Answer:
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answer : The treatment had an effect on the growth of the trees
Explanation:
H0 : No significant effect
Ha : At least one treatment has a significant effect
To confirm the above hypothesis ( conduct one way ANOVA )
From the one way Anova
( X represents each group; from control to Fertilizer and irrigation )
∑ X1^2 = 0.2338 , ∑ X2^2 = 1.8284, ∑X3^2 = 0.1982, ∑X4^2 = 11.8492
Also : ∑ N = 5 + 5 + 5 + 5 = 20 where N = number of trees in each group
also K represents number of groups i.e. = 4
Next calculate the sum of squares ( calculated using online tools )
i) Between sum of squares: value = 4.6824
ii) Within sum of squares : value = 4.3572
Next step : Calculate the degrees ( DF)
i) Between degrees of freedom
Dfb = k - 1 where K = 4 therefore Dfb = 3
ii) within degrees of freedom
Dfw = N - k , where N = 20 and K = 4 therefore Dfw = 16
Next : calculate mean sum of squares
i) Between mean sum of squares
= 4.6824 / 3 = 1.5608
ii) within mean sum of squares
= 4.3572 / 16 = 0.2723
Calculate the F-statistic value
[tex]F_{calculated }[/tex] = Between mean sum of squares / within mean sum of squares
= 1.5608 / 0.2723 = 5.73
determine the F critical value
Fcritical = [tex]F_{3,16,0.05}[/tex] = 3.2389 ( obtained from F table )
since [tex]F_{calculated } > F_{critical}[/tex] hence we can say The treatment had an effect on the growth of the trees i.e. Reject H0
