A research scientist studied the effects of various treatments on the growth of poplar trees. He decided to treat a large sample of saplings with a fertilizer, continual irrigation, irrigation and fertilizer and a control group (non-continual irrigation and no fertilization). He placed one quarter of all saplings into each of the groups and then measured their mass differential after 3 weeks. Did the treatment have an effect on the growth of the trees

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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

= Between mean sum of squares  / within mean sum of squares

= 1.5608 / 0.2723 = 5.73

determine the F critical value

Fcritical = = 3.2389  ( obtained from F table )

since   hence we can say The treatment had an effect on the growth of the trees i.e. Reject H0

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