not exist enough statistical evidence to show that               
   Fertilizers have an impact on mamey production              
   that is               
     Fertilizers have no impact on mamey production              
                  
B)   Proportion of variance explained              

η2 = 0.24096  

η2 = 0.241             (Rounding to 3 decimals)

Proportion of variance explained